{ "Tag": [], "Problem": "Fie $\\ a\\geq\\ 2$ un numar natural.Sa se arate ca multimea {$\\ a^2+a-1,a^3+a^2-1,...,a^{n+1}+a^n-1...$} contine o submultime infinita de numere relativ prime doua cate doua.\r\n $\\text{ Mircea Becheanu, Bucuresti}$", "Solution_1": "frumoasa problema.\r\nsper sa fie buna solutia asta...\r\n[hide=\"solutie\"]presupun ca am construit o multime $A=\\left\\{n_{1},n_{2},\\ldots,n_{k}\\right\\}$, astfel incat numerele $a^{n_{i}+1}+a^{n_{i}}-1$ sa fie doua cate doua prime intre ele pentru $i=\\overline{1,k}$. \nnotam cu $P_{i}=\\left\\{p\\mbox{ prim }\\Big| ~ p|(a^{n_{i}+1}+a^{n_{i}}-1)\\right\\}$, si cu \n$Q=\\bigcup_{i=\\overline{1,k}}P_{i}$. multimea $Q$ contine doar numere prime.\npentru inceput observam ca $P_{i}\\cap P_{j}=\\emptyset$ pt $i\\neq j$.\nnotam \n$s=\\prod_{q\\in Q} (q-1)$. \nPresupunem ca exista $p\\in Q$ astfel incat $a^{s+1}+a^{s}-1 \\equiv 0 (\\bmod ~ p)$. Cum $(p-1)|s$, rezulta ca $a^{s}\\equiv 1 (\\bmod ~ p)$, deci $a^{s+1}\\equiv 0(\\bmod ~ p)$, adica $a\\equiv 0(\\bmod ~ p)$. deoarece $p\\in Q$, rezulta ca $p\\in P_{r}$, cu $1\\leq r\\leq k$. atunci, rezulta ca $a^{r+1}+a^{r}-1\\equiv 0(\\bmod ~ p)$, dar cum \n$a\\equiv 0(\\bmod ~ p)$, rezulta $-1\\equiv 0(\\bmod ~ p)$, contradictie.\nalegem $n_{k+1}=s$, si pentru $j=\\overline{1,k+1}$ notam $X_{j}=a^{n_{j}+1}+a^{n_{j}}-1$.\nInitial numerele $X_{1},X_{2},\\ldots,X_{k}$ erau relativ prime (din ipoteza de inductie), iar $X_{k+1}$ este relativ cu $X_{j}$ pentru orice $j=\\overline{1,k}$, deoarece $X_{k+1}$ nu are factori primi ce se afla in $Q$, iar toti factorii primi ai lui $X_{j}$ sunt in $Q$.\ndeci $X_{1},X_{2},\\ldots,X_{k+1}$ sunt prime intre ele. acest fapt ne spune ca putem gasi o infinitate de numere, aplicand algoritmul de mai sus.[/hide]" } { "Tag": [ "logarithms", "limit", "function", "calculus", "derivative", "algebra", "binomial theorem" ], "Problem": "lim x->0 (n^x-1)/x = ln(n)\r\n\r\nthanks", "Solution_1": "The tricky thing about this problem comes as a result of its purpose. In particular, you can't differentiate the exponential while solving this, because this limit needs to be calculated before the exponential can be differentiated.\r\n\r\nSay $ n^x \\equal{} e^{\\ln x}$, then the problem becomes $ \\lim_{x \\to 0} \\frac {e^{x \\ln n}}{x \\ln n} \\equal{} 1$, or equivalently that $ \\lim_{x \\to 0} \\frac {e^x \\minus{} 1}{x} \\equal{} 1$\r\n\r\nFrom here, it depends on how you've defined the function $ e^x$.\r\n\r\n- If you've [i]defined[/i] it in terms of its power series, then you can just substitute directly.\r\n- If you've [i]defined[/i] it as the inverse of the natural logarithm, then you can express this statement equivalently as $ \\text{exp} '(0) \\equal{} 1$, and use the derivative of the logarithm to help you out*.\r\n- If you've [i]defined[/i] it as $ e^x \\equal{} \\lim_{n \\to \\infty} \\left(1 \\plus{} \\frac {x}{n} \\right)^n$, then you can use the binomial theorem and assert that the terms you want to drop out will drop out.\r\n\r\nHave I missed any definitions? I guess the point is, for some of these functions, there are many interesting relations that they satisfy. We have to pick one way to define them, and then show that all the other forms are equivalent.\r\n\r\n* - This obviously implies that you've defined the natural logarithm first somehow, and works depending on how you have done so.", "Solution_2": "oh.my english is't good.i can't translate your text well.but if you let u=$ n^x \\minus{} 1$ then using $ \\lim_{x\\rightarrow 0}ln(u \\plus{} 1)^{\\frac {1}{u}}$=1.is it easier ,athough liking your answer?" } { "Tag": [ "floor function" ], "Problem": "How many multiples of 11 are there between 100 and 1000?", "Solution_1": "basically there are 900 numbers between 100 and 1000.\r\nSo to find how many numbers that are divisible by 11, you do 900/11, which \r\nis 81.81818181.....\r\n\r\nYou round down since its asking numbers between 100 and 1000.\r\n\r\nSo your answer is $ \\mathfrak{81}$.", "Solution_2": "$ 110, 121, ...990$\r\n\r\nDivide by 11.\r\n\r\n$ 10, 11, ..., 90$\r\n\r\nClearly, there are $ 90\\minus{}9 \\equal{} \\boxed{81}$ multiples.\r\n\r\n\r\nOR\r\n\r\n$ \\lfloor{\\frac{1000}{11}\\rfloor} \\minus{} \\lfloor{\\frac{100}{11}\\rfloor}$\r\n\r\n$ 90 \\minus{} 9 \\equal{} \\boxed{81}$", "Solution_3": "The smallest # multiple of 11 is 110 in the 3digit the largest 990\n\n990-110/11 +1 =81 \n\n[hide=\"MULTIPLES ARE AWSOME\"]81 IS THE ANSWER YOU SEEK : )[/hide][/quote]\n\n[size=200][color=#FF80BF][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH IT IS GREATLY APPRECIATED : )[/u][/i][/b][/color][/size]\n\n[size=150][color=#8000FF][b][i][u]I HAVE 99.9898989898989899898989898% CONFIDENCE THAT IM CORRECT [/u][/i][/b][/color][/size]\n\n[size=50]I KNOW THIS IS SO SMALL BUT WILL YOU POST ME IF IM WRONG : ) THANKS!![/size]\n\n\n :starwars: :starwars: :starwars: :starwars:" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "$\\lim_{x\\to 0} \\frac{(x^2+2006)\\sqrt[7]{1-2x}-2006}{x}$", "Solution_1": "$\\lim_{x\\to 0} \\frac{(x^2+2006)\\sqrt[7]{1-2x}-2006}{x}$\r\n$=\\lim_{x\\to 0} {x\\sqrt[7]{1-2x}+2006(\\sqrt[7]{1-2x}-1)}{x}$\r\n$=...$" } { "Tag": [], "Problem": "Twelve students are to be divided among Mr. Mirus's and Ms. Batty's classes. No teacher is to have more than 8 students. How many different groups of students could be in Mr. Mirus's class?", "Solution_1": "We can't have any more than 8 students in each class, so these are our possible ways:\r\n\r\n8,4\r\n7,5\r\n6,6\r\n5,7\r\n4,8\r\n\r\nThere are $ \\boxed{5}$ different ways.", "Solution_2": "Yes, izzy, this is what I thought. But you can have a different 8 people or 7 people. For example, a class could be like (random names):\r\n\r\nAbigail\r\nBrendan\r\nConnie\r\nDon\r\nJulia\r\nSarah\r\n\r\nor\r\n\r\nAvery\r\nBrianna\r\nConnie\r\nDon\r\nJohnny\r\nSarah\r\n\r\nso how many different ways are possible?", "Solution_3": "But when I typed in 5, it was counted as correct...hmm...", "Solution_4": "[quote=\"GameBot\"]Twelve students are to be divided among Mr. Mirus's and Ms. Batty's classes. No teacher is to have more than 8 students. How many different [b]groups[/b] of students could be in Mr. Mirus's class?[/quote]\r\n\r\n\r\nread the question ernie--groups means amount not certain ppl", "Solution_5": "The most a class can have is 8 students, so we find that these are the possible answers...(the first number is one class, the second number is the second class)\r\n\r\n8,4\r\n7,5\r\n6,6\r\n5,7\r\n4,8\r\n\r\nThere are $ \\boxed{5}$ different ways.", "Solution_6": "Oh, duh...\r\n\r\nAnd mz, i almost never read. :P", "Solution_7": "lol.\r\ntry reading once in a while; you'll like it :P", "Solution_8": "It [b]is[/b] a bit helpful - but Ernie's a fan of only reading the last line, and skimming it, too. :P \r\n\r\n(haha, i just realized that in three consecutive posts, we all used to :P guy.)", "Solution_9": "now it's 4 posts! :P \r\nernie, ernie, ernie.....", "Solution_10": ":P :P :P \r\n\r\nlol ernie no problem i do that a lot\r\n\r\n--i messed up my cd round like that--i read 8000miles as 8000mph", "Solution_11": "[quote=\"isabella2296\"]It [b]is[/b] a bit helpful - but Ernie's a fan of only reading the last line, and skimming it, too. :P \n\n(haha, i just realized that in three consecutive posts, we all used to :P guy.)[/quote]\r\n\r\nIzzy is right. :P (just to keep it going)", "Solution_12": ":P shes right a lot.", "Solution_13": "this has seriously disintegrated to spam stop posting here for no reason pplz\r\n\r\n\r\n :P", "Solution_14": "Heed mz's advice. We shall all stop spamming.\r\n\r\n\r\nThis forum should get a mod...hmm...I wouldn't mind being a mod!!! :P", "Solution_15": "OOOHHH!!!! ME TOO!!!!\r\n\r\n\r\nthx izzy :P", "Solution_16": "[quote=\"isabella2296\"]Heed mz's advice. We shall all stop spamming.\n\n\nThis forum should get a mod...hmm...I wouldn't mind being a mod!!! :P[/quote]\r\nwait so regular users can be mods???", "Solution_17": "[quote=\"myyellowducky82\"][quote=\"isabella2296\"]Heed mz's advice. We shall all stop spamming.\n\n\nThis forum should get a mod...hmm...I wouldn't mind being a mod!!! :P[/quote]\nwait so regular users can be mods???[/quote]\r\n\r\n\r\nNICE GOING YOU BROKE THE CHAIN!!!! :P \r\n\r\n\r\nyes they can: admins i think appoint you or somethin like that. All mods, i think started as regular users", "Solution_18": "Just curious: do the admins just choose people to be mods, or do you have to ask or something?", "Solution_19": "i think they appoint people." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Solve this equation:\r\n $\\sqrt{x+\\sqrt{x}}-\\sqrt{x-\\sqrt{x}}=\\frac{3}{2}\\sqrt{\\frac{x}{x+\\sqrt{x}}}$", "Solution_1": "$x>0$ and then $x\\geq \\sqrt{x}$ gives $x\\geq 1$.\r\nMultiplying with $\\sqrt{x+\\sqrt{x}}$ gives $x+\\sqrt{x}-\\sqrt{x^{2}-x}={3\\over 2}\\sqrt{x}$.\r\nDividing with $\\sqrt{x}$ gives $\\sqrt{x}-\\sqrt{x-1}={1\\over 2}$ \r\nand then $\\sqrt{x}+\\sqrt{x-1}={1\\over \\sqrt{x}-\\sqrt{x-1}}=2$.\r\nThe LHS is strictly increasing with unique solution $x={25\\over 16}$." } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "Let $ABC$ ($\\angle{ACB}=90$) be right angle triangle. Consider a variable point $D$ on side $AC$. Circle with center $O$ through $D$ and tangents $AB$ at $A$.Prove that circumcircle of $\\triangle{AOD}$ through a fixed point (not $A$) when $D$ is vary.", "Solution_1": "probelm is wrong.\r\n\r\nwhat are B and C for?\r\nnotice that all $ \\triangle AOD$ are homothec with center $ A$ so their circumcircles are tangent to each other.hence they only pass through $ A$." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "In each cell of $ 2007 x 2007$ table there is an odd integer. Denote by $ Z_{i}$ the sum of the numbers in $ i\\minus{}$th row and by $ S_{j}$ the sum of the numbers in $ j\\minus{}$ th column.\r\nLet $ A\\equal{}\\prod_{i\\equal{}1}^{2007}Z_{i}$ and $ B\\equal{}\\prod_{j\\equal{}1}^{2007}S_{j}$.\r\nShow that $ A\\plus{}B$ can not be equal to zero.", "Solution_1": "See here http://www.mathlinks.ro/Forum/viewtopic.php?t=149440 ." } { "Tag": [], "Problem": "If $ f(x)\\equal{}\\frac{x(x\\minus{}1)}{2}$, then $ f(x\\plus{}2)$ equals:\r\n$ \\textbf{(A)}\\ f(x)\\plus{}f(2) \\qquad\\textbf{(B)}\\ (x\\plus{}2)f(x) \\qquad\\textbf{(C)}\\ x(x\\plus{}2)f(x) \\qquad\\textbf{(D)}\\ \\frac{xf(x)}{x\\plus{}2}\\\\\r\n\\textbf{(E)}\\ \\frac{(x\\plus{}2)f(x\\plus{}1)}{x}$", "Solution_1": "[hide]\nLooking at each of the options individually and very quickly testing each out,\n\n$ \\boxed{\\text{E}}$\n[/hide]", "Solution_2": "I think there would definitely be a direct method to find the answer, but I can't figure out what that method is either. :P", "Solution_3": "[hide=\"Click for solution\"]\n$ f(x\\plus{}2)\\equal{}\\frac{(x\\plus{}2)(x\\plus{}2\\minus{}1)}{2}$ and $ f(x\\plus{}1)\\equal{}\\frac{(x\\plus{}1)(x\\plus{}1\\minus{}1)}{2}$. So $ \\frac{x\\plus{}1)}{2}\\equal{}\\frac{f(x\\plus{}1)}{x} \\implies f(x\\plus{}2)\\equal{}\\frac{(x\\plus{}2)f(x\\plus{}1)}{x}$, or $ \\boxed{\\textbf{(E)}}$.\n[/hide]" } { "Tag": [ "inequalities", "floor function" ], "Problem": "Let $ p(n)$ denote the number of unrestricted partitions of $ n$. Show that $ p(n)\\geq 2^{\\lfloor \\sqrt{n} \\rfloor }$ for all $ n\\geq 2$.", "Solution_1": "[hide=\"Solution\"]\nClearly, it works for $ n \\equal{} 2$ as there are $ 2$ partitions of $ 2$ (namely $ 2$ and $ 1 \\plus{} 1$). Then, it suffices to prove that $ p(n)\\ge 2^{\\sqrt {n}}$ for when $ n$ is a perfect square. This is because $ p(n) > p(n \\minus{} 1)$, but unless $ n$ is a perfect square, the floor of $ \\sqrt {n}$ is equal to that of $ \\sqrt {n \\minus{} 1}$. Now, let $ n \\equal{} k^2$. Consider the set $ \\{ 1, 2, 3, \\cdots , k \\}$. There are $ 2^k \\equal{} 2^{\\sqrt {n}}$ such subsets of these. Notice that the maximal such sum of any subset is $ \\frac {k^2 \\plus{} k}{2}$. To each subset, we will add one more number. Let this number be $ l_1$ for subset $ S_1$. Say that the sum of the elements of $ S_1$ is $ a_1$. Then, we define $ l_1$ to be $ n \\minus{} a_1 \\equal{} k^2 \\minus{} a_1\\ge k^2 \\minus{} (\\frac {k^2 \\plus{} k}{2}) \\equal{} \\frac {k^2 \\minus{} k}{2} > k$ when $ k > 3$. We quickly test the possibilities for $ k \\equal{} 2, 3$. These yield that $ n \\equal{} 4, 9$. For $ n \\equal{} 4$, there are $ 5$ partions (namely $ 1 \\plus{} 1 \\plus{} 1 \\plus{} 1$, $ 1 \\plus{} 2 \\plus{} 1$, $ 2 \\plus{} 2$, $ 4$, and $ 1 \\plus{} 3$.) For $ n \\equal{} 9$, there are at least $ 8$ possibilities (such as $ 9$, $ 8 \\plus{} 1$, $ 7 \\plus{} 2$, $ 6 \\plus{} 3$, $ 5 \\plus{} 4$, $ 1 \\plus{} 1 \\plus{} 7$, $ 1 \\plus{} 1 \\plus{} 1 \\plus{} 6$, and $ 1 \\plus{} 2 \\plus{} 6$). Now, we only need to consider $ k > 3$, so $ l > k$ for all subsets $ S_i$. Notice as there are $ 2^k$ subsets, there are $ 2^k$ such partions created (notice that when we add $ l$, a partition is made because of the definition of $ l$.) We now show that these subsets are unique. Notice that every two distinct subsets differ in an element, say $ e \\le k$. Let subset $ S_i$ have $ e$ and subset $ S_j$ not. Then, we add $ l_i$ to $ S_i$ and $ l_j$ to $ S_j$. Since $ l_i, l_j > k$, $ S_i$ still will have $ e$, but $ S_j$ still will not. Then, these two subsets are still distinct after adding the specified element. We have thus created $ 2^k$ subsets (that can be turned into partitions of $ n$), so $ 2^k\\le p(n)$. \n[/hide]" } { "Tag": [ "LaTeX", "AMC" ], "Problem": "Which of the following is equal to $ 1\\plus{}\\frac{1}{1\\plus{}\\frac{1}{1\\plus{}1}}$?\r\n\r\n$ \\textbf{(A)}\\ \\frac{5}{4} \\qquad\r\n\\textbf{(B)}\\ \\frac{3}{2} \\qquad\r\n\\textbf{(C)}\\ \\frac{5}{3} \\qquad\r\n\\textbf{(D)}\\ 2 \\qquad\r\n\\textbf{(E)}\\ 3$", "Solution_1": "We have $ 1 \\plus{} \\frac {1}{1 \\plus{} \\frac {1}{2}} \\equal{} 1 \\plus{} \\frac {1}{\\frac {3}{2}} \\equal{} 1 \\plus{} \\frac {2}{3} \\equal{} \\frac {5}{3}$.\r\nSo the answer is $ \\text{C}$.", "Solution_2": "Also stuff in that form always have consecutive Fibonacci numbers on the top and bottom.\r\nA,D,E are eliminated. \r\nB is easily wrong and 1+1/2 is not 2.", "Solution_3": "[b]\n\n 1+ 1/1+1/1+1= 5/3 C[/code][/b]", "Solution_4": "hello, why don't you use $ \\text{\\LaTeX}$?\r\nSonnhard.", "Solution_5": "...and parentheses.", "Solution_6": "And not revive the thread for no apparent reason.. But he is a new member though :|", "Solution_7": "[quote=Complex_Ninja]And not revive the thread for no apparent reason.. But he is a new member though :|[/quote]\n\nbut i like reviving threads for no apparent reason! :)\n\n[hide=Solution]\n$\\frac{1}{1+1}=\\frac12$\n\n$1+\\frac12=\\frac32$\n\n$\\frac{1}{\\frac32}=\\frac23$\n\n$1+\\frac23=\\boxed{\\frac53}$\n[/hide]", "Solution_8": "[quote=RedFireTruck][quote=Complex_Ninja]And not revive the thread for no apparent reason.. But he is a new member though :|[/quote]\n\nbut i like reviving threads for no apparent reason! :)\n[/quote]\nGo to fun factory then", "Solution_9": "$ 1+\\frac{1}{1+\\frac{1}{1+1}}= 1+\\frac{1}{1+\\frac{1}{2}}= 1+\\frac{1}{\\frac{3}{2}}=1+\\frac23=\\frac53$ like everyone else." } { "Tag": [], "Problem": "The increasing sequence $ 1,4,5,16,17...$ consists of all positive integers which are powers of $ 4$ or sums of distinct powers of $ 4$(cannot use a power of $ 4$more than once-so you cant have $ 4^{1} \\plus{} 4^{1} \\equal{} 8$on this list). Find the 50th term of this sequence.", "Solution_1": "[hide=\"Solution\"]\nIn base 4, for each digit, we may have either a 0 or a 1.\n\n$ a_n$ strictly increases with $ n$, hence if we write $ n$ in binary and $ a_n$ in base 4, the digits will be the same.\n\n$ 50 \\equal{} 110010_2 \\implies \\boxed{a_{50} \\equal{} 110010_4 \\iff a_{50} \\equal{} 1284}$\n[/hide]", "Solution_2": "So this would work with any rational number greater than 2 right?\r\n\r\nIf this was a sequence of 3s, 5s, or even something like 4.5, you could apply t hat right?\r\n\r\nBut if it were like 1.5 it wouldnt work because 1.5<2 and it would not be strictly increasing?\r\n\r\n\r\n(because in order to do that thing with base 2, woundnt you assume $ 4^{n} > 4^{n \\minus{} 1} \\plus{} 4^{n \\minus{} 2} \\plus{} .... \\plus{} 4 \\plus{} 1$ for integer n>0)", "Solution_3": "The important part of the solution is a bijection between the indices of the sequence written in base $ 2$ and the sequence itself written in base $ 4$ (in particular, the digits used are exactly the same). The reason this bijection works is because ordering both sets gives the same sequence of binary digits. A similar idea holds for any [b]real[/b] number greater than $ 2$ (why would you restrict to rationals?).\r\n\r\nEdit: You need to be more specific when you say \"it would not be strictly increasing.\" The problem with numbers less than $ 2$ is that the ordering of the corresponding sequence is different: in particular, $ 2$ is the smallest real number $ a$ that satisfies\r\n\r\n$ a^n > a^{n\\minus{}1} \\plus{} a^{n\\minus{}2} \\plus{} ... \\plus{} a \\plus{} 1$\r\n\r\nfor all $ n$." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "In a chess tournament each player plays every other player once. A player gets 1 point for a win, 0.5 point for a draw and 0 for a loss. Both men and women played in the tournament and each player scored the same total of points against women as against men. Show that the total number of players must be a square.", "Solution_1": "Let $k$ be the number of men and $m$ the number of women, with $n=k+m$. The sum of the scores of man-against-man matches is $\\binom{k}{2}$, so that is also the total points that men scored against women. Analogously, women score $\\binom{m}{2}$ in total against men. But $km$ opposite-sex matches are played, so\r\n\\[ \\binom{k}{2}+\\binom{m}{2}=km \\\\ k^2-k+m^2-m=2km \\\\ (k-m)^2=k+m=n, \\]\r\nwhich is what we wanted to prove." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Very coll, very funny problem. I got very surprised and a bit angry when I saw the solution of this problem:\r\nShow that $ \\prod_{1\\le i < j\\le 2009}(i^\\dfrac{1}{i} - j^\\dfrac{1}{j})\\in \\mathbb{Q}$.", "Solution_1": "Look at $ i \\equal{} 2$ and $ j \\equal{} 4$. :)", "Solution_2": "Yes, correct :!: I think this problem is very funny, because can make people loose hours in it, without concluding anything relevant.", "Solution_3": "This reminds me of a joke problem I encountered in middle school: compute the product\r\n\r\n$ (n \\minus{} a)(n \\minus{} b)(n \\minus{} c) ... (n \\minus{} z)$.\r\n\r\n:P" } { "Tag": [ "inequalities", "geometry", "circumcircle", "angle bisector", "geometry proposed" ], "Problem": "In a triangle $ABC$ denote the length $l_a$ of the $A$-angle-bisector\r\nand the length $m_a$ of the A-median a.s.o. Prove that:\r\n\r\n$1.\\blacktriangleright \\boxed {a^2+b^2+c^2=1\\Longrightarrow a^2m_a+b^2m_b+c^2m_c\\le \\frac 12}\\ .$\r\n\r\n$2.\\blacktriangleright \\boxed {\\ a^2l_a+b^2l_b+c^2l_c\\le 2p^2\\sqrt{2Rr}\\ }\\ .$", "Solution_1": "I added a new geometrical inequality at the previous message !", "Solution_2": "Nice problem !\r\n\r\n$\\displaystyle (a^2m_a+b^2m_b+c^2m_c)^2\\leq (a^2+b^2+c^2)(a^2m_{a}^2+b^2m_{b}^2+c^2m_{c}^2)\\leq \\frac{1}{4}$\r\nIt is enough to prove right inequality\r\n$\\displaystyle (a^2m_{a}^2+b^2m_{b}^2+c^2m_{c}^2)\\leq \\frac{1}{4}$ or $\\sum a^2(\\frac{b^2+c^2}{2}-\\frac{a^2}{4})\\leq \\frac{1}{4}$\r\n\r\nEqualent to $4\\sum a^2b^2-\\sum a^4\\leq 1=(a^2+b^2+c^2)^2$ and this lead us to \r\n\r\n$2\\sum a^2b^2 \\leq 2\\sum a^4$ q.e.d\r\n\r\nP.S I understood how you composed second one:\r\n\r\nI think you used that $\\displaystyle \\frac{h_a}{l_a}\\geq \\sqrt{\\frac{2r}{R}}$ \r\nlook http://www.mathlinks.ro/Forum/viewtopic.php?t=18137\r\n\r\nFrom here $\\displaystyle p\\sqrt{2Rr}\\geq a\\cdot l_a$ sum them and we are done.\r\n\r\nI think we can also change our problem to\r\n\r\n$\\displaystyle a^2l_a+b^2l_b+c^2l_c \\leq min(2p^2\\sqrt{2Rr},\\frac{4}{3\\sqrt{3}}p^3)$", "Solution_3": "[color=darkred][b]Lemma.[/b] Let $ABC$ be a triangle. Denote the points $D\\in (BC)$ and the second intersection $E$ of the line $AD$ with the circumcircle $w$ of the triangle $ABC$. Then for any point $M\\in (AE)$ there is the following inequality: $\\boxed {a\\cdot AD\\le \\frac{c^2\\cdot DC+b^2\\cdot DB}{2\\sqrt{p(M)}}}$, where $p(M)$ is the power of the point $M$ w.r.t. the circle $w$.[/color]\r\n[color=darkblue][b]Proof.[/b] Apply the Stewart's theorem to the cevian-line $AD$ in the triangle $ABC$: $a\\cdot AD^2+a\\cdot DB\\cdot DC=$ $c^2\\cdot DC+b^2\\cdot DB$.But $p(D)=$ $DB\\cdot DC=$ $DA\\cdot DE$. Thus, $AE=AD+DE=$ $AD+\\frac{DB\\cdot DC}{AD}=$ $\\frac{AD^2+DB\\cdot DC}{AD}=$ $\\frac{a\\cdot AD^2+a\\cdot DB\\cdot DC}{a\\cdot AD}=$ $\\frac{c^2\\cdot DC+b^2\\cdot DB}{a\\cdot AD}$ and $AE=AM+ME\\ge 2\\sqrt{MA\\cdot ME}=$ $2\\sqrt{p(M)}$. Therefore, $\\frac{c^2\\cdot DC+b^2\\cdot DB}{a\\cdot AD}\\ge 2\\sqrt{p(M)}$, i.e. $a\\cdot AD\\le \\frac{c^2\\cdot DC+b^2\\cdot DB}{2\\sqrt{p(M)}}\\ .$\n\n[u]Two particular cases.[/u]\n\n$1.\\blacktriangleright M: =I\\Longrightarrow \\frac{DB}{c}=\\frac{DC}{b}=\\frac{a}{b+c}\\Longrightarrow$ $al_a\\le \\frac{a(c^2b+b^2c)}{2(b+c)\\sqrt{2Rr}}$ $\\Longrightarrow$ $l_a\\le \\frac{bc}{2\\sqrt{2Rr}}$ $\\Longrightarrow$ $l_a\\le \\frac{2Rh_a}{2\\sqrt{2Rr}}$ $\\Longrightarrow$ $\\boxed {\\frac{h_a}{l_a}\\ge \\sqrt{\\frac{2r}{R}}}$ (a well-known inequality). Therefore, $\\sum a^2l_a\\le \\frac{abc}{2\\sqrt{2Rr}}\\cdot \\sum a=2p^2\\sqrt{2Rr}\\ .$\n\n$2.\\blacktriangleright M: =G$ and $a^2+b^2+c^2=1\\Longrightarrow DB=DC=\\frac a2\\ ,\\ p(G)=\\frac{a^2+b^2+c^2}{9}\\Longrightarrow$\n$am_a\\le \\frac 34 a(b^2+c^2)\\Longrightarrow \\sum a^2m_a\\le \\frac 34\\sum a^2(b^2+c^2)=$$\\frac 32\\sum (ab)^2\\le \\frac 32\\cdot \\frac 13\\left(\\sum a^2\\right)^2=\\frac 12\\ .$[/color]" } { "Tag": [ "Alcumus" ], "Problem": "In how many different ways can 12 dimes be divided into three\npiles with an odd number of dimes in each pile?", "Solution_1": "No matter how many combinations you try, you will not get $ 3$ piles, with odd numbers of coins, to combine to $ 12$ total coins. An odd number plus an odd number plus an odd number ALWAYS equals another odd number. Since $ 12$ is even, the answer is $ \\boxed{0}$, since it is impossible to make an even number with the restrictions given.", "Solution_2": "[hide=\"what I thought\"]This is similar to what Alcumus said, but I know that an odd number plus an odd number is an even number, and an even number plus an odd number is an odd number. Therefore, three odd numbers add up to make an odd number. 12 is an even number, so the answer is 0.[/hide]", "Solution_3": "[hide=\"My Solution\"]Note that any odd number can be written in the form $2p+1$. \n\nLet the $1st$ pile contain $2a+1$ dimes, the $2nd$ pile contain $2b+1$ dimes, and the $3rd$ pile contain $2c+1$ dimes.\n\nWe know that the number of dimes in the three piles must add up to 12:\n\n$2a+1+2b+1+2c+1=12$\n\n$2a+2b+2c+1+1+1=12$\n\n$2(a+b+c)+3=12$\n\n$2(a+b+c)=9$.\n\n$a+b+c=\\frac{9}{2}$.\n\nBecause $a, b,$ and $c$ must be whole numbers, this equation is impossible.\n\nTherefore, there is no way that 12 dimes can be divided into three piles with an odd numer of dimes.\n\n[hide=\"answer\"]There are $\\boxed{0}$ ways for 12 dimes to be divided into three piles, each containing an odd number of dimes.[/hide][/hide]", "Solution_4": "We use generating functions:[hide = Solution]$\\left (\\frac{x}{1-x^2}\\right )^3$ is the product of the three generating functions, and it doesn't contain a $x^{12}$ term. So, the answer is $0$[/hide]", "Solution_5": "[hide=Sol]Odd+Odd+Odd = Odd.\n12 \u2260 Odd.\nAns = 0[/hide]", "Solution_6": "troll question lol\n\n[hide=sol]3 odd numbers are always odd.[/hide]", "Solution_7": "[hide]Three odd numbers can never equal an even number. \n\nExample:\n$3+5+7=15$[/hide]" } { "Tag": [ "trigonometry" ], "Problem": "Can anyone please prove this identity, its been bothering me for a while now.\r\n\r\ntan(x-y) + tan(y) / 1 - tan(x-y) * tan(y) = tan(x)", "Solution_1": "[quote=\"flyover\"]Can anyone please prove this identity, its been bothering me for a while now.\n\ntan(x-y) + tan(y) / 1 - tan(x-y) * tan(y) = tan(x)[/quote]Uh, just let $ x\\equal{}y\\plus{}z$ and you'll get the angle addition formula for tan... unless you want to prove $ \\tan(\\alpha\\plus{}\\beta)\\equal{}\\frac{\\tan\\alpha\\plus{}\\tan\\beta}{1\\minus{}\\tan\\alpha\\tan\\beta}$... :roll:", "Solution_2": "[quote=\"flyover\"]Can anyone please prove this identity, its been bothering me for a while now.\n\ntan(x-y) + tan(y) / 1 - tan(x-y) * tan(y) = tan(x)[/quote]\r\n[hide]\\[ \\frac{\\tan(x\\minus{}y)\\plus{}\\tan(y)}{1\\minus{}\\tan(x\\minus{}y)\\cdot\\tan(y)}\\equal{}\\tan((x\\minus{}y)\\plus{}(y))\\equal{}\\tan(x)\\][/hide]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "$a,b,c \\ge 0$ prove:\r\n$\\sum_{cyc}\\sqrt{a^4+a^2b^2+b^4}\\ge \\sum_{cyc}a\\sqrt{2a^2+bc}$", "Solution_1": "[quote=\"amir2\"]$a,b,c \\ge 0$ prove:\n$\\sum_{cyc}\\sqrt{a^4+a^2b^2+b^4}\\ge \\sum_{cyc}a\\sqrt{2a^2+bc}$[/quote]\r\n\r\nAlready discussed, i think....\r\n\r\n$a^4+a^2b^2+b^4\\geq \\frac{3}{4}(a^2+b^2)^2$" } { "Tag": [ "Putnam" ], "Problem": "Uitati aici o problema de Mortici, care eu consider ca e gresita, intrucat am gasit un contraexemplu.\r\n\r\nFiind date cinci puncte P1, P2, P3, P4, P5 in planul xOy, avand coordonatele intregi, demonstrati ca exista 1 <= i \\parallel y-x\\parallel ={\\sqrt{\\sum_{t=1}^{n}(y_{t}-x_{t})^{2}}}\\geq \\sqrt{(y_{i}-x_{i})^{2}}=|y_{i}-x_{i}|$.\r\nNow suppose that $y_{i}\\in [a_{i},b_{i}]$. We proceed in cases.\r\nIn the first case, assume $x_{i} |y_{i}-x_{i}|=y_{i}-x_{i}$, so $a_{i}-x_{i}>y_{i}-x_{i}$, or $y_{i}b_{i}$. Then $\\delta=|x_{i}-b_{i}|=x_{i}-b_{i}$, and $\\delta>|y_{i}-x_{i}|=x_{i}-y_{i}$. So $x_{i}-b_{i}>x_{i}-y_{i}$, or $y_{i}>b_{i}$. Once again, contradiction.\r\nThus $y_{i}\\notin [a_{i},b_{i}]$, from which we deduce $y \\in R^{n}-R$, and in turn $B(x,\\delta) \\subset R^{n}-R$. $x$ was arbitrary, so this proves that $R^{n}-R$ is open.\r\nTherefore $R$ is closed. [b]QED[/b]", "Solution_1": "It's fine and neat except the $\\le$ sign should be $\\ge$. I do not think there is a shorter way of proving the statement if one has to write down every detail as you did.", "Solution_2": "the inequality sign is fixed :) \r\n\r\nso when you're proving that a set is closed (in an undergraduate level topology course), then you almost always resort to proving that the compliment is open?", "Solution_3": "Sometimes one may prefer to prove that it contains all its boundary points instead but, if you look at how it is done and require each step to be brought to the very basic level, you usually find that it is still a proof of the openness of the complement in some disguised form, which is not surprising because that is how the closed sets are defined. So, the answer to your last question is \"Yes\". :)", "Solution_4": "compl[b]e[/b]ment\r\n\r\nYou could also use the fact that unions of open sets are open. Since $\\mathbb R^{n}\\setminus R=\\bigcup_{i=1}^{n}\\{x\\in\\mathbb R^{n}\\colon x_{i}b_{i}\\}$, this reduces the problem to showing that an \"open half-space\" is actually open. :) Not much of a simplification, actually.", "Solution_5": "Couldn't you use the fact that if x is in the complement of the interval then the coordinates of x belong to the respective complements of the components of the interval (which we know to be open) and therefore x belongs to an open set?", "Solution_6": "But the point $(1,5)$ lies outside of the square $[0,2]\\times [0,2]$ even though the first coordinate, $1$, is within $[0,2]$.", "Solution_7": "I think the most important thing is what is open according your assumption. Open balls? Open retangles?", "Solution_8": "[quote=\"Victor96\"]I think the most important thing is what is open according your assumption. Open balls? Open retangles?[/quote]\r\n\r\nThe definition I'm using:\r\nA set $S$ is open if for every $a \\in S$ there exists a $\\delta>0$ such that the ball $B(a,\\delta)$ is entirely contained in $S$.", "Solution_9": "Oh, then I think mlok's decomposition has offered a very good way of simplifying the problem because you need only to bound your radius using the information of one coordinate." } { "Tag": [ "integration", "algebra", "polynomial", "calculus", "real analysis", "real analysis unsolved" ], "Problem": "For natural number n, consider: $ H_k \\equal{} \\frac{(\\minus{}1)^{n\\minus{}k}}{n\\cdot k!(n\\minus{}k)!}\\int_0^n\\frac{q(q\\minus{}1)\\cdots(q\\minus{}n)}{q\\minus{}k}\\ ,\\ \\ k\\equal{}0,1,\\ldots,n.$\r\n\r\nProve that $ H_k$ is positive and satisfies: $ H_k\\equal{}H_{n\\minus{}k},\\ H_0\\plus{}H_1\\plus{}\\cdots\\plus{}H_n\\equal{}1.$", "Solution_1": "Just consider Lagrange's interpolation polynomial for $ f(x)\\equal{}\\frac{1}{n}$ on segment $ [0,n]$ with centers 0, 1, 2, ..., $ n$. Then $ H_{0}\\plus{}H_{1}\\plus{}\\cdots\\plus{}H_{n}\\equal{}1$ is a fact that $ \\int_0^nf(x)dx\\equal{}1$.", "Solution_2": "Don't you use Newton-Cotes formula?\r\nSo we have to consider the formula of error estimates afterwards. But the original problem precedes this theorem in my textbook. Any hints for a natural solution without numerical analysis?", "Solution_3": "What \"error estimates\"? We get the exact identity for the case $ f(x)\\equal{}1/n$ , since $ 1/n$ is a zero-degree polynomial.", "Solution_4": "[quote]Prove that $ H_k$ is positive and satisfies: $ H_k \\equal{} H_{n \\minus{} k}$[/quote]\r\n\r\nHello, I'm new here.\r\nAny ideas in proving this side? I tried expanding the product but failed.", "Solution_5": "It was an obvious part of the problem. You just need to change $ q\\to n\\minus{}q$ in the integral (dunno how to say it in english properly)." } { "Tag": [], "Problem": "What product will be obtained when 2-(3-cyclohexenyl)acetic acid is treated with $ I_2/NaHCO_3$?", "Solution_1": "Is it alpha iodonation by any chance?? :huh:", "Solution_2": "And what is \"alpha iodonation\"?", "Solution_3": "look out for methyl carbonyl chemrock....\r\n\r\ni think because...\r\nit is similar top iodoform i.e. iodine + a base..", "Solution_4": "Excuse me it is acetic acid :huh: :huh: :huh:", "Solution_5": "Both previous answers are wrong.", "Solution_6": "sorry carcul I didn't mean acetic acid!\r\nI told mr.valerium that it is acetic acid and not acetone.", "Solution_7": "you get a lactone. the double bond goes and O attaches to one of the carbons(Halo lactonisation)", "Solution_8": "Good that might be right I never thought of that :o :o", "Solution_9": "That's right: iodo-lactonization is the answer." } { "Tag": [ "rotation", "geometry", "3D geometry" ], "Problem": "When a right triangle is rotated about one leg, the volume of the cone produced is $800 \\pi$ $\\text{cm}^3$. When the triangle is rotated about the other leg, the volume of the cone produced is $1920 \\pi$ $\\text{cm}^3$. What is the length (in cm) of the hypotenuse of the triangle?", "Solution_1": "[hide=\"Solution\"]Let the legs of the right triangle be $x$ and $y$. We're trying to find $\\sqrt{x^2+y^2}$. Applying the formula for a cone's volume, $\\frac 13 xy^2=800$ and $\\frac 13 x^2y=1920$. To find $x$ and $y$, we multiply the equations together to obtain $(\\frac 13 xy^2)(\\frac 13 x^2y)=\\frac 19 x^3y^3=(800)(1920)$, so that $xy=240$. Therefore, $x=24$ and $y=10$, and the hypotenuse is $\\boxed{026}$. [/hide]" } { "Tag": [], "Problem": "The frequency distribution shows the number of CD's that students in a math class listened to in their leisure time last week. Find the mode of the data.", "Solution_1": "There isn't any data listed here.", "Solution_2": "none here either. check over their behind that tree please.", "Solution_3": "There should be a distribution with 1 mark for 0, 2 for 1, 1 for 2, continuing...the answer is 3." } { "Tag": [ "inequalities", "absolute value" ], "Problem": "The non-negative real numbers $a, b, c, d$ add up to $1$. Prove the inequality\r\n\r\n$\\left| ab - cd \\right| \\leq \\frac{1}{4}$.", "Solution_1": "[quote=\"Andreas\"]The non-negative real numbers $a, b, c, d$ add up to $1$. Prove the inequality\n\n$\\left| ab - cd \\right| \\leq \\frac{1}{4}$.[/quote]\r\nyou mean $a,b,c,d<1$ or $a+b+c+d<1$ ?", "Solution_2": "I think this works:\r\n\r\nAssume the opposite. Let's say $ab-cd > \\frac{1}{4}$.\r\n\r\nClearly, $ab > \\frac{1}{4}$, which implies $\\sqrt{ab} > \\sqrt{\\frac{1}{4}}$.\r\n\r\nBy AM-GM, $\\frac{a+b}{2} \\ge \\sqrt{ab}$. The maximum of $a+b$ is $1$.\r\n\r\nSo we have $\\frac{1}{2} \\ge \\sqrt{ab} > \\sqrt{\\frac{1}{4}}$.\r\n\r\nSo, $\\frac{1}{2} > \\frac{1}{2}$. Contradiction.\r\n\r\nIf we remove the absolute value signs, we have also $cd - ab > \\frac{1}{4}$, which leads back to the argument above.\r\n\r\nThe case for equality is when $a = .5$, $b=.5$, $c=0$, $d=0$, or $a=0$, $b=0$, $c=.5$, $d=.5$." } { "Tag": [ "inequalities" ], "Problem": "What is the coefficient of the term containing $x^{6}$ in the expansion of $(4x^{2}-\\frac{y}{2})^{9}$?", "Solution_1": "[hide]To get an $x^{6}$ from an $x^{2}$, you need it multiplied three times. There fore, we need to simplify\n$\\binom{9}{3}(4x^{2})^{3}(-\\frac{y}{2})^{6}$ which is\n$84*64x^{6}*(\\frac{y^{6}}{64})$ so I think the answer is 5376?[/hide]\r\n\r\nI don't know, it could be wrong.", "Solution_2": "[quote=\"xpmath\"][hide]To get an $x^{6}$ from an $x^{2}$, you need it multiplied three times. There fore, we need to simplify\n$\\binom{9}{3}(4x^{2})^{3}(-\\frac{y}{2})^{6}$ which is\n$84*64x^{6}*(\\frac{y^{6}}{64})$ so I think the answer is 5376?[/hide]\n\nI don't know, it could be wrong.[/quote]\r\n$(4x^{2})^{3}=4096x^{6}$. I think you typed it in wrong or something because you got the right answer, but the wrong number.", "Solution_3": "Um... ${(4x^{2})^{3}=(4*x^{2})^{3}=(4^{3}*(x^{2})^{3}})=64x^{6}$... I think", "Solution_4": "[quote=\"xpmath\"]Um... $(4x^{2})^{3}=(4*x^{2})^{3}=(4^{3}*x^{2^{3}})=64x^{6}$... I think[/quote]\r\n\r\nThe third inequality is written incorrectly. $(x^{2})^{3}= x^{6}\\neq x^{2^{3}}= x^{8}$.", "Solution_5": "[hide]\nEach term is of the form $4^{n}\\cdot x^{2n}\\cdot y^{9-n}\\cdot .5^{9-n}\\cdot (-1)^{9-n}\\cdot \\binom{9}{n}$.\nn=3 is what we want, this turns into $\\binom{9}{3}=84$[/hide]", "Solution_6": "[quote=\"diophantient\"][hide]\nEach term is of the form $4^{n}\\cdot x^{2n}\\cdot y^{9-n}\\cdot .5^{9-n}\\cdot (-1)^{9-n}\\cdot \\binom{9}{n}$.\nn=3 is what we want, this turns into $\\binom{9}{3}=84$[/hide][/quote]\r\nBut don't forget about the 4, as it will still be a coeifficient", "Solution_7": "I got the coefficient as 84y^6", "Solution_8": "The coefficient is just $84$. $y$ is considered a variable the same as $x$, so it is not part of the coefficient.", "Solution_9": "[quote=\"xpmath\"][quote=\"diophantient\"][hide]\nEach term is of the form $4^{n}\\cdot x^{2n}\\cdot y^{9-n}\\cdot .5^{9-n}\\cdot (-1)^{9-n}\\cdot \\binom{9}{n}$.\nn=3 is what we want, this turns into $\\binom{9}{3}=84$[/hide][/quote]\nBut don't forget about the 4, as it will still be a coeifficient[/quote]\r\nThe $4^{3}$ is cancelled out by the $.5^{9-3}$", "Solution_10": "Whoops, didn't see that, sorry" } { "Tag": [ "puzzles" ], "Problem": "Find the next number in the sequence\r\n\r\n1,2,2,3,2,4,2,4,3,4,2,6,2,4,4", "Solution_1": "Ive thought about this for a while and I couldn't get it. Could you post a hint that doesn't give away too much?" } { "Tag": [], "Problem": "OK, try this one - I like it and I have a solution.\r\n\r\nA word of length n that consists only of the digits \"0\" and \"1\" is called a bit-string of length n. (For example, 01101 is a bit-string of length 5.) Consider the sequence s(1), s(2), ... of bit-strings of length n > 1 which is obtained as follows : \r\n(1) s(1) is the bit-string 00...01, consisting of n-1 zeros and a \"one\" ; \r\n(2) s(k+1) is obtained as follows : \r\n(a) Remove the digit on the left of s(k). This gives a bit-string t of length n-1. \r\n(b) Examine whether the bit-string t1 (length n) is already in the set\r\n{s(1), s(2), ..., s(k)}. \r\nIf this is the case, then s(k+1) = t1. If this is not the case then s(k+1) = t0. \r\n\r\nFor example, if n = 3 we get : \r\ns(1) = 001 -> s(2) = 011 -> s(3) = 111 -> s(4) = 110 -> s(5) = 101 -> \r\ns(6) = 010 - > s(7) = 100 -> s(8) = 000 -> s(9) = 000 -> ... \r\n\r\nSuppose N = 2^n. \r\nProve that the bit-strings s(1), s(2), ..., s(N) of length n are all different.", "Solution_1": "Nobody wants to try this ? :?", "Solution_2": "I have a nice and simple proof that 00...0 is the only one that can repeat, but I'm not sure how to go about showing that it must occur first at step s(2^n). I guess I haven't given it enough thought yet.\r\n\r\nBTW, you might want to edit your statement of the procedure for determining s(k+1). Its sort of the opposite of what you want (ie if t1 is in the set then s(k+1) should be t0, not t1). It rather confused me to begin with.", "Solution_3": "A soln for this can be found on Mathlinks :lol: I couldn't solve it on my own completely, so Arne helped me (he gave me a BIG hint).", "Solution_4": "The official solution can be found at\r\n\r\nhttp://www.mathlinks.ro/phpBB/viewtopic.php?t=836." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove that if $a,b,c,d\\geq0$, then \\[\\frac a{\\sqrt{a^{2}+b^{2}+c^{2}}}+\\frac b{\\sqrt{b^{2}+c^{2}+d^{2}}}+\\frac c{\\sqrt{c^{2}+d^{2}+a^{2}}}+\\frac d{\\sqrt{d^{2}+a^{2}+b^{2}}}\\leq\\frac{4\\sqrt3}3.\\]", "Solution_1": "Arqady, Vasc, have you got solution for it?", "Solution_2": "The following estimate is nice and easy\r\n\\[\\sum_{\\rm cyclic}\\frac a{\\sqrt{b^{2}+c^{2}+d^{2}}}\\geq2. \\]", "Solution_3": "[quote=\"pvthuan\"]If $a,b,c,d\\geq0$, then\n$\\sqrt{\\frac a{a+b+c}}+\\sqrt{\\frac b{b+c+d}}+\\sqrt{\\frac c{c+d+a}}+\\sqrt{\\frac d{d+a+b}}\\leq\\frac4{\\sqrt3}$.\n[/quote]\r\n$\\sum\\sqrt{\\frac a{a+b+c}}$\r\n$\\leq\\frac 2{3\\sqrt3}\\sum\\frac{8a^{2}+b^{2}+c^{2}+8d^{2}+24ab+24ac+26ad+4bc+6bd+6cd}{3(a^{2}+b^{2}+c^{2}+d^{2})+10(ab+bc+ac+bc+bd+cd)}=\\frac4{\\sqrt3}$.", "Solution_4": "[quote=\"pvthuan\"]The following estimate is nice and easy\n\\[\\sum_{\\rm cyclic}\\frac a{\\sqrt{b^{2}+c^{2}+d^{2}}}\\geq2. \\]\n[/quote]\r\n\\[\\sum_{\\rm cyclic}\\frac a{\\sqrt{b^{2}+c^{2}+d^{2}}}\\]\r\n\\[=\\sum_{\\rm cyclic}\\frac{a^{2}}{a\\sqrt{b^{2}+c^{2}+d^{2}}}\\]\r\n\\[\\ge \\sum_{\\rm cyclic}\\frac{2a^{2}}{a^{2}+b^{2}+c^{2}+d^{2}}=2\\]", "Solution_5": "What is the last inequality from?", "Solution_6": "AM-GM inequality :wink: \r\nWe have $a\\sqrt{b^{2}+c^{2}+d^{2}}=\\sqrt{a^{2}(b^{2}+c^{2}+d^{2})}\\leq \\frac{a^{2}+b^{2}+c^{2}+d^{2}}{2}$", "Solution_7": "But doesn't that mean that equality case is when $a=\\sqrt{b^{2}+c^{2}+d^{2}}$ and so on?\r\n\r\nif so\r\n\r\n$\\sum{\\frac{a}{\\sqrt{b^{2}+c^{2}+d^{2}}}}=4$\r\n\r\nwhat am I doing wrong...?", "Solution_8": "[quote=\"pvthuan\"]Prove that if $ a,b,c,d\\geq0$, then\n\\[ \\frac a{\\sqrt {a^{2} \\plus{} b^{2} \\plus{} c^{2}}} \\plus{} \\frac b{\\sqrt {b^{2} \\plus{} c^{2} \\plus{} d^{2}}} \\plus{} \\frac c{\\sqrt {c^{2} \\plus{} d^{2} \\plus{} a^{2}}} \\plus{} \\frac d{\\sqrt {d^{2} \\plus{} a^{2} \\plus{} b^{2}}}\\leq\\frac {4\\sqrt3}3.\n\\]\n[/quote]\r\nI have an ugly proof. :(", "Solution_9": "[quote=\"shfdfzhjj\"][quote=\"pvthuan\"]Prove that if $ a,b,c,d\\geq0$, then\n\\[ \\frac a{\\sqrt {a^{2} \\plus{} b^{2} \\plus{} c^{2}}} \\plus{} \\frac b{\\sqrt {b^{2} \\plus{} c^{2} \\plus{} d^{2}}} \\plus{} \\frac c{\\sqrt {c^{2} \\plus{} d^{2} \\plus{} a^{2}}} \\plus{} \\frac d{\\sqrt {d^{2} \\plus{} a^{2} \\plus{} b^{2}}}\\leq\\frac {4\\sqrt3}3.\n\\]\n[/quote]\nI have an ugly proof. :([/quote]\r\nYes,I have a proof,But i think it is quite nice(I know it from Can hang 2007,but i find the solution independly)", "Solution_10": "[quote=\"Allnames\"][quote=\"shfdfzhjj\"][quote=\"pvthuan\"]Prove that if $ a,b,c,d\\geq0$, then\n\\[ \\frac a{\\sqrt {a^{2} \\plus{} b^{2} \\plus{} c^{2}}} \\plus{} \\frac b{\\sqrt {b^{2} \\plus{} c^{2} \\plus{} d^{2}}} \\plus{} \\frac c{\\sqrt {c^{2} \\plus{} d^{2} \\plus{} a^{2}}} \\plus{} \\frac d{\\sqrt {d^{2} \\plus{} a^{2} \\plus{} b^{2}}}\\leq\\frac {4\\sqrt3}3.\n\\]\n[/quote]\nI have an ugly proof. :([/quote]\nYes,I have a proof,But i think it is quite nice(I know it from Can hang 2007,but i find the solution independly)[/quote]\r\n\r\nCan you show it? :maybe:" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "[i]Let $ p$ be a fix odd prime .The sequence $ \\{ x_n \\}$ was defined by : $ x_{n \\plus{} 2} \\ \\equal{} \\ 4x_{n \\plus{} 1} \\ \\minus{} \\ x_n \\ \\forall \\ n \\ \\in \\ \\mathbb{N}$\n\n Choose $ x_0 \\ ; \\ x_1$ such that :\n\n For all positive integer $ k$ , there exists an index $ i$ such that : $ 4p^2 \\ \\minus{} \\ 8p \\plus{} 1 | x_i \\ \\minus{} \\ (2p)^k$ \n \n( $ \\mathbb{N}$ is the set of all nonegative integer )[/i]", "Solution_1": "[hide]\nFirst, lets say that $ 4p^{2}\\ \\minus{}\\ 8p\\plus{}1 | x_{k\\minus{}2}\\ \\minus{}\\ (2p)^{k\\minus{}2}$ and $ 4p^{2}\\ \\minus{}\\ 8p\\plus{}1 | x_{k\\minus{}1}\\ \\minus{}\\ (2p)^{k\\minus{}1}$ and let $ 4p^{2}\\ \\minus{}\\ 8p\\plus{}1\\equal{}A$.\n\nWe have that $ x_{k\\minus{}2}\\equiv(2p)^{k\\minus{}2} (mod A)$ and $ x_{k\\minus{}1}\\equiv(2p)^{k\\minus{}1} (mod A)$. \n\nThen,\n $ x_{k}\\equiv 4x_{k\\minus{}1}\\minus{}x_{k\\minus{}2}\\equiv 4(2p)^{k\\minus{}1}\\minus{}(2p)^{k\\minus{}2}\\equiv(2)^{k\\plus{}1}(p)^{k\\minus{}1}\\minus{}(2)^{k\\minus{}2}(p)^{k\\minus{}2}\n\\equiv(2p)^{k\\minus{}2}(8p\\minus{}1)\\equiv(2p)^{k\\minus{}2}(4p^2\\minus{}8p\\plus{}1\\plus{}8p\\minus{}1)\\equiv(2p)^{k\\minus{}2}(4p^2)\\equiv(2p)^{k} (mod A)$\n\nThus if we choose $ x_{0}, x_{1}$ such that $ 4p^{2}\\ \\minus{}\\ 8p\\plus{}1 | x_{0}\\ \\minus{}\\ 1$ and $ 4p^{2}\\ \\minus{}\\ 8p\\plus{}1 | x_{1}\\ \\minus{}\\ (2p)$ then we will have a sequence that satisfies the requirements. Letting $ x_{0}\\equal{}1$ and $ x_{1}\\equal{}2p$ works and we are done.\n[/hide]", "Solution_2": "[quote=\"Bacteria\"]\nFirst, lets say that $ 4p^{2}\\ \\minus{} \\ 8p \\plus{} 1 | x_{k \\minus{} 2}\\ \\minus{} \\ (2p)^{k \\minus{} 2}$ and $ 4p^{2}\\ \\minus{} \\ 8p \\plus{} 1 | x_{k \\minus{} 1}\\ \\minus{} \\ (2p)^{k \\minus{} 1}$ and let $ 4p^{2}\\ \\minus{} \\ 8p \\plus{} 1 \\equal{} A$.\n\nWe have that $ x_{k \\minus{} 2}\\equiv(2p)^{k \\minus{} 2} (mod A)$ and $ x_{k \\minus{} 1}\\equiv(2p)^{k \\minus{} 1} (mod A)$. \n\nThen,\n $ x_{k}\\equiv 4x_{k \\minus{} 1} \\minus{} x_{k \\minus{} 2}\\equiv 4(2p)^{k \\minus{} 1} \\minus{} (2p)^{k \\minus{} 2}\\equiv(2)^{k \\plus{} 1}(p)^{k \\minus{} 1} \\minus{} (2)^{k \\minus{} 2}(p)^{k \\minus{} 2} \\equiv(2p)^{k \\minus{} 2}(8p \\minus{} 1)\\equiv(2p)^{k \\minus{} 2}(4p^2 \\minus{} 8p \\plus{} 1 \\plus{} 8p \\minus{} 1)\\equiv(2p)^{k \\minus{} 2}(4p^2)\\equiv(2p)^{k} (mod A)$\n\nThus if we choose $ x_{0}, x_{1}$ such that $ 4p^{2}\\ \\minus{} \\ 8p \\plus{} 1 | x_{0}\\ \\minus{} \\ 1$ and $ 4p^{2}\\ \\minus{} \\ 8p \\plus{} 1 | x_{1}\\ \\minus{} \\ (2p)$ then we will have a sequence that satisfies the requirements. Letting $ x_{0} \\equal{} 1$ and $ x_{1} \\equal{} 2p$ works and we are done.\n[/quote]\r\n\r\n Quite nice :lol:" } { "Tag": [], "Problem": "How many different combinations of 5 dollar bills and 2 dollar bills can be used to make a total of 17 dollars? Order doesn't matter in this problem. \r\n\r\na) 2\r\nb) 3\r\nc) 4\r\nd) 5\r\ne) 6", "Solution_1": "[hide]\nYou nedd an even number of dollars left after you subtract the five dollar bills. Thus you can only subtract an odd number of five dollar bills. It must be <17, so the number of 5 dollar bills can only be 1,3.\nThe number of corresponging 2 dollar bills is 6 and 2.\nThus the answer is\n$(1,6)$\n$(3,1)$\n[/hide]", "Solution_2": "[hide]2,answer A. I won't bother with a solution as biffandoc's already got one and it's kinda obvious.[/hide]", "Solution_3": "[hide]Two. Three fives and one two, and six twos and one five.[/hide]", "Solution_4": "[hide]The combinations are\n3 5s 1 2\n\n1 5 6 2s\n\nSo there are 2 combinations\n\n$A$[/hide]", "Solution_5": "[hide]$(3,5) (1,2)$ and\n$(1,5) (6,2)$ so the answer is A[/hide]", "Solution_6": "[hide]There are two ways[/hide] what if order matters?" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ a,b,c$ be positive numbers satisfying $ a,b,c \\geq 1$ .Find the maximum of express\r\n $ P\\equal{} \\frac{(1\\plus{}a)(1\\plus{}b)(1\\plus{}c)}{abc\\plus{}1}$.", "Solution_1": "We have \r\n$ P\\leq \\frac{(a\\plus{}1)(b\\plus{}1)(c\\plus{}1)\\plus{}(a\\minus{}1)(b\\minus{}1)(c\\minus{}1)}{abc\\plus{}1}$\r\n$ \\equal{}\\frac{2(abc\\plus{}a\\plus{}b\\plus{}c)}{abc\\plus{}1}\\equal{}2\\plus{}\\frac{2(a\\plus{}b\\plus{}c\\minus{}1)}{abc\\plus{}1}$\r\nBecause $ x,y\\geq 1\\Rightarrow (x\\minus{}1)(y\\minus{}1)\\geq 0\\Rightarrow xy\\plus{}1\\geq x\\plus{}y$ then $ abc\\plus{}1\\geq ab\\plus{}c\\geq a\\plus{}b\\plus{}c\\minus{}1$\r\nThen $ P\\leq 4$", "Solution_2": "Let $ a,b,c$ be positive numbers satisfying $ a,b,c \\geq 1$ .Prove that \n $$(a+1)(b+1)(c+1)\\leq 4(abc+1)$$\n\nLet a, b, and c be positive real numbers. [url=https://artofproblemsolving.com/community/c6h1179230p5701994]Prove that [/url]\n$$(a+1)(a+b)(b+c)(c+16)\\ge81abc$$\n[url=https://artofproblemsolving.com/community/c6h866p2820]h[/url]", "Solution_3": "[quote=hoangclub]Let $ a,b,c$ be positive numbers satisfying $ a,b,c \\geq 1$ .Find the maximum of express\n $ P\\equal{} \\frac{(1\\plus{}a)(1\\plus{}b)(1\\plus{}c)}{abc\\plus{}1}$.[/quote]\n\nWe will prove that $P\\leq 4$, or $(\\frac{a+1}{2})(\\frac{b+1}{2})(\\frac{c+1}{2})\\leq \\frac{abc+1}{2}$\nFirstly, we have $(\\frac{x+1}{2})(\\frac{y+1}{2})=\\frac{xy+1}{2}-\\frac{(x-1)(y-1)}{4}\\leq \\frac{xy+1}{2}\\ \\ \\forall x,y\\geq 1$\n$\\Rightarrow LHS\\leq (\\frac{ab+1}{2})(\\frac{c+1}{2})\\leq \\frac{abc+1}{2}$ ($Q.E.D$)\n$P_{max}=4$ at $a=b=c=1$. $\\square$", "Solution_4": "$(ab-1)(c-1) \\geq 0 \\implies abc+1 \\geq ab+c \\implies a+b+c+ab+bc+ca \\leq 3abc+3 \\implies \\frac{(a+1)(b+1)(c+1)}{abc+1} \\leq 4$", "Solution_5": "To Prove : $(a+1)(b+1)(c+1) \\leq 4(abc+1) \\hspace{0.2 cm} ( a,b,c \\geq 1)$\n[hide = Solution]\nWe know that $(ab-1)(c-1)+(bc-1)(a-1)+(ca-1)(b-1) \\geq 0$\n$ab+bc+ca+a+b+c \\leq 3abc + 3$\n$abc+ab+bc+ca+a+b+c+1 \\leq 4abc+4$\n$(a+1)(b+1)(c+1) \\leq 4(abc+1)$\n\n\n[/hide]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "(a)Find a number divisible by $ 2$ and $ 9$ which has exactly $ 14$ divisors .\r\n(b)Find a number divisible by $ 2$ and $ 9$ which has exactly $ 15$ divisors .\r\n(c)Find a number divisible by $ 2$ and $ 9$ which has exactly $ 17$ divisors .", "Solution_1": "(a) The only option is $ 2\\cdot 3^6 \\equal{} 1458$.\r\n(b) The two options are $ 2^2 \\cdot 3^4 \\equal{} 324$ and $ 2^4 \\cdot 3^2 \\equal{} 144$.\r\n(c) Such a number does not exist. :(" } { "Tag": [ "MATHCOUNTS", "induction", "calculus", "function", "algebra", "polynomial", "Gauss" ], "Problem": "This is a challenge based on the 2000 National Mathcounts Masters Round:\r\n\r\nThe sum of the first n positive integers is given by the formula n(n+1)/2.\r\nThe sum of the first n squares is given by n(n+1)(2n+1)/6.\r\n\r\n1) Find a formula for the sum of the first n cubes. How did you derive this?\r\n2) \" first n 4th powers. How did you derive this?\r\n3) Give a method of finding a formula for the sum of the first n xth powers.", "Solution_1": "I am too lazy to do it myself right now, but I think you can prove all three with induction (well, at least the first two definitely).\r\n\r\nFierytycoon", "Solution_2": "Sure, you can prove that a formula works using induction (in fact, the n(n+1)/2 example is the classic example of induction), but can you make your own formula? Can you come up with some method?", "Solution_3": "Since this isn't particularly easy, I should point out that there's a nice section in [u]Concrete Mathematics[/u] by Ronald Graham, Donald Knuth, and Oren Patashnik on this topic. It has to do with Bernoulli numbers if you want to do that, but you can solve individual cases with finite calculus, generating functions, or various other methods. By the way, if you're looking for a terrific (but extremely challenging) math book, check out Graham, Knuth, Patashnik. For some reason, a friend of mine had this book and didn't like it, and so he gave it to me. I wonder what other math books he has that he doesn't like...", "Solution_4": "I've actually worked out a method long ago for extracting the explicit formula of the sums. But it's _heavy_ in calculation. Want me to demonstrate for some simple cases?", "Solution_5": "Sure, Mindspa. And I have a method that's pretty easy to understand (I think it falls under the \"generating functions\" category. If anyone wants, I'll demonstrate also.", "Solution_6": "Yes. there is a pretty simple recursive way to find the formulas for the sum of nth powers. But to jump to a particular formala for n non-recursively, I think you need a generating function that involves Bernoulli numbers.", "Solution_7": "I say it again, it's heavy on calculation really brute force, no ingeniuity, I remember doing this method a long time ago when I searched for a general formula, when I didn't even know calculus.\r\n\r\nLet's make a factored table first for the first eightterms in the sum S^4_n = 1^4 + ... + n^4,\r\n\r\n{1,1},{2,17},{3,2*7^2},{4,2*3*59},{5,11*89},{6,5^2*7*13},{7,7*2^2*167},{8,2^2*3*17*43}\r\n\r\nWe know from formulas for 1,2,3 that they contained a factor of n. So in order to be able to get a factor n in this formula (wishful thinking) we need to have S^4_n = n*q(n)/30 (S^4_n needs to be multiplied by 30 for it to be able to be divisible by n as seen in the table) for some polynomial q(n), we now build a similar table for q(n) = 30*S^4_n/n with seven terms,\r\n\r\n{1,2*3*5},{2,3*5*17},{3,2^2*5*7^2},{4,3^2*5*59},{5,2*3*11*89},{6,5^3*7*13},{7,2^3*3*5*167}.\r\n\r\nNow again we do some wishful thinking and assume that S^4_n is divisible by (n+1) thus q(n) = p(n)(n+1) for some polynomial p(n). A table for p(n) = q(n)/(n+1) (the job now is easier because we only delete factors).\r\n\r\n{1,3*5},{2,5*17},{3,5*7^2},{4,3^2*59},{5,11*89},{6,5^3*13},{7,3*5*167}. \r\n\r\nSome wishful thinking yet again gives us that it must contain a factor of (2n + 1) because we can clearly spot the \"sequence\" 3,5,7,9,11,13,15 in the factors. Thus p(n) = r(n)(2n+1) for some polynomial r(n) = p(n)/(2n+1).\r\n\r\n{1,5},{2,17},{3,5*7}={3,35},{4,59},{5,89},{6,5^3}={6,125},{7,167}. \r\n\r\nWe now see that the factors look somewhat an arithmetic sequence. Let's \"guess\" the difference r(n) - r(n-1) = (n-1)*6 + 6 = 6n thus r(n) = r(n-1) + 6n, wishful thinking again makes us want to be able to divide r(n) by 6, because the and it looks from the table as we can succed by adding 1 to each of the factors. We get (r(n)+1)/6 = m(n) for some polynomial m(n) without writing out the table for m(n) we see after writing it out that it is the sequence 1,3,6,10,... which is the n(n+1)/2 thus m(n) = n(n+1)/2.\r\n\r\nUsing all the polynomials we get m(n) = n(n+1)/2, thus r(n) = 6n(n+1)/2 - 1 why p(n) = (3n(n+1)-1)(2n+1) and q(n) = (3n(n+1)-1)(2n+1)(n+1) and thus S^4_n = n(n+1)(3n(n+1)-1)(2n+1)/30 for the nth term. I said it was heavy on calculation :(.", "Solution_8": "Yo G. Dig the avatar man..... Ramanujan is definitely my hero when it comes to contemporary mathematicians... Some of the stuff he's come up with, you can only come up with when you are on pot, crack, or both. For example, \r\n\r\n1/:pi: = ((:sqrt: 8 )/9801):Sigma: ((4n)!(1103+26390n))/(((n!)^4)396^(4n)) which is infinite sum.\r\n\r\nAnd in the famous words of my little bro, \"Damn dude, thas messed up...\"\r\n\r\nPeace,\r\nIW.OAV", "Solution_9": "[quote=\"Idiot_without_a_village\"]Some of the stuff he's come up with, you can only come up with when you are on pot, crack, or both.[/quote]\r\n\r\nHigh on mathematics...", "Solution_10": "[color=cyan]I believe this is the method that gauss was talking about, and it doesn't rely on the amount of wishful thinking required by MindSpa's solution. You need the formulas for the sums of the powers of the positive integers up to (m - 2) in order to get (m - 1).\n\nLet S(n,m) = :Sigma: i^m = 1^m + 2^m + ... + n^m\nWrite each i^m as ((i-1) + 1)^m and expand using the binomial theorem. Then, regroup terms according to their powers. Subtract all the mth powers, which will leave you with an n^m on the left side. Substitute in for all the sums of smaller powers. Subtract. Divide. And you will have your answer. It's pretty nice up until about the 4th powers. After that, I wouldn't want to touch it without a computer-algebra system or something that will keep me from writing everything out.\n\nYou can also do it a different way: you can calculate any of the base values easily. You also \"know\" that the sum of the mth powers will give you an (m+1)st degree polynomial. So the first (m+2) values uniquely determine it. So solve a system of linear equations for the coefficients.\n\nThe first way I like a lot better, and I think in general involves less calculation.[/color]", "Solution_11": "This is the way I work out sums of powers easily (though not any specific one, this is recursive), its probably the same as JBLs one here but it doesn't look exactly the same..\r\nAll we do is expand i^n - (i-1)^n. This gives us a whole bunch of terms..\r\nNow just add that up from i=0 to whatever. This gives i^n on the LHS, and a constant multiplied by each of the sums of smaller powers on the RHS, along with the one we are looking for.\r\nie, to find the sum of squares we get i^3 - (i-1)^3 = 3i^2 -3i + 1\r\n\r\nSo i^3 = 3(sum of squares) - 3i(i+1)/2 + i\r\nAnd then rearrange that to get the sum of squares.", "Solution_12": "[color=cyan]Yeah, your way is basically the same as the first way I did it, but it looks a lot nicer if you sum it up afterwards (like you did) instead of before-hand (like I did).[/color]", "Solution_13": "Here's an even faster way to produce them recursively. Take the polynomial f_k(n) for the sum of the first n kth powers. To get f_(k+1)(n) for the sum of the first n (k+1)th powers, multiply the previous formula by k and then integrate with respect to n. Then take that and add an n term that makes the sum of the coefficients equal to 1.\r\n\r\nFor example, to find the sum of the first n cubes from the forumula for the sum of the first n squares:\r\n\r\nMultiply f_2(n) = n^2/2 + n/2 by 2. Then integrate with respect to n to get n^3/3 + n^2/2. Now add n/6 to make the sum of the coefficients (i.e. 1/3 + 1/2 + 1/6) = 1. So f_3(n) = n^3/3 + n^2/2 + n/6.\r\n\r\nSimilarly, multipling this by 3 and integrating you get n^4/4 + n^3/2 + n^2/4. The sum of it's coefficients is already 1, so you don't need to add anything. Therefore f_4(n) = n^4/4 + n^3/2 + n^2/4.\r\n\r\nYou can keep doing this recursively to get the higher and higher Sum of Powers formulas.", "Solution_14": "Yeah I've heard of that way as well. Never thought about why it works though, the other method is probably quite a lot easier to remember, though the integral method is quite a bit faster. Probably best to remember both.." } { "Tag": [ "trigonometry", "inequalities", "function", "geometry", "circumcircle", "cyclic quadrilateral", "geometry proposed" ], "Problem": "In a cyclic quadrilateral $ABCD$, $AB=a$, $BC=b$, $CD=c$, $\\angle ABC = 120^\\circ$ and $\\angle ABD = 30^\\circ$. Prove that\r\n\r\n(1) $c \\ge a + b$; \r\n\r\n(2) $|\\sqrt{c + a} - \\sqrt{c + b} | = \\sqrt{c - a - b}$.", "Solution_1": "Since $ABCD$ is a cyclic quadrilateral, we have that $\\angle ACB=30$ and $\\angle CDA=180-\\angle ABC=60$, then we have that $AC=\\frac{c\\sqrt3}{2}$.\r\n\r\nNow, by the cosine law in $\\triangle ABC$ we have that\r\n$a^2+ab+b^2=\\frac{3c^2}{4} \\ \\ \\ (1)$.\r\n\r\nThen, $a)$ follows from $(1)$ and the inequality $a^2+ab+b^2\\geq\\frac{3(a+b)^2}{4}$.\r\n\r\nAnd we have $b)$ follows from $|\\sqrt{c+a}-\\sqrt{c+b}|=\\sqrt{c-a-b}\\Leftrightarrow (\\sqrt{c+a}-\\sqrt{c+b})^2=c-a-b \\\\\\\\\\Leftrightarrow 2a+2b+c=2\\sqrt{(c+a)(c+b)} \\Leftrightarrow 4a^2+4ab+4b^2=3c^2$\r\nwhich is clearly equivalent to $(1)$.", "Solution_2": "Hello,\r\n The first part of the problem can be done differently using Jensen's inequality. The problem reduces to showing that sin x + siny <=1 if x + y =60. Here, sinx is a concave function. Hence, Jensen's inequality for concave functions gives the result.\r\nSanjith", "Solution_3": "[hide=\"A elementary proof (similarly rather with the Campos-solution) for this nice and easy problem.\"]I suppose w.l.o.g. the circumradius $R=1$. Thus, $AD\\equiv d=1$, $CD=c=2$, $AC\\equiv e=\\sqrt 3$, $AB=a$, $BC=b$, $BD\\equiv f=\\sqrt{4-b^2}\\ .$\nFrom the [i]Ptolemeu's theorem[/i] $ac+bd=ef$ results $\\boxed {\\ a^2+ab+b^2=3\\ }\\ .$\nI remark $3=a^2+ab+b^2\\ge3ab\\Longrightarrow \\boxed {\\ ab\\le 1\\ }\\ .$\nTherefore, the proposed problem is equivalently with the following conditioned inequality: $a>0$, $b>0$, $a+b<\\sqrt 3$, $a^2+ab+b^2=3\\Longrightarrow$\n$a+b\\le 2\\ \\ \\wedge\\ \\ \\left|\\sqrt{2+a}-\\sqrt{2+b}\\right|=\\sqrt{2-a-b}\\ \\ (1)\\ .$\n$1^{\\circ}\\blacktriangleright ab\\le 1\\ \\wedge\\ a^2+ab+b^2=3\\Longrightarrow a^2+2ab+b^2\\le 4\\Longrightarrow a+b\\le 2\\ .$\n$2^{\\circ}\\blacktriangleright(1)\\Longleftrightarrow4+a+b-2\\sqrt{4+2(a+b)+ab}=2-a-b\\Longleftrightarrow$\n$a+b+1=\\sqrt{4+2(a+b)+ab}\\Longleftrightarrow(a+b)^2+2(a+b)+1=4+2(a+b)+ab\\Longleftrightarrow$\n$a^2+2ab+b^2+1=4+ab\\Longleftrightarrow a^2+ab+b^2=3$, what is true.[/hide]", "Solution_4": "Exactly!! Ptolemys theorem gives the shortest , simplest, direct and best proof!! No need for dirty trigno!! I did this in the paper!!!", "Solution_5": "Exactly I did the same thing though I did not assume R=1. :P" } { "Tag": [ "probability" ], "Problem": "Suppose you are living in a city where all the \r\nresidents have been allotted a three digit number between $ 100$ and \r\n$ 999$ by a numbering machine which generates three digit numbers \r\nrandomly.\r\n\r\nNow, on the centinery 'foundation day' of the city, there was a huge \r\ngathering at the city esplanade where all residents including you \r\nare to be present and where the given three digit numbers of the \r\nresidents were checked at the entrance gate. it was announced by the \r\ncity municipality that the first person who will be having a number \r\nwhich has been recorded earlier with another resident while \r\nentering, that is the first person who has a repeat number will be \r\nthe winner of a luxury car.\r\n\r\nWhen do you think your chance of winning better ? if you're the \r\n$ 13^{th}$, $ 31^{st}$, $ 251^{st}$, $ 501^{st}$ visitor ?", "Solution_1": "If the $ 501^{st}$ visitor goes, does this mean that no one else won before him?", "Solution_2": "I think it means he is the 501st in line, i.e. in calculating the probability, you must account for the possibility that someone before him won.", "Solution_3": "[quote=\"Brut3Forc3\"]I think it means he is the 501st in line, i.e. in calculating the probability, you must account for the possibility that someone before him won.[/quote]\r\n\r\nyes, thats correct.", "Solution_4": "[hide]In order for the $ 13^{th}$ person to win, the first twelve must get different numbers. The first person gets a number. The second person now has a $ \\frac{899}{900}$ chance of not winning. Continue for eleven iterations to get $ \\frac{899\\cdot898\\cdot...889}{900^{11}}$. Let that be $ a$ (for referencing). The $ 13^{th}$ person now must choose from twelve numbers, so we multiply to get $ \\frac{a}{75}$ as the probability.\n\nFor the $ 31^{st}$ visitor to win, the first thirty cannot win. We continue our expression of $ a$ by multiplying by $ a\\cdot \\frac{888\\cdot887\\cdot...871}{900^{18}}$. For the next number, there is a $ \\frac{1}{30}$ chance that the person will win. This is a greater probability than $ \\frac{a}{75}$, so we will let this be represented as $ \\frac{b}{30}$.\n\nFor the $ 251^{st}$ visitor, we have $ b\\cdot \\frac{250}{900}\\cdot \\frac{888\\cdot887\\cdot...651}{900^{238}}$. I think this is MUCH smaller than the probability for the $ 31^{st}$ person to win.\n\nAnd that logic holds for the $ 501^{st}$ person as well.\n\nTherefore, the $ 31^{st}$ person will most likely win of these four.\n\nI don't know how to do it, but there is a way to maximize the chances of winning. Earlier is better, but there needs to be enough numbers in the pool to choose from.\n\n\n[/hide]" } { "Tag": [], "Problem": "\u0388\u03c3\u03c4\u03c9 $ n$ \u03b8\u03b5\u03c4\u03b9\u03ba\u03cc\u03c2 \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03bf\u03c2 \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf\u03c2 \u03c4\u03bf\u03c5 $ 2$ \u03ba\u03b9\u03b1 \u03ad\u03c3\u03c4\u03c9 $ f$ \u03ad\u03bd\u03b1 \u03c0\u03bf\u03bb\u03c5\u03ce\u03bd\u03c5\u03bc\u03bf \u03b2\u03b1\u03b8\u03bc\u03bf\u03cd \u03cc\u03c7\u03b9 \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf\u03c5 \u03c4\u03bf\u03c5 $ n\\minus{}2$. \u0395\u03ac\u03bd $ a_{1},a_{2},...,a_{n}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03bf\u03af \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03af \u03ba\u03b1\u03b9 $ p(t)\\equal{}\\prod^{n}_{i\\equal{}1}(t\\minus{}a_{i})$ \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 $ \\sum^{n}_{i\\equal{}1}\\frac{f(a_{i})}{p^{'}(a_{i})}\\equal{}0$.", "Solution_1": "\u0391\u03c0\u03cc \u03c4\u03bf Lagrange Interpolation formula \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $ f(x) \\equal{} \\displaystyle\\sum_{i \\equal{} 1}^{n}f(a_i)\\displaystyle\\prod_{j\\neq i}\\frac {x \\minus{} a_j}{a_i \\minus{} a_j}$ \u03ba\u03b1\u03b9 \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae $ deg(f)\\le n \\minus{} 2$ \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03bf \u03c3\u03c5\u03bd\u03c4\u03b5\u03bb\u03b5\u03c3\u03c4\u03ae\u03c2 \u03c4\u03bf\u03c5 $ x^{n \\minus{} 1}$\u03c4\u03bf\u03c5 $ f(x)$ \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 $ \\sum^{n}_{i \\equal{} 1}\\frac {f(a_{i})}{p^{'}(a_{i})}$ \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 0." } { "Tag": [ "limit", "trigonometry", "function", "set theory" ], "Problem": "Not sure where this should go, but what do you guys make of this?\r\n\r\nhttp://mathworld.wolfram.com/DiagonalParadox.html", "Solution_1": "No mater what the value of n is the line will move 1 unit north total and 1 unit east total. \r\nTherefore the line will always be 2 units long.", "Solution_2": "Yes, but as the sizes of the up and downs approaches 0, the line approaches that of the diagonal. However, the distance traveled for the up and down route is always 2 while the diagonal is always $\\sqrt{2}$.\r\n\r\nI'm thinking that as the up and down approaches 0, the line would become the diagonal, but they have different lengths. Can anyone expand for me why my reasoning is incorrect, because obviously, it is, and it has not created a huge impact on mathematics that is felt everywhere (even Russels Paradox at least inspired some people to create a whole new set theory)\r\n\r\nI know that the lines have to move up and down only, but at such small sizes, wouldn't it be like a limit?\r\n\r\n\r\nUpon thinking about this a bit more, no matter how small the steps get, there is always a diagonal that is $\\sqrt{2}$ the size of the step--which was at first confusing because of my idea of a limit.\r\n\r\nHowever, if my liit idea really is correct, the steps become infinitely small, but then, the diagonal really won't go any distance because the steps are so small--and I'm rambling now so I'll go to sleep.\r\n\r\nHope some of you can help me shed some light on this :D", "Solution_3": "Here's what I make of it.\r\nA limit of something as it approaches a value is not necessarily that something evaluated at that value. \r\n\\[\\displaystyle\\lim_{x\\to 0}\\frac{\\sin x}{x}=1\\] does not imply \\[\\frac{\\sin 0}{0}=1\\]\r\nThe value of the limit is merely an indication of the function's behavior as the function approaches a value not in the function's domain. In the case of the diagonal, taking the limit of the steps are a bad indication of how long the diagonal because they will always be perpendicular while the diagonal is a straight line. As the length of the individual steps approaches 0, the steps still do not behave like the diagonal. Therefore, the limit of the length of the staircase as the length of the individual steps approaches 0 is not the same as the length of the diagonal.", "Solution_4": "It actually doesn't approach the diagonal of a square. It just looks that way since it gets really small, but no matter how many steps you take you'll never look closely and see anything that resembles a straight line. That's the intuitive answer, and the mathematical one is obvious: it's always a total of twice as long.", "Solution_5": "I think it's like this:\r\n\r\nsquare with length a, diagonal d\r\n\r\nd_n is the n-th length of the \"step\" diagonal. \r\n\r\nd_n = sum(1 to n) of (a_n) where a_n = a/n\r\n\r\nd_n = a * sum(1 to n) of (1/n)\r\n\r\nso to get the length of the diagonal we take the limit for n -> infinity. \r\nthis gives\r\n\r\nd_step = limit(n->infinity) a * sum (1 to n) of (1/n)\r\n\r\nbut sum(1 to infinity) of 1/n diverges!!", "Solution_6": "But, no matter how many subdivisions you pick, the total length of the stepwise diagonal must be 2. How can this be reconciled with the fact that the sum (2a/n from n=1 to infinity) diverges as n->infinity? \r\n\r\nOr, taking a finite case, let's say n=1000; then the sum from n=1 to 1000 of 1/n is 7.485..., not 2. Is the statement that the total length of the stepwise diagonal must be 2 incorrect? How can the line be any longer than 2 units while only travelling up and right through the square, without reversing direction or exceeding the bounds of the square?\r\n\r\nedit: The reason for this is that sum is wrong, the sum is not from n=1 to infinity at all, but is simply a/n where n=constant added to itself n times! So the sum is 2a/n times n which obviously equals 2a or the length of the stepwise diagonal." } { "Tag": [], "Problem": "for how many integer pairs (x,y) exist where $ \\sqrt{1984}\\equal{}\\sqrt{x}\\plus{}\\sqrt{y}$ and $ 07, then $ord_{2}(n!)\\ge ord_{5}(n!)+5$. It give, that four ends digits before 0 ($n!=a_{1}a_{2}...a_{k+1}a_{k+2}a_{k+3a_{k+4}00...0}$) is number ($a_{k+1}a_{k+2}a_{k+3}a_{k+4}$) divides 16. But $16\\not |2004$.", "Solution_5": "But the problem does not imply that the last four non-zero digits are $2,0,0,4$..." } { "Tag": [ "geometry", "projective geometry", "geometry proposed" ], "Problem": "Let $a$, $b$ and $l$ be (independent) lines in the projective plane and let $P$, $Q$ and $R$ be fixed points on $l$. Let $\\Delta ABC$ be a triangle such that $A \\in a$, $B \\in b$, $P \\in BC$, $Q \\in CA$ and $R \\in AB$. Show that all possible points $C$ lie on a fixed line. Projective geometry again.", "Solution_1": "Consider a central proection which put line l to infty.Then angles $(A'B',a')$ , $(A'B',b')$ , $(A'B' , A'C')$ , $(A'C' , B'C')$-constant. The result follows from gomothety in point $D'=a' \\cap b'$.Also locus of C is a line passing throug a point D." } { "Tag": [], "Problem": "Find formulae of sequence in term of $ k$\r\n\r\n$ \\mathbf{(i)}$\r\n\r\n$ a_{1}=3$\r\n$ a_{k+1}=a_{k}^{2}-2$\r\n\r\n$ \\mathbf{(ii)}$\r\n\r\n$ b_{1}=1$\r\n$ 4b_{k+1}=b_{k}^{2}+4$", "Solution_1": "The first is in the OEIS, http://www.research.att.com/~njas/sequences/A001566 . In order to figure out what the sequence is, I suggest comparing the first few values to the [hide]Lucas numbers.[/hide]", "Solution_2": "what are lucas numbers?", "Solution_3": "Well, that rather defeated the purpose of my hide tags :maybe: \r\n\r\nThe Lucas numbers are a Fibonacci-type sequence that begins $ L_{0}= 2, L_{1}= 1$. So the first few terms are $ 2, 1, 3, 4, 7, 11, \\ldots$." } { "Tag": [ "AMC", "AIME", "USAMTS", "ceiling function", "modular arithmetic", "AIME I" ], "Problem": "The sequence 3, 15, 24, 48, ... is those multiples of 3 which are one less than a square. Find the remainder when the 1994th term is divided by 1000. \r\n\r\nI looked at the Kalva's solution but I have different idea.. I wonder if there are any other ways solving it.\r\n\r\nNote: I just wrote what I had in my paper.. It's not edited so it's not like nice USAMTS solution I try to write. :D \r\n\r\n[hide=\"My answer\"]\nI decided to write the question in math term. So, we are looking for multiples of 3 that is in form of $k^2-1$. \n\nNote that $k^2-1 = (k+1)(k-1)$. So I decided to plug in some numbers that make it multiple of 3.\n\n\\[a_1 = (2+1)(2-1)\\\\\na_2 = (4+1)(4-1)\\\\\na_3 = (5+1)(5-1)\\\\\na_4 = (7+1)(7-1)\\\\\na_5 = (8+1)(8-1)\\\\\na_6 = (10+1)(10-1)\\\\\na_7 = (11+1)(11-1)\\\\\na_8 = (13+1)(13-1)\\\\\na_9 = (14+1)(14-1)\\\\\na_{10} = (16+1)(16-1)\\\\\na_{11} = (17+1)(17-1)\\\\\na_{12} = (19+1)(19-1)\\]\n\n(I just wrote out some terms to make sure I'm seeing the pattern right).\n\nSo, $a_n = (m+1)(m-1)$ is how it's expressed. From this, I developed formula with $a_8$ (I picked that one.. Seem pretty confident number).\n\nLet $Q$ to be positive integer then the value $m$ and the value of $n$ in $a_n$ can be written as $13+3Q$ and $8+2Q$ respectively.\n\nIn our problem, $8+2Q = 1994$ so $Q = 993$. Putting in $13+3Q=m$ tells us that $m=2992$.\n\nSo.. $a_{1994} = (2992+1)(2992-1)$.. You can multiply it.. I won't..\n\nSince $2993 \\equiv -7 \\mod 1000$ and $2991 \\equiv -9 \\mod 1000$, the product will be $63 \\mod 1000$. Therefore, the answer is $\\fbox{063}$. [/hide]", "Solution_1": "$n = k^2 - 1 = (k-1)(k+1)$\r\n\r\nIf $3|(k-1)(k+1)$, then $3|k-1$ or $3|k+1$.\r\n\r\nThe values of $k$ which work are in the form of $3n-1$ and $3n+1$ where $n$ is a positive integer.\r\n\r\nFor the nth term, the corresponding value for $k$ is given by $\\frac{3\\lceil\\frac{n}{2}\\rceil}{2} - 1$ when $n$ is odd and $\\frac{3\\lceil\\frac{n}{2}\\rceil}{2} + 1$ when $n$ is even.\r\n\r\n$k=3(1994/2) + 1 = 2992$. \r\n$2992^2 - 1 = 992^2 - 1 = (-8)^2 - 1 = 63\\pmod{1000}$.", "Solution_2": "I don't know what Kalva did, but I found the 1995th positive integer that is not a multiple of 3 (not 1994th because 1 doesn't count), squared it, and subtracted 1, since every integer that isn't a multiple of 3 will have a square that is 1 mod 3. Of course on the actual AIME I probably wouldn't calculate the entire square, just the last three digits, but I had a calculator handy. Your way of finding the residue mod 1000 is much smarter, of course.", "Solution_3": "I basically did what WarpedKlown1335 did at the start...\r\nYou know that 3 divides k-1 or k+1. So then the sequence of possible k goes\r\n2 4 5 7 8 10 11 ...\r\nSo then the 1994th k must be \r\n$4+3(996)$\r\nSo the 1994th term is\r\n$2992^2-1=(3000-8)^2-1=3000^2-16*3000+64-1=63(mod1000)$" } { "Tag": [], "Problem": "1. What are the dimensions of the shaded square inside the 3-4-5 right angle triangle?\r\n\r\n2. Find the position of the square that has the maximum area.\r\n\r\n3. Can you generalise for an x-y-z right angle triangle?\r\n\r\n4. What would be the dimensions of the square in terms of x, y, and z?", "Solution_1": "1, 4\r\n[hide]The square splits the entire triangle into a bunch of similar triangles. If the side of the square is $s$, then the little right triangle at the top has legs of $s$ and $3 - s$. Since it is similar to the largest triangle, we form the proportion $\\frac {3-s}{s} = \\frac {3}{4}$ and solving for $s$ we get $\\frac {12}{7}$. If we have an $x, y, z$ right triangle, we just substitute $x$ and $y$ in for $3$ and $4$ in the equations and we get that the side of the square is $\\frac {xy}{x+y}$.[/hide]", "Solution_2": "[hide=\"2\"]\nas it is drawn below\n[/hide]\n\n[hide=\"3\"]\ncall z the hypotenuse, x the shorter side, and s the side of the square.\n\n$\\displaystyle \\frac{x-a}{a} = \\frac{a}{y-a}$\n$\\displaystyle a^2 = xy - a(x+y) + a^2$\n$\\displaystyle a(x+y) = xy$\n$\\displaystyle a = \\frac{xy}{x+y}$\n[/hide]\r\n\r\nEDIT: Ignore number 3. I don't know if what i did for number 3 is generalization or not....(realized that after looking at tarquin's number [b]4[/b]", "Solution_3": "[hide=\"1\"]\n144/49\n[/hide]" } { "Tag": [ "Euler", "algorithm", "unsolved" ], "Problem": "Noah has to fit 8 species of animals into 4 cages of the Arc. He planes to put two species of animal in each cage. It turns out that, for each species of animal, there are at most 3 other species with which it cannot share a cage. Prove that there is a way to assign the animals to the cages so that each species shares a cage with a compatible species.", "Solution_1": "It is obvious we can remove the \"at most\" condition and suppose that there are 4 other compatible species for each specie.\r\n\r\nConsider the graph of $K_8$ with 3 edges at each node removed so that in total we remove 12 edges. The degree of each node is even, so we can find an euler circuit through this graph. Label the nodes A..H. The circuit looks something like ABCB... where we write each letter 2 times. Taking every two adjacent letters (where the first and last are considered adjacent) , we find we have 16 'adjacencies'.\r\n\r\nEvery adjacent pair of letters is a legal pair. Since we never use the same edge twice in a circuit, we have all 16 legal pairs. Since we use all edges in a circuit, the following method is guaranteed to work:\r\n\r\nwe take our 'circuit string' and pick any pair in it, and erase all the letters that the pair uses. We repeat this 3 more times. Each stage, we are guaranteed to find a pair, because the string necessarily contains all pairs formed of unused letters. The 4 pairs are our solution.\r\n\r\nExample:\r\n\r\nAEBFCGDHCEDFAGBH; pick AE\r\nBFCGDHC DF GBH; pick GB\r\nFC DHC DF H; pick DF\r\nC HC H; pick HC and done.", "Solution_2": "[quote=\"Singular\"]\n\n Each stage, we are guaranteed to find a pair, because the string necessarily contains all pairs formed of unused letters. The 4 pairs are our solution.\n\n[/quote]\r\n\r\nWhy is that? I don't understand. Can anybody explain?", "Solution_3": "Sorry for the reply to an old topic. :oops: \r\n\r\nIsn't the problem a direct conseguence of Dilworth's Theorem?", "Solution_4": "I think we can generalize it into $2n$ and $n$ ccases,where $3$ is replaced with $n-1$.", "Solution_5": "[quote=\"Singular\"]It is obvious we can remove the \"at most\" condition and suppose that there are 4 other compatible species for each specie.\n\nConsider the graph of $K_8$ with 3 edges at each node removed so that in total we remove 12 edges. The degree of each node is even, so we can find an euler circuit through this graph. Label the nodes A..H. The circuit looks something like ABCB... where we write each letter 2 times. Taking every two adjacent letters (where the first and last are considered adjacent) , we find we have 16 'adjacencies'.\n\nEvery adjacent pair of letters is a legal pair. Since we never use the same edge twice in a circuit, we have all 16 legal pairs. Since we use all edges in a circuit, the following method is guaranteed to work:\n\nwe take our 'circuit string' and pick any pair in it, and erase all the letters that the pair uses. We repeat this 3 more times. Each stage, we are guaranteed to find a pair, because the string necessarily contains all pairs formed of unused letters. The 4 pairs are our solution.\n\nExample:\n\nAEBFCGDHCEDFAGBH; pick AE\nBFCGDHC DF GBH; pick GB\nFC DHC DF H; pick DF\nC HC H; pick HC and done.[/quote]\n\nI also find this a bit too optimistic\n\nTake the following circuit for eg:\n\n4-5-1-2-5-3-1-6-7-8-2-7-6-4-3-8\n\nwe can remove 1-2 first, then 3-4, then we are left with\n\n5 5 6-7-8 7-6 8\n\nwhere 5 can never be paired with anything\n\nIt appears this algorithm is not at all intuitive and will need at least a 1-ply look-ahead to work properly,", "Solution_6": "actually 6 and 7 doesn't satisfy the problem requirement in my previous example, a correct example would be:\n\n4-5-1-2-5-3-1-6-7-8-2-7-3-4-6-8\n\nagain remove 12, then 34, and 5 is left hanging" } { "Tag": [ "group theory", "abstract algebra", "algebra", "polynomial", "superior algebra", "superior algebra unsolved" ], "Problem": "Hi!\r\n\r\nCan anyone give a hint for this problem:\r\n\r\nLet K be a field, g an element of Aut(K)(=group of ring automorphisms of K) of infinite order.\r\nIf F is the fixed field of the cyclic subgroup (of Aut(K)) generated by g, and K/F is an algebraic extension,\r\nshow that K/F is a normal extension.", "Solution_1": "Take $x \\in K$, and let $P$ be its minimal polynomial (which exists since $K/F$ is algebraic). If $\\alpha$ is a root of $P$, then $P(\\alpha)=0 \\Rightarrow P(g^{k}(\\alpha))=0$ for every integer $k$ (since $g^{k}$ fixes the coefficients of $P$ by hypothesis). Since $P$ has only finitely many roots and since $$ has infinite order, we have $g^{k}(\\alpha)=g^{m}(\\alpha)$ for some $n \\neq m$. But then $\\alpha=g^{k-m}(\\alpha) \\in K$, and the problem is solved." } { "Tag": [ "function", "algebra", "domain", "polynomial", "calculus", "integration", "functional equation" ], "Problem": "Can anyone of you help me how to prove that\r\n1) a function is injective, surjective and bijective?\r\n2) a function is constant, continuous and monotonous?\r\nby giving me an examples?\r\n\r\nAnother thing what is the differences between solving the functional equation that has real numbers and quotient number? I don't quite understand the notes our teacher has given us.", "Solution_1": "For 2,\r\n\r\nA constant function is a function in which the y value is a constant for whatever x value you \"plug\" into it.\r\nEx: y=3\r\n\r\nA continuous function is a function which you can \"draw\" without lifting up your pencil, such as y=x, y=x^2, y=2^x, y=lnx, y=sinx, etc....\r\n\r\nI know that a monotonically increasing function is a function in which for all pairs x_1f(x_2) for a monotonically decreasing function.\r\n\r\nI am not sure if I answered the last one correctly.\r\n\r\n-interesting_move", "Solution_2": "Constant means f(x) = C for a constant C, for all x (or put another way, f(x) = f(y) for all x, y).\r\nThe formal definition is continuous is that for any a, the limit of f(x) as x goes to a is f(a). If you don't need to do it formally, interesting_move has it right.\r\nMonotonous means either f(x) >= f(y) for all x>=y, or f(x)<=f(y) for all x<=y.\r\n\r\nInjective means one to one; ie, if f(x) = f(y) then x=y.\r\nSurjective means onto; ie, for all y in the range of the function, you can find an x with f(x)=y.\r\nBijective means injective and surjective.\r\n\r\nI'm not quite sure what you mean by quotient number, but I'll take a guess that you mean Q, ie the rational numbers. A function is something which takes a value, and outputs another value. If you have a function from the reals, that means f(any real number) exists. If you restrict things to the rationals, there is no such thing as f(sqrt(3)) for example - f is only defined for rational numbers. Often you can prove something is true for rational numbers then use something like continuity to prove it for all reals - like the famous Cauchy equations.", "Solution_3": "[quote=\"TripleM\"]Injective means one to one; ie, if f(x) = f(y) then x=y.\n[/quote]\r\n\r\nBut I found out that it has another explanation which confuses with this one.\r\n\r\nIt states that, \r\n f : A -> B\r\n if both a1 and a2 are elements of A, then a1 is not equal to a2,\r\n which implies that f(a1) is not equal to f(a2).\r\n\r\n\r\nYour say that f(x) = f(y) and this makes me confused.", "Solution_4": "[quote=\"ken_himura1984\"][quote=\"TripleM\"]Injective means one to one; ie, if f(x) = f(y) then x=y.\n[/quote]\n\nBut I found out that it has another explanation which confuses with this one.\n\nIt states that, \n f : A -> B\n if both a1 and a2 are elements of A, then a1 is not equal to a2,\n which implies that f(a1) is not equal to f(a2).\n\n\nYour say that f(x) = f(y) and this makes me confused.[/quote]\r\n\r\nTake the contrapositive of what he said, e.g, p => q is logically equivalent to ~q => ~p.\r\n\r\nSo, MMM defined injective as: f(x)=f(y) implies x=y.\r\n\r\nThe contrapositive is x != y implies f(x) != f(y), which is essentially what you wrote.", "Solution_5": "[quote=\"TripleM\"]The formal definition is continuous is that for any a, the limit of f(x) as x goes to a is f(a). If you don't need to do it formally, interesting_move has it right.[/quote]\r\n\r\nI don't understand what you mean by this sentences. Can you simplify it?\r\n\r\nAnd also, can U please help me hoe to prove the function below?\r\n\r\n how to prove \r\n\r\n1) f(an) = nf(a) \r\n\r\n2) f(a/n) = f(a)/n \r\n\r\n3) f(a(m/n)) = f(a)(m/n) \r\n where m and n are elements of N", "Solution_6": "[quote=\"ken_himura1984\"][quote=\"TripleM\"]The formal definition is continuous is that for any a, the limit of f(x) as x goes to a is f(a). If you don't need to do it formally, interesting_move has it right.[/quote]\n\nI don't understand what you mean by this sentences. Can you simplify it?\n\nAnd also, can U please help me hoe to prove the function below?\n\n how to prove \n\n1) f(an) = nf(a) \n\n2) f(a/n) = f(a)/n \n\n3) f(a(m/n)) = f(a)(m/n) \n where m and n are elements of N[/quote]\n\nHi, \n\nFirst understand that f(x) is a general representation of any relation between the variable x and another variable y. The former is called an independent variable and the latter, the dependent variable. The set of all possible values ,the independent variable assumes ,is said to be the domain while that of the dependent variable is called the range of the function\n\nexamples of function :\n\ny = sin(x)\ny = ax^2 + bx +c\n\netc. \n\nNote, that in the above examples y has been explicitly defined in terms of x.\n\nWe can also have analytically defined functions, that is functions that are not defined by a single relationship over the entire domain of definition .\n\nExamples\n\n\nf(x) = x-1 for x <= 1\n \n = sin(x) for 1 < x <2\n\n = log(x) for 2<= x\n\n\nWith this background ,let's try to answer ur queries.\n\n[quote] \nHow to prove\n\n1) f(an) = nf(a) \n\n2) f(a/n) = f(a)/n \n\n3) f(a(m/n)) = f(a)(m/n) \n where m and n are elements of N[/quote]\r\n\r\nU can prove the above only if the function f(x) is defined in a PARTICULAR way.\r\n\r\nFor example if f(x) = x (the Identity function), u can prove all of the above identities.\r\n\r\nIf f(x) = x for all real values of x,\r\n\r\nf(an) = an = na= nf(a) { since f(a) = a}\r\n\r\nsimilarly f(a/n) = a/n = f(a)/n \r\n\r\nI leave the proof of the third to u .\r\n\r\n\r\n\r\n--------------------------------------------------------------------------------------------\r\n\r\nSince u r new to functions , it's almost certain that u haven't been introduced to the concept of limits of functions. Once you understand the concept of limits, u 'll appreciate the more formal and analytical definition of continuous functions based on Limits. As of now, just visualise any function whose graph can be plotted without any breaks as a continuous function,as defined in some of the earlier replies to your message. \r\n\r\nSome examples of continuous functions could be y=x,y=2,y=sin(x),y=some polynomial in x etc.\r\n\r\nTo give u an idea of Discontinous functions,consider the greatest integer function\r\n\r\ny= [x] where [a] denotes the largest integer not exceeding a\r\n\r\nEx. [3.5] = 3\r\n \r\n [2.3]=2\r\n\r\n [-1.3]= -2\r\n\r\n [ 3] = 3\r\n\r\nSo for any integer [x] = x\r\n\r\nNow let's try to construct the graph of [x].\r\n\r\nlet's take the case of 0<= x <1\r\n\r\nhere [x] = 0 ,i.e., it's constant over the entire range 0 to 1\r\n\r\nsimilarly [x] =1 for 1<= x <2\r\n\r\nand [x] =2 for 2<= x <3\r\n\r\nand so on and so forth.\r\n\r\n\r\nIf you plot these points on a graph sheet, u 'll get a series of steps with breaks at integral values of x. In other words, the function [x] is discontinuous at integers.", "Solution_7": "santosh_blr thanks for your explanation. Now at least what function is all about.\r\n\r\nWe all know surjective function means \"onto\" but there is also a \"into\" function too. What does into means ?", "Solution_8": "[quote=\"ken_himura1984\"]santosh_blr thanks for your explanation. Now at least what function is all about.\n\nWe all know surjective function means \"onto\" but there is also a \"into\" function too. What does into means ?[/quote]\r\n\r\nConsider the association \r\n\r\n f : A-> B ( A= domain, B= Co-domain}\r\n\r\nWhere A = {1,2,3} and B= {4,5}\r\n\r\n Also let the function be a constant one with value 5\r\n\r\nThen f(1) = f(2)=f(3) = 5\r\n\r\nSo the range of f ,denoted by ran(f ), = {5} a SUBSET of B = {4,5}\r\n\r\n[b]Definition [/b]: When the range of a function (In this case {5} )is a PROPER subset of its Co-domain ( in this case, it's B), we say the function is injective\r\n\r\n\r\n\r\n\r\n", "Solution_9": "I got confused the way they solve this question.\r\n\r\n[b]Let R be the real numbers. Find all functions f:R -> R such that for all real numbers x and y,\n f(xf(y)+x) = xy+f(x)[/b]\r\n\r\nThe solution should be: \r\nTaking x = 1, y = -1-f(1) and letting a = f(y)+1, we get \r\n \r\n f(a) = f(f(y)+1) = y+f(1) = -1\r\n\r\nPutting y = a, and letting b = f(0), we get \r\n\r\n b = f(xf(a)+x) = ax+f(x),\r\n\r\nso, f(x) = -ax+b. Putting this into the equation, we have \r\n\r\n a^2xy-abx-ax+b = xy-ax+b,\r\n\r\nEquating coefficients, we get a = +/-1 and b = 0, so f(x) = x or f(x) = x. We can easily check both are solutions.\r\n\r\n\r\nMy question is\r\n1) from the 1st step, why (what is the purpose) we let y = -1-f(1) and a = f(y)+1\r\n\r\n2) cAN U help me to get the equation \" a^2xy-abx-ax+b = xy-ax+b,\"\r\n by substitution. I can't seem to get that equation for a long period.", "Solution_10": "I'll try to explain the answer in a way that you are likely to actually come up with as you solve it - remember, when writing down the solution, you don't always write it down in the same order as you tried when solving it.\r\n\r\n[quote=\"ken_himura1984\"]I got confused the way they solve this question.\n\nMy question is\n1) from the 1st step, why (what is the purpose) we let y = -1-f(1) and a = f(y)+1\n\n2) cAN U help me to get the equation \" a^2xy-abx-ax+b = xy-ax+b,\"\n by substitution. I can't seem to get that equation for a long period.[/quote]\r\n\r\nThe question says that equation f(xf(y)+x) = xy+f(x) is true for *any* x and y. You can substitute any x and y you want, and that equation will be true. Usually when trying to solve these equations, you find a particular substitution that will make things a lot easier - which often involves making part of the equation 0 to cancel things out.\r\n\r\nIn this case you could think: It would be nice if we could have most of the left hand side cancel out. If we can find some y with f(y) = -1, then we have almost solved the problem - since then for that fixed y we have f(0) = xy + f(x), f(x) = f(0) - xy and that gives us the only possible solution. \r\n\r\nNow that we know we should look to see if f(y) = -1 for some y (which may not be possible, but its a good start). Now we start playing around with some values for x or y. Lets put x=1. \r\nWe know that f(f(y) + 1) = y + f(1) for all y. Hey - now we can choose y = 1-f(1) and we have f(something) = -1, which is exactly what we want! Now all you have to do is put all this information together.\r\n\r\nWe call a that value with f(a) = -1, and we call f(0) = b. From the first step we did, we have f(x) = b - ax.\r\n\r\nf(xf(y)+x) = xy+f(x)\r\nb - a(xf(y) + x) = xy + (b-ax)\r\nb - a(x(b-ay) + x) = xy + b - ax\r\nb - abx + xya^2 - ax = xy + b - ax\r\n-abx + xya^2 = xy\r\n\r\nRemember this is true for all x and y - so the coefficient of y must be the same on each side. So xa^2 = x, so a^2 = 1. Then we have -abx = 0, so ab = 0 (we can divide by x since this is true for EVERY x), so b = 0.\r\n\r\nSo f(x) = x or f(x) = -x. They both work." } { "Tag": [ "linear algebra", "matrix", "number theory", "greatest common divisor", "calculus", "integration", "linear algebra unsolved" ], "Problem": "Are you sure about the wording ? Try the diagonal matrix with $x$ on its main diagonal such that $x^3=x+1$ ?", "Solution_1": "Yes, why? You are trying to say that the problem is wrong? :huh:", "Solution_2": "Consider $a$ a real number such that $a^3=a+1$ with a calculator you have\r\n\r\n$a\\sim 1.3247...$ \r\n\r\nTake $A=diag(a,a,a,a)$ diagonal matrix we have $A^3=A+I$ \r\n\r\nBut $det(A^4-I)=(a^4-1)^4\\sim ((1.3247)^4-1)^4\\sim 18.697 \\neq 1$", "Solution_3": "Sorry, I have made a stupid mistake. It's not $M_{4}(\\mathbb{R})$ it is $M_{4}(\\mathbb{Z})$.\r\nI must have been drunk when I typed this problem. :oops:\r\n\r\nP.S. I have edited the initial problem. :)", "Solution_4": "A is invertble \r\n\r\n\r\ngcd of $x^3-x-1$ and $x^4-1$ is 1 \r\n\r\n$A^4-I$ is invertible since its entries are in Z\r\n\r\n$det(A^4-I)=\\pm1$", "Solution_5": "This is a Romanian forum (ie Romanian speakers!). Moved to the College Playground.", "Solution_6": "Let $A\\in M_{4}(\\mathbb{Z})$ such that $A^{3}=A+I_{4}.$ Prove that $\\det(A^{4}-I_{4})=1.$", "Solution_7": "you know, there are a lot of invertible matrices, with integral entries, such that $|det(A)| \\neq 1$ ..", "Solution_8": "[quote=\"alekk\"]you know, there are a lot of invertible matrices, with integral entries, such that $|det(A)| \\neq 1$ ..[/quote]Not invertible in $\\mathcal{M}_n(\\mathbb{Z})$ :P", "Solution_9": "Well, that didn't work very well. I was trying to merge two topics. For what it's worth, Moubinool's post #7 should be at the top - that's the problem statement." } { "Tag": [ "quadratics", "number theory unsolved", "number theory" ], "Problem": "Problem)\r\nProve that for any arbitrary positive integer $ n$ which is not the square of an integer, \r\nthere exists a prime number $ p$ such that $ n$ is a quadratic nonresidue modulo p. \r\n\r\n\r\n\r\nComment) It's just my quess, so I don't know if it is true.\r\nAlso I guess that it will need Dirichlet's theorem. If you can, please don't use Dirichlet's theorem", "Solution_1": "i think that there is a generalisation,\r\nfor $ m>1$\r\n$ \\forall p\\in\\mathbb{P},\\exists k\\in\\{0,1,...,p\\minus{}1\\}: \\ n\\equiv k^{m}(mod\\ p) \\Longrightarrow \\exists A\\in\\mathbb{N}: \\ n\\equal{}A^m$", "Solution_2": "how can we prove it?\r\n\r\nIt must be generalization of IMO shortlist 2007 N2, maybe?", "Solution_3": "for your exemple $ m\\equal{}2$\r\ni think that so for $ n\\in\\mathbb{N}$ $ \\exists h: \\ h^2x\\plus{}1$ have a primp divisor p of that form $ p\\equal{}4k\\plus{}3$ (i am not sur)\r\nso $ h^2x$ is not a quadratic nonresidue modulo $ p$ ==> $ x$ is not a quadratic nonresidue modulo $ p$", "Solution_4": "See http://www.mathlinks.ro/viewtopic.php?t=64322 for a proof for the case $ m \\equal{} 2$ without Dirichlet's theorem.\r\n\r\nThe generalisation is wrong, e.g. $ 16$ is a $ 8$-th power modulo all primes.\r\nBut with some adaption, it is still correct:\r\nLet $ m$ be a natural number and $ a$ some integer. Assume that $ a$ is a $ m$-th modulo all but finitely many primes.\r\na) if $ 8 \\nmid m$, then $ a$ is the $ m$-th power of an integer.\r\nb) if $ 8|m$, then $ a$ is the $ \\frac m2$-th power of an integer.\r\n\r\nThis is sometimes called Grunwalds theorem and there is some funny story around this: Grunwald even \"proved\" (in 1933) what aviateurpilot conjectured: if something is a $ m$-th power modulo all primes, then it's a $ m$-th power. This was \"reproven\" 1942 by Whaples, and it needed six more years untill Wang found the easy counterexample I gave above and has shown the correct version.\r\n\r\nSee some texts on Class Field Theory to see some proofs." } { "Tag": [], "Problem": "Source: AHSME 1973 #14\r\n\r\nEach valve A, B, and C, when open, releases water into a tank at its own constant rate. With all three valves open the tank fills in 1 hour, with only valves A and C open it takes 1.5 hours, and with only valves B and C open it takes 2 hours. The number of hours required with only valves A and B open is\r\n$\\text{(A)} \\ 1.1 \\qquad \\text{(B)} \\ 1.15 \\qquad \\text{(C)} \\ 1.2 \\qquad \\text{(D)} \\ 1.25 \\qquad \\text{(E)} \\ 1.75$\r\n\r\n[hide=\"Solution\"]\n$(C)$\nLet $A$, $B$, and $C$ denote each respective valve's rate.\nThus,\n\\[ A+B+C = 1 \\]\n\\[ (A+C)1.5 = 1 \\implies A+C = 2/3 \\]\n\\[ (B+C)2 = 1 \\implies B+C = 1/2 \\]\nWe can quickly substitute each of the second and third equations into the first to find that $A = 1/2$ and $B = 1/3$, so together they will take $1/(1/2 + 1/3) = 6/5 = 1.2$ hours.\n[/hide]", "Solution_1": "Classic question... \r\n\r\n[hide]Let the amount of time it takes to fill a tank be $a$, $b$, and $c$ for valves A, B, and C respectively. Let x be the amount of time it takes for valves A and B to fill the tank. We then have \n\n1/a+1/c=1/1.5=2/3\n1/b+1/c=1/2\n1/a+1/b=1/x \n\nAdd: 2(1/a+1/b+1/c)=7/6+1/x \n\nsince 1/a+1/b+1/c=1/1=1, \n\n1/x+7/6=2, and x=6/5, or 1.2 hours. [/hide]" } { "Tag": [ "logarithms", "combinatorics unsolved", "combinatorics" ], "Problem": "We have n coins, and at least one of them is bad. The bad coins weigh the same, and so do the good coins. The bad coins, however, are lighter. \r\n\r\nFind the number of bad coins in O((logn) 2 ) using a balance scale", "Solution_1": "how much times should we use the balancE", "Solution_2": "As stated, $ C \\cdot \\log_{2}(n)$ times at most, $ C$ constant." } { "Tag": [ "LaTeX" ], "Problem": "\u039a\u03b1\u03bb\u03b7\u03c3\u03c0\u03ad\u03c1\u03b1 \u03c3\u03b5 \u03cc\u03bb\u03bf\u03c5\u03c2!!\r\n\u039c\u03af\u03b1 \u03b2\u03bf\u03ae\u03b8\u03b5\u03b9\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7....\r\n\r\n[img]http://img142.imageshack.us/img142/9162/92231888xt2.jpg[/img]\r\n\r\n\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03c0\u03bf\u03bb\u03cd", "Solution_1": "\u0391\u03bd \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9, \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03b5 Latex \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b4\u03b9\u03b1\u03b2\u03ac\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c0\u03b9\u03bf \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1.\r\n\r\n1. \r\n\u0393\u03b9\u03b1 $ n \\equal{} 2$ \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b4\u03c5\u03bd\u03b1\u03c4\u03ac \u03c5\u03c0\u03bf\u03c3\u03cd\u03bd\u03bf\u03bb\u03b1 \u03c0\u03bb\u03ae\u03b8\u03bf\u03c5\u03c2 $ 1 \\equal{} \\frac {2(2 \\minus{} 1)}{2}$\r\n\u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03b3\u03b9\u03b1 $ n \\equal{} k$ \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $ \\frac {k(k \\minus{} 1)}{2}$ \u03c5\u03c0\u03bf\u03c3\u03cd\u03bd\u03bf\u03bb\u03b1\r\n\u0393\u03b9\u03b1 $ n \\equal{} k \\plus{} 1$ \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 k \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03b1 \u03c5\u03c0\u03bf\u03c3\u03cd\u03bd\u03bf\u03bb\u03b1 \u03b1\u03c0\u03bf $ n \\equal{} k$ (\u03cc\u03bb\u03b1 \u03c4\u03b1 \u03c5\u03c0\u03bf\u03c3\u03cd\u03bd\u03bf\u03bb\u03b1 \u03c0\u03bf\u03c5 \u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03af\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03ad\u03bd\u03b1\u03bd \u03b1\u03c0\u03cc \u03c4\u03bf\u03c5\u03c2 \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf\u03c5\u03c2 k \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd\u03c2 \u03c0\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03c4\u03bf\u03bd \u03ba\u03b1\u03b9\u03bd\u03bf\u03cd\u03c1\u03b9\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc) \u03ac\u03c1\u03b1 \u03c4\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c4\u03c9\u03bd \u03c5\u03c0\u03bf\u03c3\u03c5\u03bd\u03cc\u03bb\u03c9\u03bd \u03b5\u03af\u03bd\u03b1\u03b9:\r\n$ \\frac {k(k \\minus{} 1)}{2} \\plus{} k \\equal{} \\frac {k(k \\plus{} 1)}{2}$\r\n\r\n2.\u0391.\r\n\u0395\u03b4\u03ce \u03b8\u03b1 \u03b5\u03c6\u03b1\u03c1\u03bc\u03bf\u03c3\u03b8\u03b5\u03af \u03bf \u03b1\u03c1\u03b9\u03c3\u03bc\u03cc\u03c2 \u03b3\u03b9\u03b1 2 \u03c4\u03cd\u03c0\u03bf\u03c5\u03c2 \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03ac \u03b8\u03b1 \u03b3\u03b5\u03bd\u03b9\u03ba\u03b5\u03c5\u03c4\u03b5\u03af \u03b3\u03b9\u03b1 n\r\n(\u0395\u03c6\u03b1\u03c1\u03bc\u03cc\u03b6\u03bf\u03c5\u03bc\u03b5 \u03b5\u03c6\u03b1\u03c1\u03bc\u03bf\u03b3\u03ae \u03c3\u03c4\u03bf\u03c5\u03c2 \u03c6\u03c5\u03c3\u03b9\u03ba\u03bf\u03cd\u03c2 \u03ba\u03b1\u03b9 \u03cc\u03c7\u03b9 \u03c3\u03c4\u03b7 \u03b4\u03bf\u03bc\u03ae \u03c4\u03c9\u03bd \u03c4\u03cd\u03c0\u03c9\u03bd, \u03b3\u03b9\u03b1\u03c4\u03af \u03b7 \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03b1 \u03b4\u03b5 \u03b4\u03af\u03b4\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03b5\u03ba\u03c6\u03ce\u03bd\u03b7\u03c3\u03b7)\r\n\r\n2.\u0392.\r\n\u0395\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03bb\u03ae \u03b5\u03c6\u03b1\u03c1\u03bc\u03bf\u03b3\u03ae \u03c4\u03bf\u03c5 2.\u0391.", "Solution_2": "\u03a3\u03b5 \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03c0\u03bf\u03bb\u03cd \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b1\u03c0\u03ac\u03bd\u03c4\u03b7\u03c3\u03b7 \u03ba\u03b1\u03b9\r\n\u03c3\u03c5\u03b3\u03bd\u03ce\u03bc\u03b7 \u03b3\u03b9\u03b1 \u03b1\u03c5\u03c4\u03ae \u03c4\u03b7 \u03bc\u03bf\u03c1\u03c6\u03ae!\r\n\u0391\u03bb\u03bb\u03ac \u03b1\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03bc\u03bf\u03c5 \u03c0\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b1 \u03ac\u03bb\u03bb\u03b1\r\n2 \u03b5\u03c1\u03c9\u03c4\u03ae\u03bc\u03b1\u03c4\u03b1 \u03bb\u03af\u03b3\u03bf \u03c0\u03b9\u03bf \u03b1\u03bd\u03b1\u03bb\u03c5\u03c4\u03b9\u03ba\u03ac \u03cc\u03c0\u03c9\u03c2 \u03c4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf :lol: .\r\n\u0391\u03bd \u03c0\u03ac\u03bb\u03b9 \u03cc\u03c7\u03b9 \u03b4\u03b5\u03bd \u03c0\u03b5\u03b9\u03c1\u03ac\u03b6\u03b5\u03b9.\u03a3\u03b5 \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03ba\u03b1\u03b9 \u03c0\u03ac\u03bb\u03b9.", "Solution_3": "\u0394\u03b5\u03bd \u03be\u03ad\u03c1\u03c9 \u03c0\u03ce\u03c2 \u03bd\u03b1 \u03b5\u03b9\u03c3\u03ac\u03b3\u03c9 \u03c4\u03bf \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03bf \u03c3\u03c5\u03bc\u03b2\u03bf\u03bb\u03b9\u03c3\u03bc\u03cc.", "Solution_4": "[quote=\"maniopas\"]\u0394\u03b5\u03bd \u03be\u03ad\u03c1\u03c9 \u03c0\u03ce\u03c2 \u03bd\u03b1 \u03b5\u03b9\u03c3\u03ac\u03b3\u03c9 \u03c4\u03bf \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03bf \u03c3\u03c5\u03bc\u03b2\u03bf\u03bb\u03b9\u03c3\u03bc\u03cc.[/quote]\r\n\u039c\u03ae\u03c0\u03c9\u03c2 \u03cc\u03bc\u03c9\u03c2 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03bc\u03b1\u03c2 \u03b4\u03ce\u03c3\u03b5\u03b9\u03c2 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03b7 \u03b2\u03bf\u03ae\u03b8\u03b5\u03b9\u03b1;", "Solution_5": "2.\u0391.\r\n\u0395\u03bb\u03bb\u03b5\u03af\u03c8\u03b7 \u03c3\u03c5\u03bc\u03b2\u03bf\u03bb\u03b9\u03c3\u03bc\u03bf\u03cd \u03b8\u03b1 \u03b3\u03c1\u03ac\u03c8\u03c9 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf \u03b3\u03b9\u03b1 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c9\u03c2 \u03b5\u03be\u03ae\u03c2:\r\n$ (Y_1 and ... and Y_n) \\equal{} not(notY_1 or ... or notY_n)$\r\n\r\n\u0391\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7:\r\n\u0393\u03b9\u03b1 $ n \\equal{} 1$ \u03b5\u03af\u03bd\u03b1\u03b9 $ Y_1 \\equal{} not(notY_1)$, \u03c0\u03bf\u03c5 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9\r\n\u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1 $ n \\equal{} k$ \u03c4\u03cc\u03c4\u03b5\r\n\u03b3\u03b9\u03b1 $ n \\equal{} k \\plus{} 1$ \u03b5\u03af\u03bd\u03b1\u03b9 $ (Y_1 and ... and Y_{k \\plus{} 1}) \\equal{} not(notY_1 or ... or notY_{k \\plus{} 1}) < \\equal{} >$\r\n$ (Y_1 and ... and Y_k and Y_{k \\plus{} 1}) \\equal{} not(notY_1 or ... or notY_k ornotY_{k \\plus{} 1}) < \\equal{} >$\r\n$ (Y_1 and ... and Y_k) and (Y_{k \\plus{} 1}) \\equal{} not(notY_1 or ... or notY_k ornotY_{k \\plus{} 1}) < \\equal{} >$\r\n$ not(notY_1 or ... or notY_n) and not(notY_{k \\plus{} 1}) \\equal{} not(notY_1 or ... or notY_k or notY_{k \\plus{} 1}) < \\equal{} >$\r\n$ not[ not(notY_1 or ... or notY_n) and not(notY_{k \\plus{} 1}) ] \\equal{} not[not(notY_1 or ... or notY_k or notY_{k \\plus{} 1})] < \\equal{} >$\r\n$ (notY_1 or ... or notY_n) or (notY_{k \\plus{} 1}) \\equal{} (notY_1 or ... or notY_k or notY_{k \\plus{} 1})$, \u03c0\u03bf\u03c5 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9\r\n\u03ac\u03c1\u03b1 \u03b1\u03bd \u03b7 \u03c0\u03b1\u03c1\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1 $ n \\equal{} k$,\u03b8\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 $ n \\equal{} k \\plus{} 1$\r\n\r\n\u03a3\u03b7\u03bc\u03b5\u03af\u03c9\u03c3\u03b7:\r\n\u0393\u03b9\u03b1 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03b8\u03b5\u03ce\u03c1\u03b7\u03c3\u03b1 \u03c9\u03c2 \u03b4\u03b5\u03b4\u03bf\u03bc\u03ad\u03bd\u03b7 \u03c4\u03b7\u03bd \u03c0\u03c1\u03cc\u03c4\u03b1\u03c3\u03b7 \u03cc\u03c4\u03b9 $ not(X and Y) \\equal{} notX or notY$ \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd \u03b1\u03bd\u03c4\u03b9\u03bc\u03b5\u03c4\u03b1\u03b8\u03b5\u03c4\u03b9\u03ba\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c4\u03b7\u03c2 no\u03c4.\r\n\r\n\u03a4\u03bf 2.\u0392 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03bb\u03ae \u03b5\u03c6\u03b1\u03c1\u03bc\u03bf\u03b3\u03ae \u03c4\u03bf\u03c5 2.\u0391 (\u03bc\u03b5 \u03b1\u03bd\u03c4\u03b9\u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7)\r\n\u0394\u03b5\u03bd \u03c4\u03bf \u03b3\u03c1\u03ac\u03c6\u03c9 \u03b3\u03b9\u03b1\u03c4\u03af \u03bc\u03bf\u03c5 \"\u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03bf \u03c0\u03ac\u03c4\u03bf\u03c2\" \u03bc\u03b5 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03c3\u03c5\u03bc\u03b2\u03bf\u03bb\u03b9\u03b1\u03c3\u03bc\u03cc.", "Solution_6": "[quote=\"maniopas\"]2.\u0391.\n\u0395\u03bb\u03bb\u03b5\u03af\u03c8\u03b7 \u03c3\u03c5\u03bc\u03b2\u03bf\u03bb\u03b9\u03c3\u03bc\u03bf\u03cd \u03b8\u03b1 \u03b3\u03c1\u03ac\u03c8\u03c9 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf \u03b3\u03b9\u03b1 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c9\u03c2 \u03b5\u03be\u03ae\u03c2:\n$ (Y_1 and ... and Y_n) \\equal{} not(notY_1 or ... or notY_n)$\n\n\u0391\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7:\n\u0393\u03b9\u03b1 $ n \\equal{} 1$ \u03b5\u03af\u03bd\u03b1\u03b9 $ Y_1 \\equal{} not(notY_1)$, \u03c0\u03bf\u03c5 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9\n\u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1 $ n \\equal{} k$ \u03c4\u03cc\u03c4\u03b5\n\u03b3\u03b9\u03b1 $ n \\equal{} k \\plus{} 1$ \u03b5\u03af\u03bd\u03b1\u03b9 $ (Y_1 and ... and Y_{k \\plus{} 1}) \\equal{} not(notY_1 or ... or notY_{k \\plus{} 1}) < \\equal{} >$\n$ (Y_1 and ... and Y_k and Y_{k \\plus{} 1}) \\equal{} not(notY_1 or ... or notY_k ornotY_{k \\plus{} 1}) < \\equal{} >$\n$ (Y_1 and ... and Y_k) and (Y_{k \\plus{} 1}) \\equal{} not(notY_1 or ... or notY_k ornotY_{k \\plus{} 1}) < \\equal{} >$\n$ not(notY_1 or ... or notY_n) and not(notY_{k \\plus{} 1}) \\equal{} not(notY_1 or ... or notY_k or notY_{k \\plus{} 1}) < \\equal{} >$\n$ not[ not(notY_1 or ... or notY_n) and not(notY_{k \\plus{} 1}) ] \\equal{} not[not(notY_1 or ... or notY_k or notY_{k \\plus{} 1})] < \\equal{} >$\n$ (notY_1 or ... or notY_n) or (notY_{k \\plus{} 1}) \\equal{} (notY_1 or ... or notY_k or notY_{k \\plus{} 1})$, \u03c0\u03bf\u03c5 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9\n\u03ac\u03c1\u03b1 \u03b1\u03bd \u03b7 \u03c0\u03b1\u03c1\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1 $ n \\equal{} k$,\u03b8\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 $ n \\equal{} k \\plus{} 1$\n\n\u03a3\u03b7\u03bc\u03b5\u03af\u03c9\u03c3\u03b7:\n\u0393\u03b9\u03b1 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03b8\u03b5\u03ce\u03c1\u03b7\u03c3\u03b1 \u03c9\u03c2 \u03b4\u03b5\u03b4\u03bf\u03bc\u03ad\u03bd\u03b7 \u03c4\u03b7\u03bd \u03c0\u03c1\u03cc\u03c4\u03b1\u03c3\u03b7 \u03cc\u03c4\u03b9 $ not(X and Y) \\equal{} notX or notY$ \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd \u03b1\u03bd\u03c4\u03b9\u03bc\u03b5\u03c4\u03b1\u03b8\u03b5\u03c4\u03b9\u03ba\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c4\u03b7\u03c2 no\u03c4.\n\n\u03a4\u03bf 2.\u0392 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03bb\u03ae \u03b5\u03c6\u03b1\u03c1\u03bc\u03bf\u03b3\u03ae \u03c4\u03bf\u03c5 2.\u0391 (\u03bc\u03b5 \u03b1\u03bd\u03c4\u03b9\u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7)\n\u0394\u03b5\u03bd \u03c4\u03bf \u03b3\u03c1\u03ac\u03c6\u03c9 \u03b3\u03b9\u03b1\u03c4\u03af \u03bc\u03bf\u03c5 \"\u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03bf \u03c0\u03ac\u03c4\u03bf\u03c2\" \u03bc\u03b5 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03c3\u03c5\u03bc\u03b2\u03bf\u03bb\u03b9\u03b1\u03c3\u03bc\u03cc.[/quote]\r\n\u03a3' \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03c0\u03bf\u03bb\u03cd \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b2\u03bf\u03ae\u03b8\u03b5\u03b9\u03ac \u03c3\u03bf\u03c5" } { "Tag": [], "Problem": "Bill has a bucket of [b]R[/b]ed, [b]W[/b]hite and [b]B[/b]lue marbles.\r\n\r\n[b]R[/b]ed marbles = 2 more than [b]B[/b]lue marbles\r\n\r\n[b]W[/b]hite marbles = twice the [b]R[/b]ed marbles\r\n\r\nWhich could be the total # of marbles? It's multiple choice: 24, 25, 26, 27, 28\r\n\r\nThe answer is 26\r\n\r\nThere are 7 Red, 5 Blue and 14 White\r\n\r\nR = 2 + B\r\n\r\nW = 2R\r\n\r\nB = R - 2\r\n\r\n26 = (R-2) + (2R) + 2 + B\r\n\r\n26 = (7 - 2) + (2*7) + 2 + 5\r\n\r\n26 = 5 + 14 + 2 + 5", "Solution_1": "what are you asking us?\r\ni have verified your answer by the way...", "Solution_2": "[quote=\"stevenmeow\"]what are you asking us?\ni have verified your answer by the way...[/quote]\r\n\r\nI guess that I'm asking what the equation(s) is/are.\r\n\r\nI got the answer by plugging in the possibilities and trial and error.\r\n\r\nI'd like to know how to setup the problem in case I don't have multiple choice in the future.", "Solution_3": "ok ill start similarly to you\r\n\r\nred=r, blue=b, white=w\r\nr=2+b\r\nw=2r=2(2+b)=4+2b\r\nadd all of them together\r\n\r\ntotal=b+r+w\r\n=b+(2+b)+(4+2b)\r\n=4b+6\r\nthe total must be in the form 4b+6\r\nwhere b is a whole number\r\nin other words, the number b must be in the series\r\n6, 10, 14, 18, 22, 26, 30...\r\nthat adds 4 each term", "Solution_4": "[quote=\"stevenmeow\"]ok ill start similarly to you\n\nred=r, blue=b, white=w\nr=2+b\nw=2r=2(2+b)=4+2b\nadd all of them together\n\ntotal=b+r+w\n=b+(2+b)+(4+2b)\n=4b+6\nthe total must be in the form 4b+6\nwhere b is a whole number\nin other words, the number b must be in the series\n6, 10, 14, 18, 22, 26, 30...\nthat adds 4 each term[/quote]\r\n\r\nThank you\r\n\r\n:)" } { "Tag": [ "geometry", "parallelogram", "geometry unsolved" ], "Problem": "Consider 5 points A, B, C, D, and E are drawn so that ABCD and BCED are parallelogram and cyclic quadrilateral. Let l be line passing through A. Suppose l intersect segment CD and line BC at F and G so that EF = EG = EC. Prove that l bisects anlge DAB.", "Solution_1": "http://www.mathlinks.ro/viewtopic.php?p=893744#893744" } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "Two Points $C$ and $D$ with constant distance move on a semicircle with diameter $AB$. Lines $AC$ and $BD$ intersect at $E$, lines $AD$ and $BC$ intersect at $F$.\r\nProve that the quadrilateral $AEBF$ has constant area!", "Solution_1": "Ok, I'll post my solution:\r\n\r\n[hide]\n\nObviously $E$ is the orthocenter of the triangle $ABF$ so $EF\\perp AB$. Hence the claim that $AEBF$ has constant area is equivalent to $EF$ being constant. \nNow obviously the quadrilateral $CEDF$ is cyclic and the diameter of its circumcircle is $EF$. But $CD$ is fixed, hence $\\angle CAD$ is fixed and hence $\\angle CFD$ and $\\angle CED$ is fixed. Thus $EF$ is fixed.\nqed\n\n[/hide]" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Let $X$ be a fixed point on bisector $AA'$ of $\\triangle ABC$. $B'=BX\\cap AC,C'=CX\\cap AB, P=A'B'\\cap CC',Q=A'C'\\cap BB'$. Prove that $\\angle PAC=\\angle QAB$", "Solution_1": "[quote=\"malinger\"]Let $X$ be a fixed point on bisector $AA'$ of $\\triangle ABC$. $B'=BX\\cap AC,C'=CX\\cap AB, P=A'B'\\cap CC',Q=A'C'\\cap BB'$. Prove that $\\angle PAC=\\angle QAB$[/quote]\r\n\r\nOne direction of http://www.mathlinks.ro/Forum/viewtopic.php?t=21297 .\r\n\r\n darij" } { "Tag": [ "calculus", "integration", "trigonometry", "logarithms", "calculus computations" ], "Problem": "Calculate:\r\n1. $ I_1 \\equal{} \\int_{\\frac { \\minus{} \\pi}{2}}^{\\frac {\\pi}{2}}\\frac {1 \\plus{} sinx}{1 \\plus{} cosx}e^{x} dx$\r\n\r\n2. $ I_2 \\equal{} \\int_{0}^{\\frac {\\pi}{3}}ln(1 \\plus{} \\sqrt {3} tgx) dx$", "Solution_1": "hello, a simple calculation shows that \r\n$ e^x\\frac {1 \\plus{} \\sin(x)}{1 \\plus{} \\cos(x)} \\equal{} \\frac {1}{2}e^x\\left(\\tan\\left(\\frac {1}{2}x\\right) \\plus{} 1\\right)^2$.It follows\r\n$ \\int \\frac {1}{2}e^x\\left(\\tan\\left(\\frac {1}{2}x\\right) \\plus{} 1\\right)^2dx \\equal{} e^x\\tan\\left(\\frac {1}{2}x\\right) \\plus{} C$. Using this, so your integral gives\r\n$ e^{\\frac {\\pi}{2}} \\plus{} e^{ \\minus{} \\frac {\\pi}{2}}$.\r\nSonnhard.", "Solution_2": "[quote=\"Dr Sonnhard Graubner\"]hello, a simple calculation shows that \n$ e^x\\frac {1 \\plus{} \\sin(x)}{1 \\plus{} \\cos(x)} \\equal{} \\frac {1}{2}\\left(\\tan\\left(\\frac {1}{2}x\\right) \\plus{} 1\\right)^2$.It follows\n$ \\int \\frac {1}{2}\\left(\\tan\\left(\\frac {1}{2}x\\right) \\plus{} 1\\right)^2dx \\equal{} e^x\\tan\\left(\\frac {1}{2}x\\right) \\plus{} C$. Using this, so your integral gives\n$ e^{\\frac {\\pi}{2}} \\plus{} e^{ \\minus{} \\frac {\\pi}{2}}$.\nSonnhard.[/quote]\r\n\r\nI don't see how $ e^x\\frac {1 \\plus{} \\sin(x)}{1 \\plus{} \\cos(x)} \\equal{} \\frac {1}{2}\\left(\\tan\\left(\\frac {1}{2}x\\right) \\plus{} 1\\right)^2$ and $ \\int \\frac {1}{2}\\left(\\tan\\left(\\frac {1}{2}x\\right) \\plus{} 1\\right)^2dx \\equal{} e^x\\tan\\left(\\frac {1}{2}x\\right) \\plus{} C$.", "Solution_3": "Hello, sorry, it was a typo it must be:\r\n$ e^x\\frac{1\\plus{}\\sin(x)}{1\\plus{}\\cos(x)}\\equal{}\\frac{1}{2}e^x\\left(\\tan\\left(\\frac{1}{2}x\\right)\\plus{}1\\right)^2$\r\nand\r\n$ \\int\\frac{1}{2}e^x\\left(\\tan\\left(\\frac{1}{2}x\\right)\\plus{}1\\right)^2dx\\equal{}e^x\\tan\\left(\\frac{1}{2}x\\right)\\plus{}C$.\r\nSonnhard.", "Solution_4": "OK, but may I know how you come up with the identity $ e^x\\frac{1\\plus{}\\sin(x)}{1\\plus{}\\cos(x)}\\equal{}\\frac{1}{2}e^x\\left(\\tan\\left(\\frac{1}{2}x\\right)\\plus{}1\\right)^2$ (from sketch)? Thanks.", "Solution_5": "hello, this is easy,\r\n$ \\frac{1}{2}\\left(\\tan\\left(\\frac{1}{2}x\\right)\\plus{}1\\right)^2\\equal{}\r\n\\frac{1}{2}\\left(\\frac{\\sin\\left(\\frac{1}{2}x\\right)\\plus{}\\cos\\left(\\frac{1}{2}x\\right)}{\\cos\\left(\\frac{1}{2}x\\right)}\\right)^2$.\r\nSonnhard.", "Solution_6": "[quote=\"borislav_mirchev\"]Calculate:\n2. $ I_2 = \\int_{0}^{\\frac {\\pi}{3}}ln(1 + \\sqrt {3} tgx) dx$[/quote]\r\n$ \\frac {\\pi}{3}ln2$\r\nSolution...\r\n[hide]\n${ I_2 = \\int_{0}^{\\frac {\\pi}{3}}ln(1 + \\sqrt {3} \\frac {sinx}{cosx}) dx = \\int_{0}^{\\frac {\\pi}{3}}ln(\\frac {cosx + \\sqrt {3}sinx}{cosx}) dx = \\int_{0}^{\\frac {\\pi}{3}}ln(2cos( x - \\pi/3)) dx - \\int_{0}^{\\frac {\\pi}{3}}ln(cosx) dx = \\int_{\\frac { - \\pi}{3}}^{0}}ln(cosx) dx - \\int_{0}^{\\frac {\\pi}{3}^}ln(cosx) dx + \\int_{0}^{\\frac {\\pi}{3}}(ln 2) dx = \\int_{0}^{\\frac {\\pi}{3}}(ln 2) dx = \\frac {\\pi}{3}ln2$[/hide]", "Solution_7": "Kirill, are you sure about your answer?", "Solution_8": "Kirill's answer seems to be correct:\r\n\r\n$ I_2\\equal{}\\int_{0}^{\\frac {\\pi}{3}}\\log(1 \\plus{} \\sqrt {3} \\tan x) dx\\equal{}$\r\n\r\n$ \\equal{}\\int_{0}^{\\frac {\\pi}{3}}\\log(1 \\plus{} \\sqrt {3} \\tan \\left(\\frac{\\pi}{3}\\minus{}x\\right) dx\\equal{}$\r\n\r\n$ \\equal{}\\int_{0}^{\\frac{\\pi}{3}}\\log\\left[1\\plus{}\\frac{3 \\cos x\\minus{}\\sqrt{3} \\sin x}{cos x\\plus{}\\sqrt{3} \\text sin x}\\right]dx\\equal{}$\r\n\r\n$ \\equal{}\\int_{0}^{\\frac{\\pi}{3}}\\log\\left[\\frac{4 \\cos x}{\\cos x\\plus{}\\sqrt{3} \\sin x}\\right]\\equal{}$\r\n\r\n$ \\equal{}\\minus{}I_2\\plus{}\\int_{0}^{\\frac{\\pi}{3}}2\\log 2 dx$\r\n\r\nHence, $ I_2\\equal{}\\frac{\\pi}{3}\\log 2$.\r\n\r\n($ \\log\\equal{}\\ln$)", "Solution_9": "It is OK. I had a mistake." } { "Tag": [ "integration", "real analysis", "real analysis unsolved" ], "Problem": "Prove that $ \\boxed {\\int_0^{\\infty}\\frac {1}{\\left(x^4 \\plus{} 1^2\\right)\\left(x^4 \\plus{} 2^2\\right)\\ldots\\left(x^4 \\plus{} n^2\\right)}\\ \\mathrm {dx} \\equal{} \\frac {\\pi}{2}\\cdot\\sum_{k \\equal{} 1}^n\\frac {( \\minus{} 1)^{k \\minus{} 1}\\cdot\\sqrt {2k}}{(n \\minus{} k)!\\cdot(n \\plus{} k)!}\\ }$ .\r\nSee http://www.mathlinks.ro/viewtopic.php?t=189621", "Solution_1": "The residue theorem yields the result almost immediately but it would be interesting to see if someone could come up with a real variable proof." } { "Tag": [ "algebra", "polynomial", "Vieta", "algebra unsolved" ], "Problem": "I'm looking for the most beautiful proof of identity (if there exist any beautiful):\r\n\r\n$ a^3\\plus{}b^3\\plus{}c^3\\minus{}3abc\\equal{}(a\\plus{}b\\plus{}c)(a^2\\plus{}b^2\\plus{}c^2\\minus{}ab\\minus{}ac\\minus{}bc)$\r\n\r\nThanks in advance! :)", "Solution_1": "Compute the determinant of \r\n\r\n$ \\left[ \\begin{array}{ccc} a & b & c \\\\\r\nc & a & b \\\\\r\nb & c & a \\end{array} \\right],$\r\n\r\nfirst by expansion (LHS), and then by computing its eigenvalues (RHS).", "Solution_2": "First proof: Expand the RHS?\r\n\r\nSecond proof:\r\n\r\nLook at the LHS as a polynomial of $ a$. Take $ a\\equal{}\\minus{}(b\\plus{}c)$, then the LHS is 0, so $ a\\plus{}b\\plus{}c$ divides $ P$. As every term on the LHS is of degree 3, we have that $ a^3\\plus{}b^3\\plus{}c^3\\equal{}(a\\plus{}b\\plus{}c)(xa^2\\plus{}xb^2\\plus{}xc^2\\plus{}yab\\plus{}ybc\\plus{}yca)$, the coefficients of the squares are the same because of symmetry. \r\n\r\nNow we have $ a^3\\equal{}xa^3$ if we look just at the $ a^3$ term, so $ x\\equal{}1$. Similarly, $ y\\equal{}\\minus{}1$.", "Solution_3": "Bugi, I've asked this because I don't like expansion proof of this... \r\nThanks for nice proofs. \r\n :)", "Solution_4": "I have answered the proof before, but I can't find it.\r\nThere were at least 4 proofs.\r\nFor example, one of them is using\r\n\\[ a^3 \\plus{} b^3 \\equal{} (a \\plus{} b)^3 \\minus{} 3ab(a \\plus{} b)\\]\r\nor Vieta or Complex number.\r\n\r\n$ \\boxed{(a \\plus{} b \\plus{} c)(a \\plus{} b\\omega \\plus{} c\\omega ^ 2)(a \\plus{} b\\omega ^ 2 \\plus{} c\\omega) \\equal{} a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} 3abc}$ for $ \\omega \\equal{} \\frac { \\minus{} 1\\pm \\sqrt {3}i}{2}$", "Solution_5": "Thanks kunny. Nice proofs.", "Solution_6": "You are welcome. :) \r\n\r\nI have just found the solution [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22252]here[/url]\r\nPlease take a look in #8 & #9 post." } { "Tag": [], "Problem": "What is the sum of outer angles of any concave quadrilateral?360?Any proof?", "Solution_1": "[hide]It is 360 degrees. If the quadrilateral is $ABCD$ and $\\angle BCD>180^\\circ$, then draw in $\\overline{AC}$ to split the quadrilateral into two triangles.\n\nThe sum of the angles of the quadrilateral is the sum of the angles of the triangles, which adds to 360 degrees.[/hide]", "Solution_2": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=397070&highlight=#397070]Once is enough[/url] :!:", "Solution_3": "Sorry,I could not find my first message" } { "Tag": [], "Problem": "One more than the reciprocal of a particular number is $ \\frac{7}{3}$. What is the original number expressed as a common fraction?", "Solution_1": "so $ \\frac73$ is one more, so $ \\frac43$ is the number, reciprocol of a number is $ \\frac43$ so the real number is $ \\frac34$", "Solution_2": "So its just 7/3-3/3=4/3 then we flip it to get 3/4." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "$ \\forall m,n \\in \\mathbb Z^\\plus{}$ , prove that :\r\n\r\n$ \\frac{(3n)!\\cdot(3m)!}{(m!)\\cdot (n!)\\cdot (m\\plus{}n)!} \\in \\mathbb Z$.\r\n\r\n :wink:", "Solution_1": "[quote=\"Hidden Scofield\"]$ \\forall m,n \\in \\mathbb Z^ \\plus{}$ , prove that :\n\n$ \\frac {(3n)!\\cdot(3m)!}{(m!)\\cdot (n!)\\cdot (m \\plus{} n)!} \\in \\mathbb Z$.\n\n :wink:[/quote]\r\n\r\nI will instead show that $ \\frac {(2n)! \\cdot (2m)!}{(m!) \\cdot (n!) \\cdot (m \\plus{} n)!} \\in \\mathbb Z$.\r\n\r\nWe just have to show that $ e_p(2n!) \\plus{} e_p(2m!) \\ge e_p(m!) \\plus{} e_p(n!) \\plus{} e_p((m \\plus{} n)!)$\r\n\r\nWhich is the same as $ \\left[\\frac{2n}{p^r}\\right] \\plus{} \\left[\\frac{2m}{p^r}\\right] \\ge \\left[\\frac{m}{p^r}\\right] \\plus{} \\left[\\frac{n}{p^r}\\right] \\plus{} \\left[\\frac{m \\plus{}n}{p^r}\\right]$ For every prime p and factor r.\r\n\r\nWe'll set $ a \\equal{} \\frac{n}{p^r}$ and $ b \\equal{} \\frac{m}{p^r}$ and then we get:\r\n\r\n$ [2a] \\plus{} [2b] \\ge [a] \\plus{} [b] \\plus{} [a\\plus{}b]$\r\n\r\nIt is easy to see that $ [2a] \\equal{} 2[a] \\plus{} [\\{2a\\}], [2b] \\equal{} 2[b] \\plus{} [\\{2b\\}], [a\\plus{}b] \\equal{} [a] \\plus{} [b] \\plus{} [\\{a\\plus{}b\\}]$\r\nAnd then we just have to show that:\r\n$ [\\{2a\\}] \\plus{} [\\{2b\\}] \\ge [\\{a\\}\\plus{}\\{b\\}]$\r\n$ RHS < 2$.. If $ RHS \\equal{} 0$ it is obviously correct. But if $ RHS \\equal{} 1$ then we have $ {a} \\ge \\frac{1}{2}$ or $ {b} \\ge \\frac{1}{2}$ which means that $ LHS \\ge 1$.\r\n\r\nThus $ \\frac {(2n)! \\cdot (2m)!}{(m!) \\cdot (n!) \\cdot (m \\plus{} n)!} \\in \\mathbb Z$.", "Solution_2": "What is e_p(2n)! ?? :huh:", "Solution_3": "[quote=\"Hidden Scofield\"]What is e_p(2n)! ?? :huh:[/quote]\r\n$ e_p(x)$ is the largest integer so $ p^{e_p(x)}|x$, where $ p$ is a prime and $ x$ an integer.\r\nAnd we know that $ e_p(n!) \\equal{} \\left[\\frac{n}{p}\\right] \\plus{} \\left[\\frac{n}{p^2}\\right] \\plus{} \\left[\\frac{n}{p^3}\\right] \\plus{} \\dots$ were $ [x]$ denotes the largest integer $ \\le x$", "Solution_4": "Great ;\r\n\r\nThank you :D" } { "Tag": [ "FTW" ], "Problem": "What is your favorite:\r\n\r\n1. Bach Prelude and Fugue?\r\n\r\n2. Chopin Etude?\r\n\r\n3. Beethhoven Piano Sonata?", "Solution_1": "You know, people wrote piano music [i]after[/i] the 19th century, too :) What about Ginastera? Prokofiev? Keith Emerson?", "Solution_2": "Well I thought I would ask about MY favourite composers, and unfortunately, I don't much like Ginastera or Prokofiev. Debussy is probably the latest I can go before I find that music takes a turn for the worst. However saying that, I do like a bit of Hindemith, Berg, Tavener and a few lesser known Scottish composers. I doubt many people would have an opinion on which movement of Ludas Tonalis is the best (my favourite is the Preludium). So I decided to restrict myself to composers that people are sure to have an opinion on. I used to attend a specialist music school, so I am not ignorant of any type of music (except from perhaps rap and hip-hop etc.). I think we have pushed he realms of tonality (or lack of tonality) as far it will go. I certainly don't enjoy listening to a solo violinist playing in quarter tone rows. And the fact the the tone rows are inverted and retrograded does not make me appreciate it any more.", "Solution_3": "Bach - A-flat Major, especially the fugue, one of the darker and slower of the major fugues in WTC.\r\n\r\nChopin - Op. 10, No. 4 (Torrent), really love Richter's recording in particular.\r\n\r\nBeethoven - Though I really love the both second movement of Appassionata and the first of Waldstein, there's no complete sonata that I enjoy more than Op. 110.\r\n\r\nAs far the others mentioned...\r\n\r\nReally love Yefim Bronfman's recordings of Prokofiev Sonatas, 3rd one's probably my favorite of his piano pieces. Horowitz does a pretty good job with them as well.\r\n\r\nAnd for Ginastera, I'd probably pick the Suite de Danzas Criollas, though the second movement of Danzas Argentinas is one of the few pieces I really enjoy playing and listening to.\r\n\r\nI'm probably not familiar enough Ludus Tonalis as a whole to pick a favorite section, though I like the twelth fugue...", "Solution_4": "Obviously Chopin Etude!!!", "Solution_5": "1. Aufschwung- Schumann\r\n\r\n2. Waltz/Valse- Chopin\r\n\r\n3. Tarantella- Liszt\r\n\r\n4. La Campanella- Liszt\r\n\r\n5. Sonata- Chopin\r\n\r\n6. Sonata- Bartok\r\n\r\n7. Images- Debussy\r\n\r\n8. Sonata (especially 'Moonlight' and 'Pathetique' and no. 32)- Beethoven\r\n\r\n9. Ballade- Chopin\r\n\r\n10. Italian Concerto- Bach\r\n\r\nTop 10 favourite!", "Solution_6": "[quote=\"Pul de Algodoncito\"]Obviously Chopin Etude!!![/quote]\r\n\r\nMeh... I think they're superficial.", "Solution_7": "But what about GOOGOSH? What are your opinions on him? Do you know him? Who is he, anyone knows?", "Solution_8": "For Sonata I prefer that of Beethoven (especially op.111) and Schubert. In my opinion Beethoven's Sonata is a bit better than Mozart's because it have more pulse and idea(don't know whether this word is correctly used)\r\n\r\nFor Etude Liszt is the best of all, not only technically but also musically(my favorites are no 3, 8, 11)\r\n\r\nChopin's ballade are brilliant also.", "Solution_9": "umm.....appasionata i think thats how you spell it", "Solution_10": "[quote=\"Identity\"][quote=\"Pul de Algodoncito\"]Obviously Chopin Etude!!![/quote]\n\nMeh... I think they're superficial.[/quote]\r\n\r\nAmen.\r\n\r\nThe Beethoven Sonatas are, in my opinion, the deepest and most extraordinary works for the piano every written. Particularly, the last four sonatas (106, 109, 110, 111) illustrates Beethoven's stronghold on music.\r\n\r\nIn terms of sound, though, sometimes I prefer Mozart, but Beethoven, in his early works (see Op. 10, No. 3!), invokes Mozart's style almost as interestingly as Mozart himself.\r\n\r\nOn another note, has anyone ever played any Messiaen? :P", "Solution_11": "Chopin etudes aren't superficial...unless you can't play them, then they start sounding pretty superficial.\r\n\r\nop.10 no.3: Sure, kinda sounds like mozart's K576 sonata (I'm not sure if the number is right, I think it was D major). \r\n\r\nMessiaen, Ew.", "Solution_12": "Mozart...ew...", "Solution_13": "Me likes counterpoint :) ... very mathematical.", "Solution_14": "Any of the Chopin polonaises, Rachmanioff, Mendelssohn, Copland, and Brahms hungarians. I used to like Mozart, now I loathe his music.", "Solution_15": "The thing with Mozart, IMO, is that since elementary school you learn how great he is, and Beethoven. Then later on, some people discover that those 2 aren't the only composers...yea...and the conclusion that Mozart's music is inferior is quickly reached. Anyways I say that because my general observation is that among people who play piano, people who are noobs like Mozart a lot (I suspect it's because EVERYONE knows just how wonderful he is, only because that's what books say), then people who are more sophisticated in the learning of the art of piano hate Mozart, since they like Liszt, Rachmaninoff, Chopin, Prokofiev, etc. more. Speaking of books, I've never seen a single book (art or history) that praises any romantic era music, or at least not as much as classical era music. My school history textbook has 2 whole pages on Mozart, while Liszt, Rach were neglected, and Chopin had one line saying \"He wrote pieces based on Polish music.\" :(", "Solution_16": "Vingt Regards sur l'Enfant Jesus FTW", "Solution_17": "beethoven sonata and bach preludes and fugues", "Solution_18": "[quote=\"CatalystOfNostalgia\"]Vingt Regards sur l'Enfant Jesus FTW[/quote]\r\nNO", "Solution_19": "[quote=\"kyyuanmathcount\"][quote=\"CatalystOfNostalgia\"]Vingt Regards sur l'Enfant Jesus FTW[/quote]\nNO[/quote]\r\nlol we've hammered this to death already, don't you think?", "Solution_20": "my favourite piano music at the moment is the lonely man from hulk, and the entertainer." } { "Tag": [ "MATHCOUNTS", "calculus" ], "Problem": "find the sum:\r\n\r\n1/(1x3)+1/(3x5)+1/(5x7)+...+1/(199x201)", "Solution_1": "erm...whass with the title? [hide]sigh...i didn't learn how to do these sort of problems till i was in calculus...don't ask[/hide]", "Solution_2": "lol, what does calculus have to do wit it?", "Solution_3": "Well, people are usually taught how to solve this sort of problem in calculus. You don't need to wait that long though.", "Solution_4": "People are usually taught partial fractions in calculus.", "Solution_5": "ah, well i dont know whats in calculus. any takers for the problem though?", "Solution_6": "this is one of the things you either know how to do or you don't. at someone's request, i have provided a hint.\n\n[hide]this is a branch of something called decomposition by partial fractions. don't be intimidated by that big name, it just means breaking something up into parts. For example, if we are finding 1/2(x-1)-1/2(x+1), we would find a common denominator and add and whatnot, giving us 1/(x-1)(x+1). In decomposition, we do this backwards. We say HEY! we can write 1/(x-1)(x+1) as A/(x-1)+B/(x+1), since those were the factors of the denominator. You can then solve for A and B, etc. I will explain this later after we solve this particular problem. I might also explain why it's useful in calculus, if you reaaaaaaaaaaally want to know. For now, suffice it to know the general idea.[/hide]", "Solution_7": "Well, with mystic having said that, unless I'm having computational problems, I get\n\n\n\n[hide]\n\n100/201\n\n[/hide]", "Solution_8": "I believe that is right. I'll tell you the rest tomorrow.", "Solution_9": "Whooo... this problem was in a Chinese 5th grade olympiad book my mom had.", "Solution_10": "got the same as southie. i always wondered what partial fractions were useful for..." } { "Tag": [ "inequalities", "geometry", "geometric transformation", "homothety", "geometry proposed" ], "Problem": "Given two intersecting circles $ c_1$ and $ c_2$.We draw a circle $ c_3$ such that tangent to $ c_1$ and $ c_2$ (such that $ c_1$ and $ c_2$ be in the $ c_3$).$ c_1$ and $ c_2$ intersects in points A,B.$ c_3$ tangent to $ c_1$ in point B and tangemt to $ c_2$ in point C. Are A,B and C collinear?prove your notation. :lol:", "Solution_1": "First,Jensen Notation should be defined bot proved .\r\nSecond you used Letter [b]b[/b] two times but your meaning is clear.\r\nthird ,the answer to your problem is [b]No[/b].\r\n[edit]go to post 4.", "Solution_2": "Yes i knew that,in my sketch they were.so thank you so much Arman!You helped me so much. :D", "Solution_3": "no Your problem is not always true.\r\ndear lomos I have read Grobber sulotion but Grobber didn't mention it or I am erroring.\r\nWould U say how is it ,true ?", "Solution_4": "Mathx ,you are right .\r\nI was wrong ,I dont know what i was thinking about.\r\nNo they are not collinear.\r\nHere is a simple prove :\r\nProblem:let $C_1$ be a circle with center $O_1$. Let $A,B$ be two point on its premeter.\r\nLet $C_2$ be a circle with center $O_2 \\in O_1A$,tangent to $C_1$.\r\nLet $AB$ intersects $C_2$ in $F$.\r\nIf the above three points be collinear , then all circles like $C_3$ with center $O_3$ on $O_1B$ and tangent to $C_1$ shall passes through $F$ but this is a contradiction ,Since $O_3B$ is not always equal to $O_3F$.", "Solution_5": "When they are collinear? :?:", "Solution_6": "when that line pass trough homothecy center" } { "Tag": [ "ratio" ], "Problem": "Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides $ 7$ times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to his distance from the stadium?\r\n\t\r\n$ \\textbf{(A)}\\ \\frac {2}{3}\\qquad \\textbf{(B)}\\ \\frac {3}{4}\\qquad \\textbf{(C)}\\ \\frac {4}{5}\\qquad \\textbf{(D)}\\ \\frac {5}{6}\\qquad \\textbf{(E)}\\ \\frac {6}{7}$", "Solution_1": "[hide] Say he walks x km/hr, his house is h kilometers away from him, and the stadius is s kilometers away from him. Thus, it will take him $\\frac{s}{x}$ hours to get to the stadium from where he is standing. Now, if he wanted to go back home, it would take him $\\frac{h}{x}$ hours. If he wanted to go to the stadium to his house on his bicylce, it would take him $\\frac{h+s}{7x}$ hours since he rides 7 times as fast. Thus,\n\\[\\frac{h+s}{7x}+\\frac{h}{x}=\\frac{s}{x}\\]\nMultiplying both sides by 7x, we get that\n\\[h+s+7h=7s\\Rightarrow 8h=6s\\Rightarrow \\frac{h}{s}=\\frac{3}{4}\\]\nSorry about that, I edited from C to B. \nTherefore, the answer is B. [/hide]", "Solution_2": "[hide]Let $a$ be the distance between him and his house, and $b$ be the distance between him and the stadium.\n\n$b=a+\\frac{a+b}7$\n\n$7b=7a+a+b$\n\n$6b=8a$\n\n$\\frac{a}b=\\frac34\\Rightarrow\\boxed{\\text{C}}$[/hide]", "Solution_3": "There's a typo in your solutions or the problem: [hide]3/4 is B, not C :P :wink: [/hide]", "Solution_4": "this problem was problem 13, not 9. :blush:", "Solution_5": "[quote=\"now a ranger\"]this problem was problem 13, not 9. :blush:[/quote]It was 13 on the AMC 10, and 9 on the AMC 12 :wink:", "Solution_6": "This was on 10A\r\nAnswer is [b]3/4[/b]" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "algebra theorems", "algebra" ], "Problem": "What are some of the main theorems or formulas needed for USAMO and other NMO's???", "Solution_1": "As i know, NMO problems are not schoolbook problems,so you don't need formulas to solve them!\r\nI admit there are some results wich can be used but simply knowing them doesn't give you the solution to the problems!" } { "Tag": [], "Problem": "After a gymnastics meet, each gymnast shook hands once with every other gymnast on every team. Afterwards, a coach came down and only shook hands with each gymnast from her own team. There were a total of 281 handshakes. What is the fewest number of handshakes the coach could have participated in?", "Solution_1": "Obviously 1?\r\nBecause if all teams have 1 competitor?", "Solution_2": "[quote=Bugi]Obviously 1?\nBecause if all teams have 1 competitor?[/quote]\n\nNo way, because it would be impossible to have 281 handshakes in total!", "Solution_3": "This is way overrated, it should be level 18 not 23", "Solution_4": "I think it's way underrated. This took me a while to solve.\n\nHowever, if the answer [hide=is]5[/hide] and we have 24 gymnasts, how would we divide them into teams? Shouldn't each team have the same amount of gymnasts?\n\n[hide=(For reference, here is the Alcumus solution.)]The number of gymnasts is some integer $n$, so that the number of gymnast-gymnast handshakes is ${n \\choose 2}$ for some $n$. Also, the coach must participate in an integer $k M$ and we have a contraction same way to step 1.", "Solution_24": "And how to solve the main problem?", "Solution_25": "[hide=Solution]\n\nAs a preliminary note, we exclude the constant solutions, $P(x)=2u^2$ for some integer $u$. Henceforth, degree of $P(x)$ is assumed to be $\\ge 1$.\n\nWe start off with the following two lemmas. For the sake of completeness, we present their proofs as well making the whole argument quite long.\n\n[i]Lemma 1.[/i] Let $a$ be a given natural number which is not a perfect square. Then there exists infinitely many primes $p$ for which $a$ is a quadratic non-residue. \n\n[i]Proof[/i] By Euler's theorem, $\\left(\\frac{ab}{p}\\right)=\\left(\\frac{a}{p}\\right)\\cdot \\left(\\frac{b}{p}\\right)$ and so, we may assume that $a$ is square-free by taking out the largest square divisor of $a$. We can write $a=2^e\\cdot p_1\\cdot \\dots \\cdot p_k$ where $p_i \\ge 3$ are primes and $e \\in \\{0,1\\}$. In case the product is empty, it shall be assumed to be equal to $1$. If $e=0$ then let $p \\equiv 1 (\\bmod \\, 4p_1\\cdot \\dots \\cdot p_{k-1})$ and $p \\equiv s (\\bmod \\, p_k)$ where $s$ is a quadratic non residue modulo $p_k$. If $e=1$ then let $p \\equiv 5 (\\bmod \\, 8)$ and $p \\equiv 1 (\\bmod \\, p_1\\cdot \\dots \\cdot p_k)$. In any case, we see that $$1=\\left(\\frac{p}{p_i}\\right)=(-1)^{\\frac{p-1}{2}\\cdot \\frac{p_i-1}{2}}\\cdot \\left(\\frac{p_i}{p}\\right)=\\left(\\frac{p_i}{p}\\right)$$ for $i=1,\\dots, k$. In the former case, however, the equality for $i=k$ is changed to $-1$ by our choice of $s$ and in the latter, as $\\left(\\frac{2}{p}\\right)=(-1)^{\\frac{p^2-1}{8}}=-1$; we conclude that $a$ is a quadratic non residue modulo $p$ by using Gauss' Quadratic Reciprocity Theorem. The existence of infinitely many such prime is guaranteed by the Chinese Remainder Theorem and the Dirichlet's Theorem on distributions of primes in an arithmetic progression. The proof is completed.\n\n---\n\nWe now present the second crucial lemma.\n\n[i]Lemma 2.[/i] Let $f$ be a non constant polynomial with integer such that $\\sqrt{f(n)}$ is an integer for all positive integers $n$. Then there exists a polynomial $g(x)$ with integer coefficients such that $f(x)=g(x)^2$. \n\n[i]Proof[/i] Let us write $f(x)$ as a product of its irreducible divisors. Clearly, we can write $$f(x)=p_1(x)\\cdot \\dots p_k(x) \\cdot \\left(R(x)\\right)^2$$ where $p_1,\\dots, p_k$ are pairwise distinct irreducible divisors of $f$ and all the polynomials $p_1,\\dots,p_k,R$ have integer coefficients. If the product $p_1\\cdot \\dots \\cdot p_k$ is a constant then it must be a square and we may conclude. Else, there exists a non-constant polynomial in the set $\\{p_1(x),\\dots,p_k(x)\\}$, which, say, is $p_1(x)$. Then, as $p_1$ being irreducible has no common factor with the product $h=p_2\\cdot \\dots \\cdot p_k$ (which might as well be empty or constant) and $p_1'$ (the derivative of $p_1$), we are assured of the existence of polynomials $A(x),B(x),C(x),D(x)$ with integer coefficients and integers $N_0,N_1 \\not= 0$ such that $$A(x)\\cdot p_1(x)+B(x)\\cdot h(x)=N_0$$ and $$C(x)\\cdot p_1(x)+D(x)\\cdot p_1'(x)=N_1$$ for all $x$ by the Euclid's Lemma. By Schur's theorem, the set of all primes dividing at least one value taken by $p_1$ over $\\mathbb{Z}$ is infinite, hence, we may choose a prime number $p>N=\\max(|N_0|,|N_1|)$ and an integer $r$ such that $p \\mid p_1(r)$. Since $p \\nmid N_0$, we see that $p \\nmid h(r)$. Now, choose $x_0=r+ip$ where $0 \\le i \\le p-1$ is an integer we shall choose later. By the Taylor's Expansion of polynomials, we have $$p_1(x_0) \\equiv p_1(r)+ip\\cdot p_1'(r) (\\bmod p^2)$$ In particular, we let $p_1(r)=p^{j_0}k_0$ and we can select $i$ such that $p^2 \\nmid p_1(x_0)$ since the polynomials $p_1$ and $p_1'$ are also coprime $p \\mid p_1(x_0),p_1'(x_0)$ will yield $p \\mid N_1$, which is false as $p>N$. Thus, $v_p(p_1(x_0))=1$ and $v_p(h(x_0))=0$. This yields that $v_p(f(x_0))$ is an odd number, a contradiction to the fact that it is a square. The result holds.\n\n---\n\nWith these two proofs finally completed, we present the main idea of the solution. \n\nNotice that if $(a+b)$ is a perfect square then $(ax^2+bx^2)=(a+b)x^2$ is also a perfect square. Define the polynomial $T(x)=P(x^2)+P((t^2-1)x^2)$ where $t$ is fixed for the moment. Then, $T$ satisfies the conditions of [i]Lemma 2.[/i] and we get that it is the square of a polynomial $Q(x)$ with integer coefficients. Let $d$ be the degree of $P$, $K$ be the leading coefficient of $P$ and $g(t)$ (not a polynomial/function etc) be the leading coefficient of $Q$. The following equation is satisfied by comparing the leading coefficients in the equation $$T(x)=Q(x)^2,$$ $$K(1+(t^2-1)^d)=g(t)^2$$ for all $t$ sufficiently large. Let $K=mc^2$ where $m$ is a square-free number or $m=1$. If $m$ is square-free and $m>1$ then we may write $$(t^2-1)^d=m\\cdot \\left(\\frac{g(t)}{mc}\\right)^2-1$$ where, for obvious reasons of divisibility, $$\\left(\\frac{g(t)}{mc}\\right)$$ is an integer. Let $p$ be a prime number and set $t \\equiv 1 (\\bmod \\, p)$,i.e., $p \\mid t^2-1$ and so $p \\mid m\\cdot \\left(\\frac{g(t)}{mc}\\right)-1$ giving that $m$ is a quadratic residue modulo $p$. However, by [i]Lemma 1.[/i] we can choose a sufficiently large prime $p$ with $\\left(\\frac{m}{p}\\right)=-1$ and we get the requested contradiction. If $m=1$ then suppose $d>1$. If $d$ is even then $1$ is a difference of sufficiently big perfect squares, and if $d$ is odd then we have $$\\left((t^2-1)^{\\frac{d-1}{2}}-1\\right)^2<\\left(\\frac{g(t)}{tc}\\right)^2<\\left((t^2-1)^{\\frac{d-1}{2}}\\right)^2$$ which is again a contradiction! Thus, we must have $d=1$. If $d=1$ then we have $P(x)=vx+w$ and if $a=1,b=t^2-1$ then we see that the equation $$z^2-vt^2=w$$ is satisfied for some integer $z$. If $v$ is a perfect square then we get that $w=0$. If not, then it is a Pell type equation, so its solutions must grow exponentially, however, they grow only linearly, as $t$ grows linearly, only to be multiplied by a constant (equal to the largest perfect square divisor of $v$) and we get a contradiction!\n\nThus, the only solutions to this are the polynomials $P(x)=k^2x$ for some integer $k$ or $P(x)=2u^2$ for some fixed integer $u$. These also satisfy our given equation and we may conclude.\n\n\n\n\n[/hide]\n\n[hide=Credits and remarks] This happens to be a joint solution with [url=https://www.artofproblemsolving.com/community/user/158287]dibyo_99[/url] at the IMO training camp in India, this May. There were other notable contributors to this as well. This is one of my favorite number theory problems! :) [/hide]", "Solution_26": "[quote=anantmudgal09]and if $d$ is odd then we have $$\\left((t^2-1)^{\\frac{d-1}{2}}-1\\right)^2<\\left(\\frac{g(t)}{tc}\\right)^2<\\left((t^2-1)^{\\frac{d-1}{2}}\\right)^2$$ which is again a contradiction! Thus, we must have $d=1$. [/quote]How can you get the left side?", "Solution_27": "[quote=anantmudgal09]\n[i]Lemma 2.[/i] Let $f$ be a non constant polynomial with integer such that $\\sqrt{f(n)}$ is an integer for all positive integers $n$. Then there exists a polynomial $g(x)$ with integer coefficients such that $f(x)=g(x)^2$.[/quote]\n\nI learned another simple way to prove this lemma from lectures on winter camp in Korea around 2018.\n\nProof) Think of $f(x+m)f(x)$ for some natural number $m$. If $m$ is sufficiently large, we can separate complex roots of $f(x)$ and $f(x+m)$ and therefore $f(x+m)$ and $f(x)$ is relatively prime. By assumption, $f(x+m)f(x)$ is even degree and the maximal coefficient is square. Therefore, $f(x+m)f(x)=h(x)^2$ for some $h(x) \\in \\mathbb{Z}[x]$. Since $f(x)$ and $f(x+m)$ is prime, you can easily see that $f(x)=g(x)^2$ for some $g(x) \\in \\mathbb{Z}[x]$.\n\nSame idea is used in Proposition mentioned in https://artofproblemsolving.com/community/u314211h1985879p14247495", "Solution_28": "Another astonishing proof for Lemma by my student.\n\nProof) Let $f(x)=a_n x^n +....+a_0$. By assumption, $a_n >0$.\n[Case 1] $n$ is odd.\nMultiply the term $(a_n x +1)$. There are infinitely many $x$ such that $a_n x+1$ is perfect square. Therefore, infinitely many $x$ such that $(a_n x+1)f(x)$ is perfect square. However, there are also $x$ such that $a_n x +1$ is not a perfect square and contradiction.\n[Case 2] $n$ is even, but $a_n$ is not a perfect square.\nMultiply the term $(a_n x^2 +1)$. And same as above. Here we used the fact that pell equation $y^2 - a_n x^2 =1$ has infinitely many solution.\n\nTherefore, $n$ is even and $a_n$ is perfect square and $f(x)=g(x)^2$ for some $g(x) \\in \\mathbb{Z}[x]$.\n\n\n", "Solution_29": "Here is a smol bren solution.\n Write the polynomial,as $x^d.P(x)$, with $P(0)$ non zero.Assume $P$ non-constant.Then assuming,$P$ is not a square ,write it as a square times a bunch of irreducibles.Consider such a irreducible ,say $h(x)$,and then by Schur ,choose a very large prime factor of the polynomial $h(x^2)$.Then using Bezout,we can conclude that if $p|h(m^2)$,then $p$ does not divide any other irreducible factor of $P$ evaluated at $m^2$.Then,again using bezout but this time with $h'(x)$,we conclude that if $p$,is large enough ,then $p$ also doesn't divide $h'(m^2)$,thus from Hensel,we can uniquely lift the root corresponding to $m^2$ mod $p$ to mod $p^2$ .Due to this unique lift ,we can find such an $n$,such that $n \\equiv m^2$ mod $p$ but $p^2$ does not divide $P(n)$.Thus we have that $v_p(P(n))$ is odd.Now choose a $b$ such that $v_p(b)=j$ is even and quite large.As $p$ is large (forgot to mention $p$ doesn't divide $P(0)$,so $p$ doesn't divide $P(b)$).\nNow,we use hensel's again.Firstly note that $n$ is a quadratic residue mod $p$.So by hensel's on the polynomial $x^2-n$,we see that the derivative is non zero mod $p$.Thus,we can find a solution to this poly mod $p^{j+1}$,such that $v_p(x^2-n)$ is exactly $j$.Thus we have found $b+n$ is a perfect square but $v_p(n^d.P(n)+b^d.P(b))$ is odd which is a contradiction.Thus,$P(x)$ is a perfect square.\n\nUsing a very similar argument as above,we can prove that $d$ is even(Basically choosing $a,b$,such that $P(a)$ has a large $v_p$,which is obv \n even,except here we choose $b$ to have small odd $v_p$,and obv $p$ doesn't divide $P(b)$.$a+b$ being a perfect square is ensured by the very similar hensel argument as above).\nSo we have that we can write $P(a)+P(b)$ as $a^{2d}Q(a)^2+b^{2d}Q(b)^2$.So if we fix $a$,we see that $a^dQ(a)$ is a part of infinitely many Pythagorean triples,which is impossible.\nObserving the left out cases(which are not that hard),we get our solutions.", "Solution_30": "As in #26, $f(ax^2)+f(bx^2)$ is always a square for a fixed pair $(a,b)$ with $a+b$ being a square, so $g(x)^2=f(ax^2)+f(bx^2)$ for some integer polynomial $g(x)$. In addition, we consider only the case $\\deg(f) \\geq 2$.\n\nNext, we consider the leading coefficient of this polynomial; it is of the form $T(a^n+b^n)$ ($T$ is the leading coefficient of $f$), which shall be a square. If $n$ is even, plug $(1,p^2-1),(1,p^2+2p)$ for some prime $p$ to see that $2T, T$ are QR's modulo $p$ for all primes $p$, so $2$ is a QR for all primes $p$, absurd, since $(\\frac{2}{p})=(-1)^{\\frac{p^2-1}{8}}$. Now we can finish the other case with $(2,2),(3,1)$ : $2^{n+1}T$ is a square, hence $T$ is a square, so $3^n+1=x^2$ is a square, so $3^n=(x-1)(x+1)$ and thus $x-1=1$ and $n=1$, contradiction.", "Solution_31": "I see pleny of solutions over here depending on this theorem about a polynomial being square too many times. I solved to problem using a totally different approach with pells equations, some estimations of the growth of some terms, and an unexpected use of Norms function at the end. I will write down the solution tomorrow.\n", "Solution_32": "Suppose $degP > 0$. If $P$ is constant so $2P(2)$ must be a square, so $P \\equiv 2k^2$ for some integer $k$. Indeed all of these work.\nAlso suppose that the leading coefficient of $P$ is positive. If it's negative then $P(2N^2)$ is negative for big $N$, but is must also be twice a perfect square, contradiction.\nFix some positive integer $t$ such that the equation $b^2 - 2a^2=t$ has a solution. \nWe know there are infinetely many big $t$'s satisfying this (perfect squares for instance).\nBy pell's equation we know that, given one solution $b_0^2 - 2a_0^2$, we can find infinetely many solutions of the form \n$$b_n + a_n \\sqrt2= (b_0 + a_0 \\sqrt2)(3+2sqrt2)^n.$$\nJust so I'll write quicker, set $ r = 3 + 2 \\sqrt2$, we'll use it at lot. \nNow we know that, as $2a_n^2 + 2a_n^2$ is the square of an integer, $2P(a_n^2)$ must be the square of an even integer, say, by definition, $4r_n^2$. So $P(a_n^2) = 2r_n^2$.\nWe also know that $b_n^2 - t + t$ is a perfect square, so $P(b_n^2 - t) = s_n^2 - P(t)$, for some integer $s_n$. But, by definition, $b_n^2 - t = 2a_n^2$, so $2r_n^2$ and $s_n^2 - P(t)$ must be equal. So we have that $s_n^2 - 2r_n^2 = P(t)$. \nBut this means that $s_n, r_n$ are pell's solutions for $x^2 - 2y^2 = P(t)$. This means that $$s_n + r_n \\sqrt2 = (s_0 + r_0 \\sqrt2)r^{t_n}$$ for some minimal solution $(s_0,r_0)$ and some integer $t_n$. As there are only finitely many minimal solutions for pell's equation, by the pigeon hole principle one of them must appear infinetely many times, varying $n$. WLOG call it $(s_0,r_0)$. \nIn other words, we can say that \n$$s_n = \\frac{s_0 + r_0 \\sqrt2}{2}r^{t_n} + \\frac{s_0 - r_0 \\sqrt2}{2}r^{-t_n}$$\nso\n$$s_n^2 = \\big( \\frac{s_0 + r_0 \\sqrt2}{2} \\big)^2 r^{2t_n} + \\big( \\frac{s_0 - r_0 \\sqrt2}{2} \\big)^2 r^{-2t_n} + 2\\frac{s_0^2 - 2r_0^2}{4} = (1+o(1))\\big( \\frac{s_0 + r_0 \\sqrt2}{2} \\big)^2 r^{2t_n}$$\nSo this is one estimate for $s_n^2$.\nLet $d = degP$ and $c$ the leading coefficient. Now we also know that \n$$s_n^2 = P(b_n^2 - t) + P(t) = (1+o(1))P(b_n^2) = (1+o(1))c.b_n^{2d}.$$\nAs $$b_n = (1+o(1))\\frac{b_0+a_0 \\sqrt2}{2}r^n,$$\n$$s_n^2 = (1+o(1))c.\\big( \\frac{b_0+a_0 \\sqrt2}{2} \\big) ^{2d}r^{2nd}$$\nTherefore the ratio between the $2$ estimates converges to $1$. So\n$$lim_{n \\to \\infty} \\frac{\\big( \\frac{s_0 + r_0 \\sqrt2}{2} \\big)^2 r^{2t_n}}{c.\\big( \\frac{b_0+a_0 \\sqrt2}{2} \\big) ^{2d}r^{2nd}} = 1.$$\nSo\n$$lim_{n \\to \\infty} \\frac{\\big( \\frac{s_0 + r_0 \\sqrt2}{2} \\big)^2}{c.\\big( \\frac{b_0+a_0 \\sqrt2}{2} \\big) ^{2d}} r^{2t_n - 2dn} = 1.$$\nSo, a constant times an integer power of $r$ converges to $1$. Due to the fact the the sequence $C.r^m$ is discrete, if it converges to $1$, it's eventually $1$. So there exists an integer $q$ such that\n$$\\frac{\\big( \\frac{s_0 + r_0 \\sqrt2}{2} \\big)^2}{c.\\big( \\frac{b_0+a_0 \\sqrt2}{2} \\big) ^{2d}} r^q = 1.$$\nRearranging things and substituting $r$ back to $3+2 \\sqrt2$, we have\n$$(s_0 + r_0 \\sqrt2)^2 (3+2 \\sqrt2)^t = c(b_0 + a_0 \\sqrt2)^{2d} 2^{2d-2}$$\nNow we are gonna make use of the Norm function defined by $N(m + n \\sqrt2) = m^2 - 2n^2$, where $m,n$ are rationals. One can easily prove that this function is well defined and is multiplicative. So we have\n$$(s_0^2 - 2r_0^2)^2 . 1^t = c^2(b_0^2 - 2a_0^2)^2 2^{4d-4}$$\nSo\n$$P(t)^2 = c^2 . t^{2d} . 2^{4d-4}$$\nAs the leading coefficient of $P$ is positive, $P$ is eventually positive, so we can take the squareroot of both sides to conclude that\n$$P(t) = c.t^d.2^{2d-2}$$\nBut notice that this is true for infinetely many big integers $t$'s, so it must be true that\n$$P(x) = x^d.c.2^{2d-2}$$\nBut remember that $c$ is the leading coefficient of $P$, so $c = c.2^{2d-2}$, so $2^{2d-2}=1$, so $d$ must be $1$, and $P(x) = cx$.\nNow, as $2P(2)$ is a square, so $4c$ is a square, so $c$ itself is a perfect square. Given that, clearly $a+b$ is a square if and only if ac+bc is a square.\nTherefore all solutions are $P(x) = k^2x$ and $ P(x) = 2k^2$, where $k$ is an arbitrary integer." } { "Tag": [ "vector", "algebra", "polynomial", "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Suppose that F is a field and all F^n are written as column vectors\r\nLet A be an nxn matrice with entry from F, and \r\nW = { B in M_nxn (F) | AB = BA}\r\nSuppose there exists a vector v in F^n such that { v, Av, A^2 v, ........ A^(n-1) n} is a basis for F^n.\r\n\r\nProve that { I, A, A^2, ...., A^(n-1)} is a basis for W.\r\n\r\n\r\nAbove is the question I get stuck. It is easy to prove that W is a subspace and { I, A, A^2, ...., A^(n-1)} is linearly independent.\r\nBut how are we going to prove that it is spanning W?\r\nAny idea?", "Solution_1": "[hide=\"Hint\"]It suffices to show that $ \\dim W\\le n$. For this purpose consider the evaluation mapping $ W\\to F^n$, $ B\\mapsto Bv$ which is injective.[/hide]", "Solution_2": "Alternately (and I'm not sure if this works), write the rows of $ B \\in W$ in terms of the basis given.", "Solution_3": "Why is the linear transformation stated above injective? \r\n\r\nIf we consider Bv = 0, I can just reach a conclusion that B is non-invertible or B=0 if B is in the span of {I, A,......, A^n-1}. :| \r\n :|", "Solution_4": "Assuming that $ Bv\\equal{}0$ can you conclude that $ B\\equal{}0$ using a basis of $ F^n$ and that $ B\\in W$?", "Solution_5": "Oh! Is that because for every vector in F^n, Bx = 0, therefore B must be equal to zero?\r\n\r\nBv = 0, BAv = 0, BA^2 v = 0 .... BA^(n-1) v =0, hence for every x in F^n, Bx = 0.", "Solution_6": "Correct.[quote=\"distantstar\"]Oh! Is that because for every vector in F^n, Bx = 0, therefore B must be equal to zero?\n\nBv = 0, [/quote]implying (together with $ B\\in W$)[quote]BAv = 0, BA^2 v = 0 .... BA^(n-1) v =0, hence for every x in F^n, Bx = 0.[/quote]And \"hence\" stems from the assumption that $ A^mv$, $ m\\equal{}0,\\ldots,n\\minus{}1$, is a Basis of $ F^n$.", "Solution_7": "See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=113276]here[/url].\r\n\r\nThe relevant argument:\r\nSuppose the minimal and characteristic polynomials of $ A$ are both equal to $ p$. Consider the algebra* $ V$ of polynomials mod $ p$, and let $ A$ act on this space by $ A(q(x))\\equal{}x\\cdot q(x)$. In the standard basis of $ V$, the matrix of $ A$ is the companion matrix associated to $ p$, and this is isomorphic to the original matrix form.\r\nNow, we show that any linear transformation $ B$ on $ V$ which commutes with $ A$ commutes with polynomial multiplication: $ B(x^k)\\equal{}B(x\\cdot x^{k\\minus{}1})\\equal{}B(A(x^{k\\minus{}1}))\\equal{}A(B(x^{k\\minus{}1}))\\equal{}x\\cdot B(x^{k\\minus{}1})\\equal{}\\cdots$\r\n$ \\equal{}x^k\\cdot B(1)$, and $ B(q(x))\\equal{}q(x)\\cdot B(1)$ by linearity. The map $ B(q(x))\\equal{}q(x)\\cdot \\lambda(x)$ is exactly $ \\lambda(A)$, and $ B$ is a polynomial in $ A$.\r\n\r\n*An algebra is both a ring and a vector space. In this case, we started with a generic $ n$-dimensional vector space and added ring structure to make things easier.", "Solution_8": "Hmm, I cannot see why you need all this formalism. All you need here is that $ B\\in W$ commutes with $ A^m$, $ m>0$, and that $ F^n$ is $ A$-cyclic (which comes very close what you intend to mean I think). The key fact is that the evaluation mapping $ W\\to F^n$ is an isomorphism.", "Solution_9": "The natural structure on $ W$ is an algebra. The natural structure on $ F^n$ is a vector space. I wouldn't call it an isomorphism without either adding structure on $ F^n$ or removing it from $ W$. I chose the former, while you chose the latter.\r\n\r\nOur arguments really are very similar; it's just that the algebra formalism lets me say $ B(q(x))\\equal{}q(x)\\cdot B(1)$ rather than $ B\\cdot (q(A)\\cdot v)\\equal{}q(A)\\cdot (B\\cdot v)$, and the finish after that is easier." } { "Tag": [], "Problem": "1+8+27+............................+8000\r\n\r\nTry to solve this. Please show all of your work.\r\n\r\n[hide=\"hint\"]Prime factorize each individual term.[/hide]\r\n\r\nSource: AoPS(not a post)", "Solution_1": "[hide]From a formula, $1^3+2^3+3^3+4^3...20^3$ is $(\\frac{20(21)}{2})^2$ which is $44100$ if i'm not mistaken.[/hide]", "Solution_2": "tatz how i did it", "Solution_3": "Same here.", "Solution_4": "if you did it the same way, its still ok to post your solution... :)", "Solution_5": "1+8+27+............................+8000= 1^3+2^3+3^3+...+20^3=(1+2+3+...+20)^2=(1/2*20*21)^2=44100.", "Solution_6": "[hide]use the formula 20*21/2^2=44100[/hide]", "Solution_7": "I did it just like the 2nd guy. :alien:" } { "Tag": [], "Problem": "Jason counts up from 1 to 9 skipping the even numbers, and then immediately counts down without skipping any of the numbers, and then back up to 9 skipping the evens, and so on, alternately counting up and down \r\n\\[ (1, 3,5,7,9,8,7,6,5,4,3,2,1,3,5,\\ldots ).\\] What is\r\nthe $ 1000^{\\text{th}}$ integer in his list?", "Solution_1": "The pattern 1,3,5,7,9,8,7,6,5,4,3,2 repeats.\r\n\r\n$ 1000\\div 12\\equal{}83 R 4$\r\n\r\nAnswer: 7", "Solution_2": "Who else took mod 13 instead of 12?\n\nwhoops 11 year bump", "Solution_3": "It's mod 12 because there are 12 elements in each cycle.\n", "Solution_4": "[quote=hashtagmath]It's mod 12 because there are 12 elements in each cycle.[/quote]\n\nnice bump lol yeah i know ", "Solution_5": "[hide=bestzack66 (#3)][url=aops.com/community/user/547702][b]bestzack66 (Profile)[/b][/url] $\\cdot$ Oct 13, 2020, 12:17 PM $\\cdot$ [aops]Y[/aops] 1 $\\cdot$ [tip=Avatar][img]https://avatar.artofproblemsolving.com/avatar_547702.jpg?t=1618853880[/img][/tip] \n ----- Who else took mod 13 instead of 12?\n\nwhoops 11 year bump $\\phantom{Userscript made by WyattB.}$ [/hide] \n \nhow is that a whoops bump" } { "Tag": [ "trigonometry", "probability", "function", "geometry", "algebra", "polynomial", "conditional probability", "\\/closed" ], "Problem": "I'm probably only going to take Intermediate Trig in the fall, and I was wondering if the administrators could post the topic outline for Intermediate Counting.\r\nAlso, will AoPS Volume 2 cover most of the material in the counting class? What about AcoPS? Thanks.", "Solution_1": "what's AcoPS?", "Solution_2": "[quote=\"ffdbzathf\"]what's AcoPS?[/quote]\r\n\r\nI'm guessing that is the book The Art and Craft of Problem Solving by Paul Zeitz, which is a recommended book for some AoPS classes.", "Solution_3": "[quote=\"1234567890\"]I'm probably only going to take Intermediate Trig in the fall, and I was wondering if the administrators could post the topic outline for Intermediate Counting.\n[/quote]LESSON I: Conditional Probability\nLESSON II: More Conditional Probability & A Constructive Approach to Counting\nLESSON III: Constructive Expectations & 1-1 Counting\nLESSON IV: Using Correspondences to Count\nLESSON V: Distributions\nLESSON VI: Recursion\nLESSON VII: The Catalan Numbers\nLESSON VIII: Combinatorial Identities\nLESSON IX: Principle of Inclusion-Exclusion (PIE)\nLESSON X: PIE as a State of Mind\nLESSON XI: Generating Functions\nLESSON XII: Using Generating Function with Partitions\n[quote]Also, will AoPS Volume 2 cover most of the material in the counting class? What about AcoPS? Thanks.[/quote]Chapters 17 and 19 of AoPS Vol 2 cover a lot of material from the class, but in the class we will cover it more thoroughly. Also, generating functions in particular are covered in significant detail in ACoPS.", "Solution_4": "Is Mathematics of Choice too easy? Like Intro level?", "Solution_5": "It goes beyond intro level. It's a good primer for our Intermediate class.", "Solution_6": "I went through it, and it looked like it covered intro stuff very thouroghly, but I am not sure about the intermediate class.", "Solution_7": "I took Intro Counting. Does that mean that Mathematics of Choice won't help much?", "Solution_8": "It will, I am sure.... there are chapters on generating functions and advanced techniques, but I don't think it will cover it fully.", "Solution_9": "Will Intermediate Trig and Complex numbers be offered in the spring?\r\n\r\nCould you post a syllabus for that as well?\r\n\r\nthanks.", "Solution_10": "intermediate trig and complex numbers will be offered in the fall, and anything that's offered in the fall is not offered in the spring (at least that's how it is this year). i am not sure if intermediate trig and complex numbers will be offered in the summer though. do any of the instructors have an idea of that yet?", "Solution_11": "We will not be considering our Summer 2006 schedule for quite some time.", "Solution_12": "Intermediate Trigonometry/Complex Numbers:\r\n\r\nLesson I: Basic Trigonometry\r\nLesson II: Trigonometric Identities I\r\nLesson III: Trigonometric Identities II\r\nLesson IV: Laws of Sines and Cosines\r\nLesson V: Geometry with Trigonometry\r\nLesson VI: Problem Day!\r\nLesson VII: Intro to Complex Numbers and the Complex Plane\r\nLesson VIII: DeMoivre\u2019s Theorem, cis, Re{z}, and Im{z}\r\nLesson IX: Exponential Form\r\nLesson X: Roots of Unity and Polynomials\r\nLesson XI: Geometry with Complex Numbers\r\nLesson XII: More Problems" } { "Tag": [ "algorithm" ], "Problem": "I'm interested to learn how it is possible to write huge precision arithmetic programs for example in C++. I know that the numbers are in a linked lists but to do the arithmetic effectively the numbers should be in base 2. But I don't see how to convert a large integer $ n$ to base 2 because I think in some cases one has to approximate $ \\operatorname{log}_2n$ more accurate than double precision. I might be wrong of course.\r\n\r\nAnother question is that how many bits one should put in one cell in the linked list so that one can do the arithmetic effectively? Or is there some other data structures than linked list that one should use in that kind of problems?", "Solution_1": "A large integer is already stored on a computer in base $ 2$, so I'm not sure what you mean. A program that receives large integer input (as a string) should read each character as a digit and parse accordingly. What is the issue?\r\n\r\nAs for your second question, a different choice of data structure (such as a stack or a queue) will make some operations faster and other operations slower, if I remember correctly. So what operations do you need?", "Solution_2": "[quote=\"t0rajir0u\"]A large integer is already stored on a computer in base $ 2$, so I'm not sure what you mean.[/quote]\nBut if I ask a user an input, he wants it to be in base 10. I can read that input in string or char array, not in longs because there might be overflow. Then if I like to compute something effectively I should convert that string to the corresponding base 2 value and do arithmetic in base 2. Then the user want the result in base 10 so I have to convert it back to base 10.\n[quote=\"t0rajir0u\"]As for your second question, a different choice of data structure (such as a stack or a queue) will make some operations faster and other operations slower, if I remember correctly. So what operations do you need?[/quote]\r\nNothing in special. I was just thinking how much it takes memory to build a linked list.", "Solution_3": "Hmm. I think what you want is a list of [url=http://en.wikipedia.org/wiki/Computational_complexity_of_mathematical_operations]computational complexities[/url]. The naive base-conversion algorithm I'm thinking of is $ O(n^2)$, which is annoying but doable.", "Solution_4": "Well, if you want to do arbitrary precision arithmetic, you can use the [url=http://gmplib.org]GNU MP Bignum Library[/url]. Using a library like GMP will allow you to avoid trying to implement arbitrary precision yourself, which could take a while and might be a bit buggy.\r\n\r\nOn the other hand, if you don't want to use a library, the Wikipedia page t0rajir0u suggested is useful.", "Solution_5": "In my opinion the best data structure to use would be an array ;). To make it easier to convert from decimal to your internal representation and back, I would use an array of ints but store in each element only values from 0 to 999 999 999. (So it would be a base-one billion representation.) You'll be wasting 2 bits of every 32-bit integer but that's not too bad. The advantage is that you can\r\n\r\n1. easily calculate the length of the array you'll need - it's number of digits / 9 (rounded up)\r\n2. easily convert string to your representation and back", "Solution_6": "In the programming languages I'm aware of, I believe the size of an array is an int, which bounds the size of an array of ints.", "Solution_7": "i think what he is asking is how big can ya get?", "Solution_8": "[quote=\"t0rajir0u\"]In the programming languages I'm aware of, I believe the size of an array is an int, which bounds the size of an array of ints.[/quote]\r\nAre you concerned that puuhikki will need an array bigger than 8GB? ;)" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "AMC 10", "trigonometry", "trig identities" ], "Problem": "This is not for any of administrators.\r\n\r\nThe discussion I held resulted out very hard for me. The reason, I found, was that most people who participated in there were in level of IMO or USAMO, or at least AIME.\r\n\r\nSo, I need volunteering advanced math chat coach.\r\n\r\nIf you are interested, please sign-up here", "Solution_1": "Uh oh, nobody has signed up...", "Solution_2": "No, Nukular did. And if he's busy, tetrahedr0n will do it. I already talked to with them.", "Solution_3": "Ah. Which one should I do (easy/hard)?", "Solution_4": "I don't really mind but I would be appreciated if you can join on the easy too :wink:", "Solution_5": "Ay. Well can you give example problems from both?", "Solution_6": "For the easy one, it's hard for me to give an example problem.. But for the level of the question, it would be like this:\r\n\r\nSimplify: (this is very similar to question on AMC10)\r\n\r\n[tex]\\frac {(2^5)^6}{(4^6)^5}[/tex]\r\n\r\nFor hard one, I really don't know but since it's going to be in high level, I'll just show you a difficult one\r\n\r\nProve that the sum of the numbers x*sin(x) (x = 2, 4, 6, ..., 180) is 90*cos(1)/sin(1). (degrees)\r\n\r\nP.S. I just got from internet, so don't ask me how to do it..", "Solution_7": "Ay, well the first one is easy...the second...um...", "Solution_8": "Hint for the second one: [hide]use either telescoping series or complex numbers[/hide]", "Solution_9": "Er..okay..I will try!\r\n\r\n[quote]Prove that the sum of the numbers x*sin(x) (x = 2, 4, 6, ..., 180) is 90*cos(1)/sin(1). (degrees)\n[/quote]\r\n\r\nOkay, still stuck. I thought a telescoping series was when the whole sum will have a partial fraction decomposition so that it telescopes into like 2 terms. Like I think AoPS2 gives the example of sum(1, :inf: of 1/n(n+1) so if you do the decomposition it is:\r\n1/n(n+1) = A/n + B/(n+1)\r\n1 = A(n+1) + Bn\r\n1 = An + A + Bn\r\n1 = A\r\nA+B = 0\r\nB=-1\r\nSo you want the sum from 1 to :inf: of 1/n - (1/(n+1))\r\nYou end up w/:\r\n1/1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4...\r\nSo the answer =1\r\nHowever now this confuses me since the limit of 1/n(n+1) as n-> :inf: seems like it should be 0. So I have confused myself.\r\nAbout this problem, I will try writing out some terms:\r\n\r\n2sin2+4sin4+6sin6+8sin8+...+180sin(180)\r\nCan factor out 2:\r\n2(sin2+sin4+sin6+...+sin180)\r\nOh wait, okay I dunno if this proves it but think of the graph of sine:\r\nSo:\r\n2( 2(sin2+sin4...+sin88) + sin90 + sin180)\r\n2( 2(sin2+sin4+...+sin88) +1+0)\r\n4(sin2+sin4+...+sin88) +2\r\nNow I am stuck again.", "Solution_10": "That trig problem is really nice except that it takes about a minute to do it if you see the trick. Someone once said here that college math doesn't help people on the USAMO, but one of my professors would definitely expect us to be able to solve this problem. (We're covering Fourier series right now, and all the problems he does in class are just like this one.) Well, this one is a bit harder than the ones in class, but it is the same idea.", "Solution_11": "Yes but can you offer a suggestion so I can see the trick?", "Solution_12": "Perhaps this is an awfully big hint, but [hide]multiply by sin(1) and ask yourself what identities you know about expressions like that[/hide].", "Solution_13": "[quote=\"JS1527\"]However now this confuses me since the limit of 1/n(n+1) as n-> :inf: seems like it should be 0. So I have confused myself.[/quote]\r\n\r\nNote that at first, you were calculating the sum of the series. Then you confused yourself by looking at the limit of the sequence. In fact, if you want the sum of those things to converge, they better get really small, or else your sum will explode.", "Solution_14": "JBL - Right. My teacher acts like the limit of an expression is the same as its sum from 1 to infinity.... So I was correct w/my example w/the partial fraction decomposition?\r\n\r\nComplexZeta - Do you want me to:\r\n\r\n[2sin2 + 4sin4 + ... + 180sin180 ]*sin1\r\n\r\nor:\r\n\r\nS*sin1 = 90*cos(1) ?", "Solution_15": "[quote=\"JS1527\"]ComplexZeta - Do you want me to:\n\n[2sin2 + 4sin4 + ... + 180sin180 ]*sin1\n\nor:\n\nS*sin1 = 90*cos(1) ?[/quote]\r\n\r\nHe wants you to take the expression\r\n[2sin2 + 4sin4 + ... + 180sin180 ]*sin1\r\nand manipulate it to show that it is equal to the expression\r\n90*cos(1).", "Solution_16": "Sorry I didn't abandon this problem, mathfanatic and complexzeta helped me get this one. Basically, the strategy is to [hide]multiply everything by sin(1) and then use sum-and-difference trig identities so that everything telescopes and you end up w/a simple expression that simplifies to 90*cos(1). [/hide] Thanks for the help.", "Solution_17": "Just for the purpose of completeness, here is a full solution:\n\n\n\n[hide]\n\n\n\n2sin(2) + 4sin(4) + ... + 180sin(180) = 1/sin(1)[2sin(1)sin(2) + 4sin(1)sin(4) + ... + 180sin(1)sin(180)]. \n\n\n\nNow we use the formula \n\n\n\nsin(a)sin(b)=1/2[cos(a-b)-cos(a+b)] \n\n\n\nto write \n\n\n\n1/sin(1)[2sin(1)sin(2) + 4sin(1)sin(4) + ... + 180sin(1)sin(180)]=1/sin(1)[cos(1)-cos(3) + 2cos(3)-2cos(5) + 3cos(5)-3cos(7) + ... + 90cos(179)-90cos(181)], \n\n\n\nwhich telescopes to\n\n\n\n1/sin(1)[cos(1) + cos(3) + cos(5) + ... + cos(177) + cos(179) -90cos(181)].\n\n\n\nBut cos(a)+cos(180-a)=0, so this simply reduces to\n\n\n\n1/sin(1)[-90cos(181)]=1/sin(1)[-90cos(179)]=1/sin(1)[90cos(1)],\n\n\n\nas desired.[/hide]", "Solution_18": "Wow you always go the extra step. I actually remember doing it a little differently, but I think it is essentially the same reasoning:\n\n[hide]\n\n2sin2sin1 + 4sin4sin1 + 6sin6sin1 + ... + 178sin178sin1 + 180sin180sin1 = 90*cos1\n\n\n\nsince sinA*sinB=1/2 [cos(A-B) - cos(A+B)]\n\n\n\n2(1/2)(cos1 -cos3) + (4)(1/2)(cos3 - cos5) + (6)(1/2)(cos7 - cos5) + ... + (178)(1/2)(cos177 - cos179) +(180)(1/2)(cos179 - cos181) = 90*cos1\n\n\n\ncos1 - cos3 +2cos3 - 2cos5 + 3cos7 - 3cos5 + ... + 89cos177 -89cos179 + 90cos179 - 90cos181 = 90*cos1\n\n\n\ntelescoping to:\n\n\n\ncos1 + cos3 + cos5 + cos7 + ... + cos179 + (-90cos181) = 90*cos1\n\n\n\nsin1cos1 + cos3sin1 + cos5sin1 + cos7sin1 + ... + cos179sin1 + \n\n-90cos181sin1 = 90*cos1sin1\n\n\n\nSince cosAsinB = 1/2 *[sin(A+B) - sin(A-B)]\n\n\n\n1/2 * (sin2 - sin0 + sin4 - sin2 + sin6 - sin4 + ... + sin180 - sin178) \n\n-90cos181sin1 =90*cos1*sin1\n\n\n\ntelescoping to:\n\n\n\n1/2 * (-sin0+sin180) -90cos181sin1 = 90*cos1*sin1\n\n-90cos181sin1 = 90*cos1*sin1\n\n-cos181sin1 = cos1sin1\n\n\n\nSince cosAsinB = 1/2 *[sin(A+B) - sin(A-B)]\n\n\n\n-1/2 * [sin182 - sin180] = 1/2 * [sin2 - sin0]\n\n-sin182 = sin2\n\n\n\nNow if you think about the graph of y=sinX, these two are equal.\n\n\n\nAy, this took way more work than Simon's but I just had to do it again so I would remember. BTW this is not the way they showed me how to do it.[/hide][/hide]" } { "Tag": [ "search", "number theory", "relatively prime" ], "Problem": "let a,b,c be 3 natural numbers such that ab,\\ a>c$. Then \r\n\r\n$BN=CL\\Longleftrightarrow \\{\\ b=c\\ \\}\\ \\vee \\ \\{\\ (b^2+c^2-a^2)[(a-b)(a-c)-2a(s-a)]=0\\ \\}.$\r\n\r\n[u]Remark.[/u] For example, $a=15,\\ b=12,\\ c=5:$\r\n$a-b=3,\\ a-c=10,\\ 2(s-a)=b+c-a=2\\Longrightarrow$\r\n$2a^2-2a(b+c)+bc=(a-b)(a-c)-2a(s-a)=0.$\r\n\r\n[u]Generally:[/u] $m,n\\in N^*,m>n;a=m(m+n),b=2mn,c=m^2-n^2$ and in this\r\ncase the $\\triangle ABC$ is obtuse ($90^{\\circ}-1,b>-1,a\u2260b, prove \r\n $\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{(n+a)(n+b)}=\\frac{1}{b-a}\\int_a^b \\frac{x^a-x^b}{1-x}dx$", "Solution_1": "This is absurd. Let $b=a+1$.\r\nThis equation asserts that \r\n$\\frac1{1+a}=\\sum_{n=1}^\\infty\\frac1{(n+a)(n+a+1)}=\\int_a^{a+1}x^a\\, dx=\\frac{(a+1)^{a+1}-a^{a+1}}{a+1}$.\r\nFor large $a$, this last expression is comparable to $a^a$, which gets very large.", "Solution_2": "Probably, the integral should be from $0$ to $1$ rather than from $a$ to $b$ ;) .", "Solution_3": "I'm sorry.It is from 0 to 1.But how to do it?", "Solution_4": "Expand $\\frac 1{1-x}=1+x+x^2+x^3+\\dots$ and integrate each term $(x^a-x^b)x^n$ separately.", "Solution_5": "Thank you!" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let a,b,c be positive numbers such that $ a^2b^2\\plus{}b^2c^2\\plus{}c^2a^2\\equal{}1$\r\nProve that\r\n $ \\frac{a^3}{b^2\\plus{}c^2}\\plus{}\\frac{b^3}{c^2\\plus{}a^2}\\plus{}\\frac{c^3}{a^2\\plus{}b^2} \\geq \\frac{3\\sqrt{3}}{2}abc(a^4\\plus{}b^4\\plus{}c^4)$", "Solution_1": "\\[ \\frac {a^3}{b^2 + c^2} + \\frac {b^3}{c^2 + a^2} + \\frac {c^3}{a^2 + b^2} \\geq kabc(a^4 + b^4 + c^4),\r\n\\]\r\n\r\n\\[{ k_{max} = \\frac {\\sqrt {3}}{2}}.\r\n\\]", "Solution_2": "[quote=\"fjwxcsl\"]\n\\[ \\frac {a^3}{b^2 + c^2} + \\frac {b^3}{c^2 + a^2} + \\frac {c^3}{a^2 + b^2} \\geq kabc(a^4 + b^4 + c^4),\n\\]\n\n\\[{ k_{max} = \\frac {\\sqrt {3}}{2}}.\n\\]\n[/quote]\r\n I don't understand you. Do you want to tell that my ineq is wrong? :(" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "In the plane is given an infinite set of points such that any 1*1 square contains finitely many points. Prove that one can find two points A,B from the set such that for every other poit X in the set, min(XA,XB)>=0.999AB.", "Solution_1": "It works for $k<1$ (here $k=0.999$). Assume there are no two such points. Then choose any two points $X_0,X_1$. We can find $X_2$ s.t. $X_2X_1 0$ for all $ \\varepsilon > 0$ such that $ \\varepsilon \\equal{} \\frac {1}{3M^3}$.\r\n\r\nWhen $ x > M$, $ \\left|\\frac {x^3 \\plus{} 1}{3x^3 \\plus{} x} \\minus{} \\frac {1}{3}\\right| < \\left|\\frac {x^3 \\plus{} 1}{3x^3} \\minus{} \\frac {1}{3}\\right| \\equal{} \\frac {1}{3x^3} < \\frac {1}{3M^3} \\equal{} \\varepsilon$. By definition of the limit, $ \\lim_{x\\rightarrow{\\infty}}\\frac {x^3 \\plus{} 1}{3x^3 \\plus{} x} \\equal{} \\frac {1}{3}$.", "Solution_4": "thanksssssssssssssssssssssss" } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "Find all polynomial $ P \\in\\mathbb R[x]$ such that there exists a unique $Q \\in \\mathbb R[x]$ with $Q(0)=0$ satisfying\n\\[x+Q(y+P(x)) = y+Q(x+P(y)),\\]\nfor all $x,y \\in \\mathbb R$.", "Solution_1": "[quote=\"hoangclub\"]Find all polynomial $ P$ $ \\in$ $ R[x]$ such that only have unique $ Q$ $ \\in$ $ R[x]$ with $ Q(0) \\equal{} 0$ satisfying:\n $ x\\plus{}Q(y\\plus{}P(x))$=$ y\\plus{}Q(x\\plus{}P(y))$for all x,y $ \\in$$ R$.[/quote]\r\n\r\nLet $ p$ an $ q$ be the degree of polynomials $ P$ and $ Q$.\r\n\r\n1) $ p\\equal{}0$. $ P(x)\\equal{}a$ and so $ x\\plus{}Q(y\\plus{}a)\\equal{}y\\plus{}Q(x\\plus{}a)$. So $ q\\equal{}1$, $ Q(x)\\equal{}cx$ and $ x\\plus{}cy\\plus{}ca\\equal{}y\\plus{}cx\\plus{}ca$ and $ c\\equal{}1$\r\n\r\n2) $ p\\equal{}1$. $ P(x)\\equal{}ax\\plus{}b$ and so $ x\\plus{}Q(y\\plus{}ax\\plus{}b)\\equal{}y\\plus{}Q(x\\plus{}ay\\plus{}b)$\r\nThen, if we put $ y\\equal{}\\minus{}P(x)$ in the original equation, we have $ x\\plus{}P(x) \\equal{} Q(x\\plus{}P(\\minus{}P(x)))$ and LHS has degree 0 (if $ a\\equal{}\\minus{}1$) or 1 and RHS degree 0 (if $ a^{2}\\equal{}1$) or q. And so :\r\n2.1) LHS has degree 0 : $ P(x)\\equal{}b\\minus{}x$ and so $ b\\equal{}Q(2b)$ and obviously $ Q(x)$ is not unique.\r\n2.2) LHS has degree 1 and $ q\\equal{}1$ : $ P(x)\\equal{}ax\\plus{}b$ with $ a\\neq\\minus{}1$ and $ Q(x)\\equal{}cx$ (since $ Q(0)\\equal{}0$). So :\r\n$ x\\plus{}c(y\\plus{}ax\\plus{}b)\\equal{}y\\plus{}c(x\\plus{}ay\\plus{}b)$ and so $ 1\\plus{}ac\\minus{}c\\equal{}0$ and so $ a\\neq 1$ and $ c\\equal{}\\frac{1}{1\\minus{}a}$\r\n\r\n3) $ p>1$\r\nThen, in $ x\\plus{}Q(y\\plus{}P(x))$=$ y\\plus{}Q(x\\plus{}P(y))$, LHS has degree in $ x$ $ \\max(1,pq)$ and RHS has degree $ q$ in $ x$ and, since $ p>1$, this is impossible.\r\n\r\nAnd the solutions are :\r\n$ P(x)\\equal{}a$ (and unique $ Q$ is $ Q(x)\\equal{}x$)\r\n$ P(x)\\equal{}ax\\plus{}b$ with $ a\\notin\\{\\minus{}1,1\\}$ (and unique $ Q$ is $ Q(x)\\equal{}\\frac{1}{1\\minus{}a}x$)" } { "Tag": [ "Putnam" ], "Problem": "Hi Everyone,\r\n\r\nOn another thread, people on MathLinks were discussing the benefits and fallbacks of starting out very early on in mathematics. One common theme among the posts was that one must enjoy what one is doing.\r\n\r\nI would like to start a new thread here. How many people started out mathematics relatively late? What made you decide to study mathematics, either at school or on your own? What attracted you to Olympiad problem solving?\r\n\r\nI was about 19 when I first discovered that I really liked mathematics. Before then I always thought that I would be a theoretical physicist. I ended falling in love with the [i]language[/i] of physics though --- mathematics. Since I was 19 and wasn't in the normal school system here in Ontario, Canada, my math skills were probably very weak compared to the math skills some of you on MathLinks have or had at the age of 19. Also not being in the normal school system, I didn't get opportunities to participate in local and provincial mathematics competitions. I recognized however that Olympiad type problems are a good means to make one learn and absorb \"elementary\" mathematics. I also see Olympiad type problems as a way to keep one's mathematical mind sharp, as one most often needs to be very creative when solving Olympiad type problems.\r\n\r\nNot having much experience with Olympiad type mathematics problems, I tried the Putnam exam three times. I didn't do very well as I one would expect, but I enjoyed the experience. Sitting down and just working on math problems for the hell of it for 6 hours was great, despite my low scores. I now subscribe to Crux Mathematicurum with Mathematical Mayhem. Many of you have probably heard about this problem-solving journal published by the Canadian Mathematical Society. It's a great resource for problems, etc. See [url]http://www.cms.math.ca/Publications/CRUXwithMAYHEM.html[/url] for more on the Journal.\r\n\r\nI have one chance left to take the Putnam exam. I hope this time I'll be much more prepared. In any case, the whole learning process about Olympiad type problems and the fact that I'm not that great at them has made me interested in how to teach Olympiad problem solving to a receptive audience. Do any of you have any advice, or alternatively stories to tell of your own experiences with Olympiad coaches or others of mathematical influence that may be of use to someone interested in how to teach Olympiad problem solving better? Can one even teach mathematical Olympiad type problem solving other than by just getting students to solve problem after problem?\r\n\r\nAnyway, I look forward to replies to this post.\r\n\r\nHave a mathematistic week. :lol: \r\n\r\nYours truly,\r\n\r\nmathfreak", "Solution_1": "Weeeee\n14 YEAR BUMMMMP" } { "Tag": [ "probability and stats" ], "Problem": "Three prisoners Mark, Luke and Thomas find in one cell. Two of they will have to be shoot to you but everyone of they does not know if dovra to die or not. Poiche at least between Luke and Thomas dovra to die Mark asks to one its guard is of the two. The guard answers Luke. To this Mark point rallegra thinking that its probabilita one to die is last from 2/3 to 1/2. She has effectively reason rallegrar?", "Solution_1": "http://en.wikipedia.org/wiki/Monty_Hall_problem" } { "Tag": [], "Problem": "Jeff has an equal number of nickels, dimes and quarters worth a total of $ \\$1.20$. Anne has one more of each type of coin than Jeff has. How many coins does Anne have?", "Solution_1": "Let $ x$ be the number of each coin Jeff has.\r\n\r\n$ 25x\\plus{}10x\\plus{}5x\\equal{}120$\r\n$ 40x\\equal{}120$\r\n$ x\\equal{}3$\r\n\r\nJeff has 3 nickels, 3 dimes, and 3 quarters.\r\n\r\nTherefore Anna has 4 nickels, 4 dimes, and 4 quarters, for a total of $ 3(4)\\equal{}\\boxed{12}$ coins." } { "Tag": [ "LaTeX", "geometry" ], "Problem": "hello\r\ni have a question concerning maple (plots to be precise)\r\nI wanted to ask it here, but didn't find a place (latex help seemed the best spot)\r\n\r\nnow is there already a place where such questions, in my case about maple commands, can be asked\r\n\r\nor, due to maple being a commercial nonfree product, that the majority probably doesn't have here, is it not an option?", "Solution_1": "So uhm forgive me for pushing it up but is there any maple related areas around here?", "Solution_2": "You may ask, but it is not certain that anyone will answer :-). Rather than asking permission to ask questions, you are probably better off just posting some of them -- if there was a really good place for them, some moderator would probably move them there. This forum is a good first guess." } { "Tag": [ "induction" ], "Problem": "Please help me perform induction on this problem:\r\n\r\nProve that if n \u2265 12 then n can be written as a sum of 4\u2019s and 5\u2019s. For example, 23 = 5 + 5 + 5 + 4 + 4 = 3 \u00b7 5 + 2 \u00b7 4. [Hint. In this case it will help to do the cases n = 12, 13, 14, and 15 separately. Then use\r\ninduction to handle n \u2265 16.]", "Solution_1": "12 = 4 + 4 + 4.\r\n\r\nAssume it holds for n. If n can be written as a sum of 4's and 5's, then replace a 4 with a 5 and you get n+1. If n only has 5's in its sum, replace 3 5's (n >= 15) with 4 4's and you get n+1.", "Solution_2": "How about this:\r\n\r\n$ n \\equal{} 4(\\minus{}n\\plus{}5t)\\plus{}5(n\\minus{}4t)$\r\n\r\nFor any integer $ t$. What we need to do is to find an integer $ t$ which gives $ \\minus{}n\\plus{}5t\\geq 0$ and $ n\\minus{}4t\\geq 0$, or equivalently $ \\frac{1}{5}n\\leq t\\leq\\frac{1}{4}n$. This can be achieved for any $ n\\geq 12$", "Solution_3": "Here is a generalisation:\r\n\r\nProve that every sufficiently large positive integer can be expressed as the sum of $ p$'s and $ (p\\plus{}1)$'s where $ p$ is a positive integer.", "Solution_4": "Thanks, so I'll try two cases for n = 5a + 4b, 1: a 5's and 0 4's, 2: a 5's and b 4's.", "Solution_5": "The more relevant generalization is the Chicken McNugget theorem. ;)" } { "Tag": [ "geometry", "geometric transformation", "number theory", "modular arithmetic" ], "Problem": "When n is divided by 5 the remainder is 2, what is the remainder when n+4 is divided by 5?\r\n\r\nThe local high school band has x musicians. Tube players make up 1/6th of the band. There are 3 more than 3 times as many drummers than tuba players. There are 4 fewer flutists than tuba players. How many musicians in the band are not tuba players, drummers, or flutists?\r\n\r\nJohn works 5 days a week, on Mondays he completes 1/2 of his week's work. On Tuesday he completes 1/4 of the remainder of his work for the week. What fraction of the week remains to be done?\r\n\r\nThere are 660 feet in one furlong, and 1/2 of a fathom in one yard. how many fathoms are there in one furlong?", "Solution_1": "[quote=\"nick42\"]1. $ 1$\n2. \n3. $ 3/8$\n4. $ 110$\n\nThat's what i think they are. can anybody double-check? :lol:[/quote]\r\n\r\n2. (5/6)x musicans in the band are not Tube players, \r\n(1/2)x -3 musicans in the band are not drummers\r\n(5/6)x+4 musicans in the band are not flutists", "Solution_2": "Explain your reasoning. :P", "Solution_3": "[quote=\"Forte\"]Explain your reasoning. :P[/quote]\r\n\r\nT - Tube players\r\nD - drummers\r\nF - flutists\r\n\r\n$ T \\equal{} \\frac {1}{6}x$\r\n\r\n$ x \\minus{} T \\equal{} x \\minus{} \\frac {1}{6}x \\equal{} \\frac {5}{6}x$\r\n\r\n\r\n$ D \\equal{} 3T \\plus{} 3 \\equal{} 3 \\cdot \\frac {1}{6}x \\plus{} 3 \\equal{} \\frac {1}{2}x \\plus{} 3$\r\n\r\n$ x \\minus{} D \\equal{} x \\minus{} (\\frac {1}{2}x \\plus{} 3) \\equal{} \\frac {1}{2}x \\minus{} 3$\r\n\r\n\r\n$ F \\equal{} T \\minus{} 4 \\equal{} \\frac {1}{6}x \\minus{} 4$\r\n\r\n$ x \\minus{} F \\equal{} x \\minus{} (\\frac {1}{6}x \\minus{} 4) \\equal{} \\frac {5}{6}x \\plus{} 4$", "Solution_4": "[quote=\"nick42\"] \n2. the number of tuba players is $ 1/6x$, drummers are $ 1/2x \\plus{} 3$, and flutists are $ 1/6x \\minus{} 4$. just subtract these numbers from $ x$ to get the answers. [/quote]\r\n\r\nStatement doesn't say that there are only Tube players, drummers and flutists in the band... there can be some singers and gitarists...", "Solution_5": "Irrelevant. That's why the value is in terms of x.\r\n\r\nTuba players are $ \\frac{1}{6}$ of the band, so the number of Tuba players is $ \\frac{1}{6}x$ regardless, and the number that are not Tuba players is $ \\frac{5}{6}x$\r\n\r\nIf we know the number of Tuba players to be $ \\frac{1}{6}x$, then 3 more than 3 times that amount is $ 3\\left(\\frac{1}{6}x\\right)\\plus{}3\\equal{}\\frac{1}{2}x\\plus{}3$ Drummers, regardless. The number that are not Drummers is then $ \\frac{1}{2}x\\minus{}3$\r\n\r\nIf the number of Tuba players is $ \\frac{1}{6}x$, then 4 less is $ \\frac{1}{6}x\\minus{}4$ Flutists, regardless. The number that are not Flutists, then, is $ \\frac{5}{6}x\\plus{}4$\r\n\r\nFor each of these proofs, it is irrelevant that there might be other instruments, because each one is based on the first value, which was given to us.", "Solution_6": "1. We write this in modulo 5. So,$ n\\equiv2(\\mod5)$. Since we can do addition in modular arithmetic, we add 4 on both sides to get $ n\\plus{}4\\equiv6\\equiv1(\\mod5)$ So, the remainder is 1." } { "Tag": [], "Problem": "How could \"change\" the view we have about cardinalities if we defined alternatively when 2 sets have the same cardinality? Are there such definitions?", "Solution_1": "1) This is not the appropriate forum for such a discussion.\r\n\r\n2) The notion of bijection is so fundamental to anything we could call cardinality that I would hesitate to call any related but distinct concept \"cardinality\" at all. On the other hand, if you restrict to the special case that you are comparing sizes of two subsets of the same set - for example, $ \\mathbb{N}$ - there are many interesting and useful notions of [url=http://en.wikipedia.org/wiki/Natural_density]density[/url]." } { "Tag": [ "Euler" ], "Problem": "Fie triunghiul $ABC$ de afixe $a,b,c$ avand acelasi modul si $a^{3}+b^{3}+c^{3}+abc=0$. Demonstrati ca centrul cercului lui Euler se afla pe cercul circumscris triunghiului $ABC$.", "Solution_1": "[hide]Un mic ajutor, centrul cercului lui Euler are afixul $\\frac{a+b+c}{2}$(Vedeti \"Matematica pentru grupele de performanta\",clasa a X-a,Vasile Pop)[/hide]", "Solution_2": "FIe $|a|=|b|=|c|=R$ si $|a+b+c|=x$ ,unde $R,x\\in \\mathbb R$ ,$R,x>0$\r\nAvem:\r\n$a^{3}+b^{3}+c^{3}+abc=(a+b+c)^{3}-3(a+b+c)(ab+bc+ca)+4abc=(a+b+c)^{3}-3(a+b+c)\\frac{R^{4}(\\overline{a}+\\overline{b}+\\overline{c})}{\\overline{abc}}+4abc=(a+b+c)^{3}-3\\frac{1}{R^{2}}x^{2}\\cdot abc+4abc=0$ \r\n$\\Rightarrow$ $abc=\\frac{-(a+b+c)^{3}}{4-3x^{2}\\frac{1}{R^{2}}}$ ,aplicand modulul obtinen\r\n$R^{3}=\\frac{x^{3}}{|4-3x^{2}\\frac{1}{R^{2}}|}$ $\\Leftrightarrow$ $x^{3}=|4R^{3}-3Rx^{2}|$ $\\Rightarrow$ $R=\\frac{x}{2}$ sau $R=x$.\r\nDar daca $R=x$ $\\Rightarrow$ $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}$ $\\Leftrightarrow$ $3(a+b)(b+c)(c+a)=0$ $\\Leftrightarrow$ $b+c=0$ \r\nDaca $b+c=0$ atunci $a^{3}+b^{3}+c^{3}+abc=a^{3}-ab^{2}=0$ $\\Leftrightarrow$ $a=b$ contradictie cu $ABC$ triunghi.\r\nDeci $R=\\frac{x}{2}$.qed" } { "Tag": [], "Problem": "How many integers in base $b$ have $n$ digits? I know the formula - $(b-1)b^{n-1}$ - but I need a reference. Surely it is somewhere!", "Solution_1": "[hide]$b^n-b^{n-1}$\nSay, 3 digits in base 10\n1000-100=900\n4 digits in base 2\n16-8=8\n[/hide]", "Solution_2": "What exactly are you asking for when you state you desire a \"reference\" ... a formal statement verifying and/or explaining the given formula ...? If it is the latter, I can proudly say I am qualified for the task! :P \r\n\r\n Take the base b expression of n digits:\r\n\r\n$(x_1)b^0 + (x_2)b^1 + (x_3)b^2 ... (x_z)b^{n-1}$\r\n\r\n In the very last term the coefficient $(x_z)$ can be filled with any of the digits from $1$ to $b-1$ a total of $(b-1)-1+1=b-1$ choices. The rest of the coefficients can be filled with any of the digits $0$ to $b-1$ a total of $(b-1)-0+1=b$ choices. Since $n-1$ of our terms can choose from $b$ different digits and only one term from $b-1$, there are $(b-1)b^{n-1}$ such numbers." } { "Tag": [ "geometry", "perimeter" ], "Problem": "1) Two runners jog around an oval path at constant rates, but in opposite directions. One runner completes a lap every 10 minutes, and the other does a lap every 12 minutes. How many minutes elapse between consecutive meetings of the runners? Express your answer as a mixed number. [hide]5 5/11[/hide]", "Solution_1": "[b]Let the perimeter of the path be 600 feet[/b].... Yes I know since i suck at math the only thing i can do is pick numbers for this kind of stuff.....\r\n\r\nthat means one runner runs 60 feet/min and the other runs 50 feet/min.\r\nyou want it to be in the form of \"[i]x[/i]feet/ minute b/c the question asks \" How many minutes elapse......\"\r\n\r\nso that means the runners separate at the rate of 110 feet/minute.\r\n\r\nsince the path is 600 feet, you have 600=110*T (from the D=RT formula)\r\nT is how many minutes elapse......\r\nYou find that T=5.555555555555555555555555555555\r\n- drawing a diagram helps", "Solution_2": "for most people, thats easier, but for people that like to get into algebra, then you can set the distance equal to x. that way helps if you are slow at multipling big numbers and stuff. and you probably want your answer as a fraction, so leaving it as 60/11 is fine i think" } { "Tag": [ "complex numbers", "algebra open", "algebra" ], "Problem": "We have two points M1(z1), M2(z2). We know that the middle of the segment [M1M2] has the afix M((z1+z2)/2).\r\n\r\nThe recyproc is true???", "Solution_1": "You make no sense. What do you mean by afix and recyproc?", "Solution_2": "This is my interpretation of the problem. There are two points, $ M_{1}$ and $ M_{2}.$ The complex numbers representing them are $ z_{1}$ and $ z_{2}.$ We know that if a point $ M$ is the midpoint of $ M_{1}M_{2},$ then the complex number (\"afix\") corresponding to $ M$ is $ \\frac{z_{1}+z_{2}}{2}.$ Does the converse (\"recyproc\") hold? That is, if the complex number of $ M$ is $ \\frac{z_{1}+z_{2}}{2},$ must $ M$ be the midpoint of $ M_{1}M_{2}$?\r\n\r\nThe answer is yes. Let $ M'$ be the midpoint of $ M_{1}M_{2}.$ Then the complex number of $ M'$ is $ \\frac{z_{1}+z_{2}}{2},$ the same as that of $ M.$ Since $ M$ and $ M'$ have the same complex number, they must be the same point. $ M'$ is the midpoint of $ M_{1}M_{2},$ so $ M$ is the midpoint of $ M_{1}M_{2}$." } { "Tag": [ "geometry", "symmetry", "ratio", "circumcircle", "trigonometry", "cyclic quadrilateral", "geometry proposed" ], "Problem": "[img]http://i198.photobucket.com/albums/aa269/ConanKudo5/geo.jpg[/img]\r\n$\\hat BPC= 90$\r\nProve that:AP+AE=PD", "Solution_1": "Let $ EF$ meet $ BC$ at $ L.$ Let $ BP, CP$ meet the incircle $ (I)$ at $ X, Y.$ $ BC$ is polar of $ D$ and $ EF$ is polar of $ A$ with respect to $ (I).$ $ A, P, D$ are collinear $ \\implies$ tangent of $ (I)$ at $ P$ goes through $ L.$ Project the circle $ (I)$ to a circle $ (I')$ and the line $ AL$ to infinity, labeling all projecterd points with primes. By symmetry, $ X'Y' \\parallel E'F' \\parallel B'C'$ $ \\implies$ $ XY$ goes through $ L.$ The diagonal intersection $ Q$ of the cyclic quadrilateral $ EFYX$ lies on the polar $ AD$ of $ L \\equiv EF \\cap XY.$ Let $ XY$ cut $ AD$ at $ Z,$ let $ PF, PE$ cut $ BC$ at $ M, N.$ Projected $ \\triangle P'M'N'$ becomes isosceles right and $ B', C'$ midpoints of $ M'D', N'D'$ $ \\implies$ cross ratios $ (D, M, B, L) \\equal{} (D', M', B', L') \\equal{} \\minus{} 1$ and $ (D, N, C, L) \\equal{} (D', N', C', L') \\equal{} \\minus{} 1$ are harmonic. Suppose $ P'B'$ intersects $ E'M'$ at $ X^*.$ These are P-, M-medians of isosceles right $ \\triangle D'P'M, \\triangle P'M'N'$ $ \\implies$ $ \\measuredangle M'P'B' \\equal{} \\measuredangle E'M'P'$ $ \\implies$ $ \\triangle M'P'B \\sim \\triangle X^*M'B'$ $ \\implies$ $ \\measuredangle P'X^*E' \\equal{} \\measuredangle M'X^*B' \\equal{} \\measuredangle P'M'B' \\equal{} \\measuredangle P'F'E'\\ ( \\equal{} 45^\\circ)$ $ \\implies$ $ X^* \\equiv X' \\in (I')$ $ \\implies$ $ E', X', M'$ are collinear, $ E, X, M$ are collinear and similarly, $ F, Y, N$ are collinear. By central projection of $ BC$ to $ AD$ from $ X,$ the cross ratio $ (D, P, Q, Z) \\equal{} (D, M, B, L) \\equal{} \\minus{} 1$ is also harmonic, $ \\frac {\\overline{DP}}{\\overline{DZ}} \\equal{} \\minus{} \\frac {\\overline{QP}}{\\overline{QZ}}.$\n\n$ \\measuredangle BPC \\equal{} 90^\\circ$ is right $\\iff$ $ XY$ is a diameter of $ (I)$ $\\iff$ $ Z$ is the midpoint of $ DP.$ Using the harmonic cross ratio $(D, P, Q, Z) \\equal{} -1,$ this is equivalent to $ \\frac {\\overline{QZ}}{\\overline{QP}} \\equal{} \\minus{} \\frac {\\overline{DZ}}{\\overline{DP}} \\equal{} \\minus{} \\frac {_1}{^2}$ and by the following lemma, to $ \\frac {\\overline{AD}}{\\overline{AP}} \\equal{} \\left(\\frac {\\overline{QD}}{\\overline{QP}}\\right)^2 \\equal{} \\left(2\\ \\frac {\\overline{QZ}}{\\overline{PQ}} \\plus{} 1\\right)^2 \\equal{} 4.$\nSince power of $ A$ to $ (I)$ is $\\overline{AE}^{2} \\equal{} \\overline{AP} \\cdot \\overline{AD} \\equal{} 4\\ \\overline{AP}^{2},$ this is further equivalent to $ \\overline{AE} \\equal{} 2\\ \\overline{AP}$ $\\iff$ $ \\overline{AE} \\plus{} \\overline{AP} \\equal{} 3\\ \\overline{AP} \\equal{} \\overline{AD} \\minus{} \\overline{AP} \\equal{} \\overline{PD}.$\n\n[color=#0000FF][b]Lemma:[/b][/color] Let $ (I)$ be a circle with diameter $ XY$ and $ Z$ arbitrary point on $ XY.$ Perpendicular to $ XY$ at $ Z$ cuts the circle $ (I)$ at $ P, D.$ Let $ Q \\in PZ$ be arbitrary, let $ XQ, YQ$ cut $ (I)$ again at $ F, E,$ respectively, and let tangents to $ (I)$ at $ E, F$ meet at $ A.$\nThen $ AQ \\perp XY$ and $ \\frac {\\overline{AD}}{\\overline{AP}} \\equal{} \\left(\\frac {\\overline{QD}}{\\overline{QP}}\\right)^2$\n\nLet $ XE, YF$ meet at $ T.$ $ Q$ is orthocenter of $ \\triangle XYT,$ $ TQ \\perp XY.$ By Pasacal theorem for degenerated cyclic hexagon $ XEEFFY,$ the intersections $ T, Q, A$ are collinear $ \\implies$ $ AQ \\perp XY.$ Circumcircle of the right angled kite $ IEAF$ is the 9-point circle $ (K)$ of the $ \\triangle TXY$ $ \\implies$ $ A$ is the midpoint of $ TQ,$ the circumcenter of right angled quadrilateral $ TEQF,$ let $ (A)$ be its circumcircle. As the angles $ \\measuredangle AEI \\equal{} \\measuredangle AFI$ are right, the circles $ (I) \\perp (A)$ are perpendicular. $ XY \\perp TQ$ are perpendicular diameters of the perpendicular circles $ (I), (A)$ and $ (I)$ cuts the line $ TQ$ at $ D, P.$ It follows that the points $ D, P$ and the circle $ (A)$ are coaxal and consequently, powers of $ Q \\in (A)$ to the points $ D, P$ are in the ratio $ \\frac {\\overline{QD}^2}{\\overline{QP}^2} \\equal{} \\frac {\\overline{AD}}{\\overline{AP}}.$", "Solution_2": "[color=darkblue][b]Very nice problem ![/b]\n\n[u][b]Proof.[/b][/u] Suppose w.l.o.g. that $b>c$. Denote $\\{\\begin{array}{c}K\\in BC\\ ,\\ AK\\perp BC\\\\\\ L\\in EF\\cap BC\\\\\\ S\\in LI\\cap AD\\\\\\ \\phi=m(\\widehat{DLI})\\end{array}$. From the well-known property $LI\\perp AD$ (see the above [b][u]Yetti's proof[/u][/b])\n\nobtain $L\\in PP$ and $\\sin\\phi=\\frac{SD}{LD}=\\frac{KD}{AD}$. Prove easily that $KD=\\frac{(b-c)(p-a)}{a}$, $LB=\\frac{a(p-b)}{b-c}$, $LD=\\frac{2(p-b)(p-c)}{b-c}$.\n\n$1\\blacktriangleright$ $\\{\\begin{array}{c}\\boxed{PA+AE=PD}\\\\\\\\ [AP\\cdot AD=(p-a)^{2}]\\end{array}\\|$ $\\Longleftrightarrow$ $\\{\\begin{array}{c}PD-PA=p-a\\\\\\ PA\\cdot (PA+PD)=(p-a)^{2}\\end{array}\\|$ $\\Longleftrightarrow$ $PA\\cdot (PA+PD)=(PD-PA)^{2}$ $\\Longleftrightarrow$\n\n$PD=3\\cdot PA$ $\\Longleftrightarrow$ ${\\begin{array}{c}AD=2(p-a)\\\\\\ (\\sin\\phi =\\frac{KD}{AD})\\end{array}\\|}$ $\\Longleftrightarrow$ $\\boxed{\\sin\\phi=\\frac{b-c}{2a}}$.\n\n$2\\blacktriangleright$ Because the division $(B,C,D,L)$ is harmonically obtain : $\\boxed{PB\\perp PC}$ $\\Longleftrightarrow$ $\\widehat{BPD}\\equiv\\widehat{BPL}$ $\\Longleftrightarrow$\n\n$\\frac{PD}{PL}=\\frac{BD}{BL}$ $\\Longleftrightarrow$ $2\\cdot \\frac{DS}{LD}=\\frac{BD}{BL}$ $\\Longleftrightarrow$ $2\\sin\\phi=\\frac{(p-b)(b-c)}{a(p-b)}$ $\\Longleftrightarrow$ $\\boxed{\\sin\\phi =\\frac{b-c}{2a}}$.[/color]", "Solution_3": "[color=darkblue][b][size=134][u]Remark.[/u][/size]\n\nThe incircle $C(I,r)$ of an acute triangle $ABC$ is tangent (internally) to the circle with the diameter $[BC]$ if and only if $\\boxed{a+h_{a}=b+c}$\n\nand in this case the common tangent point of these circles belongs to the $A$- Gergonne's line if and only if $a=4r$.[/b][/color]" } { "Tag": [ "trigonometry", "geometry unsolved", "geometry" ], "Problem": "Solve the equation:\r\n$sinx+sin2x+sin3x=1$", "Solution_1": "\\[\\begin{array}{l}\\sin (3x) = 3\\sin (x)-4\\sin^{3}x \\\\ \\sin (2x) = 2\\sin (x)\\cos (x) \\\\ \\end{array}\\]\r\n\r\nPlug in and convert cos to sin and have fun.", "Solution_2": "[quote=\"Davidd\"]\n\\[\\begin{array}{l}\\sin (3x) = 3\\sin (x)-4\\sin^{3}x \\\\ \\sin (2x) = 2\\sin (x)\\cos (x) \\\\ \\end{array}\\]\nPlug in and convert cos to sin and have fun.[/quote]\r\n :lol: did u try solving it that way?", "Solution_3": "Here's what you could do:\r\n$\\sin x+\\sin 2x+\\sin 2x \\cos x+\\cos 2x \\sin x=\\sin 2x(1+\\cos x)+2\\sin x \\cos^{2}x= \\sin 2x (1+\\cos 2x)$\r\nSo:\r\n$\\sin 2x (1+2 \\cos x)=1$\r\nform here its trivial. :D" } { "Tag": [ "AMC", "AIME" ], "Problem": "Suppose that $\\mid x_{i}\\mid <1$ for all $i=1,2,3\\dots,n$.\r\n\r\nSuppose further that $\\mid x_{1}\\mid +\\mid x_{2}\\mid +\\mid x_{3}\\mid +\\dots+\\mid x_{n}\\mid =19+\\mid x_{1}+x_{2}+x_{3}\\dots+x_{n}\\mid $.\r\n\r\nWhat is the smallest possible value of $n$?", "Solution_1": "[quote=\"Andreas\"][hide]We have that $19+\\mid x_{1}+x_{2}+\\cdot\\cdot\\cdot+x_{n}\\mid < n$.\nObviously $-1 < x_{i}< 1$. The smallest value of $n$ occurs when for example all the $x_{i}$ are equal to $0$.\nThen $n = 20$.[/hide][/quote]\n[hide]Well, looking at the LHS, each of the $\\mid x_{i}\\mid $ terms is less than 1: $\\mid x_{1}\\mid <1$, $\\mid x_{2}\\mid <1$, etc. If we add all these together, we get $\\mid x_{1}\\mid +\\mid x_{2}\\mid +\\dots+\\mid x_{n}\\mid 0$ and $ abc \\equal{}1$. Prove that: $ \\frac{a\\plus{}b\\plus{}c}{3} \\geq\\ \\sqrt[10]{\\frac{a^3\\plus{}b^3\\plus{}c^3}{3}}$\r\nIt's a hard problem and I remember that there wasn't any \"nice proof\" for it. \r\nNow, Am-Gm solved it :)", "Solution_1": "Will your way work for the following:\r\n$ \\sqrt{\\frac{a\\plus{}b\\plus{}c}{3}} \\ge \\sqrt[19]{\\frac{a^3\\plus{}b^3\\plus{}c^3}{3}}$\r\nwith the same condition? ;)", "Solution_2": "With $ k\\equal{}10$, Am-Gm solved it easily. But indeed, with $ k \\equal{} 9.5$, my way is not effect :oops: \r\n[quote=\"can_hang2007\"]Will your way work for the following:\n$ \\sqrt {\\frac {a \\plus{} b \\plus{} c}{3}} \\ge \\sqrt [19]{\\frac {a^3 \\plus{} b^3 \\plus{} c^3}{3}}$\nwith the same condition? ;)[/quote]", "Solution_3": "Thank you, nguoivn! I proved it by $ uvw,$ but my proof is very ugly.\r\nI posted this inequality here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=230715", "Solution_4": "We have the following lemma:\r\n[i][b]Lemma. [/b]If $ c \\ge \\max\\{a,b\\} \\ge 0$ and $ t\\equal{}\\frac{a\\plus{}b}{2},$ then we have\n$ a^2b^2(a^3\\plus{}b^3\\plus{}c^3) \\le t^4(2t^3\\plus{}c^3).$[/i]\r\nUsing this lemma, we can easily solve the following problem:\r\n[i][b]Problem. [/b]Let $ a,b,c$ be positive real numbers such that $ abc\\equal{}1$. Determine the least constant $ k>0$ such that the following inequality holds:\n$ \\frac{a\\plus{}b\\plus{}c}{3} \\ge \\sqrt[k]{\\frac{a^3\\plus{}b^3\\plus{}c^3}{3}}.$[/i]\r\n\r\nIn my opinion, the mixing variables method is the best way for this problem (Certainly, except EV of Vasc). I would like to dedidated the lemma to arqady as you're the author of this nice problem. :)", "Solution_5": "[quote=\"nguoivn\"]Given $ a, b, c > 0$ and $ abc \\equal{} 1$. Prove that: $ \\frac {a \\plus{} b \\plus{} c}{3} \\geq\\ \\sqrt [10]{\\frac {a^3 \\plus{} b^3 \\plus{} c^3}{3}}$\nIt's a hard problem and I remember that there wasn't any \"nice proof\" for it. \nNow, Am-Gm solved it :)[/quote]\r\n@arqady: And here is my proof with $ k \\equal{} 10$. I hope that you'll like it :)" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Prove that for any positive reals $a_i,0\\le i \\le 8$\r\n$\\displaystyle\\sum_{i=1}^8\\left(\\frac{a_i}{a_i+a_{i+1}}\\right)^3\\ge 1,$ $a_9=a_1$", "Solution_1": "Put $x_i=\\frac{a_{i+1}}{a_i}$ Then $\\prod x_i=1$ and we obtain\r\n$\\sum\\frac1{(1+x_i)^3}\\ge1$ which is a special case of [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=202141#p202141]the first one of the nice ineqs given here.[/url]" } { "Tag": [ "inequalities", "linear algebra", "linear algebra unsolved" ], "Problem": "Let $A$ be a real matrix. Suppose that $|a_{ij}| \\leq M$ for some M > 0. \r\nProve the inequality \r\n\\[ \\det A \\leq n^{\\frac{n}{2}}M^n. \\]\r\n\r\nShow that this estimate cannot be improved for $n = 2^m$.", "Solution_1": "You just have to prove that if all the coefficients satisfy $|a_{i,j}|<1$ then $|det(A)| \\leq n^{\\frac{n}{2}}$. And this is easy (Hadamard inequality) if you see a determinant as a volume.", "Solution_2": "[quote=\"alexilic\"]\nShow that this estimate cannot be improved for $n = 2^m$.[/quote]Hadamard matrices", "Solution_3": "I think it is an open problem to know for wich $n$ this inequality can be improved.", "Solution_4": "Another proof for hadamard ineq is using Gram-Schmidt method of orthonormal basis in n dimensions.\r\n[this is not the place to expand this...]\r\nFor the cases when equality holds it may be helpful the following\r\n\r\n[url=http://mathworld.wolfram.com/HadamardsMaximumDeterminantProblem.html]link[/url]\r\n[url=http://www.answers.com/main/ntquery;jsessionid=330owgv4aw4tu?method=4&dsid=2222&dekey=Hadamard+matrix&gwp=8&curtab=2222_1&sbid=lc08a&linktext=Hadamard%20matrix]link2[/url]" } { "Tag": [ "function", "algebra", "partial fractions", "complex analysis" ], "Problem": "When you're creating a Laurent series, does the annulus have to be just around one pole, or could it be around multiple poles.\r\n\r\nFor example, let $ f(x) \\equal{} \\frac{1}{x(x^2\\plus{}1)}$. This has simple poles at $ 0$, $ i$, and $ \\minus{}i$. You can have a Laurent series in the region $ 0<|z|<1$, centered at $ 0$. Do you also get a Laurent series (a different one) for $ 1<|z|<\\infty$?\r\n\r\nCan you make a Laurent series centered at a point that is not itself a pole, but for which the annulus contains poles. For example, for the function above, could you make a Laurent series around $ 1$ for $ \\sqrt{2}<|z|<\\infty$?", "Solution_1": "I think I've figured out the first part of my question.\r\n\r\nFor $ 0 < |z| < 1$, we have\r\n\\[ f(z) \\equal{} \\frac {1}{z}\\left(1 \\minus{} z^2 \\plus{} z^4 \\minus{} z^6 \\plus{} \\cdots\\right)\r\n\\]\r\nFor $ 1 < |z| < \\infty$, the series looks like\r\n\\[ \\frac {1}{z^3} \\minus{} \\frac {1}{z^5} \\plus{} \\frac {1}{z^7} \\minus{} \\frac {1}{z^9} \\plus{} \\cdots\r\n\\]\r\nThe first was easy. For the second, I used the Residue Theorem to compute the coefficients. Is this the typical way to do it?\r\n\r\nAlso, why are there infinitely many negative terms in the second series even though there are no essential singularities?", "Solution_2": "You get a Laurent series for each annulus between poles. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=681920#681920]here[/url] for an example and a method for rational functions. Each term in the partial fractions expansion has a series at zero and/or a series at $ \\infty$; the various Laurent series combine these based on which series converge on each region." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $\\large a;b;c$ are real numbers. Prove that:\r\n$\\large a(a+b)^5+b(b+c)^5+c(c+a)^5 \\geq 0$ :lol:", "Solution_1": "[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=a%28a%2Bb%29%5E5&t=55723]www.mathlinks.ro/Forum/viewtopic.php?highlight=a%28a%2Bb%29%5E5&t=55723[/url]" } { "Tag": [ "\\/closed" ], "Problem": "Many of you may have noticed that you once again have access to past class materials. This will only be available for a couple of days, so please be sure to print out and save materials today or tomorrow. This is the only time we will make these past materials available.", "Solution_1": "Ummm...I DIDN'T! Are they still here? And how can I access them?", "Solution_2": "Sing - this was for people who were in classes long ago. You can access the files to classes you are in currently by clicking on My Classes.", "Solution_3": "Oh, haha, I guess it was wishful thinking to have FREE transcripts. Thanks", "Solution_4": "Check out the old Math Jams - plenty of free transcripts there!" } { "Tag": [ "AMC", "AIME", "geometry", "3D geometry" ], "Problem": "I'm currently working on a Mock State (sorry, no team :(). PM me if you would like to take it, so I can get a better sense to see if I can turn this into a competition. :)", "Solution_1": "Put me in. :D :D :D", "Solution_2": "Competitors:\r\n\r\naleph0\r\nAIME15\r\ndnkywin\r\ngauss1181\r\n\r\nAnd [b]PM me[/b], cough cough...", "Solution_3": "Ill take it too if theres pm reminders.", "Solution_4": "Competitors:\r\n\r\naleph0\r\nAIME15\r\nAIME_is_hard\r\nBOGTRO\r\nBugi\r\ndnkywin\r\nFantasyLover\r\ngauss1181\r\nhighflyer_10\r\nmewto55555\r\nshentang?\r\n\r\nAnd [b]PM me[/b], cough cough (mewto and gauss)...", "Solution_5": "i'll take it", "Solution_6": "READ THE POST!\r\n\r\nPM him.", "Solution_7": "i thought this was to get an idea....so if it is then it would have to depend on school work for me. So I might do it if I have time.", "Solution_8": "Competitors:\r\n\r\n007math\r\naleph0\r\nAIME15\r\nAIME_is_hard\r\nBOGTRO\r\nBugi\r\ndnkywin\r\nFantasyLover\r\ngauss1181\r\nhighflyer_10\r\nmewto55555\r\nshentang?\r\n\r\nAnd [b]PM me[/b], cough cough (mewto and gauss)...\r\n\r\nOk, this will become a competition.", "Solution_9": "Is it just gonna be like a 1 round competition, or like elimination?", "Solution_10": "A 1-rounder. There might be a countdown round if enough people take it. And it is due to come out next week.\r\n\r\nCompetitors:\r\n\r\n007math\r\naleph0\r\nAIME15\r\nAIME_is_hard\r\nandrewliu\r\nBOGTRO\r\nBugi\r\ndnkywin\r\nFantasyLover\r\ngauss1181\r\nhighflyer_10\r\nisabella2296\r\nmewto55555\r\nshentang?", "Solution_11": "Progress:\r\n\r\n30 sprint questions, typed\r\n8 target questions, need to be typed\r\n30 sprint solutions, 20 need to be typed\r\n2 target solutions\r\n\r\nHopefully, it will be out by Sunday.", "Solution_12": "What are sprint and target???", "Solution_13": "Sprint: 30 questions, 45 minutes (usually 40, but I'm giving more, you'll see why)\r\n\r\nTarget: 4 pairs of questions, 6 minutes each\r\n\r\nThese are the two main individual rounds that form the basis of MathCounts.", "Solution_14": "007math \r\naleph0 \r\nAIME15 \r\nAIME_is_hard \r\nandrewliu \r\nBOGTRO \r\nBugi \r\ndnkywin \r\nFantasyLover \r\ngauss1181 \r\nhighflyer_10 \r\nisabella2296 \r\nmewto55555 \r\nshentang\r\nDojo\r\n\r\npm'd", "Solution_15": "I borrowed a template from mr frost about a year or so ago, then modified it a bit", "Solution_16": "[quote=\"5849206328x\"]I borrowed a template from mr frost about a year or so ago, then modified it a bit[/quote]\r\n\r\nMay I have that template? :roll:", "Solution_17": "[quote=\"ernie\"][quote=\"5849206328x\"]I borrowed a template from mr frost about a year or so ago, then modified it a bit[/quote]\n\nMay I have that template? :roll:[/quote]\r\n\r\nWhathe aid--but to me. :roll: :D", "Solution_18": "May I have the template? :roll:", "Solution_19": "you probably should pm me if you want it, I prefer not to have that comment reduce this thread to spam...", "Solution_20": "#1 technically isn't a cube :P", "Solution_21": "How would you know? You're on your honor. :P", "Solution_22": "I HAVE RECEIVED ONLY ONE SET OF ANSWERS SO FAR. PLEASE SUBMIT. THANK YOU, THAT IS ALL.\r\n\r\noops, caps :lol:", "Solution_23": "bump bump\r\n\r\nplz submit...", "Solution_24": "Extending the deadline to January 7, 8 pm EST.", "Solution_25": "[quote=\"ernie\"]Extending the deadline to January 7, 8 pm EST.[/quote]\r\nGood idea. That way I have more winter break time to do USAMTS. :D", "Solution_26": "Can I join now?", "Solution_27": "No problem!\r\n\r\nI need answers from:\r\n\r\n#5-1Mole\r\n007math\r\naleph0\r\nAIME15\r\nAIME_is_hard\r\nandrewliu\r\nBOGTRO\r\nBugi\r\nDiscrete_Math\r\ndnkywin\r\nDojo\r\nFantasyLover\r\ngauss1181\r\nhighflyer_10\r\nisabella2296\r\nmewto55555\r\nshentang\r\ntheprodigy\r\nxpmath\r\n\r\nDeadline: January 10, 8 pm EST", "Solution_28": "ok, since this is failing, i will send out PM reminders\r\n\r\ndeadline (re-extended): jan 12, 8 pm est", "Solution_29": "I've now already submitted my answers. :)" } { "Tag": [ "quadratics", "induction", "algebra", "polynomial", "limit", "quadratic formula" ], "Problem": "Let $ f(x) \\equal{} x^2 \\plus{} 2007x \\plus{} 1$. Prove that for every positive integer $ n$, the equation $ \\underbrace{f(f(\\ldots(f}_{n\\ {\\rm times}}(x))\\ldots)) \\equal{} 0$ has at least one real solution.", "Solution_1": "Solution 1 :\r\n[u] Result [/u]\r\nConsider the equation :\r\n$ f(x) = a$ where $ a\\geq \\frac { - 2007^2 + 4}{4}$ has solution $ x_1,x_2$ then \r\n${ x_1 = \\max\\{x_1,x_2}\\geq \\frac { - 2007}{2}$\r\nProof : From $ x_1 + x_2 = - 2007$\r\nWe prove by induction $ f_{n}x = a$ has at least real solution if $ a > \\frac { - 2007^2 + 4}{4}$\r\nSuppose it is true for $ n$ \r\nWe prove it is true for $ n + 1$\r\n$ f_{n + 1}(x) = f_n(x)^2 + 2007f_{n}(x) + 1$\r\nIf $ f_n(x_n) = x_1$ then $ x_n$ is a root of $ f_{n + 1}(x) = 0$\r\nBut we have $ x_1 > \\frac { - 2007^2 + 4}{4}$ so the equation has solution .Call it is $ x_{11}\\geq \\frac{-2007}{4}$\r\nContine consider equation $ f_{n-1}=x_{11}$ ..\r\nIt has solution. \r\nOur induction claim.\r\nApply this result our problem was be claim.\r\nI know that have other solution so please contine discuss it .", "Solution_2": "[quote=\"cyshine\"]Let $ f(x) = x^2 + 2007x + 1$. Prove that for every positive integer $ n$, the equation $ \\underbrace{f(f(\\ldots(f}_{n\\ {\\rm times}}(x))\\ldots)) = 0$ has at least one real solution.[/quote]\r\n\r\nThe equation $ f(x)=x$ has two solutions $ a0$\r\n\r\nAnd so $ g(x)$ has at least one real root in $ (b,+\\infty)$", "Solution_3": "Define $ a\\equal{}\\minus{}\\frac{2007}{2}\\,,\\,b\\equal{}\\minus{}a^2\\plus{}1$ and the intervals $ D\\equal{}[a,\\infty[\\,,\\,W\\equal{}[b,\\infty[$, then we have $ 0\\in D\\subset W$ and $ f(x)\\equal{}(x\\minus{}a)^2\\plus{}b$, i.e. $ f(\\mathbb{R})\\equal{}f(D)\\equal{}W$ and therefore $ f(\\ldots f(\\ldots (\\mathbb{R})\\ldots )\\equal{}W$", "Solution_4": "$ f(x)\\equal{}q$ has solution if $ q>1\\minus{}\\frac{2007^2}4$. One solution for $ f(x)\\equal{}q$ is $ x_1\\equal{}\\frac{\\sqrt s \\minus{}2007}2>\\minus{}2007>1\\minus{}\\frac{2007^2}4$. So $ f(x)\\equal{}x_1$ has a root $ x_2$ such that $ f(x)\\equal{}x_2$ has a root $ x_3$..., and if we assume $ q\\equal{}0$, we can do this process indefinitely, which is what we had to prove.", "Solution_5": "Here is general problem (can solve with my method )\r\nFind then number of solution of the equation : $ f_{n}(x)\\equal{}0$", "Solution_6": "I think that this works.\n\n[b]Solution[/b]\n\nLet $Q(x)=\\underbrace{f(f(\\ldots(f}_{n\\ {\\rm times}}(x))\\ldots))$.\n\nWe have that\n\\[x \\rightarrow \\infty \\quad \\Rightarrow \\quad f(x) \\rightarrow \\infty\\]\nTherefore we also have that\n\\[x \\rightarrow \\infty \\quad \\Rightarrow \\quad Q(x) \\rightarrow \\infty\\]\nHence there exists arbitrarily large value $\\alpha$ such that $Q(\\alpha)>0$.\n\nLet $\\beta$ denote a root of the equation $f(x)=x$. The quadratic formula yields that $\\beta < 0$. This yields that $Q(\\beta)=\\beta<0$.\n\nBy the intermediate value theorem, there exists $x$ between $\\alpha$ and $\\beta$ such that $Q(x)=0$.", "Solution_7": "i have done a solution\nif anything is wrong, please rectify it\nwe have$f(x)=x(x+2007)+1$ a parabola\nwhere $f'(x)>0$ when $x>-\\frac{2007}{2}$, $f'(x)<0$ when $x<-\\frac{2007}{2}$\nso $f(x)$ is increasing in $[-\\frac{2007}{2},\\infty)$ and decreasing in $(-\\infty,-\\frac{2007}{2}]$\nnow $f(0)=1, f(f(0))=f(1)>f(0)=1, f(f(f(0)))>f(1)>f(0)=1$\nso $f^n(0)>1>0$ for all n in $N$\nagain\n$f(-2006)=-2006+1=-2005, f(f(-2006))=f(-2005)0$. Next, let $\\alpha$ be the negative root of $x^2+2006x+1=0$. Thus, \n\\[f^n(\\alpha)=\\alpha\\]\nfor all $n$, and so since $f^n(\\alpha)<0 -1 $\n[color=#f00]Claim: [/color] We can now use induction to make the claim that \nIf $r > -1$ satisfies $f^{n}(r) = 0$ for some positive integer $n$, then there exists a real $x > -1$ such that $f(x) = x^2+2007x+1 = r.$ \n[i]Proof.[/i] One $x$ that satisfies $x^2 + 2007x + 1-r = 0$ is $x = \\frac{-2007 + \\sqrt{2007^2 +4r-4}}{2},$ which is obviously real and greater than -1 for any $r > -1.$ \nHence if we use the first claim, this second claim shows that because one of the roots to $f(x)$ is greater than -1, there will always be a real root greater than -1 to $f^{n}(x)$ for all positive integers $n.$ Hence $r > -1 \\ge \\frac{4 - 2007^2}{4},$ which is what we wanted to show and we're done. " } { "Tag": [ "rotation", "analytic geometry", "LaTeX", "AMC" ], "Problem": "Triangle $ OAB$ has $ O \\equal{} (0,0)$, $ B \\equal{} (5,0)$, and $ A$ in the first quadrant. In addition, $ \\angle{ABO} \\equal{} 90^\\circ$ and $ \\angle{AOB} \\equal{} 30^\\circ$. Suppose that $ \\overline{OA}$ is rotated $ 90^\\circ$ counterclockwise about $ O$. What are the coordinates of the image of $ A$?\r\n\r\n$ \\textbf{(A)}\\ \\left( \\minus{} \\frac {10}{3}\\sqrt {3},5\\right) \\qquad \\textbf{(B)}\\ \\left( \\minus{} \\frac {5}{3}\\sqrt {3},5\\right) \\qquad \\textbf{(C)}\\ \\left(\\sqrt {3},5\\right) \\qquad \\textbf{(D)}\\ \\left(\\frac {5}{3}\\sqrt {3},5\\right) \\\\ \\textbf{(E)}\\ \\left(\\frac {10}{3}\\sqrt {3},5\\right)$", "Solution_1": "Answer:\r\n[hide]This forms a 30,60, 90 triangle.... \njust draw it out with the angles and you'll find that point A is at:\n(5,5/rt(3)) or (5,5rt(3)/3)\nBy rotating it 90 degrees counterclockwise, you move point to a point with (-,+) coordinates.. \nso you already know the answer is A or B.\nBy rotating Point A around point O, you're basically moving Point A to a point that is as far from point O as Point A is from Point O...\n\nby looking, you can tell that B is just as far from Point O as Point A is as from Point O\n\nAns:B[/hide]", "Solution_2": "[hide]call any point on the negative x-axis, $ D$. So when you rotate $ OA$ by $ 90$ degrees, the angle $ A^1 OD(\\mbox{a the image of a it is meant to be a prime but well okay}) \\equal{} 180 \\minus{} 30 \\minus{} 90 \\equal{} 60$. and since $ OA \\equal{} \\frac {10}{\\sqrt {3}}$, the x coordinate and y coordinate are part of the 30-60-90 triangle and hence the coordinates are $ (\\frac { \\minus{} 5\\sqrt {3}}{3}, 5) \\longrightarrow \\fbox{\\mbox{B}}$[/hide]\r\n\r\nhad some LaTeX issues sorry." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "I was trying to look at the problem and solution at a book and i can't understand some lines on the solution.\r\nProblem:\r\n\r\nShow by $\\epsilon-\\delta$ definition such that $lim_{x\\to2}x^{2}+5 = 9$.\r\n \r\nWe want $x^{2}+5 = 9$ to be less than $\\epsilon$\r\n$|x^{2}+5-9| = |x^{2}-4|$ \r\n$=|(x+2)(x-2)| < \\delta|x+2|$ \r\n\r\nWe therefore cannot choose $\\frac{\\delta}{|x+2|}$ since $\\delta$ has to be constant\r\n\r\nNote that $|x-2| < 1$ implies that $x \\in (1,3)$\r\n[b]QUESTION[/b]: Where did 1 came from in $|x-2| < 1$ . Do we always use 1?\r\n\r\nHence if $|x-2| < 1$, we must have $|x+2|<5$\r\nSo that $|x+2\\parallel x-2|<\\delta|x+2|<5\\delta$\r\n\r\nProof:\r\n\r\nLet $\\epsilon > 0$\r\nchoose $\\delta = \\frac{\\epsilon}{5}$\r\n...\r\n...", "Solution_1": "[quote]QUESTION: Where did 1 came from in $|x-2| < 1 .$ Do we always use 1?[/quote]\r\nYou have to do something like that, to estimate the less-critical $|x+2|$ factor. But using 1 is just arbitrary. It's a convenient positive number. Do we always use 1? Let's just ask this: if I asked you to name a convenient positive number, what would you say?", "Solution_2": "But can we use 1 all the time? I think there are cases that you cannot use 1", "Solution_3": "Of course there could be cases where you might need something else. This isn't a rule - it's a convenience. Keep your eyes open.", "Solution_4": "We therefore cannot choose $\\frac{\\delta}{|x+2|}$ since has to be constant \r\n\r\nI am just wondering, where did $\\frac{\\delta}{|x+2|}$ came from? Isn't that suppose to be $\\frac{\\epsilon}{|x+2|}$\r\n\r\nSorry I'm so dumb at this." } { "Tag": [ "MATHCOUNTS" ], "Problem": "What is the greatest hole number that MUST be a factor of the sum of any four consecutive whole numbers? (number 5)\r\n\r\nJohn, Mike, and Chantel will divide a pile of pennies amongst themselves using the following process:\r\n\r\nThe number of pennies is counted:\r\nIf the number of pennies is even, Mike will get half the pile.\r\nIf the number of pennies is odd, one penny is given to Chantel, and then John will get half the pennies remaining in the pile.\r\n\r\nIf the number of pennies in the pile is 2005, how many pennies will Mike end up with?\r\nNumber eight. ^^^", "Solution_1": "it said consecutive odd whole numbers for number five right?\r\nguess and check and you get 8.", "Solution_2": "5. (the way hwenterprise stated it)\r\n[hide]The sum of 4 consecutive whole numbers can be expressed as $x+(x+1)+(x+2)+(x+3)$. You can simplify this to get $4x+6$. The greatest number that $4x+6$ is always divisible by is $2$.[/hide]\n\nEdit: The actual question on the test said \"What is the greatest whole number that MUST be a factor of the sum of any four consecutive positive [b]odd[/b] numbers?\n\n5. (the way it was stated on the MATHCOUNTS chapter test)\n[hide]The sum of 4 consecutive odd numbers can be expressed as $x+(x+2)+(x+4)+(x+6)$, which can be simplified to $4x+12$. Note that x must always be odd, so $4x=4 (mod 8)$. $12$ is also $4 (mod 8)$. $4+4=0(mod8)$, so $4x+12$ must always be divisible by 8.[/hide]", "Solution_3": "number 8 is just a long process. Just keep doing it over and over and over again.", "Solution_4": "For #5, I plugged in values and got it right. (I got all 8 in chapter target)", "Solution_5": "For #8:\r\nThe question begins with 2005 pennies in the pile. This is odd, so Chantel gets 1 and John gets 1002. There are now 1002 pennies in the pile.\r\n\r\n1002 is even, so Mike getrs 501 pennies. There are now 501 pennies in the pile.\r\n\r\n501 is odd, so Chantel gets 1 penny (total 2) and John gets 250 pennies (total 1252). There are now 250 pennies in the pile.\r\n\r\n250 is even, so Mike gets 125 pennies (total 626). The pile now has 125 pennies.\r\n\r\n125 is odd, so Chantel gets 1 (3 total) and John gets 62. There are 62 left.\r\n\r\n62 is even, so Mike gets 31 (total 657). There are 31 left.\r\n\r\n31 is odd, so Chantel gets 1 (total 4) and John gets 15. There are 15 left.\r\n\r\n15 is odd, so Chantel gets 1 and John gets 7. There are 7 left.\r\n\r\n7 is odd, so Chantel gets 1 and John gets 3. There are 3 left.\r\n\r\n3 is odd, so Chantel gets 1 and John gets 1. There is one penny left. This then goes to Chantel. Mike has [b]657[/b] pennies.", "Solution_6": "I feel sorry for chantel" } { "Tag": [ "logarithms", "function", "linear algebra", "matrix", "linear algebra open" ], "Problem": "Let $A \\in M_n(\\mathbb{C})$. What conditions must we impose on A such that there exists $B \\in M_n(\\mathbb{C})$ with the property that exp(B)=A.\r\n\r\nRecall that $exp(X)=\\sum_{k=0}^{\\infty}\\frac{X^k}{k!}$ for $X\\in M_n(\\mathbb{C})$.", "Solution_1": "There is a theorem:\"spectral mapping theorem\" (applied in this case, because it is more generally)\r\nlet $ \\sigma=\\{a\\in C|A-aI_n \\mbox{is not invertible} \\} $ and f a function C-analitique such that we can define f(A). Then \r\n$ f(\\sigma(A))= \\sigma(f(A)). $", "Solution_2": "[quote=\"amfulger\"]Let $A \\in M_n(\\mathbb{C})$. What conditions must we impose on A such that there exists $B \\in M_n(\\mathbb{C})$ with the property that exp(B)=A.\n[/quote]\r\n\r\nAs caugusts did, let $\\sigma (A)=\\{ a\\in\\mathbb{C} |A-a\\cdot I_n \\ is\\ not\\ invertible\\}$\r\n\r\nI'll prove that a matrix B such as exp(B)=A exists iff $0\\ not\\ \\in \\sigma (A)$.\r\nIt is easy to see, as caugust said, that $\\sigma (\\exp(B))=\\exp(\\sigma (B))$. \r\n\r\n$ \\Rightarrow $ tIf a matrix B exists such as exp(B)=A, then $\\sigma (A)=\\exp(\\sigma (B)) \\ not \\ \\ni 0$ \r\n\r\n$ \\Leftarrow $ What if $0\\ not\\ \\in \\sigma(A)$?\r\n\r\n$\\exp(U\\cdot B\\cdot U^{-1})=\\sum_{k=0}^{\\infty} \\frac{(U\\cdot B\\cdot U^{-1})^k}{k!}=U\\cdot \\exp(B)\\cdot U^{-1}=U\\cdot A\\cdot U^{-1}$\r\nThis goes to prove that it is enough to solve the problem for matrices A in Jordan canonical form.\r\nIt is also easy to see that it is enough to solve the problem for matrices A that are actually Jordan cells.\r\n\r\nDenote by $E=(e_{i,j})$, $e_{i,j}=\\{ \\begin {array}{cc} 1,&if\\ j=i+1 \\\\ 0,& if j\\neq i+1\\end{array}$. \r\nIt is easy to see that $E^k=(e_{i,j}^k)$, $e_{i,j}^k=\\{ \\begin {array}{cc} 1,&if\\ j=i+k \\\\ 0,& if j\\neq i+k\\end{array}$, for $k\\geq 0$. Also for $k\\geq n,\\ E^k=0_n$.\r\n\r\nSince A is a Jordan cell and $0\\ not \\in \\sigma(A)$ ,$A=e^x \\cdot I_n+E$ for some complex number x. I'll prove that there exists a matrix B of the form $B=x\\cdot I_n+\\sum_{k=1}^{n-1} a_k\\cdot E^k$ such that exp(B)=A, $a_k\\in \\mathbb{C},\\ for\\ 1\\leq k \\leq n-1$ .\r\n\r\nLets solve the equation $\\exp(x\\cdot I_n+\\sum_{k=1}^{n-1} a_k\\cdot E^k)=e^x\\cdot I_n+E$.\r\nIt is easy to see that if XY=YX, then exp(X+Y)=exp(X)exp(Y).\r\nLet $T=\\sum_{k=1}^{n-1} a_k\\cdot E^k$. T is superior triangular with all the elements of the diagonal 0, so $\\sigma(T)=\\{0\\}$. So T is nilpotent, so $T^n=0_n$.\r\n$\\exp(B)=\\exp(x\\cdot I_n+T)=\\exp(x\\cdot I_n)\\cdot \\exp(T)=e^x\\cdot I_n\\cdot ( \\sum_{k=0}^{n-1}\\frac{T^k}{k!})$.\r\nSo we have to solve the equation $\\sum_{k=1}^{n-1} \\frac{( \\sum_{i=1}^{n-1} a_i\\cdot E^i)^k}{k!}=e^{-x}\\cdot E$.\r\nA quick inductive argument shows that this equation is always solvable.", "Solution_3": "Statement: If $0 \\ not\\in \\sigma(A)$ then for all $m \\in \\mathbb{N},\\ m\\geq 1$ there exists a matrix $C_m$ such that $C_{m}^{m}=A$.\r\n\r\nProof: If $0 \\ not\\in \\sigma(A)$ then there exists B such as exp(B)=A. Take $C_m=exp(\\frac{B}{m})$. It is easy to see that $C_{m}^{m}=A$.\r\n---------------------------------------------------------------------------------------------\r\nOpen questions: If for all $m \\in \\mathbb{N}^*$ there exists $C_m$ such as $C_{m}^{m}=A$, does it follow that 0 is not in $\\sigma (A)$? Does it follow that there exists a matrix B such as exp(B)=A?" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Prove that for every two natural numbers like $a$ and $n$ we have $a^{n}=a^{n-\\phi (n)}(\\mod n)$.", "Solution_1": "Quite easy.But we should let $(a,n)=1$.", "Solution_2": "[quote=\"Hawk Tiger\"]Quite easy.But we should let $(a,n)=1$.[/quote]\r\nI don't think so... Not must be $(a,n)=1$\r\n(Is it right grammer? :blush: )", "Solution_3": "[quote=\"mathematica\"]Prove that for every two natural numbers like $a$ and $n$ we have $a^{n}=a^{n-\\phi (n)}(\\mod n)$.[/quote]\r\nif $n-\\phi (n)=0$ ,$a=n$,it is not true.(Am I right?)", "Solution_4": "It's always true and doesn't require any conditions.\r\n\r\n@Hawk Tiger: wh do not have $n-\\varphi(n)=0$ very often.", "Solution_5": "$(a,n)=1$ is not needed, just prove that $n-\\varphi(n)\\geq \\beta$ for $p^\\beta \\parallel n$ ($p$ prime) which is very obvious :)", "Solution_6": "Sorry.I consider it as $a^{k}=a^{k-\\varphi (n)}$ :blush: \r\nThanks." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Let A,B,C,D be a convex quadrilateral. Points K,L,M,N are the midpoints of AB,BC,CD,DA respectively. KM intersects AC and BD at P and Q respectively. LN intersects AC and BD at R and S respectively. Prove that if AP x PC=BD x DQ then $ \\frac {PR}{QS} \\equal{} \\frac{SD \\minus{} BQ}{RC \\minus{} AP}$", "Solution_1": "you have got a typo: $ AP \\cdot PC \\equal{} BQ \\cdot QD$." } { "Tag": [ "rotation" ], "Problem": "Last year, Laura rotated the tires on her car so that all 5, including the spare, received equal wear. Laura drove 20,000 miles last year. How many miles of wear did she put on each tire during the year?", "Solution_1": "[color=00FFFF][code]The answer is 16000[/code][/color][/b]", "Solution_2": "at any given time, there are 4 wheels on a car, so the total miles on the wheels she has must be $ 20,000\\cdot4\\equal{}80,000$. however, these $ 80,000$ miles are spread out over five wheels equally, giving us $ \\frac{80,000}5\\equal{}\\boxed{16,000}$, as prakhar23 said before me." } { "Tag": [ "vector" ], "Problem": "Find a vector with a direction opposite to v = 2i - 5j with thalf the length of v.", "Solution_1": "$-\\frac{1}{2}(2i-5j)$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c,d > 0$ . Prove that :\r\n\r\n$ (\\frac {a \\plus{} b}{a \\plus{} b \\plus{} c})^{2} \\plus{} (\\frac {b \\plus{} c}{b \\plus{} c \\plus{} d})^{2} \\plus{} (\\frac {c \\plus{} d}{c \\plus{} d \\plus{} a})^{2} \\plus{} (\\frac {d \\plus{} a}{d \\plus{} a \\plus{} b})^{2}\\geq\\frac {16}{9}$", "Solution_1": "the inequality is wrong.try a=b=c=2 ,d=4 and the inequality is reversed" } { "Tag": [ "function", "algebra", "domain", "complex analysis", "complex numbers", "complex analysis unsolved" ], "Problem": "I have gotten through most part of the question but i am totally blocked at this part, hope someone can help...\r\n\r\nLet f be a non-constant entire function such that |f(z)|=1 for all complex numbers z satisfying |z|=1. Show that for any complex number y from the open unit disc, there exist a complex number z (from the open unit disc) such that f(z)=y.\r\n\r\nI have proven the following statements which was given in the lecturer's hint. \r\n\r\n(i) The above function f maps the open unit disc to the open unit disc.\r\n(ii) g(z)= (z-a)/(1-a_bar*z) where |a|<1, is an analytic function mapping the unit disc to the unit disc.\r\n(iii) Let h be an analytic non-constant function in the bounded domain D and continuous up to and including the boundary of D. If |f(z)| is constant on the boundary of D then h must have at least one zero in D.\r\n\r\nRemarks: I strongly believed that schwarz's lemma needs to be used in the question. I have been trying to show, but to no success, that there exist some non-zero z in the open unit disc such that |f(z)|=|z|, which will allow me to complete the question.\r\n\r\nHope someone can help :roll:", "Solution_1": "let $ w \\in\\mathbb{D}\\equal{}\\{ z: |z|<1\\}$ and consider $ F\\equal{}f(z)\\minus{}w$ and then use the Rouche theorem", "Solution_2": "Cobmine iii and ii to get the hypothesis of Schwarz Lemma.\r\n\r\niii says that there is a zero of $ f$ inside the disk. You need to move the zero to zero by composing with the function in ii, and then check that new composition of functions hasn't messed up mapping the disk to the disk.\r\n\r\nSuppose $ f(a)\\equal{}0$. Consider $ h(z)\\equal{}f\\left(\\frac{z\\plus{}a}{1\\plus{}\\overline{a}z}\\right)$. Then $ h(0)\\equal{}f(a)\\equal{}0$, and it still maps the disk to the disk, so Schwarz applies to $ h$.", "Solution_3": "See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=277820]this topic.[/url]\r\n\r\nI didn't specifically use Schwarz there, but I used arguments from the same family as the proof of Schwarz. But then fedja simplified it down from there. Go read his proof on the link." } { "Tag": [ "MATHCOUNTS", "ARML", "trigonometry", "HMMT", "Duke", "college", "geometry" ], "Problem": "I thought it would be nice to add a little organization to this forum. And as it stands, there aren't too many posts in this forum that it would be that tough of a task to maintain :D If you have any suggestions to make this post better, I'd love to hear them.\r\n\r\n\\[ \\text{\\Large Practice Problem Lists} \\]\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21038#]Mathcounts Practice[/url] - MCrawford\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20985]ARML Practice Problems[/url] - MCrawford\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21104]ARML Practice[/url] - MCrawford\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=36545]Two Trigonometry Problems[/url] - gauss202\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46375]Problem Marathon[/url] - Iversonfan2005\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61308]Some HMMT Problems[/url] - Iversonfan2005\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61776]Some Problems from the Duke Math Meet[/url] - gauss202\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=69252]Some Problem Shobhit Made Up[/url] - Iversonfan2005\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35645]Collection of Geometry Problems[/url] - gauss202\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=74217]Random AMC Type Problems[/url] - Iversonfan2005\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=74354]ARML NT Practice Problems[/url] - joml88\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=432882#p432882]NCML[/url] Problems - prybarczyk\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75721]ARML Practice Problems[/url] - joml88\r\n\r\n\\[ \\text{\\Large Lessons/Reference Material} \\]\r\n\\[ \\text{Number Theory} \\]\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75061]Divisibility Rules[/url] - joml88\r\n\r\n\\[ \\text{Geometry} \\]\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75469]Triangle Area[/url] - joml88\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=434617#p434617]Circles[/url] -Iversonfan2005\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75573]Reference for Geometry Formulas[/url] - Iversonfan2005\r\n\r\n\\[ \\text{Algebra} \\]\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75738]Factoring[/url] - joml88", "Solution_1": "sweet :coolspeak: , it's amazing how much activity this forum's gotten in the last 2 weeks :roll:\r\n\r\nI have a question, how exactly do we make pdfs, because it would be nice if we can compile all the problems into a single pdf.. :) , if someone can tell me, i can do it", "Solution_2": "[quote=\"Iversonfan2005\"]I have a question, how exactly do we make pdfs, because it would be nice if we can compile all the problems into a single pdf.. :) , if someone can tell me, i can do it[/quote]\r\n\r\n\r\nWell, I can think of two options:\r\n\r\n(A) You can use TeXnic Center and type up all of the questions separately (well, you can just copy, paste, and do a little bit of editing to make it come out a little neater). This requires you to have TeXnice Center downloaded on your computer and hooked up and everything which is good thing to have anyways.\r\n\r\nor\r\n\r\n(B) You can download CutePDF or some other program that converts files to pdf's. This would allow you to take regular documents go to File > Print, change the printer to CutePDF and instead of actually printing something out on paper, it will output a PDF file on your computer of the document. So basically, you could copy and paste everything into word and use CutePDF.\r\n\r\nOption A would tend to be more time consuming. Particularly, downloading all the stuff to get TeXnic Center working on your computer can be a hassle but I think it's well worth it if you plan on $\\text{\\TeX}$ing up documents every once in a while. The plus side is that you'll have TeXnic Center working :) and you'll have more control over what the document looks like.\r\n\r\nOption B is more practical and quicker for this specific task.", "Solution_3": "[quote=\"joml88\"]I thought it would be nice to add a little organization to this forum. And as it stands, there aren't too many posts in this forum that it would be that tough of a task to maintain [/quote]\r\n\r\nLOL :rotfl: :rotfl: , I think there's been so much activity on this forum now that you should just title it $\\text{Lessons and References}$ and just put links for the lessons and stuff and probably the document files of questions posted." } { "Tag": [ "geometry", "geometric sequence" ], "Problem": "If $ b$ is positive, what is the value of $ b$ in the geometric sequence $ 9, a, 4, b$? Express your answer as a common fraction.", "Solution_1": "$ 4 \\equal{} 9r^2$\r\n\r\n$ r^2 \\equal{} \\frac{4}{9}$\r\n\r\n$ r \\equal{} \\frac{2}{3}$\r\n\r\n$ b \\equal{} 4r \\equal{} 4(\\frac{2}{3}) \\equal{} \\boxed{\\frac{8}{3}}$" } { "Tag": [ "probability", "LaTeX", "probability and stats" ], "Problem": "We all know that P ( a U b ) = P(a) + P(b) - P(a int b)\r\n\r\nProve the addition principle for n events..\r\nIt comes in the form of alternate positive and negative summations with 2,3, ... n terms at a time.", "Solution_1": "Usually this is called the \"Principle of Inclusion-Exclusion,\" frequently abbreviated \"PIE\" on the forums. What have you tried?\r\n\r\nAlso, [[LaTeX]] is equipped on the forums -- read http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 and the first link in this sentence for an introduction.", "Solution_2": "Yea i know tht its calld principle of incusion-exclusion..n its very often used in probability..\r\n\r\nthts y its posted here..", "Solution_3": "You have to prove something for a general $ n$ case. What does that suggest to you?" } { "Tag": [], "Problem": "I would like to meet some penpals who go to competitions and who are willing to chare their experiances and problems with me. \r\nIf someone knows of some free on-line preparation program, please let me know!\r\nmy e-mail: nada011@yubc.net \r\nplease write,\r\nregards,\r\nTijana", "Solution_1": "[quote=\"tijana_kovacevic\"]If someone knows of some free on-line preparation program[/quote]\r\n\r\nWhy don't you just try the many nice problems that Arne, JBL, grobber and others freely and nicely give us ??" } { "Tag": [], "Problem": "Prove that the equation $ \\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c}\\plus{}\\frac{1}{abc} \\equal{} \\frac {12}{a \\plus{} b \\plus{} c}$ has infinitely many solutions $ (a,b,c)$ in natural numbers.", "Solution_1": "This is a very strange problem. Clearly if $ (a,b,c)$ is a solution, then so is $ (ka,kb,kc)$ where $ k\\in\\mathbb{Z}^\\plus{}$. Hence the problem reduces to finding one solution.\r\n\r\nFrom expanding, we see that $ (b\\plus{}c)|11abc$. If we scale the variables sufficiently, we would have a solution such that $ b\\plus{}c|bc$. From here, we can write $ b\\equal{}r(s\\plus{}t)s$, and $ c\\equal{}r(s\\plus{}t)t$. Back in the original equation, this gives\r\n$ a\\equal{}\\frac{r}{2}\\cdot(\\minus{} s^2 \\plus{} 9 s t \\minus{} rt^2 \\pm \\sqrt{s^4 \\minus{} 22 s^3 t \\plus{} 75 s^2 t^2 \\minus{} 22 s t^3 \\plus{} t^4})$\r\n\r\nwhich is certainly promising...but we still restated the problem in the sense that we need that expression to be a square...\r\n\r\nI don't have a nice way to do this. I have checked $ 1\\le s\\le 200$, $ 1\\le t\\le 2000$ and not gotten a solution. I wonder what the solution is...I have tried root flipping but it is a quartic...", "Solution_2": "I think that you can work as here \r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?p=118691#118691", "Solution_3": "I'm very sorry, Altheman, for the mistake in my post.the problem is OK now." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "x $ >$ y $ >$ z $ \\geq$ 0, prove that:\r\n\r\n$ 8$ $ \\times$ $ ($ $ x^3$+$ y^3$+$ z^3$ $ )^2$ $ \\geq$ $ 9$ $ (x^2 \\plus{} yz)$ $ (y^2 \\plus{} xz)$ $ (z^2 \\plus{} xy)$", "Solution_1": "$ 9(x^2\\plus{}yz)(y^2\\plus{}zx)(z^2\\plus{}yx)\\leq9(\\frac{x^2\\plus{}y^2\\plus{}z^2\\plus{}xz\\plus{}yz\\plus{}yx}{3})^3\\leq9(\\frac{2(x^2\\plus{}y^2\\plus{}z^2)}{3})^3\\equal{}9*8[(\\frac{x^2\\plus{}y^2\\plus{}z^2}{3})^\\frac{1}{2}]^6\\leq9*8[(\\frac{x^3\\plus{}y^3\\plus{}z^3}{3})^\\frac{1}{3}]^6\\equal{}8(x^3\\plus{}y^3\\plus{}z^3)^2$", "Solution_2": "$ (x^3\\plus{}y^3\\plus{}z^3)(x^3\\plus{}y^3\\plus{}z^3)(1\\plus{}1\\plus{}1)\\geq(\\sum x^2)^3$" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $x;y;z$ be positive real numbers.Prove that:\r\n$x^{2}(y+z)+y^{2}(x+z)+z^{2}(x+y)\\ge 6xyz$[/QUOTE]", "Solution_1": "Use AM-GM inequality", "Solution_2": "[quote=\"chien than\"]Let $x;y;z$ be positive real numbers.Prove that:\n$x^{2}(y+z)+y^{2}(x+z)+z^{2}(x+y)\\ge 6xyz$[/quote][/quote]\nok\nsolution S.O.S\n$xy(x+y)+yz(y+z)+zx(z+x)\\ge 6xyz$\nwith\nx/y+y/x=(x-y)^2/xy[/quote]", "Solution_3": "auhongan_au, you do nothing with SOS?\r\ndid you just use Cauchy?", "Solution_4": "Can anyone tell me what the s.o.s method is ?", "Solution_5": "[quote=\"mathubu\"]Can anyone tell me what the s.o.s method is ?[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127", "Solution_6": "we can writ it like this, $x(y^{2}+z^{2})+y(z^{2}+x^{2})+z(x^{2}+y^{2}) \\geq 6xyz$ wich becom easy by the fact $x^{2}+y^{2}\\geq 2xy$", "Solution_7": "yeah that's right. :lol:", "Solution_8": "Why do we discuss this kind of inequalities?\r\nIt's obvious for\r\n$x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)\\geqslant3\\sqrt[3]{x^{2}\\cdot2\\sqrt{yz}\\cdot y^{2}\\cdot2\\sqrt{zx}\\cdot z^{2}\\cdot2\\sqrt{xy}}=6xyz$" } { "Tag": [ "articles", "geometry", "geometric transformation" ], "Problem": "I have the following problem - I enjoy browsing news from the world, reading about politics, social affairs, economy, science etc., but I can' find a decent news site on the net that would satisfy my expectations. Portals in my country (Poland) are poor, they tend to write too much about Polish politics (which is getting more and more indigestible) and anyway, they are full of tabloid-style crap. Sites like NYTimes.com etc. would make a good reading if they allowed filtering information - but they're too chaotic , messy and full of useless stuff (also, their layout makes it hard to concentrate and find valuable news among the informational litter). I enjoy the style of writing about world affairs that \"The Economist\" presents, so a news site designed in a similar style would be OK. Any suggestions?", "Solution_1": "http://www.theonion.com ", "Solution_2": "You can read the Onion if you want, but keep in mind that it is satire, not actual news. I am not sure if Treething was trying to make a commentary on something or not, but I just want you to be aware of what you are getting into.", "Solution_3": "[url]http://news.bbc.co.uk/[/url] is about the best site I know.\r\n\r\n\r\nThe problem -of course - is....how many languages do you speak. My experience is that that can be a huge barrier (I often compare articles on the internet).", "Solution_4": "Well you can translate it, but the automatic translations aren't very good...still though you can get a good amount of the information from it.\r\n\r\n[url]http://babelfish.altavista.com/[/url]", "Solution_5": "Timesonline is quite a good one:\r\n[url]http://www.timesonline.co.uk/tol/news/[/url]" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Let $ n$ be a positive integer. Prove that for all real $ x,y$, \r\n$ (nx)^n\\minus{}((n\\minus{}1)x\\plus{}y)^n\\plus{}((n\\minus{}2)x\\plus{}2y)^n\\minus{}\\dots\\plus{}(\\minus{}1)^{n\\minus{}1}(x\\plus{}(n\\minus{}1)y)^n\\plus{}(\\minus{}1)^n (ny)^n$\r\n$ \\equal{}n!(x\\minus{}y)^n$", "Solution_1": "for $ (x,y)\\equal{}(1,0)$, the equation becomes $ n!\\equal{}n^n\\minus{}(n\\minus{}1)^n\\plus{}(n\\minus{}2)^n\\minus{}\\cdots$, which is false. the correct formula is $ n!\\equal{}n^n\\minus{}\\binom{n}{1}(n\\minus{}1)^n\\plus{}\\binom{n}{2}(n\\minus{}2)^n\\minus{}\\cdots$. you're missing the necessary coefficients.", "Solution_2": "Sorry about that :blush: \r\n\r\nThe identity i meant was \r\n\\[ \\binom{n}{0}(nx)^{n}\\minus{}\\binom{n}{1}((n\\minus{}1)x\\plus{}y)^{n}\\plus{}\\binom{n}{2}((n\\minus{}2)x\\plus{}2y)^{n}\\minus{}\\dots\\]\r\n\\[ \\plus{}(\\minus{}1)^{n\\minus{}1}\\binom{n}{n\\minus{}1}(x\\plus{}(n\\minus{}1)y)^{n}\\plus{}(\\minus{}1)^{n}\\binom{n}{n}(ny)^{n}\\]\r\n\\[ \\equal{}n!(x\\minus{}y)^n\\]", "Solution_3": "corresponding coefficients are equal by a straightforward application of PIE.", "Solution_4": "What is PIE?", "Solution_5": "Principle of Inclusion-Exclusion", "Solution_6": "ok thanks :)" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c \\in (0,\\infty)$ such that $ b^20 we have a(y+z)/(b+c)+b(z+x)/(a+c)+c(x+y)/(a+b)>=sqrt(3(xy+yz+zx).\r\n Prove this inequality. It is trivial to see that it is stronger than Crux problem. And I have a 3-4 lines solution to this one.", "Solution_1": "Do not try this one! I didn't posted in the correct place. It is extraordinary difficult.", "Solution_2": "Wonderful, fantastic and beautiful inequality. It is really one of the best I have ever seen. I like a lot inequalities like the FAMOUS CRUX INEQUALITY and this is as nice as it. \r\nI hope you will post solution as soon as possible.\r\n\r\nThank you very much, Harazi.", "Solution_3": "I promised a short solution. I hope it will be at most 4 lines:\r\n Take x+y+z=1. From Cauchy we have sum ax/(b+c)+sqrt(3 sum xy)<=sqrt[(sum x^2)*(sum (a/(b+c))^2)]+sqrt( sum xy*3/4)+sqrt(sum xy*3/4)<=sqrt(( sum x^2+2 sum xy)*(3/2+sum (a/(b+c)^2))<=sum a/(b+c) since it reduces to sum ab/(c+a)(c+b)>=3/4, trivial. So, 4 lines.", "Solution_4": "Fantastic solution! You virtuously manage with Cauchy. Congratulations!\r\n\r\nNamdung", "Solution_5": "I am really astonished.\r\n\r\nThis is a SUPER FANTASTIC solution!!!!!" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "$ (a \\plus{} b \\plus{} c)(b\\plus{}a\\minus{}c)(a\\plus{}c\\minus{}b)(b\\plus{}c\\minus{}a)(\\frac {1}{a^4} \\plus{} \\frac {1}{ b^4} \\plus{} \\frac {1}{c^4} )\\ge \\frac{9}{4}$\r\n\\]\r\nfor all $ a$ , $ b$ , $ c$ are the lenghts of a triangle", "Solution_1": "[quote=\"ABRORBEK\"]$ (a \\plus{} b \\plus{} c)(b \\plus{} a \\minus{} c)(a \\plus{} c \\minus{} b)(b \\plus{} c \\minus{} a)(\\frac {1}{a^4} \\plus{} \\frac {1}{ b^4} \\plus{} \\frac {1}{c^4} )\\ge \\frac {9}{4}$\n\\]\nfor all $ a$ , $ b$ , $ c$ are the lenghts of a triangle[/quote]\r\nTry $ a\\equal{}1.99,b\\equal{}c\\equal{}1$." } { "Tag": [ "LaTeX" ], "Problem": "a,b are distinct postive integers s.t. $ a^2\\plus{}ab\\plus{}b^2 | ab(a\\plus{}b)$\r\nProve that $ |a\\minus{}b| > {(ab)}^{1/3}$", "Solution_1": "a^2 + ab + b^2 | ab(a+b)\r\n\r\nab(a+b) = 0 (mod a^2 +ab +b^2)\r\n\r\na^2b + ab^2 = 0 (mod a^2 +ab +b^2)\r\n\r\n3a^2b + 3ab^2 = 0 (mod a^2 +ab +b^2)\r\n\r\n-3a^2b + 3ab^2 = -6a^2B (mod a^2 +ab +b^2)\r\n\r\n(a-b)(a^2 +ab +b^2) -3a^2b + 3ab^2 = -6a^2B (mod a^2 +ab +b^2)\r\n\r\na^3 - b^3 -3a^2b + 3ab^2 = -6a^2B (mod a^2 +ab +b^2)\r\n\r\n\r\n(a-b)^3 = -6a * ab (mod a^2 +ab +b^2)\r\n\r\nsince |-6a| is bigger than 1\r\n\r\n|a-b| < (ab)^(1/3)\r\n\r\n\r\nsorry it is kind of messy, i dont know how to use latex :(", "Solution_2": "can someone help me check my solution plz \r\ni am not sure if it is right\r\n\r\nthanks", "Solution_3": "[quote=\"lgd0612\"]a^2 + ab + b^2 | ab(a+b)\n\n(a-b)^3 = -6a * ab (mod a^2 +ab +b^2)\n\nsince |-6a| is bigger than 1\n\n|a-b| < (ab)^(1/3)\n\n\nsorry it is kind of messy, i dont know how to use latex :([/quote]\r\n\r\nThe transition here is incorrect, i.e. you cannot go from the equivalence to the inequality.", "Solution_4": "hmm\r\nok from this step \r\n\r\n(a-b)^3 = -6a * ab (mod a^2 +ab +b^2) \r\n\r\ni can write it in anther form \r\n(a-b)^3 = K(a^2 +ab+ b^2) - 6a*ab and K is a non negitive integer \r\nsince a,b are both positive, K(a^2 +ab+ b^2) is also a non negitive integer\r\n\r\nif K is positive then \r\n(a-b)^3 > -6a*ab\r\n(b-a)^3 > 6a*ab\r\nb-a > (6a)^(1/3)*ab^(1/3) \r\nb-a > ab^(1/3) \r\n|a-b|> ab^(1/3)\r\n\r\n\r\nthe only case that i need to worry about is when K = 0, (a-b)^3 is at its minimum value: -6a*ab\r\nso i have \r\n(a-b)^3 = -6a*ab\r\na-b = (-6a)^(1/3)*ab^(1/3) \r\nb-a = 6a^(1/3)*ab^(1/3)\r\n\r\nnow i know that 6a^(1/3) > 1 (a is a positive integer)\r\nso\r\nb-a > ab^(1/3)\r\n|a-b| > ab^(1/3)", "Solution_5": "From $ \\textrm{[Russia 2001]}$ and the solution was:\r\nSet $ g\\equal{}\\gcd(a,b)$ and write $ a\\equal{}xg$, and $ b\\equal{}yg$ with $ \\gcd(x,y)\\equal{}1$.\r\nThen:\r\n$ \\frac{ab(a\\plus{}b)}{a^2\\plus{}ab\\plus{}b^2}\\equal{}\\frac{xy(x\\plus{}y)g}{x^2\\plus{}xy\\plus{}y^2}$ is an integer.\r\nnote that $ \\gcd(x^2 \\plus{} xy\\plus{} y^2,x) \\equal{} \\gcd (y^2 , x) \\equal{} 1$.\r\nSimilary: $ \\gcd(x^2\\plus{}xy\\plus{}y^2,y)\\equal{}1$. Because $ \\gcd(x\\plus{}y,y)\\equal{}1$, we have\r\n$ \\gcd (x^2 \\plus{} xy \\plus{} y^2 , x\\plus{}y) \\equal{} \\gcd(y^2 , x\\plus{}y) \\equal{} 1$\r\nSo we have:\r\n$ x^2 \\plus{} xy \\plus{} y^2|g$\r\nimplying that $ g \\ge x^2 \\plus{} xy \\plus{} y^2$. Therefore,\r\n$ |a\\minus{}b|^3$\r\n$ \\equal{} |g(x\\minus{}y)|^3 \\equal{} g^2|x\\minus{}y|^3 g$\r\n$ \\ge g^2 (1)(x^2 \\plus{} xy\\plus{} y^2)$\r\n$ > g^2 xy$\r\n$ \\equal{}ab$" } { "Tag": [ "induction", "number theory proposed", "number theory" ], "Problem": "Is there a sequence $ a_1,a_2,...,a_{100}$ of positive integers such that $ a_{i\\minus{}1}lcm(a_i,a_{i\\plus{}1})$ , for $ i\\equal{}2$ to $ 99$ ?", "Solution_1": "Assume that it was true for a sequence of $ k$ integers $ a_1,a_2,...,a_k$ (the proposition). (When k=3 you can easily find an example)\r\nThen we can choose a large enough $ m$ and a large enough prime $ p < ma_1$ which coprime with $ ma_1$\r\nso that $ lcm(p,ma_1) > lcm(ma_1,ma_2)$ because it's equivalent to $ pa_1 > lcm(a_1,a_2)$ and obviously $ L.H.S.$ can be made artitrary large.\r\n\r\nNow from our assumption, $ lcm(a_1,a_2) > lcm(a_2,a_3) > .... > lcm(a_{k \\minus{} 1}, a_{k})$\r\n$ \\Leftrightarrow lcm(ma_1,ma_2) > lcm(ma_2,ma_3) > .... > lcm(ma_{k \\minus{} 1}, ma_{k})$\r\n\r\nSo $ p,ma_1,ma_2,...., ma_k$ will be a sequence of $ k \\plus{} 1$ integers that satisfies the condition.\r\n\r\nBy the principle of mathematical induction, the proposition is true for all $ k \\ge 3$, so it's true when $ k \\equal{} 100$, done." } { "Tag": [ "probability" ], "Problem": "Spinners A and B are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' is even?\r\n\r\n[img]http://img142.imageshack.us/img142/3076/spinnersls3.gif[/img]\r\n\r\n$\\text{(A) }\\frac{1}{4}\\qquad\\text{(B) }\\frac{1}{3}\\qquad\\text{(C) }\\frac{1}{2}\\qquad\\text{(D) }\\frac{2}{3}\\qquad\\text{(E) }\\frac{3}{4}$", "Solution_1": "[hide]There are 12 total possibilities.\n\nIf the first spinner is 2 or 4, the product will always be even. If it is 1 or 3, only multiplying by 2 gives an even number.\n\nThe probability is $\\frac{8}{12},$ or $\\frac{2}{3}.$ The answer is $\\text{D}.$[/hide]", "Solution_2": "[hide=\"answer\"]D-The product of two #s is even if both aren't odd. The odds of both being odd are 1/2*2/3=1/3. So, the probability of an even # is 1-1/3=2/3. [/hide]", "Solution_3": "[hide]casework (longer) way:\n\nto have an even product...\n\ncase 1 both spinners give even numbers. 1/2 * 1/3 = 1/6\ncase 2 1st spinner even, 2nd odd. 1/2 * 2/3 = 1/3\ncase 3 1st spinner odd, 2nd even. 1/2 * 1/3 = 1/6\n\n1/6 + 1/3 + 1/6 = 2/3. D[/hide]", "Solution_4": "[hide]\nD=2/3\n\nI just looked at the possibilities:12 (4 x 3), and figured half already had even numbers, so there were 6. For the other numbers on the spinner with 4 numbers, they each had 1 even possibility, making 2 extra. So, 2 + 6 = 8. But, we have to remember its asking the probability. So, there are 8 out of 12, making 8/12. We put that in lowest terms to get 2/3 (divide the numerator and denominator by 4).\n\n[/hide]\r\n\r\n-jorian", "Solution_5": "[hide=\"solution\"]The probability that the product of the numbers is NOT even is 1/2*2/3=1/3. The probability that the product is even is 2/3. D[/hide]" } { "Tag": [ "vector", "induction", "superior algebra", "superior algebra unsolved" ], "Problem": "How many ordered sequences of $ k$ vectors in $ \\mathbb{F}_q^n$ span? ($ \\mathbb{F}_q$ denotes finite field of $ q$ elements).", "Solution_1": "Let $ V$ be a sub-space of dimension $ 0\\leq m\\leq n$ and for $ k\\in\\mathbb{N}$ let $ A_{k,m}(V)$ denote the set $ \\left\\{\\left.(v_1,\\ldots,v_k)\\in(\\mathbb{F}_q^n)^k\\,\\right|\\,V+ =\\mathbb{F}_q^n\\right\\}$, i.e. the set of ordered vector sequences, which together with $ V$ will span $ \\mathbb{F}_q^n$. Obviously, the number of elements of $ A_{k,m}(V)$ does not depend on $ V$. Therefore, we will write $ a_{k,m}$ for the number of such ordered sequences.\r\n\r\nFor $ k=1$ we get: $ a_{1,m}=\\left\\{\\begin{array}{ll}0 & \\mbox{if }m1$ then we observe the following construction principle.\r\n \r\n$ \\begin{array}{ll}(v_1,\\ldots,v_k)\\in A_{k,m}(V)\\Longleftrightarrow & v_1\\in V,(v_2,\\ldots,v_k)\\in A_{k-1,m}(V)\\mbox{ or }\\\\[0.2cm] & v_1\\not\\in V,(v_2,\\ldots,v_k)\\in A_{k-1,m+1}(V\\oplus )\\end{array}$\r\n\r\nThis implies the recursion: $ a_{k,m}=q^m\\cdot a_{k-1,m}+(q^n-q^m)\\cdot a_{k-1,m+1}$ and using straightforward induction by $ k$ we can verify:\r\n\r\n$ a_{k,m}=\\left\\{\\begin{array}{ll}q^{kn}\\cdot\\prod_{i=1}^{n-m}\\left(1-q^{-(k+1-i)}\\right) & \\mbox{if } k+m\\geq n\\\\[0.2cm]\r\n0 & \\mbox{otherwise}\r\n\\end{array}\\right.$\r\n\r\napplying the usual convention $ 1=\\prod_{\\emptyset}\\ldots$\r\n\r\nThe problem asked for the number of ordered vector sequences of length $ k$, which span $ \\mathbb{F}_q^n$. This number is given by $ a_{k,0}$." } { "Tag": [ "function", "calculus", "derivative", "integration", "geometry" ], "Problem": "this is homework as a source but ive already done it, just thought it was quite nice.\r\n\r\nA uniform rope of mass per unit length $\\lambda$ is coiled on a table. One end is pulled straight up with constant velocity $v$. Find the force exerted on the rope as a function of the height, $y$.\r\n\r\nFind the total work done on the rope in lifting the end to a height $y$.\r\n\r\nCompare (and explain the result) the instantaneous power needed to lift the rope with the rate of change of total mechanical energy (ie. potential plus kinetic).", "Solution_1": "Hey, there's something wrong with this problem :), I can't get the right answer!", "Solution_2": "it is quite a nice problem I think. hint..\r\n[hide]\nconsider the rate of change of momentum. [/hide]\n\nmy solution to the first part:\n\n[hide]there will be two reasons a force is required; to change the momentum of the rope from zero to some constant value when lifting it, and to keep the rope already lifted from falling. The second part is quite simple..\n\nthe time can be expressed by t = y/v, simple distance/speed\n\nafter a time t, there will be a length y in the air, giving a total mass of by (im using b for $\\lambda$ because its easier!). The weight of this is gby, and so this is the force needed to keep the rope moving with constant velocity.\n\nthe second force is the force needed to change the velocity of the rope from 0 to v. the force is the time derivative of the momentum.\n\nafter a time t, \n\nthe rate of change of momentum\n\nafter a time t, there will be y = tv of the rope in the air. now have a small look at time t+dt, and there will be now v(t+dt) in the air; so v dt extra rope will be in the air. the mass of this little bit of rope is bv dt, and so the momentum is bv^2 dt, so bv^2 is the instantaneous change on momentum=force. (a bit of messy calculus but I think its allowed!)\n\nso the total force is F = bv^2 + gby[/hide]", "Solution_3": "That's exactly what I did. The problem is that this can't be right! If you integrate the force you'll get the energy all messed up - you'll do more work than the rope gets energy.", "Solution_4": "Ok the rest of the problem, \r\n\r\n[hide]ok so work done is \n\n$W = \\int^{y}_{o} F dx$\n$W = \\int^{y}_{o} F bv^{2} + gby dx$\n$W = [xbv^{2} + xgby]^{y}_{0}$\n$W = ybv^{2} + bvy^{2}$\n\nPower \n\n$P = \\frac{dW}{dt} = bv^{3} + gbtv^{2}$\n\nAnd the mechanical energy is the potential energy plus the kinetic energy. The kinetic energy is given by:\n\n$T = \\frac{mv^{2}}{2}$\n$T = \\frac{byv^{2}}{2}$\n\nAnd the potential energy can be found by modelling the entire length of rope, mass by, and length y, as a particle of mass m halfway up the rope. So this is:\n\n$U = mgh$\n$U = \\frac{bgy^{2}}{2}$\n\nSo the total mechanical energy is the sum,\n\n$E = U + T = \\frac{bv^{2}t^{2}g + bv^{3}t}{2}$\n\nusing $y = vt$\n\nSo $E = \\frac{Pt}{2}$... I guess this is like saying is you draw a graph of power against time, the energy will be the area underneath...[/hide]", "Solution_5": "Two things:\r\n[hide=\"#1\"]Dude, how can you integrate the work like that?! You wrote:\n\n$W=\\int_0^HF(y)\\text{d}x$\n\nbut it should really be this:\n\n$W=\\int_0^HF(y)\\text{d}y$ !!!\n\nAnd that's different. The result should be:\n\n$W=Hbv^2+\\frac{1}{2}H^2bv$.[/hide]\n[hide=\"#2\"]You say that $E=\\frac{Pt}{2}$. If you apply my correction from #1, it turns out a little different but anyway.... shouldn't be $E=Pt$??? What happens to the rest of the work done? Shouldn't $E=W$?[/hide]" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra solved" ], "Problem": "$R$ is a ring. The product of any two non-zero left ideals of $R$ is not zero. If $RxR=0$ for some $x$. Show that $x=0$.", "Solution_1": "Let $I=Rx$. Then $I$ is a left ideal and $II=0$, so $I=0$. \r\nLet $J=$, the additive subgroup generated by $x$. Then $RJ=0$, so $J$ is a left ideal.\r\n$I=Rx=0 \\Rightarrow x^2=0$, so $JJ=0$, hence $J=0$ and $x=0$.", "Solution_2": "[quote=\"amfulger\"]Let $J=$, the additive subgroup generated by $x$. [/quote]\r\nNice solution. I never thought about that." } { "Tag": [ "calculus", "LaTeX", "real analysis", "real analysis unsolved" ], "Problem": "This is a case study problem on my textbook Applied Calculus 4th ed. Waner/Costenoble ISBN: 9780495384281.\r\n\r\nabout reducing sulfer emission. If you have this book it's on page 285 and 2 page long problem.\r\n\r\nNow i am not sure if topic like this is appropriate in here, please advise me if i am in a wrong place.\r\n\r\nAnyway if you have this textbook i would appreciate any help on on exercise 3~8.\r\n\r\nThank you\r\n\r\n[img]http://ecx.images-amazon.com/images/I/51h%2BiSzavGL._AA240_.jpg[/img]", "Solution_1": "It's rather unlikely that anyone is going to have the same textbook as you. Your options are to either scan the page or to do your own homework.", "Solution_2": "Actually, you have another option: choose one of the problems (or a similar problem), type it into the forum (using [[LaTeX]] -- see that link or [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]this one[/url]), tell us what you've done and where you're stuck, and someone will probably come along and give you some hints." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "find all function with reel numbers wish satisfay \r\n$f(x-1)+f(x+1)=2f(x)+x$", "Solution_1": "[quote=\"arkhammedos\"]find all function with reel numbers wish satisfay \n$f(x-1)+f(x+1)=2f(x)+x$[/quote]\r\nLet $f(x)=g(x)+\\frac{x^{3}}{6}$, then $g(x+1)+g(x-1)=2g(x)$.\r\nIt give $g(x)=g(\\{x\\})+[x](g(1+\\{x\\})-g(\\{x\\}))$, were g(x) is any function in $[0,2)$.", "Solution_2": "$h(x)=f(x)-f(x-1)$\r\nthen $h(x)=h(x-1)+2(x-1)$\r\nso, if $p\\le x 0$. If $ S$ has mean curvature $ H \\ge a$ at each point for some real number $ a$, show that $ ar \\leq 1$", "Solution_1": "If $ z\\equal{}f(x,y)$ is the equation, then $ 2H$ is the divergence of the field $ \\vec G: \\equal{}\\frac{\\nabla f}{\\sqrt{1\\plus{}|\\nabla f|^2}}$. If $ H\\ge a$ everywhere, then by the divergence theorem the flux of $ \\vec G$ across the boundary is at least $ 2a\\pi r^2$. On the other hand, the flux must less than $ 2\\pi r$ because $ |\\vec G|<1$.\r\nhttp://en.wikipedia.org/wiki/Mean_curvature" } { "Tag": [ "email", "summer program", "Mathcamp" ], "Problem": "Lets say i send my online application and quiz solutions on April 27 but i have to send my recommendations by snail mail... will I still get \"full consideration\" if the envelope the recommendations are postmarked on the 27th?", "Solution_1": "[quote=\"dangvh\"]Lets say i send my online application and quiz solutions on April 27 but i have to send my recommendations by snail mail... will I still get \"full consideration\" if the envelope the recommendations are postmarked on the 27th?[/quote]\r\n\r\nYour recommendations should be sent so that they are [i]received[/i] by the 27th.", "Solution_2": "Same thing for solutions? I dont understand the def of postmark for snailmail.", "Solution_3": "[quote=\"mcalderbank\"]Same thing for solutions? I dont understand the def of postmark for snailmail.[/quote]\r\n\r\nYes, same for solutions -- they should be received by April 27, not postmarked April 27.\r\n(The post office stamps the date on each letter on the day they receive it.) As an alternative, you can fax your solutions to (617) 812-6339.", "Solution_4": "I'm still finishing up on my solutions and putting the finals details on the application (essay, background, and other stuff as well).\r\n\r\nExactly how much less consideration do you give late applications? And are they eligible for merit scholarships and financial aid still?", "Solution_5": "[quote=\"MithsApprentice\"]Exactly how much less consideration do you give late applications? And are they eligible for merit scholarships and financial aid still?[/quote]\r\n\r\nHowever many new applicants we have slots for, we'll admit that many people from the on-time applications. Late applications will be considered for the waitlist alongside the remaining applications. Yes, late applications are eligible for merit scholarships and financial aid.", "Solution_6": "Hmm...I have sent my rec and quiz solutions a little while ago, and they definitely should have arrived by now. However, the deadline is getting dangerously close, and the status is still set to \"Awaiting receipt\". Is that just because the page hasn't been updated yet, or should I consider faxing my solutions?", "Solution_7": "[quote=\"aurelie\"]Hmm...I have sent my rec and quiz solutions a little while ago, and they definitely should have arrived by now. However, the deadline is getting dangerously close, and the status is still set to \"Awaiting receipt\". Is that just because the page hasn't been updated yet, or should I consider faxing my solutions?[/quote]\r\n\r\nMost likely the page hasn't been updated yet. (Mira, who receives the mail at the Cambridge, MA address, is away at the moment.)", "Solution_8": "My teachers finally got around to writing me my recommendations... but I was planning originally going to snail mail them... would it be ok if i just scanned them and posted the file on my application?\r\n\r\nedit- now that i think of it i could scan them and email them to you if that's ok", "Solution_9": "[quote=\"dangvh\"]My teachers finally got around to writing me my recommendations... but I was planning originally going to snail mail them... would it be ok if i just scanned them and posted the file on my application?\n\nedit- now that i think of it i could scan them and email them to you if that's ok[/quote]\r\n\r\nthe problem with that idea is, your letters may be confidential and your teachers may be expecting you not to read them; so this would be OK only if your teachers write to us to let us know that you have their permission to do this. as an alternative, your teachers could email us the letters (either the original file, or a scanned copy).", "Solution_10": "When you talk about faxing in your qualifying quiz, how can you change your option on the online app so that Mathcamp knows you are submitting by fax. The app only has mail and web.", "Solution_11": "[quote=\"bubala\"]When you talk about faxing in your qualifying quiz, how can you change your option on the online app so that Mathcamp knows you are submitting by fax. The app only has mail and web.[/quote]\r\n\r\nIf you'll fax, select that you're submitting by mail.", "Solution_12": "umm i got my 2 required recommendations and they will send via email... but i still have one that is in an envelope... could i send that one too? It will arrive late but i just want to know that if another recommendation arrives later would my application be considered late.", "Solution_13": "[quote=\"dangvh\"]umm i got my 2 required recommendations and they will send via email... but i still have one that is in an envelope... could i send that one too? It will arrive late but i just want to know that if another recommendation arrives later would my application be considered late.[/quote]\r\n\r\nIf 2 recs are in by the deadline, that's fine.", "Solution_14": "thanks!", "Solution_15": "What time in what time zone does it have to be submitted by? I was unable to find it on the Mathcamp website. I know I shouldn't be procrastinating long enough to have to ask this, but I have a couple things left to tweek.", "Solution_16": "[quote=\"picounts\"]What time in what time zone does it have to be submitted by? I was unable to find it on the Mathcamp website. I know I shouldn't be procrastinating long enough to have to ask this, but I have a couple things left to tweek.[/quote]\r\n\r\nYour home time zone is fine, whatever that may be. We're not [i]such[/i] sticklers. (-:", "Solution_17": "It still says that the recommendation that I sent in by mail was not received. Is that page up to date?", "Solution_18": "Hehe...they received mine Tim. :P", "Solution_19": "What should I do if one of my recommenders said she sent the recommendation online today, but it says \"awaiting receipt\" on the site. I really need to have an on-time application.", "Solution_20": "[quote=\"picounts\"]It still says that the recommendation that I sent in by mail was not received. Is that page up to date?[/quote]\r\n\r\nIt is now up to date, so if it's not marked as received then we don't have it yet. But don't worry -- we're not evil -- there is always slack on rec letters arriving by mail when the rest is on time.", "Solution_21": "My solutions arrived on the 28th. Does that mean my applications is considered late? Everything else was in on time.", "Solution_22": "[quote=\"The Original Pi Guy\"]My solutions arrived on the 28th. Does that mean my applications is considered late? Everything else was in on time.[/quote]\r\n\r\nBecause of the computer difficulties, we're counting everything that arrived on the 28th as on time.", "Solution_23": "[quote]Because of the computer difficulties, we're counting everything that arrived on the 28th as on time.[/quote]\r\n\r\nso, what is the final count of applications then? I assume it went above 209?", "Solution_24": "[quote=\"lfm\"][quote]Because of the computer difficulties, we're counting everything that arrived on the 28th as on time.[/quote]\n\nso, what is the final count of applications then? I assume it went above 209?[/quote]\r\n\r\nwe've had a few more applications, but not many. i think the total is 219 now, but a handful of those may now be late applications." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "geometry" ], "Problem": "I already have AOPS vol. 2 and I am pretty sure I will qualify for the USAMO, but my only proof experience is with the in-book vol.2 proofs and the USAMTS. My goal is to qualify for Red MOP, and I'm not in WOOT.", "Solution_1": "wait. you are confident that you can qualify for USAMO just after completing Aops vol 2? sweet!\r\n\r\nafter that, i guess you could do art and craft of problem solving by paul zeitz for higher level, stepping into olympiad level problems.\r\nthen problem-solving strategies by william engle\r\nfor specific practice in particular areas, do the ...for the training of the USA IMO books\r\nthey're really good. i have 104 NT and it is awesome!", "Solution_2": "Engel and Zeitz are both good resources. I would also recommend [i]Mathematical Olympiad Treasures[/i] by Titu Andreescu.", "Solution_3": "You could also look through old USAMO tests to practice.", "Solution_4": "Just do a ton of #1s, #2s and #4s from past USAMOs until you feel pretty confident in your ability to solve the majority of them. If you can solve at least two problems, you should be able to easily make Red.", "Solution_5": "[quote=\"t0rajir0u\"]Engel and Zeitz are both good resources. I would also recommend [i]Mathematical Olympiad Treasures[/i] by Titu Andreescu.[/quote]\r\n\r\nIsn't it called [i]Mathematical Olympiad Challenges[/i]? Perhaps I'm thinking of a different book...", "Solution_6": "[quote=\"Temperal\"][quote=\"t0rajir0u\"]Engel and Zeitz are both good resources. I would also recommend [i]Mathematical Olympiad Treasures[/i] by Titu Andreescu.[/quote]\n\nIsn't it called [i]Mathematical Olympiad Challenges[/i]? Perhaps I'm thinking of a different book...[/quote]\r\n\r\nI believe there are two different books, one called Treasures and one called Challenges. I haven't seen Challenges, but I do know that Treasures book is pretty good.", "Solution_7": "The material in \"Mathematical Olympiad N\", where N $ \\in \\{Treasures, Challenges\\}$, is generally pretty specialized. While the problems are excellent and numerous, don't expect to find Pompeiu's theorem on every contest you take.", "Solution_8": "I think you can also order the 21st Century Contests CD through the AMC Store. There are past real competitions from USAMO that you can download, print out, and practice as many times as you want to. Or you can print out the USAMO Competition Booklet/Pamphlet, which has all of the actual problems and solutions.", "Solution_9": "Thanks for the suggestions!", "Solution_10": "There are also a lot of Olympiad problems in the [url=http://www.artofproblemsolving.com/Forum/resources.php]AoPS contests[/url] section" } { "Tag": [ "calculus", "derivative", "LaTeX" ], "Problem": "Hi, can anyone tell me how to make a case differentation like this with $\\LaTeX$?\r\n\r\nThanks", "Solution_1": "Use the cases environment (which requires the AMS package)\r\n$S(x,a,b): = \\begin{cases} 0 & x \\leq a \\\\ 2*\\left(\\frac{a-x}{b-a}\\right)^2 & a \\leq x \\leq \\frac{a+b}{2} \\\\ 1-2*\\left(\\frac{b-x}{b-a}\\right)^2 & \\frac{a+b}{2} \\leq x \\leq b \\\\ 1 & x \\geq b \\end{cases}$", "Solution_2": "Thank you!!" } { "Tag": [ "ratio", "geometric sequence", "arithmetic sequence" ], "Problem": "Prove that this has an infinite number of solution\r\na^3+b^3=c^2\r\nand \r\nx^3+r^3=ghj\r\nwhere g, h,j are in geometric progress \r\nand\r\nsame problem as above but g, h,j are in arithmetic progression.", "Solution_1": "For number 1, $ (a,b,c) \\equal{} (m(m^3 \\plus{} n^3),n(m^3 \\plus{} n^3),(m^3 \\plus{} n^3)^2)$ works for any $ a,b,c$.\r\n\r\nFor number 2, let $ g,h,j$ be in geometric progression with common ratio $ \\alpha$, so $ \\alpha \\equal{} j/h \\equal{} h/g\\implies gj \\equal{} h^2\\implies ghj \\equal{} h^3$. Then this turns into $ x^3 \\plus{} r^3 \\equal{} h^3$, which by Fermat's Last Theorem has no nontrivial solutions, so either $ x \\equal{} 0$ and $ r \\equal{} h$, $ r \\equal{} 0$ and $ x \\equal{} h$, or $ h\\equal{}0$ and $ x\\equal{}\\minus{}r$.", "Solution_2": "Here is a family for #1 (may be encompassed in math154's though): $ (a,b,c)\\equal{}(x^{2}, 2x^{2}, 3x^{3})$.", "Solution_3": "thx but could anyone solve the third question", "Solution_4": "It would certainly help if you posted all the questions together, but if you want to be cheap you can just let the arithmetic progression be $ h,h,h$ which gives infinitely many solutions... (it becomes the same as number 2).\r\n\r\nOtherwise you can let $ (r,x,g,h,j) \\equal{} (r,3r,r,4r,7r)$ motivated by\r\n\\[ (x \\plus{} r)((x \\plus{} r)^2 \\minus{} 3xr) \\equal{} h(h^2 \\minus{} d^2),\\]where $ d$ is the common difference.", "Solution_5": "Prove: if they have no solutions or an infinite amount of solution\r\n1. a^3+b^3=3d where d is a prime greater than 3\r\n2. a^3+b^3=2^c\r\n3. a^3+b^3=3^c\r\n4. a^3+b^3=p^c where p is a prime greater than 3\r\n\r\nThis is the rest of the problem.", "Solution_6": "Oh yeah and a>B and all unknowns are postivie integers", "Solution_7": "@Math154.\r\n\r\nI think we have to take your second option because if we take the first we get that:\r\n\r\n$ x^{3}\\plus{}r^{3}\\equal{}h^{3}$\r\n\r\nThis is false by Fermat's Last Theorem.", "Solution_8": "I did not realize that it was positive integers at the time of that post..." } { "Tag": [ "function", "algebra", "partial fractions", "geometric series" ], "Problem": "Does anyone have any general tips for dealing with problems involving functions defined recursively? I know this question is somewhat general, but I am not very good at these types of problems.", "Solution_1": "I always find guessing to be useful ... \r\n\r\nactually, that isn't that far off the mark. Recursive functions are very often write-outable, so just put down the first many terms and see if you can find a pattern. Things I look for tend to be the differences between terms and other possibly identifying characteristics, like prime factorizations of terms.", "Solution_2": "Guessing is definitely a good technique. However, for recursions, you'll often end up with something that you'd never guess. Thus I feel that generating functions are the way to go. Provided that you can sum a variety of infinite series (not always an easy task!), you should be able to solve most recursions that come up on math contests. Having said that, try guessing or whatever else hits you first before you resort to techniques that are way too powerful for the problem you're solving.", "Solution_3": "If you have specific questions, advice is easier to give.", "Solution_4": "Thanks for the advice guys. Though Simon, what do you mean by summing infinite series using generating functions? I thought (simple) generating functions were only used for counting problems.", "Solution_5": "To tell you the truth, I used generating functions for recursions long before I started using them for counting problems. Here's the method in action for a very simple recursion because I'm just lazy.\r\n\r\nProblem: f(0)=0, f(n+1)=2f(n)+1. Find f(n).\r\n\r\nSolution: Let's multiply both sides of the recursion by xn to get f(n+1)xn=2f(n)xn+xn. Now sum from n=0 to infinity, and call :Sigma: f(n)xn from n=0 to infinity F(x). Note that the left sides is (F(x)-f(0))/x=F(x)/x. The right side is 2F(x)+1/(1-x). Thus F(x)/x=2F(x)+1/(1-x), or F(x)((1-2x)/x)=1/(1-x), or F(x)=x/(1-x)(1-2x). Now we expand this out by partial fractions and (hopefully!) get F(x)= :Sigma: (2n-1)x^n from n=0 to infinity. Thus f(n)=2n-1.", "Solution_6": "Cool! I don't think I've ever seen that type of problem solved that way before.\r\n\r\nJust one minor detail:\r\n\r\n[quote=\"ComplexZeta\"]Now we expand this out by partial fractions and (hopefully!) get F(x)= :Sigma: (2n-1)x^n from n=0 to infinity. Thus f(n)=2n-1.[/quote]\r\n\r\nWhen we expand it out, we actually get your sum from n=1 to infinity (since, after some algebra, F(x) = Sum from n=0 to infinity x^(n+1) * (2^(n+1) - 1)). Shouldn't we need to take this into account and subtract off the expression when n=0?", "Solution_7": "Ah, but f(0)=0, so it doesn't matter. (How do you like the way I wriggled out of that mess?)", "Solution_8": "Here's a handout entitled [url=http://compgeom.cs.uiuc.edu/~jeffe/teaching/373/notes/recurrences.pdf]Notes on Solving Recurrence Relations[/url] that I found useful.\r\n\r\n--Dan", "Solution_9": "[quote]Ah, but f(0)=0, so it doesn't matter. (How do you like the way I wriggled out of that mess?)[/quote]\r\n\r\n:D\r\n\r\nBy the way, the notes are interesting petra.", "Solution_10": "Hmm try this one:\r\n\r\nFind an explicit formula for a[size=75]n[/size] if a[size=75]n+2[/size] = (a[size=75]n[/size] + a[size=75]n+1[/size])/2.", "Solution_11": "2a_n+2=a_n+a_n+1\r\ncharacteristic equation-->2x^2-x-1-->2x^2-2x+x-1-->(2x+1)(x-1) \r\nhow bout you tell me a_0 or something?", "Solution_12": "Use a0=1 and a1=2. (I just made those up, but why not?) Also, try to do it with generating functions. It's much more universal than characteristic equations (i.e. you can solve way more recursions).", "Solution_13": "A(-1/2)^0 +B=1->A+B=1\r\n-A/2+B=2\r\n3A/2=-1, A=-2/3, B=5/3\r\n\r\na_n = (-2/3)(-1/2)^n + 5/3\r\n\r\nhow would i use generating functions to do this?", "Solution_14": "Okay, here goes nothing. Our terms are all an. Let \r\nf(x) = a0x0 + a1x1 + a2x2 + a3x3 + ...\r\n\r\nUsing our identity, we get that\r\nf(x) = a0x0 + a1x1 + (a0 + a1)x2/2 + (a1 + a2)x3/2 + (a2 + a3)x4/2 + ...\r\n\r\nAnd, recombining terms, we get\r\nf(x) = a0x0 + a1x1 + 1/2*(a0x2 + a1x3 + ...) + 1/2*(a1x2 + a2x3 + ...)\r\n\r\nwhich we can express in the form\r\n\r\nf(x) = a0x0 + a1x1 + x2f(x)/2 + x*(f(x) - a0)/2\r\n\r\nand combining terms with f(x) on the left, we get\r\n\r\nf(x)*(1 - x/2 - x2/2) = a0 + a1x - a0x/2 = x*(a1 - a0/2) - a0\r\n\r\nor\r\n\r\nf(x) * (x2 + x - 2) = (a0 - 2a1)x + 2a0\r\n\r\nor\r\n\r\nf(x) = ((a0 - 2a1)x + 2a0)/(x2 + x - 2)\r\n\r\n\r\nNow, I'm stopping here, but to finish it off, you need to split that into two fractions (one over (x - 1), the other over (x + 2) which we get from factoring (x2 + x - 2)). Each of these will have a form similar to that for the summation of a geometric series, so you simply need to express them as the geometric series that they are and then add them together to find out what ever coefficient is.\r\n\r\n(By the way, you can do the characteristic equation method for general first and second terms, too. It just gets messy, as this does. This would be much prettier if I had done it for given values of a0 and a1." } { "Tag": [ "trigonometry", "complex numbers", "geometry unsolved", "geometry" ], "Problem": "For a complex number $ z\\equal{}x\\plus{}iy$ we denote by $ P(z)$ the corresponding point $ (x,y)$ in the plane. Suppose $ z_1,z_2,z_3,z_4,z_5,\\alpha$ are nonzero complex numbers such that:\r\n$ (i)$ $ P(z_1),...,P(z_5)$ are vertices of a complex pentagon $ Q$ containing the origin $ O$ in its interior, and\r\n$ (ii)$ $ P(\\alpha z_1),...,P(\\alpha z_5)$ are all inside $ Q$.\r\nIf $ \\alpha\\equal{}p\\plus{}iq$ $ (p,q \\in \\mathbb{R})$, prove that $ p^2\\plus{}q^2 \\le 1$ and $ p\\plus{}q \\tan \\frac{\\pi}{5} \\le 1$.", "Solution_1": "I think I have the first part...but I'm not very good at complex numbers so if I made a mistake, please correct me!\r\n\r\nLet $ r = \\max(|z_1|,|z_2|,|z_3|,|z_4|,|z_5|)$ so that $ P(z_1), P(z_2),...,P(z_5)$(and therefore $ Q$) lie inside the region $ |z|\\le r$.\r\nSince we are given that $ P(\\alpha z_1),P(\\alpha z_2),...,P(\\alpha z_5)$ lie inside $ Q$ these points must also lie inside the region $ |z| \\le r$.\r\nHence for all $ 0\\le i \\le 5$,\r\n$ |\\alpha z_i|=|\\alpha\\parallel{}z_i| \\le r \\Rightarrow |\\alpha|\\le \\min\\left(\\frac{r}{|z_1|},\\frac{r}{|z_2|},\\frac{r}{|z_3|},\\frac{r}{|z_4|},\\frac{r}{|z_5|}\\right) = 1$." } { "Tag": [ "real analysis", "induction", "real analysis unsolved" ], "Problem": "Good Day, Everyone. \r\n\r\n[quote=\"W. Rudin\"]Construct a Borel set $E\\subset R^1$ such that $0 < m(E\\cap I)0\r\n 2) a(n)---> : :inf: \r\n Prove that {b(n)} is dense in [0,1] , where b(n)= a(1)+a(2)+...+a(n).[/list][/quote][/b][/u]", "Solution_1": "This has already been discussed (quite a few times, I might add :)).", "Solution_2": "Actually,I think it should look like this: \r\n\r\nGiven that $x_n\\to\\infty$ and $x_{n+1}-x_n\\to 0$,show that $\\{x_n\\}$ is dense in [$0,1$].", "Solution_3": "the real problem is to show thet {b(n)} is dense in [0,1]" } { "Tag": [ "calculus", "integration", "trigonometry", "LaTeX", "calculus computations" ], "Problem": "hey all could you solve this integral pls \r\n\\int(cos(x)/(5+4cos(x)))dx", "Solution_1": "[quote=\"Javkhaa\"]hey all could you solve this integral pls \n\\int(cos(x)/(5+4cos(x)))dx[/quote]\r\n\r\nThis type of integrals is typically solved with the substitution $ y\\equal{}\\tan\\frac{x}{2}$ to transform it into the integral of a rational function. Try it!", "Solution_2": "[quote=\"Javkhaa\"]\n$ \\int \\frac{\\cos x}{5+4\\cos x}dx$\n\n[/quote]\r\n\r\nPut $ \\$$ signs around your $ \\text{\\LaTeX}$ code.\r\n\r\nSubstitute $ t=\\tan \\frac{x}{2}$." } { "Tag": [ "geometry solved", "geometry" ], "Problem": "Two circles intersect at points A and B. An arbitrary line through B intersects the first circle again at C and the second circle again at D. The tangents to the first circle at C and the second at D intersect at M . Through the intersection of AM and CD, there passes a line parallel to CM and intersecting AC at K. Prove that BK is tangent to the second circle.", "Solution_1": "Really, not very hard.\r\nLet the intersection between AM and CD be E. Then the most important thing is to notice A,D,M,C and A,B,E,K are both concyclic. The remaining is just some simple angle chasing.", "Solution_2": "I've tried to prove that A,D,M,C and A,B,E,K are both concyclic and finish the solution but I couldn't. Could you describe the angle chasing?", "Solution_3": "Okay, here is a detailed solution. I will work with directed angles modulo 180.\r\n\r\nDenoting by E the point of intersection of the lines AM and CD, we have EK || CM.\r\n\r\nSince the line CM is the tangent to the circle through the points A, B and C at the point C, we have < (CM; BC) = < CAB by the tangent-chord angle theorem. In other words, < (CM; CD) = < (CA; AB). Similarly, < (DM; CD) = < (DA; AB). Thus,\r\n\r\n < CMD = < (CM; DM) = < (CM; CD) - < (DM; CD) = < (CA; AB) - < (DA; AB) = < (CA; DA) = < CAD,\r\n\r\nand it follows that the points A, D, M, C are concyclic. This yields < ACM = < ADM. On the other hand, since EK || CM, we have < AKE = < ACM, and thus < AKE = < ADM. But since the line DM is the tangent to the circle through the points A, B and D at the point D, we have < (AD; DM) = < ABD after the tangent-chord angle theorem. In other words, < ADM = < ABD. Hence, < AKE = < ADM = < ABD = < ABE, and this yields that the points A, B, E, K are concyclic. This entails < ABK = < AEK. Since EK || CM, we have < AEK = < AMC. And since the points A, D, M, C are concyclic, we have < AMC = < ADC. Thus,\r\n\r\n < ABK = < AEK = < AMC = < ADC = < ADB.\r\n\r\nIn other words, < (AB; BK) = < ADB. By the converse of the tangent-chord angle theorem, this yields that the line BK is the tangent to the circle through the points A, B and D at the point B.\r\n\r\n$\\blacksquare$\r\n\r\n Darij" } { "Tag": [ "floor function", "combinatorics unsolved", "combinatorics" ], "Problem": "Written the numbers from 1 to 100 on the cards. If there doesn\u2019t exist two consecutive numbers on any chosen 5-cards amoung them then these 5-cards will be called as \u201cgood\u201d . Find how many \u201cgood\u201d verifiying this condition can be found ?", "Solution_1": "Is the question, \"In how many ways can one choose 5 numbers from $ \\{1, 2, \\ldots, 100\\}$ such that no two consecutive numbers are chosen?\" It's a standard exercise and the answer is $ \\binom{100 \\minus{} 5 \\plus{} 1}{5}$.", "Solution_2": "In general, for $ N\\equal{}\\{1,\\dots,n\\}$ and $ k\\leq \\lfloor \\frac{n}{2} \\rfloor$, the number of k-subset so that $ a_i\\minus{}a_j\\neq 1\\quad \\forall a_i,a_j\\in N$ is given by $ n\\plus{}k\\minus{}1\\choose k$\r\n\r\nProof: Arrange k elements in ascending order. There have to be atleast k-1 elements, one each to be kept between 1st and the 2nd,2nd and the 3rd and so on upto (k-1)th and the kth. Rest of the n-2k+1 can be kept in k possible places (now including the space left and right of 1 and k respectively). This can be done in $ n\\plus{}2k\\minus{}1\\plus{}k\\choose k$ ways." } { "Tag": [ "number theory", "relatively prime" ], "Problem": "When .2006, with 006 repeating, is expressed as a fraction whose numerator and denominator are relatively prime, what is the sum of the numerator and denominator?\r\n\r\nI made 10000x = 2006. 006006 etc. and 10x= 2.006006 etc. and I subtracted, and got:\r\n\r\n9900x=2004, and then 167/825, which would give a sum of 825+167=992. What did I do wrong?", "Solution_1": "[quote=\"ckck\"]When .2006, with 006 repeating, is expressed as a fraction whose numerator and denominator are relatively prime, what is the sum of the numerator and denominator?\n\nI made 10000x = 2006. 006006 etc. and 10x= 2.006006 etc. and I subtracted, and got:\n\n9900x=2004, and then 167/825, which would give a sum of 825+167=992. What did I do wrong?[/quote]\r\n$ 10000\\minus{}10 \\equal{} 9990$, not $ 9900$.\r\nSo, $ 9990x\\equal{}2004 \\rightarrow x\\equal{}\\dfrac {2004}{9990}\\equal{}\\dfrac {334}{1665}$, and your answer is $ \\boxed{1999}$." } { "Tag": [], "Problem": "if electronegativity differnce among the atoms in a covalent bond increase it's bond strength increases", "Solution_1": "If the electronegativity difference of the atoms in a bond increase, then the bond caracter changes from covalent to ionic. So we are to compare covalent versus ionic bonds, perhaps by comparing bond dissociation enthalpies with lattice enthalpies.", "Solution_2": "well actually i am seraching for a explaination of the order of acididty in case of halogen acids", "Solution_3": "For the halogen acids HX, although bond polarity decreases down the halogen group, the strenght of the bond between hydrogen and the halogen also decreases, and so the acids are increasingly stronger. Bond strenght decreases because the bond is formed by superposition of an H 1s orbital with an sp3 halogen orbital: however, as the halogen size increases the orbital combination becomes increasingly poor due to a incresing size difference - for iodine for example, the orbitals in question are H 1s and I \"5sp3\" - note the big size difference between the orbitals. Is is so expected that HI is the strongest acid, which is indeed the case. On the other hand, although the electronegativity difference in HF is the greastest (and so it would be expected for this acid to be the strongest) its covalent bond is the strongest because orbital overlap in this case is very efficient (due to the very similar atoms sizes): because of that, HF is only an weak acid in water.", "Solution_4": "A somewhat similar argument can be used to explain the anomalous behaviour of halogens regarding reactivity and orientation in aromatic electrophilic substitutions.", "Solution_5": "thanks :D" } { "Tag": [ "geometry", "incenter", "ratio", "Euler", "geometry unsolved" ], "Problem": "In acute angled triangle $ABC$, $BC=a$,$CA=b$,$AB=c$, and $a>b>c$. $I,O,H$ are the incentre, circumcentre and orthocentre of $\\triangle{ABC}$ respectively. Point $D \\in BC$, $E \\in CA$ and $AE=BD$, $CD+CE=AB$. Let the intersectionf of $BE$ and $AD$ be $K$. Prove that $KH \\parallel IO$ and $KH = 2IO$.", "Solution_1": "Let $CK$ meet $AB$ at $F$ then from Seva teorem \r\n$\\frac{AE}{EC}\\frac{CD}{DB}\\frac{BF}{FA}=1 \\Longrightarrow AE=BD \\Longrightarrow\\frac{CD}{EC}=\\frac{FA}{FB}\\Longrightarrow \\frac{CD+CE}{EC}=\\frac{AB}{FB}$ $\\Longrightarrow EC=FB$ and $AF=CD$\r\nSo $AF=CD=p-b, BF=CE=p-a, BD=AE=p-c$\r\nLet $A_1,B_1,C_1$ be the feet of altitudes from $A,B,C$, respectively.Let $K_1,K_2$ be feet of perpendiculars from $K$to the side $BC$ and to the altitude $AA_1$ and let $S,T$ and $R$ be the feet of perpendiculars from $I$ and $O$ to the side $BC$, and foot of altitude $O$ to $IS$, respectively.\r\nFrom the Seva teorem we get $\\frac{AD}{KD}=\\frac{p}{p-a}=\\frac{AA_1}{KK_1} \\Longrightarrow KK_1=\\frac{p-a}{p}AA_1$\r\n$|HK_2|=|HA_1-KK_1|=|AA_1-AH-AA_1\\frac{p-a}{p}|=|\\frac{a}{p}AA_1-AH|=|\\frac{2S}{p}-AH|=|2r-2OT|$ \r\n\r\n$\\Longrightarrow HK_2=2IR$ $(1)$\r\n$KK_2=A_1K_1=A_1D-K_1D=A_1D-A_1D \\frac{KD}{AD}=A_1D-A_1D\\frac{p-a}{p}=\\frac{a}{p}A_1D=\\frac{a}{p}(CA_1-CD)=\\frac{a}{p}(\\frac{a^2+b^2-c^2}{2a}-(p-b))=b-c$\r\n$\\Longrightarrow KK_2=2(p-c-\\frac{a}{2})=2ST$\r\n$\\Longrightarrow KK_2=2OR$$(2)$\r\nUsing $(1)$,$(2)$ and $\\angle KK_2H=\\angle ORI=\\frac{\\pi}{2}$\r\nwe get $\\bigtriangleup KK_2H \\sim \\bigtriangleup ORI$ with ratio $2$. \r\nSo we get $KH \\Vert OI$ and $KH=2OI$", "Solution_2": "We have $(p-a)\\overrightarrow{KA}+(p-b)\\overrightarrow{KB}+(p-c)\\overrightarrow{KC}=\\overrightarrow{0}$, $\\overrightarrow{OA}+\\overrightarrow{OB}+\\overrightarrow{OC}=\\overrightarrow{OH}$, $a\\overrightarrow{IA}+b\\overrightarrow{IB}+c\\overrightarrow{IC}=\\overrightarrow{0}$. Now, it is easy ! :D", "Solution_3": "It's very easy problem.We have $ K $ is Nagel point of the triangle $ ABC $ and by Nagel line theorem $ K,I,G $ are collinear and $ GK=2IG $, on the other hand $ H,G,O $ are collinear ( Euler line ) and $ HO=2OG $ ,so the triangle $ IGO $ and the triangle $ HGK $ similar, so $ IO $ parallel to $ HK $ and $ HK=2IO $ .", "Solution_4": "Where $ G $ is centroid of the trianle $ ABC $." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Let $P$ be an interior point of equilateral triangle ABC such that $\\angle{PAB}+\\angle{PBC}+\\angle{PCA}=90^{\\circ}$ . Find all $P$.", "Solution_1": "See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=54016]Locus (nice, nice, ...)[/url]." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Let triangle $ ABC, A\\minus{}$excircle tagent $ AB,AC$ at $ M,N$ respectively,$ B,C\\minus{}$excircle tagent $ BC$ at $ P,Q$ respectively. Prove that $ PM,QN,AH$ concurent (here $ AH$ be altitude of triangle $ ABC$)", "Solution_1": "See [b]vittasko[/b]'s lemma: http://www.mathlinks.ro/Forum/viewtopic.php?t=164092" } { "Tag": [ "summer program", "MathPath", "puzzles" ], "Problem": "I'm not sure if this has been posted before, but I like this problem a lot.\r\n\r\nYou have just found out that you have a deadly disease. Fortunately, your doctor has the cure. He/she gives you two bottles of pills: one labled A and one labled B. The pills in each bottle are exactly identical. The doctor tells you to take one pill from each bottle for the next 15 days. If you do this successfully, you live. If you fail, you die.\r\n\r\nThe first day is fine. You take your pills as planned and all goes well. The second day though, you put one A pill into your palm and when you get a B pill, two pills fall from the bottle into you palm. You don't which pills are which. What do you do to make certain that you live?\r\n\r\nGood Luck!", "Solution_1": "Seen this one before.\r\n\r\n[hide]\nTake a pill from A.\n\nLay all four of them so that they are not close to each other.\n\nCut each one exactly in half.\n\nTake one from each of the 4 piles and you have 2 1/2s A and 2 1/2s B. Remember you have to separate the pills before you cut them. Then you have to eat the remainder tomorrow.\n\nQED[/hide]", "Solution_2": "[hide=\"I think the solution is...\"]You crush those three pills and another A pill together, and then eat half of the mixture that day, and the next half the next day.[/hide]", "Solution_3": "[hide=\"Or...\"]Just throw those pills away and eat new ones[/hide]", "Solution_4": "You can't throw those out because you only have 15 of each and the doctor doesn't have any more of them. But the other solutions work. Good job! \r\n\r\nAnother solution I've seen:\r\n\r\n[hide]Add another A pill and dissolve all four pills in water. Then drink half the mixture today and half tomorrow. Eat the other pills as planned. (This only works if the pills dissolve in water)[/hide]\r\n\r\nI personally prefer mathgeniuse's solution, but anything that works is fine too.", "Solution_5": "Was this a MathPath problem of the day? It sounds really familiar.\r\n\r\nUse the Banach-Tarski Paradox to duplicate yourself some more pills!!! :roll:", "Solution_6": "Yeah this was and yes it is familiar." } { "Tag": [ "geometry", "geometric transformation", "reflection", "geometry proposed" ], "Problem": "ABC is a triangle.And the circumcenference of ABC is circleO. H is the orthocenter.\r\nGiven point P in the plane.AP,BP,CP meets the Circle O at A1,B1,C1.\r\nThe reflection of A1 with respect to the midpoint of BC is A2.B2,C2 similarly defined.\r\nProve that H,A2,B2,C2 are concyclic.\r\nAlso if we define A3,B3,C3 as the reflection point of A1 with respect to BC.\r\nWe also have H,A3,B3,C3 concyclic.", "Solution_1": "Dear Mathlinkers,\r\nsee the discussion : http://www.mathlinks.ro/Forum/viewtopic.php?t=303466\r\nSincerely\r\nJean-Louis" } { "Tag": [], "Problem": "is a good author. Last year, when I had more time than I do now, I used to read many of his books. One of his best is False Memory.", "Solution_1": "Oh yeah, Dean Koontz rocks. He is one of the few authors even close to being as good as Stephen King. Watchers and Phantoms are his best books.", "Solution_2": "Personally I have never read Dean Koontz, but Pennywise here is a good friend of mine in person, so I thought I'd post and get him to elaborate on why they are so \"good\".:p", "Solution_3": "Dean Koontz is awesome because a.) his books are just as much humorous as they are creepy; he makes fun of all the horror cliches b.) the only fictional character even close to being as cool as Pennywise is Einstien, the dog, or The Outsider, both of which are in Watchers; finally, c.) he writes with a unique style, and doesn't fall into the classic horror stereotypes." } { "Tag": [], "Problem": "For a positive integer $r$, the sequence $(a_n)$ is defined by $a_1 = 1$ and $a_{n+1} = \\frac{n a_n + 2(n + 1)^{2r}}{n + 2}$ for $n \\geq 1$.\r\n\r\nProve that each $a_n$ is a positive integer, and find the $n$\u2019s for which $a_n$ is even.", "Solution_1": "[quote=\"M4RI0\"] Prove that each $a_n$ is a positive integer, and find the $n$\u2019s for which $a_n$ is even.[/quote]\r\n\r\num, what about $n=1$ and $r=2$? i'm getting $a_2=\\frac{17}{3}$...", "Solution_2": "Hm, we have $a_2= \\frac{1+8r}3$. But this isn't an integer for all $r$. Maybe you're missing some piece of information? Or maybe I'm misunderstanding the question?\r\n\r\nedit: Jason beat me to it :P", "Solution_3": "OK. It's corrected now. :D" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$a,b,c,d >0$.Prove that:\r\n $27(a^4+b^4+c^4+d^4-4abcd) \\geq (a+b+c+d)^4-4^4abcd$\r\nConjecture:\r\n$x_{1},x_{2},..,x_{n} >0$ and $n \\in N \\geq 3$.Prove that:\r\n $(n-1)^{(n-1)}(x_{1}^n+x_{2}^n+...+x_{n}^n-nx_{1}x_{2}...x_{n}) \\geq (x_{1}+x_{2}+...+x_{n})^n-n^nx_{1}x_{2}...x_{n}$", "Solution_1": "The (n-1)EV transforms this one (even your conjecture) to a two variable inequality!\r\n :)", "Solution_2": "sorry what is the (n-1)EV?" } { "Tag": [ "geometry", "geometric transformation", "geometry proposed" ], "Problem": "Let m be a fixed line and A, B are points lying on different halfplane divided by m such that AB is not perpendicular to m. C and D moves in m such that the angle CBD = CAD. Call H the orthocenter of CDB. (S) is a circle passing A and H and intersects HC, HD respectively at M,N. Through H, we draw a line paralel to m and cut (S) at I. Let K be the intersection of AI and MN.\r\n a) Prove that K lies on m.\r\n b) Now A and B are fixed. Let J, L be the orthocenter of KCM and KDN. Prove that J, L lies on a fixed line.", "Solution_1": "Wow! I found this on the $10$'th page! :D I guess people didn't want to read it. It is a bit long, isn't it? :)\r\n\r\nIt's very rich in properties, so I'll state many of them without proof, because it's all just a big angle chase :).\r\n\r\n[b]a)[/b] $ACHD$ is a cyclic quadrilateral. $AMN$ is obtained from $ACD$ by a spiral similarity (I'm not sure we'll use this :)). Let $K'=AI\\cap CD,\\ K''=MN\\cap CD$. We have $\\angle ADK'=\\angle AHM,\\angle AKD=\\angle AMH$, so $\\angle HAD=\\angle MAK'$. We have thus found $AHD,AMK'$ to be similar $(*)$.\r\n\r\nWe can now use $(\\#)$ to find $\\angle MK''C=\\angle MAC$, so $MK''AC$ is cyclic. From here we find $\\angle AK''N=\\angle ACH,\\ \\angle ANK''=\\angle AHC$, so, again, $\\angle MAK''=\\angle HAD$, getting $AHD,AMK''$ similar $(**)$.\r\n\r\n$(*),(**)\\Rightarrow K'=K''=K$.\r\n\r\n[b]b)[/b] If we assume for now that $C,D$ are also fixed, the locus of $J$ (let's just talk about $J$) is easily seen to be a line passing through $B$. We need now to show that these lines when $C,D$ move on $m$ are, in fact, one and the same. Its not hard to prove that the symmetric of $A$ in $m$ also belongs to this locus no matter how $C,D$ move, so when $A,B$ are fixed, the locus of $J$ is the line passing through $B$ and the symmetric of $A$ wrt $m$ (which are different, according to the hypothesis).", "Solution_2": "[quote=\"grobber\"]Wow! I found this on the $10$'th page![/quote]\r\nWhat do you mean?", "Solution_3": "He probably means the 10'th page of the \"Geometry Proposed & Own Problems\" topics list.\r\n\r\n Darij", "Solution_4": ":lol: :lol: :lol:", "Solution_5": "That is indeed what I mean. I was just surprised that we ignored it for so long :).", "Solution_6": "There are NUMEROUS problems, which we ignore for a long time. :D" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$ a,b,c>0 $ show that this inequality holds:\r\n\\[ \\frac{2a}{a^2+bc} +\\frac{2b}{b^2+ac} + \\frac{2c}{c^2+bc} \\leq \\frac{a}{bc}+\\frac{b}{ac} +\\frac{c}{ba} \\]", "Solution_1": "$2a^2 + b^2 + c^2 \\ge 4a\\sqrt{bc}$; add up the similar inequalities to get $a^2 + b^2 + c^2 \\ge a\\sqrt{bc} + b\\sqrt{ac} + c\\sqrt{ab}$. Thus $\\frac{a}{bc} + \\frac{b}{ac} + \\frac{c}{ab} = \\frac{a^2 + b^2 + c^2}{abc} \\ge \\frac{a\\sqrt{bc} + b\\sqrt{ac} + c\\sqrt{ab}}{abc} = \\frac{a}{a\\sqrt{bc}} + \\frac{b}{b\\sqrt{ac}} + \\frac{c}{c\\sqrt{ab}} \\ge \\frac{2a}{a^2 + bc} + \\frac{2b}{b^2 + ac} + \\frac{2c}{c^2 + ab}$", "Solution_2": "WLOG, assume abc=1. Observe that \\[\\frac{2a}{a^2+bc}=\\frac{2a^2}{a^3+1}\\le\\frac{1}{2}(a+1)\\]The result follows immediately from $a^2+b^2+c^2\\ge a+b+c$ and $a^2+b^2+c^2\\ge 3$.", "Solution_3": "Thank you, both of you but mecrazywong can you explain you solution ? :lol:", "Solution_4": "[quote=\"jensen\"]Thank you, both of you but mecrazywong can you explain you solution ? :lol:[/quote]\r\nWhat do you want him to explain? :?", "Solution_5": "$\\frac{2a^2}{a^3+1}\\le\\frac{1}{2}(a+1)\\Leftrightarrow(a-1)^2(a^2+3a+1)\\ge0$.\r\n$a^2+b^2+c^2\\ge(a+b+c)^2/3\\ge a+b+c$." } { "Tag": [ "inequalities", "quadratics", "inequalities proposed" ], "Problem": "Prove that for all reals a,b,c we have $ a^4+b^4+c^4+a^3b+b^3c+c^3a\\geq 2ab^3+2bc^3+2ca^3 $. A very nice one.", "Solution_1": "assuming that a is the smallest and writing $b=a+x,c=a+y$ doing some algebra, the inequality is immediately clear(the hardest part is showing that $x^4+x^3-2x+1\\geq 0$ for all positive x(actually i'm not completely sure about it, but setting $c=0$ in the original equation that one is implied and inequalities in one variable aren't that hard)).\r\n\r\nPeter", "Solution_2": "I have also a proof\r\nIf a.b>0 we can suppose c>=0 also because else we get a rather easy inequality\r\n|Now use ab^3+bc^3+ca^3-a^3b-b^3c-c^3a|=|(a-b)(b-c)(c-a)|(*)\r\nSo we must prove sum{a^4+3b^4-4ab^3}/4>=*\r\nOr sum{(a-b)^2(a^2+2ab+3b^2}/4>=(*)\r\nNow if we increase each of a,b,c by x we can see the RHS increases less than LHS.\r\nSo we can suppose one of a,b,c is zero and then it's easy", "Solution_3": "i just had problems whether my solution actually works for all real a(i thought they were positive), but actually i get a quadratic in a with discriminant $\\Delta=-3(x^3-3x^2y+y^3)^2\\leq 0$, so the solution works.\r\n\r\nPeter", "Solution_4": "Jura,[tex] a^3b+b^3c+c^3a-ab^3-bc^3-ca^3=(a+b+c)(a-b)(b-c)(a-c) [/tex] :)", "Solution_5": "Following Peter Scholze I have also solved it but, after realizing a \r\n\r\ndeep similarity with INEQ3, I am now going to think of this problem :\r\n\r\n\" Which inequality between INEQ3 and this one is stronger ? \"\r\n\r\nHas anyone an answer?\r\n\r\nThank you very much.", "Solution_6": "These inequalities are\r\n$\\#3(a^4+b^4+c^4)+3(a^3b+b^3c+c^3a)\\geq 6(ab^3+bc^3+ca^3)$\r\nand\r\n$2(a^4+b^4+c^4)+4(a^2b^2+b^2c^2+c^2a^2)\\geq 6(ab^3+bc^3+ca^3)$.\r\n\r\nFor a, b, c positive, the second is stronger than the first, because\r\n$LHS1\\geq LHS2$.\r\n\r\nFor a, b, c negative we cannot say one is stronger than the other." } { "Tag": [ "trigonometry", "inequalities unsolved", "inequalities" ], "Problem": "$ 00}{\\frac{1}{2r}{\\int_{x-r}^{x+r}{f(t)dt}}}$. Prove that \r\n a) if $f$ is continuous, then $M$ is too.\r\n b) The set of all real numbers $x$ such that $M(x)>1$ can be covered with a family of intervals the sum of lengths of which is at most $5\\int_{-\\infty}^{\\infty}{f(t)dt}$. Thus, the set of $x$ for which $M(x)=\\infty$ is of measure 0.", "Solution_1": "Harazi, could you please tell what was the year when this problem was proposed in the ENS exam.\r\n\r\nI found it very interesting.\r\n\r\nI'll try to cover the second part. My idea for the first one seems a bit unclear.\r\n\r\nThis average functiion is called Hardy-Littlewood Maximal function. The second point of this problem can be easily done using a version of Hardy-Littlewood Maximal theorem:\r\n$|\\{ Mf \\geq \\epsilon \\}| \\leq \\frac{2}{\\epsilon} ||f||_1$ (*) . Here is the reference http://planetmath.org/encyclopedia/HardyLittlewoodMaximalTheorem.html\r\n\r\nYou can also find a proof here http://www.math.ntnu.no/~eugenia/TMA4225/lecturenotes/lecture15.pdf.\r\n\r\n*. From (*) if we tend $\\epsilon \\to \\infty$ we get the result that $| \\{ Mf = \\infty \\} | = 0$.\r\n\r\n**.To get the result $\\{ x : M(x)>1 \\}$ can be covered with intervals of summary length $\\leq 5||f||_1$, i think that in order to use (*) we can notice that $M(x)$ is measurable and even more $\\{ x: M(x) > a \\}$ is open.\r\n\r\nThe measurability is clear: $M(x)$ being the supremum of average functions(which are in our case continious) is lower semicontinious and the set $\\{ x: M(x) > a \\}$ is measurable.\r\n\r\nTo get a stronger version that $\\{ x: M(x) > a \\}$ is open, we can define $A_r(x) = \\frac{1}{2r} \\int_{x-r}^{x+r} f(t)dt$, and $M(x)=\\sup_{r>0} A_r(x)$, so for any fixed $x$ and $n$ there exists $r_n$ such that $A_{r_n}(x)>M(x)-\\frac{1}{n}$. Thus\r\n\r\n$\\{ x: M(x)>a\\}= \\bigcup_{n} \\bigcap_{r \\in \\mathbb{Q}} \\{ x: A_r(x)> a-\\frac{1}{n} \\}$ which is open as well as each set $\\{ x: A_r(x)> a-\\frac{1}{n} \\}$ is open because $A_r(x)$ is continuous.\r\n\r\nSo, the set $\\{ x : M(x)>1\\}$ is open and can be represented as a union of intervals which summary leghts is due to (*) $\\leq 5||f||_1$.\r\n\r\nSorry if it is unclear.", "Solution_2": "Suppose we replace $\\mathbb{R}$ by $\\mathbb{R}^n$ and the average over the interval $(x-r,x+r)$ by the average over the ball of radius $r$ centered at $x.$ Then replace the $5$ in part (b) by $5^n.$\r\n\r\nThe result is still true, basically in the way that Eugene outlines." } { "Tag": [ "logarithms", "real analysis", "real analysis theorems" ], "Problem": "If $ |z+1|< \\sqrt{2}\\; ,\\; $ then\r\n\\[\\pi +\\; 4\\; \\arctan{z}+ 2\\; \\ln{\\frac{1-2z-z^2}{1+z^2}}= \\]\r\n\\[=4\\sum\\limits_{k=0}^{\\infty} \\frac{(1+z)^{8k+1}}{8k+1}\\frac{1}{16^k}\\; -\\; 2\\sum\\limits_{k=0}^{\\infty}\\frac{(1+z)^{8k+4}}{8k+4}\\frac{1}{16^k}\\; -\r\n\\sum\\limits_{k=0}^{\\infty} \\frac{(1+z)^{8k+5}}{8k+5}\\frac{1}{16^k}\\; -\\; \\]\\[- \\sum\\limits_{k=0}^{\\infty}\r\n\\frac{(1+z)^{8k+6}}{8k+6}\\frac{1}{16^k}\\; \\; .\r\n\\]\r\n\r\n\r\nFor $ z=0 $ we find remarkable formula. Which one ?", "Solution_1": "$\\pi$ from Baily-Borwein-Plouffe 1995", "Solution_2": "It's OK ! Yes , for z=0 is the so-called BBP-formula which enable us to find the n-$ {}^{th}$ hexa-decimal of $ \\pi $ , without calculation the first $ n-1 .$ Regards,Alex" } { "Tag": [ "calculus", "derivative", "algebra", "binomial theorem" ], "Problem": "Factorise $(a+b)^7-a^7-b^7$ without expanding $(a+b)^7$.", "Solution_1": "$(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$", "Solution_2": "Method is the thing that counts,answer is fine :)", "Solution_3": "Can you post a solution? I'm confused.", "Solution_4": "Yeah, how do you go about doing this? It's obviously the exapansion of $(a+b)^7$ without the $a^7$ and $b^7$ that are on either end, but how do you find what that is without just using the binomial theorem?", "Solution_5": "[quote=\"Pocket Sand\"]Can you post a solution? I'm confused.[/quote]\r\n\r\nI'm in a hurry,all i can say is see symetry or something.the given thing is a symmetrical expression,so the factors have to be something like $a+b(Aa^2+ab+Bb^2)$(put a=0,b=0....) which are both symmetrical in their respective degrees.....will xplain later :)", "Solution_6": "FWIW: \r\n\r\nQuick way to see:\r\n[hide]$a=0$ makes the expression $0$ so ====> $a$ is a factor\n$b=0$ makes the expression $0$ so ===> $b$ is a factor \n$a=-b$ makes the expression $0$ so ===> $(a+b)$ is a factor \n$a=b\\omega$ or $a=b\\omega^2$ makes the expression $0$ so ===> $(a-b\\omega)(a-b\\omega^2) = (a^2+ab+b^2)$ is a factor \n[hide]Here $\\omega$ is cuberoot of unity and notice $1+\\omega+\\omega^2 = 0$\nYou notice that the whole expression is $b^7((1+\\omega)^7 - 1 - \\omega^7)$ comes out to be zero because $1+\\omega = -\\omega^2$ \nand you get $-\\omega^{14} - \\omega^7 -1 = -\\omega^2-\\omega -1 = 0$[/hide]\nYou can see that its double root.. too so last expression should be a saure term. \n\n[hide](Easiest to see this requires calculus , the derivative would be zero too when $a=b\\omega$ etc..[/hide]\n\n[hide](If you don't see its a double roots right away, you can also find the remaing term as it will only of a second degree ... etc..) [/hide]\n\nSo you use $(a+b)^7-a^7-b^7 = k a b (a+b) (a^2+ab+b^2)^2$ \nSince both sides are now of equal degrees , you can find the value of $k$ by using, say $a=b=1$ which will give $k=7$ [/hide]\r\n\r\nHope this helps.", "Solution_7": "Gyan, that's a very interesting method, I've never thought of factoring that way before.", "Solution_8": "Yeah. Plugging in values is a very good way to begin a factorization problem.\r\n\r\nAlso, you can tell immdiately by the $-a^7-b^7$ that $ab$ is a factor.", "Solution_9": "[quote=\"Gyan\"]FWIW: \n\nQuick way to see:\n[hide]$a=0$ makes the expression $0$ so ====> $a$ is a factor\n$b=0$ makes the expression $0$ so ===> $b$ is a factor \n$a=-b$ makes the expression $0$ so ===> $(a+b)$ is a factor \n$a=b\\omega$ or $a=b\\omega^2$ makes the expression $0$ so ===> $(a-b\\omega)(a-b\\omega^2) = (a^2+ab+b^2)$ is a factor \n[hide]Here $\\omega$ is cuberoot of unity and notice $1+\\omega+\\omega^2 = 0$\nYou notice that the whole expression is $b^7((1+\\omega)^7 - 1 - \\omega^7)$ comes out to be zero because $1+\\omega = -\\omega^2$ \nand you get $-\\omega^{14} - \\omega^7 -1 = -\\omega^2-\\omega -1 = 0$[/hide]\nYou can see that its double root.. too so last expression should be a saure term. \n\n[hide](Easiest to see this requires calculus , the derivative would be zero too when $a=b\\omega$ etc..[/hide]\n\n[hide](If you don't see its a double roots right away, you can also find the remaing term as it will only of a second degree ... etc..) [/hide]\n\nSo you use $(a+b)^7-a^7-b^7 = k a b (a+b) (a^2+ab+b^2)^2$ \nSince both sides are now of equal degrees , you can find the value of $k$ by using, say $a=b=1$ which will give $k=7$ [/hide]\n\nHope this helps.[/quote]\r\n\r\nThat's a better explaination than i can give. :D It is a handy technique.\r\n\r\nThough gyan,wouldn't $a^2$ and $b^2$ have some coefficients so we write it as $Aa^2+ab+Bb^2$ and eventually A and B come out to be 1?" } { "Tag": [ "trigonometry", "inequalities", "inequalities unsolved" ], "Problem": "For $a,b,c,d$ positive reals prove the ineq\r\n\r\n$\\sqrt{(a^2-ab+b^2)(c^2-cd+d^2)}+\\sqrt{(a^2+ad+d^2)(c^2+cb+b^2)} \\geq(a+c)(b+d) $", "Solution_1": "There is something wrong with your inequality. a=b=c=d gives a contradiction.\r\nPlease correct your fault because I have a rather nice idea what to do with the left side and I would like to post it here.\r\n\r\n\r\nMisha", "Solution_2": "Sorry, I will correct it just now.", "Solution_3": "[quote=\"manlio\"]For $a,b,c,d$ positive reals prove the ineq\n\n$\\sqrt{(a^2-ab+b^2)(c^2-cd+d^2)}+\\sqrt{(a^2+ad+d^2)(c^2+cb+b^2)} \\geq(a+c)(b+d) $[/quote]\r\n\r\nThat's t$\\infty$ c$\\infty$l!!\r\n\r\nLet ABP be a triangle with AP = a, BP = b and < APB = 60. Let C be a point on the extension of the side AP beyound the point P such that CP = c, and let D be a point on the extension of the side BP beyound the point P such that DP = d. Then, < BPC = 120, < CPD = 60 and < DPA = 120.\r\n\r\nIn the triangle APB, the Cosine Law yields\r\n\r\n$AB=\\sqrt{a^{2}+b^{2}-2ab\\cos \\measuredangle APB}=\\sqrt{a^{2}+b^{2}-2ab\\cos 60^{\\circ }}$\r\n$=\\sqrt{a^{2}+b^{2}-2ab\\cdot \\frac{1}{2}}=\\sqrt{a^{2}-ab+b^{2}}$.\r\n\r\nIn the triangle BPC, the Cosine Law yields\r\n\r\n$BC=\\sqrt{b^{2}+c^{2}-2bc\\cos \\measuredangle BPC}=\\sqrt{b^{2}+c^{2}-2bc\\cos 120^{\\circ }}$\r\n$=\\sqrt{b^{2}+c^{2}-2bc\\cdot \\left( -\\frac{1}{2}\\right) }=\\sqrt{b^{2}+bc+c^{2}}$.\r\n\r\nSimilarly,\r\n\r\n$CD=\\sqrt{c^{2}-cd+d^{2}}$\r\n\r\nand\r\n\r\n$DA=\\sqrt{d^{2}+da+a^{2}}$.\r\n\r\nAlso, AC = AP + CP = a + c and BD = BP + DP = b + d.\r\n\r\nBy the Ptolemy inequality, applied to the quadrilateral ABCD, we have\r\n\r\n$AB\\cdot CD+BC\\cdot DA\\geq AC\\cdot BD$;\r\n\r\nthus,\r\n\r\n$\\sqrt{a^{2}-ab+b^{2}}\\cdot \\sqrt{c^{2}-cd+d^{2}}+\\sqrt{b^{2}+bc+c^{2}}\\cdot \\sqrt{d^{2}+da+a^{2}}\\geq \\left( a+c\\right) \\cdot \\left( b+d\\right) $,\r\n\r\ncompleting the proof.\r\n\r\nSorry, Misha, for posting this - I guess your idea was the same...\r\n\r\n Darij", "Solution_4": "It's exactly what I wanted to do! I love inequalities with geometric interpretations.\r\n\r\nMisha" } { "Tag": [ "calculus", "integration", "abstract algebra", "function", "Ring Theory", "algebra", "domain" ], "Problem": "Hi!\r\n\r\nI really don't quite see, why the statement below is correct, or is it just a \"typo\" ?.\r\nLet X be an integral noetherian scheme. Also, suppose X is Dedekind, that is, an irreducible, locally noetherian, normal scheme of dimension 1 or 0.\r\n\r\n[b]Then for every open subset of X, its corresponding ring of functions is a Dedekind ring.[/b]\r\n\r\nThis is clear for affine open subsets, but what about more general open sets?. Any thoughts?. Anyone out there who could prove this or give a counterexample?.\r\n\r\nNote: The author of the book allows Dedekind rings to be 1 or 0-dimensional.", "Solution_1": "If X is 0-dimensional, then X is a singleton and the claim is trivial. Now let X be 1-dimensional. Note that for every open subset $ U$ it holds $ \\Gamma(U,\\mathcal O_X)\\equal{}\\cap_{x\\in U}\\mathcal O_{X,x}$ where the intersection takes place in the function field of $ X$, the stalk of $ \\mathcal O_X$ at the generic point of $ X$. Then use the fact that the intersection of normal subrings of a field $ K$ is normal again (w.r.t. $ K$).", "Solution_2": "[quote=\"Third Edition\"]If X is 0-dimensional, then X is a singleton and the claim is trivial. Now let X be 1-dimensional. Note that for every open subset $ U$ it holds $ \\Gamma(U,\\mathcal O_X) \\equal{} \\cap_{x\\in U}\\mathcal O_{X,x}$ where the intersection takes place in the function field of $ X$, the stalk of $ \\mathcal O_X$ at the generic point of $ X$. Then use the fact that the intersection of normal subrings of a field $ K$ is normal again (w.r.t. $ K$).[/quote]\r\n\r\nWell, yes, i understand from the argument in the textbook that $ \\Gamma(U,\\mathcal O_X)$ is an integral domain, and integrally closed and of dimension 0 or 1. [b]Why is it a Noetherian ring?.[/b]\r\n\r\nOf course, $ U$ has a finite open affine cover by spectra of Noetherian rings, but then what?.\r\n\r\nI must confess i haven't thought too much about this, and it may well be quite trivial :oops: .\r\n\r\nWould you enlighten me?. :)\r\n\r\nThanks anyway!.", "Solution_3": "Are you working over a base field $ k$? If so then take a projective normal model $ \\bar{X}$ of (the function field) 0f $ X$. Note that every (non-void) subset $ U$ of $ X$ is the complement of finitely many points of $ \\bar{X}$ which implies that $ U$ is affine." } { "Tag": [ "inequalities", "trigonometry", "symmetry", "triangle inequality", "trig identities", "Law of Cosines", "algebra unsolved" ], "Problem": "let $a$ and $b$ be two positive reel number \r\nsolve in the set of reel number the equoition \r\n$\\sqrt{x^{2}+a^{2}-ax\\sqrt{3}}+\\sqrt{y^{2}+b^{2}-by\\sqrt{3}}+\\sqrt{x^{2}+y^{2}-xy\\sqrt{3}}=\\sqrt{a^{2}+b^{2}}$\r\n$x$ et $y$ sont les nombres inconu \r\ngood luck :roll:", "Solution_1": "Hint:\r\nUse the inequality to estimate left hand:\r\n$\\sqrt{a^{2}+b^{2}}+\\sqrt{c^{2}+d^{2}}+\\sqrt{e^{2}+f^{2}}\\geq\\sqrt{(a+c+e)^{2}+(b+d+f)^{2}}$", "Solution_2": "$x^{2}+a^{2}-\\sqrt{3}ax=x^{2}+a^{2}-2ax\\cos 30^\\circ.$", "Solution_3": "Kunny's hint gives it away.\r\nConsider a right triangle with leg lengths $b$ and $a$ with $a$ in the horizontal direction and $b$ vertical. Construct the lines which make an angle of $30^{\\circ}$ with the $x$ and $y$ (or $a$ and $b$) axes, respectively. (In a clockwise order going around the origin, one encounters first line $a$ then two lines then line $b$)\r\nPlace points $X$, $Y$, at a distance $x$,$y$ from the origin on the first and second lines (going counterclockwise) respectively, all in the first quadrant. We claim that the given problem is equivalent to being given in the diagram that $XA+XY+YB=AB$; this is clear by considering the law of cosines condition since $XA^{2}=a^{2}+x^{2}-2ax\\cos 30^{\\circ}$. Thus, by the triangle inequality, the points $X,Y$ lie on $AB$.\r\n\r\nTHe rest is basic stuff... we realize setup similar right triangles and painlessly find that $a=m(a+b)$ for some scale factor $m$ then $x=2mb$ so $x=\\frac{2ab}{a+b}$ and by symmetry, the same goes for $y$, which is interesting since these two both are the harmonic mean of $a$ and $b$.\r\n\r\nA really nice but basic problem (well at least after kunny's tip)", "Solution_4": "[quote=\"me@home\"]so $x=\\frac{2ab}{a+b}$ and by symmetry, the same goes for $y$[/quote]\r\nThe correct solution is $(x,y)=(\\frac{2ab}{a+\\sqrt{3}b},\\frac{2ab}{\\sqrt{3}a+b})$.\r\nIf you look at the figure you see $a>b\\Rightarrow x>y$.\r\n\r\n\r\nTake $L$ the perpendicular point of $X$ on $OA$ ($O$ be the origin with $\\overline{OA}=a,\\overline{OB}=b,\\overline{OX}=x,\\overline{OY}=y$), \r\nthen $BOXLA$ is an intercept theorem figure with ${\\overline{XL}\\over\\overline{BO}}={\\overline{LA}\\over\\overline{OA}}$.\r\n$\\angle XOL=30^\\circ$ gives $\\overline{XL}={x\\over 2}$ and $\\overline{OL}={\\sqrt{3}x\\over 2}$.\r\nSo ${{x\\over 2}\\over b}={a-{\\sqrt{3}x\\over 2}\\over a}$ and $x=\\frac{2ab}{a+\\sqrt{3}b}$, and symmetric for $y$." } { "Tag": [ "function", "algebra", "domain", "algebra unsolved" ], "Problem": "Find $ f: R\\minus{}>R\\plus{}$ such that:\r\n$ f(x)^2\\equal{}1\\plus{}xf(x\\plus{}1)$", "Solution_1": "Hmm...if I got it correctly,the problems you proposed is:\r\nfind all $ f: \\mathbb{R}\\rightarrow\\mathbb{R}^{ \\plus{} }$,such that...\r\nsubstitution $ x \\equal{} 0$ implies that $ f(0) \\equal{} 1$, and $ x \\equal{} \\minus{} 1$ implies that $ f( \\minus{} 1)^2 \\equal{} 0$,therefore $ f( \\minus{} 1) \\equal{} 0$,contradiction.\r\n[i]Added after reading Allnames reply below:[/i]\r\nWell...then show us the full version.\r\nI suspect you misunderstood what I meant.In fact,I gave a correct solution for the given problem,I had emphasized the condition before I posted a solution.I guess misconceptions appeared due to incompleteness of the statement,furthermore,I am starting to doubt about the domain of $ f$,whereas image seems to be correct.I would rather think of a case,where domain is all negative integers.", "Solution_2": "It is not enough to solve it Erken\r\nAlthough it is from a book ,it is not full version of this problem,I have seen it in a notebook but with other condition (I remember $ f$ is continious function)\r\nAt all I think we cant kill it without any condition :wink:", "Solution_3": "It is enough to solve,don't have any other condition all names!" } { "Tag": [ "limit", "number theory open", "number theory" ], "Problem": "This is related to one of the problems on the forum, which has been unsolved for a long time, but forget about that. Here's what I was wondering:\r\n\r\nGiven $r$ irrational, is it true that $\\lim\\inf_{n\\rightarrow \\infty}2^n\\{nr\\}=\\infty$?\r\n\r\nIn other words: given $k>0$, is it true that there are only finitely many $n$ s.t. $2^n\\{nr\\}1$ we have $ f^'(x)=\\frac{1}{(x+1)\\ln(x)}-\\frac{\\ln(x+1)}{x\\ln(x)^2}<0 \\Leftrightarrow x\\ln(x)<(x+1)\\ln(x+1)$ because $ h(x)=x\\ln(x)$ ist for $ x>1$ strictly monotonously increasing, further we have $ \\lim_{x\\to 1+0}f(x)=+\\infty$ and $ \\lim_{x\\to \\infty}f(x)=1$, so our equation has no real solution.\r\nSonnhard.", "Solution_2": "[color=darkblue]Yes! Thank your post. :lol: \n\nThis problem can be done this way:\n\n+) If $ 0 < x < 1 \\Longrightarrow \\log_{x}{(x \\plus{} 1)} < \\log_{x}{1} \\equal{} 0$, but $ \\log_{2008}{\\frac {3}{2}} > \\log_{2008}{1} \\equal{} 0 \\Longrightarrow$ equation have no root.\n\n+) If $ x > 1 \\Longrightarrow x \\plus{} 1 > x$ $ \\Longrightarrow \\log_{x}{(x \\plus{} 1)} > \\log_{x}{x} \\equal{} 1$ $ \\equal{} \\log_{2008}{2008} > \\log_{2008}{\\frac {3}{2}} \\Longrightarrow$. equation have no root.\n\nHence equation have no root.\n\nThanhnam2902! :P [/color]" } { "Tag": [ "geometry", "rectangle", "conics", "ellipse" ], "Problem": "Bob and Ed are standing inside a simple figure at $(-3,0)$ and $(3,0)$ respectively. Whenever Bob throws a ball at the walls of the figure, it bounces back and hits Ed (assume that gravity, air resistence, etc. are not in play), traveling a distance of 10 units. Find the area enclosed by this figure.", "Solution_1": "Umm.. isn't there infinite solutions? As long as he keeps throwing it at (0,x) you can make as big as a rectangle you want.", "Solution_2": "he's assuming that you know the rule of the foci for an ellipse. now he's saying find the area of the elipse with foci at those points, wich is trivial...", "Solution_3": "Darn you.\r\n\r\nThis is an interesting property of ellipses, which is that bouncing a \"ball\" off the sides from one focus will just hit the other focus, and the distance traveled will be constant. However I don't think I gave enough information in the original problem. edited now.", "Solution_4": "[hide]The walls are in the shape of an elipse with the foci at $(-3,0)$ and $(3,0)$. The center is at $(0,0)$\nSince the ball travels $10$ units, the length of the major axis is $10$. In an elipse, the square of major axis equals the sum of the squares of the minor axis and the distance between the foci. Using that fact, the length of the minor axis is $\\sqrt{10^{2}-6^{2}}=8$. \nTherefore, the area of the elipse is $\\pi\\cdot\\frac{10}{2}\\cdot\\frac{8}{2}=\\boxed{20\\pi}$[/hide]", "Solution_5": "very intresting problem.... did you make it?", "Solution_6": "It's a pretty classic problem. I first saw it on the HMMT." } { "Tag": [ "inequalities", "rotation", "triangle inequality", "superior algebra", "superior algebra unsolved" ], "Problem": "Let x,y,z,t belongs to R^n where d(x,y)=||x-y||.\r\n\r\nShow that(Ptolemy's inequality):\r\nd(x,y)d(z,t)<=d(x,z)d(y,t)+d(x,t)d(y,z)Thanks anyone in advance.", "Solution_1": "Not the best possible topic match:\r\n\r\n1. For $ n\\equal{}2$, you have the classical inequality.\r\n2. For $ n\\equal{}3$, you have a tetrahedron. Take one side. It bounds 2 triangles. Rotate one of them until it lies in the same plane as the other one and such that the common side is a diagonal in the newly formed quadrilateral. Draw the other diagonal, consider the intersection point and join it to the other vertex that is not already joined to. Now apply the inequality in case $ n\\equal{}2$ and the triangle inequality in some triangle with a vertex in the intersection of the diagonals.\r\n3. 4 points span a 3-dimensional space, so case $ n>3$ reduces to $ n\\equal{}3$." } { "Tag": [ "inequalities", "logarithms", "Euler" ], "Problem": "Prove or disprove:\r\n$ 1\\plus{}\\frac{1}{2}\\plus{}\\frac{1}{3}\\plus{}...\\plus{}\\frac{1}{n}<4$", "Solution_1": "Depends on n doesnt it. If n tends to inf , sum is also inf.", "Solution_2": "$ \\sum_{k\\equal{}1}^{\\infty} \\frac{1}{k}\\equal{}\\infty$", "Solution_3": "Can you tell me for which $ n$ the expression is $ \\ge 4$?", "Solution_4": "We can use $ \\log n \\approx 4\\minus{} \\gamma$, where $ \\gamma$ is the [url=http://en.wikipedia.org/wiki/Euler-Mascheroni_constant]Euler-Mascheroni constant[/url]. Then we get $ n \\approx e^{4 \\minus{} \\gamma} \\approx 30.65 \\ldots$\r\n\r\n$ n \\equal{} 30 \\implies \\sum \\equal{} \\frac {9304682830147}{2329089562800} < 4$\r\n\r\n$ \\boxed{n \\equal{} 31} \\implies \\sum \\equal{} \\frac {290774257297357}{72201776446800} > 4$" } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "B is a point on the circle (S)S. A is a point (distinct from B) on the tangent to (S) at B. C is a point not on (S) such that the line segment AC meets the circle (S) at two distinct points. (S') is a circle which touches AC at C and (S) at D, where B and D are on opposite sides of the line AC. Show that the circumcenter of (BCD) lies on the circumcircle (ABC).", "Solution_1": "Denote s,s' are center of circle S and S'\r\nLet SB S'C meet at G, then G A C B are on one circle,\r\n\u2220COB=\u2220CS'O+\u2220OSB,\u2220S'GS+\u2220CS'O+\u2220OSB=\u2220CAB+\u2220CS'O+\u2220OSB=\u2220CAB+\u2220COB=180\r\nSo OCAB lie on one circle,where O is circumcenter of triangle BCD" } { "Tag": [ "function", "quadratics" ], "Problem": "I'm sure everyone knows of the dollar question which is\r\n\r\n[u]How many ways can you make a dollar with change?[/u]\r\n\r\nSo the way to solve it using generating functions, and finding the coefficient of $x^{100}$. \r\n\r\n[b]How do you find the coefficient of $x^{100}$ without multiplying it all out?[/b]\r\n\r\nThis link [url]http://www.cut-the-knot.org/ctk/GeneratingFunctions.shtml#Graham[/url] shows it but it does not really explain it (or I'm missing something). \r\n\r\nPlease do tell.", "Solution_1": "[quote=\"SM4RT\"]I'm sure everyone knows of the dollar question which is\n\n[u]How many ways can you make a dollar with change?[/u]\n\nSo the way to solve it using generating functions, and finding the coefficient of $x^{100}$. \n\n[b]How do you find the coefficient of $x^{100}$ without multiplying it all out?[/b]\n\nThis link [url]http://www.cut-the-knot.org/ctk/GeneratingFunctions.shtml#Graham[/url] shows it but it does not really explain it (or I'm missing something). \n\nPlease do tell.[/quote]\r\n\r\numm unless im not mistaken the coefficiant of $x^{100}$ is $1$? unless im missing something which im sure i am :D", "Solution_2": "He means the coefficient of $x^{100}$ in the generating function of\r\n\r\n$P(x) = \\frac{1}{(1-x)(1-x^{5})(1-x^{10})(1-x^{25})(1-x^{50})}$", "Solution_3": "Yes but how would you get it, like they did\r\n\r\n$\\binom{6}{4}+45\\binom{5}{4}+52\\binom{4}{4}+\\binom{3}{4}=15+225+52+0=\\boxed{292}$ \r\n\r\nHow do they derive that?", "Solution_4": ":maybe: Anyone?!?!?", "Solution_5": "Yes, I'm also curios about where the 4's in the binomial coefficients come from", "Solution_6": "I've wondered this too. If someone knows, it could help a lot in future problems.", "Solution_7": "They give a reference to pages 344-346 of Concrete Mathematics. I had a quick look through, it looks like they set up a recursive equation which turns out to have a quadratic solution. C[z] = in-terms-of-z. Then they used binomial expansion to expand the $\\sqrt{..z..}$ part and it goes on from there.\r\n\r\nSo yeah, do have a look through the book, it's a very good one :)", "Solution_8": "Sorry to bring up an old topic, but I'm curious about this. Could you perhaps elaborate some on what he did?", "Solution_9": "You could multiply the denominator out and then use the extended Binomial Theorem." } { "Tag": [ "MATHCOUNTS", "email", "Duke", "college" ], "Problem": "It was n problem of the week :roll:\r\n\r\nBTW where did you get these states?", "Solution_1": "I got them from my coach who got them each year at competition.", "Solution_2": "Do you think you could scan me a copy of a couple? XD", "Solution_3": "If you give me them, ill give you 5 years of chapters, 7 years of states and 9 years of nationals :hopeful:", "Solution_4": "[quote=\"biffanddoc\"]If you give me them, ill give you 5 years of chapters, 7 years of states and 9 years of nationals :hopeful:[/quote]\r\n\r\nno need to give me any because I am not gonna compete next year. My coach has all the competitions from 1995 (i think) to 2005. I'll get them from her and scan them. I am just gonna do states right now. I'll just post them here when i get them into my computer.", "Solution_5": "Youre my hero :)", "Solution_6": "And mine as well.", "Solution_7": "Klebian, you should probably talk to mr. boyd about getting other states. Or request them on the coaches forum of mathcounts.org", "Solution_8": "Hey I want them too.", "Solution_9": "[quote=\"biffanddoc\"]Klebian, you should probably talk to mr. boyd about getting other states. Or request them on the coaches forum of mathcounts.org[/quote]\r\nI have no real way of contacting Mr. Boyd other than through AoPS, which probably wouldn't work, or by asking a student of his. As of now I have asked a student, but he's never on on AIM. XD", "Solution_10": "Please tell me which years' competitions you want. I'll do them them weekend. :)", "Solution_11": "For states,\r\npre-1998\r\nFor nats,\r\npre-1990,1996,1997,2001,2002,2004", "Solution_12": "I'll PM you 01, 02 nationals.\r\nCan you get me 1999 states Biff?\r\nAnd the rest of the states/nats would apply to me also. :)", "Solution_13": "Sure. I FOUND IT LAST NIGHT :wow: :wow:\r\n\r\nI had taken it to NY the week before and it was in my suitcase (and I made sure that I didn't take along any compeition sets :roll:)", "Solution_14": "I have 1989 -1991 Nationals and States, but sadly, I have no scanner so I can't really bargain for tests :'(", "Solution_15": "Sorry for the double post, but can I get a copy of all those State Tests as well?", "Solution_16": "oooh more mathcounts tests? I have only done back to 2001 on the state level. my coach has more but she's out of town and state for me is on saturday, so i'd love to have a copy!!! and national level tests? I've never had a copy! if that's okay with you guys thanks. \r\nsorry but I probably cannot contribute any unique competitions myself.", "Solution_17": "can I get a copy of those tests too? sorry, i can't contribute any competitions either.", "Solution_18": "Wow! i didnt know that so many people wanted these tests. sorry to disappoint you all, but my scanner just broke, so i can't really post them here. sorry again. :(", "Solution_19": "[quote=\"piano1321\"]Wow! i didnt know that so many people wanted these tests. sorry to disappoint you all, but my scanner just broke, so i can't really post them here. sorry again. :([/quote]\r\n\r\nMust remember to buy you a scanner then.\r\nOh please please let me make nats. ;)", "Solution_20": "Can people send me everything they have pre 1999?\r\nMy email is mcpreet@gmail.com.\r\nIf anyone wants I can give them a gmail invite.", "Solution_21": "I would post everything I have here, but it's against AoPS policy. Last time I did it, it got deleted immediately. So here's what I'm going to do: Make a website, and post everything on it. Besides MathCounts, I also have a GIG of other math competitions' PDF's, most unavailable online right now.\r\n\r\nBelow, is a list of everything I need. Please email anything on that list to mysmartmouth@gmail.com.\r\nIf you need anything that I have, email me and I will be glad to send you anything you need for free.\r\n\r\n\r\n[u]I Need:[/u]\r\nAll Nationals (I have 2003, but low quality)\r\nPre-2001 Chapter & State\r\nPre-2001 Handbooks\r\n[u]Also...[/u]\r\n2006 Duke Math meet (I have a low quality copy)\r\nALL AMC 8's\r\nALL AMC 10's (except for 2000 & 2001)", "Solution_22": "Check out my signature." } { "Tag": [ "real analysis", "function", "limit", "real analysis unsolved" ], "Problem": "Suppose $ f$ is Lebesgue measurable on $ (0,1)$ and not essentially bounded. Is it true that for every positive function $ g$ on $ (0,\\infty)$ such that $ \\lim_{p\\to\\infty}g(p)\\equal{}\\infty$ one can find an $ f$ such that $ \\lim_{p\\to\\infty}\\|f\\|_p\\equal{}\\infty$ but $ \\|f\\|_p\\leq g(p)$ for all sufficiently large $ p$?", "Solution_1": "Yes. The key point is that you can increase $ \\|f\\|_n$ without changing $ \\|f\\|_k$, $ k\\equal{}1,\\dots,n\\minus{}1$ by much, by adding a very thin and tall peak to $ f$. This can be repeated indefinitely.", "Solution_2": "Yes, it works, thanks. Any ideas about the convexity for Lebesgue measure in the other problem?", "Solution_3": "[quote=\"mlok\"]Yes. The key point is that you can increase $ \\|f\\|_n$ without changing $ \\|f\\|_k$, $ k \\equal{} 1,\\dots,n \\minus{} 1$ by much, by adding a very thin and tall peak to $ f$. This can be repeated indefinitely.[/quote]\r\n\r\nI don't know how.Can somebody tell me how to use this \"key point\" to consruct $ f$?Thank you in advance." } { "Tag": [], "Problem": "Its been so long since I posted here. :) \r\nDetermine all five-digit numbers with the property that, when the first digit and the last digit are both deleted, the resulting number is $\\frac{2}{215}$ of the original number.", "Solution_1": "[hide=\"Solution\"]If the first digit is $a$, the last $b$ and the middle part $x$, then\n\n$10000a+b+10x={215\\over 2}x\\iff 10000a+b={195\\over 2}x\\iff x=\\frac{2(10000a+b)}{3\\cdot 5\\cdot 13}$\n\nHence $10000a+b$ is divisible by $3$ and $5$, giving $30000,60000,90000,10005,40005,70005$ as candidates. Checking shows that only $70005$ is divisible by $13$, hence $x={2\\cdot 70005\\over 195}=718$. The only such number is $77185$.[/hide]" } { "Tag": [ "trigonometry", "AMC", "USA(J)MO", "USAMO" ], "Problem": "Someone broke a calculator ( :oops_sign: ) so that the only keys functioning are the $\\sin(x)$, $\\cos(x)$, $\\tan(x)$, $\\sin^{-1}(x)$, $\\cos^{-1}(x)$, and $\\tan^{-1}(x)$. Given any rational number $q$, show that you can hit a finite number of buttons so that you get the number $q$ on the panel. Assume the calculator is EXACT no matter what, everything is in radians, and the initial display is $0$. \r\n\r\n\r\nExample. \r\n$q=1$\r\n$\\cos^{-1}(0) = 1$", "Solution_1": "This is a USAMO-problem and has been posted and solved[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=136808#p136808]here[/url].", "Solution_2": "Thank you very much. :P", "Solution_3": "Since it's been already discussed, it's been locked. :)" } { "Tag": [ "geometry", "MATHCOUNTS", "percent", "AMC 8", "number theory", "AMC 10" ], "Problem": "What do you think you'll get?\r\n\r\nI'll probably get a 23,24, or 25. \r\n\r\n...and why is everyone saying they have to practice so much?, I mean, they're REALLY easy!\r\n-jorian", "Solution_1": "25 PLZZZ! :D \r\n\r\nI certainly hope so. I haven't prepared at all, but it seems really easy. I have been practicing geometry since mid-october till now, devoted some time to geometry each day. And I have asked lots of people, and on AOL many problems from geometry. so yeah... \r\n\r\ni duno. i want a 25 really bad , and I think it is really do-able\r\n\r\nI gota watch my silly mistakes, 1st time taking it", "Solution_2": "Its really not that hard to get a perfect...its just that some kids mess up on the last few\r\n :( \r\n-jorian", "Solution_3": "I know. I am in geometry, so I will be utterly disappointed with anything but a 25. \r\n\r\nIt's my 1st and only time taking it, so I won't ever get another chance. Unless I fail 8th grade, which I don't plan on doing. :lol: \r\n\r\nThe only thing I am worried about is if luck will be on my side. Time on my side, questions on my side, luck... \r\n\r\n\r\nYeah... I don't feel the pressure, but this score.. will define me for the rest of my life... well... at least this year. till vestavia that is", "Solution_4": "Yeah, points off usually come from stupid mistakes or trick questions. You have plenty of time, so you really just want to take the test slowly and carefully...", "Solution_5": "its gonna be fun this year gazing around after i finish while others are doing problems like mad dogs :maybe: \r\n-jorian :ninja:", "Solution_6": "so usually there is nothing that is not easy on amc 8? just trick questions and careless mistakes that usually prevent people from getting a 25?\r\n\r\nI hope I won't make any mistakes or get tricked. *crosses fingers*", "Solution_7": "as long as you know how/when to apply methods like pyth., ull be fine\r\n-jorian", "Solution_8": "this website is cool. I found it today, and I will work on it as much as possible tomorrow when I get home from school to get ready for the day ahead. http://www.unl.edu/amc/mathclub/04,1-problems.html", "Solution_9": "Ya. when....\r\n\r\n[hide]\nLoki, Moe, Nick and Ott are good friends. Ott had\nno money, but the others did. Moe gave Ott one-\u00affth\nof his money, Loki gave Ott one-fourth of his money\nand Nick gave Ott one-third of his money. Each gave\nOtt the same amount of money. What fractional part\nof the group's money does Ott now have?\n(A) 1/10 (B) 1/4 (C) 1/3 (D) 2/5 (E) 1/2\n[/hide]\r\n\r\nis the hardest problem on the test (last question, '02), u know its easy!\r\n-jorian", "Solution_10": "haha i thought the hardest one was 2004- *25. It was worded totally confusing.", "Solution_11": "yes, and the image was amazingly discrete!\r\n-jorian", "Solution_12": "[quote=\"jhredsox\"]Ya. when....\n\n[hide]\nLoki, Moe, Nick and Ott are good friends. Ott had\nno money, but the others did. Moe gave Ott one-\u00affth\nof his money, Loki gave Ott one-fourth of his money\nand Nick gave Ott one-third of his money. Each gave\nOtt the same amount of money. What fractional part\nof the group's money does Ott now have?\n(A) 1/10 (B) 1/4 (C) 1/3 (D) 2/5 (E) 1/2\n[/hide]\n\nis the hardest problem on the test (last question, '02), u know its easy!\n-jorian[/quote]\r\nNumber 25 isn't always the hardest.", "Solution_13": "i know, i was just trying to make a point :blush: \r\n\r\n-jorian\r\n\r\n@bpms: when do u go to school if ur on at this time?", "Solution_14": "I'm pretty sure I will ace the AMC 8. On the practice tests, I got 25, 25, 25, 23, 19. On the 23 and 19, I screwed up on reading. I hope I get 25!", "Solution_15": "He may not be in your time zone.", "Solution_16": "[quote=\"jhredsox\"]i know, i was just trying to make a point :blush: \n\n-jorian\n\n@bpms: when do u go to school if ur on at this time?[/quote]\r\n\r\nIt depends when you take it. I take around 9 in my math class.", "Solution_17": "Cool. Thanks for the site. Anyway, I won't ace the AMC 8. :(", "Solution_18": "i have a chance. I hope luck is on my side. Because I get 24's on practice very frequently :(", "Solution_19": "So what? Practice isn't what you normally get on the real test. Unfortunately... :(", "Solution_20": "You really don't need to get worried about getting a 25 on AMC 8. Shoot for higher goals, like a good score on AMC 10. AMC 8 is definitely nothing to worry about.", "Solution_21": "The practice tests basically shows what score you get -3. Not all the time but most of the time.", "Solution_22": "Its more like -4. :wink:", "Solution_23": "is that really something reasonable anirudh?\r\n\r\ni dont think one or two points makes a difference", "Solution_24": "wait so if you get like a 5 on a practice test, you will get a 2, or an 8 on next test approximately?\r\n\r\nive never heard of that be4", "Solution_25": "i think what they're pointing out is not to trust your practice tests :D \r\n\r\nand no, u guys are being way too optimistic considering 6 of u guys ALREADY think ur getting a perfect score\r\n\r\nwell guess what, at best, 1 of u will :wink: \r\n-jorian", "Solution_26": "I recently did two AMC 8's, and on both I got 25. Hopefully, this time I'll get the same. :)", "Solution_27": "I do better under the pressure...", "Solution_28": "how is that possible\r\n\r\nis it because you don't have the motivation with no time pressure?\r\n-jorian", "Solution_29": "[quote=\"jhredsox\"]how is that possible\n\nis it because you don't have the motivation with no time pressure?\n-jorian[/quote]\r\nEveryone is different.", "Solution_30": "[quote=\"jhredsox\"]yes, and the image was amazingly discrete!\n-jorian[/quote]\r\n\r\nDo you know what discrete means?", "Solution_31": "discrete- kinda apart\r\n\r\nanyway I really wish I can get a 25. and i work well under pressure also", "Solution_32": "[quote=\"now a ranger\"]discrete- kinda apart\n\nanyway I really wish I can get a 25. and i work well under pressure also[/quote]\r\n\r\npfft I don't\r\n\r\nNational practice scores last year (approximate)\r\n\r\n40\r\n40\r\n40\r\n38\r\n46\r\n40\r\n40\r\n35\r\n31 (2005)\r\n40\r\n40\r\n40\r\n42\r\n39\r\n\r\nNational Score:25", "Solution_33": "25, but i won't be shocked or upset if I don't. I know that if I miss something, it'll be due to reading or some small mistake. I *could* spend a great deal of time checking it and whatnot, but why bother? It doesn't mean anything.", "Solution_34": "to me it sorta does , but then again it doesnt , cuz i havent studied, if i thot i twas important i would have studied a long time ago. Instead I spend time on geometry and other math stuff and number theory and aops", "Solution_35": "you should focus more on amc 10, which is more of a mathcounts level. AMC 8 doesn't really require studying for. It will just naturally come to you as you study for other competitions.", "Solution_36": "25 I hope. More likely towards 24 but with a stroke of luck, the 1/4 percent or the 4 leaf clover or magic, or god, or other extraneous things which you probably don't care about so I will continue on, I CAN DO IT! \r\n\"Emphasize added\"\r\n :D", "Solution_37": "http://www.unl.edu/amc/mathclub/04,1-problems.html\r\n\r\nis very nice\r\n\r\nnow a ranger\r\nthank you", "Solution_38": "Ug. I feel sick right now. This my first competition while sick. This sucks so bad. I'll probably know how to do all the problems, or maybe all but 1 in case something special occurs, but I might have a ton of errors.", "Solution_39": "Ikh, I make so many stupid mistakes on AMC's. On math club practices, I ended up finishing in 12 minutes and I would make 1 or 2 mistakes. If I tried taking about the whole time, I would just end up making the same mistake when checking.", "Solution_40": "[quote=\"bpms\"]Ug. I feel sick right now. This my first competition while sick. This sucks so bad. I'll probably know how to do all the problems, or maybe all but 1 in case something special occurs, but I might have a ton of errors.[/quote]\r\n\r\nBeing sick is an advantage. I was sick at chapters and got a 44 last year, without any practices. Being sick makes you delusionally paranoid about how many you got right, so you will check and check and check without getting bored.", "Solution_41": "[quote=\"13375P34K43V312\"][quote=\"bpms\"]Ug. I feel sick right now. This my first competition while sick. This sucks so bad. I'll probably know how to do all the problems, or maybe all but 1 in case something special occurs, but I might have a ton of errors.[/quote]\n\nBeing sick is an advantage. I was sick at chapters and got a 44 last year, without any practices. Being sick makes you delusionally paranoid about how many you got right, so you will check and check and check without getting bored.[/quote]\r\nMAybe for you, but focusing is impossible right now. Also, um a 44 would be bad for me at chapters, just because I set high goals for myself.", "Solution_42": "[quote=\"13375P34K43V312\"][quote=\"bpms\"]Ug. I feel sick right now. This my first competition while sick. This sucks so bad. I'll probably know how to do all the problems, or maybe all but 1 in case something special occurs, but I might have a ton of errors.[/quote]\n\nBeing sick is an advantage. I was sick at chapters and got a 44 last year, without any practices. Being sick makes you delusionally paranoid about how many you got right, so you will check and check and check without getting bored.[/quote]\r\n\r\nBeing sick is a big disavantage for me in math but a big advantage for me during sports. :D", "Solution_43": "I was sick at states(it came on right before the Target) so I was still able to do well on sprint. I only got 39, but still was 2nd. I'd like to know how that happened. Just so oyu know, I found it to be a HUGELY bad thing.", "Solution_44": "[color=blue]kyyuanmathcount: took out this post. see the no discussion of AMC 8 thread everyone. [/color]" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Given that $a_1, a_2, ..., a_m$ are nonzero integers.\r\nFor every $k = 0,1,...,m$ ($n -:inf:, then f(x) -> :inf:.", "Solution_3": "yeh i meant a relative maximum, i.e. the highest possible positive value in this case. I know its somewhere b/w f(10.5) and f(11.5), but i would like an exact value (and how to do it).", "Solution_4": "You'd break out the calculus. It'd look a little like this:\r\n\r\nf(x) = (16-x)(x^2)\r\nf(x) = 16x^2 - x^3\r\nf'(x) = 32x - 3x^2 (took the derivative)\r\nFind the 0s of the derivative with the quadratic formula\r\n-32:pm:32 / -6 = 0 or 32/3\r\n\r\nIt's a local max iff the sign of the derivative changes from + to -.\r\n\r\nf'(-0.1) = -3.23\r\nf'(0.1) = 3.17\r\nSo f(0) is a min, not a max\r\n\r\nf'(10) = 20\r\nf'(11) = -11\r\nso f(32/3) is a max.\r\n\r\nOf course, this post is meaningless if you don't know calculus... and you probably wouldn't have asked if you did.", "Solution_5": "Expand the expression to get 16x2 - x3. Take the derivative to get 32x - 3x2. Set this equal to 0 and solve for x. You should get x = 0 or x = 32/3. 0 is the local minimum, and 32/3 the local maximum.\r\n\r\nConfuted beat me to it. ;)", "Solution_6": "If you don't know calculus, you can also use the arithmetic-geometric mean inequality. You said that you wanted to restrict to x :ge: 0. We can also restrict to x :le: 16, since otherwise the expression is negative. Now apply the AM-GM inequality to the three numbers 32 - 2x, x, and x. Their arithmetic mean is 32/3. So, by the AM-GM inequality, (32 - 2x) x2 :le: (32/3)3. Now divide by 2 to get an upper bound on your expression (16 - x) x2. Equality occurs in AM-GM when the numbers are equal. Solving 32 - 2x = x gives x = 32/3 as previously observed.", "Solution_7": "doh.... i dont know calculus, but i do know some of derivatives from my prechal book. So i definitely should have gotten that. Hmm....", "Solution_8": "[quote=\"jelyman\"]doh.... i dont know calculus, but i do know some of derivatives from my prechal book. So i definitely should have gotten that. Hmm....[/quote]\r\nNaw, that's in chapter 8 of my calculus book. Of course, intuitively it makes sense, but it doesn't surprise me that you didn't think to do it that way.", "Solution_9": "actually, this kind of thing is handled in my precalc book.", "Solution_10": "Yeah, my precalculus text has an introduction to calculus at the end." } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "quadratics", "linear algebra unsolved" ], "Problem": "I am looking for a problem of positive definiteness of a 3 by 3 matrix $A=A^{T}$ for the form $Q(x,y,z)=[x,y,z]^{T}Q \\left[\\matrix{x\\cr y\\cr z}\\right]$\r\n\r\non the set of $\\{x,y,z~\\vert~xz=y^{2}\\}$\r\n\r\nAre there any standrad result for this?", "Solution_1": "I guess one possible solution is by Lagrangian Multiplier Method\r\nWe form the Lagrangian \r\n$L(x,\\lambda)=Q(x,y,z)+\\lambda (xz-y^{2})$\r\nand follow the standard procedure :)", "Solution_2": "If we follow the bchui's idea:\r\nWe seek the extrema of $X^{T}AX$ on the compact \r\n$K=\\{X\\mid{X^{T}}UX=0,{X^{T}}X=1\\}$ where \\[U=\\begin{pmatrix}0&0&-1\\\\0&2&0\\\\-1&0&0\\end{pmatrix}\\].\r\n$\\partial{K}$ is empty therefore it's sufficient to use lagrange's formula.\r\nThere exist $\\lambda,\\mu\\in\\mathbb{R}$ such that\r\nfor all $H$: $X^{T}AH+\\lambda{X^{T}}H+\\mu{X^{T}UH}=0$ or \r\n(1): $AX+\\lambda{X}+\\mu{UX}=0$.\r\nThus we have 5 equations (3 in (1) and the 2 equations which define $K$) in 5 unknowns ${x,y,z,\\lambda,\\mu}$.\r\nThe condition for positive definiteness is: if $X$ is a solution of the system then $X^{T}AX$>0. (1) implies that $X^{T}AX=-\\lambda$. Thus for each solution $\\lambda$ must be <0.\r\nRemark: necessarily $det(A+\\lambda{id}+\\mu{U})=0$.\r\nExample: Let \\[A=\\begin{pmatrix}126&102&33\\\\102&-16&-50\\\\33&-50&-124\\end{pmatrix}\\].\r\nBy a Maple calculation we find 2 solutions (max and min if there don't exist any other solution):$\\{\\lambda\\approx-173.9,\\mu=\\approx-29.3\\},\\{\\lambda\\approx134.6,\\mu\\approx62\\}$. \r\nThus $A$ is not >0 over $K$.", "Solution_3": "There is also another approach. We need to check that the expression $ax^{2}+bx^{z}+cz^{2}+(dx+fz)y$ is non-negative for all $x,y,z$ such that $xz=y^{2}$. Since $y$ can be replaced by $-y$, we see that $ax^{2}+bxz+cz^{2}$ must be non-negative (i.e., we must have $a,c\\ge 0$, $b^{2}\\le 4ac$ and the fourth degree expression $(ax^{2}+bxz+cz^{2})^{2}-(dx+fz)^{2}xz$ should be non-negative for all $x,z$ with $xz\\ge 0$. Introducing a new variable $t=x/z$, we see that the latter requirement is equivalent to non-negativity of a forth degree polynomial on $[0,+\\infty)$, which can be checked by Sturm test. The advantage of this approach is that it can be done by hand and doesn't require any floating point computations if all coefficients are integer.", "Solution_4": "[quote=\"loup blanc\"]If we follow the bchui's idea:\nWe seek the extrema of $X^{T}AX$ on the compact \n$K=\\{X\\mid{X^{T}}UX=0,{X^{T}}X=1\\}$ where .......[/quote]\r\n\r\nThat's what I have obtained already. The problem is: \r\nFor a quadratic optimization problem with quadratic constraint, such as \r\n\r\nMinimize $Q(x)=x^{T}Ax$\r\nwith constraint $x^{T}G_{1}x=1,~x^{T}G_{2}x=0$\r\n\r\nDo we have a general solution or any general solution for certain particular cases?\r\n\r\n\r\nNote also that $AX+\\lambda X+\\mu UX=0$ involves a quadratic term. Can we possibly solve such equations explicitly?", "Solution_5": "We seek the extrema of $x^{T}Ax$ when $x$ goes through $K=\\{x\\mid{x^{T}Ux=0}$ and $x^{T}Vx=1\\}$. I want to be sure that $K$ is a non empty compact; so I choose $V$ symmetric >0 and $U$ symmetric $\\not>0$ and $\\not<0$; here I assume that $V=id$.Again we have $Ax+\\lambda{x}+\\mu{Ux}=0$. We solve this 5-5 system with the symbolic calculation of Maple $via$ the Gr\u00f6bner basis.Then the $\\lambda$ are the real roots of a polynomial of degree 6. Let $\\lambda_{1}$ and $\\lambda_{2}$ the min and the inf of these real roots.The request max is $-\\lambda_{1}$ and the request min is $-\\lambda_{2}$.\r\nExample: \\[A=\\begin{pmatrix}-88&31&-16\\\\31&-100&-34\\\\-16&-34&51\\end{pmatrix},U=\\begin{pmatrix}126&102&33\\\\102&-16&-50\\\\33&-50&-124\\end{pmatrix}\\]\r\nThe polynomial in $\\lambda$'s is:2473506906946$\\lambda^{6}$-758144102940511$\\lambda^{5}$+98688563506650912$\\lambda^{4}$-6373227166814110795$\\lambda^{3}$+33017456342608831768$\\lambda^{2}$+14937125574310179954480$\\lambda$-221426066491786238260800.\r\nThe real roots are:$\\approx$ -40.3,15.6,98.3,118.4\r\nThus the request maxima are $\\approx$ 40.3 and -118.4" } { "Tag": [ "function", "calculus", "algebra", "functional equation", "calculus computations" ], "Problem": "If $ f(x+y) = f(x) + f(y) -xy -1 $ $ \\forall x,y \\in \\mathbb{R} $, and $ f(1) = 1 $, find the number of solutions of $ f(x) = n$ where $ n \\in \\mathbb{N} $. Here $x \\in \\mathbb{R} $[hide](Probably this part os wrong -- not sure!!)[/hide]", "Solution_1": "Let $f(x)=g(x)+1-\\frac{x^2}{2}$. Then $g(x+y)=g(x)+g(y)$. The solutions of this functional equation are far too varied to give a definite answer to your question. We can show that $f(x)=1+\\frac x2-\\frac{x^2}2$ for rational $x$, but not necessarily for all $x$.\r\n\r\nIt is possible for $f$ to take a particular value on an uncountable dense set, or to avoid it completely.", "Solution_2": "So the question is wrong?? I though so?? What if $x$ in $f(x)=n $ is an integer???", "Solution_3": "As was said by jmerry, you can show that, for every $q\\in\\mathbb{Q}$ the following holds:\r\n\\[f(q)=1+\\frac{q}{2}-\\frac{q^2}{2}\\]\r\nIf you allow $x$ to vary only among the integers, you are searching $x\\in\\mathbb{Z}$ such that\r\n\\[1+\\frac{x}{2}-\\frac{x^2}{2}\\in\\mathbb{N}\\]\r\nNow, it's easy to see that the only integer solutions are $-1,0,1,2$.\r\n\r\nThe impossibility to give an answer when you let $x$ vary among the reals is related to the non continuous solutions of Cauchy's equation (I think this is how it's called) : $f(x+y)=f(x)+f(y)$.\r\n\r\nBtw, if you ask your $f$ to be continuous, the problem can be solved (in the same way, with the same solution) with $x\\in\\mathbb{R}$.\r\n\r\nEhr, now that I think of it...this isn't properly a Calculus subject, is it?" } { "Tag": [ "LaTeX", "inequalities unsolved", "inequalities" ], "Problem": "Find the minimum of $E= \\frac{xy}{z}+ \\frac{yz}{x}+ \\frac{zx}{y}$, if $x,y,z>0 $ and $x^2+y^2+z^2=1$.\r\n\r\nmade by Kurliandcik L.D.\r\n\r\ncheers! :D :D :cool:", "Solution_1": "I think you mean $x^2+y^2+z^2=1$. So\r\n\\[ \\left(\\frac {xy}z+\\frac{yz}x+ \\frac{xz}y\\right)^2=\\sum \\left (\\frac {xy}z\\right )^2+2(x^2+y^2+z^2). \\] \r\nBut $ \\displaystyle \\sum \\left (\\frac {xy}z\\right )^2\\geq x^2+y^2+z^2.$, so \\[ \\left(\\frac {xy}z+\\frac {yz}x+\\frac {xz}y\\right )^2 \\geq 3(x^2+y^2+z^2)=3. \\] \r\n\\[\\Leftrightarrow \\frac {xy}z+\\frac {yz}x+\\frac {xz}y \\geq \\sqrt 3. \\] \r\nThe equal occurs when $ x=y=z= \\displaystyle \\frac {\\sqrt 3}3.$\r\n\r\n[Admin Edit: Made Latex compliant]", "Solution_2": "Yes I meant x^2+y^2+z^2=1.\r\n\r\nSorry about that! :blush:" } { "Tag": [], "Problem": "A palindrome is a number that is the same forwards and backwards. For example, 212 and 21466412 are palindromes. Consider a palindrome with $n$ digits where $n$ is even. Find a formula, in terms of $n$, of how many $n$-digit palindromes exist.", "Solution_1": "[quote=\"ragnarok23\"]A palindrome is a number that is the same forwards and backwards. For example, 212 and 21466412 are palindromes. Consider a palindrome with $n$ digits where $n$ is even. Find a formula, in terms of $n$, of how many $n$-digit palindromes exist.[/quote]\r\n[hide]For the first and last digits they have to be even. So there are four choices for them since they can't be 0. There are n-2 remaining digits. For even values of n there are because there are ten values for each pair of digits and each pair is coordinated. $(10^{\\frac{n-2}{2}})(4)$. For odds, there are $(10^{(\\frac{n-3}{2})})(40)$ which is equal to $(10^{\\frac{n-1}{2}})(4)$ numbers.\nIf you want a generalization that's too bad :P although there probably is one using the floor function.[/hide]", "Solution_2": "[quote=\"bpms\"][quote=\"ragnarok23\"]A palindrome is a number that is the same forwards and backwards. For example, 212 and 21466412 are palindromes. Consider a palindrome with $n$ digits where $n$ is even. Find a formula, in terms of $n$, of how many $n$-digit palindromes exist.[/quote]\n[hide]For the first and last digits they have to be even. So there are four choices for them since they can't be 0. There are n-2 remaining digits. For even values of n there are because there are ten values for each pair of digits and each pair is coordinated. $(10^{\\frac{n-2}{2}})(4)$. For odds, there are $(10^{(\\frac{n-3}{2})})(40)$ which is equal to $(10^{\\frac{n-1}{2}})(4)$ numbers.\nIf you want a generalization that's too bad :P although there probably is one using the floor function.[/hide][/quote]\r\nI didn't say the actualy number was even, I simply said the number had an even number of digits. :wink:", "Solution_3": "[quote=\"ragnarok23\"][quote=\"bpms\"][quote=\"ragnarok23\"]A palindrome is a number that is the same forwards and backwards. For example, 212 and 21466412 are palindromes. Consider a palindrome with $n$ digits where $n$ is even. Find a formula, in terms of $n$, of how many $n$-digit palindromes exist.[/quote]\n[hide]For the first and last digits they have to be even. So there are four choices for them since they can't be 0. There are n-2 remaining digits. For even values of n there are because there are ten values for each pair of digits and each pair is coordinated. $(10^{\\frac{n-2}{2}})(4)$. For odds, there are $(10^{(\\frac{n-3}{2})})(40)$ which is equal to $(10^{\\frac{n-1}{2}})(4)$ numbers.\nIf you want a generalization that's too bad :P although there probably is one using the floor function.[/hide][/quote]\nI didn't say the actualy number was even, I simply said the number had an even number of digits. :wink:[/quote]\nHeh, I misread that. This should be easier by a lot. Using the same logic...\n[hide]Each digit pairs up with one other so there are $(9)(10^{\\frac{n-2}{2}})$ since the first digit can't be 0.[/hide]" } { "Tag": [ "calculus" ], "Problem": "I'm currently taking an AP US History course. I've read that if you take an AP class you should undoubtedly take the national exam; the reason being that it shows admissions people that you really know the course material. If you're getting an A in the course, is this really necessary? It seems like a lot of money considering all you would be doing is reinforcing your teachers opinion of your performance, unless of course admissions people really do want to see that you take the test.", "Solution_1": "Take the test.\r\n\r\nThere are several things at stake here. The first is that it's actual college credit. I know that different colleges have different policies, but at an awful lot of places, it counts. For us, we have an general education requirement for U.S. History, and that AP exam satisfies the requirement. The course title and grade on your high school transcript would not satisfy the requirement.\r\n\r\nThe college credit frees you up to move forward. If you don't take that U.S. History course during your freshman year, what do you do? College campuses are full of choices, only your own time is limited and you can't take everything. Perhaps the history of a different part of the world? Or an interesting literature course? Or accelerate your math and science courses?\r\n\r\nThe second is that you telling me you got an A in the course is mostly worth a \"so what?\" from me. I'm in advising, not admissions, but look at it from the advisor's side, and bring it back to my field. If you tell me you took a calculus course in high school but not the AP exam, then I don't know the standards of your course, and we won't recognize that course. After all, many - maybe most - of the students who took a high school precalculus course aren't really ready to start with our Calculus I. I do know the standards of the AP exam.\r\n\r\nGiving \"weighted GPA\" credit for AP and honors courses is quite widespread. Admissions people know that you have an incentive to take the course just for that reason - but, again, what is the level of that course and what are its standards? The AP exam may only sample your accomplishments, but it is a national standard to weigh that course against. \r\n\r\nBecause so many AP exams are taken in the senior year, with the results known long after admissions decisions are made, it's hard for those doing admissions to find a way to account for them. But I'm guessing that you're not taking that U.S. History course in your senior year. More likely, you're a junior or a sophomore.\r\n\r\nOh, and you said, \"It seems like a lot of money.\" It's a lot less money than one course's portion of a semester worth of college tuition and/or fees, even at a low-cost state university like mine.", "Solution_2": "Kent Merryfield has said it very well. I will simply reemphasize the point that the test fee (and one day out of your life) is a small price to pay to take the test and get recognition of your work during the course. Knowing ahead of time that you will take the test will also help you concentrate during the course and very likely learn more than you would if you know that you will avoid the test. And after all, isn't learning the most important goal here? Having the test score in hand, as the first reply pointed out, makes cooler learning goals more achievable, and that's worth a lot more than what the test costs.", "Solution_3": "Thanks Kent & Tokenadult. I definitely apreciate it, and I'll most likely end up taking the exam.\r\n\r\nBy the way, yes Kent, I am a sophomore.", "Solution_4": "the really shows how well u know the subject. \r\n\r\nand at my school, u get 5 pts added to your final average if u take the test" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "How do you solve this problem?\r\n$N(m)={4m\\choose 0}+{4m\\choose 4}+{4m\\choose 8}+\\cdots+{4m\\choose 4m}$, where m is a positive integer. Find the smallest m where 81 evenly divides N(m).", "Solution_1": "I'm thinking Pascal's triangle. (of course, there are other equivalent or similar ways of looking at it)", "Solution_2": "I don't quite see how that helps... :?", "Solution_3": "I advise Moderators to move it to the Number Theory Section.", "Solution_4": "[quote=\"jen7\"]How do you solve this problem?\n$N(m)={4m\\choose 0}+{4m\\choose 4}+{4m\\choose 8}+\\cdots+{4m\\choose 4m}$, where m is a positive integer. Find the smallest m where 81 evenly divides N(m).[/quote] Note that $4N(m)=(1+1)^{4m}+(1+i)^{4m}+(1-1)^{4m}+(1-i)^{4m}=2^{2m+1}(2^{2m-1}+(-1)^m)$.\r\nTherefore, you are to find minimal $m$ s.t. $81\\mid 2^{2m-1}+(-1)^m$.\r\nMore hints?" } { "Tag": [ "geometry", "circumcircle", "angle bisector", "geometry unsolved" ], "Problem": "(F.Nilov, A.Zaslavsky) Let $ CC_0$ be a median of triangle $ ABC$; the perpendicular bisectors to $ AC$ and $ BC$ intersect $ CC_0$ in points $ A_c$, $ B_c$; $ C_1$ is the common point of $ AA_c$ and $ BB_c$. Points $ A_1$, $ B_1$ are defined similarly. Prove that circle $ A_1B_1C_1$ passes through the circumcenter of triangle $ ABC$.", "Solution_1": "see http://www.mathlinks.ro/Forum/viewtopic.php?t=223753\r\nsin\u2220ABB1/ sin\u2220CBB1 * sin\u2220CAA1/ sin\u2220BAA1 * sin\u2220BCC1/ sin\u2220ACC1=\r\nsin\u2220B0BC/ sin\u2220B0BA * sin\u2220BAA0/ sin\u2220CAA0 * sin\u2220ACC0/sin\u2220BCC0 =\r\nB0A/B0C * BA0/CA0 * BC0/AC0 =1\r\nWe have AA1,BB1,CC1 are concurrent,denote the intersection is D\r\nWe can acquire O,D,C1,B1 are concyclic and OP is diameter \r\nO,A1,B1,D are concyclic and OP is diameter\r\nSo A1,B1,C1,D,O are concyclic with OP its diameter", "Solution_2": "As it can be seen from above proof, CC1 is symmedian in triangle ABC and angle bisector of angle AC1B. Knowing that the symmedian passes through the common point of the tangents to (ABC) at A and B, we get OC1 and CC1 perpendicular, hence OC1 is angle bisector as well. Inverting about O so that the circle O transforms to it self, the circle (OAB) is transformed into AB, hence the image of C1 is AB\u2229OC1=C2, similarly getting A2 and B2.\r\nTaking into account the above statements, we get C_2A/C_2B = (b/a)^2 and other 2 similar relations which, multiplied, show that A2, B2 and C2 are collinear, the conclusion follows (a,b and c being the side lengths of triangle ABC).\r\n\r\nBest regards,\r\nsunken rock", "Solution_3": "Can someone solve this problem using complex numbers ?\r\nI tried to, but I failed. \r\nAt least give a hint or something.", "Solution_4": "By USAMO 2008/2, we know that $C_1$ is the midpoint of the chord intercepted by the $C$-symmedian in the circumcircle of triangle $ABC$. Now, inverting about $O$ maps $C_1$ to the intersection point of the tangent to the circle at $C$ and line $AB$. Thus, inversion maps the circumcircle of $A_1B_1C_1$ to the trilinear polar of the symmedian point $K$ of triangle $ABC$ and so, points $A_1,B_1,C_1,O$ are con cyclic." } { "Tag": [ "geometry", "geometric transformation", "rotation", "factorial" ], "Problem": "In how many ways can 5 people be seated around a round table? (Two seatings are considered the same if one is a rotation of the other.)", "Solution_1": "What does 5!/5=4!=24 mean? What does 5! mean?", "Solution_2": "We have $ 5\\cdot4\\cdot3\\cdot2\\cdot1$ choices but there are $ 5$ rotations. Thus, we have $ 4\\cdot3\\cdot2\\cdot1 \\equal{} 24$.\r\n\r\nchubbyg-- $ n!\\equal{}(n)(n\\minus{}1)(n\\minus{}2)(n\\minus{}3)...(2)(1)$. Basically, it is the multiplication of all integers less than or equal to itself until 1 is reached. It is called a [url=http://www.artofproblemsolving.com/Wiki/index.php/Factorial]factorial[/url]. Try google if you need more information.", "Solution_3": "cant you just do (n-1)! instead of (n!)/n? It is much simpler. so for this problem it would be (5-1)! or 4!", "Solution_4": "[quote=\"slayer42\"]cant you just do (n-1)! instead of (n!)/n? It is much simpler. so for this problem it would be (5-1)! or 4![/quote]\r\n\r\nyes\r\n$ (n\\minus{}1)!$ is the same as $ \\frac{n!}{n}$ <--can you figure out why?\r\n\r\nso, since these two are the same, you can do the problem either way", "Solution_5": " $\\dfrac{5!}{5} = 4! = \\boxed{24}$ \n5! ways to place everyone but divide by 5 because of rotational symmetry." } { "Tag": [ "LaTeX" ], "Problem": "does anyone know how to simplify:\r\n$ \\frac {(3^{2008})^2 \\minus{} (3^{2006})^2}{(3^{2007})^2 \\minus{} (3^{2005})^2}$\r\nThanks in advance :D\r\nEDIT: ALL LATEX IS CORRECTED", "Solution_1": "what does 3^2 008 mean? does it mean 3^2008? if yes,\r\n\r\nit simplifies to (3^4016-3^4012)/(3^4014-3^4010\r\nfactor it out so (3^4010)(3^6-3^2)/(3^4010)(3^4-1)\r\nso you get 3^6-3^2/3^4-1\r\n\r\n729-9/81-1\r\n\r\n9.", "Solution_2": "[quote=\"neurosurgeon1118\"]does anyone know how to simplify:\n$ \\frac {(3^{2008})^2 \\minus{} (3^{2006})^2}{(3^{2007})^2 \\minus{} (3^{2005})^2}$ \nThanks in advance :D\n______\nBOGTRO: HA HA fixed in this quote.[/quote]\r\n\r\nRewriting, we get\r\n\r\n$ \\frac{3^{4010}(3^6\\minus{}3^2)}{3^{4010}(3^4\\minus{}1)}\\equal{}\\frac{720}{80}\\equal{}\\boxed{9}$.", "Solution_3": "yeah sorry but for some reason i don't understand why the Latex considers the 3^2008 as 3^2 008", "Solution_4": "You realize that this is an AMC 10A 2008 problem, right?", "Solution_5": "Yes I do... Considering I took it", "Solution_6": "Well sorry but i was wondering if anyone had any idea on how to number 25 if u have the test then can you do it and post a solution.", "Solution_7": "What's #25? On the 10A?\r\n\r\nIf so that's totally off topic. \r\n\r\nAnyway, solutions would be up. I got that by realizing that there\r\n\r\nis a $ \\sqrt{7}$, so... Multi choice helped there\r\n\r\nOf course, I still failed.", "Solution_8": "First of all sorry for my ignorance of not looking it up\r\nanyways i scanned the contest pamphlet so i hope this helps:\r\nWait for some reason i cant attach images on aops anymore so im just going to link it to my website\r\nhttp://neurosurgeon1118.web.officelive.com/Documents/AMC%2025.JPG", "Solution_9": "[url]http://isabio.org/amc/2008AMC1012A/AMC10A2008_25_En/[/url]" } { "Tag": [ "ratio", "geometry", "Pythagorean Theorem" ], "Problem": "The diagram consists of three nested squares. Find the ratio of the area of the smallest square to the area of the largest square. Express your answer as a common fraction.\n[asy]draw((0,0)--(7,0)--(7,7)--(0,7)--cycle);\ndraw((3,0)--(7,3)--(4,7)--(0,4)--cycle);\ndraw((1.2,2.4)--(4.6,1.2)--(5.8,4.6)--(2.4,5.8)--cycle);\nlabel(\"4\",(2,7),N);\nlabel(\"3\",(0,5.5),W);\nlabel(\"2\",(.6,3.2),.5ENE);[/asy]", "Solution_1": "[quote=\"GameBot\"]The diagram consists of three nested squares. Find the ratio of the area of the smallest square to the area of the largest square. Express your answer as a common fraction.\n[asy]draw((0,0)--(7,0)--(7,7)--(0,7)--cycle);\ndraw((3,0)--(7,3)--(4,7)--(0,4)--cycle);\ndraw((1.2,2.4)--(4.6,1.2)--(5.8,4.6)--(2.4,5.8)--cycle);\nlabel(\"4\",(2,7),N);\nlabel(\"3\",(0,5.5),W);\nlabel(\"2\",(.6,3.2),.5ENE);[/asy][/quote]\r\n\r\nThe largest square has side lenght 7, the second square has side length 5, the smallest square has side lenght $ \\sqrt{11}$.\r\n\r\n$ \\frac{11}{49}$.", "Solution_2": "Wait, are you sure of that? Because by the Pythagorean Theorem, we get $ \\sqrt{3^2 \\plus{} 2^2}$ or $ \\sqrt{13}$, giving an area of $ 13$ not $ 11$...", "Solution_3": "[quote=\"ernie\"]Wait, are you sure of that? Because by the Pythagorean Theorem, we get $ \\sqrt {3^2 \\plus{} 2^2}$ or $ \\sqrt {13}$, giving an area of $ 13$ not $ 11$...[/quote]\r\n\r\nTypo. Whatever." } { "Tag": [], "Problem": "$ P$ and $ Q$ are whole numbers such that $ 0 < P < 10$ and $ 0 < Q < 10$. How many common\nfractions $ \\frac{P}{Q}$ exist if $ \\frac{1}{2} < \\frac{P}{Q} < 1$?", "Solution_1": "First you find the possible values for Q: 1, 2... 8, 9.\r\nFor these values:\r\nQ is equal to\r\n1 - P is not possible\r\n2- P is not possible\r\n3- 2/3\r\n4- 3/4\r\n5- 3/5, 4/5\r\n6- 4/6, 5/6\r\n7- 4/7, 5/7, 6/7\r\n8- 5/8, 6/8, 7/8\r\n9- 5/9, 6/9, 7/9, 8/9\r\nThere are 16.\r\nBUT, the question asks for COMMON fractions. So we list only the simplified here:\r\n2/3, 3/4, 3/5, 4/5, 5/6, 4/7, 5/7, 6/7, 5/8, 7/8, 5/9, 7/9, 8/9\r\nSo there are 13.\r\nAnswer: 13" } { "Tag": [ "geometry", "3D geometry", "sphere", "topology", "superior algebra", "superior algebra unsolved" ], "Problem": "Is it possible to define a multiplication so that the following is a group?\r\n(1) open unit interval (-1, 1)\r\n(2) open unit disc\r\n(3) open unit sphere in $R^n$", "Solution_1": "(1) Yes\r\n(2) Yes\r\n(3) Yes\r\nJust take some bijection $b$ of the desired set to $\\mathbb{R}$ (or $\\mathbb{C}$ or somewhat else), then define $x*y : = b^{-1}\\left( b(x)+b(y) \\right)$.", "Solution_2": "It's more interesting to ask whether these sets are [i]topological[/i] groups as subspaces of the according euclidean space. (3) works exactly for $n=1,2,4$. But I think a lot of algebraic topology is necessary to proove that.", "Solution_3": "For (1): I don't think normal multiplikation works (or what is normal\u00bf).\r\n\r\nPS: hi Martin, welcome here :)", "Solution_4": "@Daniel: Ach, ich hab die inversen Elemente vergessen! Wie hast du mich eigentlich erkannt?", "Solution_5": "[quote=\"-oo-\"]@Daniel: Ach, ich hab die inversen Elemente vergessen! Wie hast du mich eigentlich erkannt?[/quote]\r\n@Martin: only english outside of the the national forums please :) \r\nWell I could break that rule now since I doubt that anyone else will understand it even in English, but well: You used that name in the \"Schw\u00e4tz\" and sometimes also on the MP ;) (and mods can see the E-Mail addresses).\r\n\r\n@all: he just said that he forgot the inverses and asked how I identified him." } { "Tag": [ "group theory", "abstract algebra", "vector", "superior algebra", "superior algebra unsolved" ], "Problem": "[size=150][color=red][b]Problem.[/b][/color] If each of two [color=blue]groups[/color] is isomorphic to a subgroup of the other, are the [color=blue]groups[/color] necessarily isomorphic?\n[/size]\r\nI look forward to reading your responses one of these days. Happy solving to you [b]all[/b]! :wink: \r\n\r\nP.S. Last month's problem can be found [url=http://www.mathlinks.ro/viewtopic.php?t=256626][color=orange]here.[/color][/url]", "Solution_1": "I remembered doing this way back in the day, and luckily I still have Artin sitting on my shelf to fill in the details. :)\r\n[hide=\"Solution\"]The answer is \"no\": if $ F$ is the free group generated by $ x, y$ then the elements $ u \\equal{} x^2, v \\equal{} xy$ and $ w \\equal{} y^2$ generate a subgroup of $ F$ isomorphic to the free group on 3 generators. Thus the free group on 3 generators is isomorphic to a subgroup of the free group on 2 generators and vice-versa, but the two groups are not isomorphic.[/hide]", "Solution_2": "[hide=\"Another solution\"]\nSet $ G : \\equal{} S_{\\mathbb{N}}, \\ H: \\equal{} S_{\\mathbb{N}}\\times \\mathbb{Z}_{2}.$\nThen $ G < H$ and $ H\\cong \\langle S_{\\mathbb{N}\\setminus\\{1,2\\}}, (1\\ 2)\\rangle < G,$ but $ G\\not \\cong H,$ because $ \\# Z(G) \\equal{} 1\\neq 2 \\equal{} \\# Z(H).$[/hide]", "Solution_3": "Another example (I won't bother tex-ing it - it's not really necessary for this) - take G=RxZ, H=R. Using Zorn's lemma one proves that RxR and R are isomorphic as they are both continuum cardinality Q vector spaces (the existence of a basis is where Zorn is needed). It follows that RxZ is a subgroup of R, and it is obvious that R is a subgroup of RxZ. To see they are not isomorphic, note that one is divisible (R) whilst the other isn't (as Z certainly isn't).\r\nThe problem can be generalized to other categories in an obvious manner - it's fun to see where it holds. A particularly fun one (in my opinion) is for fields, enjoy!", "Solution_4": "Well done, [b]guys[/b]!\r\n\r\nI didn't expect anything less from you... :wink:" } { "Tag": [ "AMC", "AIME", "percent", "AIME I" ], "Problem": "Uh, looking at past AIMEs and AMCs (and Altheman's Mock AIME), I see questions like \"Bob goes to starbuck from 2-4 and stays for 30 minutes. Jon goes to starbuck from 1-3 and stays for 30 minutes.\" How do you do these types of questions?", "Solution_1": "You have to use a geometric approach, since there are infinitely many possibilities.\r\n\r\nTry reading this excerpt from Intro to Counting & Probability. It has some problems like the one you mentioned.\r\n\r\n[url=http://www.artofproblemsolving.com/Books/IntroCount/exc2.pdf]Click Here[/url]", "Solution_2": "Thank you, I get it now. :)", "Solution_3": "Don't know if this is one of the problems that you said you saw already...\r\nbut here's an example of an AIME problem that uses the geometric method:\r\n\r\n[size=150]AIME 1998 #9[/size]\r\nTwo people arrive at a cafe independently at random times between $9 AM$ and $10 AM$ and each stay for $m$ minutes. What is $m$ if there is a $40$ percent chance that they are in the cafe together at some moment?", "Solution_4": "if you understand the process, try my problem now, it is a step above the normal ones :)", "Solution_5": "AHHHH....Adunakhor....you almost confused me...I didn't get an integer answer and I started checking my answer...then I went to the contests section and found out that it's in the form $a-b\\sqrt{c}$...:P...Altheman, what is your problem? you mean the question in your mock AIME?", "Solution_6": "[quote=\"Zellex\"]AHHHH....Adunakhor....you almost confused me...I didn't get an integer answer and I started checking my answer...then I went to the contests section and found out that it's in the form $a-b\\sqrt{c}$...:P...Altheman, what is your problem? you mean the question in your mock AIME?[/quote]\r\n\r\nyes!", "Solution_7": "[quote]AHHHH....Adunakhor....you almost confused me...I didn't get an integer answer and I started checking my answer...[/quote]\r\n\r\nHeh, sorry :?. I took this problem from Kalva which didn't include that part.", "Solution_8": "so how would u do that problem...i am not really sure.\r\ni know that you have to draw part of a circle on a graph, but dont really understand how you lable the axies." } { "Tag": [], "Problem": "3 circles, touching one another externally, share a common tangent. If the radii of the two larger circles are 36 and 9 respectively, find the radius of the smallest circle.", "Solution_1": "[hide=\"Answer\"]$4$\n\nSee http://www.mathlinks.ro/Forum/viewtopic.php?t=64305[/hide]" } { "Tag": [ "floor function", "search", "number theory", "number theory proposed" ], "Problem": "Prove that the exponent of number 2 in the expansion of $n!$ is $\\sum_{j=1}^{\\infty}\\left\\lfloor{n \\over 2^{j}}\\right\\rfloor$.", "Solution_1": "Similary the exponent of number $p$ in the expansion of $n!$ is $\\sum_{k=1}^{\\infty}[\\frac{n}{p^{k}}]$. You can find in some text book on number theory :wink:", "Solution_2": "Can you post a solution? I don't have many number theory books.", "Solution_3": "you can search on lookforbook.com\r\nwrite:number theory.", "Solution_4": "How many numbers in the set $\\{1,2,\\cdots,n\\}$ is divisible by $2$? Of course $\\left\\lfloor{n \\over 2^{1}}\\right\\rfloor$.\r\nHow many numbers in the set $\\{1,2,\\cdots,n\\}$ is divisible by $2^{2}$? Of course $\\left\\lfloor{n \\over 2^{2}}\\right\\rfloor$.", "Solution_5": "[quote=\"Lovasz\"]Can you post a solution? I don't have many number theory books.[/quote]\r\n[url=http://modular.math.washington.edu/scans/papers/hardy/]Theory of Numbers by Hardy and Wright[/url]" } { "Tag": [ "function", "analytic geometry", "geometry", "conics", "parabola", "calculus", "integration" ], "Problem": "Given the function $ f(x) \\equal{} x^2 \\minus{} C$ for some constant $ C$, find the vertices of a square (coordinates in terms of $ C$) centered at the origin with sides parallel to the axes such that the area above the curve and in the square is equal to the area below the curve and in the square.\r\n\r\nFor example, when $ C \\equal{} \\frac13$, we have the square with vertices $ (\\pm1,\\pm1);(\\pm1,\\mp1)$.\r\n\r\n[asy]import graph;\nunitsize(x=1cm);\nreal f(real x){return x^2-0.333333333;}\nxlimits( -1.5, 1.5);\nylimits( -3, 1);\ndraw(graph(f,-3,3));\nxaxis(Ticks(\"%\",extend=true));\nyaxis(Ticks(\"%\",extend=true));\ndot((0,0));\ndraw((1,1)--(1,-1)--(-1,-1)--(-1,1)--cycle);[/asy]", "Solution_1": "Well won't you have two cases? One when the parabola is enitrely contained within the square, and the one when it is not? Am I being unclear?", "Solution_2": "Hm, slightly. Tell me what you would do as a first step.", "Solution_3": "Well we have two cases (I think?) one described in your first post and the other where its range exceeds the square's mximum y value. To eases integration we could move it up by a perhaps?", "Solution_4": "No, $ a$ makes the algebra harder. I assume $ a$ is a half sidelength? Try a different transformation.\r\n\r\n[hide=\"Hint\"]\nShift up by $ C$. Now divide into two cases and solve.[/hide]", "Solution_5": "Hm do we have to adjust when it hits [b]exactly[/b] on the squares vertices?", "Solution_6": "No. Either case will then be inclusive.", "Solution_7": "Gah we will probably have to use a bunch of calc for this. So we get a shift up by C. So that makes the square (a+c,a+c) and so on?", "Solution_8": "Yes. Now, see if you can simplify the integration.", "Solution_9": "Wait a minute! if we shoft like that the x-coordinate remians a right? gah that screws up a bit. and how u use latex for integration?", "Solution_10": "I think it's\r\n[code]\\int_a^bf(x)\\,dx[/code] to produce $ \\int_a^bf(x)\\,dx$.", "Solution_11": "Hm if we flip the graph over, then we get a nice upside-down parabola which would be ez to integrate. This will probably be a nasty cubic relating a and c then.", "Solution_12": "Well, the parabola is a quadratic $ \\minus{} x^2 \\plus{} a\\plus{}c$. All that remains is to find the bounds of the integral.", "Solution_13": "well in the case described in the first post it is 0-a", "Solution_14": "Yes; what about the second case?", "Solution_15": "Well if the top point is a+c, and the graph is x^2 now it goes from 0-sqrt(a+c). Too lazy to integrate.", "Solution_16": "Bah. Practicing $ \\text{\\LaTeX}$:\r\n\r\n$ \\int^{\\sqrt {a + c}}_{ - \\sqrt {a + c}}x^2\\ dx = \\left(\\frac {x^3}3\\right)\\bigg|^{\\sqrt {a + c}}_{ - \\sqrt {a + c}}$\r\n\r\n$ = \\frac23(a + c)\\sqrt {a + c} = 2a^2$.\r\n\r\nThus, $ (a + c)^3 = 3a^2$, and the rest follows. Blah." } { "Tag": [ "trigonometry", "function", "parameterization" ], "Problem": "How do I make a function in terms of x and y when I have a parametric functions involving trigs and logs? :?:", "Solution_1": "It depends on how complex it is, but just use the inverse functions.", "Solution_2": "Usually when you have the parametric functions $ x(t),y(t)$ in terms of trigs and logs, use their inverse functions (and paying attention to their domain/range <---this is very important) and do the algebra to get the relation $ f(x,y)\\equal{}0$. But, the point is... why would you want to do that? Parametric functions allow you to see the behavior of a curve in terms of a controlled parameter $ t$. If you are trying to graph it, a graphing calculator with parametric graphing capabilities should be sufficient." } { "Tag": [ "limit", "function", "parameterization", "complex analysis", "algebra", "functional equation", "complex analysis unsolved" ], "Problem": "Prove there is no such sequences $\\{a_{n}\\}_{n=1}^{\\infty},\\{b_{n}\\}_{n=1}^{\\infty}$ in $\\mathbb{C}$satisfy the following conditions:\r\n(1)\r\n\\[\\lim_{n\\to \\infty}\\sqrt[n]{|a_{n}|}=\\lim_{n\\to \\infty}\\sqrt[n]{|b_{n}|}=0 \\]\r\n(2)\r\n\\[b_{0}a^{2}_{0}=b_{0}a_{0}+1 \\]\r\n\r\n\\[b_{1}a^{2}_{0}+b_{0}(a_{1}a_{0}+a_{0}a_{1})=b_{1}a_{0}+b_{0}a_{1}\\]\r\n\r\n\\[b_{2}a^{2}_{0}+b_{1}(a_{1}a_{0}+a_{0}a_{1})+b_{0}(a_{2}a_{0}+a^{2}_{1}+a_{0}a_{2})=b_{2}a_{0}+b_{1}a_{1}+b_{0}a_{2}\\]\r\n\r\n\\[\\cdots \\]\r\n\r\n\\[b_{n}a^{2}_{0}+b_{n-1}(a_{1}a_{0}+a_{0}a_{1})+\\cdots+b_{0}(a_{n}a_{0}+a_{n-1}a_{1}+\\cdots+a_{0}a_{n})=b_{n}a_{0}+b_{n-1}a_{1}+\\cdots+b_{0}a_{n}\\]\r\n\r\n\\[\\cdots \\]", "Solution_1": "Translating into the language of power series:\r\n$f(x)=\\sum a_{n}x^{n}$ and $g(x)=\\sum b_{n}x^{n}$ have infinite radius of convergence.\r\n$f(x)^{2}g(x)=f(x)g(x)+1$\r\n\r\nAll right, that makes $f$ and $g$ entire functions. Since $f(z)(f(z)-1)g(z)=1$, $f(z)$ is never zero or 1. By the Picard theorem, $f$ is constant. Then $g$ is also constant. The only solutions have $a_{n}=b_{n}=0$ for $n\\ge1$.", "Solution_2": "Let $f(z)=\\sum_{n=0}^{\\infty}a_{n}z^{n}$ and $g(z)=\\sum_{n=0}^{\\infty}b_{n}z^{n}.$\r\n\r\nThe limit involving roots is the statement that these series have infinite radii of convergence - in other words, $f$ and $g$ are entire.\r\n\r\nNow, let's take the chain of identities and make them the coefficients of power series:\r\n\r\n$\\sum_{n=0}^{\\infty}\\left(\\sum_{k=0}^{n}b_{n-k}\\sum_{j=0}^{k}a_{k-j}a_{j}\\right)z^{n}=1+\\sum_{n=0}^{\\infty}\\left(\\sum_{k =0}^{n}a_{n-k}b_{k}\\right)z^{n}$\r\n\r\nNow let's translate this into something we can see:\r\n\r\n$(f(z))^{2}g(z)=1+f(z)g(z)$\r\n\r\nSo that's what the question comes to: can we find entire functions that satisfy that identity?\r\n\r\nThe identity can be rewritten as:\r\n\r\n$f(z)(f(z)-1)g(z)=1.$\r\n\r\nIn order for that to be true for all $z\\in\\mathbb{C},$ we can never have $f(z)=0$ or $f(z)=1.$ That flies in the face of the \"little\" Picard theorem: the range of a nonconstant entire function can omit at most one value. Therefore, the identity cannot be satisfied, except by constant functions.\r\n\r\n[color=green][Ah - took too long to type it. jmerry beat me to it.][/color]", "Solution_3": "Oh,thank you two :) .This is problem is just come out when I want to prove the Picard's Theorem.And in fact it can be show is equicvalent to Picard's theorem.So I want to know a proof without Picard's theorem,or more elemtrary way.Then there will be an alternative way to prove the Picard's theorem.", "Solution_4": "What happens if you just run the chain of identities? Can we find formulas for $a_{1},a_{2},\\dots,b_{1},b_{2},\\dots$ in terms of $a_{0}$ and $b_{0}?$", "Solution_5": "If you try that, be aware that radius of convergence matters- there are many local solutions to the functional equation, and you'd need infinitely many parameters to make it work.\r\n\r\nThis is not the way to an easier proof of the Picard theorem.", "Solution_6": "[quote=\"Kent Merryfield\"]What happens if you just run the chain of identities? Can we find formulas for $a_{1},a_{2},\\dots,b_{1},b_{2},\\dots$ in terms of $a_{0}$ and $b_{0}?$[/quote]\r\nSorry,I forget the constant entire functions.\r\n\r\n\r\nIn fact $b_{n}$ can expressed by $a_{n}$ but also seems tredious. :blush:" } { "Tag": [ "parameterization" ], "Problem": "We may say a process is adiabatic if it occurs fast enough such that no heat is exchanged from the system.But we also say a process is quasi-static when it occurs very slow.\r\n\r\nThen how can PV^gamma = c hold for a fast occuring adiabatic process because a polytropic process should be quasistatic as wiki says http://en.wikipedia.org/wiki/Polytropic_process .\r\n\r\nAlso why is it true that only quasistatic process can be a polytropic process.?", "Solution_1": "I wanna say just briefly:it is slow enough to be considered that pressure and other parameters are the same everywhere in the thermal system and it is ,simultaneously, fast enough to be considered that no heat exchange could be occured.Inside the system heat exchange is very fast,because every kind of heat transfer takes place,but the heat transfer with outside mostly by conduction through some material." } { "Tag": [], "Problem": "How do you guys learn how to write pre-olympiad problems? Many of you say source: myself. I'd like to come up with some new problems. Does anyone have any tips?", "Solution_1": "Inspiration for my problems (though they may or may not be difficult enough for this forum) comes from randomly examining an interesting subject. Try to think of a potential trick to solving a problem first, and go from there. Also, you can use ideas you have seen in other places.", "Solution_2": "I don't know about Pre-Olympiad but I think just making the problems is same for all levels.\r\n\r\nEven though Pre-Olympiad problems are harder to create, if you have solved many problems in that level, it shouldn't be too hard for you to create one. It doesn't mean that it's easy but not totally impossible neither.\r\n\r\nI can't make Pre-Olympiad problems but I'm pretty sure I'll be able to do them in future when I have solved many harder problems. There isn't really technique way on how to make a problem. Just like problem solving, you create more problems and you become better problem maker.\r\n\r\nAnd speaking of my own experience, it's true and sometimes it's even more fun than problem solving :lol:!\r\n\r\nSincerely,\r\n\r\nSilverfalcon", "Solution_3": "Well... depends what you expect. Just a good problem is not something you seek for. Well, actually it is.\r\n\r\nI sougth for days on my very first problem. Which was incredibly bad. In the beginning it was incredible work for no results. But that way you open your mind, and after weeks I got my first good problem, and since then I hardly ever put time in seeking... I just had the idea at once. And sometimes I have some quite nice ones :)\r\n\r\nThere are 2 important things in problem writing:\r\n1) finding a problem\r\n2) finding an elegant way to solve it\r\n\r\nHow to make a problem harder? often for pre-olympiad level it is sufficient to think of a good solution and write a problem around it. I don't know for olympiad level, I'll wait until some more problems of mine get olympiad-approved :)" } { "Tag": [ "analytic geometry", "AMC", "AIME", "LaTeX", "floor function", "modular arithmetic" ], "Problem": "Bob lives at $(0, 0)$ and works at $(6, 6)$. Every day, to get to work, he walks down $12$ blocks, each block either increasing his $x$-coordinate or his $y$-coordinate. How many ways are there for him to get to work if the intersections $(2, 1)$ and $(2, 4)$ are closed off?\r\n\r\nWhat number do you think this would be on an AIME?", "Solution_1": "[quote=\"miyomiyo\"]Bob lives at $(0, 0)$ and works at $(6, 6)$. Every day, to get to work, he walks down $12$ blocks, each block either increasing his $x$-coordinate or his $y$-coordinate. How many ways are there for him to get to work if the intersections $(2, 1)$ and $(2, 4)$ are closed off?\n\nWhat number do you think this would be on an AIME?[/quote]\r\nUnless I'm missing something, probably number 3-5.\r\nMaybe I'm trivializing it though.", "Solution_2": "What's your solution?", "Solution_3": "[hide=\"solution\"]\nThe total number of ways from $(0,0)$ to $(6,6)$ is $\\binom{12}{6}$\nor\n$\\frac{12!}{6!6!}={11*7*3*2^{2}}$\n\nThe total number of ways that pass through $(2,1)$ is $\\binom{3}{2}* \\binom{9}{4}$\nor\n$\\frac{3!}{2!1!}* \\frac{9!}{4!5!}={7*3^{3}*2}$\n\nLikewise, the total number of ways that pass through $(2,4)$ is\n$\\binom{6}{2}*\\binom{6}{4}$\n\n$\\frac{6!}{2!4!}* \\frac{6!}{2!4!}={3^{2}*5^{2}}$\n\nThe number of paths that pass through BOTH $(2,1)$ and $(2,4)$ is $\\binom{3}{2}* \\binom{6}{2}$\n$\\frac{3!}{2!1!}* \\frac{6!}{2!4!}={3^{2}*5}$\n\nby the inclusion exclusion principle, the paths that don't pass through those intersections are\n\n${11*7*3*2^{2}}-{7*3^{3}*2}-{3^{2}*5^{2}}+{3^{2}*5}=$\n$(7*3*2)(22-9)-(3^{2}*5)(5-1) =$\n$(2*3)[ 7*13-2*3*5 ] =$\n$(6)(91-30) = (6)(61) = 366$\n\nplease check the arithmetic. I could have made a mistake.\n[/hide]\r\n\r\noh and how do you use combinatoric notation in latex?\r\n\r\nEDIT:\r\nthank you Hokkage", "Solution_4": "I did the same as chickendude's solution. Didn't do the arithmetic out though. Combinatorial form is \\binom{n}{k} in LaTeX. For AIME difficulty, I'd say around a 4.", "Solution_5": "my solution was the same as chickendude's but i got 360 instead. Could someone check his arithmetic please?", "Solution_6": "366 is correct.", "Solution_7": "Another way to do it: the number of ways to get to a certain intersection is the the sum of the number of ways to get to the intersections below it and to the left.", "Solution_8": "[quote=\"Phelpedo\"]Another way to do it: the number of ways to get to a certain intersection is the the sum of the number of ways to get to the intersections below it and to the left.[/quote]\r\n\r\nThis method is called \"block walking\", and is useful for smaller situations in which a lot of annoying casework would get in the way of a quick combinatorical solution.\r\n\r\nLike:\r\n\r\nFind the number of lattice paths of minimal distance from $(0,0)$ to $(8,8)$ if the squares with coordinates $( 2 \\pm 1, 3 \\pm 1)$ and $(6 \\pm 1, 5 \\pm 1)$ are blocked off due to a parade.", "Solution_9": "I'd put this as a 2 or a 3 because it is so easy to brute force by block-walking. If you want to make it later to force combinatorics use, make the numbers larger or something.", "Solution_10": "Yeah, but then it's not an AIME answer. The biggest you can go that's under 1000 is (6, 7) which is easy to block-walk.", "Solution_11": "You can always do stuff like find $\\lfloor x/1000 \\rfloor$.", "Solution_12": "Yeah, I know....\r\n\r\nHow's this. He works at (12, 12), (3, 3) and (5, 6) are closed. Find the remainder when the number of ways is divided by 1000.", "Solution_13": "how could he work a (3,3) and go to work through (5,6)?", "Solution_14": "[i]Hmm.. what if:[/i]\r\n\r\nA man works at the location $(7,7)$ and lives at the origin. He can only travel along the lattice grid. Each change by $\\pm 1$ of either of the x- or y-coordinate takes the man $1$ minute to complete. If the number of ways in which he can reach his work, if he has at most $20$ minutes to get there, is $N$, find the remainder when $N$ is divided by $1000$.", "Solution_15": "[hide=\"Casewise\"]\nClassify directions as U, D, L, and R for up, down, left, and right, respectively.\n\nHe can either go directly (7 ups and 7 rights), or take 1, 2, or 3 detours, in the form of an additional LR pair or UD pair.\n\nIf he were to go:\n\n1. Directly, $\\frac{14!}{7! 7!}$\n\n2. Add one pair (either LR or UD), $2 \\frac{16!}{7! 8! 1!}$\n\n3. Add two of the same pair (either LR or UD), $2 \\frac{18!}{7! 9! 2!}$\n\n4. Add one of each pair, $\\frac{18!}{8! 8! 1! 1!}$\n\n5. Add three of one pair (either LR or UD), $2 \\frac{20!}{7! 10! 3!}$\n\n6. Add two of one pair and one of the other (either two LR's or two UD's), $2 \\frac{20!}{8! 9! 2! 1!}$\n\nTotal:\n\n$\\frac{14!}{7! 7!}+2 \\frac{16!}{7! 8! 1!}+2 \\frac{18!}{7! 9! 2!}+\\frac{18!}{8! 8! 1! 1!}+2 \\frac{20!}{7! 10! 3!}+2 \\frac{20!}{8! 9! 2! 1!}$\n\n$\\equiv 432+920+640+220+440+400 \\equiv \\boxed{052 \\; \\pmod{1000}}$\n[/hide]", "Solution_16": "[quote=\"The Zuton Force\"][hide=\"Casewise\"]\nClassify directions as U, D, L, and R for up, down, left, and right, respectively.\n\nHe can either go directly (7 ups and 7 rights), or take 1, 2, or 3 detours, in the form of an additional LR pair or UD pair.\n\nIf he were to go:\n\n1. Directly, $\\frac{14!}{7! 7!}$\n\n2. Add one pair (either LR or UD), $2 \\frac{16!}{7! 8! 1!}$\n\n3. Add two of the same pair (either LR or UD), $2 \\frac{18!}{7! 9! 2!}$\n\n4. Add one of each pair, $\\frac{18!}{8! 8! 1! 1!}$\n\n5. Add three of one pair (either LR or UD), $2 \\frac{20!}{7! 10! 3!}$\n\n6. Add two of one pair and one of the other (either two LR's or two UD's), $2 \\frac{20!}{8! 9! 2! 1!}$\n\nTotal:\n\n$\\frac{14!}{7! 7!}+2 \\frac{16!}{7! 8! 1!}+2 \\frac{18!}{7! 9! 2!}+\\frac{18!}{8! 8! 1! 1!}+2 \\frac{20!}{7! 10! 3!}+2 \\frac{20!}{8! 9! 2! 1!}$\n\n$\\equiv 432+920+640+220+440+400 \\equiv \\boxed{052 \\; \\pmod{1000}}$\n[/hide][/quote]\r\n\r\nIs there a way to solve the general problem without a load of casework? Since it would get pretty ugly if he had to travel $14$ units and had, say, two hours in which to get there.", "Solution_17": "Well, I mean, it can clearly be generalized into a summation form, but taking it modulo 1000 will be, if I'm not missing something blatantly obvious, the standing problem.\r\n\r\nLike, say he had $120$ minutes to go. Even the case in which he takes the full 2 hours with 53 UD routes has $42906253386216936243966766644196440$ possibilities... and that's just one of the $\\frac{(53)(54)}{2}=1431$ cases.", "Solution_18": "Hmm... It seems like a good problem, but to be an AIME problem it would need a non-brute force solution.", "Solution_19": "[quote=\"The Zuton Force\"][hide=\"Casewise\"]\nClassify directions as U, D, L, and R for up, down, left, and right, respectively.\n\nHe can either go directly (7 ups and 7 rights), or take 1, 2, or 3 detours, in the form of an additional LR pair or UD pair.\n\nIf he were to go:\n\n1. Directly, $\\frac{14!}{7! 7!}$\n\n2. Add one pair (either LR or UD), $2 \\frac{16!}{7! 8! 1!}$\n\n3. Add two of the same pair (either LR or UD), $2 \\frac{18!}{7! 9! 2!}$\n\n4. Add one of each pair, $\\frac{18!}{8! 8! 1! 1!}$\n\n5. Add three of one pair (either LR or UD), $2 \\frac{20!}{7! 10! 3!}$\n\n6. Add two of one pair and one of the other (either two LR's or two UD's), $2 \\frac{20!}{8! 9! 2! 1!}$\n\nTotal:\n\n$\\frac{14!}{7! 7!}+2 \\frac{16!}{7! 8! 1!}+2 \\frac{18!}{7! 9! 2!}+\\frac{18!}{8! 8! 1! 1!}+2 \\frac{20!}{7! 10! 3!}+2 \\frac{20!}{8! 9! 2! 1!}$\n\n$\\equiv 432+920+640+220+440+400 \\equiv \\boxed{052 \\; \\pmod{1000}}$\n[/hide][/quote]\r\n\r\nI do not think that is correct becaue you have not accounted for the fact that you have repeated ALOT of paths (for example in some 16 minute paths, they would have already reached the office by the 14th minute so that path cannot be achieved but you hvae counted that into your answer) (for example for a 14 minute path you can have NNNNNNNEEEEEEE but for a 16 minute path you can have NNNNNNNEEEEEEEWE, well that path cannot be done because you would have already reached the office by the 14th minute but you still incorperated that into your 16 minute path counting)." } { "Tag": [ "Putnam", "inequalities", "function", "calculus", "derivative", "AMC" ], "Problem": "Prove for non-negative real numbers a1, a2, ..., an and b1, b2, ..., bn,\r\n\r\n(a1a2...an)1/n + (b1b2...bn)1/n :le: ((a1 + b1)(a2 + b2)...(an + bn))1/n", "Solution_1": "The inequality clearly holds if one of a1 .... an or b1....bn = 0 so suppose not.\n\n\n\nThen we can divide everything by GM(a1 ... an) and set x_n = b_n/a_n and find that the original inequality is equivelent to\n\n\n\n1 < = [(1 + x_1)(1 + x_2) ... (1 + x_n)]^(1/n) - (x_1*x_2 ... x_n)^(1/n).\n\n\n\nThis isn't rigorous but suppose x = x_2 = x_2 ... = x_n. Then considering RHS as a function of x_1 by differentiation we find that absolute minimum occurs at x_1 = x. Similarly for other variables. I haven't taken multivariable, so I don't know how to convert this into a tight argument.\n\n\n\nAnother way is to replace x_1, x_2 with x_1' and x_2' such that GM(x_k's) remains constant while GM(1 + x_k's) decreases. With a little bit of argument this implies that inequality must hold[hide][/hide]" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "if , (x-y){f(x+y)}-(x+y){f(x-y} =2xy(x^2 - y^2)\r\n\r\nthen find f(2)", "Solution_1": "Not enough information: $ f(x)\\equal{}\\frac{x^3}{2}\\plus{}Cx$ is a solution for any constant $ C$.", "Solution_2": "how did you get this?please give the solution sir...", "Solution_3": "Let $ g(t)\\equal{}f(t)/t$. In terms of $ g$ the equation simplifies to $ g(x\\plus{}y)\\minus{}g(x\\minus{}y)\\equal{}2xy$. Dividing by $ y$ and letting $ y\\to 0$, get $ g'(x)\\equal{}x$. Hence $ g(x)\\equal{}x^2/2\\plus{}C$.", "Solution_4": "[quote=\"mlok\"]Let $ g(t) \\equal{} f(t)/t$. In terms of $ g$ the equation simplifies to $ g(x \\plus{} y) \\minus{} g(x \\minus{} y) \\equal{} 2xy$. Dividing by $ y$ and letting $ y\\to 0$, get $ g'(x) \\equal{} x$. Hence $ g(x) \\equal{} x^2/2 \\plus{} C$.[/quote]\r\nI is not continuous . But with this you can find out it .Please try again.", "Solution_5": "what is \"I\" here , sir?", "Solution_6": "Oh I am sorry. This function is not continuous.", "Solution_7": "What are you talking about? mlok gave a (valid) family of continuous solutions to the given equation. There's no claim that these are exhaustive, just that these are enough to demonstrate that all solutions need not pass through a given point $ (2, f(2))$.", "Solution_8": "This solution to find out :$ f(x)\\equal{}\\frac{x^3}{2}\\plus{}Cx$ is not true. \r\nBecause from problem we don't know do there exist $ g'(x)$ or not.", "Solution_9": "Did you not read what I wrote? The function $ f(x) \\equal{} x^3/2 \\plus{} cx$ [i]is[/i] a solution of the given equation for every constant $ c$. (Plug it in and check.) mlok made no claim that this family was exhaustive, because for the question mlok was answering (to compute the value $ f(2)$) it was sufficient to demonstrate this family. (It shows that $ f(2)$ is not uniquely determined by the given equation.)\r\n\r\nTelling someone that they didn't solve a question that wasn't asked and that they weren't trying to solve is not helpful. If you actually were going to provide a complete solution, that would be interesting.", "Solution_10": "[quote=\"mlok\"]Not enough information: $ f(x) \\equal{} \\frac {x^3}{2} \\plus{} Cx$ is a solution for any constant $ C$.[/quote]\r\nOh sorry i think he wrote solution. \r\nHere is my solution.\r\n$ h(x) \\equal{} f(x) \\minus{} \\frac {x^3}{2} \\minus{} Cx$\r\nWhere $ C \\equal{} f(1) \\minus{} \\frac {1}{2}$\r\nSo $ h(1) \\equal{} 0$\r\nApply this to the equation:\r\n$ (x \\minus{} y)h(x \\plus{} y) \\equal{} (x \\minus{} y)h(x \\plus{} y)$\r\n$ \\Longrightarrow xh(y) \\equal{} yh(x)$\r\nLet $ x \\equal{} 1$ then $ h(y)\\equiv 0$\r\nImply that $ f(x) \\equal{} \\frac {x^3}{2} \\plus{} Cx$ where $ C \\equal{} f(1) \\minus{} \\frac {1}{2}$\r\nThe solution only use property of this function not necessary exist $ g'(x)$." } { "Tag": [ "function", "calculus", "derivative", "limit", "calculus computations" ], "Problem": "Let's say I have two functions, g(x) and f(x). How can I find and prove that g(x)~f(x) at a certain x=c? Would I find the derivative of both functions, set them equal to each other, then solve for x, or something like that?", "Solution_1": "[quote=\"epkid08\"]Let's say I have two functions, g(x) and f(x). How can I find and prove that g(x)~f(x) at a certain x=c? Would I find the derivative of both functions, set them equal to each other, then solve for x, or something like that?[/quote]\r\n\r\n$ f(x)\\underbrace{\\sim}_{x\\to{c}}g(x)\\implies\\lim_{x\\to{c}}\\frac{f(x)}{g(x)}\\equal{}1$", "Solution_2": "[quote=\"Mathstud28\"][quote=\"epkid08\"]Let's say I have two functions, g(x) and f(x). How can I find and prove that g(x)~f(x) at a certain x=c? Would I find the derivative of both functions, set them equal to each other, then solve for x, or something like that?[/quote]\n\n$ f(x)\\underbrace{\\sim}_{x\\to{c}}g(x)\\implies\\lim_{x\\to{c}}\\frac {f(x)}{g(x)} \\equal{} 1$[/quote]\r\n\r\nYes, but I would also like to prove that x=c is the only x that would make g(x)~f(x) true. What would I do to prove this?" } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "How about $ \\int \\frac{1}{x^{2008}\\plus{}x}dx$", "Solution_1": "[hide=\"Hint\"]\n\\[ \\frac{1}{x^{2008} \\plus{} x} \\equal{} \\frac{1}{x} \\minus{} \\frac{x^{2006}}{x^{2007} \\plus{} 1}\\]\n[/hide]" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "What is the last digit of $2005^{2005}$ ?", "Solution_1": "I think that it's $3$ but my prove is to comlicated to post it here, if you are interested i can e-mail you :) :) :) :D :D :D", "Solution_2": "YES E -MAIL ME \r\n\r\n :lol: [color=red]mathematicien@naseej.com[/color]", "Solution_3": "The last digit of $2005^{2005}$ is $3$??? Am I missing something? Or did I see the problem wrongly?", "Solution_4": "Last digit seen from which side\u00bf\r\nFrom left to right: it's obviously $3$ and the first digit is $7$ then.", "Solution_5": ":?: :?: :?: \r\nThe last digit IS 5!! and last digit by convention refers to the units digit, so there's no issue of reading from the left or right...", "Solution_6": "[quote=\"dblues\"]:?: :?: :?: \nThe last digit IS 5!! and last digit by convention refers to the units digit, so there's no issue of reading from the left or right...[/quote]\r\nI'm agree with you :? ;)", "Solution_7": "no it's a $3$ !!!", "Solution_8": "I must agree with M@re and Rodolphe: after much tedious calculation and a little advanced numbe theory, I conclude the last digit is a 3. The fact that the last digit of 2005^k is a 5 for small values of k is just a fluke.", "Solution_9": "Deleted because I'm stupid.", "Solution_10": "No, no.... Its certanly 3, i was solving that problem for two days, and then, when i was so close to give up it just came out to me..... 3\r\n\r\nAnd i have to agree with MysticTerminator about small values for k in $2005^k$, it is only a fluke", "Solution_11": "how can you don't see that's 3 !!", "Solution_12": "hey, M@re(or MysticTerminator) , can you post your solution plz? ;)", "Solution_13": "It is too difficult. It would take long time to type", "Solution_14": "Hey guys, I know that it's very interesting research subject here, but I was browsing the forum and I saw some other problems- it is true, a little bit easier- that wait for solutions, so why don't you leave a little bit this magnificient problem and try to concentrate on those problems.", "Solution_15": "dear friends\r\nexplain your solutions :mad: [size=150]ok ok ok [/size]", "Solution_16": "Solution is so complicated and long ... Are you sure you want it ?", "Solution_17": "It also contains a clever leap of logic which I initially failed to make. :)", "Solution_18": "Unfortunately, the proof does not fit my margin; I cannot post it here.", "Solution_19": "Oh, sorry for my stupid mistake, its obviously $3$ :blush: :roll: \r\nBut I was unable to solve the problem for $2006^{2006}$, any ideas\u00bf", "Solution_20": "Ok, guys, sorry, but the topic is closed. There are too many interesting ideas on 1 m^2 of posts." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "this is the problem, i couldn't solve:\r\ngiven prime p, find all solutions iz Z: xyz=p(x+y+z).\r\nthanx", "Solution_1": "we have to find all possible solutions $(x, y, z)$ to this equation, right ?", "Solution_2": "Please also consider this problem: http://www.mathlinks.ro/viewtopic.php?p=2406#2406 :)", "Solution_3": "Yes you have to find all triples (x, y, z).\r\nPlease, help me about this problem..." } { "Tag": [ "geometry", "perimeter", "inradius", "incenter", "Pythagorean Theorem", "AMC" ], "Problem": "How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to 3 times their perimeters?\r\n\r\n(A) 6\r\n(B) 7\r\n(C) 8\r\n(D) 10\r\n(E) 12", "Solution_1": "I posted this solution somewhere else, but since this is my favorite problem of the year, I am going to repost\r\n\r\n$Area=semi perimeter*inradius$\r\nwhich means r=6\r\n\r\nAssume the side lengths of the legs are x and y, then the hypotenuse will be $(x-6)+(y-6)=x+y-12$\r\n\r\n$x^{2}+y^{2}=(x+y-12)^{2}$\r\n$x^{2}+y^{2}=x^{2}+y^{2}+144+2(xy-12x-12y)$\r\n$0=144+2(xy-12x-12y)$\r\n$0=72+xy-12x-12y$\r\n$72=xy-12x-12y+144$\r\n$72=(x-12)(y-12)$\r\n\r\nlet $x>y$\r\n(x-8,y-8)=(72,1),(36,2),(24,3),(18,4),(12,6),(9,8)\r\nwhich gives 6 possible triangles", "Solution_2": "Good one :P . I didn't know the formula $Area = Semiperimeter \\times Radius$.", "Solution_3": "[quote=\"Allan Z\"]\nAssume the side lengths of the legs are x and y, then the hypotenuse will be $(x-6)+(y-6)=x+y-12$\n[/quote]\r\n\r\nCan you explain that part? I get $c=xy/6-x-y$", "Solution_4": "The key is that if $r$ is the radius then there exist positive reals $p, q$ such that\r\n\r\n$x = p+r$\r\n$y = q+r$\r\n$z = p+q$\r\n\r\nThese are the three side lengths that the altitudes from the incenter divide the sides into. We conclude that\r\n\r\n$z = x+y-2r = x+y-12$\r\n\r\nCombining this result with the Pythagorean Theorem gets us a nice solution.", "Solution_5": "I put a,b, and c as sides of the right triangle with c as the hypotenuse\r\n\r\nAfter squaring the two equations I had\r\nI got\r\n[hide]\na+b-2=ab/4\n[/hide]\r\nbut I realized I wrote down the wrong equation\r\nso I'm wrong\r\nand\r\nI don't feel like doing it again\r\nso I think the others are right\r\n\r\nI did if the area=perimeter\r\ninstead of 3area=perimeter", "Solution_6": "I was trying the test earlier today, and I have a semi-new solution.\r\n\r\n[hide] At the start, we know that $ ab\\equal{}6(a\\plus{}b\\plus{}c)$ and $ a^{2}\\plus{}b^{2}\\equal{}c^{2}$. Double the first and add it to the second equation to get \\[ (a\\plus{}b)^{2}\\equal{}c^{2}\\plus{}12(a\\plus{}b\\plus{}c)\\] \\[ (a\\plus{}b)^{2}\\minus{}c^{2}\\equal{}12(a\\plus{}b\\plus{}c)\\] \\[ (a\\plus{}b\\minus{}c)(a\\plus{}b\\plus{}c)\\equal{}12(a\\plus{}b\\plus{}c)\\] Because a, b, and c are positive, we can cross out $ (a\\plus{}b\\plus{}c)$ from both sides getting $ a\\plus{}b\\minus{}12\\equal{}c$. This can now be put directly into what Allen Z said to finish the problem.[/hide]" } { "Tag": [], "Problem": "Albatross has an income of 15.51 dollars for every 7 hours he works. If so how much money does he make in a week.\r\n\r\nP.S. Don't ask about the name Albatross it's a cool name :P .\r\n\r\nP.P.S If you are clearly above getting started I'm going to have to ask you NOT to answer unless people are cleary stumped and it has been taking a long time to solve. Because geez do you get self-satisyfaction in knowing you can solve problems intended for 12 year olds? ;)", "Solution_1": "Does he have an 8 hour work day or a 24 hour work day?", "Solution_2": "24 Hour. :lol:", "Solution_3": "[hide]lets see....\nthere is 7*24=168 hours in a week \n168/7=24 \n15.51*24=372.24\nso...... \"Albatross\" gets paid a measily $372.24 for his hard toiled 168 hours of work[/hide]", "Solution_4": "Does he still get paid if he falls asleep on the job :P ? I don't think anybody can stay awake for 168 hours :P .", "Solution_5": "The problem depends on where albatross lives. Assuming he lives in the gopher state and is employed by a large ($500,000 annual receipts or more) company, he gets higher wages after the first 48 hours.\r\n\r\nAssuming he is employed by a small company, then Albatross' best move would be to sue the compnay for having an hourly wage less than 4.90. Same for the large company.\r\n\r\nOr maybe Albatross isn't employed by a company, in which none of the above apply." } { "Tag": [ "Gamebot" ], "Problem": "What positive integer value of $ x$ will satisfy the equation $ (x^2 \\minus{} 6x \\plus{} 9)^4 \\equal{} 5^{16}$?", "Solution_1": "This equation seems quite scary with the huge exponents, so let's not solve right away yet and instead try to simplify a little. \n\nIn the left side, \n$ (x^2\\minus{}6x\\plus{}9)^4$\n$ \\equal{}((x\\minus{}3)^2)^4$ (since $ a^2\\minus{}ab\\plus{}b^2\\equal{}(a\\minus{}b)^2$)\n$ \\equal{}(x\\minus{}3)^8$. (multiplying exponents according to power of power property)\n\nOur exponent is $ 8$. Notice that that is half of the exponent on the right side, $ 16$! \n\nWe can rewrite the right side as $ 5^{16}\\equal{}(5^2)^8$ due to the power of power property, and now we take the eighth root of both sides of the equation, eliminating the large exponent!\n\nNow, it's just a case of simple solving-for-$ x$:\n$ x\\minus{}3\\equal{}5^2$\n$ x\\minus{}3\\equal{}25$\n$ x\\equal{}\\fbox{28}$", "Solution_2": "[hide=Solution]We can start by taking the fourth root from $ (x^2 - 6x + 9)^4 = 5^{16}$, yielding with $ (x^2 - 6x + 9)= 5^{4}$. So, we get that $x^2 - 6x = 616$. So, $x(x-6)=616$. Prime factorizing $616$, we get $ x=\\fbox{28}$ [/hide]", "Solution_3": "you can say $(x^2-6x+9)^4 = (x-3)^8 = 5^{16}$, and this gives x-3 = 25, so x = 28", "Solution_4": "[quote=PenguinMoosey]you can say $(x^2-6x+9)^4 = (x-3)^8 = 5^{16}$, and this gives x-3 = 25, so x = 28[/quote]\n\nNice!, didn't think about it that way!", "Solution_5": "[quote=PenguinMoosey]you can say $(x^2-6x+9)^4 = (x-3)^8 = 5^{16}$, and this gives x-3 = 25, so x = 28[/quote]\n\nthat's really straightfoward! :D", "Solution_6": "[quote=MWZ][quote=PenguinMoosey]you can say $(x^2-6x+9)^4 = (x-3)^8 = 5^{16}$, and this gives x-3 = 25, so x = 28[/quote]\n\nthat's really straightfoward! :D[/quote]\n\nNo, I didn't think of factoring $x^2-6x+9$ as a quadratic. It occured to me to do prime factorization.", "Solution_7": "Wait can\u2019t u take 4th root?", "Solution_8": "probably negatives or something but idk", "Solution_9": "[quote=dragoon]Wait can\u2019t u take 4th root?[/quote]\n\nProbably better to factor $(x^2-6x+9)^4$ as $(x-3)^8$.", "Solution_10": "[quote=naren_pr][quote=dragoon]Wait can\u2019t u take 4th root?[/quote]\n\nProbably better to factor $(x^2-6x+9)^4$ as $(x-3)^8$.[/quote]\n\nI mean:\n$(x^2-6x+9)^4=5^{16} \\implies (x^2-6x+9)=5^4 \\implies (x-3)=\\pm 5^2 \\implies x\\in \\{28, -22\\}$\n\nWhy is no one noticing the solution $-22$? Like I get positive int and all but...", "Solution_11": "[quote=firebolt360][quote=naren_pr][quote=dragoon]Wait can\u2019t u take 4th root?[/quote]\n\nProbably better to factor $(x^2-6x+9)^4$ as $(x-3)^8$.[/quote]\n\nI mean:\n$(x^2-6x+9)^4=5^{16} \\implies (x^2-6x+9)=5^4 \\implies (x-3)=\\pm 5^2 \\implies x\\in \\{28, -22\\}$\n\nWhy is no one noticing the solution $-22$? Like I get positive int and all but...[/quote]\n\nEveryone notices it but no one states it for the reason you gave :)", "Solution_12": "[quote=firebolt360][quote=naren_pr][quote=dragoon]Wait can\u2019t u take 4th root?[/quote]\n\nProbably better to factor $(x^2-6x+9)^4$ as $(x-3)^8$.[/quote]\n\nI mean:\n$(x^2-6x+9)^4=5^{16} \\implies (x^2-6x+9)=5^4 \\implies (x-3)=\\pm 5^2 \\implies x\\in \\{28, -22\\}$\n\nWhy is no one noticing the solution $-22$? Like I get positive int and all but...[/quote]\n\nThat is how the question is worded. They said positive integer, so we have to go with the flow." } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "Find all polynomials $ P(x)$ that satisfy equation: $ P(x\\plus{}1)\\equal{}P(x)\\plus{}2x\\plus{}1$ for all reals $ x$.", "Solution_1": "we take $ Q(x)=P(x)-x^2$\r\nwe have $ Q(n+1)=P(n+1)-n^2-2n-1=P(x)-n^2=Q(n)$ so $ Q(x)=c$ for some $ c\\in\\mathbb{R}$\r\nthen $ \\boxed{P(x)=x^2+c,\\ c\\in\\mathbb{R}}$", "Solution_2": "[quote=\"aviateurpilot\"]we take $ Q(x) = P(x) - x^2$\nwe have $ Q(n + 1) = P(n + 1) - n^2 - 2n - 1 = P(x) - n^2 = Q(n)$ so $ Q(x) = c$ for some $ c\\in\\mathbb{R}$\nthen $ \\boxed{P(x) = x^2 + c,\\ c\\in\\mathbb{R}}$[/quote]\r\nwhat is Q(x+1)=Q(x) -> Q(x)=c", "Solution_3": "[quote=\"ABRORBEK\"][quote=\"aviateurpilot\"]we take $ Q(x) = P(x) - x^2$\nwe have $ Q(n + 1) = P(n + 1) - n^2 - 2n - 1 = P(x) - n^2 = Q(n)$ so $ Q(x) = c$ for some $ c\\in\\mathbb{R}$\nthen $ \\boxed{P(x) = x^2 + c,\\ c\\in\\mathbb{R}}$[/quote]\nwhat is Q(x+1)=Q(x) -> Q(x)=c[/quote]\r\n\r\n$ Q(n)=Q(n-1)=Q(n-2)=...=Q(1)=Q(0)$\r\nwe take $ c=Q(0)$\r\nthen equation $ Q(x)=c$ have an infinity of solution then $ \\forall x: \\ Q(x)=c$ because $ Q$ is polynom", "Solution_4": "[quote=\"aviateurpilot\"][quote=\"ABRORBEK\"][quote=\"aviateurpilot\"]we take $ Q(x) = P(x) - x^2$\nwe have $ Q(n + 1) = P(n + 1) - n^2 - 2n - 1 = P(x) - n^2 = Q(n)$ so $ Q(x) = c$ for some $ c\\in\\mathbb{R}$\nthen $ \\boxed{P(x) = x^2 + c,\\ c\\in\\mathbb{R}}$[/quote]\nwhat is Q(x+1)=Q(x) -> Q(x)=c[/quote]\n\n$ Q(n) = Q(n - 1) = Q(n - 2) = ... = Q(1) = Q(0)$\nwe take $ c = Q(0)$\nthen equation $ Q(x) = c$ have an infinity of solution then $ \\forall x: \\ Q(x) = c$ because $ Q$ is polynom[/quote]IT IS TRY $ n$ not integer $ n$ real number", "Solution_5": "A polynomial equation can't have infinitely many solutions unless the polynomial is a constant. That is because an $ n$th degree polynomial has no more than $ n$ roots, and $ n$ is finite.", "Solution_6": "[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=211560]The method of finite differences[/url] settles the issue. $ \\Delta P(0) \\equal{} 1, \\Delta^2 P(0) \\equal{} 2$ immediately gives $ P(x) \\equal{} P(0) {x \\choose 0} \\plus{} {x \\choose 1} \\plus{} 2{x \\choose 2} \\equal{} P(0) \\plus{} x^2$ for every integer $ x$, hence everywhere." } { "Tag": [ "inequalities", "trigonometry", "geometry", "perimeter", "geometry solved" ], "Problem": "Let ABC -- triangle such that ABc* \\sqrt (ac)/(a+c) \\leq \\sqrt (p-a)(p-c) <=>\r\na*c 3 \\leq (p-a)(p-c)*(a+c) 2 .(1)\r\n\r\nDenote p-a=A and so on . We have cA>B.\r\n(1) comes to (B+C)(A+B) 3 \\leq AC(A+2B+C) 2.\r\nI will use a instead of A and so on.\r\n(b+c)(a 3+3a 2b+3ab 2 +b 3) \\leq ac(a 2 +4b 2 +c 2 +2ac+4ab+4bc).\r\nThis comes by computations to a 3b+3a 2b 2 +3ab 3+b 4+b 3c \\leq ac 3+abbc+aabc+4abcc+2aacc.\r\nThis is b(a+b) 3+b 3c \\leq ac(cc+ab+4bc+2ac)+ab 2 c.\r\n\r\nWe have c>a>b, we have bbbc3 \\leq ac(c 2 +ab+4bc+2ac).\r\nSince c>a we have\r\nac(cc+ab+4bc+2ac)>a 3 (3a+5b).\r\n\r\nSo it suffices to show b(a+b) 3 \\leq a 3(3a+5b). This come by computations to \r\n3a 2 b 2 +3ab 3+b 4 \\leq 3a 4+4a 3b.\r\nSince a>b it suffices to show\r\na 2 b 2 +ab 3 \\leq a 4+a 3b <=> (a+b)(a 3-b 3) \\geq 0, which is true since a>b.\r\n\r\n----------------------------------------------------------------------------------\r\nSorry about this monster. I know what you will say :\"Ah ce scarbos\" :D .\r\nI just really wanted to finish the proof of the first problem I sove in 2004 :D .", "Solution_2": "[quote=\"Fedor Bakharev\"]Let $ABC$ be a triangle such that $AB\\leq BC\\leq CA$, $w_{b}$ is the angle-bisector from $B$ to $CA$, $h_{c}$ is the altitude from $C$ to $AB$. Prove that $w_{b}\\leq h_{c}$.[/quote]\r\nLet we show a refinement of the original inequality :\r\n\r\n$c\\leq a\\leq b\\Longrightarrow h_{c}\\geq r_{a}\\geq w_{b}$,\r\n\r\nwhere $r_{a}$ is the radius of excircle.\r\n[hide=\"Proof.\"]\n$h_{c}-r_{a}=\\frac{2\\Delta}{c}-\\frac{\\Delta}{s-a}$\n\n$=\\frac{\\Delta[2(s-a)-c]}{c(s-a)}$\n\n$=\\frac{\\Delta(b-a)}{c(s-a)}\\geq0$;\n\n$r_{a}-w_{b}=\\frac{\\Delta}{s-a}-\\frac{2ca}{c+a}\\cos{\\frac{B}{2}}$\n\n$=\\sqrt{\\frac{s(s-b)(s-c)}{s-a}}-\\frac{2\\sqrt{cas(s-b)}}{c+a}$\n\n$=\\sqrt{s(s-b)}\\left(\\sqrt{\\frac{s-c}{s-a}}-\\frac{2\\sqrt{ca}}{c+a}\\right)$\n\n$=\\frac{(a-c)[a^{2}+5ca+ab+c(c+a-b)]\\sqrt{s(s-b)}}{2(c+a)^{2}\\sqrt{(s-c)(s-a)}+4(s-a)(c+a)\\sqrt{ca}}\\geq0$,\n\nwhere $s$ is the semi-perimeter and $\\Delta$ is the area of $ABC$. [/hide]" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find all real positive solutions (if any) to \n\\begin{align*}\nx^3+y^3+z^3 &= x+y+z, \\mbox{ and} \\\\\nx^2+y^2+z^2 &= xyz. \n\\end{align*}", "Solution_1": "Well as we have $x,y,z>0$ we have also \\[ \\sqrt[3]{\\frac{x^3+y^3+z^3}{3}}\\geq \\frac{x+y+z}{3} \\] so from the first equation \\[ \\sqrt[3]{\\frac{x+y+z}{3}}\\geq \\frac{x+y+z}{3} \\] and as $x+y+z>0$ we obtained $x+y+z \\leq 3$.\r\nNow we have also \\[ \\sqrt{\\frac{x^2+y^2+z^2}{3}}\\geq \\sqrt[3]{xyz} \\] and thus we may get $xyz \\geq 27$ but this is contradiction cos' we would have \\[ 1 \\geq \\frac{x+y+z}{3} \\geq \\sqrt[3]{xyz} \\geq 3 \\] this means that our system of equation has no solutions in positive real numbers :)", "Solution_2": "http://www.mathlinks.ro/Forum/viewtopic.php?t=5389", "Solution_3": "Multiply the first equation by the square of the second, you get\r\n$ (x^3\\plus{}y^3\\plus{}z^3)(x^2\\plus{}y^2\\plus{}z^2)^2\\equal{}(x\\plus{}y\\plus{}z)x^2y^2z^2$\r\nBut the left side contains every term on the right side and more, hence there is no solution.", "Solution_4": "$x+y+z=x^3+y^3+z^3 \\geq 3xyz=3(x^2+y^2+z^2) \\geq (x+y+z)^2$\n\n$\\implies x+y+z \\leq 1 \\implies 0 \\leq x,y,z \\leq 1 \\implies x^3 \\leq x, y^3 \\leq y, z^3 \\leq z \\implies x^3+y^3+z^3 \\leq x+y+z$. \n\nEquality occurs only at $x=y=z=1 \\implies x+y+z=3$, contradiction.", "Solution_5": "Jastrzab's is the most classy solution....\n\nMaths is the doctor of science..\nSayantan...", "Solution_6": "[quote=\"shobber\"]Find all real positive solutions (if any) to \\[ \n\\begin{align*}\nx^3+y^3+z^3 &= x+y+z, \\mbox{ and} \\\\\nx^2+y^2+z^2 &= xyz. \n\\end{align*}. \\][/quote]\nDividing the first equation by the second we get:\n$\\frac{\\sum x}{\\sum x^2}=\\sum \\frac{x^2}{yz}\\ge \\frac{\\left(\\sum x\\right)^2}{\\sum xy}$ [b][Cauchy-Schwarz][/b]\n$\\Rightarrow \\frac{\\sum x}{\\sum x^2}\\ge \\frac{\\left(\\sum x\\right)^2}{\\sum xy}\\Rightarrow \\sum xy\\ge \\left(\\sum x\\right)\\left(\\sum x^2\\right)$\n$\\ge \\left(\\sum x\\right)\\left(\\sum xy\\right)\\Rightarrow \\sum x\\le 1$. But\n$\\sum x=\\sum x^3$, so that $xyz\\le \\tfrac{1}{3}\\left(\\sum x^3\\right)\\le \\tfrac{1}{3}$ [b][AM-GM][/b]\nHowever, $xyz=\\sum x^2\\ge 3\\sqrt[3]{x^2y^2z^2}\\Rightarrow xyz\\ge 27$, absurd!\nThus, $\\exists$ no solutions in $x,y,z>0$ to the system. $\\Box$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$p$ is a positive number. For given $p$ find the greatest $C$ for which the following inequality stands:\r\n$x^{2}+y^{2}+pxy \\geq C(x+y)^{2}$\r\nwhere $x$ and $y$ are nonnegative numbers.", "Solution_1": "Take $x=y=1$ then $2+p\\geq 4C\\iff \\frac12+\\frac{p}{4}\\geq C$\r\nNow $C=\\frac12+\\frac{p}{4}$ is sufficient; indeed, \\[x^{2}+y^{2}+pxy \\geq \\left(\\frac12+\\frac{p}{4}\\right)(x+y)^{2}\\iff (x^{2}+y^{2})\\left(\\frac12-\\frac{p}{4}\\right)+(\\frac{p}{2}-1)xy\\geq 0\\\\ \\iff \\left(\\frac12-\\frac{p}{4}\\right)(x-y)^{2}\\geq 0\\]\r\nwhich is true for $p\\leq 2$." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Let n>1 be an integer, and let S be a subset of {1,2,...,n} such that S neither contains two coprime elements, nor does it contain two elements, one of which divides the other. What is the maximum possible number of elements of S ?", "Solution_1": "the integer nearest to n/4" } { "Tag": [ "algebra", "polynomial", "vector", "linear algebra", "linear algebra unsolved" ], "Problem": "Let $A$ be a $C$-commutative algebra of dimension $n$ over $C$ and assume that 0 and 1 are its only idempotents. Prove that $A$ is generated by a finite number of nilpotent elements.", "Solution_1": "Is there something missing, or wrong, or have I gone crazy? :? What if this algebra is $\\mathbb C$ itself, for instance? It's true that $0$ and $1$ are the only idempotent elements, but there's only one nilpotent element, $0$, and it certainly doesn't generate $\\mathbb C$..\r\n\r\nI'd also like to know what \"to generate\" means here, in the context of algebras. For instance, do you assume that $1$ and $X$ generate $K[X]$, because all elements of $K[X]$ are polynomials in $1$ and $X$, or do you mean that the elements have to generate the algebra as a vector space?", "Solution_2": "Oh, I am terribly sorry, I forgot to add what I understand by generates. Actually, we say that the family $x_1,..,x_k$ generates the algebra $A$ if every element of $A$ is a linear combination of $1_A$ and some of $x_1,...,x_k$. Again, I apologise.", "Solution_3": "As usual, identify $a\\in \\mathcal A$ with the endomorphism $x\\mapsto ax$ represented in a basis of $\\mathcal A$ over $\\mathbb C$. What we get is an embedding of $\\mathcal A$ as a subalgebra of $\\mathcal M_n(\\mathbb C)$. What we must prove can be rephrased thus:\r\n\r\nGiven a subalgebra $\\mathcal A$ of $\\mathcal M_n(\\mathbb C)$ with no idempotent elements except for $0,1$ ($1=I_n$ here; in general, things like $A-\\lambda,\\ A\\in\\mathcal A$ mean $A-\\lambda I_n$), prove that $\\mathcal A$ is generated by $1$ and several nilpotent elements. \r\n\r\nIn order to prove this, it suffices to show that for every element $A\\in\\mathcal A$ there is a scalar $\\lambda$ such that $A-\\lambda$ is nilpotent. Every $A\\in\\mathcal A$ can be written uniquely as $A=D_a+N_a$, where $D_a,N_a$ are diagonalizable and nilpotent respectively, and they commute with one another. Moreover, they're polynomials in $A$, so they belong to $\\mathcal A$. Suppose now that $D_a$ is not a scalar matrix. It must then have $k\\ge 2$ distinct eigenvalues $\\lambda_1,\\ldots,\\lambda_k$, and it can be written as $D_a=\\sum_{i=1}^k\\lambda_iP_i$, where $P_i$ is the projection onto the space of $\\lambda_i$-eigenvectors associated to $A$. $P_i$ can be written as some scalar times $\\prod_{j\\ne i}(D_a-\\lambda_j)$, so it too belongs to $\\mathcal A$, but then $P_i$ would be an idempotent element of $\\mathcal A$ different from $0,1$, so we have a contradiction. This means that $D_a=\\lambda$, and we have $A-\\lambda=N_a$, which is nilpotent, and this is exactly what we wanted." } { "Tag": [ "LaTeX" ], "Problem": "Hi I'm new to the forum & looking for help displaying the algebraic logic to get from one equivelant formula to next.\r\n\r\n1) Starting: Chrg = Pmt + Cont + Woff\r\n2) Chrg/Chrg = Pmt/Chrg + Cont/Chrg + Woff/Chrg\r\n3) ??\r\n4) End: Chrg/Chrg = Pmt/(Chrg-Cont) + Woff/(Chrg -Cont)\r\n\r\nNeed help on showing the logic needed between step 2 & 4? \r\n\r\nThanx in advance", "Solution_1": "What are these 'symbols' :maybe: \r\n Can anyone put this in LaTeX", "Solution_2": "I think each \"string\" of 4 letters is a different variable. (i.e. $ Chrg\\equal{}a, Pmt\\equal{}b\\cdots$)" } { "Tag": [ "trigonometry", "geometry", "circumcircle", "symmetry", "geometry unsolved" ], "Problem": "the inscribed circle of ABC is tangent to each side at D,E,F, AD meets the inscribed circle at X,take DY=AX in AD,YB and YC meets the circle at P,Q.prove:FP,EQ,AD are parallel.", "Solution_1": "[color=darkblue]I have another nice problem. \n\nThe inscribed circle of $ \\triangle ABC$ is tangent to each side $ BC, CA, AB$ at $ D,E,F$ respectively. Let $ AD$ meets the inscribed circle at $ X$, take $ DY \\equal{} AX$ with $ Y \\in AD$\nLet $ BY$ meet $ AC$ at $ P$, $ CY$ meet $ AB$ at $ Q$. Prove that $ FP, EQ, AD$ are concurrent.[/color]\r\n\r\n[color=darkblue]What do you think? Is it true?[/color]", "Solution_2": "Draw the line which passes through $ E$ and parallel to $ AD$, it cuts $ (I)$ at $ Q$, similarly $ FP\\parallel AD$. We will prove $ DY \\equal{} AX$ and complete the proof.\r\nApplying the law of sine we have:\r\n$ \\frac {YD}{DC} \\equal{} \\frac {YQ}{QC}\\cdot\\frac {\\sin \\angle YDQ}{\\sin \\angle QDC} \\equal{} \\frac {AE}{EC}\\cdot\\frac {\\sin \\angle AEX}{\\sin \\angle AXE}$\r\n$ \\Leftrightarrow YD \\equal{} AE\\cdot\\frac {AX}{AE} \\equal{} AX$\r\nWe are done!", "Solution_3": "[quote=\"thanhnam2902\"][color=darkblue]I have another nice problem. \n\nThe inscribed circle of $ \\triangle ABC$ is tangent to each side $ BC, CA, AB$ at $ D,E,F$ respectively. Let $ AD$ meets the inscribed circle at $ X$, take $ DY \\equal{} AX$ with $ Y \\in AD$\nLet $ BY$ meet $ AC$ at $ P$, $ CY$ meet $ AB$ at $ Q$. Prove that $ FP, EQ, AD$ are concurrent.[/color]\n\n[color=darkblue]What do you think? Is it true?[/color][/quote]\r\n\r\nThis is obviously true by apply Pappus's theorem to 6 points BQFEPC (because $ AD,BE,CF$ are concurrent at Gergonne point).And you don't need the statement $ DY\\equal{}AX$ in this problem.", "Solution_4": "[quote=\"livetolove212\"]Draw the line which passes through $ E$ and parallel to $ AD$, it cuts $ (I)$ at $ Q$, similarly $ FP\\parallel AD$. We will prove $ DY \\equal{} AX$ and complete the proof.\nApplying the law of sine we have:\n$ \\frac {YD}{DC} \\equal{} \\frac {YQ}{QC}\\cdot\\frac {\\sin \\angle YDQ}{\\sin \\angle QDC} \\equal{} \\frac {AE}{EC}\\cdot\\frac {\\sin \\angle AEX}{\\sin \\angle AXE}$\n$ \\Leftrightarrow YD \\equal{} AE\\cdot\\frac {AX}{AE} \\equal{} AX$\nWe are done![/quote]\r\n\r\ntypo: $ \\frac {YD}{DC} \\equal{} \\frac {YQ}{QC} \\cdot \\frac {\\sin \\angle QDC}{\\sin \\angle YDQ}$", "Solution_5": "Let te parallel to $ AD$ through $ E$ cut $ (I)$ at $ Q$. Let $ CQ$ intersect $ AD$ and $ (I)$ at $ Y$ and $ Q'$, respectively. Let the tangent line to $ (I)$ at $ Q$ intersect $ AD$ and $ AC$ at $ Y'$ and $ M$, respectively. We will see that $ AX \\equal{} DY$, which solves the problem.\r\n\r\nProjecting from $ C$ we get $ (E,Q,D,Q') \\equal{} (E,Q',D,Q)$. This implies $ (E,Q,D,Q') \\equal{} \\minus{}1$.\r\n\r\nNow, consider the projection from $ Q$ which maps the incircle into line $ AD$. We get $ \\minus{}1 \\equal{} (E,Q,D,Q') \\equal{} (\\infty ,Y',D,Y) \\equal{} DY/DY'$. This implies $ DY \\equal{} Y'D$.\r\n\r\nAs $ QME$ is isosceles, $ Y'MA$ is isosceles too. By simmetry, we get $ Y'D \\equal{} AX$. Then $ DY \\equal{} AX$, which is what we wanted to prove.", "Solution_6": "Dear Mithril,\r\nare you sure of some path of your proof?\r\nFor example: (E,Q,D,Q')=(E,Q',D,Q)?\r\nIf you have some time can you explain again your third line...\r\nSincerely\r\nJean-Louis", "Solution_7": "Dear Mithril,\r\nyour idea is very nice and helpfull.\r\nI come up with your notation and some explications:\r\nthe quadrilateral EQDQ' is harmonic; in consequence, the tangent to the incircle at Q' meet the tangent \u00e0 Q on DE at a point N.\r\nLet X be the meetpoint of QQ' and DE.\r\n(E, D, X, N) is harmonic , so is the pencil with origin at Q\u2026\r\nthe rest follows\r\nSincerely\r\nJean-Louis", "Solution_8": "Dear Mithril,\r\nyour idea conduct to a converse wich solve the initial problem of Bye.\r\nSincerely\r\nJean-Louis", "Solution_9": "$ (E,Q,D,Q') \\equal{} (E,Q',D,Q)$ because if you project from $ C$, mapping the incircle into itself, $ E$ and $ D$ are fixed points, $ Q$ goes to $ Q'$ and $ Q'$ goes to $ Q$. This means $ EQDQ'$ is harmonic.\r\n\r\nThen, project from $ Q$ and send $ E$ to the infinity, and $ Q$, $ D$, $ Q'$ to $ Y'$, $ D$ and $ Y$, respectively.\r\n\r\n[quote=\"jayme\"]\n(E, D, X, N) is harmonic , so is the pencil with origin at Q\u2026\n[/quote]\r\nWhat do you mean here? I don't think you need to project to $ ED$ before.", "Solution_10": "Based on the same idea as in [b][size=100]mithril's[/size][/b] solution, let us to try an alternative approach.\r\n\r\n[b][size=100][color=DarkBlue]LEMMA. - A triangle $ \\bigtriangleup ABC$ is given and let $ D$ be, the point of intersection of its circumcircle $ (O),$ from the $ A$-symmedian. We draw a line through vertex $ B$ and parallel to the line segment $ CD,$ which intersects $ AD$ at point $ E,$ the circle $ (O)$ at point $ F,$ and the tangent line of $ (O)$ at vertex $ C,$ at point $ Z.$ Prove that $ BE \\equal{} FZ.$[/color][/size][/b]\r\n\r\n[b][size=100]PROOF OF THE LEMMA.[/size][/b] - We draw the tangent line of $ (O)$ at point $ D,$ which intersects the line segment $ CZ,$ at point so be it $ K$ and we denote the points $ P\\equiv BC\\cap KD$ and $ Q\\equiv BC\\cap AD.$\r\n\r\nThe tangent line of $ (O)$ at vertex $ B$ of $ \\bigtriangleup ABC,$ passes through the point $ A'\\equiv AD\\cap CK$ as well.\r\n\r\n$ \\bullet$ Because of the line segment $ BC,$ as the polar of $ A'$ with respect to $ (O),$ passes through the point $ P,$ we have that the polar of $ P$ with respect to $ (O),$ passes through the point $ A'.$\r\n\r\nSo, we conclude that the line segment $ A'D,$ connecting the point $ A'$ with the point $ D,$ as the tangency point of $ (O),$ from the line through $ P,$ is the polar of $ P$ with respect to $ (O)$ and then, we have that the points $ P,\\ B,\\ Q,\\ C,$ are in harmonic conjugation.\r\n\r\nThat is, the pencil $ D.PBQC$ is also harmonic and because of $ BE\\parallel DC,$ we conclude that $ BE \\equal{} BE'$ $ ,(1)$ where $ E'\\equiv DP\\cap BE.$\r\n\r\nBut, it is easy to show that $ BE' \\equal{} FZ$ $ ,(2)$ because of the symmetry with respect to the line segment $ KO.$\r\n\r\nFrom $ (1),\\ (2)$ $ \\Longrightarrow$ $ BE \\equal{} FZ$ and the proof of the [b][size=100]Lemma[/size][/b] is completed.\r\n\r\n$ \\bullet$ Return now in the proposed problem, we draw the line through the point $ F$ and parallel to $ AD,$ which intersects the incircle $ (I)$ of the given triangle $ \\bigtriangleup ABC,$ at point so be it $ P$ and we denote the points $ Y\\equiv AD\\cap BP$ and $ B'\\equiv (I)\\cap BP.$\r\n\r\nThe line segment $ B'B,$ is the $ B'$-symmedian of $ \\bigtriangleup B'DF$ and because of $ AD\\parallel FP,$ where $ P\\equiv B'B\\cap (I),$ based on the above [b][size=100]Lemma[/size][/b], we conclude that $ DY \\equal{} AX.$\r\n\r\nBy the same way, we can prove that the line segment $ CQ,$ where $ Q$ is the point of intersection of $ (I),$ from the line through the point $ E$ and parallel to $ AD,$ intersects $ AD$ at the same point $ Y$ and the proof of the proposed problem is completed.\r\n\r\nKostas Vittas." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c \\ge 0$ such that $ a\\plus{}b\\plus{}c\\equal{}3$. Prove that\r\n\\[ \\frac{a^2}{2a^2\\minus{}3a\\plus{}3}\\plus{}\\frac{b^2}{2b^2\\minus{}3b\\plus{}3}\\plus{}\\frac{c^2}{2c^2\\minus{}3c\\plus{}3} \\le \\frac{3}{2}\\]\r\n:)", "Solution_1": "[quote=\"can_hang2007\"]Let $ a,b,c \\ge 0$ such that $ a \\plus{} b \\plus{} c \\equal{} 3$. Prove that\n\\[ \\frac {a^2}{2a^2 \\minus{} 3a \\plus{} 3} \\plus{} \\frac {b^2}{2b^2 \\minus{} 3b \\plus{} 3} \\plus{} \\frac {c^2}{2c^2 \\minus{} 3c \\plus{} 3} \\le \\frac {3}{2}\n\\]\n:)[/quote] \r\nI think the following more general statement holds:\r\n\r\nLet $ a_1,a_2,...a_n$ be nonnegative real numbers such that $ a_1\\plus{}a_2\\plus{}...\\plus{}a_n\\equal{}n$. If $ p\\le \\frac n{2(n\\minus{}1)}$ and $ q\\ge 0$, then\r\n\r\n$ \\frac {a_1\\minus{}1}{(a_1\\minus{}p)^2\\plus{}q}\\plus{} \\frac {a_2\\minus{}1}{(a_2\\minus{}p)^2\\plus{}q}\\plus{}...\\plus{} \\frac {a_n\\minus{}1}{(a_n\\minus{}p)^2\\plus{}q} \\le 0$.\r\n\r\nIn the case $ p\\equal{}\\frac n{2(n\\minus{}1)}$, equality holds for $ a_1\\equal{}a_2\\equal{}...\\equal{}a_n$, and again for $ a_1\\equal{}0$ and $ a_2\\equal{}...\\equal{}a_n$, or any cyclic permutation.", "Solution_2": "Yes, it can be rewrite:\r\n$ \\sum\\ \\frac{(2b\\plus{}2c\\minus{}a)^2}{15a^2\\plus{}(2b\\plus{}2c\\minus{}a)^2} \\geq\\ \\frac{9}{8}$\r\nUsing Cauchy Schwarts:\r\n$ LHS \\geq\\ \\frac{[\\sum\\ (2b\\plus{}2c\\minus{}a)^2]^2}{\\sum\\ (2b\\plus{}2c\\minus{}a)^2[15a^2\\plus{}(2b\\plus{}2c\\minus{}a)^2]} \\geq\\ \\frac{9}{8}$\r\n(obviously with Schur).\r\nIt also canhang's proof for my similar problem:\r\n$ \\sum\\ \\frac{a^2}{2a^2\\plus{}(b\\plus{}c\\minus{}a)^2} \\leq\\ 1$\r\nOf course, with my problem, we can prove easily with Am-Gm :) \r\nNotice that with 2 ineqs above, equality occurs when $ a\\equal{}b\\equal{}c$ or $ a\\equal{}b;c\\equal{}0$", "Solution_3": "[quote=\"Vasc\"][quote=\"can_hang2007\"]Let $ a,b,c \\ge 0$ such that $ a \\plus{} b \\plus{} c \\equal{} 3$. Prove that\n\\[ \\frac {a^2}{2a^2 \\minus{} 3a \\plus{} 3} \\plus{} \\frac {b^2}{2b^2 \\minus{} 3b \\plus{} 3} \\plus{} \\frac {c^2}{2c^2 \\minus{} 3c \\plus{} 3} \\le \\frac {3}{2}\n\\]\n:)[/quote] \nI think the following more general statement holds:\n\nLet $ a_1,a_2,...a_n$ be nonnegative real numbers such that $ a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_n \\equal{} n$. If $ p\\le \\frac n{2(n \\minus{} 1)}$ and $ q\\ge 0$, then\n\n$ \\frac {a_1 \\minus{} 1}{(a_1 \\minus{} p)^2 \\plus{} q} \\plus{} \\frac {a_2 \\minus{} 1}{(a_2 \\minus{} p)^2 \\plus{} q} \\plus{} ... \\plus{} \\frac {a_n \\minus{} 1}{(a_n \\minus{} p)^2 \\plus{} q} \\le 0$.\n\nIn the case $ p \\equal{} \\frac n{2(n \\minus{} 1)}$, equality holds for $ a_1 \\equal{} a_2 \\equal{} ... \\equal{} a_n$, and again for $ a_1 \\equal{} 0$ and $ a_2 \\equal{} ... \\equal{} a_n$, or any cyclic permutation.[/quote]\r\nNice, VasC. Let me tell about the source of my inequality. Indeed, it is just my inspiration from your inequality: If $ a,b,c \\ge 0,a\\plus{}b\\plus{}c\\equal{}3$, then\r\n\\[ a^{\\frac{3}{2a}}\\plus{}b^{\\frac{3}{2b}}\\plus{}c^{\\frac{3}{2c}} \\le 3 \\qquad (1)\\]\r\nIn case $ a,b,c$ are the side lengths of a triangle, we can use Bernoulli's Inequality:\r\n\\[ \\left( \\frac{1}{a}\\right)^{\\frac{3}{2a}} \\ge 1\\plus{}\\frac{3}{2a}\\left( \\frac{1}{a} \\minus{}1\\right) \\equal{}\\frac{2a^2\\minus{}3a\\plus{}3}{2a^2}\\]\r\nor\r\n\\[ a^{\\frac{3}{2a}} \\le \\frac{2a^2}{2a^2\\minus{}3a\\plus{}3}\\]\r\nThus, we only need to prove\r\n\\[ \\frac{a^2}{2a^2\\minus{}3a\\plus{}3} \\plus{}\\frac{b^2}{2b^2\\minus{}3b\\plus{}3} \\plus{}\\frac{c^2}{2c^2\\minus{}3c\\plus{}3} \\le \\frac{3}{2} \\qquad (2)\\]\r\nAfter working on (2), I see that it holds for all $ a,b,c \\ge 0$. That's how I came with this inequality. :blush: :)\r\n\r\n@nguoivn: Em co nghi den cach khac ko dung Cauchy Schwarz ko? Vi thuc su anh post bai nay de gioi thieu ky thuat khac. :)", "Solution_4": "Nice your idea Can_Hang. :lol: \r\n\r\n[quote=\"Vasc\"] Let $ a_1,a_2,...a_n$ be nonnegative real numbers such that $ a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_n \\equal{} n$. If $ p\\le \\frac n{2(n \\minus{} 1)}$ and $ q\\ge 0$, then\n\n$ \\frac {a_1 \\minus{} 1}{(a_1 \\minus{} p)^2 \\plus{} q} \\plus{} \\frac {a_2 \\minus{} 1}{(a_2 \\minus{} p)^2 \\plus{} q} \\plus{} ... \\plus{} \\frac {a_n \\minus{} 1}{(a_n \\minus{} p)^2 \\plus{} q} \\le 0$.\n\nIn the case $ p \\equal{} \\frac n{2(n \\minus{} 1)}$, equality holds for $ a_1 \\equal{} a_2 \\equal{} ... \\equal{} a_n$, and again for $ a_1 \\equal{} 0$ and $ a_2 \\equal{} ... \\equal{} a_n$, or any cyclic permutation.[/quote]\r\nActually, the condition $ q\\ge \\frac {n(3n \\minus{} 4)}{4(n \\minus{} 1)^2}$ is needed, too. We can reformulate this statement as follows:\r\n\r\nLet $ a_1,a_2,...a_n \\ge \\frac { \\minus{} n}{n \\minus{} 2}$ such that $ a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_n \\equal{} n$. If $ r\\ge \\frac {n(3n \\minus{} 4)}{(n \\minus{} 2)^2}$ , then\r\n\r\n$ \\frac {1 \\minus{} a_1 }{r \\plus{} a_1^2} \\plus{} \\frac {1 \\minus{} a_2 }{r \\plus{} a_2^2} \\plus{} ... \\plus{} \\frac {1 \\minus{} a_n }{r \\plus{} a_n^2} \\ge 0$.", "Solution_5": "@Can_hang: Ko biet y' tuong? nay` co' dc ko anh :\r\nAssume that $ c \\equal{}$ min$ {a, b, c}$. => $ 3\\geq\\ a \\plus{} b \\geq\\ 2 ; \\frac {9}{4} \\geq\\ ab \\geq\\ 0.$\r\nWe can prove: \r\n$ \\frac {a^2}{2a^2 \\minus{} 3a \\plus{} 3} \\plus{} \\frac {b^2}{2b^2 \\minus{} 3b \\plus{} 3} \\leq\\ \\frac {a \\plus{} b}{2}$ (let $ a \\plus{} b \\equal{} S$; $ ab \\equal{} P$...)\r\nWe only need to prove:\r\n$ \\frac {a \\plus{} b}{2} \\plus{} \\frac {c^2}{2c^2 \\minus{} 3c \\plus{} 3} \\le \\frac {3}{2}$\r\n<=> $ \\frac {c(c \\minus{} 1)(2c \\minus{} 3)}{2c^2 \\minus{} 3c \\plus{} 3} \\geq\\ 0$ (always trues with $ c\\equal{}$ min $ {a, b, c}$.)\r\n :maybe:", "Solution_6": "[quote=\"nguoivn\"]@Can_hang: Ko biet y' tuong? nay` co' dc ko anh :\nAssume that $ c \\equal{}$ min$ {a, b, c}$. => $ 3\\geq\\ a \\plus{} b \\geq\\ 2 ; \\frac {9}{4} \\geq\\ ab \\geq\\ 0.$\nWe can prove: \n$ \\frac {a^2}{2a^2 \\minus{} 3a \\plus{} 3} \\plus{} \\frac {b^2}{2b^2 \\minus{} 3b \\plus{} 3} \\leq\\ \\frac {a \\plus{} b}{2}$ (let $ a \\plus{} b \\equal{} S$; $ ab \\equal{} P$...)\n[/quote]\r\nThis inequality does not hold, nguoivn. Try $ a\\equal{}1.1,b\\equal{}c\\equal{}0.95$ ;)" } { "Tag": [ "trigonometry", "ratio", "number theory", "relatively prime", "AMC" ], "Problem": "Triangle $ABC$ has a right angle at $C,$ $AC = 3$ and $BC = 4$. Triangle $ABD$ has a right angle at $A$ and $AD = 12$. Points $C$ and $D$ are on opposite sides of $\\overline{AB}$. The line through $D$ parallel to $\\overline{AC}$ meets $\\overline{CB}$ extended at $E$. If\r\n\r\n\\[ \\frac{DE}{DB} = \\frac{m}{n} \\]\r\n\r\nwhere $m$ and $n$ are relatively prime positive integers, then $m+n =$\r\n\r\n$\\text{(A)} \\ 25 \\qquad \\text{(B)} \\ 128 \\qquad \\text{(C)} \\ 153 \\qquad \\text{(D)} \\ 243 \\qquad \\text{(E)} \\ 256$", "Solution_1": "edit: better picture\r\n\r\nfirst use Pythagorean to find all lengths (the 5 and the 13).\r\ndraw $AG$ parallel to $CE$ and $BF$ parallel to $AC$ and $DE$. Then we see $GE = 3$ and $AF = 4$ because of the rectangles. Label $DG = x$ and $FG = y$ then we get $x^2 + (y+4)^2 = 12^2$ and $(x+3)^2 + y^2 = 13^2$ from the right triangles. At this point bust out your TI and graph the two equation in terms of $y$ (ignore the negative radical since lengths must be positive). Then you get $x = 9.6$ and $y = 3.2$. so bust out your TI again and find $(9.6+3)/13$ in fraction and adding top and bottom we get $128$. answer is $B$.", "Solution_2": "If you say ${\\angle{ABC}=\\theta}$ and extend $CA$ past $A$, then the angle there is also $\\theta$, and then really it's just ratios and pythagorean theorem.", "Solution_3": "[quote=\"K81o7\"]If you say ${\\angle{ABC}=\\theta}$ and extend $CA$ past $A$, then the angle there is also $\\theta$, and then really it's just ratios and pythagorean theorem.[/quote]\r\n\r\nNow that I think of it, that's probably better because there are no systems of equations of powers of two. But with TI that really is no problem at all..." } { "Tag": [], "Problem": "Water weighs 62.4 pounds per cubic foot. What is the weight of the water in a full, right, cylindrical tank with a 7-foot diameter and height of 2 feet? Use 22/7 as an approximation for \u03c0. Express your answer to the nearest whole number", "Solution_1": "$ \\mathrm{weight} \\equal{} \\left (\\frac{22}{7} \\right )\\left (\\frac{7}{2} \\right )^{2}2 \\hspace{1mm}\\mathrm{ft}^{3}\\times \\frac{62.4 \\hspace{1mm} \\mathrm{pounds}}{\\mathrm{ft}^{3}}\\equal{}\\boxed{4805 \\hspace{1mm} \\mathrm{pounds}}$" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "Solve the equation\r\n\r\nx*((y')^2+1)=y^4\r\n\r\nThanx", "Solution_1": "hello, have someone solved this equation?\nSonnhard." } { "Tag": [], "Problem": "\u0393\u03b5\u03b9\u03b1 \u03c3\u03b5 \u03bf\u03bb\u03bf\u03c5\u03c2. \u03bc\u03b9\u03b1 \u03ba\u03b1\u03bb\u03b7 (\u03b8\u03b5\u03c9\u03c1\u03b7\u03bc\u03b1): \u03b1\u03bd p \u03c0\u03c1\u03c9\u03c4\u03bf\u03c2 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03b7\u03c2 4\u03ba+1 \u03bd\u03b1 \u03b4\u03b5\u03b9\u03be\u03b5\u03c4\u03b5 \u03bf\u03c4\u03b9 \u03c5\u03c0\u03b1\u03c1\u03c7\u03bf\u03c5\u03bd 00 \u03c9\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9 \u03bd\u03b1 \u03b3\u03c1\u03b1\u03c6\u03b5\u03b9 \u03c9\u03c2 \u03b1\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03b4\u03c5\u03bf \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03c9\u03bd \u03bc\u03b5 \u03b4\u03c5\u03bf \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03bf\u03c5\u03c2 \u03c4\u03c1\u03bf\u03c0\u03bf\u03c5\u03c2 \u03c0\u03c7 p=a^2+b^2=c^2+d^2 , \u03c4\u03bf\u03c4\u03b5 p = \u03c3\u03c5\u03bd\u03b8\u03b5\u03c4\u03bf\u03c2 :ewpu:", "Solution_1": "\u039a\u03b1\u03bb\u03b5\u03c2.!..... :lol:", "Solution_2": "\u038c\u03c3\u03bf\u03bd \u03b1\u03c6\u03bf\u03c1\u03ac \u03c4\u03b7\u03bd \u03c0\u03c1\u03ce\u03c4\u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7, \u03b5\u03af\u03bd\u03b1\u03b9 \u03b3\u03bd\u03c9\u03c3\u03c4\u03ae \u03c9\u03c2 \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \u03c4\u03bf\u03c5 Fermat (\u03ba\u03b1\u03b9 \u03b1\u03c5\u03c4\u03ae... :wink: ) \u03ba\u03b1\u03b9 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03bd\u03cd\u03b5\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03b1\u03c1\u03ba\u03b5\u03c4\u03bf\u03cd\u03c2 \u03c4\u03c1\u03cc\u03c0\u03bf\u03c5\u03c2. \u0391\u03bd \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03c4\u03ce\u03c1\u03b1 \u03b8\u03ad\u03bb\u03b5\u03b9 \u03bd\u03b1 \u03b3\u03c1\u03ac\u03c8\u03c9 \u03bc\u03b9\u03b1 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7,\u03b1\u03c2 \u03bc\u03bf\u03c5 \u03c4\u03bf \u03c0\u03b5\u03b9. :wink: \r\n\r\n\u0397 \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7 \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03b1\u03c0\u03b5\u03c5\u03b8\u03b5\u03af\u03b1\u03c2 \u03b1\u03c0\u03cc \u03c4\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1,\u03b1\u03c6\u03bf\u03cd \u03b1\u03bd $p$ \u03c0\u03c1\u03ce\u03c4\u03bf\u03c2 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 $4k+1$ \u03c4\u03cc\u03c4\u03b5 \u03b3\u03c1\u03ac\u03c6\u03b5\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03bc\u03bf\u03bd\u03b1\u03b4\u03b9\u03ba\u03cc \u03c4\u03c1\u03cc\u03c0\u03bf \u03c9\u03c2 \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03b4\u03cd\u03bf \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03ce\u03bd\u03c9\u03bd,\u03b5\u03bd\u03ce \u03b1\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 $4k+3$ \u03c4\u03cc\u03c4\u03b5 \u03b1\u03c0\u03bb\u03ac \u03b4\u03b5\u03bd \u03b3\u03c1\u03ac\u03c6\u03b5\u03c4\u03b1\u03b9 \u03c9\u03c2 \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03b4\u03cd\u03bf \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03ce\u03bd\u03c9\u03bd :P . \u0391\u03bd \u03c4\u03ce\u03c1\u03b1 $p=2$ \u03c4\u03cc\u03c4\u03b5 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 \u03bf $p$ \u03b4\u03b5\u03bd \u03b3\u03c1\u03ac\u03c6\u03b5\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03b4\u03cd\u03bf \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03c4\u03c1\u03cc\u03c0\u03bf\u03c5\u03c2 \u03c9\u03c2 \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03b4\u03cd\u03bf \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03ce\u03bd\u03c9\u03bd. \u0386\u03c1\u03b1, \u03bf $p$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03cd\u03bd\u03b8\u03b5\u03c4\u03bf\u03c2.\r\n\r\nY.\u0393. \u039a\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03bc\u03bf\u03c5 \u03c0\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03cc\u03c4\u03b9 \u03b8\u03ad\u03bb\u03b5\u03b9 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c3\u03c4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b5\u03b3\u03ac\u03bb\u03b7 \u03ba\u03b1\u03b9 \u03b2\u03b1\u03c1\u03b9\u03ad\u03bc\u03b1\u03b9... :P :lol: . \r\n\r\nStelios..." } { "Tag": [ "function", "topology", "real analysis", "real analysis unsolved" ], "Problem": "Can anyone help me with this problem? \r\n\r\nLet $S$ be a subset of the metric space $E$ with the property that each point of $S^{c}$ (complement of S) is a cluster point of $S$. Let $E'$ be a complete metric space and $f: S \\to E'$ a uniformly continuous function. Prove that $f$ can be extended to a continuous function from $E$ into $E'$ in one and only one way, and that this extended function is also uniformly continuous. \r\n\r\nI don't get the initial definition: how can some cluster point of $S$ also be in $S^{c}$?\r\n\r\nCan anyone give some pointers?", "Solution_1": "[quote=\"teacupontheridge\"] I don't get the initial definition: how can some cluster point of $S$ also be in $S^{c}$?\n[/quote] \r\nLet $(\\mathbb{R}, d)$ be a metric space with $d$ as the usual distance function, i.e. if $x, y \\in \\mathbb{R},$ then $d(x,y) = |x-y|.$ Now, let $S =$the set of all rational numbers $\\subset \\mathbb{R}.$ Then, $S^{c}=$ the set of all irrationals. Clearly, each point of $S^{c}$ is a cluster point of $S.$\r\n\r\nMore importantly, a cluster point of $S$ need not necessarily be in $S.$", "Solution_2": "Thanks, but can anyone give pointers for showing that $f$ can be extended to a continuous function from $E$ into $E'$ in one and only one way?", "Solution_3": "hmmm... i think there's a thm as such: $f: X\\to Y$ is continuous at $x$ if and only if for every sequence $\\{x_{n}\\}$ of $X$ such that $x_{n}\\to x$, $f(x_{n})\\to f(x)$.\r\n\r\nthis should help q a bit." } { "Tag": [ "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "find all function $f:R->R$ such that:\r\nfor every real series $\\sum u_n$, if $\\sum u_n$ converges then $\\sum f(u_n)$ converges", "Solution_1": "This has been posted before. Only functions $f(x)$ which are locally equal to $ax$ near 0 for some $a$ work.", "Solution_2": "I remember I posted it a long time ago but I've never seen a proof for it ..", "Solution_3": "But if you have a proof for it, then you must be happy, isn't it? Here are the main steps:\r\n 1) Prove that it is odd on a neighbourhood of 0. If not, consider a sequence of the form $ a_0, -a_0, a_0, -a_0,...,a_1,-a_1,a_1,-a_1,...$ where $a_n$ is decreasing to 0 and verifies $ |f(a_n)+f(-a_n)|>0$ and take each group $a_i,-a_i,a_i,-a_i,...$ such that $a_i, -a_i$ appears $N_i$ times and $N_i |f(a_i)+f(-a_i)|>1$. \r\n 2) Prove that f is continuous in 0. This is trivial.\r\n 3) Prove that f is additive on a neighbourhood of 0. If not, consider sequences of the form $ a_0+b_0,-a_0,-b_0,a_0+b_0,-a_0,-b_0,... $ of the same type as in 1), use the result in 1) and conclude.", "Solution_4": "harazi ?\r\n\r\nWhy is the continuity in 0 trivial ?", "Solution_5": "Okay, I found out for the continuity.\r\n\r\nI have additiveness on a neighboorhood of zero and continuity in zero.\r\n\r\nThe only thing that I miss is to prove that f is continuous on some neighboorhood of zero, and then it's over.\r\n\r\nI know there is a theorem which states that if f is additive on R and contiuous in one point, then it is linear on all R, but a can't figure out how to modify this theorem so that it works on the neighborhood...", "Solution_6": "If $x$ is in the neighborhood, then $f_x(t)=f(x+t)=f(x)+f(t)$ is continuous at $t=0$, so $f$ is continuous at $x$.", "Solution_7": "Actually, even if it is ``trivial'', I do have a problem for the continuity in zero :blush: :mad:.\r\n\r\nCould anyone give a bit more details ?", "Solution_8": "If $f$ is not continuous in 0 you can find a decreasing sequence $a_n$ which tends to 0 such that $|f(a_n)-f(0)|\\geq \\epsilon$.But since $f(0)=0$ you have $|f(a_n)|\\geq \\epsilon$.Now choose the serie with the following terms:$a_1,-a_2,....,a_{2n-1},-a_{2n},...$and conclude.", "Solution_9": "Sorry in advance for the stupid question:\r\n[quote=\"jmerry\"]Only functions $f(x)$ which are locally equal to $ax$ near 0 for some $a$ work.[/quote]\r\nDo you mean $\\lim_{x \\rightarrow 0} \\frac{f ( x )}{x} = a$ ?", "Solution_10": "no, it means here that there exists some $\\epsilon > 0$ such that $f(x)=ax$ on $]- \\epsilon; + \\epsilon[$", "Solution_11": "$\\epsilon$-neighborhood ;) ? Ok, right ! Thanks." } { "Tag": [ "ratio", "MATHCOUNTS", "AMC", "AIME", "USA(J)MO", "USAMO", "geometry" ], "Problem": "What's your dumbest stupid mistake?\r\n\r\nMine would have to be getting the correct answer but changing the answer when I transferred it from my scrap paper to my answer sheet.", "Solution_1": "My most dumbest stupid mistake (that i can remember) was question 10 the amc 10a. i still can't believe i got the wrong answer!!!. :mad:", "Solution_2": "my dumbest mistake happened in a previous competition back when I was in middle school.\r\n\r\nI bubbled in the right answer for the wrong problem! :blush:", "Solution_3": "7*7=7 :o", "Solution_4": "On a similar note, I said on a number of occasions 16*5 = 90", "Solution_5": "Chapter CountDown Last Year\r\n\r\nFinal Round\r\n\r\nSudden Victory\r\n\r\nQuestion: 5-x=8\r\n\r\nAnswer: x= -8", "Solution_6": "Last year my friends and I were quizzing each other before State Mathcounts.\r\nChris asks me what the ratio of the shortest diagonal in a regular hexagon to the longest diagonal is.\r\nI automatically say sqrt3/2.\r\nIn Sudden Victory in the countdown the question is what is the ratio of the shortest diagonal in a regular hexagon to the longest diagonal.\r\nI buzz in and say sqrt3/3.", "Solution_7": "Ouch JMI, sorry about that. I didn't have the luxury of going to a school that offered MathCounts and hadn't heard about it until 5 months ago.\r\n\r\nThe question asks for A:B, I gave B:A, where A and B are correct.\r\n\r\nI forgot the 3s in the expansion of (a+b)^3.\r\n\r\nAt my school, we have someone who makes stupid mistakes all the time. We now call the act of making such mistakes \"pulling a (person's name here)\".", "Solution_8": "[quote=\"mcalderbank\"]On a similar note, I said on a number of occasions 16*5 = 90[/quote]\r\n\r\nWow, I do that all the time.. \r\n\r\nOnce in a test I found one solution to an equation, went on to prove there were no more by assuming that x/x = 0, and concluded by saying there was no solutions at all..\r\n\r\nOr when I publicly said on this forum that some particular number ending with 5 was a prime..", "Solution_9": "Hmm...does state MC count?\r\n\r\nNumber 1...10 - 2 = 6\r\n\r\nWent from like 3rd to 10th place...", "Solution_10": "AHSME 1999 #4 (paraphrased)\r\nFind the sum of all primes less than 100 which leave a remainder of 1 when divided by 4 and a remainder of 4 when divided by 5.\r\n\r\nMy thought process: Okay, that's 9 mod 20. 9 is not prime, 29 is prime, 49 is not prime, 69 is prime, 89 is prime. 29+69+89=187. That answer was available.\r\n\r\nI checked it later, and went through the same thoughts to verify my answer.\r\n\r\n\r\n\r\n\r\n\r\nI got the other 29 questions right.", "Solution_11": "4(x2 +4x+4)\r\n= 4x2 +4x+4", "Solution_12": "24/3=3. Gah!\r\n\r\nAnd 27-8=15.\r\n\r\nBoth mistakes cost me dearly. :(", "Solution_13": "ok well there's probably a stupider one, but the previous one is the last stupid mistake i remember... :blush: :D ;)", "Solution_14": "MathCounts. When I read my answer out loud at countdown, I had the right answer on paper, but read the wrong thing", "Solution_15": "In 8th grade:\r\n21 --> 4 because the bottom line on the 2 went too far and crossed the 1, then on the next line I wrote a 4.\r\n\r\nMore recently: on my calculus tests, there are calculator and non-calculator parts. On the calculator part, one problem asked to find a definite integral of the function ln(xe^x) (before we had learned how to integrate ln(u)), and i spent about 10 minutes trying to find the antiderivative bfore giving up. That was the only problem on the test I got wrong.", "Solution_16": "Trying to eat a whole jar of pickles in one day...not really\r\n\r\nSaying that 91 was the largest integer prime less than 100 :( forgot 97, and $7*13$ :rotfl:", "Solution_17": "$x+6=24$, $x=4$. :dry:", "Solution_18": "saying that pi/6 rad is 60 degrees (ok its not that bad)\r\n\r\nbut it was a costly mistake, on this year's AIME I", "Solution_19": "School math test (thankfully not state mc):\r\n\r\n6+2=7\r\n\r\n:cursing::wallbash:", "Solution_20": "My best one (on AIME, though I caught it): 0*n=n", "Solution_21": "Got 20th instead of 8th because of:\r\n\r\nI thought an angle equaled 49 when it was complementary to it (misread)", "Solution_22": "Countdown round:\r\nNot paying attention and saying the same wrong answer as the one my opponent said 20 seconds earlier :oops: . Didn't even know until my dad told me, which was after the awards ceremony.\r\n\r\nLuckily, it didn't cost me much.", "Solution_23": "3*3 = 6, I argued with someone for 2 minutes over that once :blush:", "Solution_24": "i think i put that 0.75=25% in a school math test :blush:", "Solution_25": "my dumbest mistake is when i do my homework fast and misread a $\\times$ sign for a $+$ sign", "Solution_26": "last geometry test-I did 16*4=16 for like 10 problems.", "Solution_27": "My dumbest mistake as of now would be tied between misbubbling two problems(got them right, bubbled them wrong), reading a 90 element subset as a three element one (wonder how i pulled that one off), and thinking .012>.123, all of which were made on the AIME this year, so my score right now is just below the predicted cutoff for USAMO.", "Solution_28": "3*9=18 twice", "Solution_29": "Mine would have to be... joining AoPS." } { "Tag": [ "inequalities", "trigonometry", "inequalities proposed" ], "Problem": "[quote=\"Virgil Nicula\"]Let $ABC$ be a triangle. Define $2p=a+b+c$, the circumradius $R$, the inradius $r$,\nthe A-exinradius $r_a$ and the lengths $m_a$, $l_a$, $h_a$ of the median, angle-bisector and altitude respectively from the vertex $A$.\n\n\\begin{eqnarray} & \\Longrightarrow & \\left(A-90^{\\circ}\\right)\\cdot\\left[m_a-R(1+\\cos a)\\right]\\ge 0\\\\\\ & \\Longrightarrow & \\frac{a(p-a)}{\\sqrt{2Rr}}\\le \\frac{b+c}{2}\\cos \\frac A2\\le m_a\\\\\\ & \\Longrightarrow & \\sqrt{p(p-a)}\\le m_a\\ \\ \\wedge\\ \\ \\frac{b^2+c^2}{4R}\\le m_a\\\\ & \\Longrightarrow & \\sqrt{p(p-a)}\\le \\frac{a(p-a)}{\\sqrt{2Rr}}\\Longleftrightarrow (a-b)(a-c)\\le 0\\\\ & \\Longrightarrow & \\boxed {(4R+r)\\sqrt{\\frac{2r}{R}}\\le m_a+m_b+m_c}\\le 4R+r\\\\ & \\Longrightarrow & 3\\le \\boxed {\\frac 12\\cdot \\sum \\frac{b+c}{a}\\le \\sum \\frac{m_a}{h_a}}\\le 1+\\frac Rr\\\\ & \\Longrightarrow & 3\\le \\boxed {\\frac{a^3+b^3+c^3}{abc}\\le \\sum \\frac{m_a}{r_a}}\\\\\\ & \\Longrightarrow & \\boxed {3r\\le \\frac{am_a+bm_b+cm_c}{a+b+c}}\\le R+r\\\\ & \\Longrightarrow & \\frac{a+b+c}{2R}\\le \\boxed {\\frac{3}{2R}\\cdot \\frac{a^2+b^2+c^2}{a+b+c}\\le \\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}}\\\\ & \\Longrightarrow & \\frac pR\\le \\boxed {\\frac{p}{\\sqrt{2Rr}}\\le \\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}} \\end{eqnarray}[/quote][b][u]Remark 1.[/u][/b] The non-enclosed inequalities are evidently or well-known and I shall prove them further.\r\n\r\n[b][u]Remark 2.[/u][/b] Prove easily that $\\frac{a(p-a)}{\\sqrt{2Rr}}=(1+\\cos A)\\sqrt{2Rr}\\le m_a$. \r\n\r\n[b][u]Remark 3.[/u][/b] We can prove the inequality $(0)$ with two remarkable identities :\r\n$1^{\\circ} \\blacktriangleright\\ \\boxed {(p-b)(p-c)(b-c)^2+p(p-a)(b+c-2a)^2=4\\left[bcm^2_a-2ap(p-a)^2\\right]}\\Longrightarrow$\r\n$\\boxed {m_a\\ge\\frac{a(p-a)}{\\sqrt{2Rr}}}$ (with equality if and only if $a=b=c$).\r\n\r\n$2^{\\circ} \\blacktriangleright\\ \\boxed {16S^2+\\left(b^2-c^2\\right)^2=4a^2m^2_a}\\Longleftrightarrow a^2(4Rm_a)^2=4(4RS)^2+4R^2\\left(b^2-c^2\\right)^2\\Longleftrightarrow$\r\n$a^2(4Rm_a)^2=4(abc)^2+4R^2\\left(b^2-c^2\\right)^2\\Longleftrightarrow (4Rm_a)^2=4b^2c^2+4R^2\\left(\\frac{b^2-c^2}{a}\\right)^2\\Longleftrightarrow$\r\n$(4Rm_a)^2=\\left(b^2+c^2\\right)^2-\\left(b^2-c^2\\right)^2+4R^2\\left(\\frac{b^2-c^2}{a}\\right)^2\\Longleftrightarrow$\r\n$(4aRm_a)^2=\\left[a\\left(b^2+c^2\\right)\\right]^2+\\left(4R^2-a^2\\right)\\left(b^2-c^2\\right)^2\\Longrightarrow \\boxed {4Rm_a\\ge b^2+c^2}$\r\n\r\n(with equality if and only if $b=c\\ \\vee \\ A=90^{\\circ}$).\r\n\r\n$3^{\\circ} \\blacktriangleright$ From the identity $(1^{\\circ})$ we obtain $4bcm^2_a\\ge 8ap(p-a)^2+p(p-a)(b+c-2a)^2\\Longleftrightarrow$\r\n\r\n$\\frac{4bcm^2_a}{p(p-a)}\\ge (b+c)^2\\Longleftrightarrow \\boxed {m_a\\ge\\frac{b+c}{2}\\cos \\frac A2}$ (with equality if and only if $b=c$).\r\n\r\n[b][u]Remark 4.[/u][/b] $\\boxed {\\frac{a(p-a)}{\\sqrt{2Rr}}\\le \\frac{b+c}{2}\\cos \\frac A2}\\Longleftrightarrow 2a(p-a)\\le (b+c)\\sqrt{2Rr}\\cos \\frac A2\\Longleftrightarrow$\r\n\r\n$2a(p-a)\\le (b+c)\\sqrt{\\frac{2Rrp(p-a)}{bc}}\\Longleftrightarrow 2a(p-a)\\ge (b+c)\\sqrt{\\frac{2Rpra(p-a)}{abc}}\\Longleftrightarrow$\r\n\r\n$\\Longleftrightarrow 2a(p-a)\\le (b+c)\\sqrt{\\frac{a(p-a)}{2}}\\Longleftrightarrow$ $2\\sqrt{a(b+c-a)}\\le b+c\\Longleftrightarrow (b+c-2a)^2\\ge 0$\r\n\r\n(with equality if and only if $b+c=2a$).\r\n\r\n[b][u]Remark 5.[/u][/b] $\\boxed {1\\ge\\cos \\frac{B-C}{2}\\ge \\sqrt{\\frac{2r}{R}}}\\Longleftrightarrow 2\\sin \\frac{B+C}{2}\\cos \\frac{B-C}{2}\\ge 2\\sqrt{\\frac{2r}{R}}\\cos \\frac A2\\Longleftrightarrow$\r\n\r\n$\\sin B+\\sin C\\ge \\sqrt{\\frac{8rp(p-a)}{Rbc}}\\Longleftrightarrow b+c\\ge \\sqrt{\\frac{32Rpr(p-a)}{bc}}\\Longleftrightarrow$\r\n\r\n$b+c\\ge \\sqrt{\\frac{32Rpra(p-a)}{4Rpr}}\\Longleftrightarrow b+c\\ge 2\\sqrt{a(b+c-a)}\\Longleftrightarrow (b+c-2a)^2\\ge 0\\ .$\r\n\r\n(with equality if and only if $b+c=2a$).\r\n\r\n[b]Another (geometrical) proof.[/b] Denote $w=C(O,R)$, $A_1\\in (AI\\cap w$, $A_2\\in (A_1O\\cap w$, i.e. $A_1A_2=2R$, $A_2A\\perp A_1A$. Therefore, $\\cos \\frac{B-C}{2}=$ $\\frac{A_1A}{A_1A_2}=$ $\\frac{IA+IA_1}{2R}\\ge$ $\\frac{2\\sqrt{IA\\cdot IA_1}}{2R}=$ $\\frac{\\sqrt{2Rr}}{R}=$ ${\\sqrt{\\frac{2r}{R}}\\ .}$", "Solution_1": "Ineq:\r\n$\\Large \\frac{1}{2}\\sum \\frac{b+c}{a}\\le \\sum \\frac{m_a}{h_a}$ is real hardy.\r\nCan you post your solution,thank :P .", "Solution_2": "Maybe later. The all above non-enclosed inequalities have a common spring: the primitive inequality $(0)$.", "Solution_3": "is it true that \r\n$\\sum \\frac{1}{\\sqrt{p-a}} \\sum \\sqrt{p-a} \\geq \\sum \\frac{1}{a} \\sum a$\r\nbecause if it is we have a proof of three by using $m_a \\geq \\sqrt{p(p-a)}$" } { "Tag": [ "\\/closed" ], "Problem": "Looking at some quotes about IMO 2003, I see guests posting....... how's that?", "Solution_1": "[quote=\"Jos\u00e9\"]Looking at some quotes about IMO 2003, I see guests posting....... how's that?[/quote]\r\nWhat's wrong with that? Why can't guests have some place to post?", "Solution_2": "There might be some sections in the forum where this is allowed", "Solution_3": "Also, that was almost three years ago. Mathlinks was different then.", "Solution_4": "[quote=\"shobber\"][quote=\"Jos\u00e9\"]Looking at some quotes about IMO 2003, I see guests posting....... how's that?[/quote]\nWhat's wrong with that? Why can't guests have some place to post?[/quote]\r\n\r\nNothing wrong. But if you want to post, why don't you create an account?????\r\n\r\nAnd why in 2003 you could and now you can not?", "Solution_5": "Maybe it was like that so people could post without going through the account-making process and then people started to spam too much so they changed the rule.", "Solution_6": "[quote=\"amcavoy\"]Maybe it was like that so people could post without going through the account-making process and then people started to spam too much so they changed the rule.[/quote]\r\n\r\nYes, that would be a good reason then" } { "Tag": [ "geometry", "3D geometry", "summer program", "MathPath" ], "Problem": "a 4x4x4 in cube originally built from 1x1x1 in cubes is cut into exactly 29 cubes with integer edge leghths and no material left over. how any of the 29 cubes are 2x2x2 cubes?", "Solution_1": "[quote=\"espark52\"]a 4x4x4 in cube originally built from 1x1x1 in cubes is cut into exactly 29 cubes with integer edge leghths and no material left over. how any of the 29 cubes are 2x2x2 cubes?[/quote]\r\n[hide=\"hint\"]systems[/hide]", "Solution_2": "[hide]Um....... I think the answer is 5. [/hide]", "Solution_3": "[hide]Let $x$ be the number of 1x1x1 cubes and $y$ be the number of 2x2x2 cubes. Then we have\n\n$x+y=29$\n$x+8y=64$\n\nThis gives $y=5$ and $x=24$, so the answer is $5$.[/hide]", "Solution_4": "yes, the answer is five :lol:", "Solution_5": "I figured out a solution for the 4x4x4 all by myself at MathPath!!!!\r\n\r\nOf course, it only worked about 1/4 of the time cuz of the parity case which I couldn't figure out...", "Solution_6": "Another way to think about it is this:\r\n[hide]If you had all 2x2x2 cubes, you'd have 8 cubes right?\nSo then for every 2x2x2 cube that you take off, you make 8 1x1x1 cubes so you add 7 every time. Start with 8 cubes (8 2x2x2's), then 15 cubes (7 2x2x2's), 22 cubes (6 2x2x2's) and finally 29 cubes (5 2x2x2's and 24 1x1x1's)[/hide]" } { "Tag": [ "search", "LaTeX" ], "Problem": "How can I make a square with shadow?", "Solution_1": "Search here: http://mirror.hmc.edu/ctan/info/symbols/comprehensive/symbols-letter.pdf\r\n\r\nIf you can't find it here, it probably doesn't exist.", "Solution_2": "With the fancybox package, you can type something like \\shadowbox{hello}. I don't know if that's what you were looking for." } { "Tag": [ "inequalities", "LaTeX", "inequalities unsolved" ], "Problem": "prove that for every x1,x2,...,xn are positive real , we have:\r\n$ \\sum cos^2(x_i \\minus{} x_j) \\geq \\frac {n(n \\minus{} 2)}{4}$\r\nplz help :(", "Solution_1": "[quote=\"long14893\"]prove that for every x1,x2,...,xn are positive real , we have:\n$ \\sum cos^2(x_i \\minus{} x_j) \\geq \\frac {n(n \\minus{} 2)}{4}$\nplz help :([/quote]\r\nHello....no one's gonna help me ??? :((", "Solution_2": "could you explain the question properly? :) \r\n\r\nand use late3x preferably\r\nfor subscripts use eg: u_i in dollar signs = $ u_i$", "Solution_3": "sr ... but I think t used Latex :-/ ?\r\nabout my question....what makes u hard to understand it :-/?\r\nFor example , if n=3 , we need to prove:\r\n$ cos^2(x_1\\minus{}x_2) \\plus{} cos^2(x_2\\minus{}x_3) \\plus{} cos^2(x_3\\minus{}x_1)^2 \\geq 3/4$\r\nplz help me :| i really need the solution of this question >\"<", "Solution_4": "[quote=\"long14893\"]prove that for every x1,x2,...,xn are positive real , we have:\n$ \\sum cos^2(x_i \\minus{} x_j) \\geq \\frac {n(n \\minus{} 2)}{4}$\nplz help :([/quote]\r\n It should be \r\n\r\n$ \\sum_{1\\leq i < j \\leq n } cos^2(x_j \\minus{} x_i) \\geq \\frac {n(n \\minus{} 2)}{4}$ instead of $ \\sum cos^2(x_i \\minus{} x_j) \\geq \\frac {n(n \\minus{} 2)}{4}$ :)", "Solution_5": "[quote=\"HTA\"][quote=\"long14893\"]prove that for every x1,x2,...,xn are positive real , we have:\n$ \\sum cos^2(x_i \\minus{} x_j) \\geq \\frac {n(n \\minus{} 2)}{4}$\nplz help :([/quote]\n It should be \n\n$ \\sum_{1\\leq i < j \\leq n } cos^2(x_j \\minus{} x_i) \\geq \\frac {n(n \\minus{} 2)}{4}$ instead of $ \\sum cos^2(x_i \\minus{} x_j) \\geq \\frac {n(n \\minus{} 2)}{4}$ :)[/quote]\r\nyes ... that's exactly what I wanted to say .... cuz I forget how to use Latex to write like that :D\r\n\r\n[hide] l\u00e0m ra ch\u01b0a ... n\u1ea3n qu\u00e1 :(( t\u01b0\u1edfng n\u00f3 t\u1ed1t b\u1ee5ng =)) [/hide]", "Solution_6": "[quote=\"long14893\"][quote=\"HTA\"][quote=\"long14893\"]prove that for every x1,x2,...,xn are positive real , we have:\n$ \\sum cos^2(x_i \\minus{} x_j) \\geq \\frac {n(n \\minus{} 2)}{4}$\nplz help :([/quote]\n It should be \n\n$ \\sum_{1\\leq i < j \\leq n } cos^2(x_j \\minus{} x_i) \\geq \\frac {n(n \\minus{} 2)}{4}$ instead of $ \\sum cos^2(x_i \\minus{} x_j) \\geq \\frac {n(n \\minus{} 2)}{4}$ :)[/quote]\nyes ... that's exactly what I wanted to say .... cuz I forget how to use Latex to write like that :D\n\n[hide] l\u00e0m ra ch\u01b0a ... n\u1ea3n qu\u00e1 :(( t\u01b0\u1edfng n\u00f3 t\u1ed1t b\u1ee5ng =)) [/hide][/quote]\r\nok , here is my solution :\r\nFirstly , notice that :\r\n$ \\sum_{i,j \\equal{} 1}^{n} cos^2(x_i \\minus{} x_j) \\equal{} 2 \\sum_{1 \\leq i,j\\leq n}^{n}cos^2(x_i \\minus{} x_j) \\plus{} n$\r\nHence , we need to prove \r\n$ \\sum_{i,j \\equal{} 1}^{n} cos^2(x_i \\minus{} x_j) \\minus{} n \\geq \\frac {n(n \\minus{} 2)}{2}$\r\n$ \\sum_{i,j \\equal{} 1}^{n}(\\frac {1 \\plus{} cos(2x_i \\minus{} 2x_j)}{2}) \\minus{} n \\geq \\frac {n(n \\minus{} 2)}{2}$\r\n$ \\sum_{i,j \\equal{} 1}^{n}(1 \\plus{} cos(2x_i \\minus{} 2x_j)) \\minus{} 2n \\geq n^2 \\minus{} 2n$\r\n$ \\sum_{i,j \\equal{} 1}^{n}cos(2x_i \\minus{} 2x_j) \\geq 0$\r\n$ (\\sum_{i \\equal{} 1}^{n}cos2x_i)^2 \\plus{} (\\sum_{j \\equal{} 1}^{n}sin2x_j)^2 \\geq 0$\r\nwhich is true :)" } { "Tag": [ "pigeonhole principle", "combinatorics unsolved", "combinatorics" ], "Problem": "Let $ A$ be a subset of $ \\{ 0,1,2,...,1997 \\}$ containing more than $ 1000$ elements. Prove that either $ A$ contains a power of $ 2$ (that is, a number of the form $ 2^k$ with $ k\\equal{}0,1,2,...)$ or there exist two distinct elements $ a,b \\in A$ such that $ a\\plus{}b$ is a power of $ 2$.", "Solution_1": "[quote=\"moldovan\"]Let $ A$ be a subset of $ \\{ 0,1,2,...,1997 \\}$ containing more than $ 1000$ elements. Prove that either $ A$ contains a power of $ 2$ (that is, a number of the form $ 2^k$ with $ k \\equal{} 0,1,2,...)$ or there exist two distinct elements $ a,b \\in A$ such that $ a \\plus{} b$ is a power of $ 2$.[/quote]\r\n\r\nAssume that we can find a counterexample $ A$ , which should be a subset of $ \\{0,1,2,...,1997\\}$ \\ $ \\{2,4,8,...,1024\\}$\r\n\r\nThen divide $ \\{0,1,2,...,1997\\}$ \\ $ \\{2,4,8,...,1024\\}$ into $ \\{1997,51\\}, \\{1996,52\\}, \\{1995,53 \\},...,\\{1025,1023 \\},$\r\n$ \\{50,14\\}, \\{49,15\\}, \\{48\\}, \\{47,17\\}, ..., \\{33,31\\},$\r\n$ \\{13,3\\}, \\{12\\},\\{11,5\\},...,\\{9,7\\}, \\{1\\}, \\{0\\}$ 998 groups.\r\n\r\nSo by pigeonhole principle, we can find two distinct elements of $ A$ in one group and so sum of the two elements is a power of $ 2$." } { "Tag": [ "inequalities", "induction" ], "Problem": "Show that\r\n\r\n$ n\\left(1 \\plus{} \\frac {1}{2} \\plus{} \\frac {1}{3} \\plus{} ... \\plus{} \\frac {1}{n}\\right) \\geq \\left(n \\plus{} 1\\right)\\left(\\frac {1}{2} \\plus{} \\frac {1}{3} \\plus{} ... \\plus{} \\frac {1}{n \\plus{} 1}\\right)$\r\n\r\nfor all natural numbers $ \\,\\,n \\geq 1$\r\n\r\n\r\nSee i have made progress, i just dont know how to finnish it off. This is what i did.\r\n\r\n$ n \\plus{} n\\left( \\frac {1}{2} \\plus{} \\frac {1}{3} \\plus{} ... \\plus{} \\frac {1}{n}\\right) \\geq \\left(n \\plus{} 1\\right)\\left(\\frac {1}{2} \\plus{} \\frac {1}{3} \\plus{} ... \\plus{} \\frac {1}{n}\\right) \\plus{} \\frac {n \\plus{} 1}{n \\plus{} 1}$\r\n\r\n$ n \\plus{} n\\left( \\frac {1}{2} \\plus{} \\frac {1}{3} \\plus{} ... \\plus{} \\frac {1}{n}\\right) \\geq n\\left(\\frac {1}{2} \\plus{} \\frac {1}{3} \\plus{} ... \\plus{} \\frac {1}{n}\\right) \\plus{} 1\\left(\\frac {1}{2} \\plus{} \\frac {1}{3} \\plus{} ... \\plus{} \\frac {1}{n}\\right) \\plus{} 1$\r\n\r\n(Subtracting $ n\\left( \\frac {1}{2} \\plus{} \\frac {1}{3} \\plus{} ... \\plus{} \\frac {1}{n}\\right)$ from both sides)\r\n\r\n$ n \\geq 1 \\plus{} \\frac {1}{2} \\plus{} \\frac {1}{3} \\plus{} ... \\plus{} \\frac {1}{n}$\r\n\r\nHow do i prove this? What do i do to finnish my proof?", "Solution_1": "To get to:\r\n\r\n$ n\\geq 1+\\frac{1}{2}+\\frac{1}{3}+...+\\frac{1}{n}$\r\n\r\nYou can express $ n$ as the sum of $ n$ one's: $ n=\\underbrace{1+\\cdots+1}_{n}$.\r\n\r\nAnd for each $ n\\ge k\\ge 1$ : $ 1\\ge \\dfrac{1}{k}$,(equality in $ k=1$) you add all of those up and get the result you want.\r\n\r\nIn other words\r\n\r\n$ \\sum_{k=1}^n 1 \\ge \\sum_{k=1}^n \\dfrac{1}{k}$\r\n\r\nExample: $ n=2$,\r\n\r\n\\begin{eqnarray*}\r\n1&\\ge& 1\\\\\r\n1 &>& \\dfrac{1}{2}\r\n\\end{eqnarray*}\r\n\r\n$ 1+1\\ge 1+\\dfrac{1}{2}\\implies 2\\ge 1+\\dfrac{1}{2}$", "Solution_2": "Since the opposite is not true, the equation you have written is. If $ n \\equal{} 2$, then we would need $ 1 \\plus{} \\frac {1}{2}$ to be greater than or equal to $ 2$. But it is not. Finally, since the integers increase by $ 1$ on the LHS, and on the RHS, the fractions increase by $ \\frac {1}{n}$, we are actually trying to prove $ n$ is greater than or equal to $ \\frac {1}{n}$ for every natural number $ n$. But this is always true.\r\n\r\nI hope you understood my proof.", "Solution_3": "Thanks.\r\n\r\n\r\nBut if you writing a proof for a contest what would you write to finnish it off?\r\n\r\nie is it satisfactory to state that this is obviously true?", "Solution_4": "No; prove by induction. $ 1\\ge 1/1$; $ 1>1/k$ for all $ k$ (which is what you add to both sides when you go from $ k\\minus{}1$ to $ k$), hence the inequality is true.", "Solution_5": "I'm sorry; I'm not very good at writing proofs. I thought you didn't know how to solve it and that it was a casual question, so I just posted a proof for it.", "Solution_6": "Another way is induction:\r\n\r\n1. It's true for $ n \\equal{} 1$.\r\n\r\n2. Assume it's true for $ n$\r\n$ n\\ge 1 \\plus{} \\dfrac{1}{2} \\plus{} \\cdots \\plus{} \\dfrac{1}{n}$ $ (1)$\r\n\r\n3. Prove it's true for $ n \\plus{} 1$:\r\n\r\n$ n \\plus{} 1\\ge 1 \\plus{} \\dfrac{1}{2} \\plus{} \\cdots \\plus{} \\dfrac{1}{n} \\plus{} \\dfrac{1}{n \\plus{} 1}$ $ (2)$\r\n\r\nSubtract $ 1$ from $ 2$.\r\n\r\n$ 1\\ge \\dfrac{1}{n \\plus{} 1}$\r\n\r\n$ n \\plus{} 1\\ge 1$ Which is true for all natural numbers.\r\n\r\nEdit: oops didn't read Temperal's post", "Solution_7": "I considered induction and tried it but i just didnt think properly ;(\r\n\r\nThanks heaps :lol:", "Solution_8": "Most of the time induction gets pretty messy!!!!!!!!! :(", "Solution_9": "[quote=\"tonypr\"]\nSubtract $ 1$ from $ 2$.\n[/quote]\r\n\r\nNup, you can't do that !!\r\n\r\nFortunately it's fixable. Just add $ 1 \\ge \\frac{1}{1\\plus{}n}$ to $ (1)$ to get $ (2)$" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Why A contains at least two elements denoted by red line ?\r\n\r\nThanks", "Solution_1": "If $ A$ contains just one point(for example $ a$) then our sequence $ x_n$ converges to $ a$ and we are done :)", "Solution_2": "[quote=\"leshik\"]If $ A$ contains just one point(for example $ a$) then our sequence $ x_n$ converges to $ a$ and we are done :)[/quote]\r\n\r\nThanks.\r\n\r\nAlthough I might had thought of it , but I didn't.\r\n\r\nWhen read books of math , we need to discuss and trigger off mutually.", "Solution_3": "in (i), why $ f(\\ell) \\neq \\ell$ ? I see. if $ f(\\ell) \\equal{} \\ell$, we will get contradition.\r\n\r\nThen in (ii) , why $ [\\ell_{\\_}\\plus{}f(\\ell_{\\_})]/2\\in A$ ?" } { "Tag": [ "inequalities", "AMC", "USA(J)MO", "USAMO", "inequalities proposed" ], "Problem": "Let 0 3$ so $ (k^2 \\minus{} 3)^2 < k^4 \\minus{} 6k^2 \\plus{} 4k \\minus{} 3 < (k^2 \\minus{} 2)^2$ is a contradiction. If $ k \\equal{} 1$ or $ k \\equal{} 2$ so $ m^2 < 0$ , $ k \\equal{} 3$ $ \\Longrightarrow$ $ m \\equal{} 6$ and $ p \\equal{} 19$ , $ q \\equal{} 17$ is only solution.\r\n$ Remark$: If $ q$ is not a prime number, the problem is also true($ p,q >1$).The only solution is $ p \\equal{} 19$, $ q \\equal{} 17$.", "Solution_2": "http://www.mathlinks.ro/Forum/viewtopic.php?p=224998#224998", "Solution_3": "I think $ p\\equal{}19$ and $ q\\equal{}7$ : :wink:", "Solution_4": "Sorry :blush: $ 343\\equal{}7^3$ :lol:" } { "Tag": [ "trigonometry", "inequalities", "inequalities unsolved" ], "Problem": "1.x,y>0.Prove\r\n$\\frac{x}{\\sqrt{x^{2}+3y^{2}}}+\\frac{y}{\\sqrt{y^{2}+3x^{2}}}+\\frac{2xy}{\\sqrt{x^{2}+3y^{2}}\\sqrt{y^{2}+3x^{2}}}\\le\\frac{3}{2}$\r\n2.a,b,c>0.Prove\r\n$(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})(\\frac{1}{1+a}+\\frac{1}{1+b}+\\frac{1}{1+c})\\ge\\frac{9}{1+abc}$", "Solution_1": "1) We can put $\\cos A =\\frac{x}{\\sqrt{x^2+3y^2}}$ ; \r\n $\\cos B = \\frac{y}{\\sqrt{y^2+3x^2}}$ ; \r\n $\\cos C =\\frac{2xy}{\\sqrt{(x^2+3y^2)(y^2+3x^2)}}$ \r\n because $\\cos^2A+\\cos^2A+\\cos^2A+2{\\cosh}{\\cosh}{\\cosh} =1$ \r\n 2) Use the well-known ineq :\r\n $\\frac{1}{a(1+b)}+\\frac{1}{b(1+c)}+\\frac{1}{c(1+a)} \\geq \\frac{3}{1+abc}$", "Solution_2": "$cos^2(A)+cos^2(B)+cos^2(C)=-1$? :rotfl:", "Solution_3": "[quote=\"tkhtn\"]$cos^2(A)+cos^2(B)+cos^2(C)=-1$? :rotfl:[/quote]\r\nthere may be a small mistake,I think nttu may mean $cos^2(A)+cos^2(B)+cos^2(C)+2\\cos A \\cos B \\cos C=1$\r\n :)", "Solution_4": "I think the first inequality can be solved by Cauchy.If anyone can do it, please post here", "Solution_5": "To nttu:For the second inequality, can you post all your solution in details?", "Solution_6": "The second ineq is very old and easy!\r\nPlease read ineq agains.", "Solution_7": "Please solve the first ineq by Cauchy,Bunhiacopki,Schwarz,AM-GM or Chebusev" } { "Tag": [ "induction", "trigonometry", "limit", "calculus" ], "Problem": "Let $a_{1},a_{2},...$ be positive numbers such that $a_{n+1}=a_{n}^{2}-2(n=1,2,...)$. Prove that $a_{n}\\geq 2$ for all $n\\geq 1$.", "Solution_1": "[hide=\"complete proof\"]\nIf $a_{i}\\geq 2$, then $a_{i}^{2}\\geq4$ and $a_{i}^{2}-2 = a_{i+1}\\geq 2$.\n\nSo, we just have to show that $a_{1}\\geq 2$. Suppose $a_{1}< 2$. Then $a_{1}^{2}-a_{1}-2 = (a_{1}-2)(a_{1}-1)$\nSuppose $11\\forall n$ :wink:", "Solution_5": "so basically he was saying that it wasn't proven that $a_{i}$ will ever go below 1?", "Solution_6": "Because if $a_{i}<1$ then $a_{i+1}<0$, contradiction!", "Solution_7": "[hide]\nIf $a_{1}\\geq 2$, then $a_{n}$ will always be $\\geq 2$ by induction (as $a_{k}\\geq 2\\Rightarrow a_{k+1}=a_{k}^{2}-2\\geq 2$).\n\nSuppose that $a_{1}<2$ but $a_{n}$ is positive for all $n$. It follows by induction that $a_{n}<2$ for all $n$, so we can write\n\\[a_{n}=2\\cos \\alpha_{n}\\]\nwhere $0<\\alpha_{n}<\\frac{\\pi}{2}$. But we have\n\\[2\\cos \\alpha_{n+1}=a_{n+1}=a_{n}^{2}-2=4\\cos^{2}\\alpha_{n}-2=2\\cos 2\\alpha_{n}\\]\nso $\\alpha_{n+1}=2\\alpha_{n}$. Let $\\theta=\\alpha_{1}$; then it follows that\n\\[a_{n}=2\\cos 2^{n-1}\\theta\\]\nfor all $n$. Since $\\theta>0$ and $2^{n-1}$ gets arbitrarily large for large enough $n$, there will be some $M$ such that $2^{M-1}\\theta \\geq \\frac{\\pi}{2}$. Consider the least such $M$. Then because $2^{M-2}\\theta<\\frac{\\pi}{2}$, $2^{M-1}\\theta <\\pi$. It follows that\n\\[\\frac{\\pi}{2}\\leq 2^{M-1}\\theta<\\pi\\Rightarrow a_{M}\\leq 0\\]\nbut this is a contradiction!\n[/hide]", "Solution_8": "Me too :wink: \r\n[quote=\"scorpius119\"][hide]\nIf $a_{1}\\geq 2$, then $a_{n}$ will always be $\\geq 2$ by induction (as $a_{k}\\geq 2\\Rightarrow a_{k+1}=a_{k}^{2}-2\\geq 2$).\n\nSuppose that $a_{1}<2$ but $a_{n}$ is positive for all $n$. It follows by induction that $a_{n}<2$ for all $n$, so we can write\n\\[a_{n}=2\\cos \\alpha_{n}\\]\nwhere $0<\\alpha_{n}<\\frac{\\pi}{2}$. But we have\n\\[2\\cos \\alpha_{n+1}=a_{n+1}=a_{n}^{2}-2=4\\cos^{2}\\alpha_{n}-2=2\\cos 2\\alpha_{n}\\]\nso $\\alpha_{n+1}=2\\alpha_{n}$. Let $\\theta=\\alpha_{1}$; then it follows that\n\\[a_{n}=2\\cos 2^{n-1}\\theta \\]\nfor all $n$. Since $\\theta>0$ and $2^{n-1}$ gets arbitrarily large for large enough $n$, there will be some $M$ such that $2^{M-1}\\theta \\geq \\frac{\\pi}{2}$. Consider the least such $M$. Then because $2^{M-2}\\theta<\\frac{\\pi}{2}$, $2^{M-1}\\theta <\\pi$. It follows that\n\\[\\frac{\\pi}{2}\\leq 2^{M-1}\\theta<\\pi\\Rightarrow a_{M}\\leq 0 \\]\nbut this is a contradiction!\n[/hide][/quote]\r\nThis is solution of my friends:\r\nIf $1a_{n+1}\\forall n\\geq 1$. Now from above we have $(a_{n})$ convergen. Contradiction! (Because if $L=\\lim_{n\\to\\infty}a_{n}$ then $L=-1$ or $2$).", "Solution_9": "[quote=\"JBL\"][quote=\"Hamster1800\"]Then, $a_{1}^{2}-a_{1}-2 < 0 \\implies a_{1}^{2}-20$ and $min\\left\\{a^{2}-\\frac{8}{a^{4}},b^{2}-\\frac{8}{b^{4}},c^{2}-\\frac{8}{c^{4}}\\right\\}=A-\\frac{8}{A^{2}}$\r\nThus: $A+A+A \\leq a^{2}+b^{2}+c^{2}\\leq 6+min\\left\\{a^{2}-\\frac{8}{a^{4}},b^{2}-\\frac{8}{b^{4}},c^{2}-\\frac{8}{c^{4}}\\right\\}= 6+A-\\frac{8}{A^{2}}$,\r\nAnd by $AM\\geq GM$ : $6\\geq A+A+\\frac{8}{A^{2}}\\geq 3\\sqrt[3]{A \\times A \\times \\frac{8}{A^{2}}}= 6$. As a consequence $2=A=a^{2}=b^{2}=c^{2}$ and again the answers are as Mimoide said..." } { "Tag": [ "probability", "search" ], "Problem": "Hi all! :)\r\n\r\nOne choose a seven digits password at random.\r\nFind the following probabilities :\r\n1) All the digits are distincts\r\n2) The product of those digits is divisible by 2.\r\n3) The digits form :\r\na/ a strictly increasing sequence.\r\nb/ an increasing sequence.\r\n\r\nEnjoy! ;)", "Solution_1": "1) [hide]There are 10 choices for the first digit, 9 choices for the second such that the second is not the same as the first, 8 for third, and so on down to 4 choices for the 7th. (10*9*8*7*6*5*4)/10^7=189/3125\n[/hide]\n2) [hide]We must have at least 1 of the 7 digits a multiple of 2. There are 5 such digits to choose from. We may count the number of passwords whose product is not a multiple of 2 and subtract this from the total to find the number whose product is a multiple of 2. There are 5^7 passwords with a product not divisible by 2, since there are 5 digits not multiples of 2 to choose from and 7 positions possible. The probability of getting one of these is 5^7/10^7=1/2^7=1/128, so the probability that you get one whose product is divisible by 2 is 1-1/128=127/128.[/hide]\n\n3a) [hide]I'm guessing that a strictly increasing sequence means that the common difference is not 0, but there is a common difference. I shall have the solution for the no-common-difference also.\n\nIf we have a common difference of 1, we can have the first term of the sequence any digit from 0 to 3 inclusive, and since each other member is equal to 1 more then the previous member, there are a total of 4 possible successful passwords. Therefore there is 4/10^7=1/2500000 chance of these. (You cannot have a common difference of any other positive real as those differences do not allow for our limited stock of digits)\n\nIf we are strictly increasing, we can have as the first term any digit from 0 to 3 inclusive. There is only 1 sequence for first digit=3 where all increase. There are 7 such for 2. There are 29 such for 1. There are (I think) 67 such for 0. There are a sum of 100 such, giving 100/10^7=1/100000[/hide]\n\n3b) [hide]Assuming one can have a common difference of 0: There are 10 with difference 0 and 4 with difference 1, for 14/10^7=7/1000000\n\nWith variable difference between terms, there are 1 for first digit 9, 7 for 8, 63 for first of 7, 189 for 6, 441 for 5, 903 for 4, 1695 for 3, 2937 for 2, 4789 for 1, and 7920 for 0. At least I hope so, because I've wasted a lot of time otherwise...There are thus 18945 passwords that fulfill our conditions. 18945/10^7=3789/2000000[/hide]\r\n\r\nDid I get some of it right at least? :( I probably messed up somewhere...", "Solution_2": "I think I have got different answers for 3(a) and (b) .\r\n\r\n3(a) [hide]\nFor strictly increasing sequence we just have to select sets of $7$ different numbers. They can be arranged in only one way in strictly increasing order.\nThus the probability is $\\frac{\\binom{10}{7}}{10^7}$\n[/hide]\n\n3(b) [hide]\nFor increasing sequence, we can divide the job as selecting 7 different numbers, 6 different numbers, 5 different numbers,\u2026,1 number.\nNow they have to be arranged in increasing order where numbers can be repeated to get a total of $7$ digits.\nSo, probability of obtaining such a sequence is\n\\[ \\frac{\\binom{7}{7}\\binom{10}{7}+\\binom{7}{6}\\binom{10}{6}+\\cdot\\cdot\\cdot+ \\binom{7}{1}\\binom{10}{1}}{10^7} \\]\n[/hide]\r\n\r\nSomeone can give a better a solution? :oops:", "Solution_3": "Yes another approach\r\n\r\n3(a) \r\n\r\n[hide]The number of such number is also given by the number of solution of \n\n$a+b+c+d+e+f+g \\leq 9$ (seventh digit is $\\leq 9$) with $0 < a,b,c,d,e,f,g \\leq 9$\nOr $a'+b'+c'+d'+e'+f'+g'+h = 2$ with $0 \\leq a',b',c',d',e',f',g',h$\nThis is a \"combinaison with repetitions\" an the number of solutions is $\\binom{9}{7} = 36$\n[/hide]\n3(b)\n\n[hide]Same thing, $a+b+c+d+e+f+g \\leq 9$ with $0 < a, 0 \\leq b,c,d,e,f,g \\leq 9$\n\nThis gives $\\binom{17}{7} = 6435$ numbers[/hide]", "Solution_4": "And the winner is... T\u00b5t\u00b5! :first: \r\n\r\n(at least, according to my own results :P)\r\n\r\nApart from that, solafidefarms, your other results are good! :)\r\n\r\nVarun, I think you misunderstood the questions...", "Solution_5": "Well well well, after I tried to search a bit more about this problem..\r\nI've come to the conclusion that you were right for 3/a/ Varun!\r\n\r\nAnd, t\u00b5t\u00b5, your idea (combinations with repetitions) was good, but, you've made a little error ;). The 3/b/ should be : ${16} \\choose {7}$.\r\n\r\nBut, once again, this is just my work..\r\n\r\n\r\nps: sorry for double-posting..", "Solution_6": "What is the difference between increasing and strictly increasing? :?", "Solution_7": "Read solafidefarms' post for an explanation ;)" } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "The question is:\r\nFind all continuous functions satisfying:\r\n$ f(x + y) = f(x) + f(y) + xy(x + y) ; x, y \\in \\mathbb R.$\r\n\r\n[hide=\"What I did is:\"]\nLet $ y = x$.\nThen,\n$ f(2x) = 2f(x) + x^2(2x)$\n$ 2f'(2x) = 2f'(x) + 3x^2$\n$ 2f''(2x) = f''(x) + 6x$\n$ 4f'''(2x) = f'''(x) + 6$\n\nTherefore:\n$ 4f'''(0) = f'''(0) + 6$\n$ f'''(0) = 2$\nNow integrating from $ 0$ to $ x$, I got $ f(x) = \\frac {x^3}{3} + cx^2$\nThe correct answer is ${ \\color{red}{f(x) = \\frac {x^3}{3} + cx}}$ where $ \\color{red}c = f'(0)$\n[b]Where did I possibly go wrong???[/b] :huh: \n\n[/hide]", "Solution_1": "See http://goiit.com/posts/list/algebra-functional-equation-91610.htm#425429", "Solution_2": "[quote=\"hsbhatt\"]See http://goiit.com/posts/list/algebra-functional-equation-91610.htm#425429[/quote]\r\nThank you. As stated by other members of [hide=\"forum\"] I accidentally posted the same thing thrice![/hide], solving it by $ Cauchy's$ $ Eq^n$ is also a quick and convenient trick.\r\n\r\nThank You for the assistance.\r\n\r\n-$ Rajiv$ $ Patki.$" } { "Tag": [ "trigonometry" ], "Problem": "x,y,z reale astfel incat x+y+z=pi/4\r\n\r\ndemonstrati ca (sinx+cosx)(siny+cosy)(sinz+cosz)=cos2x+cos2y+cos2z", "Solution_1": "oups... scuze pt eroare\r\n\r\ndemonstrati ca sqrt2 * (sinx+cosx)(siny+cosy)(sinz+cosz)=cos2x+cos2y+cos2z\r\n\r\n\r\nsi mai pun una aici:\r\n\r\ndemonstrati ca orice numar natural se poate scrie ca o suma de termeni distincti ai sirului lui Fibbonacci .", "Solution_2": "a doua iese direct prin inductie\r\nfie $F_{k}\\leq n < F_{k+1}$. atunci exista inegalitatile :\r\n$0 \\leq n-F_{k}< F_{k+1}-F_{k}= F_{k-1}$.\r\ninsa numarul $n-F_{k}$ se scrie ca o astfel de suma, si toti termenii sunt distincti, si mai mici decat $F_{k-1}$. adaugand $F_{k}$ la suma, proprietatea se pastreaza.\r\nverficarea pentru cazurile mici e imediata.", "Solution_3": "parca era o teorema problema 2, zeckendorf sau ceva in genu.\r\n\r\ncat despre prima problema , iese cu complexe si cu calcule multe, si anume:\r\nluam $z_{1}=\\cos{x}+i \\sin{x}$ etc, si avem $z_{1}z_{2}z_{3}={\\frac{\\sqrt{2}}{2}+i \\frac{\\sqrt{2}}{2}}$, si tre sa demonstram ca $\\frac{\\sqrt{2}}{8z_{1}z_{2}z_{3}}\\cdot \\prod{(z_{1}^{2}+1-i(z_{1}^{2}-1)))}= \\sum{\\frac{z_{1}^{4}+1}{2z_{1}}}$ , asta in caz ca nu ai ce face :)", "Solution_4": "E mai usor deoarece $2x+2y+2z=\\frac{\\pi}{2}$ avem $\\cos 2z=\\sin{2(x+y)}=2\\sin{(x+y)}\\cos{(x+y)}$ acum $\\cos 2x+\\cos 2y=2\\cos{(x+y)}\\cos{(x-y)}$ de unde $\\cos 2x+\\cos 2y+\\cos 2z=2\\cos{(x+y)}(\\cos{(x-y)}+cos{\\left(\\frac{\\pi}{2}-x-y\\right)}) =4 cos{(x+y)}\\cos{(y+z)}cos{(z+x)}=4\\prod_{cyc}\\cos\\left(\\frac{\\pi}{4}-x\\right)= \\sqrt{2}\\prod_{cyc}(\\cos x+\\sin x)$ :lol:", "Solution_5": "[b][u]Lemma (well-known or prove easily).[/u][/b]\r\n\r\n$\\{\\begin{array}{c}\\{\\alpha ,\\beta ,\\gamma \\}\\subset\\mathrm R\\\\\\ \\alpha+\\beta+\\gamma=\\pi\\end{array}\\|\\Longrightarrow\\{\\begin{array}{c}\\sin 2\\alpha+\\sin 2\\beta+\\sin 2\\gamma=4\\sin\\alpha\\sin\\beta\\sin\\gamma\\\\\\\\ \\cos 2\\alpha+\\cos 2\\beta+\\cos 2\\gamma =-4\\cos\\alpha\\cos\\beta\\cos\\gamma-1\\end{array}$\r\n\r\n[b][u]Particular cases.[/u][/b]\r\n\r\n$\\{\\begin{array}{c}x+y+z=\\frac{\\pi}{4}\\\\\\ \\alpha : =\\frac{\\pi}{4}+x\\ ,\\ \\beta : =\\frac{\\pi}{4}\\ ,\\ \\gamma : =\\frac{\\pi}{4}+z\\\\\\ (\\alpha+\\beta+\\gamma =\\pi )\\end{array}\\|\\Longrightarrow\\{\\begin{array}{c}\\sum \\cos 2x=\\sqrt 2\\cdot\\prod (\\cos x+\\sin x)\\\\\\\\ \\sum\\sin 2x=1+\\sqrt 2\\cdot\\prod (\\cos x-\\sin x)\\end{array}$" } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "For positive real numbers $ a_1, ... , a_n, b_1, ... , b_n$ ($ n \\geq 3$) with the $ b_i$'s pairwise distinct, denote $ S \\equal{} a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_n$ and $ T \\equal{} b_1b_2 ... b_n$.\r\nLet $ f(x) \\equal{} (x \\minus{} b_1)(x \\minus{} b_2) ... (x \\minus{} b_n)\\sum_{j \\equal{} 1}^n\\frac {a_j}{x \\minus{} b_j}$. \r\nFind the number of distinct real zeroes of the polynomial $ f(x)$.", "Solution_1": "Anybody? :)", "Solution_2": "It is obviosly. Consider \\[ f(b_j)\\equal{}a_j\\prod_{i\\not \\equal{}j}(b_i\\minus{}b_j)\\]\r\nIf $ b_10 \\forall j$ $ f(b_j)(\\minus{}1)^{n\\minus{}j}>0$, therefore they had n-1 roots $ b_1= x+y+z ,which it clearly true.", "Solution_4": "\\[ \\frac {a^2 \\plus{} bc}{b \\plus{} c} \\plus{} \\frac {b^2 \\plus{} ca}{c \\plus{} a} \\plus{} \\frac {c^2 \\plus{} ab}{a \\plus{} b}\\,\\geq\\,a \\plus{} b \\plus{} c \\longleftrightarrow \\sum_{cyclic}x(a \\minus{} b)(a \\minus{} c) \\geq 0\r\n\\]where $ x \\equal{} \\frac {1}{b \\plus{} c},y \\equal{} \\frac {1}{c \\plus{} a},z \\equal{} \\frac {1}{a \\plus{} b}$\r\nWLOG $ a \\ge b \\ge c$. $ \\longrightarrow x \\ge y$ \r\nfrom Vornicu-Schur; $ \\sum_{cyclic}x(a \\minus{} b)(a \\minus{} c) \\geq 0$\r\nthus,\r\n\\[ \\frac {a^2 \\plus{} bc}{b \\plus{} c} \\plus{} \\frac {b^2 \\plus{} ca}{c \\plus{} a} \\plus{} \\frac {c^2 \\plus{} ab}{a \\plus{} b}\\,\\geq\\,a \\plus{} b \\plus{} c\r\n\\]", "Solution_5": "The following stronger inequality is true too.\r\nLet $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\\plus{}ac\\plus{}bc\\neq0.$ Prove that:\r\n\\[ \\frac {a^2 \\plus{} bc}{b \\plus{} c} \\plus{} \\frac {b^2 \\plus{} ca}{c \\plus{} a} \\plus{} \\frac {c^2 \\plus{} ab}{a \\plus{} b}\\geq\\frac{2(a^2\\plus{}b^2\\plus{}c^2)\\plus{}ab\\plus{}ac\\plus{}bc}{a\\plus{}b\\plus{}c}\\]", "Solution_6": "[quote=\"arqady\"]The following stronger inequality is true too.\nLet $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \\plus{} ac \\plus{} bc\\neq0.$ Prove that:\n\\[ \\frac {a^2 \\plus{} bc}{b \\plus{} c} \\plus{} \\frac {b^2 \\plus{} ca}{c \\plus{} a} \\plus{} \\frac {c^2 \\plus{} ab}{a \\plus{} b}\\geq\\frac {2(a^2 \\plus{} b^2 \\plus{} c^2) \\plus{} ab \\plus{} ac \\plus{} bc}{a \\plus{} b \\plus{} c}\n\\]\n[/quote]\r\nThe inequality is equivalent to\r\n\r\n$ \\sum \\frac{a^3\\plus{}abc}{b\\plus{}c} \\geq \\sum a^2,$\r\n\r\n$ \\Leftrightarrow \\sum \\frac{a(a\\minus{}b)(a\\minus{}c)}{b\\plus{}c} \\geq 0,$\r\n\r\nWLOG we can assume that $ a \\geq b \\geq c,$\r\n\r\nwe have $ \\frac{c(c\\minus{}a)(c\\minus{}b)}{b\\plus{}c} \\geq 0,$\r\n\r\nand $ \\frac{a(a\\minus{}b)(a\\minus{}c)}{b\\plus{}c} \\geq \\frac{b(a\\minus{}b)(b\\minus{}c)}{a\\plus{}c}$ because it's equivalent to $ (a\\minus{}b)(a^2\\plus{}b^2\\minus{}ab\\minus{}c^2) \\geq 0,$\r\n\r\nand the proof is completed.", "Solution_7": "[quote=\"rachid\"][quote=\"arqady\"]The following stronger inequality is true too.\nLet $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \\plus{} ac \\plus{} bc\\neq0.$ Prove that:\n\\[ \\frac {a^2 \\plus{} bc}{b \\plus{} c} \\plus{} \\frac {b^2 \\plus{} ca}{c \\plus{} a} \\plus{} \\frac {c^2 \\plus{} ab}{a \\plus{} b}\\geq\\frac {2(a^2 \\plus{} b^2 \\plus{} c^2) \\plus{} ab \\plus{} ac \\plus{} bc}{a \\plus{} b \\plus{} c}\n\\]\n[/quote]\nThe inequality is equivalent to\n\n$ \\sum \\frac {a^3 \\plus{} abc}{b \\plus{} c} \\geq \\sum a^2,$\n\n$ \\Leftrightarrow \\sum \\frac {a(a \\minus{} b)(a \\minus{} c)}{b \\plus{} c} \\geq 0,$\n\nWLOG we can assume that $ a \\geq b \\geq c,$\n\nwe have $ \\frac {c(c \\minus{} a)(c \\minus{} b)}{b \\plus{} c} \\geq 0,$\n\nand $ \\frac {a(a \\minus{} b)(a \\minus{} c)}{b \\plus{} c} \\geq \\frac {b(a \\minus{} b)(b \\minus{} c)}{a \\plus{} c}$ because it's equivalent to $ (a \\minus{} b)(a^2 \\plus{} b^2 \\minus{} ab \\minus{} c^2) \\geq 0,$\n\nand the proof is completed.[/quote]\r\n\r\nVery nice proof. How about the even stronger one\r\n\r\n$ \\frac {a^2 \\plus{} bc}{b \\plus{} c}\\plus{}\\frac {b^2 \\plus{} ca}{c \\plus{} a}\\plus{}\\frac {c^2 \\plus{} ab}{a \\plus{} b}\\geq\\frac {2(a^2 \\plus{} b^2 \\plus{} c^2) \\plus{}ab\\plus{}ac\\plus{}bc}{a\\plus{}b\\plus{}c} \\plus{} \\frac {27}{8} \\frac {(ab\\plus{}bc\\plus{}ca) \\sum (a^2(a \\minus{} b)(a \\minus{} c))} {(a \\plus{} b \\plus{} c)^4}$" } { "Tag": [ "trigonometry" ], "Problem": "Find $\\sum_{k=0}^\\infty \\frac{\\tan(k \\sqrt2)}{2^{k}}$, or prove that it diverges. (The argument is in radians.)", "Solution_1": "$\\tan{k\\sqrt{2}}$ is in RADIANS???!!! :what?:", "Solution_2": "Whoa! Scary. But I'm pretty sure that there's a not-too-hard way to do this...", "Solution_3": "I've gotten it down to:\r\n\r\n[hide] \n$\\frac{1}{i}\\sum_{k=0}^{\\infty}\\left(\\left(\\frac{e^{\\sqrt{2}i}}{2}\\right)^{k}-\\left(\\frac{e^{-\\sqrt{2}i}}{2}\\right)^{k}\\right)\\left(\\frac{1}{e^{ik\\sqrt{2}}+e^{-ik\\sqrt{2}}}\\right)$\n[/hide]\r\n\r\nAnd now I'm stuck. It's the factor on the right that gets me. If it weren't there, I could separate it into geometric sequences.", "Solution_4": "Sorry to bump, but it would be neat if someone could solve this." } { "Tag": [ "induction" ], "Problem": "Prove that $\\forall n\\in\\mathbb{N}$:\r\n(1) $\\prod_{k=1}^n{(n+k)}=2^n\\prod_{k=1}^n{(2k-1)}$\r\n(2) $\\left(\\frac 23\\right)^n<\\frac4{n(n+1)}$\r\nby using mathematical induction.", "Solution_1": "Well, I don't think they're pre-olympiad level.", "Solution_2": "No one can do them? :?", "Solution_3": "This is very easy and probably would rather go in Getting Started ?\r\n\r\n\r\nAnyway note that $1/(n+1) * (2n+1)(2n+2) = 2(2n+1)$ and prove that $2/3 * \\frac 4{n(n+1)} <\\frac 4{(n+2)(n+1)}$", "Solution_4": "[quote=\"t\u00b5t\u00b5\"]This is very easy and probably would rather go in Getting Started ?\n[/quote]\r\nI have posted it in rather getting started or intermediate but someone moved it to this area.", "Solution_5": "[quote=\"t\u00b5t\u00b5\"]This is very easy and probably would rather go in Getting Started ?[/quote]\r\nNo proof in Getting Started ;) \r\nSee the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32837]guideline[/url] (especially post #2).\r\n\r\nSo I think this is appropriate for Intermediate or Pre-olympiad.", "Solution_6": "[quote=\"t\u00b5t\u00b5\"]This is very easy and probably would rather go in Getting Started ?\n\n\nAnyway note that $1/(n+1) * (2n+1)(2n+2) = 2(2n+1)$ and prove that $2/3 * \\frac 4{n(n+1)} <\\frac 4{(n+2)(n+1)}$[/quote]This explains (2) is not true $\\forall n\\in\\mathbb{N}$ :) Counterexample: let $n=1$.", "Solution_7": "who can do the second?" } { "Tag": [ "modular arithmetic", "number theory", "relatively prime" ], "Problem": "Find the number of solutions of $x^{2}\\equiv x\\pmod{m}$ for any positive integer $m$.", "Solution_1": "EDIT: Ok, I really have no clue about anything in NT I guess.", "Solution_2": "[hide]\n\\[x^{2}\\equiv x\\Rightarrow m|x(x-1)\\]\n[i]Let $k$ be the number of distinct prime factors of $m$.[/i] Since $x$ and $x-1$ are relatively prime, for each prime $p$ dividing $m$, either all factors of $p$ divide $x$ or all divide $x-1$. Since there are two choices for each prime, there are $\\boxed{2^{k}}$ solutions, and the Chinese Remainder Theorem guarantees that they all work.\n[/hide]" } { "Tag": [ "Mafia", "FTW", "probability" ], "Problem": "THE EPIC SPACE WAR:\r\n\r\nII:\r\n\r\nREVENGE OF THE CRONOSIANS: (What? I had to think of a name)\r\n\r\nAfter the great battle, where many heroes like EggyLv.999, Spaceguy524 and others were defeated, the survivors fled to Titan, the largest moon of Saturn. (Also known as Cronos) However, on titan, there are strange beings that feed on the air particles. These are known as Cronosians and feared by all.\r\n\r\nSignups available:\r\n\r\n1.\r\n2.\r\n3.\r\n4.\r\n5.\r\n6.\r\n7.\r\n8.\r\n9.\r\n10.\r\n11.\r\n12.\r\n13.\r\n\r\nExtra Game Mod:\r\n\r\nPlease copy and paste list then type your username:\r\n\r\nLike this:\r\n\r\n\r\n1. My username here!\r\n2.\r\n3.\r\n4.\r\n5.\r\n6.\r\n7.\r\n8.\r\n9.\r\n10.\r\n11.\r\n12.\r\n13.\r\n\r\nExtra Game Mod:\r\n\r\nSHOPPING:\r\n\r\nrefer to this anytime needed, copy and paste or quote if you will. \r\n\r\nSHIPS: \r\n\r\nClass 1\r\n\r\n\r\nAverage fighter - 1000 dollars \r\n\r\nGood fighter - 2000 dollars \r\n\r\nGreat fighter! - 3000 dollars\r\n\r\n\r\nClass 2\r\n\r\n\r\nAverage cruiser - 10,000 dollars \r\n\r\nGood cruiser - 20,000 dollars \r\n\r\nGreat Cruiser! - 35,000 dollars\r\n\r\n\r\nClass 3\r\n\r\n\r\nAverage Battle cruiser - 100,000 dollars \r\n\r\nGood battle cruiser - 210,000 dollars \r\n\r\nGreat Battle cruiser - 400,000 dollars\r\n\r\n\r\nSHIELD SHOP: \r\n\r\nshield - 1000 dollars \r\n\r\ngood shield - 10,000 dollars \r\n\r\ngreat shield! - 100,000 dollars \r\n\r\nWEAPON SHOP: note: all weapons never run out of ammo \r\n\r\nlaser - 500 dollars \r\n\r\nMass Driver - 10,000 dollars \r\n\r\nMagnetic accelerator - 15,000 dollars \r\n\r\nEnergy Pulse - 30,000 dollars \r\n\r\nMissile - 10,000 dollars \r\n\r\nProtius Missiles - 50,000 dollars \r\n\r\nAntimatter pulse - 500,000 dollars (OMG!) \r\n\r\nUPDATE: all events are now 1000 dollars each\r\n\r\nThis is almost like EPIC SPACE WAR I; the only thing is that in this one, you can fight each other or aliens, AND BE WARNED, ALIENS HAVE SHIPS TOO!\r\n\r\nGood luck, and don\u2019t spam this thread!\r\n\r\n\r\nAnd this time, there's no sudden ends to the game. Phew. :D", "Solution_1": "1. spaceguy524\r\n2.\r\n3.\r\n4.\r\n5.\r\n6.\r\n7.\r\n8.\r\n9.\r\n10.\r\n11.\r\n12.\r\n13.\r\n\r\n@tcjy: PM me when it is my turn.", "Solution_2": "1. spaceguy524\r\n2. Chompy\r\n3.\r\n4.\r\n5.\r\n6.\r\n7.\r\n8.\r\n9.\r\n10.\r\n11.\r\n12.\r\n13. \r\n\r\nI'm bored", "Solution_3": "1. spaceguy524 \r\n2. Chompy \r\n3. tennis123\r\n4. \r\n5. \r\n6. \r\n7. \r\n8. \r\n9. \r\n10. \r\n11. \r\n12. \r\n13.", "Solution_4": "1. spaceguy524\r\n2. Chompy\r\n3. tennis123\r\n4. EggyLv.999\r\n5.\r\n6.\r\n7.\r\n8.\r\n9.\r\n10.\r\n11.\r\n12.\r\n13.", "Solution_5": "1. spaceguy524\r\n2. Chompy\r\n3. tennis123\r\n4. EggyLv.999\r\n5. cyberspace\r\n6.\r\n7.\r\n8.\r\n9.\r\n10.\r\n11.\r\n12.\r\n13.", "Solution_6": "[quote=\"spaceguy524\"]\n\n@tcjy: PM me when it is my turn.[/quote]\r\n\r\nsame for me.", "Solution_7": "1. spaceguy524 \r\n2. Chompy \r\n3. tennis123 \r\n4. EggyLv.999 \r\n5. cyberspace \r\n6. Bluecarneal\r\n7. \r\n8. \r\n9. \r\n10. \r\n11. \r\n12.", "Solution_8": "[quote=\"spaceguy524\"]@tcjy: PM me when it is my turn.[/quote]\r\n\r\nsame here\r\n\r\n\r\n1. spaceguy524 \r\n2. Chompy \r\n3. tennis123 \r\n4. EggyLv.999 \r\n5. cyberspace \r\n6. Bluecarneal \r\n7. Zheng\r\n8. \r\n9. \r\n10. \r\n11. \r\n12.\r\n13.", "Solution_9": "1. spaceguy524\r\n2. Chompy\r\n3. tennis123\r\n4. EggyLv.999\r\n5. cyberspace\r\n6. Bluecarneal\r\n7. Zheng\r\n8. AIME15\r\n9.\r\n10.\r\n11.\r\n12.\r\n13.\r\n\r\nPerhaps you could PM anyone when it's their turn instead of random people requesting it :P", "Solution_10": "1. spaceguy524 \r\n2. Chompy \r\n3. tennis123 \r\n4. EggyLv.999 \r\n5. cyberspace \r\n6. Bluecarneal \r\n7. Zheng \r\n8. AIME15 \r\n9. louiethechewy123\r\n10. \r\n11. \r\n12. \r\n13. \r\n\r\nThe veteran fighter, louiethechewy123 is back from a mission Epic Space War I !!!", "Solution_11": "If not enough people join can we play with less people?", "Solution_12": "Well, 13 is a small number. :P\r\n\r\nWell, that's compared to Mafia games anyway...\r\n\r\nIt won't take that long for the game to fill up.", "Solution_13": "why 13? that is my favorite number", "Solution_14": "If no one signs up in a week will this game be dead?\r\nPlease sign up, people!!!", "Solution_15": "Whoops, forgot about that. :oops: \r\n[b]Unattack[/b]\r\n[b]Attack abcak[/b]", "Solution_16": "Could I please join? Though dont p.m. me please, and if you could just skip my turn starting Friday for a week plz.? :maybe:", "Solution_17": "Why don't you want to be PM'd? It's the only way the mod can inform you that it's your turn.", "Solution_18": "Because, I dont always check them, and I check my posts on a normal daily basis.Basically, I dont really use p.m.s. :lol: So may I join or not?", "Solution_19": "Ask spaceguy.\r\n\r\nI'm no mod, just the creator lolz :rotfl: :rotfl: :rotfl:", "Solution_20": "Yes you may join Butler. \r\n\r\nwestiepaw-92\r\nabcak-92\r\nzheng-92\r\nEggy-90\r\ncyberspace-92\r\ntennis-78\r\n\r\nabcak, choose somebody.", "Solution_21": "I counter.\r\n\r\nYou jerk.", "Solution_22": "[quote=\"abcak\"]I counter.\n\nYou jerk.[/quote]\r\n\r\nHuh?\r\n\r\nme=:?:", "Solution_23": "I attack westiepaw.", "Solution_24": "Bump.\r\n\r\nDon't want this to die, its a good game", "Solution_25": "can i replace someone?\r\n\r\npm me when it's my turn", "Solution_26": "join LEGEND OF THE BLACK DRAGON", "Solution_27": "Um..... \"BUMP\"", "Solution_28": "MODS. PLEASE. LOCK. THIS. THREAD.", "Solution_29": "If there's another advertisement here, Then this should get locked.\r\n\r\nWell, I guess it's dead. So better create another. SO LONG TESWII!!!" } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Show that if $ p$ is an odd prime,then the congruence $ x^4 \\equiv \\minus{}1 (modp)$ has a solution iff $ p \\equiv 1(mod 8)$", "Solution_1": "This is to prove that x^4 +1 / ( 8k+1) .\r\nConsider two cases i) x is even x =2m \r\nThen we get that T.P. 16*m^4 +1 / 8k+1 \r\nWe know that 16*m^4 +1 is congruent to 1(mod 8) also 8k+1 is congruent to 1(mod 8) .\r\nHence 16*m^4 +1 / 8k+1 .\r\nSimilarly consider x =2m+1. and we get similar congruent conditions.\r\nHowever i am not sure about the following statement if a is congruent to m(mod c) and b is also congruent to m(mod c) . Then either a|b or b|a .\r\nPlease do check that result alone.", "Solution_2": "[quote=\"314 Einstien\"]This is to prove that x^4 +1 / ( 8k+1) .\nConsider two cases i) x is even x =2m \nThen we get that T.P. 16*m^4 +1 / 8k+1 \nWe know that 16*m^4 +1 is congruent to 1(mod 8) also 8k+1 is congruent to 1(mod 8) .\nHence 16*m^4 +1 / 8k+1 .\nSimilarly consider x =2m+1. and we get similar congruent conditions.\nHowever i am not sure about the following statement if a is congruent to m(mod c) and b is also congruent to m(mod c) . Then either a|b or b|a .\nPlease do check that result alone.[/quote]\n\nCasually has come across this post. The very crazy text. But the problem is quite reasonable and is interesting for beginning. For solution use the primitive root modulo $p$.", "Solution_3": "$x^8\\equiv 1 (mod p), x^4 \\neq 1(mod p)$ so order of $x$ modulo $p$ is 8 so $8 | p-1$ =>\n$QED$", "Solution_4": "[quote=\"Ovchinnikov Denis\"]$x^8\\equiv 1 (mod p), x^4 \\neq 1(mod p)$ so order of $x$ modulo $p$ is 8 so $8 | p-1$ =>\n$QED$[/quote]\n\nIt's correct. But we need to prove inverse statement: if $p \\equiv 1 \\pmod{8}$, then the congruence $x^4 \\equiv -1 \\pmod{p}$ is solvable. The simplest way to prove it -- use primitive root modulo $p$." } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let R be a group ring AG, where A is a ring, and G is a finite group. Show that R is self-injective iff A is.", "Solution_1": "[quote=\"flowers\"]Let R be a group ring AG, where A is a ring, and G is a finite group. Show that R is self-injective iff A is.[/quote]\r\n\r\nAG is a free A-module, so a flat A-module, so the restriction of an injective AG-module is an injective A-module. In particular, if AG is AG-injective, then AG is A-injective, and then so is its A-direct summand, A.\r\n\r\nAG is a free AG-module, so a projective AG-module, so if A is A-injective, then Hom_A(AG,A) is AG-injective, but viewing AG as Hom_AG(AG,AG) you get that AG is isomorphic to Hom_A(AG,A) (for instance by Nakayama reciprocity, or just element pushing).\r\n\r\nThe \"injective producing lemma\" is very useful for this. It is 3.5 p. 62 in Lam's Lectures on Modules and Rings. If P is an (R,S)-bimodule that is R-flat and M is an injective S-module, then Hom_S(P,M) is an injective R-module. This allows for some basic change of rings.\r\n\r\n\r\nThis is theorem 4 of:\r\n\r\nConnell, Ian G. \"On the group ring.\"\r\nCanad. J. Math. 15 (1963) 650-685.\r\nhttp://www.ams.org/mathscinet-getitem?mr=153705\r\nhttp://www.cms.math.ca/cjm/v15/p650\r\n\r\nThis is also a part of the result of a paper of Renault, though the main point there was the necessity that G be finite:\r\n\r\nRenault, Guy. \"Sur les anneaux de groupes.\"\r\nC. R. Acad. Sci. Paris S\u00e9r. A-B 273 (1971) A84-A87.\r\nhttp://www.ams.org/mathscinet-getitem?mr=288189" } { "Tag": [ "geometry", "ratio", "inequalities", "geometric inequality", "areas", "IMO", "IMO 1966" ], "Problem": "Let $ ABC$ be a triangle, and let $ P$, $ Q$, $ R$ be three points in the interiors of the sides $ BC$, $ CA$, $ AB$ of this triangle. Prove that the area of at least one of the three triangles $ AQR$, $ BRP$, $ CPQ$ is less than or equal to one quarter of the area of triangle $ ABC$.\r\n\r\n[i]Alternative formulation:[/i] Let $ ABC$ be a triangle, and let $ P$, $ Q$, $ R$ be three points on the segments $ BC$, $ CA$, $ AB$, respectively. Prove that\r\n\r\n$ \\min\\left\\{\\left|AQR\\right|,\\left|BRP\\right|,\\left|CPQ\\right|\\right\\}\\leq\\frac14\\cdot\\left|ABC\\right|$,\r\n\r\nwhere the abbreviation $ \\left|P_1P_2P_3\\right|$ denotes the (non-directed) area of an arbitrary triangle $ P_1P_2P_3$.", "Solution_1": "I do not think the problem is very difficult :\r\n\r\nWlog, we may assume that [ABC] = 1.\r\n\r\nLet A', B' and C' be the midpoints of the sides BC, CA and AB respectively.\r\nIf R is on the segment AC' and Q is on the segment AB', then clearly [AQR] \\leq [AB'C'] = 1/4, so we are done in that case.\r\nWe have similar conclusion in the case where two of the points P,Q,R lies on the half-sides from the common vertex.\r\n\r\nThus, we may assume that R is on the segment BC', Q is on AB' and P is on CA' (the case R on AC', Q on CB' and P on BA' is similar).\r\nLet's consider R and P to be fixed. From ouir choices of the points, the line parallel to PR through B' meets the side AB in a point which belongs to the segment AC' (look about the respective slope). Thus, for any point in the segment AB', the distance from this point to the line PR is at least the distance from B' to the line PR.\r\nIt follows that [PQR] \\geq [B'PR].\r\nWith a similar reasoning, we prove that :\r\n[PQR] \\geq [B'PR] \\geq [B'C'P] \\geq [A'B'C'] = 1/4.\r\n\r\nSince [AQR]+[BRP]+[CPQ]+[RPQ] = [ABC] = 1, we deduce that\r\n[AQR]+[BRP]+[CRQ] \\leq 3/4, from which the result follows easily.\r\n\r\nPierre.", "Solution_2": "Pierre gave too complicated solution ;)\r\nLet $P$, $Q$, $R$ divide sides in ratios $a:1-a$, $b:1-b$ and $c:1-c$ respectively, then if we suppose contrary it will mean that $a(1-b)>1/4$, $b(1-c)>1/4$, $c(1-a)>1/4$. Multiplying them we obtain $abc(1-a)(1-b)(1-c)>1/64$, though $x(1-x)\\leq 1/4$ for all $x\\in[0,1]$. Contradiction. :D", "Solution_3": "[quote=\"Myth\"]Pierre gave too complicated solution ;)[/quote]\r\n\r\n'Too complicated' is my third name. :D \r\n('Too long' is my second)\r\n\r\n[Moderator edit: Another solution to this problem is to conclude it from the stronger problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=5179 .]\r\n\r\nPierre.", "Solution_4": "We note that the equality is attained if P,Q,R are the middle points of the sides. Now we define $\\overline{AR}=a$, $\\overline{BR}=b$, $\\overline{BP}=c$, $\\overline{CP}=d$, $\\overline{CQ}=e$, $\\overline{AQ}=f$. Now we suppose that $(\\triangle ARQ)>\\frac{(\\triangle ABC)}{4}$, $(\\triangle BRP)>\\frac{(\\triangle ABC)}{4}$, $(\\triangle CPQ)>\\frac{(\\triangle ABC)}{4}$\n\nNow we have\n\n$\\frac{af\\cdot sen\\angle BAC}{2}>\\frac{\\frac{(a+b)(e+f)\\cdot sen\\angle BAC}{2}}{4}$\n\n$\\frac{bc\\cdot sen\\angle ABC}{2}>\\frac{\\frac{(a+b)(c+d)\\cdot sen\\angle ABC}{2}}{4}$\n\n$\\frac{de\\cdot sen\\angle BCA}{2}>\\frac{\\frac{(c+d)(e+f)\\cdot sen\\angle BCA}{2}}{4}$\n\nSimplifying and adding the inequalities we have\n\n$abcdef>\\frac{(a+b)^2(c+d)^2(e+f)^2}{64}$\n\nthat is a contradiction, because for AM-GM over abd, afd, bcd, cdf, abe, aef, cbe, cfe we have\n\n$\\frac{abd+afd+bcd+cdf+abe+aef+cbe+cfe}{8}\\ge \\sqrt[8]{a^4b^4c^4d^4e^4f^4}$\n\n$\\frac{(a+b)^2(c+d)^2(e+f)^2}{64}\\ge abcdef$\n\nwe finished :))" } { "Tag": [], "Problem": "Find the number of two-digit positive integers whose digits total 7.", "Solution_1": "16\r\n25\r\n34\r\n43\r\n52\r\n61\r\n70\r\n\r\nThere are 7.", "Solution_2": "None of the digits can be larger than 7, and the units cannot be 0, so the answer is 7", "Solution_3": "[quote=\"AIME15\"]None of the digits can be larger than 7, and the units cannot be 0, so the answer is 7[/quote]\r\nyou mean that the ten's digit can't be 0" } { "Tag": [ "AMC" ], "Problem": "How many positive integer multiples of 1001 can be expressed in the form 10^j - 10^i, where i and j are integers and 0 :le: i < j :le: 99?", "Solution_1": "My solution ...\n\n\n\n[hide]When j=i+6, we have\n\n\n\n10^j - 10^i = (10^i)(10^6-1) =(10^i)(10^3-1)(10^3+1)\n\n\n\n(e.g. 999,999 = 999 x 1001; 9,999,990 = 9,990 x 1001 etc.)\n\n\n\nWith the constraints 0 <= i <= j=i+6 <= 99 we have 0 <= i <= 93, so there are 94 instances.\n\n\n\nThen, when j=i+12, we have\n\n\n\n10^j - 10^i = (10^i)(10^12-1) =(10^i)(10^9-10^6+10^3-1)(10^3+1)\n\n\n\n(e.g. 999,999,999,999 = 999,000,999 x 1001)\n\n\n\nand we have 0<=i<=87, so 88 instances.\n\n\n\nWe can continue the pattern up to j=i+96, which gives 4 instances. The total number of instances is \n\n\n\nSum (k=0 to 15) [6k+4] = 8x98 = 784 instances[/hide]" } { "Tag": [], "Problem": "Does anyone know what scores on the verbal(60 questions) and reading comprehesion(40) tests for the SSAT will give a 90+ percentile?\r\n\r\nSorry, i know this doesnt belong here, but i dont know where it should go.", "Solution_1": "What is the SSAT? Do you mean SAT? By definition, percentile depends on how others do, so it varies from year to year. I think somewhere around a 670 gets you close.", "Solution_2": "The SSATs are the standardized test for private and some other secondary high schools. It's 3 sections just like the SAT, and scores for each section are from 250 to 350 points. \r\n\r\n@OP, the scores range every year, it just depends how hard the test is. Also you should think in terms of raw score, since like the SAT there's a penalty for wrong answers." } { "Tag": [ "probability", "ratio", "geometric series" ], "Problem": "Erica flips a fair two-sided coin. NAte then flips the coin and wins if it matches Erica's flip. If it doesn't match, then Noah flips the coin and wins if it matches Nate's flip. If Noah doesn't win, then Erica flips and wins if it matchs Noah's flip. The game continues until there is a winner. What is the probability that Nate wins the game?", "Solution_1": "[hide]I think that the probability of Nate winning can be expressed by $\\displaystyle \\frac{\\frac12}{\\frac78}=\\boxed{\\frac47}$[/hide]", "Solution_2": "[hide]\nNot quite sure about this but,\nThe probability that he wins on the first \"set\" of flips is\n$\\frac{1}{2}*\\frac{1}{2} = \\frac{1}{4}$\nOn the second\n$\\frac{1}{2}^5 = \\frac{1}{32}$\nOn the third\n$\\frac{1}{2}^8 = \\frac{1}{256}$\n\nSo it looks like an infinite sum of geometric series with first term $\\frac{1}{4}$ and ratio $\\frac{1}{8}$, so\n$\\frac{\\frac{1}{4}}{\\frac{7}{8}} = \\frac{2}{7}$[/hide]", "Solution_3": "I think that the first term of the infinite series is $\\displaystyle \\frac12$ though. Because it doesnt matter what Erica flips... so the chance for Nate to win on his first flip is $\\displaystyle 1*\\frac12$... so I think that the sum of the infinite series would be $\\displaystyle \\frac{\\frac12}{\\frac78}=\\frac47$", "Solution_4": "Ahhh your right." } { "Tag": [ "LaTeX" ], "Problem": "hey can anyone help me?\r\n\r\nI tried making a new document on TeXnic (Hello world! because im just starting LaTeX) and when I try to compile it, nothing happens. I'm wondering if it is because the bar at the top is blank, while it should say \"LaTeX => PDF\" (I think) but I can't type anything in there and the dropdown doesn't do anything.\r\n\r\nany suggestions?\r\nthx", "Solution_1": "You probably didn't tell TexnicCenter where to find the latex files when you installed it. [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Help#I.27ve_upgraded_MiKTeX_and_now_TeXnicCenter_won.27t_work]This[/url] will take you through it (you do the same as if you had upgraded MiKTeX)." } { "Tag": [ "MATHCOUNTS", "search" ], "Problem": "What is the best way to prepare for Mathcounts?\r\nWork on the 2010 handbook problems or work on Past mathcounts school and chapter problems?\r\nI like to do both but I am short on time.\r\nOur Mathcounts school test will be next week, hopefully I can make to the chapter (my school is a tough school, lots of competitions). The chapter is next month.\r\n\r\nIf I have to make choice on focusing on 2010 handbook problems or past School/Chapter problems, which one should I choose?", "Solution_1": "Search the forums for Mathcounts preparation. This question has been asked countless times before.", "Solution_2": "Hmm... the optimal strategy would be to use the handbook to learn how to solve new kinds of problems, and then focus on speed and accuracy by doing old tests, but since you're low on time, I'd say it's best to practice with old tests and review what you can. I hope this helps!", "Solution_3": "Yes, I agree with mathluver that doing old tests would be best in your situation.\r\nGood luck! :)" } { "Tag": [ "function", "Euler", "inequalities", "number theory proposed", "number theory" ], "Problem": "Hey guys, \r\ni would like to give u guys this problem . \r\nwell i got this result when i was doing some research on eulers phi function. \r\nits: \r\nphi(1)+phi(2)+phi(3)+phi(4)+.......+phi(k)> (1+2+3+4+......+k)/2 \r\nfor all k>4 \r\nguys its a very nice result as the difference becomes very narrow and then suddenly widens and then again narrows and so on........ \r\ntry proving this!!!! \r\nKarthik", "Solution_1": "For $ k = 5,6,7,8,9,10$ the inequality can be verified instantly. Assume that $ k\\geq 11$.\r\n\r\nNote that $ \\varphi(1) + \\varphi(2) + \\dots + \\varphi(k)$ counts the number of pairs of integers $ (m,n)$ such that $ 1\\leq m\\leq n\\leq k$ and $ \\gcd(m,n) = 1$. \r\nTherefore, $ 2(\\varphi(1) + \\varphi(2) + \\dots + \\varphi(k)) - 1$ counts the number of pairs $ (m,n)$ such that $ 1\\leq m,\\, n\\leq k$ and $ \\gcd(m,n) = 1$.\r\n\r\nThe total number of pairs $ (m,n)$ with $ 1\\leq m, n\\leq k$ is $ k^2$. Therefore, the number of pairs with $ \\gcd(m,n) = 1$ equal $ k^2$ minus the number of pairs with $ \\gcd(m,n) > 1$. It is clear that for prime $ p$ the number of pairs with $ p|\\gcd(m,n)$ does not exceed $ (k/p)^2$. Hence,\r\n\\[ 2(\\varphi(1) + \\varphi(2) + \\dots + \\varphi(k)) \\geq k^2 - \\sum_{p\\,\\mbox{is prime}} \\frac {k^2}{p^2} = k^2(1 - P(2))\r\n\\]\r\nwhere $ P(\\cdot)$ is the [url=http://mathworld.wolfram.com/PrimeZetaFunction.html]prime zeta function[/url]. Since $ P(2)\\approx 0.452247 < 0.453$, we have\r\n\\[ 2(\\varphi(1) + \\varphi(2) + \\dots + \\varphi(k)) \\geq k^2(1 - P(2)) > 0.547 k^2 > 1 + 2 + \\dots + k,\\quad(\\mbox{for}\\, k\\geq 11)\r\n\\]\r\nimplying that\r\n\\[ \\varphi(1) + \\varphi(2) + \\dots + \\varphi(k) > \\frac {1 + 2 + \\dots + k}{2}.\r\n\\]" } { "Tag": [ "geometry", "incenter", "circumcircle", "Euler", "rhombus", "exterior angle", "angle bisector" ], "Problem": "A, B and C are the feet of the interior bisectors of a triangle EFG. \nShow that the triangle ABC is a right triangle iff one of the angles of EFG is 120\u00b0.", "Solution_1": "$Lemma:$ For a triangle ABC ($BC>AC$) let the interior angle bisectors AD and CF and the exterior angle bisector BE, then D, E and F are collinear. $Prove$ By Menelaus, occur iff and only if $\\frac{BD.CE.AF}{DC.EA.FB}=1$, by angle bisector theorem we have $\\frac{AB.BC.AC}{AC.AB.BC}=1$, which is true.\nSuppose X (some banach space) is riemann integrable, does that mean that || f ||_x is riemann integrable as well?", "Solution_1": "$f$ is Riemann integrable on $[a,b]$ iff for any $\\epsilon>0$ the interval $[a,b]$ can be partitioned into subintervals $I_{k}$ such that $\\sum_{k}\\mathrm{osc}_{I_{k}}(f)|I_{k}|<\\epsilon$. (Here $\\mathrm{osc}_{I}(f)=\\sup_{x,y\\in I}\\parallel f(x)-f(y)\\parallel$. Post-composition of $f$ with any Lipschitz map preserves this property.\r\n\r\n[i]Most pointless remark of the week:[/i] the notion \"Riemann integrable\" makes sense for maps into an abstract metric space, when no reasonable definition of \"Riemann integral\" is available.", "Solution_2": "I'm not that sure: unlike $\\mathbb R$, in infinite-dimensional Banach spaces you can have a limit of Riemann sums with large sums of oscillations times lengths. Let, for instance, $X$ be a Hilbert space with an orthonormal basis $\\{e_{t}\\}_{t\\in\\mathbb R}$ and let $f(t)=e_{t}$. Then the norm of each Riemann sum for the interval $[0,1]$ and any partition does not exceed the square root of the length of the longest interval in the partition, so this terrible function is Riemann integrable to $0$ if one understands Riemann integrability of a vector-valued function as existence of a limit of Riemann sums. This conclusion will remain true if you change $e_{t}$ to $2e_{t}$ here or there but in this way you can get $\\|f\\|$ even not Lebesgue measurable! I should confess that I'm not completely sure what is the commonly accepted definition of Riemann integrability of vector-valued functions, i.e., whether it is just existence of a limit of Riemann sums or the oscillation condition cited by mlok. My point was only that those two are no longer equivalent.", "Solution_3": "Now I think it's the existence of Riemann sums... Indeed, [url=http://www.zentralblatt-math.org/zmath/en/advanced/?q=an:0764.28008&format=complete]this paper[/url] concerns, among other things, the spaces $X$ such that every $X$-valued Riemann integrable function is a.e. continuous. There is also [url=http://www.worldscibooks.com/mathematics/5905.html]a recent book[/url] on Riemann-style integrals of Banach space-valued functions.", "Solution_4": "thanks guys. it was recently suggested to me that a function such as this maybe a suitable counterexample:\r\n$f: [0, 1] \\rightarrow \\ell^{2}$ such that $\\int_{0}^{1}f = 0$ but $\\parallel f \\parallel = \\chi_{Q}$ and therefore not riemann integrable.", "Solution_5": "This works\r\n$\\{r_{n}\\}_{n=1}^\\infty = \\mathbb{Q}\\cap [0, 1]$ and $\\{e_{n}\\}_{n=1}^\\infty$ a collection of \"orthonormal basis\" of infinite dimension. i.e. $e_{1}= (1, 0, 0, ...), e_{2}= (0, 1, 0, 0, ...)$\r\n\r\n$f : [0, 1] \\rightarrow \\ell^{2}$ such that, $f(x = r_{n}) = e_{n}$ and$f(x = otherwise) = 0$" } { "Tag": [ "logarithms" ], "Problem": "Log base 2 (64) = x\r\nSo thats 2 to the x = log 64? but what does x equal", "Solution_1": "$\\log_{2} 64 = x$ $\\Longrightarrow$ $2^x = 2^6$ $\\Longrightarrow$ $x = 6$.", "Solution_2": "You just guessed that how did you get it?", "Solution_3": "[hide]We have $\\log_{2}64=x \\Longrightarrow 2^x=64 \\Longrightarrow 2^6=64, x=\\boxed{\\boxed{6}}$[/hide]", "Solution_4": "As shown by Andreas and ragnarok23, when you get to the step $2^x=64$, you're basically saying \"2 to the what power equals 64?\" Since $2^6=64$, 6 is the answer.", "Solution_5": "I don't understand.It is too easy", "Solution_6": "Hes writing $64$ as $2^y$. So then you can equate the exponents.", "Solution_7": "lol its the way the aops volume 1 book teaches it :P", "Solution_8": "[quote=\"lillyiggie\"]Log base 2 (64) = x\nSo thats 2 to the x = log 64? but what does x equal[/quote]\r\n\r\n[hide]$log_{2} 64=x \\Longrightarrow 2^x=64 \\Longrightarrow x=6$[/hide]", "Solution_9": "[quote=\"Jos\u00e9\"]$log_{2} 64=x \\Longrightarrow 2^x=64 \\Longrightarrow x=6$[/quote]\r\n :spam:", "Solution_10": "[quote=\"Andreas\"][quote=\"Jos\u00e9\"]$log_{2} 64=x \\Longrightarrow 2^x=64 \\Longrightarrow x=6$[/quote]\n :spam:[/quote]\r\nif he copied your answer then its spam, if he came up with that on his own it isn't spam.", "Solution_11": "[quote=\"math92\"][quote=\"Andreas\"][quote=\"Jos\u00e9\"]$log_{2} 64=x \\Longrightarrow 2^x=64 \\Longrightarrow x=6$[/quote]\n :spam:[/quote]\nif he copied your answer then its spam, if he came up with that on his own it isn't spam.[/quote]\r\n\r\nI did it by my own" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "A= [6 4]\r\n [-2 -1] \r\n\r\n \r\n\r\n\r\nB=[2 -3]\r\n [4 4]\r\n\r\n\r\n\r\ntr((AB)^-T)= ? Can you give any hints of what to do with the negative transpose[/list]", "Solution_1": "That notation makes no sense to me. No one is interested in anything called a \"negative transpose\". On the other hand the inverse of a transpose - which is also the transpose of an inverse - is often an interesting item. Do you perhaps mean $ ((AB)^{\\minus{}1})^T,$ which is the same as $ ((AB)^T)^{\\minus{}1}?$\r\n\r\nAlso, that \"tr\" you have in front. That would usually mean the trace of a matrix. Is that what you intend?", "Solution_2": "well yes the tr stands for trace of a matrix ive been trying to do this matrix and this is what a did I multiply A.B and then do the inverse of AB is that corrrect? After doing that I did the transpose right?\r\n\r\n((AB))= [28 -2]\r\n [-8 2]\r\n\r\n\r\n\r\n((AB^-1))=[7/10 -1/20]\r\n [-1/5 1/20] \r\n\r\nis this correct?" } { "Tag": [], "Problem": "A and B can build a wall in 24 hrs. After A worked 7 hrs, B helped A and both together finished the rest of the work in 20 hours. How long does it takes to each one to make the work?", "Solution_1": "Let $a$ be the amount of time in hours it takes A to build the wall and $b$ be the amount of time in hours it takes B to build the wall.\r\n\r\nSo in 1 hour, A can do $\\frac{1}{a}$ of it, and in 1 hour, B can do $\\frac{1}{b}$ of it. I hope this helps you get started.", "Solution_2": "[quote=\"lingomaniac88\"]Let $a$ be the amount of time in hours it takes A to build the wall and $b$ be the amount of time in hours it takes B to build the wall.\n\nSo in 1 hour, A can do $\\frac{1}{a}$ of it, and in 1 hour, B can do $\\frac{1}{b}$ of it. I hope this helps you get started.[/quote]\r\n\r\nusing this you get:\r\n[hide]\n$\\frac{1}{a}+\\frac{1}{b}= \\frac{1}{24}$\nsince they worked together for $20$ hrs, they worked on\n$\\frac{20}{24}$ or $\\frac{5}{6}$ of the job.\nthat means $a$ did $\\frac{1}{6}$ in $7$ hrs, or $\\frac{1}{42}$\nin one hour. thus $a$ can do the job in 42 hrs.\nuse $\\frac{1}{a}+\\frac{1}{b}= \\frac{1}{24}$ to solve for\n$\\frac{1}{b}$ which is $\\frac{1}{56}$ so $b$ does the\njob in 56 hrs\n[/hide]", "Solution_3": "Okay...excellent replies. I did not know that a fractional equation set up was needed to answer this question.\r\n\r\nIs there a formula or equation for questions dealing with working alone and working together?", "Solution_4": "[quote=\"Interval\"]Okay...excellent replies. I did not know that a fractional equation set up was needed to answer this question.\n\nIs there a formula or equation for questions dealing with working alone and working together?[/quote]\r\nNot really. You just need to know when to use these kinds of fractions in these kinds of problems. Try this. A pipe can drain a tank in 8 hours. Another pipe can fill the same tank in 6 hours. John who was watching the tank accidentally left both pipes on. How many hours will it take for the tank to be filled with water?" } { "Tag": [ "inequalities", "geometry", "trigonometry", "geometry proposed" ], "Problem": "(A.Abdullayev, 9--11) Prove that the triangle having sides $ a$, $ b$, $ c$ and area $ S$ satisfies the inequality\r\n\\[ a^2\\plus{}b^2\\plus{}c^2\\minus{}\\frac12(|a\\minus{}b|\\plus{}|b\\minus{}c|\\plus{}|c\\minus{}a|)^2\\geq 4\\sqrt3 S.\\]", "Solution_1": "See also http://www.mathlinks.ro/Forum/viewtopic.php?t=199673 (with not quite the nicest solution).\r\n\r\n dg", "Solution_2": "Let $ a \\leq b \\leq c$. Hence, we need to show that\r\n\\[ b^2 \\minus{} a^2 \\minus{} c^2 \\plus{} 4ac \\geq 4\\sqrt {3}S\\,.\\]\r\nHowever, since $ 2ac\\cos(B) \\equal{} a^2 \\plus{} c^2 \\minus{} b^2$ and $ 2S \\equal{} ac\\sin(B)$, the inequality is equivalent to\r\n\\[ ac\\big(2 \\minus{} \\cos(B)\\big) \\geq \\sqrt {3}ac\\sin(B)\\,.\\]\r\nThus, it suffices to show\r\n\\[ \\cos(B) \\plus{} \\sqrt {3}\\sin(B) \\leq 2\\,,\\]\r\nbut this can be done using Cauchy-Schwarz Ineq, or expanding $ \\cos\\left(B \\minus{} 60^\\circ\\right)$. In addition, the equality holds iff the second largest angle is equal to $ 60^\\circ$, which, in this case, is $ \\angle{B}$." } { "Tag": [], "Problem": "If 1/2x +1/2(1/2x + 1/2(1/2x +1/2(1/2x + ... = y, \r\nthen x = ?", "Solution_1": "[hide]$y=\\frac{1}{2x}+\\frac{1}{2}\\left(\\frac{1}{2x}+\\frac{1}{2}\\left(\\frac{1}{2x}\\ldots\\right)\\right)$\n\n$2\\left(y-\\frac{1}{2x}\\right)=\\frac{1}{2x}+\\frac{1}{2}\\left(\\frac{1}{2x}+\\frac{1}{2}\\left(\\frac{1}{2x}\\ldots\\right)\\right)$\n\n$y=2\\left(y-\\frac{1}{2x}\\right)$\n\n$y=2y-\\frac{1}{x}$\n\n$x=\\frac{1}{y}$[/hide]", "Solution_2": "Can someone move all of this guy's problems to a middle school forum?" } { "Tag": [ "search" ], "Problem": "I know this might not be the right place to ask for help on computer, but it is related to USAMTS. My story is here: half an hour ago I tried to transfer my Round 4 solutions to my new computer, but for some freak reasons all the related files disappeared in both my old and new computers. Afterwards I searched through the disks and found nothing, and later discovered that my recycle bin setting was \"Always Delete Completely\". So it's very likely that I deleted my files accidentally and, due to my dumb automatical setting, permanently. I don't really want to re-type everything again since I'm so poor at LaTeX. If anyone can tell me how to recover files deleted completely like this, I would greatly appreciate. Thanks in advance.", "Solution_1": "I know there's some kind of computer program that can recover files that have been \"permanently\" deleted, but since it costs money, it might not be the best option.", "Solution_2": "You could go to http://www.download.com, search on \"lost files\", and download a file-recovery program for a free trial. That trial should last long enough to recover the accidentally erased files.\r\n\r\nHowever, there is a risky paradox involved. Any downloaded file is written to supposedly empty space on your disk, such as the space that contains your lost file. If you download a file-recovery program, it could overwrite the lost file that you want it to recover. It is a risk you have to take, unless you can download it onto your new computer and use it to remotely scan your old computer.\r\n\r\nErin Schram", "Solution_3": "To both of you, thanks for your helpful response. I've just re-typed my solutions for this round - it's not really a good experience - since it's my only choice. Some of my friends also introduced to me the software you mentioned, but considering its cost and my inability to use these modern technology, I decided to write it again... :huh: hopefully I will not make this mistake again. Anyways thank you very much. :)", "Solution_4": "For future reference:\r\n\r\n[quote=\"ZzZzZzZzZzZz\"]...but since it costs money, it might not be the best option.[/quote]\r\n\r\nThat's not true: there are several free file-recovery programs. You can find some browsing the Internet for a while. Scan the program for viruses to avoid running a fake spyware program. To avoid (or at least minimize) writing to the disk you need to recover from, install the program to a flash disk." } { "Tag": [], "Problem": "Veronica High School has 85 seniors, each of whom plays on at least one of the school's three varsity sports teams: football, baseball, and lacrosse. It so happens that 74 are on the football team; 26 are on teh baseball team; 17 are on both the football and lacrosse teams; 18 are on both the baseball and football teams; and 13 are on both the baseball and lacrosse teams. Compute the number of seniors playing all three sports given that twice this number are the members of the lacrosse team.", "Solution_1": "[hide]\nI get 11.\nI think the formula for this is a+b+c-ab-ac-bc+abc, where ab is someone doing first two, etc.\n74+26+2x-18-17-13+x=85\n100+3x-48=85\n52+3x=85\n3x=33\nx=11\n[/hide]" } { "Tag": [ "quadratics", "number theory", "greatest common divisor", "relatively prime" ], "Problem": "Please write your solution and as the instructions say if necessary to \"prove\" your solutions:\r\nThanks in Advance :D \r\n \r\nheres the link for the problems, please note that you need acrobat or some form of pdf. reader to view the file\r\n[url]http://www.math.ksu.edu/%7Esoibel/2002-7-8.pdf[/url]", "Solution_1": "in my opinion this should be moved down to intermediate\r\n[hide=\"1\"]\nAssume we can find such squares.\n$ 4m \\plus{} 3 \\equal{} x^2 \\plus{} y^2$\nOne of the ones on the right must be even, the other odd. WLOG x is even. x=2a.\n$ 4m \\plus{} 3 \\equal{} 4a^2 \\plus{} y^2$\nRearrange.\n$ 4(a^2 \\minus{} m) \\equal{} 3 \\minus{} y^2$\nConsider the equation mod 4.\n$ 0\\equiv 3 \\minus{} y^2 \\mod 4$\nBut quadratic residue mod 4 is always 0 or 1, so this equation has no solutions.\n\n[/hide]\n\nFor 2, couldn't you theoretically choose all 1001 even numbers, of which no 2 are relatively prime?\n\n[hide=\"3\"]\n$ \\frac {(100!)^{50}}{\\binom{100}{1}\\binom{100}{2}\\cdots}$\nprobably unhelpful, but it looks like it might go somewhere\n[/hide]", "Solution_2": "[hide=\"4\"]Place four points such that it forms a square with side length 1. Then place a point over any one side (outside the square) that it forms an equilateral triangle of side 1 with two of the square's vertices. The final point goes 1 unit \"below\" it in the square, or such that it is \"above\" the opposite side of the square and forms an equilateral triangle.\nGood job if you actually understand that.[/hide]", "Solution_3": "That doesn't work for question 4, since the point above is only next to 2 other points.\r\n\r\n[hide=\"question 1\"]\n\n$ k^2$ and $ n^2$ are congruent to either 0 or 1 mod 4.\n\nTherefore, $ k^2 \\plus{} n^2$ is congruent to 0,1, or to mod 4, for all integers k and m.\n\nBut 4m+3 is congruent to 3 mod 4, so it can't be of the form k^2+m^2.\n\n[/hide]\n\n[hide=\" question 2\"]\n\nThis question is simply wrong, just take 2,4,6,8,...,2002.\n\nThe question is really supposed to be 1002 numbers\n\nIn this case, pair off the numbers (1,2) (3,4) (5,6) ... (2001,2002)\n\nThere are 1001 pairs, so 2 numbers lie in the same pair by pigeonhole. 2 consecutive integers have gcd 1, since\n\n(k+1)-k=1.\n\nTherefore, since there is at least 2 numbers which make such a pair, we are done, since their gcd is 1.\n\n[/hide]\n\n[hide=\"question 3\"]\n\n$ 1!2!3!...49!51!52!...100! \\equal{} (100^{1}*99^{2}*98^{3}*...*51^{50})*(49^{51}*48^{52}*...*1^{99})$\n\nwe may remove any terms with even powers, since they dont affect whether the number is a square or not.\n\n$ (100*98*96*...*52)*(49*47*...*1) \\equal{} 2^{25}*(50*49*...*26)*(49*47*...*1)$\n\nremoving all of the doubled terms again\n\n$ 2^{25}*(50*48*...*26)*(25*23*...*1) \\equal{} 2^{25}*2^{13}*(25*24*23*...*13)*(25*23*...*1)$\n\nremoving all of the doubled terms again\n\n$ 2^{(25 \\plus{} 13)}*(24*22*20*18*16*14)*(11*9*7*5*3*1) \\equal{} 2^{38}*(24*22*20*18*16*14)*(11*9*7*5*3*1)$\n\nremoving the 2^ term, which is a square,\n\n$ (24*22*20*18*16*14)*(11*9*7*5*3*1) \\equal{} 2^{6}*(12*11*10*9*8*7)*(11*9*7*5*3*1)$\n\nremoving all of the doubled terms again, and the 2^ term,\n\n$ (12*10*8)*(5*3*1) \\equal{} 2*2*3*5*2*2*2*2*5*3*1 \\equal{} 2^{6}*3^{2}*5^{2}$\n\nwhich is a perfect square.\n\n[/hide]\n\n[hide=\"question 4\"]\ntake an equilateral triangle with side length one. There are 3 points.\n\nTake another equilateral triangle that's exactly one cm higher than the first one, with sidelength one, and the same orientation. Details are easy to fill out from here.\n\n[/hide]", "Solution_4": "sorry but i could'nt understand number 3, can someone please simplify the wording or is it just my lack of understanding into what is happening, I could'nt read much more than into the 2nd line until i started getting confused.", "Solution_5": "[quote=\"rofler\"]That doesn't work for question 4, since the point above is only next to 2 other points.[/quote]\r\nI'm pretty sure it touches three points.... Actually, my configuration is exactly what you did, except worded differently. (And horribly.)" } { "Tag": [ "SFFT" ], "Problem": "So my competition is on the 21st, today is the 14th, (I live in Washington state btw) and I want to know if you guys know of any formulas or math tricks that might help out there, like SFFT, Balls and Urns,etc.\r\n\r\nEdit: Oh, and what score would I need on written to get into Countdown (top 10) and what score to get top 4", "Solution_1": "Wow.. Washington? That's where Darryl Wu lives!! I think... :maybe: But your score would really have to be at least 43+ or something. But at least you have countdown. I can't really help you with tricks, because you probably know more than me. :blush:", "Solution_2": "43? Unless this year is easier then normal, you definitely do not need to score that high. Darryl Wu will probably get like perfect but their are still 3 spot open. Just practice a lot of problems and you will naturally learn tricks. I learned balls and urns on my own.", "Solution_3": "Would a score of mid to upper 30s get me to countdown?\r\n\r\nI can usually do good at Countdown, I got 6th or 7 seat and beat everyone up to 2nd seat at chapter.[/quote]", "Solution_4": "hmm...probably don't need 43, mid-upper 30s should get u in cd but u have to get top 4 through the countdown unless u get 40+\r\n\r\nand unless u were in darryl wu's chapter u know u're going to have an extra guy in front of u...so only 2 spots :P", "Solution_5": "I'm sorry. I meant top 4. But there should be some REALLY good people in Washington. Don't worry Romeo. I bet your one of them! :wink: But just listen to the scores other people tell you. I probably don't have as much experience because I am a 7th grader. :(" } { "Tag": [ "geometry", "perimeter", "geometric transformation", "rotation" ], "Problem": "1) What is the radius of the circle? (pic below)\r\n\r\n2) HEPTAGO is a star polygon with its 7 vertices equally spaced around a circle. If Sam walks along the sides of the star polygon and turns at each vertex, how many degrees will she turn in coming back to her original position? (pic below)", "Solution_1": "[hide]use the formula A=rs. Since its an isosceles triangle, we have a 5-12-13 triangle. So the height is 12. Quickly finding the area of the triangle which is 60, and the semi-perimeter is 18, we can find the radius. The radius is $\\frac{10}{3}$.[/hide]", "Solution_2": "3) In 1736, the cousin of the famous Gabriel Fahrenehit, Otto Fahrenheit, invented his angle measuring scale. No rotation was 32 and half a rotation was 212. When Otto Fahrenheit measured the angles of a hexagon with his not-so-famous Fahrenheit protractor and then added them what was the sum of the angles?\r\n\r\n4) The net (2-d drawing of a 3-d shape) of a Green house is given. The 20 by 20 square is the floor of the building. IF the air system circulates 8 cubic feet per second, how many minutes will it take to circulate the entire volume of the building (answer to the nearest minute) (pic below)\r\n\r\n[hide]Answers:\nAccording to the answer key...here they are...but I don't think they are correct\n\n1) 7 $\\frac{1}{24}$\n2) 1080\n3) 912\n4) 29\n[/hide]", "Solution_3": "[quote=\"ckck\"][hide]use the formula A=rs. Since its an isosceles triangle, we have a 5-12-13 triangle. So the height is 12. Quickly finding the area of the triangle which is 60, and the semi-perimeter is 18, we can find the radius. The radius is $\\frac{10}{3}$.[/hide][/quote]\n\nno, you found the radius of an inscribed circle. \n\n[hide]using the height of the triangle being 12, as ckck correctly explained, we can do one of two things.\n\n1. use the formula for radius of a circle circumscibed around a triangle with sides a, b, and c.\n\n$r=\\frac{abc}{4(area)}$\nso $\\frac{13\\cdot13\\cdot10}{4\\cdot60}=\\boxed{\\frac{169}{24}}$\n\nor 2. we can be creative.\n\ndraw a radius from the center of the circle, which lies on the altitude of the triangle to one of the base angles. We then have a right triangle with legs 5, 12-r, and r. Using pythagorean, we gets $169-24r+r^{2}=r^{2}$\n\nSo 169=24r and $r=\\boxed{\\frac{169}{24}}$\n\nEDIT: yah, as the answer key says $\\frac{169}{24}=7\\frac{1}{24}$[/hide]", "Solution_4": "[quote=\"Walk Around The River\"][quote=\"ckck\"][hide]use the formula A=rs. Since its an isosceles triangle, we have a 5-12-13 triangle. So the height is 12. Quickly finding the area of the triangle which is 60, and the semi-perimeter is 18, we can find the radius. The radius is $\\frac{10}{3}$.[/hide][/quote]\n\nno, you found the radius of an inscribed circle. \n\n[hide]using the height of the triangle being 12, as ckck correctly explained, we can do one of two things.\n\n1. use the formula for radius of a circle circumscibed around a triangle with sides a, b, and c.\n\n$r=\\frac{abc}{4(area)}$\nso $\\frac{13\\cdot13\\cdot10}{4\\cdot60}=\\boxed{\\frac{169}{24}}$\n\nor 2. we can be creative.\n\ndraw a radius from the center of the circle, which lies on the altitude of the triangle to one of the base angles. We then have a right triangle with legs 5, 12-r, and r. Using pythagorean, we gets $169-24r+r^{2}=r^{2}$\n\nSo 169=24r and $r=\\boxed{\\frac{169}{24}}$\n\nEDIT: yah, as the answer key says $\\frac{169}{24}=7\\frac{1}{24}$[/hide][/quote]\r\nOops, my bad. I thought the circle was inscribed. Well, I was too tired to see it. Oh well...", "Solution_5": "[quote=\"ckck\"][quote=\"Walk Around The River\"][quote=\"ckck\"][hide]use the formula A=rs. Since its an isosceles triangle, we have a 5-12-13 triangle. So the height is 12. Quickly finding the area of the triangle which is 60, and the semi-perimeter is 18, we can find the radius. The radius is $\\frac{10}{3}$.[/hide][/quote]\n\nno, you found the radius of an inscribed circle. \n\n[hide]using the height of the triangle being 12, as ckck correctly explained, we can do one of two things.\n\n1. use the formula for radius of a circle circumscibed around a triangle with sides a, b, and c.\n\n$r=\\frac{abc}{4(area)}$\nso $\\frac{13\\cdot13\\cdot10}{4\\cdot60}=\\boxed{\\frac{169}{24}}$\n\nor 2. we can be creative.\n\ndraw a radius from the center of the circle, which lies on the altitude of the triangle to one of the base angles. We then have a right triangle with legs 5, 12-r, and r. Using pythagorean, we gets $169-24r+r^{2}=r^{2}$\n\nSo 169=24r and $r=\\boxed{\\frac{169}{24}}$\n\nEDIT: yah, as the answer key says $\\frac{169}{24}=7\\frac{1}{24}$[/hide][/quote]\nOops, my bad. I thought the circle was inscribed. Well, I was too tired to see it. Oh well...[/quote]\r\n\r\nWithout looking at the diagram you knew there was a 5-12-13 triangle inside a circle? Wow!", "Solution_6": "Ah, thanks Walk Around the River! I'm still confused on the other questions though :huh:", "Solution_7": "[quote=\"Phoenix257\"]1) What is the radius of the circle? (pic below)\n\n2) HEPTAGO is a star polygon with its 7 vertices equally spaced around a circle. If Sam walks along the sides of the star polygon and turns at each vertex, how many degrees will she turn in coming back to her original position? (pic below)[/quote]\r\n[hide=\"1, a lower tech method\"]Ok, using pythag the altitude has length 12.\nNow draw the radius from circles center to any of the vertices.\nCall this length x.\n$(12-x)^{2}+25=x^{2}$\n$144-24x+25=0$\n$169=24x$\n$x=\\frac{169}{24}$[/hide]", "Solution_8": "Can anyone get the second question ? (the star one)", "Solution_9": "[hide=\"#2\"]Since all vertices are located on the circle and are equally spaced, each \"outside\" angle must be congruent to one another. Therefore, if we were to find the measure of one angle, suppose $PTA$, we could find the total number of degrees she turns by multiplying by seven. We notice that since P, A, and T are all on the circle, the angle $PTA$ is $\\frac{1}{2}$ of the central arc $PA$. The measure of this arc is $\\frac{360}{7}$. Therefore the angle $PTA=\\frac{360}{14}$ and the total [b]$\\frac{360}{7}$[/b].[/hide]", "Solution_10": "I was thinking that too but somehow the answer is [hide]1080[/hide] :huh:", "Solution_11": "EDIT: i didn't read the question right :oops: \r\ni think you can solve it by assuming that you get a regular heptagon in the middle, which is a perfectly okay assumption", "Solution_12": "[hide]A possible solution to the second question...\n\nMark the intersection of GA and HE as $X$. The angles that are like $\\angle GXH$ are equal to $\\frac{arc GH+arc EA}{2}= \\frac{360}{7}\\cdot \\frac{3}{2}$. There are seven of these angles, so they add up to 540 degrees. Angles like $\\angle PTA$ are half arc $PA = \\frac{360}{7 \\cdot 2}$. There are 7 of these angles so they add up to 180 degrees. Together, that's $540+180 = 720$. Maybe the answer key is wrong?\n\n[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "If $ a,b,c$ are nonnegative real numbers, then\r\n\r\n$ 2\\sum a^6\\plus{}16\\sum a^3b^3 \\ge 9\\sum a^2b^2(a^2\\plus{}b^2)$.", "Solution_1": "[quote=\"Vasc\"]If $ a,b,c$ are nonnegative real numbers, then\n\n$ 2\\sum a^6 \\plus{} 16\\sum a^3b^3 \\ge 9\\sum a^2b^2(a^2 \\plus{} b^2)$.[/quote]\r\nWe have\r\n\\[ a^6\\plus{}b^6\\plus{}16a^3b^3\\minus{}9a^2b^2(a^2\\plus{}b^2) \\equal{}(a^3\\minus{}b^3)^2 \\minus{}9a^2b^2(a\\minus{}b)^2 \\equal{}(a\\minus{}b)^2[(a^2\\plus{}ab\\plus{}b^2)^2\\minus{}9a^2b^2] \\ge (a\\minus{}b)^2[9\\cdot a^2b^2\\minus{}9a^2b^2]\\equal{}0\\]\r\nHence\r\n\\[ 2\\sum a^6\\plus{}16\\sum a^3b^3\\minus{}9\\sum a^2b^2(a^2\\plus{}b^2) \\equal{}\\sum [a^6\\plus{}b^6\\plus{}16a^3b^3\\minus{}9a^2b^2(a^2\\plus{}b^2)] \\ge 0\\]\r\nWe have done. :)", "Solution_2": "A stronger inequality is the following:\r\n\r\n$ \\sum a^6 \\plus{} 8\\sum a^3b^3\\plus{}2abc(\\sum a)(\\sum ab) \\ge 5\\sum a^2b^2(a^2 \\plus{} b^2)$.", "Solution_3": "[quote=\"Vasc\"]A stronger inequality is the following:\n\n$ \\sum a^6 \\plus{} 8\\sum a^3b^3 \\plus{} 2abc(\\sum a)(\\sum ab) \\ge 5\\sum a^2b^2(a^2 \\plus{} b^2)$.[/quote]\r\n\r\nIs there any mistake, Vasc.", "Solution_4": "Of course. This is corrected statement.\r\n\r\nA stronger inequality is the following:\r\n\r\n$ \\sum a^6 \\plus{} 8\\sum a^3b^3 \\plus{} 2abc(\\sum a)(\\sum ab) \\ge 5(\\sum a^2)(\\sum a^2b^2)$." } { "Tag": [ "geometry", "geometric transformation", "reflection", "calculus", "trigonometry", "Pythagorean Theorem" ], "Problem": "A man is 6 miles east and 5 miles south of his home. He is also 3 miles north of a river, which is 8 miles south of his home. What's the least amount of miles he may travel to fetch water from the river and then return home?", "Solution_1": "[hide]\nDistance from him and river : 3mi\nDistance from river and home : 10mi by pythagorean theorem\nTotal distance : 13mi\n[/hide]", "Solution_2": "[hide]12.58529852..........[/hide]", "Solution_3": "[quote=\"mathgeek2006\"][hide]12.58529852..........[/hide][/quote]\r\n\r\nHow did you get that?", "Solution_4": "[hide] the answer is ten (there is a really special way to do this involving using the river as a mirror[/hide]", "Solution_5": "Good job you guys, a lot of critical thinking involved. I'll post the correct answer tomorrow... :)", "Solution_6": "This is a good one ,I tried using a lot of laws of reflection [hide]\n12.77 miles[/hide]", "Solution_7": "[hide]10[/hide]", "Solution_8": "[hide]\nI got 12.529964...\n\nRiver is a mirror, reflect home, make right triangle with legs 6 and 11. Find hypotenuse.\n[/hide]", "Solution_9": "[hide]First he travels 3 miles south to the river. then he goes along the hypotenuse of the triangle (with sides 6 and 8) which is 10 miles. so he travels 13 miles.[/hide]", "Solution_10": "does someone want to give in the final answer", "Solution_11": "Hey everybody:\r\n\r\n Using Calculus I got an answer of 12.53 miles.\r\n Hamster1800 had the best solution.", "Solution_12": "Calculus has no place in this forum and is utterly unnecessary to solve this problem.\r\n\r\nIf you post an answer, please explain how you got your answer. This thread contains many guesses -- mostly wrong. But without explanations, nobody can help you correct your mistakes and nobody can learn from what you did right. Please post explanations, not just answers!\r\n\r\nThe man must travel 6 miles west to get home. He must travel 3 miles south and then 8 miles north to get home. Using the river as a mirror, the man's minimum travel distance is the hypotenuse of a right triangle with legs of 6 and 11.\r\n\r\nSo, assuming the man can access the river (flowing east-west) at any point along its length, the minimum distance is $\\sqrt{6^2+11^2}=\\sqrt{157}\\approx 12.53$.", "Solution_13": "I got an answer of thirteen because he has to go 3 miles down and then then travel across the hypothenuses which is ten.", "Solution_14": "People are trying to minimize distance to the river, which is incorrect. While this is not terrible, it is not the minimum SUM of the distanced from him to the river to the house. What we want, is for him to hit the river at the angle, so that when he leaves the river, he goes at the same angle, but on the other sides. some higher level math is nice here to show why it works to reflect the house, but let's look at it like this.\r\n\r\nWhen you reflect the house over the river, it moves down and you can create a line to the contact point on the river. Now, instead of being 8 miles north of the river, the house is 8 miles south of the river. Obviously, any point of contact on the river will be \"on the way\" to his house, so we want the shortest distance: a straight line. This straight line will intersect the river. This point of intersection is where he should get water from the river.", "Solution_15": "Does any one want to give in an answer or Am i correct at 12.7", "Solution_16": "It has already been given. You just put the guy beneath the river, and draw a straight line to his house. Legs of 6 and 11, make the hypotenuse and the answer $\\sqrt{6^2+11^2}=\\sqrt{157}\\approx12.53$", "Solution_17": "Yes, the answer is approximately 12.53. Can calculus be used to solve any problem? Just curious.", "Solution_18": "[quote=\"nat mc\"]It has already been given. You just put the guy beneath the river, and draw a straight line to his house. Legs of 6 and 11, make the hypotenuse and the answer $\\sqrt{6^2+11^2}=\\sqrt{157}\\approx12.53$[/quote]\r\n\r\nOh that explains it very well. :) :10:", "Solution_19": "4everwise -- Calculus can't be used to solve [i]any[/i] problem, but it is powerful when dealing with areas under curves, rates, and particularly in this kind of problem: optimization (that is, maximizing or minimizing quantities).\r\n\r\nMost math competition students I know despise calculus simply because it is less elegant or insightful than a wiser approach with more basic math concepts. As seen here, the reflection using the river as a mirror is an excellent approach, and calculus (while giving the same answer as this approach), is just brute force. A computer could do the calculus approach, but could not conceive of the mirror idea by itself.", "Solution_20": "Hi everyone:\r\n\r\n The only reason I used calculus was this: I believe that the best way to check an answer to a problem is by solving it a different way. I'm a strong believer in being able to solve a problem with more than one solution. I agree (and I mentioned this in my last post on this topic) that Hamster1800 had the best solution. It was creative\r\nyet simple to understand. I only used calculus as a check.", "Solution_21": "I completely agree with you! I'm not bashing your solution :) . The spirit of checking a problem multiple ways is good, but I dont want people who haven't heard of calculus to start giving up on optimization problems and what not... and also not to assume that just because a problem seems like it requires calculus that calculus is the only way to solve it.\r\n\r\nso many interesting examples of Archimedes solving areas under curves exactly without using calculus..", "Solution_22": "Let's not curve the topic into something several years ahead of this forum! :D \r\nI have never learned calculus, neither precalculus. I should start studying that :lol:", "Solution_23": "Don't you just hate it when a problem goes into triginometry and calculus stuff? I always get lost. :( \r\n\r\nAnyway, shouldn't the answer be 13 how ccy did it? I read the explanation but I don't quite understand what you mean by mirror and a load of other things. Please explain.", "Solution_24": "I'm learning pre-calculus next school year and didn't know what it was for until now. :( \r\n\r\nAnyway, let's get this thread back on track." } { "Tag": [], "Problem": "hey all.\r\ni jus wrote FT1 today at home and wanted to know how every1 else has done so i can check wat a \"decent\" mark is. \r\n\r\nSo ppl, plz post ur marks if youve written it too!\r\n\r\nI got 215 (120 + 95)\r\n\r\nFYI it was out of 522 (270 + 252) for all those who've not written it. :)", "Solution_1": "[quote=\"hash_include\"]hey all.\ni jus wrote FT1 today at home and wanted to know how every1 else has done so i can check wat a \"decent\" mark is. \n\nSo ppl, plz post ur marks if youve written it too!\n\nI got 215 (120 + 95)\n\nFYI it was out of 522 (270 + 252) for all those who've not written it. :)[/quote]\r\n\r\nThala!! \r\n\r\nWell, not sure about these\r\n\r\nChappli - 252\r\nPrasad - 246\r\nShreyas - 557\r\n\r\nNo da ,thats my no of posts.", "Solution_2": "haven't written it ...not even thinking of writing it :P", "Solution_3": "rock got 162 in paper 1 and it seems 118 in paper2 making it 280!!! :D \r\nMe too not written! :D", "Solution_4": "Havent written it yet. Was in the process of cleaning my room and now all my FIITJEE papers are missing. Must be the work of a over neatness concious mother of mine :D \r\n\r\nBut hell, my room actually looks organized...\r\n\r\n\r\nI HATE IT\r\n\r\nP.S - Sorry about the rant, but yeah, will post here once I take the test", "Solution_5": "I got 256, Prasad 242.. You swapped the units digit ,shre...", "Solution_6": "Yes I got 161 in paper 1 and 120 in paper 2 so totally 281:\r\nOfcourse central marks in all of them and felt chemistry was easy out of the three :D :D", "Solution_7": "what's your sp(l)it da :D", "Solution_8": "Chemistry:108\r\nPhysics:86\r\nMaths:87\r\nCentral marks in all of them :D :D :D", "Solution_9": "gr8 da :D", "Solution_10": "I thought paper-2 was really good and I spent 1hr and 15mins. in that paper for maths and still ended up getting 33 which was awful and then rolled over chemistry da. :D" } { "Tag": [ "geometry", "3D geometry", "probability" ], "Problem": "A fair coin and a 6-sided number cube with numbers 1 through 6 on its faces are simultaneously tossed. What is the probability that a ``head'' and an even number appear, respectively, on the top faces of the coin and the cube?", "Solution_1": "Each of these independent events comes up with $ \\frac{1}{2}$ probability, so the answer is with multiplication.\r\n\r\n$ \\frac{1}{2} \\cdot \\frac{1}{2} \\equal{} \\frac{1}{4}$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Sent to pvthuan, I hope you'll like this\r\nProve that for all x,y >0, we always have\r\n$\\frac{4}{(x+1)^{4}}+\\frac{4}{(y+1)^{4}}+\\frac{x^{2}y^{2}}{x^{2}y^{2}+1}\\ge 1$", "Solution_1": "Yes, I like that one. Please come one and send me your results.", "Solution_2": "Yes, It is my result when I solve the ChinaTST 2005. I'm really like the not homogeneous inequalities. i'll sent you the tex file next day about my work. I hope you'll enjoy it!", "Solution_3": "[quote=\"The soul of rock\"]Sent to pvthuan, I hope you'll like this\nProve that for all x,y >0, we always have\n$\\frac{4}{(x+1)^{4}}+\\frac{4}{(y+1)^{4}}+\\frac{x^{2}y^{2}}{x^{2}y^{2}+1}\\ge 1$[/quote]\nIt's true for all reals $x\\neq-1$ and $y\\neq-1$. :wink:", "Solution_4": "It's just a corollary of the known inequality: If $a$ and $b$ are positive real numbers, then \\[\\frac{1}{(1+a)^2}+\\frac{1}{(1+b)^2} \\ge \\frac{1}{1+ab}.\\]", "Solution_5": "And how can you deal with the last one?", "Solution_6": "[quote=\"TheIronChancellor\"]And how can you deal with the last one?[/quote]\nWith the most natural way: \n$\\frac{1}{(1+a)^{2}}+\\frac{1}{(1+b)^{2}}-\\frac{1}{1+ab}=\\frac{ab(a-b)^2+(ab-1)^2}{(a+1)^2(b+1)^2(ab+1)}$.", "Solution_7": "Thanks :D" } { "Tag": [], "Problem": "Consider a standard twelve-hour clock whose hour and minute hands move continuously. Let $m$ be an integer, with $1 \\leq m \\leq 720$. At precisely $m$ minutes after 12:00, the angle made by the hour hand and minute hand is exactly $1^\\circ$.\r\nDetermine all possible values of $m$.", "Solution_1": "[hide]The hour hand moves 1/12th the speed of the minute hand. $1/6+5x$ where x is the hour 12:00=0 1:00=1,... would form one degree with the hour hand if the hour hand did not move. However, the hour hand does so an infinite series is set up. Everytime the minute hand catches up to form one degree if the hour didn't move, the hour hand moves at 1/12th the minute hand's speed and prevents it. This means that the expression above is multiplied by $12/11$. $2/11+60/11x$. 60 is 5 mod 11, so x=4 works. The resulting term means that at 22 minutes past 4:00, the minute hand and hour hand forms a one degree angle. That's 262 minutes after 12:00.[/hide]", "Solution_2": "[hide=\"solution\"]consider the angles starting at 12:00, going clockwise, after $m$ minutes, the minute hand will make an angle of $6m$, and the hour hand will make an angle of $\\frac{m}{2}$, we then have that $|6m-\\frac{m}{2}|=\\pm 1+360k$ ${\\{m,k}\\} \\subset \\mathbb{Z}$\n\\begin{eqnarray} 11m=\\pm 2+720k\\\\ 11m=\\pm 2+5k+11*65*k\\\\ 11(m-65k)=\\pm 2+5k \\end{eqnarray}\nwe get the solutions: $k=4$, and $m-65k=2$ or $\\boxed{m=262}$ or\n$k=7$, $m-65k=3$ then $\\boxed{m=458}$\nby CRT these are the unique solutions mod 55 (the numbers being k and m-65k) if we add 55, the minutes are not in the given range so that is all the solutions[/hide]" } { "Tag": [ "limit" ], "Problem": "Evaluate :\r\n\\[ S_n \\equal{} \\arctan{\\frac {1}{2}} \\plus{} \\arctan{\\frac {1}{2 \\cdot 2^2}} \\plus{} \\dots \\plus{} \\arctan{\\frac {1}{2 \\cdot n^2}}\r\n\\]\r\n.\r\n\r\nIs it true that : \r\n\r\n$ \\arctan{(2k \\plus{} 1)} \\minus{} \\arctan{(2k \\minus{} 1)} \\equal{} \\arctan{\\frac {1}{2 \\cdot k^2}}$ ?? If yes , prove it :)", "Solution_1": "I think it is true, but I feel like I'm missing something in my proof:\r\n\r\n$ \\arctan(2k\\plus{}1)\\equal{}a$, $ \\arctan(2k\\minus{}1)\\equal{}b$\r\n\r\n$ \\tan(a\\minus{}b)\\equal{}\\frac{\\tan(a)\\minus{}\\tan(b)}{1\\plus{}\\tan(a)\\tan(b)}\\equal{}\\frac{(2k\\plus{}1)\\minus{}(2k\\minus{}1)}{1\\plus{}(2k\\plus{}1)(2k\\minus{}1)}\\equal{}\\frac{1}{2k^2}$\r\n\r\nNow just take the $ \\arctan$ of both sides.\r\n\r\n... well after this the evaluation is almost............ trivial :P \r\n\r\nInteresting: $ \\lim_{n\\rightarrow \\infty}{S_n}$??\r\n\r\n\r\n\r\n :)", "Solution_2": "What is the motivation for identity used?", "Solution_3": "Well, in terms of arguments, it's given by\r\n\r\n$ \\frac {1 \\plus{} (2k \\plus{} 1)i}{1 \\plus{} (2k \\minus{} 1)i} \\equal{} \\frac {(1 \\plus{} (2k \\plus{} 1)i)(1 \\minus{} (2k \\minus{} 1)i)}{4k^2 \\minus{} 4k} \\equal{} \\frac {2k^2 \\plus{} i}{2k^2 \\minus{} 2k}$\r\n\r\nThese sort of identities should be expected; they appear, for example, in [url=http://en.wikipedia.org/wiki/Machin-like_formula]Machin-like formulae[/url]." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "For any positive real numbers $a,\\ b$ and $c$,\r\n$\\frac{a}{a+b}+\\frac{b}{b+c}+\\frac{c}{c+a} \\ge \\frac{81abc}{2(a+b+c)^3}$", "Solution_1": "[quote=\"ductrung\"]For any positive real numbers $a,\\ b$ and $c$,\n$\\frac{a}{a+b}+\\frac{b}{b+c}+\\frac{c}{c+a} \\ge \\frac{81abc}{2(a+b+c)^3}$[/quote]\r\n\r\nIn [b]\"Toantuoitho[\\b] magazine in Vietnam\r\nit stronger ..", "Solution_2": "We can make $x=\\frac{a}{a+b+c}$, $y=\\frac{b}{a+b+c}$ and $z=\\frac{c}{a+b+c}$. And now we have to prove that (In this passage we must use the fact that $x+y+z=1$)\r\n\r\n$\\frac{x}{x+y}+\\frac{y}{y+z}+\\frac{z}{z+x} \\geq \\frac{81xyz}{2}$.\r\n\r\nBut we know that:\r\n\r\n$\\frac{x}{x+y}+\\frac{y}{y+z}+\\frac{z}{z+x} \\geq 3\\frac{\\sqrt[3]{xyz}}{\\sqrt[3]{(x+y)(y+z)(z+x)}}$.\r\n\r\nLet $P=\\sqrt[3]{xyz}$.\r\n\r\nWe can observe too that $\\frac{(x+y)+(y+z)+(z+x)}{3} \\geq \\sqrt[3]{(x+y)(y+z)(z+x)}$, and so:\r\n$\\frac{3}{2} \\leq \\frac{1}{\\sqrt[3]{(x+y)(y+z)(z+x)}}$.\r\n\r\nAnd now we have that:\r\n\r\n$\\frac{x}{x+y}+\\frac{y}{y+z}+\\frac{z}{z+x} \\geq \\frac{9}{2}P$ and we\u00b4ll have $\\frac{9}{2}P \\geq \\frac{81}{2}P^3$, if and only if $\\frac{1}{3}\\geq P$, and this last ineq is true because $\\frac{x+y+z}{3} \\geq \\sqrt[3]{xyz} = P$", "Solution_3": "[quote=\"ductrung\"]For any positive real numbers $a,\\ b$ and $c$,\n$\\frac{a}{a+b}+\\frac{b}{b+c}+\\frac{c}{c+a} \\ge \\frac{81abc}{2(a+b+c)^3}$[/quote]\r\n\r\nI think the stronger ineq holds, too:\r\n\r\n$\\frac{a}{a+b}+\\frac{b}{b+c}+\\frac{c}{c+a} \\ge \\frac{27abc}{2(a+b+c)(ab+bc+ca)}$\r\n\r\n\r\nHappy new year :)", "Solution_4": "[quote=\"socrates\"][quote=\"ductrung\"]For any positive real numbers $a,\\ b$ and $c$,\n$\\frac{a}{a+b}+\\frac{b}{b+c}+\\frac{c}{c+a} \\ge \\frac{81abc}{2(a+b+c)^3}$[/quote]\n\nI think the stronger ineq holds, too:\n\n$\\frac{a}{a+b}+\\frac{b}{b+c}+\\frac{c}{c+a} \\ge \\frac{27abc}{2(a+b+c)(ab+bc+ca)}$\n\n\nHappy new year :)[/quote]\r\n\r\n[i]Unless[/i] my computer program is buggy, I don't think it's true. (Funny that it's exactly the original ineq I attempted until I found a counter example using computer.) Otherwise, please be so kind to post a solution.\r\n\r\nHappy New Year !!!", "Solution_5": "I have another nice solution. \r\n\r\n$\\sum_{cyclic} \\frac{a}{a+b} \\geq \\frac{81abc}{2(a+b+c)^3}$\r\n\r\n$\\Leftrightarrow (\\sum_{cyclic} a+b ) ( \\sum_{cyclic} \\frac{a}{a+b} )\\geq \\frac{81abc}{(a+b+c)^2}$\r\n\r\nHere, by $Cauchy-Swartz$ $Ineq$, we get $(\\sum_{cyclic} a+b ) ( \\sum_{cyclic} \\frac{a}{a+b} )\\geq (\\sum_{cyclic} \\sqrt{a})^2$\r\n\r\nTherefore, It is enough to prove that $(a+b+c)^2 (\\sqrt{a}+\\sqrt{b}+\\sqrt{c})^2 \\geq 81 abc$. \r\n\r\nThis is self-evident by using $AM-GM$. \r\n\r\n$Q.E.D.$", "Solution_6": "[quote=\"ductrung\"]\n[i]Unless[/i] my computer program is buggy, I don't think it's true. (Funny that it's exactly the original ineq I attempted until I found a counter example using computer.) Otherwise, please be so kind to post a solution.\n\nHappy New Year !!![/quote]\r\n\r\n\r\nYour computer's statement [i]must[/i] be true...\r\n\r\nAnyway here goes my solution, please check its validity... :P \r\n\r\n$\\frac{a}{a+b}+\\frac{b}{b+c}+\\frac{c}{c+a} = \\frac{ac}{ac+bc}+\\frac{ab}{ab+ac}+\\frac{bc}{bc+ab}\\geq \\frac{(\\sqrt{ac}+\\sqrt{ab}+\\sqrt{bc})^2}{2ab+2bc+2ca}$\r\n\r\nSo it remains to show that \r\n\r\n$(\\sqrt{ac}+\\sqrt{ab}+\\sqrt{bc})^2 \\cdot (a+b+c)\\geq 27abc$\r\n\r\nwhich is true due to AM-GM :\r\n\r\n$(\\sqrt{ac}+\\sqrt{ab}+\\sqrt{bc})^2\\geq (3\\sqrt[3]{abc})^2$\r\n\r\nas well as\r\n\r\n$a+b+c\\geq 3\\sqrt[3]{abc}$\r\n\r\n :)", "Solution_7": "[quote=\"socrates\"]\nYour computer's statement [i]must[/i] be true...\n[/quote]\r\n\r\nHello socrates,\r\n\r\nSo simple and totally valid. I regret deeply that I discarded it so quickly (such an easy problem) 'cause the computer says so - it is a rounding error! Anyway, please excuse me and thanks for your reply." } { "Tag": [ "ARML", "AMC", "USA(J)MO", "USAMO", "Stanford", "college", "analytic geometry" ], "Problem": "Hi everyone,\r\ni'm working on getting a bunch of information out about ARML tonight including practice schedules, how to pay, etc., but i need to pass this on right away. You can go to http://www.southwest.com right now and buy a flight to Vegas for ARML for 117 bucks if you want to fly. If you're worried about being on the team, i can assure you that i will accept you onto the team if you want to go ahead and purchase a flight. Remember, you will still need to pay the registration fee, but for those of you who were planning to fly, this may be your best chance to get a good deal. Sale ends at midnight apparently, so don't waste any time. :) Again, my flight details are Southwest flight 215 from SJC-LAS on May 28 and Southwest flight 270 from LAS-SJC on May 30, and you need to arrange to be on the same flight as me if you're flying..\r\n\r\n-Tim", "Solution_1": "Is it 117 round trip?", "Solution_2": "yeah, i canceled my old flight and bought a new one at this rate! :)", "Solution_3": "Er, my mom just checked on southwest.com, and the lowest rates she can find are 92 and 99 (there and back, respectively), which after tax amounts to over 200 bucks. Did you get some sort of discount, and when did you change the flight?\r\n\r\nEDIT: Found the promotion. Booked the flight :)", "Solution_4": "Hi everyone,\r\n\r\nThis letter is for everyone - students, parents, and coaches - planning to go to ARML this year. First, if you haven\u2019t yet filled out an application form, do it now. http://mathleague.org/armlapp.php. There is still time. This goes for every person who is going to ARML regardless of age \u2013 I filled one out for myself just an hour ago. As for organizing the teams, it looks like we have the interest and the transportation capacity to take four full teams to ARML this year, so my goal will be to do just that. A and B teams have been chosen according to the formula I outlined at last Saturday\u2019s contest, and they are listed below alphabetically. Anyone else who wants to go to ARML may as well, subject of course to space constraints as described in the next paragraph. I mentioned earlier that the cost of the trip would be $ \\$$250; I\u2019m going to knock that down to $ \\$$245, because assuming you went to three tryouts you\u2019ve already paid $ \\$$30, and $ \\$$30 off last year\u2019s cost brings us to $ \\$$245. Of course, I promised at the beginning of the year that ALL your tryout entry fees would come back to you; so if you attended more than three tryouts feel free to take an additional $ \\$$10 off the cost for each additional tryout you went to (eg. if you went to six tryouts your cost is only $ \\$$215)..\r\n\r\nIf you want to secure a spot on this year\u2019s ARML team, you need to bring a check for the amount of your registration fee - payable to Great Plains Math League - to the practice on May 2 (see below). I may be able to accept payments next week assuming there is still space on one of our teams. If on Saturday we have more interest in our lower two teams than I can accommodate, I will accept students in order of their ARML selection indices as defined last weekend. If those two teams remain unfilled after Saturday, I will accept students onto those teams in the order their checks are received. If students selected for the A or B team are unable to participate, I will move students up from lower teams based on some objective combination of selection index and USAMO performance..\r\n\r\nYou will also need to submit a notarized permission slip, which can be found at http://mathleague.org/permission.pdf. You have two options to get this turned in: you can hand it to me with your check at the practice on May 2, or you can mail your permission slip to PO Box 111, Sparta WI 54656. I will not collect any permission slips personally after this Saturday\u2019s practice. Permission slips don\u2019t have to be in until May 20th, but please get yours in as soon as possible, because the person checking the PO Box in Wisconsin will be bugging you incessantly about your permission slip until it shows up. You will also be bugged in the same manner if you filled out an application form but haven\u2019t committed to going or not going to ARML by next week.. :)\r\n\r\nBecause the vast majority of you indicated on your application forms that leaving early on Thursday would not be a problem, I anticipate having the bus leave from Stanford at 2pm on Thursday May 28th, heading straight for Las Vegas. If this is impossible for you, let me know when you submit your check and I will hold your check until a firm decision has been made regarding departure time. I will be glad to return any such checks in the event this presents an irreconcilable schedule conflict; aside from that, we will be unable to issue refunds if you end up not being able to go to ARML. We will probably leave Las Vegas around 5pm on the 30th, putting us back to Stanford shortly after midnight. Again, if you are thinking of flying and are willing to pay the extra cost, make sure you coordinate with me so that you\u2019re on my flight..\r\n\r\nFinally, a word about practices. We\u2019ll be running afternoon practices from 2:00 to 5:00 on May 2, 10, and 24, and from 2:00 to 8:30 (with dinner served) on May 16. These will be in the math building at Stanford (you all know where that is I\u2019m sure), and we will be giving more details at the first practice. Suffice it to say practices will be a little different this year, so you\u2019ll have to show up to see what we mean..\r\n\r\n\r\nAnd now on to our alphabetical listing of A and B team qualifiers:\r\n\r\nA team \r\nArul\tVishal\r\nChang\tAlan\r\nChen\tYanping\r\nChen\tLewis\r\nChu\tTimothy\r\nLiu\tAndrew\r\nLiu\tBowei\r\nLu\tAnthony\r\nO'Dorney\tEvan\r\nSaxena\tAmrit\r\nShweh\tKevin\r\nWang\tBayley\r\nWein\tAlex\r\nYang\tKevin\r\nYang\tPatrick\r\n\r\n\r\nB team\r\nBoyle\tJohn\r\nBurks\tMichael\r\nChen\tWilliamZ\r\nchoi\tdongho\r\nDay\tCynthia\r\nGu\tAlbert\r\nHeh\tKevin\r\nHo\tTony\r\nKouevda\tNikita\r\nKulkarni\tArchit\r\nLau\tMatthew\r\nSung\tFelix\r\nSwernofsky\tJoseph\r\nYoung\tBrandon\r\nZhang\tBrian", "Solution_5": "Eh...what if we can't exactly remember how many tryouts we went to? :P", "Solution_6": "ah yes. i guess in that case you could send an email to operations at mathleague dot org and someone can look that up for you. same thing if you're not sure if you filled out your application correctly/at all..\r\n\r\nBTW i'm assuming everyone got the letter above via email as well; if you didn't and think you should have, send an email to operations and we'll try to figure out what went wrong..", "Solution_7": "For those flying, is the bus expense taken off the price or not?", "Solution_8": "^It should be, but sadly it isn't =(", "Solution_9": "I did not get the email.\r\n\r\nIn response to Silas' post below: Yes, got the new one.", "Solution_10": "I have not been receiving any emails either, and I have signed up numerous times over the past few months.", "Solution_11": "Same here, even though signed up for team and qualified for A team.", "Solution_12": "Same here. I signed up for ARML, then signed up for mailing list twice with the same email account, no emails yet.", "Solution_13": "The email was just resent to a smaller mailing list (just this year's team). Did you get it this time?", "Solution_14": "Yes, I received this email.", "Solution_15": "probably better if anyone who didn't get the email contact mathleague directly rather than posting here - we do want to keep this one non-spammy.. :)", "Solution_16": "Hey,\r\n\r\nJust wondering, for the AIME scores used in the ARML index, is the team based on last years AIME scores or this years AIME scores?\r\n\r\nThanks,\r\nTimothy Chu", "Solution_17": "[quote=\"Scrambled\"]is the team based on last years AIME scores[/quote]\r\n\r\nI wish it were :( This year's scores are used in the index." } { "Tag": [ "\\/closed" ], "Problem": "Where can I find an exhaustive list of permitted attachment types? I can't find it in FAQ.", "Solution_1": "Here you are: http://www.mathlinks.ro/Forum/faq.php?mode=attach\r\nThe 0-s there mean that there is no limit for the upload of one single file (other than your own total quota).", "Solution_2": "Thank you! :)" } { "Tag": [ "geometry", "trapezoid" ], "Problem": "number 1 solution was given to me by someone else that I talk to :lol: \r\n\r\n\r\nIn a trapezoid the bases are 5 inches and 8 inches, and the legs are 4 inches and 6 inches. If the legs are extended to meet in a point, how much must the shorter leg be extended?", "Solution_1": "Trapezoids don't have medians...", "Solution_2": "Label the length extended from the leg whose side is $4, x$. If we also extend the leg whose side is $6$ until it meets the extended length $x$, we notice that we form $2$ similar triangles. So we have \\[\\frac{x}{x+4}=\\frac{5}{8}\\Rightarrow 8x=5x+20 \\Rightarrow x=\\frac{20}{3}\\]", "Solution_3": "nice explanation :) \r\n\r\nso you could also find the extension needed for the longer leg also if you wanted", "Solution_4": "[quote=\"i_like_pie\"]Trapezoids don't have medians...[/quote]\r\n\r\nYes they do.", "Solution_5": "As he solved\r\nif we label extension of longer side as $a$\r\nthen $\\frac{5}{8}=\\frac{a}{6+a}$\r\nso $a=10$ inches[/b]" } { "Tag": [ "email" ], "Problem": "[quote]\u03a4\u039f \u03a3\u0395\u039c\u0399\u039d\u0391\u03a1\u0399\u039f \u0391\u039d\u0391\u039b\u03a5\u03a3\u0397\u03a3 \u039a\u0391\u0399 \u039a\u0391\u03a4\u0391\u03a3\u039a\u0395\u03a5\u0397\u03a3 \u039c\u0391\u0398\u0397\u039c\u0391\u03a4\u0399\u039a\u03a9\u039d \u03a0\u03a1\u039f\u0392\u039b\u0397\u039c\u0391\u03a4\u03a9\u039d \n(\u03a3\u03a5\u039d\u03a4\u039f\u039d\u0399\u03a3\u03a4\u0397\u03a3 : \u0395\u03a5\u0393\u0395\u039d\u0399\u039f\u03a3 \u0391\u0393\u0393\u0395\u039b\u039f\u03a0\u039f\u03a5\u039b\u039f\u03a3) \u0398\u0391 \u0393\u0399\u039d\u0395\u0399 \u039a\u0391\u039d\u039f\u039d\u0399\u039a\u0391 \u03a4\u0397\u039d \n\u03a0\u0391\u03a1\u0391\u03a3\u039a\u0395\u03a5\u0397 26 I\u0391\u039d\u039f\u03a5\u0391\u03a1\u0399\u039f\u03a5 2007 \n\u03a3\u03a4\u0399\u03a3 17:00, \u039a\u03a4\u0399\u03a1\u0399\u039f \u0395, 2\u03bf\u03c2 \u039f\u03a1\u039f\u03a6\u039f\u03a3, \n\u03a3\u03a4\u0397 \u0391\u0399\u0398\u039f\u03a5\u03a3\u0391 \u03a3\u0395\u039c\u0399\u039d\u0391\u03a1\u0399\u03a9\u039d \u03a4\u039f\u03a5 \u03a4\u039f\u039c\u0395\u0391 \u039c\u0391\u0398\u0397\u039c\u0391\u03a4\u0399\u039a\u03a9\u039d (\u03a3\u03a4\u0397\u039d \u03a0\u0395\u03a1\u0399\u03a0\u03a4\u03a9\u03a3\u0397 \u03a0\u039f\u03a5 \n\u0391\u03a5\u03a4\u039f \u0394\u0395\u039d \u0395\u0399\u039d\u0391\u0399 \u0395\u03a6\u0399\u039a\u03a4\u039f \u039b\u039f\u0393\u03a9 \u03a4\u03a9\u039d \u039a\u0399\u039d\u0397\u03a4\u039f\u03a0\u039f\u0399\u0397\u03a3\u0395\u03a9\u039d \u03a4\u03a9\u039d \u03a6\u039f\u0399\u03a4\u0397\u03a4\u03a9\u039d\n\u0395\u03a7\u0395\u0399 \u0395\u039e\u0391\u03a3\u03a6\u0391\u039b\u0399\u03a3\u03a4\u0395\u0399 \u0391\u039b\u039b\u0397 \u0391\u0399\u0398\u039f\u03a5\u03a3\u0391 \u03a3\u03a4\u039f \u039a\u03a4\u0399\u03a1\u0399\u039f \u0394\u0399\u039f\u0399\u039a\u0397\u03a3\u0397\u03a3). \n \n\u039f\u0399 \u03a3\u03a5\u039c\u039c\u0395\u03a4\u0395\u03a7\u039f\u039d\u03a4\u0395\u03a3 \u039a\u0391\u039b\u039f \u0398\u0391 \u0395\u0399\u039d\u0391\u0399 \u039d\u0391 \u0395\u03a1\u0398\u039f\u03a5\u039d \u039b\u0399\u0393\u039f \u039d\u03a9\u03a1\u0399\u03a4\u0395\u03a1\u0391 \u0391\u03a0\u039f \u03a4\u0399\u03a3 17:00\n\u03a0\u03a1\u039f\u039a\u0395\u0399\u039c\u0395\u039d\u039f\u03a5 \u039d\u0391 \u0395\u039e\u0391\u03a3\u03a6\u0391\u039b\u0399\u03a3\u03a4\u0395\u0399 \u039f \u039a\u0391\u039b\u03a5\u03a4\u0395\u03a1\u039f\u03a3 \u0394\u03a5\u039d\u0391\u03a4\u039f\u03a3 \u03a3\u03a5\u039d\u03a4\u039f\u039d\u0399\u03a3\u039c\u039f\u03a3.\n \n\u03a3\u0391\u039d \u0398\u0395\u039c\u0391 \u03a4\u039f\u03a5 \u03a3\u0395\u039c\u0399\u039d\u0391\u03a1\u0399\u039f\u03a5 \u0395\u03a7\u0395\u0399 \u039a\u0391\u03a4' \u0391\u03a1\u03a7\u0397\u039d \u03a0\u03a1\u039f\u03a4\u0391\u0398\u0395\u0399 \u03a4\u039f \u0398\u0395\u03a9\u03a1\u0397\u039c\u0391 \u03a4\u039f\u03a5 FERMAT (x^n + \ny^n = 1).[/quote]\r\n\u0395\u03af\u03c0\u03b1 \u03bd' \u03b1\u03bd\u03bf\u03af\u03be\u03c9 \u03bd\u03ad\u03bf topic \u03b1\u03c6\u03bf\u03cd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03ae\u03b4\u03b7 3-4 \u03ac\u03c4\u03bf\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd\u03c4\u03b1\u03b9.\u0395\u03b4\u03ce \u03b8' \u03b1\u03bd\u03b1\u03ba\u03bf\u03b9\u03bd\u03ce\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b1 \u03b8\u03ad\u03bc\u03b1\u03c4\u03b1 \u03c4\u03c9\u03bd \u03b5\u03c0\u03cc\u03bc\u03b5\u03bd\u03c9\u03bd \u03c3\u03b5\u03bc\u03b9\u03bd\u03b1\u03c1\u03af\u03c9\u03bd \u03ba\u03b1\u03b8\u03ce\u03c2 \u03c4\u03bf site \u03c4\u03bf\u03c5\u03c2 \u03b4\u03b5\u03bd \u03b1\u03bd\u03b1\u03bd\u03b5\u03ce\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c5\u03c7\u03bd\u03ac.", "Solution_1": "akougete endiaferon, to seminario einai anoikto gia olous?", "Solution_2": "\u03c6\u03c5\u03c3\u03b9\u03ba\u03b1 \u03ba\u03b1\u03b9 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b1\u03bd\u03bf\u03b9\u03ba\u03c4\u03bf \u03b3\u03b9\u03b1 \u03c4\u03bf\u03bd \u03ba\u03b1\u03b8\u03b5\u03bd\u03b1, :)", "Solution_3": "ok :) , opote tha pao", "Solution_4": "\u03b5\u03bb\u03c0\u03b9\u03b6\u03c9 \u03bf \u0398\u03b1\u03bd\u03bf\u03c2 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03bc\u03b1\u03c2 \u03c4\u03b7\u03bd \u03ba\u03b1\u03bd\u03b5\u03b9 ,,, \u03bf\u03c0\u03bf\u03b9\u03bf\u03c2 \u03b8\u03b5\u03bb\u03b5\u03b9 \u03bd\u03b1 \u03c0\u03b1\u03b5\u03b9 \u03b1\u03c2 \u03b6\u03b7\u03c4\u03b7\u03c3\u03b5\u03b9 \u03bb\u03b5\u03c0\u03c4\u03bf\u03bc\u03b5\u03c1\u03b9\u03b5\u03c2 \u03c4\u03bf\u03c5 \u03ba\u03c4\u03b9\u03c1\u03b9\u03bf\u03c5 (\u03b5\u03b9\u03bd\u03b1\u03b9 \u03b4\u03c5\u03c3\u03ba\u03bf\u03bb\u03bf \u03bd\u03b1 \u03c4\u03bf \u03b2\u03c1\u03b5\u03b9\u03c2 ) :D", "Solution_5": "\u03a0\u03bf\u03b9\u03b1 \u03b1\u03af\u03b8\u03bf\u03c5\u03c3\u03b1 \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c5\u03c4\u03ae \u03c0\u03bf\u03c5 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b7 \u03c3\u03c5\u03b6\u03ae\u03c4\u03b7\u03c3\u03b7??\r\n\u0395\u03af\u03bd\u03b1\u03b9 \u03bc\u03af\u03b1 \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03ad\u03c2 \u03c0\u03bf\u03c5 \u03b3\u03c1\u03ac\u03c6\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03b9 \u03b3\u03af\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bc\u03b1\u03b8\u03ae\u03bc\u03b1\u03c4\u03b1??", "Solution_6": "\u03b5\u03b9\u03bd\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03c3\u03c7\u03bf\u03bb\u03b7 \u03a3.\u0395\u039c\u03a6\u0395(\u03b5\u03c6\u03b1\u03c1\u03bc\u03bf\u03c3\u03bc\u03b5\u03bd\u03c9\u03bd \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03c9\u03bd \u03ba\u03b1\u03b9 \u03c6\u03c5\u03c3\u03b9\u03ba\u03c9\u03bd \u03b5\u03c0\u03b9\u03c3\u03c4\u03b7\u03bc\u03c9\u03bd\r\n\u03c4\u03bf\u03bc\u03b5\u03b1\u03c2 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03c9\u03bd -- \u03ba\u03c4\u03b7\u03c1\u03b9\u03bf \u0395.\r\n2\u03bf\u03c2 \u03bf\u03c1\u03bf\u03c6\u03bf\u03c2 \r\n\u03b1\u03b9\u03b8\u03bf\u03c5\u03c3\u03b1 \u03c3\u03b5\u03bc\u03b9\u03bd\u03b1\u03c1\u03b9\u03c9\u03bd\r\n\u03bc\u03b1\u03bb\u03bb\u03bf\u03bd \u03b4\u03b5\u03bd \u03b8\u03b1 \u03ba\u03b1\u03c4\u03b1\u03c6\u03b5\u03c1\u03c9 \u03bd\u03b1 \u03b5\u03c1\u03b8\u03c9 :( \r\n[hide]\u03b8\u03b1 \u03b4\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03b5\u03b9\u03c3\u03bf\u03b4\u03bf \u03c4\u03bf\u03c5\u03c2 \u03ba\u03b1\u03c4\u03b1\u03bb\u03b7\u03c8\u03b5\u03b9\u03b5\u03c2 \u03c4\u03b9\u03c2 \u03c3\u03c7\u03bf\u03bb\u03b7\u03c2 ! \u03b4\u03b5 \u03b8\u03b1 \u03b5\u03c7\u03b5\u03b9\u03c2 \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1 \u03bd\u03b1 \u03bc\u03c0\u03b5\u03b9\u03c2 , \u03b5\u03b9\u03bd\u03b1\u03b9 \u03ba\u03b1\u03bb\u03b1 \u03c0\u03b1\u03b9\u03b4\u03b9\u03b1 :) [/hide]", "Solution_7": ":blush: \r\n :blush: \r\n\r\n\u03b2\u03b1\u03c3\u03b9\u03ba\u03ac \u03b5\u03af\u03c7\u03b1 \u03c3\u03c4\u03b5\u03af\u03bb\u03b5\u03b9 email \u03c4\u03b7\u03bd \u03a4\u03b5\u03c4\u03ac\u03c1\u03c4\u03b7 \u03ba\u03b1\u03b9 \u03bc\u03bf\u03c5 \u03b1\u03c0\u03ac\u03bd\u03c4\u03b7\u03c3\u03b5 \u03c4\u03b7\u03bd \u03a0\u03ad\u03bc\u03c0\u03c4\u03b7 \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03c4\u03bf \u03b5\u03af\u03c7\u03b1 \u03b4\u03b5\u03b9 :( \r\n\u03b1\u03c5\u03c4\u03ae \u03c4\u03b7 \u03c6\u03bf\u03c1\u03ac \u03cc\u03bc\u03c9\u03c2 \u03b8\u03b1 \u03b5\u03af\u03bc\u03b1\u03b9 \u03b5\u03ba\u03b5\u03af!", "Solution_8": "\u03b4\u03b5\u03bd \u03c0\u03b5\u03b9\u03c1\u03b1\u03b6\u03b5\u03b9 :) \r\n@thatin: \u03b5\u03c3\u03c5 \u03b8\u03b1 \u03bc\u03b9\u03bb\u03b7\u03c3\u03b5\u03b9\u03c2 ?", "Solution_9": "[quote=\"kostas\"]\u03b5\u03bb\u03c0\u03b9\u03b6\u03c9 \u03bf \u0398\u03b1\u03bd\u03bf\u03c2 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03bc\u03b1\u03c2 \u03c4\u03b7\u03bd \u03ba\u03b1\u03bd\u03b5\u03b9 ,,, [/quote]\r\n\r\nsimpairana pws to ikseres...\r\nnai", "Solution_10": "\u03c4\u03bf \u03b7\u03be\u03b5\u03c1\u03b1 , :P", "Solution_11": "telika de tha ertheis?", "Solution_12": "\u0395\u03b3\u03ce \u03cc\u03bc\u03c9\u03c2, \u03b8\u03b1 \u03ad\u03c1\u03b8\u03c9.\u03a4\u03b7\u03bd \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03b7 \u03c6\u03bf\u03c1\u03ac,\u03c7\u03bc\u03bc,\u03b5,\u03c7\u03bc\u03bc,...,\u03b5,thatiiiiin;;;;;; :mad: \r\n[hide]\u0391\u03bb\u03ae\u03b8\u03b5\u03b9\u03b1,\u03ad\u03bd\u03b1 pm \u03c0\u03bf\u03c5 \u03c3\u03bf\u03c5 \u03ad\u03c7\u03c9 \u03c3\u03c4\u03b5\u03af\u03bb\u03b5\u03b9,\u03c4\u03bf \u03b5\u03af\u03b4\u03b5\u03c2;[/hide]", "Solution_13": "[quote=\"thatin\"]telika de tha ertheis?[/quote]\r\n\u03bf\u03c7\u03b9, \u03b1\u03bb\u03bb\u03b1 \u03bf\u03c0\u03bf\u03b9\u03bf\u03c2 \u03b8\u03b5\u03bb\u03b5\u03b9 \u03b1\u03c2 \u03c0\u03b5\u03b9 \u03c4\u03b9 \u03ba\u03b1\u03bd\u03b1\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03b5\u03b9\u03c0\u03b1\u03c4\u03b5 ,\u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03c9 :)", "Solution_14": "@Fischerman\r\n\u03cc\u03c7\u03b9 \u03b1\u03ba\u03cc\u03bc\u03b1!\r\n\u03b5, \u03c4\u03b9? \u03ba\u03ac\u03bd\u03b1\u03c4\u03b5 \u03bc\u03b9\u03b1 \u03c9\u03c1\u03b1\u03b9\u03cc\u03c4\u03b1\u03c4\u03b7 \u03b5\u03b9\u03c3\u03b1\u03b3\u03c9\u03b3\u03ae!\r\n\r\n\r\n\u03c0\u03b1\u03c1\u03bf\u03c5\u03c3\u03b9\u03ac\u03c3\u03c4\u03b7\u03ba\u03b5 \u03b7 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03c0\u03b1\u03c1\u03bf\u03c5\u03c3\u03b9\u03ac\u03c3\u03c4\u03b7\u03ba\u03b5 \u03c3\u03c4\u03bf \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf \u03ba\u03b1\u03b9 \u03bf \u0391\u03b3\u03b3\u03b5\u03bb\u03cc\u03c0\u03bf\u03c5\u03bb\u03bf\u03c2 \u03bc\u03af\u03bb\u03b7\u03c3\u03b5 \u03b3\u03b9\u03b1 \u03c4\u03c1\u03b9\u03c4\u03bf\u03b2\u03ac\u03b8\u03bc\u03b9\u03b5\u03c2 \u03ba\u03b1\u03bc\u03c0\u03cd\u03bb\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03b5\u03b9\u03b4\u03b9\u03ba\u03cc\u03c4\u03b5\u03c1\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03b9\u03c2 \u03b5\u03c6\u03b1\u03c0\u03c4\u03bf\u03bc\u03ad\u03bd\u03b5\u03c2 \u03ba\u03b1\u03bc\u03c0\u03ae\u03c2 \u03c4\u03bf\u03c5\u03c2\r\n\r\n\u03c4\u03b7\u03bd \u03b5\u03c0\u03cc\u03bc\u03b5\u03bd\u03b7 \u03c0\u03b1\u03c1\u03b1\u03c3\u03ba\u03b5\u03c5\u03ae \u03b8\u03b1 \u03b3\u03af\u03bd\u03b5\u03b9\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\naskisi gia to spiti:\r\naposeikste to avatar mou", "Solution_15": "\u03a4\u03bf \u03c3\u03b5\u03bc\u03b9\u03bd\u03ac\u03c1\u03b9\u03bf \u03b1\u03bd \u03be\u03ad\u03c1\u03b5\u03c4\u03b5 \u03b8\u03b1 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03ba\u03ac\u03b8\u03b5 \u03a0\u03b1\u03c1\u03b1\u03c3\u03ba\u03b5\u03c5\u03ae?", "Solution_16": "\u03b5\u03b9\u03bd\u03b1\u03b9 \u03ba\u03b1\u03b8\u03b5 \u03a0\u03b1\u03c1\u03b1\u03c3\u03ba\u03b5\u03c5\u03b7 \u03c4\u03b7\u03bd \u03b9\u03b4\u03b9\u03b1 \u03c9\u03c1\u03b1", "Solution_17": "\u03a4\u03b7\u03bd \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03b7 \u03a0\u03b1\u03c1\u03b1\u03c3\u03ba\u03b5\u03c5\u03ae (2/2) \u03b4\u03b5\u03bd \u03ad\u03b3\u03b9\u03bd\u03b5 \u03c4\u03bf \u03c3\u03b5\u03bc\u03b9\u03bd\u03ac\u03c1\u03b9\u03bf.", "Solution_18": "kserei kaneis an tha ginei auti tin paraskeui?", "Solution_19": "\u0394\u03b5\u03bd \u03ad\u03c7\u03bf\u03c5\u03bd \u03c3\u03c4\u03b5\u03af\u03bb\u03b5\u03b9 mail.", "Solution_20": "emena pote de mou exei erthei mail...\r\nama erthei pes mou plz\r\nexeis pm...!", "Solution_21": "\u03c3\u03c5\u03bd\u03b5\u03c7\u03b9\u03b6\u03b5\u03b9 \u03bd\u03b1 \u03b3\u03b9\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c5\u03c4\u03bf \u03c4\u03bf \u03c3\u03b5\u03bc\u03b9\u03bd\u03b1\u03c1\u03b9\u03bf ? \u03b1\u03c5\u03c4\u03b7 \u03c4\u03b7\u03bd \u03a0\u03b1\u03c1\u03b1\u03c3\u03ba\u03b5\u03c5\u03b7 \u03b8\u03b1 \u03b3\u03b9\u03bd\u03b5\u03b9?", "Solution_22": "[quote]\u0395 \u039b \u0395 \u03a5 \u0398 \u0395 \u03a1 \u039f \u03a3 \u0395 \u039c \u0399 \u039d \u0391 \u03a1 \u0399 \u039f\n\u0391\u03bd\u03ac\u03bb\u03c5\u03c3\u03b7\u03c2 \u03ba\u03b1\u03b9 \u039a\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03c5\u03ae\u03c2 \u039c\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ce\u03bd \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03ac\u03c4\u03c9\u03bd\nhttp://www.math.ntua.gr/~probsolv/ \n\n(\u03c4\u03c1\u03af\u03c4\u03bf\u03c2 \u03c7\u03c1\u03cc\u03bd\u03bf\u03c2 \u03bb\u03b5\u03b9\u03c4\u03bf\u03c5\u03c1\u03b3\u03af\u03b1\u03c2, 2007-2008)\n\u03a3\u03c5\u03bd\u03c4\u03bf\u03bd\u03b9\u03c3\u03c4\u03ae\u03c2 : \u0395\u03c5\u03b3\u03ad\u03bd\u03b9\u03bf\u03c2 \u0391\u03b3\u03b3\u03b5\u03bb\u03cc\u03c0\u03bf\u03c5\u03bb\u03bf\u03c2\n\n\n\u03a4\u03bf \u03a3\u03b5\u03bc\u03b9\u03bd\u03ac\u03c1\u03b9\u03bf \u03b1\u03c0\u03b5\u03c5\u03b8\u03cd\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b5 \u03c6\u03bf\u03b9\u03c4\u03b7\u03c4\u03ad\u03c2 \u03bf\u03c0\u03bf\u03b9\u03bf\u03c5\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03ad\u03c4\u03bf\u03c5\u03c2. \u03a3\u03c4\u03bf \u03c0\u03b1\u03c1\u03b5\u03bb\u03b8\u03cc\u03bd \u03c4\u03bf \u03ad\u03c7\u03bf\u03c5\u03bd \u03c0\u03b1\u03c1\u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03ae\u03c3\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03b4\u03b9\u03b4\u03ac\u03c3\u03ba\u03bf\u03bd\u03c4\u03b5\u03c2 \u039c\u03ad\u03c3\u03b7\u03c2 \u0395\u03ba\u03c0\u03b1\u03af\u03b4\u03b5\u03c5\u03c3\u03b7\u03c2. \u0395\u03af\u03bd\u03b1\u03b9 \u03b1\u03bd\u03bf\u03b9\u03c7\u03c4\u03cc \u03c3\u03b5 \u03c3\u03c5\u03bd\u03b1\u03b4\u03ad\u03bb\u03c6\u03bf\u03c5\u03c2 \u03c4\u03bf\u03c5 \u03a4\u03bf\u03bc\u03ad\u03b1 \u039c\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ce\u03bd \u03c4\u03bf\u03c5 \u0395\u039c\u03a0 (\u03bf\u03c1\u03b9\u03c3\u03bc\u03ad\u03bd\u03bf\u03b9 \u03ad\u03c7\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf \u03c0\u03b1\u03c1\u03b5\u03bb\u03b8\u03cc\u03bd \u03c0\u03b1\u03c1\u03bf\u03c5\u03c3\u03b9\u03ac\u03c3\u03b5\u03b9 \u03b8\u03ad\u03bc\u03b1\u03c4\u03b1). \n\u03a4\u03b1 \u03b8\u03ad\u03bc\u03b1\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03b1\u03bd\u03b1\u03c0\u03c4\u03cd\u03c3\u03c3\u03bf\u03bd\u03c4\u03b1\u03b9 (\u03b2\u03b1\u03c3\u03b9\u03ba\u03ac \u03b1\u03c0\u03cc \u03c4\u03bf\u03c5\u03c2 \u03c6\u03bf\u03b9\u03c4\u03b7\u03c4\u03ad\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03af\u03b4\u03b9\u03bf\u03c5\u03c2) \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03ba\u03c4\u03cc\u03c2 \u03cd\u03bb\u03b7\u03c2, \u03c0\u03c1\u03bf\u03c3\u03b5\u03b3\u03b3\u03af\u03c3\u03b9\u03bc\u03b1 \u03cc\u03bc\u03c9\u03c2 \u03bc\u03b5 \u03c3\u03c4\u03bf\u03b9\u03c7\u03b5\u03b9\u03ce\u03b4\u03b5\u03b9\u03c2 \u03b3\u03bd\u03ce\u03c3\u03b5\u03b9\u03c2. \u0391\u03c0\u03ce\u03c4\u03b5\u03c1\u03bf\u03c2 \u03c3\u03c4\u03cc\u03c7\u03bf\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03bf\u03c1\u03b3\u03ac\u03bd\u03c9\u03c3\u03b7 \u03b8\u03b5\u03bc\u03ac\u03c4\u03c9\u03bd \u03c3\u03b5 \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03b1-\u03b5\u03c1\u03b3\u03b1\u03c3\u03af\u03b5\u03c2, \u03bc\u03b5 \u03b1\u03bb\u03bb\u03b5\u03c0\u03ac\u03bb\u03bb\u03b7\u03bb\u03b1 \u03b2\u03ae\u03bc\u03b1\u03c4\u03b1, \u03ad\u03c4\u03c3\u03b9 \u03ce\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03ba\u03b1\u03b8\u03bf\u03b4\u03b7\u03b3\u03b5\u03af\u03c4\u03b1\u03b9 \u03bf \u03bb\u03cd\u03c4\u03b7\u03c2 \u03c3\u03c4\u03bf \u03bd\u03b1 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03b5\u03af \u03c4\u03b9\u03c2 \u03b3\u03bd\u03ce\u03c3\u03b5\u03b9\u03c2 \u03c4\u03bf\u03c5 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c3\u03b5\u03b3\u03b3\u03af\u03c3\u03b5\u03b9 \u03c4\u03bf \u03bd\u03ad\u03bf \u03ba\u03b1\u03b9 \u03ac\u03b3\u03bd\u03c9\u03c3\u03c4\u03bf. \n\u0397 \u03bb\u03ad\u03be\u03b7 \u03b5\u03bb\u03b5\u03cd\u03b8\u03b5\u03c1\u03bf \u03c5\u03c0\u03bf\u03b4\u03b7\u03bb\u03ce\u03bd\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03b1\u03bc\u03af\u03b1 \u03c5\u03c0\u03bf\u03c7\u03c1\u03ad\u03c9\u03c3\u03b7 \u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03bf\u03c7\u03ae\u03c2 \u03ba\u03b1\u03b9 \u03ba\u03b1\u03bc\u03af\u03b1 \u03c5\u03c0\u03bf\u03c7\u03c1\u03ad\u03c9\u03c3\u03b7 \u03c0\u03b1\u03c1\u03bf\u03c5\u03c3\u03af\u03b1\u03c2 (\u03b5\u03ba\u03c4\u03cc\u03c2 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03b7\u03b8\u03b9\u03ba\u03ae \u03c5\u03c0\u03bf\u03c7\u03c1\u03ad\u03c9\u03c3\u03b7 \u03bd\u03b1 \u03c0\u03b1\u03c1\u03bf\u03c5\u03c3\u03b9\u03ac\u03c3\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03c4\u03b7\u03bd \u03bf\u03bc\u03b9\u03bb\u03af\u03b1 \u03c0\u03bf\u03c5 \u03ad\u03c7\u03b5\u03b9 \u03b1\u03bd\u03b1\u03bb\u03ac\u03b2\u03b5\u03b9). \u0391\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1, \u03b4\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03ba\u03b1\u03bc\u03b9\u03ac \u03b1\u03bd\u03c4\u03b1\u03bc\u03bf\u03b9\u03b2\u03ae, \u03b2\u03b1\u03b8\u03bc\u03bf\u03bb\u03bf\u03b3\u03b9\u03ba\u03ae \u03ae \u03ac\u03bb\u03bb\u03b7, \u03c0\u03ad\u03c1\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03af\u03b7\u03c3\u03b7 \u03c4\u03b7\u03c2 \u03c0\u03b5\u03c1\u03b9\u03ad\u03c1\u03b3\u03b5\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd \u03b5\u03bd \u03b3\u03ad\u03bd\u03b5\u03b9 \u03b1\u03c0\u03cc\u03bb\u03b1\u03c5\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03b7 \u03af\u03b4\u03b9\u03b1 \u03b7 \u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03bf\u03c7\u03ae \u03c0\u03c1\u03bf\u03c3\u03c6\u03ad\u03c1\u03b5\u03b9. \n\n\n\n\u039a\u03ac\u03b8\u03b5 \u03a0\u03b1\u03c1\u03b1\u03c3\u03ba\u03b5\u03c5\u03ae, \u03ce\u03c1\u03b1 17:00\n\n\u0391\u03af\u03b8\u03bf\u03c5\u03c3\u03b1 \u03c3\u03b5\u03bc\u03b9\u03bd\u03b1\u03c1\u03af\u03c9\u03bd, \u03ba\u03c4\u03af\u03c1\u03b9\u03bf \u0395 (\u03a3\u0395\u039c\u03a6\u0395), 2\u03bf\u03c2 \u03cc\u03c1\u03bf\u03c6\u03bf\u03c2\n(\u03a0\u03bf\u03bb\u03c5\u03c4\u03b5\u03c7\u03bd\u03b5\u03b9\u03bf\u03cd\u03c0\u03bf\u03bb\u03b7, \u0396\u03c9\u03b3\u03c1\u03ac\u03c6\u03bf\u03c5)\n\n\u0388\u03bd\u03b1\u03c1\u03be\u03b7 :\t\t\u03a0\u03b1\u03c1\u03b1\u03c3\u03ba\u03b5\u03c5\u03ae 23 \u039d\u03bf\u03b5\u03bc\u03b2\u03c1\u03af\u03bf\u03c5 2007[/quote]", "Solution_23": "\u03b8\u03b1 \u03c0\u03b1\u03c2?", "Solution_24": "OXI :D .\u0392\u03c1\u03ae\u03ba\u03b1 \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03ad\u03c8\u03b1\u03c7\u03bd\u03b1 :lol: http://eclass.uoa.gr/courses/MATH206/" } { "Tag": [ "logarithms", "geometry", "complex analysis", "algebra", "polynomial", "function" ], "Problem": "Hi, I was wondering if anyone can help me with this inverse Laplace Transform using the complex inversion formula. Thanks.\r\n\r\nQuestion: Evaluate $ \\displaystyle\\mathcal{L}^{\\minus{}1}\\left\\{\\ln \\left(1\\plus{}\\frac{1}{s^2}\\right)\\right\\}$ by the inversion formula.", "Solution_1": "use a 3 keyhole contour\r\n\r\nWhat can you say about the branch-cut geometry of the holomorphic extension of $ \\text{Log} \\frac{P_n(z)}{Q_m(z)}$ to the left half-plane $ \\textnormal{Re}(z)<0$ with $ P$ and $ Q$ polynomials (of sufficient degree to make the inversion valid) when both the zeros and poles are arranged along the imaginary axis like your function? How will the function value along (horizontal) branch-cuts at each pole and zero differ depending on the order of each pole and zero?" } { "Tag": [], "Problem": "Which one of the following statement explain why the emission spectrum of atomic hydrogen is a line spectrum?\r\n\r\nA.an electron can possess any amount energy\r\nB.the electron in hydrogen atom can only occupy fixed energy levels.\r\nC.Transition of an electron from a higher energy level to a lowerenergy level emits light\r\nD.An emission spectrum consists of discrete lines which have fixed frequencies.", "Solution_1": "I would say B, but I have to review my stuff.", "Solution_2": "The correct answer is B. \r\n\r\nThe theory of fixed energy levels was constructed to answer the question as to why line spectrums were found for hydrogen (and everything else). The uniqueness of the frequency is the basis for spectral chemical analysis.\r\n\r\nWe may find new evidence that calls this theory into questions, but as far as I know it seems correct so far." } { "Tag": [], "Problem": "AB\r\n +BA\r\n ______\r\n CD4\r\nIn the correctly worked addition problem above, each letter represents a different nonzero digit. What is one possible value of the two-digit number represented above as AB ?\r\n\r\n(A) 74 (B) 31 (C) 86 (D) 47 (E) 95", "Solution_1": "[quote=\"sri340\"]AB\n +BA\n ______\n CD4\nIn the correctly worked addition problem above, each letter represents a different nonzero digit. What is one possible value of the two-digit number represented above as AB ?\n\n(A) 74 (B) 31 (C) 86 (D) 47 (E) 95[/quote]\r\n\r\nBy adding the units digits, we see that $ a \\plus{} b \\equiv 4 \\mod{10}$, this means that $ a \\plus{} b \\equal{} 4$ or $ a \\plus{} b \\equal{} 14$. Note how our sum is a 3 digit number, because of this there is no way for $ a \\plus{} b \\equal{} 4$ (try plugging in some numbers and see why). Because $ a \\ne b$, we can simply guess & check. If $ a \\equal{} 6$ and $ b \\equal{} 8$ we get $ 68 \\plus{} 86 \\equal{} 154$, since $ a \\ne b \\ne c \\ne d$, $ ab \\equal{} 86$.\r\n\r\nAlso, because you are given choices, simply add the two digits given and see if they equal $ 14$. This eliminates A, B, and D. Checking 86 and 95, we see only 86 works.", "Solution_2": "Another way is to note that \r\n$ 10a\\plus{}b\\plus{}10b\\plus{}a\\equal{}11(a\\plus{}b)$\r\nThe only number in the hundreds that is divisible by $ 11$ and ends in $ 4$ is $ 154$\r\nHence $ ab$ is $ 86$" } { "Tag": [ "counting", "distinguishability", "geometry", "geometric transformation", "rotation" ], "Problem": "One problem in the first volume of Aops asks,\r\nIn how many ways can four idential red chips and two identical white chips be arranged in a circle?\r\n\r\nI failed to solve the problem correctly, and I understood the solution the solutions manual gave me. However...I was wondering why and where my method failed.\r\n\r\nHere is my solution:\r\nWe take a simpler situation, and pretend we are ordering the chips in a line. There are six chips, so there are 6! ways the chips can be arranged. However, since the four red chips are indistinguishable, each possible arrangement is repeated 4! times. So we must divide 6! by 4!. We do the same thing with the white chips, so we have $ \\frac{6!}{4!2!}$. This equals 15, the number of possible arrangements if we were concerned with a line. I'm pretty sure I am right up to now (although this may be an unnecessary step...).\r\n\r\nHowever, we are dealing with a circle. Since rotations of the circles don't matter, we divide by n, the amount of chips on the circle...so my final solution is\r\n$ \\frac{15}6$. This is obviously wrong because it is not an integer number.\r\n\r\nOnce again, I know a different approach to the problem that will get me to the right answer, but I really want to know what is wrong with this logic. I am not looking for an alternate solution, but rather a correction.\r\n\r\nThanks", "Solution_1": "[quote=\"choidavid91\"]However, we are dealing with a circle. Since rotations of the circles don't matter, we divide by n[/quote]\r\n\r\nYou can only do that if all of the objects are distinguishable (I think).\r\n\r\nI'm not too sure about this logic myself.", "Solution_2": "When you convert a string of beads into a bracelet, you're overcounting different configurations by different amounts. I really can't find a clear way to explain precisely why this occurs, but if you list out all 15 configurations (it's not too bad), you should at least be able to see how this happens." } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "Let $f,g : \\mathbb{R}\\to\\mathbb{R}$. $f$ has Darboux property and for all $x\\in\\mathbb{R}$ \r\n\r\n$\\lim_{h \\to 0}\\frac{f(x+h)-g(x)}{h}=0$\r\n\r\nexists and is finite.\r\n\r\nProve that $f=g$ ... i just can't solve it :(", "Solution_1": "[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88958]some hints in romanian[/url]\r\n\r\n$\\lim_{h\\to0}f(x+h)=g(x)$ so $f$ has finite lateral limits which implies that it hasn't disconituities of the first order but also $f$ has Darboux property which means that it hasn't discontinuities of the second order thus $f$ is continous.", "Solution_2": "What is the Darboux property? And if the limit is 0, doesn't that imply its finite?", "Solution_3": "That $=0$ after the limit is probably a typo.\r\n\r\nThe Darboux property is also called the intermediate value property. For any value $t$ between $f(a)$ and $f(b)$ there is $c\\in(a,b)$ such that $f(c)=t$.", "Solution_4": "Yes, it's a typo :maybe: , i'm sorry.\r\n\r\nThanks for the links and indication." } { "Tag": [ "algebra", "polynomial", "function", "quadratics", "AMC", "AIME", "rational function" ], "Problem": "Assume that $x_1,x_2,\\ldots,x_7$ are real numbers such that \n\\[ \\begin{array}{r} x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7=1\\,\\,\\,\\,\\,\\,\\,\\, \\\\ 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7=12\\,\\,\\,\\,\\, \\\\ 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7=123. \\\\ \\end{array} \\] Find the value of \\[16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7.\\]", "Solution_1": "[hide]The corresponding coefficients with $x_k$ are $k^2,(k+1)^2,(k+2)^2$, and we're looking for a linear combination of equations which will yield the coefficient $(k+3)^2$ with $x_k$. By the method of undetermined coefficients we have\n\n$ak^2+b(k+1)^2+c(k+2)^2=(k+3)^2$\n\nExpanding and equating the coefficients with the corresponding powers of $k$, we get\n\n$a+b+c=1$\n$2b+4c=6$\n$b+4c=9$\n\nand the solution is $a=1, b=-3, c=3$\n\nHence the desired sum is $1\\cdot 1-3\\cdot 12+3\\cdot 123=334$[/hide]", "Solution_2": "Sorry if this is a little unclear, but:\r\n\r\n[hide=\"Solution\"]Denote the three equations, from top to bottom, as $(1)$, $(2)$, and $(3)$. Notice that $(2)-(1)=11$ and $(3)-(2)=111$ are both sequences of consecutive odd integers. Let's denote these sequences as $(4)$ and $(5)$, respectively. Now, notice that $(5)-(4)$ gives $2x_1+2x_2+\\cdots+2x_7=100$, which we can use to find the desired sequence. Our answer is $(3)+(5)+100=\\boxed{334}$.[/hide]", "Solution_3": "[hide]Let the left-hand sides of the equations be A, B, C, and D (from top to bottom).\n\nNotice that $z^2 + (z+2)^2 = 2(z^2 + 2z + 2) = 2((z+1)^2 + 1)$\n\nApplying this to the coefficients of the equations, $(A+C)/2 = B + (x_1 + x_2 + ... + x_7)$ and that $(B+D)/2 = C + (x_1 + x_2 + ... + x_7)$\n\nPlugging values of A, B, and C into the first equation, $x_1 + x_2 + ... + x_7 = 50$, so plugging this into the second equation gives:\n\n$(12+D)/2 = 123 + 50$, $\\boxed{D = 334}$[/hide]", "Solution_4": "Farenhajt, nice solution!", "Solution_5": "[quote=\"Farenhajt\"][hide]The corresponding coefficients with $x_k$ are $k^2,(k+1)^2,(k+2)^2$, and we're looking for a linear combination of equations which will yield the coefficient $(k+3)^2$ with $x_k$. By the method of undetermined coefficients we have\n\n$ak^2+b(k+1)^2+c(k+2)^2=(k+3)^2$\n\nExpanding and equating the coefficients with the corresponding powers of $k$, we get\n\n$a+b+c=1$\n$2b+4c=6$\n$b+4c=9$\n\nand the solution is $a=1, b=-3, c=3$\n\nHence the desired sum is $1\\cdot 1-3\\cdot 12+3\\cdot 123=334$[/hide][/quote]\r\n\r\ncan you explain this a little bit more?? It seems like a powerful system solving technique with patterns and I'd like to learn it", "Solution_6": "[quote=\"pkerichang\"]can you explain this a little bit more?? It seems like a powerful system solving technique with patterns and I'd like to learn it[/quote]\r\n\r\n[hide=\"Detailed explanation\"]Well, the general idea is this: We have $3$ equations with $7$ variables, hence there's no unique solution, so the idea of solving it should be discarded as too cumbersome. On the other hand, we can try to find numbers $a,b,c$ such that when we multiply the first equation by $a$, the second by $b$ and the third by $c$, and then add all that up, we get exactly the desired expression as LHS. (The \"shop\" term for the resulting equation is [i]linear combination[/i] of the given equations.)\n\nTherefore, in the first equation we'll get terms of the form $ak^2x_k$, in the second of the form $b(k+1)^2x_k$, and in the third $c(k+2)^2x_k.$ In the same time, the free terms will be $a\\cdot 1, b\\cdot 12, c\\cdot 123$.\n\nNow adding up must yield the terms of the form $(k+3)^2x_k$ for EVERY $k.$ So we write\n\n$ak^2+b(k+1)^2+c(k+2)^2=(k+3)^2$\n\nThis is where actual power technique comes handy, and it's called [i]the method of undetermined coefficients[/i] (at least I think that's the correct english term :)): We don't know what are the values of $a,b,c$, but we know that they must satisfy the above equation for every $k$. Let's expand and group the LHS and expand the RHS:\n\n\\begin{eqnarray*}ak^2+b(k^2+2k+1)+c(k^2+4k+4) &=& ak^2+bk^2+2bk+b+ck^2+4ck+4c\\\\ &=& (a+b+c)k^2+(2b+4c)k+(b+4c)\\end{eqnarray*}\n\n$(k+3)^2=k^2+6k+9$\n\nSo we get\n\n$(a+b+c)k^2+(2b+4c)k+(b+4c)=k^2+6k+9$\n\nNow the important point: These are two polynomials in $k$ and the [u]only[/u] way they could be equal for EVERY $k$ (this \"every\" is very crucial) is that all the corresponding coefficients (leading, linear and constant) are equal. Therefore we must have\n\n$a+b+c=1$\n$2b+4c=6$\n$b+4c=9$\n\nThis is the system of linear equations which isn't difficult to solve by any known method. For example, subtracting the third equation from the second we get $b=-3.$ Now substituting that into the third we get $-3+4c=9\\implies 4c=12\\implies c=3$. Then the first becomes $a-3+3=1\\implies a=1.$\n\nSo we have found $a,b,c$ such that we get what we want: If we multiply the first equation by $1$, the second by $-3$ and the third by $3$ and add that all up, we get the desired expression. So we must calculate the free term we'll obtain in the process. As we said, it must be $a\\cdot 1+b\\cdot 12+c\\cdot 123=1-36+369=334.$\n\nHope this helped :)\n\nNB. The method of undetermined coefficients is applicable in many different situations concerning polynomials (notably in calculating remainders in polynomial division and splitting a rational function into simple fractions), so having it in mind could sometimes be a life-saver.[/hide]", "Solution_7": "[quote=\"Farenhajt\"][quote=\"pkerichang\"]can you explain this a little bit more?? It seems like a powerful system solving technique with patterns and I'd like to learn it[/quote]\n\n[hide=\"Detailed explanation\"]Well, the general idea is this: We have $3$ equations with $7$ variables, hence there's no unique solution, so the idea of solving it should be discarded as too cumbersome. On the other hand, we can try to find numbers $a,b,c$ such that when we multiply the first equation by $a$, the second by $b$ and the third by $c$, and then add all that up, we get exactly the desired expression as LHS. (The \"shop\" term for the resulting equation is [i]linear combination[/i] of the given equations.)\n\nTherefore, in the first equation we'll get terms of the form $ak^2x_k$, in the second of the form $b(k+1)^2x_k$, and in the third $c(k+2)^2x_k.$ In the same time, the free terms will be $a\\cdot 1, b\\cdot 12, c\\cdot 123$.\n\nNow adding up must yield the terms of the form $(k+3)^2x_k$ for EVERY $k.$ So we write\n\n$ak^2+b(k+1)^2+c(k+2)^2=(k+3)^2$\n\nThis is where actual power technique comes handy, and it's called [i]the method of undetermined coefficients[/i] (at least I think that's the correct english term :)): We don't know what are the values of $a,b,c$, but we know that they must satisfy the above equation for every $k$. Let's expand and group the LHS and expand the RHS:\n\n\\begin{eqnarray*}ak^2+b(k^2+2k+1)+c(k^2+4k+4) &=& ak^2+bk^2+2bk+b+ck^2+4ck+4c\\\\ &=& (a+b+c)k^2+(2b+4c)k+(b+4c)\\end{eqnarray*}\n\n$(k+3)^2=k^2+6k+9$\n\nSo we get\n\n$(a+b+c)k^2+(2b+4c)k+(b+4c)=k^2+6k+9$\n\nNow the important point: These are two polynomials in $k$ and the [u]only[/u] way they could be equal for EVERY $k$ (this \"every\" is very crucial) is that all the corresponding coefficients (leading, linear and constant) are equal. Therefore we must have\n\n$a+b+c=1$\n$2b+4c=6$\n$b+4c=9$\n\nThis is the system of linear equations which isn't difficult to solve by any known method. For example, subtracting the third equation from the second we get $b=-3.$ Now substituting that into the third we get $-3+4c=9\\implies 4c=12\\implies c=3$. Then the first becomes $a-3+3=1\\implies a=1.$\n\nSo we have found $a,b,c$ such that we get what we want: If we multiply the first equation by $1$, the second by $-3$ and the third by $3$ and add that all up, we get the desired expression. So we must calculate the free term we'll obtain in the process. As we said, it must be $a\\cdot 1+b\\cdot 12+c\\cdot 123=1-36+369=334.$\n\nHope this helped :)\n\nNB. The method of undetermined coefficients is applicable in many different situations concerning polynomials (notably in calculating remainders in polynomial division and splitting a rational function into simple fractions), so having it in mind could sometimes be a life-saver.[/hide][/quote]\r\n\r\nThank you for this useful approach to solve systems of equations (the coefficient technique I know already, just not the name). Are there any circumstances where it's not applicable???", "Solution_8": "As far as I know, it can be used only for polynomials and rational functions (but of course, the scope of problems involving these is immense).", "Solution_9": "Farenhajit,could you solve this question using the coeffoicient method?I do have the solution but i just can't understand it..\r\n\r\n$1.2+2.3+3.4.....$.Find sum.\r\n\r\nNow in this ,They've equated the given thing to $A+Bn+C(n+1).....$\r\nWhy?", "Solution_10": "I'm not sure what do you mean, shady... Can you post some more details from their solution?", "Solution_11": "[quote=\"Farenhajt\"]$ak^2+b(k+1)^2+c(k+2)^2=(k+3)^2$[/quote]\r\n\r\nJust a note, at this point you can just substitute $k=0,-1,-2$ to get \\[ b+4c=9, \\] \\[ a+c=4, \\] and \\[ 4a+b=1, \\] from which you get the same solution $(1,-3,3)$. :)", "Solution_12": "Sure, but it's always better to work with variables, since that way you make sure that the method is applicable. (Remember that trigonometric sum? :))", "Solution_13": "[quote=\"Farenhajt\"]Remember that trigonometric sum? :)[/quote]\r\n\r\n:D Ok, sure.", "Solution_14": "i have a few questions.\r\n\r\nwhat are the free terms. i do not understand y u used these values.\r\n\r\nmy other question is, would you be able to use that method on this problem.\r\n\r\n$a_1+8a_2+27a_3+64a_4=1$\r\n$8a_1+27a_2+64a_3+125a_4=27$\r\n$27a_1+64a_2+12a_3+216a_4=125$\r\n$64a_1+125a_2+216a_3+343a_4=343$\r\n\r\nshow that\r\n$a_1(x+1)^3+a_2(x+2)^3+a_3(x+3)^3+a_4(x+4)^3\\equiv(2x+1)^3$", "Solution_15": "A free term is the number on the RHS of an equality. Let's say you have these two:\r\n\r\n$x=12$\r\n$y=7$\r\n\r\nWhen you multiply the first by $a$ and the second by $b$, you get\r\n\r\n$ax=12a$\r\n$by=7b$\r\n\r\nNow if you add the LHS's, you must also add the RHS's:\r\n\r\n$ax+by=12a+7b$\r\n\r\nDoes this help?\r\n\r\nAnd yes, this method can be used to solve the problem you posted, but I'll let someone else to show it (it's not too difficult).", "Solution_16": "thanks that has helped. ill will try to do the problem i posted with this method, later right now i have to head to class. :blush: :(", "Solution_17": "[quote=\"maokid7\"]i have a few questions.\n\nwhat are the free terms. i do not understand y u used these values.\n\nmy other question is, would you be able to use that method on this problem.\n\n$a_1+8a_2+27a_3+64a_4=1$\n$8a_1+27a_2+64a_3+125a_4=27$\n$27a_1+64a_2+12a_3+216a_4=125$\n$64a_1+125a_2+216a_3+343a_4=343$\n\nshow that\n$a_1(x+1)^3+a_2(x+2)^3+a_3(x+3)^3+a_4(x+4)^3\\equiv(2x+1)^3$[/quote]\r\n\r\n[hide]To show this, all you need to do is show that the two cubics are equal for at least 4 different values of $x$, because if they were not equal, their difference would be nonzero and of degree 3 or less, so it would have at most 3 roots. But finding 4 values of $x$ is easy, since the cases where $x$ is 0, 1, 2, and 3 are given.[/hide]", "Solution_18": "wow you guys sure made a big deal of an easy problem :D jk not really\r\nI just have an easy solution because when I first saw the problem I had just finished doing this problem: which is in aops vol 2\r\n[quote=\"maokid7\"]i have a few questions.\n\nwhat are the free terms. i do not understand y u used these values.\n\nmy other question is, would you be able to use that method on this problem.\n\n$a_{1}+8a_{2}+27a_{3}+64a_{4}=1$\n$8a_{1}+27a_{2}+64a_{3}+125a_{4}=27$\n$27a_{1}+64a_{2}+12a_{3}+216a_{4}=125$\n$64a_{1}+125a_{2}+216a_{3}+343a_{4}=343$\n\nshow that\n$a_{1}(x+1)^{3}+a_{2}(x+2)^{3}+a_{3}(x+3)^{3}+a_{4}(x+4)^{3}\\equiv(2x+1)^{3}$[/quote]\r\nanyway, all you do is notice that $f(x)=\\sum (x+k)^{2}*x_{k}$. F is obviously a quadratic equation. We have $f(1)=1,f(2)=12,f(3)=123$. Now look at differences: the first ones are $11,111$. The second level of differences must be a constant (because this is a quadratic), and we have $111-11=100$ must be this value. So we can easily find $f(4)$ using only basic arithmetic: $100+111=211$ is the next first level difference, so $f(4)=211+f(3)=\\boxed{334}$", "Solution_19": "[quote=\"shadysaysurspammed\"]Farenhajit,could you solve this question using the coeffoicient method?I do have the solution but i just can't understand it..\n\n$1.2+2.3+3.4.....$.Find sum.\n\nNow in this ,They've equated the given thing to $A+Bn+C(n+1).....$\nWhy?[/quote]\r\n\r\nI don't see why you need the A, B, C thing but here is a solution:\r\n\r\n$\\sum_{n=1}^{k}{n+\\frac{n+1}{10}}$\r\n$\\frac{1}{10}\\sum_{n=1}^{k}{11n+1}$\r\n$(\\frac{1}{10})(11\\binom{k+1}{2}+k)$\r\n$\\sum=\\frac{11k^{2}+13k}{20}$", "Solution_20": "[quote=\"Farenhajt\"][hide]The corresponding coefficients with $x_{k}$ are $k^{2},(k+1)^{2},(k+2)^{2}$, and we're looking for a linear combination of equations which will yield the coefficient $(k+3)^{2}$ with $x_{k}$. By the method of undetermined coefficients we have\n\n$ak^{2}+b(k+1)^{2}+c(k+2)^{2}=(k+3)^{2}$\n\nExpanding and equating the coefficients with the corresponding powers of $k$, we get\n\n$a+b+c=1$\n$2b+4c=6$\n$b+4c=9$\n\nand the solution is $a=1, b=-3, c=3$\n\nHence the desired sum is $1\\cdot 1-3\\cdot 12+3\\cdot 123=334$[/hide][/quote]\r\n\r\nUh... what did you do?", "Solution_21": "[quote=\"lingomaniac88\"][quote=\"Farenhajt\"][hide]The corresponding coefficients with $x_{k}$ are $k^{2},(k+1)^{2},(k+2)^{2}$, and we're looking for a linear combination of equations which will yield the coefficient $(k+3)^{2}$ with $x_{k}$. By the method of undetermined coefficients we have\n\n$ak^{2}+b(k+1)^{2}+c(k+2)^{2}=(k+3)^{2}$\n\nExpanding and equating the coefficients with the corresponding powers of $k$, we get\n\n$a+b+c=1$\n$2b+4c=6$\n$b+4c=9$\n\nand the solution is $a=1, b=-3, c=3$\n\nHence the desired sum is $1\\cdot 1-3\\cdot 12+3\\cdot 123=334$[/hide][/quote]\n\nUh... what did you do?[/quote]\r\n\r\nFarenhajt notes that the coefficient of every $x_{k}$ in the system is, in the first equation $k^{2}$, in the second equation $(k+1)^{2}$, and in the third equation $(k+2)^{2}$.\r\n\r\nThe goal of the problem is to find the value of the expression where every coefficient is $(k+3)^{2}$. Thus, we multiply the first equation by some real, add it to the second equation multiplied by some real, then to the third equation multiplied by some real, to get that desired $(k+3)^{2}$. \r\n\r\nTo do this we need to find reals $a, b, c$ that satisfy the equation Farenhajt specified, as coefficients of a polynomial of $k$. We equate coefficients to get his system and solve. \r\n\r\nI hope this is a little clearer.", "Solution_22": "first we know that\r\nif f(x)=ax^2+bx+c then (f(x1)-f(x2))/(x1-x2)=(f(x1+x2)-f(0))/((x1+x2)-0)\r\nit is not so difficult to prove it....\r\n\r\nand then we look at this question\r\nwe have f(x)=x1(x+1)^2+x2(x+2)^2+........x7(x+7)^2\r\nand f(0)=1 f(1)=12 f(2)=123 and we want to find f(4)\r\n\r\nthen it is easy.....", "Solution_23": "I'm sorry that I've made a mistake above \r\nthat we want to find f(3) instead of f(4)", "Solution_24": "um....DEAD TOPIC", "Solution_25": "Nobody mentioned this was an AIME problem. I think like 1987 or something like that.", "Solution_26": "Sorry for the revive, but no one has posted the following solution yet: \n[hide=Summation overkill]Rewrite all of the equations as follows:\n$\\sum_{i=1}^{7}i^2x_i = 1$\n$\\sum_{i=1}^{7}(i+1)^2x_i = 12$\n$\\sum_{i=1}^{7}(i+2)^2x_i = 123$\nAnd we would like to find the value of $\\sum_{i=1}^{7}(i+3)^2x_i$. We can rewrite the desired quantity as $\\sum_{i=1}^{7}i^2x_i + 6\\sum_{i=1}^{7}ix_i + 9\\sum_{i=1}^{7}x_i = 1 + 6\\sum_{i=1}^{7}ix_i + 9\\sum_{i=1}^{7}x_i$. The second equation rewrites to $2\\sum_{i=1}^{7}ix_i + \\sum_{i=1}^{7}x_i = 11$ and the third equation rewrites to $2\\sum_{i=1}^{7}ix_i + 2\\sum_{i=1}^{7}x_i = 61$. Solving this system of equations gives us $\\sum_{i=1}^{7}ix_i = -\\frac{39}{2}$ and $\\sum_{i=1}^{7}x_i = 50$. So our answer is $1 + 6\\left(-\\frac{39}{2}\\right) + 9(50) = \\boxed{334}$.", "Solution_27": "[hide=Solution]We want a cool solution to post on the aops forums, so we try to obliterate everything at the same time :P Thus, we want some linear combination of $x^2, (x+1)^2, (x+2)^2$ to give $(x+3)^2$. Trying random stuff and equating some coefficients gives that the first minus three times the second plus three times the third gives the desired (try it yourself if you haven't; its kinda magical!), so the answer is $1-36+369=\\boxed{334}$.[/hide]", "Solution_28": "[hide=Is this legal]\nUse finite differences\n$E_2-E_1=11$\n$E_3-E_2=111$\n$(E_3-E_2)-(E_2-E_1)=100$\nSince the second difference in a quadratic is constant, we know $(E_4-E_3)-(E_3-E_2)=100$\nSo $E_4-E_3=211$\n$E_4=334$\n[/hide]\n\nEdit: it is legal because we can say that the LHS are $E(1)$, $E(2)$, $E(3)$ of the function $E(n)=\\sum_{k=0}^6 x_k(k+n)^2$. Clearly this is quadratic wrt $n$.", "Solution_29": "Where did we get the $(x+3)^2$ from?", "Solution_30": "[quote=smartguy888]Where did we get the $(x+3)^2$ from?[/quote]\n\nI'm going to assume you're responding to my post, because its the most recent that uses that terminology.\n\nTo be honest, using $x$ there is a bit confusing.\n\n[hide=maybe this gives away some stuff so dont read it if you haven't solved this yet]The point is that the value we're looking for can be expressed as $\\sum_{i=1}^{7} (i+3)^2x_i$, so we want to use some combination of $i^2, (i+1)^2,$ and $(i+2)^2$ to create $(i+3)^2$ to obtain every term needed.[/hide]", "Solution_31": "tbh I think finite differences is overkill", "Solution_32": "Also, you automatically know that $(x+3)^2$ can be represented as a linear combination of $x^2, (x+1)^2, (x+2)^2$ because those 3 polynomials form a basis of all polynomials with degree at most $2$.", "Solution_33": "[hide=Solution]We have $a^2, (a+1)^2, (a+2)^2$ and want $(a+3)^2$, so take $x(a^2)+y(a+1)^2+z(a+2)^2=(a+3)^2$. This means $x+y+z=1, 2y+4z=6, y+4z=9$. Multiplying the last equation by $2$ gives $2y+8z=18$, so $4z=12 \\implies z=3$. This means $y=-3$ and $z=1$, so the answer is $1 \\cdot 1 - 3 \\cdot 12 + 3 \\cdot 123 = \\boxed{334}$.[/hide]" } { "Tag": [ "topology" ], "Problem": "I am getting stuck on the following problem, I have tried to use the map $ p_{*} : \\pi_1(S^n)\\to \\pi_1(\\mathbb{R}P^n), \\mbox{and} \\,\\,\\,p_{*} : H_{n}(S^n)\\to H_{n}(\\mathbb{R}P^n)$ but it seems not working :( \r\n\r\nLet $ n\\geq2$ and $ p : S^n\\to \\mathbb{R}P^n$ be the standard covering map. Prove that $ p$ is NOT nullhomotopic\r\n\r\nAny idea?", "Solution_1": "Here $ S^n$ is the universal cover for $ \\mathbb{R}P^n$, and so the covering map induces an isomorphism of $ \\pi_k$ for $ k > 1$. In particular $ p_* : \\pi_n(S^n) \\to \\pi_n(\\mathbb{R}P^n)$ is not the zero map.", "Solution_2": "thanks Nukular\r\n\r\nYour solution is great, I also figured out one solution for my own : \r\n\r\nLet $ \\tilde {p}$ be the lifting map from $ S^n \\to S^n$ of $ p : S^n\\to \\mathbb{R}P^n$, by the uniqueness of the lifting map I have $ \\tilde {p}\\equal{}Id_{S^n}$ so I have $ Id_{S^{n}}$ homotopic to constant map. This is a contradiction since $ S^n$ is not contractible", "Solution_3": "1234567a: your solution is almost the same as Nukular's. The reason you know the lifting map $ \\tilde{p}$ exists is because $ \\pi_1(S^n)\\equal{}0$. So in fact, if $ \\tilde{X}\\rightarrow X$ is a covering map, then by the lifting criterion, for $ n\\geq 2$, $ \\pi_n(\\tilde{X})\\cong\\pi_n(X)$ precisely because we can lift any map from $ S^n\\rightarrow X$ into its covering space and do all our homotopy stuff in the universal cover!" } { "Tag": [ "AMC", "AIME" ], "Problem": "I am finding following Engel book to be quite difficult. I have finished AoPS 2, can score about a 9 or 10 consistently on the AIME (provided that I don't make any careless errors), have no olympiad experience, and am taking WOOT. When providing solutions to the questions, for example, what is the motivation to find the solutions? Though the solution itself is important, how can we figure out the motivation to solve the problem?", "Solution_1": "try Problem Solving Through Problems it's nicer! I'm doing it right now :)", "Solution_2": "What chapter are you doing, specifically?\r\n\r\nThe first few are easy, but some of that NT stuff at the end blows my mind...", "Solution_3": "The Larson is much easier at the beginning, but gets hard fast :wink:", "Solution_4": "[quote=\"PenguinIntegral\"]What chapter are you doing, specifically?\n\nThe first few are easy, but some of that NT stuff at the end blows my mind...[/quote]\r\n\r\nDude... when I read the invariance chapter, i'm like what on earth is this?\r\n\r\nAlso, which is better for Olympiad preparation? Larson or PSS?", "Solution_5": "Well PSS is more difficult and longer...but if you can't do it, there's no point in bulling through it.", "Solution_6": "What exactly gives you trouble???", "Solution_7": "I'd imagine that he has to look at the solutions for most of the problems, and upon doing so the motivations for these solutions seems rather obscure.\r\n\r\n-If you have to look up solutions for most/all of the problems (especially in chapter 1), you should probably start easier...you gain very little by simply looking at the solution and a lot more by wrestling with the problem, understanding why certain things work and don't work, even if you never get the solution.\r\n\r\nI will say that Engel's solutions are rather minimal, which can be good because you still have to figure out large parts for yourself.", "Solution_8": "[quote=\"Phelpedo\"]I'd imagine that he has to look at the solutions for most of the problems, and upon doing so the motivations for these solutions seems rather obscure.\n\n-If you have to look up solutions for most/all of the problems (especially in chapter 1), you should probably start easier...you gain very little by simply looking at the solution and a lot more by wrestling with the problem, understanding why certain things work and don't work, even if you never get the solution.\n\nI will say that Engel's solutions are rather minimal, which can be good because you still have to figure out large parts for yourself.[/quote]\r\nThe motivation to decompose some large number into a sum of fourth powers still boggles me to this day. But I honestly agree with you, Karth should probably start easier. If he's in the same spot I was about a year ago and am in now where everything was/is way too hard or way too easy, I don't have any good suggestions.\r\n\r\nI have to second PSS for a large part of pre-olympiad prep. Larson and ACops skim way too much off the top.\r\n\r\nWould you mind posting some problems you are having difficulty with, Karth?" } { "Tag": [ "trigonometry", "quadratics", "algebra" ], "Problem": "the equation of a projectile can be written as $ y \\equal{} xtan\\theta \\minus{} \\frac {1}{4h}x^2(1 \\plus{} tan^2\\theta) \\; where \\frac {V^2}{2g} \\equal{} h , g > 0$\r\n(i)show that the point $ (X,Y)$can be hit by firing at two different angles $ \\theta_1$ and $ \\theta_2$ if $ X^2 < 4h(h \\minus{} Y)$\r\n(ii) show that no point ABOVE the x axis can be hit by firing at two different angles such that $ \\theta_1 < \\pi/4$ and $ \\theta_2 < \\pi/4$ \r\ni have completed part one but i have no idea how to approach part two", "Solution_1": "Treat $ \\tan \\theta$ as a variable and solve for it (it's a quadratic equation), you then get\r\n\\[ \\tan \\theta \\equal{} \\frac {2hx\\pm \\sqrt {4h^2x^2 \\minus{} x^4 \\minus{} 4hx^2y}}{x^2}\r\n\\]\r\nNow show that if $ x > 0$ then $ \\frac {2hx \\plus{} \\sqrt {4h^2x^2 \\minus{} x^4 \\minus{} 4hx^2y}}{x^2}\\geq 1$ (or it's undefined) and therefore it corresponds to $ \\theta \\geq \\frac {\\pi}{4}$." } { "Tag": [ "algebra", "functional equation", "IMO", "IMO 2003" ], "Problem": "His result for IMO 2003: 777 776. He missed perfect score by one point. That's really unlucky for him.\r\n\r\nWhat did he do wrong on problem 6? Why did he scored 6 points? Does anyone knows?", "Solution_1": "i belive that (from what he said) he didn't acutally solve the problem - it was more like a very stupid marking scheme!\r\n\r\nexample \"for not proving that ... \" -1p !!! etc.", "Solution_2": "I remember that Mihai Manea from Romania got 41/42 points .. that is pretty unlucky! :( .. this happened a few years ago.. I don't remember quite well when.. but it happened! :( ... I guess that all contestant who get 41/42 are unlucky.. one point away from perfectness! :(\r\n\r\ncheers :D", "Solution_3": "i was on the team that year (2001) and yeah mihai was good. the 6 which you get at an IMO can from a 3-4 to a 7, from my point of view. \r\n\r\nI myself got three 6-s last year, and one of them was a clear 6 (at the functional equation I forgot to mention that the solutions found verify the original equation - much as all the Romanian team did) and the other two, were 7-s ( i repeat in my opinion).", "Solution_4": "I heard that in the year Mihai Manea took 41 pnts, 4 42 pnts the organizers gave a notebook. So he didn't get one\r\n :(" } { "Tag": [ "AMC" ], "Problem": "What does the sign above BF=EC mean?", "Solution_1": "Question 11 of what? What sign?", "Solution_2": "it means arc" } { "Tag": [], "Problem": ":!: \r\n\u6211\u662f\u4e00\u540d\u4e2d\u5b66\u751f\uff0c\u662f\u4e00\u4f4d\u5de8\u65f6\u7f51\u6c11\uff0c\u4f46\u59cb\u7ec8\u6ca1\u80fd\u627e\u5230\u597d\u7684\u6570\u5b66\u8d44\u6e90\u7f51\u7ad9\r\n\u5e0c\u671bmathlinks\u53ef\u4ee5\u6539\u53d8\r\n\u8c22\u8c22\u5927\u5bb6", "Solution_1": "\u54e5\u4eec\u4f60\u5728\u54ea\u4e2a\u57ce\u5e02\u5440? \u51e0\u5e74\u7ea7?", "Solution_2": "\r\n\u56fd\u5185\u6709\u597d\u591a\u5965\u6570\u7f51\u7ad9\u8ba8\u8bba\u95ee\u9898\u548c\u53d1\u8868\u6750\u6599\u66f4\u65b9\u4fbf.\u5982\"\u6570\u5b66\u7ade\u8d5b\u8bba\u575b\",\"\u5965\u6570\u4e4b\u5bb6\u8bba\u575b\".\u53ef\u5728GOOGLE\u6253\u4e2d\u6587\u641c\u7d22.\r\n\u6211\u5728\"\u5965\u6570\u4e4b\u5bb6\"\u4e3b\u6301\u4e13\u680f,\u90a3\u513f\u4e5f\u6709\u4e0d\u5c11\u56fd\u5185\u5916\u7684\u6750\u6599,\u6b22\u8fce\u5149\u4e34.www.aoshoo.com./bbs1", "Solution_3": "[quote=\"georgew_bush\"]:!: \n\u6211\u662f\u4e00\u540d\u4e2d\u5b66\u751f\uff0c\u662f\u4e00\u4f4d\u5de8\u65f6\u7f51\u6c11\uff0c\u4f46\u59cb\u7ec8\u6ca1\u80fd\u627e\u5230\u597d\u7684\u6570\u5b66\u8d44\u6e90\u7f51\u7ad9\n\u5e0c\u671bmathlinks\u53ef\u4ee5\u6539\u53d8\n\u8c22\u8c22\u5927\u5bb6[/quote]\r\n\u4f60\u5c31\u662ftale\u5417\uff1f" } { "Tag": [ "logarithms" ], "Problem": "Find all $f : \\left( 0, \\infty \\right) \\to \\mathbb R$ such that $f(x) \\leq \\ln x$ for all $x > 0$ and $f(xy) \\leq f(x)+f(y)$ for all $x, y > 0$.", "Solution_1": "[quote=\"perfect_radio\"]Find all $f : \\left( 0, \\infty \\right) \\to \\mathbb R$ such that $f(x) \\leq \\ln x$ for all $x > 0$ and $f(xy) \\leq f(x)+f(y)$ for all $x, y > 0$.[/quote]\r\n\r\nsimple problem.\r\n\r\nf(x) <=lnx \r\nfor x=1 we have f(x) <=0\r\n\r\nf(xy) <= f(x) + f (y) <= lnx + lny; relation (1)\r\nfor x=y=1 we have: f(1) <= 2*f(1) <= 0 so rezults that f(1) = 0\r\n\r\nNow taking y=1/x in (1) we have:\r\nf(1) <= f(x) + f(1/x)<= 0;but f(1)=0 so f(x) + f(1/x) =0 so f(1/x)= -f(x) we say to that relation (2)\r\n\r\nin relation (1) we take x=y=1/x rezults:\r\n\r\nf(1/(x^2)) <= 2*f(1/x) <= ln(1/(x^2)); but from relation (2) rezults:\r\n\r\nf(1/(x^2)) <= -2*f(x) <= - ln(x^2) we multiply with (-1) so it rezults:\r\n\r\nf(1/(x^2)) >= 2*f(x) >=2*lnx so f(x) >= lnx; but the problem says that f(x) <= lnx so we finally got that f(x) = lnx; \r\n\r\n\r\n[/hide]", "Solution_2": "two things: 1) you have to always check your solution works\r\n2) there are more solutions, such as, 0.5 ln x", "Solution_3": "Nice solution, Zamfirmihai.\r\n\r\n[quote=\"Altheman\"]1) you have to always check your solution works[/quote]\nTrue.\n[quote=\"Altheman\"]2) there are more solutions, such as, 0.5 ln x[/quote]\r\nFalse.", "Solution_4": "[quote=\"perfect_radio\"]Nice solution, Zamfirmihai.\n\n[quote=\"Altheman\"]1) you have to always check your solution works[/quote]\nTrue.\n[quote=\"Altheman\"]2) there are more solutions, such as, 0.5 ln x[/quote]\nFalse.[/quote]\r\n\r\nyes, I forgot to check. \r\n\r\nmersi", "Solution_5": "uhh...? wow, am i being stupid or something?, 0.5 ln x works...", "Solution_6": "no, it doesn`t work!\r\n\r\nf(x)<=lnX =>\r\n\r\n0.5 * lnx <= lnx rezults 0 <= lnx . but if you chose x=1/e rezults\r\nlnx = -1 rezults 0 <= -1. false" } { "Tag": [ "quadratics", "search", "number theory" ], "Problem": "Has someone tell me about square residues?\r\nThank so much.", "Solution_1": "i suppose it is quadratic residues :) \r\n\r\nit is related to number theory\r\n\r\nplease check [url]http://en.wikipedia.org/wiki/Quadratic_residue[/url]", "Solution_2": "You might also want to visit the following links:\r\n1. http://mathworld.wolfram.com/QuadraticResidue.html\r\n2. http://primes.utm.edu/glossary/page.php?sort=QuadraticResidue\r\n3. http://www.mathreference.com/num-mod,res.html\r\n4. http://marauder.millersville.edu/~bikenaga/numbertheory/quadratic-residues/quadratic-residues.html\r\n5. http://www.iec.csic.es/~ismael/QRs.htm\r\n\r\nOr rather, try out your search in google: \r\n http://www.google.co.in/search?q=Quadratic+residue&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a", "Solution_3": "Thank you so much ! :lol:" } { "Tag": [ "geometry", "Euler", "ratio", "geometry proposed" ], "Problem": "Let Fb be Feuerbach point,and $ M_{c},M_{c},M_{c}$ be midpoints of sides AB,BC,CA respectively.Let \u2299I be the incircle of triangle ABC. The three sides are tangent to the circle I at D,E,F respectively.\r\nThen the Simson lines for Fb for triangle MCMaMb and DEF are the same,and the line is parallel to the Euler line of triangle DEF.\r\nLet $ H_{c},H_{c},H_{c}$ be feet of three vertices C,A,B on three sides AB,BC,CA respectively.Then Simson Line $ H_{2}H_{4}$ is parallel to Euler line GH of triangle DEF.\r\n In the figure,the Simson lines are green,and Euler line is blue.\r\n[img]http://bbs.mathchina.com/usr1PvjRWKew/5/68/CEB4C3FCC3FB_1164420633.jpg[/img]\r\nI solved this problem by using conjugate ratio [url]http://webs.uvigo.es/adg2006/talks_files/ling.pdf[/url].", "Solution_1": "Do your best then you can solve it.", "Solution_2": "Because the Feuerbach point $ F_{b}$ of triangle $ ABC$ is the Anti-Steiner of the Euler line of triangle $ DEF$ w.r.t triangle $ DEF$, thus the Simson line for $ F_{b}$ w.r.t triangle $ DEF$ is parallel to the Euler line of triangle $ DEF$." } { "Tag": [ "geometry", "AMC", "AMC 8" ], "Problem": "Right isosceles triangles are constructed on the sides of a\n3-4-5 right triangle, as shown. A capital letter represents the area of each\ntriangle. Which one of the following is true?\n\n[asy]/* AMC8 2002 #16 Problem */\ndraw((0,0)--(4,0)--(4,3)--cycle);\n\ndraw((4,3)--(-4,4)--(0,0));\ndraw((-0.15,0.1)--(0,0.25)--(.15,0.1));\n\ndraw((0,0)--(4,-4)--(4,0));\ndraw((4,0.2)--(3.8,0.2)--(3.8,-0.2)--(4,-0.2));\n\ndraw((4,0)--(7,3)--(4,3));\ndraw((4,2.8)--(4.2,2.8)--(4.2,3));\n\nlabel(scale(0.8)*\"$Z$\", (0, 3), S);\nlabel(scale(0.8)*\"$Y$\", (3,-2));\nlabel(scale(0.8)*\"$X$\", (5.5, 2.5));\nlabel(scale(0.8)*\"$W$\", (2.6,1));\nlabel(scale(0.65)*\"5\", (2,2));\nlabel(scale(0.65)*\"4\", (2.3,-0.4));\nlabel(scale(0.65)*\"3\", (4.3,1.5));[/asy]\n\n$ \\text{(A)}\\ X\\plus{}Z\\equal{}W\\plus{}Y \\qquad \\text{(B)}\\ W\\plus{}X\\equal{}Z$ \\\\ $ \\text{(C)}\\ 3X\\plus{}4Y\\equal{}5Z \\qquad \\text{(D)}\\ X\\plus{}W\\equal{}\\frac{1}{2}(Y\\plus{}Z)$ \\\\ $ \\text{(E)}\\ X\\plus{}Y\\equal{}Z$ \\\\ \\\\\nSelect the most appropriate letter.", "Solution_1": "...this is derivation of the pythagorean theorm.\r\n\r\nE ) X+Y=Z.\r\n\r\nanswer ; e", "Solution_2": "Could you please explain in further detail? This isn't my question, but I want to know the solution as well. Thanks.", "Solution_3": "im not really a teacher type... but w/e\r\n\r\nin the derivation of pythagorean theorm, there are three squares around the triangle.\r\n\r\nand in the problem, those triangles are 1/2 of the squares.\r\n\r\nin derivation of pythagorean theorm, its $ x^2\\plus{}y^2\\equal{}z^2$,\r\n\r\nand in this problem, it becomes $ \\frac{x^2}{2}\\plus{}\\frac{y^2}{2}\\equal{}\\frac{z^2}{2}$, which is also right.", "Solution_4": "also food for thought\r\nmost the time multiple choice answers are e :wink: :wink:" } { "Tag": [ "topology", "ratio", "function", "real analysis", "real analysis solved" ], "Problem": "hi all, \r\n\r\nhere is a problem that seems intuitive but rather hard (i find) to formalize:\r\nconsider the set E of all the continuous (with uniform norm) fonctions from D to S, where D is the unit disk and S the unit circle. Prove that E is arc connected.", "Solution_1": "Do you mean $\\forall\\, x,y\\in E$ $\\exists$ homotopy $H:[0,1]\\to E$ s.t. $H(0)=x$ and $H(1)=y$?", "Solution_2": "Couldn't we take the homotopy (from $f$ to $g$) $\\phi(t)(x)=$ the point dividing the positively directed arc $\\widehat {(f(x),g(x))}$ in the ratio $t:1-t$? By positively directed arc I mean the directed arc going from $f(x)$ to $g(x)$ in a counterclockwise direction.\r\n\r\nIt looks Ok to me right now, but it's 4 AM, so I'm sure you'll be understanding, should it turn out to be just another stupid mistake of mine :).", "Solution_3": "[quote=\"grobber\"]Couldn't we take the homotopy (from $f$ to $g$) $\\phi(t)(x)=$ the point dividing the positively directed arc $\\widehat {(f(x),g(x))}$ in the ratio $t:1-t$? By positively directed arc I mean the directed arc going from $f(x)$ to $g(x)$ in a counterclockwise direction.\n\nIt looks Ok to me right now, but it's 4 AM, so I'm sure you'll be understanding, should it turn out to be just another stupid mistake of mine :).[/quote]\r\nI think it doesn't work, because it is possible $H(t)\\not\\in E$. Unfortunately, I don't know how to describe appropriate example using my poor english. :(", "Solution_4": "Yeah, I think I also see a pretty obvious counterexample (it's not a concrete counterexample, but I think I can see why the construction above can fail).", "Solution_5": "Just take $y(u)\\equiv (1,0)$ and $x(u)$ as it is described on figure, $t=\\frac{1}{2}$.\r\nLet $\\epsilon>0$ be small number. You see that for $u=(w,\\epsilon)$ point $H(t)(u)$ lies in neighbourhood of $(1,0)$ and for $u=(w,-\\epsilon)$ point $H(t)(u)$ lies in neighbourhood of $(-1,0)$.\r\nOk?", "Solution_6": "Ok, I'll have another shot at it. I'm sure there will be someone to dismiss this one too, but I think that learning through mistakes is the best way of learning :D.\r\n\r\nSince path connectedness is an equivalence relation, it's enough to show that every function is path connected to the constant function. Moreover, since any two constant functions are obviously path connected, it's enough to show that $f$ is path connected to the constant function $g$ whose value is $f(0)$ (the value of $f$ in the origin; I'll use complex number language). Now take $H(t)(x)=f((1-t)x),\\ \\forall x\\in D,\\ \\forall t\\in [0,1]$.\r\n\r\nIs it totally stupid? :)", "Solution_7": "It seems correct..." } { "Tag": [ "inequalities", "trigonometry", "quadratics", "Euler", "algebra", "inequalities unsolved" ], "Problem": "Let $x$, $y$, $z$ be positive real numbers such that $x^2+y^2+z^2+2xyz=1$. Prove the inequality $x+y+z \\leq 3/2$.", "Solution_1": "Oh,I think It's very well-Known.\r\n$x=\\cos{A},y=\\cos{B},z=\\cos{C}$", "Solution_2": "Exuse me, but I don't uderstand.", "Solution_3": "if A+B+C=180 then\r\ncos^2 A+cos^2 B + cos^2 C + 2cosAcosBcosC=1\r\nand cosA+cosB+cosC<=3/2", "Solution_4": "But we have to show that if $x^2+y^2+z^2+2xyz=1$ then there exist $A$, $B$, $C$, such that $A+B+C=180^{\\circ}$ and $x=\\cos{A}$, $y=\\cos{B}$, $z=\\cos{C}$.\r\n\r\nThis is not the same. At first, I suppose, we have to prove that $x,y,z \\leq 1$. Don't you think so?", "Solution_5": "well, x,y,z <=1 are quite obvious because\r\nx^2 180\u00b0 - 90\u00b0 - 90\u00b0 > 0\u00b0. On the other hand, clearly, C < 180\u00b0. Thus, the angle C lies between 0\u00b0 and 180\u00b0. Hence, with C = 180\u00b0 - A - B, we have\n\nNow we are going to prove that z = cos C. In fact, since A is an acute angle, $\\sin A = \\sqrt{1-\\cos^2 A} = \\sqrt{1-x^2}$ (since x = cos A), and similarly, $\\sin B = \\sqrt{1-y^2}$.\n\n$\\cos C = \\cos\\left(180^{\\circ}-A-B\\right) = \\cos\\left(180^{\\circ}-\\left(A+B\\right)\\right) = -\\cos\\left(A+B\\right) = -\\left(\\cos A \\cos B - \\sin A \\sin B\\right)$\n$= \\sin A \\sin B - \\cos A \\cos B = \\sqrt{1-x^2}\\sqrt{1-y^2}-xy = \\sqrt{\\left(1-x^2\\right)\\left(1-y^2\\right)}-xy$.\n\nOn the other hand, the equation $x^2+y^2+z^2+2xyz=1$ can be rewritten as\n\n$z^2 + 2xy\\cdot z + \\left(x^2+y^2-1\\right) = 0$,\n\nwhat is a quadratic equation for the number z, and its two solutions are\n\n$z_1 = \\frac{-2xy+\\sqrt{\\left(2xy\\right)^2-4\\left(x^2+y^2-1\\right)}}{2}$;\n$z_2 = \\frac{-2xy-\\sqrt{\\left(2xy\\right)^2-4\\left(x^2+y^2-1\\right)}}{2}$.\n\nActually, $z = z_2$ is impossible since $z_2$ is negative, while z was assumed to be positive. Hence, we must have $z = z_1$. Thus,\n\n$z = \\frac{-2xy+\\sqrt{\\left(2xy\\right)^2-4\\left(x^2+y^2-1\\right)}}{2} = \\frac{-2xy+\\sqrt{4\\left(1-x^2\\right)\\left(1-y^2\\right)}}{2}$\n$ = \\frac{-2xy+2\\sqrt{\\left(1-x^2\\right)\\left(1-y^2\\right)}}{2} = \\sqrt{\\left(1-x^2\\right)\\left(1-y^2\\right)} - xy$.\n\nComparing this with the equation $\\cos C = \\sqrt{\\left(1-x^2\\right)\\left(1-y^2\\right)}-xy$ obtained above, we see that z = cos C. Also, the equation C = 180\u00b0 - A - B yields A + B + C = 180\u00b0. Furthermore, we know that the angles A and B are acute; moreover, since C is an angle between 0\u00b0 and 180\u00b0 and its cosine is positive (cos C = z, and z is positive by assumption), it is also acute, and thus, the angles A, B, C are the three angles of an acute-angled triangle. Thus, Theorem 1 is proven.\n\nTheorem 1 justifies the substitution x = cos A, y = cos B and z = cos C, with A, B, C being the angles of an acute triangle, for positive reals x, y, z satisfying $x^2+y^2+z^2+2xyz=1$.\n\nBTW, we had a discussion of the same inequality at http://www.mathlinks.ro/Forum/viewtopic.php?t=21387 .\n\n[b]PS.[/b]\n\n[quote=\"Sailor\"]Belive it or not, this property can be used to solve this [i]unsolved [/i] problem:\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=23446[/quote]\r\n\r\nHmm, do you mean the identity $\\cos^2 A + \\cos^2 B + \\cos^2 C + 2\\cos A\\cos B\\cos C = 1$ ? Well, maybe, I don't know (I have never been interested about that problem).\r\n\r\n Darij", "Solution_8": "This equality can be derived from the first cosine theorem.", "Solution_9": "I have the same idea with cosA,cosB,cosC.We use only that cosA+cosB+cosC=1+r/R=<3/2,using the Euler Inequalitie. ;)", "Solution_10": "We can use approach ABC to solve this problem\r\nWhen we use this we have:a=b\r\nThen we only have to prove this inequality with two variables such as:a and c or b and c.Easy?", "Solution_11": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=3398963#p3398963]With condition x^2+y^2+z^2+2xyz=1:[/url]\n$1< x+y+z \\leq \\frac{3}{2}.$", "Solution_12": "[quote=\"Remike\"]Let $x$, $y$, $z$ be positive real numbers such that $x^2+y^2+z^2+2xyz=1$. Prove the inequality $x+y+z \\leq 3/2$.[/quote]\n[hide=\"Non-Trigo\"]\nWe will prove that $x+y+z\\leq \\frac{3}{2}$ diven $x^2+y^2+z^2+2xyz=1$ and $x,y,z\\in R^+$\nnow, if $x\\geq y\\geq z$\nthen $3x^2+2x^3\\geq x^2+y^2+z^2+2xyz=1$\n$\\implies 3x^2+2x^3\\geq 1$ if $x \\leq \\frac{1}{2}$ then $3x^2+2x^2\\leq 1$ which is not possible hence\n$x\\geq \\frac{1}{2}$\nand $3z^2+2z^3\\leq x^2+y^2+z^2+2xyz\\implies 3z^2+2z^3\\leq 1$ if $z\\geq \\frac{1}{2}\\implies 3z^2+2z^3\\geq 1$hence $z\\leq \\frac{1}{2}$\nwhich means if $x\\geq y \\geq z$ then we can surely say that $x\\geq \\frac {1}{2}$ and $z\\leq \\frac{1}{2}$\nbut we can't say anything about $y$ i.e. it may be more that $\\frac{1}{2}$ or less than $\\frac{1}{2}$\nnow, forget about the condition $x\\geq y\\geq z$\nwe can say that two of $(x-\\frac{1}{2})$, $(y-\\frac{1}{2}$),$(z-\\frac{1}{2})$ will bear the same sign\nhence $(x-\\frac{1}{2})(y-\\frac{1}{2})\\geq 0\\implies (2x-1)(2y-1)\\geq 0\\implies 2xy+\\frac{1}{2} \\geq x+y$\nhence $2xy + \\frac{1}{2}+z \\geq x+y+z\\implies$ it is enough to prove\n$\\frac{3}{2}\\geq 2xy+\\frac{1}{2}+z$\ni.e.,$1 \\geq 2xy+z$\nwe will assume the contrary\ni.e. $2xy+z\\geq 1$\nbut $1=x^2+y^2+z^2+2xyz\\geq 2xy+z(z+2xy)$\nif $z+2xy\\geq 1\\implies z(z+2xy)\\geq z \\implies 2xy+z(z+2xy)\\geq 2xy+z\\implies 2xy+z(z+2xy)\\geq 1$(as we have assumed $z+2xy\\geq 1$\nbut we have already proved $2xy+z(z+2xy)\\leq 1$ hence $2xy+z\\geq 1$ is not possible hence $2xy+z\\leq 1$\nHence Proved. :)[/hide]" } { "Tag": [ "LaTeX" ], "Problem": "hey i wanted to type up my algebra midterm in latex in the texnic center thing but i wanted to shorten the margins so i can fit more wording on a line\r\n\r\nso does ne1 kno the latex for that? changing the margins?", "Solution_1": "I have it set to this for my document:\r\n\r\n\\textwidth 6.5in\r\n\\oddsidemargin 0in\r\n\\evensidemargin 0in\r\n\r\nwhich works pretty well, but you can change it to whatever. Those \\()sidemargin commands are for the left side and the right side apparently depends on it--at least that's how I understand it, but I don't know much about $\\LaTeX$. And the odd, even are just odd and even pages, and they should probably be the same unless you are putting together a book. Also, the \\()sidemargin commands also accept negative values, so that should help if you are making the \\textwidth bigger.", "Solution_2": "aite thanks!" } { "Tag": [ "quadratics", "Diophantine Equations" ], "Problem": "Are there integers $m$ and $n$ such that $5m^2 -6mn+7n^2 =1985$?", "Solution_1": "$ (5m \\minus{} 3n)^2 \\plus{} 26n^2 \\equal{} 5.1985$\r\nBut $ (\\frac { \\minus{} 26}{397}) \\equal{} \\minus{} 1$ imply that it has no solution.", "Solution_2": "Solving the equation w.r.t. $ m$ we obtain $ m \\equal{} \\frac {3\\pm\\sqrt {9925 \\minus{} 26n^2}}{5}$, which means $ 9925 \\minus{} 26n^2$ must be a perfect square.\r\nLet $ k^2 \\equal{} 9925 \\minus{} 26n^2$. So we have $ k^2\\equiv_{13}6$, and it's easy to see $ 6$ is not a quadratic residue modulo $ 13$, therefore the equation has no solutions on integers." } { "Tag": [], "Problem": "You are in a room with 10 people, all of whom have perfect senses of logic. Then, someone announces that they will be auctioning off a dollar. But, the person who comes in second will also have to pay what they bid. What will happen?", "Solution_1": "What about this scenario:\r\n\r\nSomeone is auctioning off a dolar. Everyone will (secretly) write down their bids. The winner gets the dollar, but everyone has to pay the amount they bid (ignore ties or something).", "Solution_2": "The first question to ask is: what happens when two bid the same\u00bf", "Solution_3": "I read about this once- more often than not, the 2 bidders will first bid to win the dollar, then, after the bidding has reached a dollar, they bid to minimize their losses (Would you rather pay a dollar and not get anything, or pay a dollar and 2 cents, and win the dollar?) The bidding can supposedly get up to 3 or 4 dollars.", "Solution_4": "Did anyone ever try to sell a dollar bill on ebay?", "Solution_5": "yep [url]http://coins.search.ebay.com/dollar_Paper-Money-US_W0QQsacatZ3412[/url]\r\n\r\nbut they seem to be like \"antique\" \"collectable\" dollars. not normal ones.", "Solution_6": "My friend bought a twenty dollar bill for 13 dollars. \r\n\r\n\"Can I just have 7 dollars transfered directly to my Paypal account?\"", "Solution_7": "[quote=\"B-tone_Party\"]I read about this once- more often than not, the 2 bidders will first bid to win the dollar, then, after the bidding has reached a dollar, they bid to minimize their losses (Would you rather pay a dollar and not get anything, or pay a dollar and 2 cents, and win the dollar?) The bidding can supposedly get up to 3 or 4 dollars.[/quote]\r\n\r\nTechnically, because of this and the fact that each person has a perfect sense of logic, no one will bid.", "Solution_8": "Or 1 person will bid 1 cent and take the dollar.", "Solution_9": "this reminds me of that one problem where you buy a magic thing that gives you everything you want, but you have to sell it to someone for less than you bought it, or else you lose all the stuff and your life stinks. and you have to sell it for a positive integer number of cents :roll: \r\n\r\nthe question is, how much would you pay for it? you don't want to pay 1 cent for it, because then you can't sell it. but you don't want to pay 2 cents for it, because no one will buy it from you for 1 cent. etc.", "Solution_10": "best bid....", "Solution_11": "Once i auctionate a 50 pesos bill because I desperately needed change. I sold it on 41.50" } { "Tag": [ "calculus", "derivative", "function", "algebra", "domain", "calculus computations" ], "Problem": "What is the difference between a global maximum and a local maximum?\r\n\r\nand\r\n\r\nName two types of critical numbers and how they differ. \r\n\r\nThanks in advance :lol:", "Solution_1": "The critical numbers are, according to the standard definition used by most math texts, either the solutions of the equation $f'(x)=0$ or the points where the first derivative of a function does not exists.\r\nEx: $f(x)=\\frac{1}{x}$, then $f'(x)=\\frac{-1}{x^{2}}$, hence the only critical point is 0 since at 0 the derivative is not well defined.\r\nNow, if $f(x)=x^{2}-3x+2$ then $3/2$ is a critical number i.e. $f'(3/2)=0$.\r\n\r\nGlobal max is the maximum value of a given function on a specific interval, however the local maximum is the maximum value of f near a point (of course the point is a critical point), not on the whole interval.\r\nDoes it make any sense?", "Solution_2": "[quote]Ex: $f(x)=\\frac{1}{x},$ then $f'(x)=\\frac{-1}{x^{2}},$ hence the only critical point is 0 [/quote]\r\nNo, zero is not a critical point for the function; instead, it is a point not in the domain of the function. I would restrict use of the word \"critical point\" to something that actually is a point - hence to something in the domain of the function.\r\n\r\n$f(x)=|x|,$ at $x=0,$ would be an example of a critical point where a derivative doesn't exist." } { "Tag": [ "MATHCOUNTS" ], "Problem": "He means are we doing ladder (Denote a(n) as the nth place finisher, a(12) competes against a(11) and the winner is w(1). Denote w(n) as the winner when a(n) first competes. a(n) competes against w(n) n>1)\r\n\r\nOr are we doing bracket (I am at a loss to express it mathematically, but it just pitches random people against each other, so that places 1 through 4 go less than the other places).\r\n\r\nAnd yes, that is a good question: which one are you doing? Also, which are they doing this year at nats? Probably bracket...", "Solution_1": "[quote=\"jb05\"]Also, which are they doing this year at nats? Probably bracket...[/quote]\r\n\r\nI think it's ladder at Nats, but with my tendency to be wrong today, it is probably bracket. :(", "Solution_2": "Actually I think it is a bracket like last year with the top 12 written round. Also, besides whether ladder or bracket for this, how does it work, how do we do it?", "Solution_3": "I like ladder better.\r\nIn 6th grade at chapter I beat 4 8th graders in a row but was eliminated\r\n before facing one of my team mates.", "Solution_4": "I have never done a ladder but it sounds cool", "Solution_5": "Bracket includes the top 12 instead of 10, so i'm for that ;)", "Solution_6": "Gives us a tiny bit better chance at nats...\r\n1/22.4 : 1/19\r\n.....", "Solution_7": "Well [i]some[/i] people don't randomly get their rank...", "Solution_8": "I used to like ladder (when I went to chapter and state and always got top two), but in the rare event that I make the countdown, I think I'll be hoping for bracket.", "Solution_9": "The thing is (unless you have byes, or double elimination) you have to have 1,2,4,8,16,32,or 64(etc.) people to do a bracket. Because if you have 12 people, that would cut down to 6, then 3. But with 3 people left, you could no longer continue the bracket (unless you give a bye to the highest rank, or do double elimination. I can't see MC giving byes, but if it is a bracket, it will be with double elimination or byes.", "Solution_10": "Mathcounts divides 12 into 8 and 4. It brings the 8 down to 4, and then does an ordinary round robin tournament (I think; correct me if I'm wrong). How is this going to work tomorrow?", "Solution_11": "[quote=\"jb05\"]Mathcounts divides 12 into 8[/quote]\r\n\r\nI believe you, but how does that work?", "Solution_12": "[quote=\"mysmartmouth\"][quote=\"jb05\"]Mathcounts divides 12 into 8[/quote]\n\nI believe you, but how does that work?[/quote]\r\n\r\nThey take ranks 5-12, have 5 v 6, 7 v 8, etc... eliminate those 8 to 4, then do it until they have 1.", "Solution_13": "Actually, it's 8v9 7v10 6v11 5v12, and then against the other four", "Solution_14": "Like this:\r\n\r\nthis took 2 min for me to make.", "Solution_15": "I think it will be like that again this year.", "Solution_16": "You can read the ladder format rules and the bracket format rules on the [url=http://www.mathcounts.org/webarticles/articlefiles/510-C&SGenInstruct.pdf]MATHCOUNTS Web Site[/url]." } { "Tag": [ "geometry", "trapezoid", "AoPSwiki" ], "Problem": "cananyone explain what is a median of a trapezoid ?", "Solution_1": "let $ s$ be the median than \r\n\r\n$ s \\equal{} \\frac {a \\plus{} c}{2}$\r\n\r\nthis is a quote from AoPSWiki:\r\n\r\n[quote]The median of a trapezoid is defined as the line connecting the midpoints of the two legs. Its length is the arithmetic mean of that of the two bases $ \\dfrac{b_1 \\plus{} b_2}{2}$. It is also parallel to the two bases.[/quote]" } { "Tag": [ "function", "parameterization", "induction", "algebra", "functional equation", "strong induction", "algebra unsolved" ], "Problem": "Let $\\mathbb{N}$ denote the set of positive integers. Find all functions $f: \\mathbb{N} \\to \\mathbb{N}$ such that \\[ f(m+n)f(m-n) = f(m^2) \\] for $m,n \\in \\mathbb{N}$.", "Solution_1": "I like this - you really just need to calculate...\r\n\r\nThe functional equation $f\\left(m-n\\right)f\\left(m+n\\right)=f\\left(m^2\\right)$ allows us to recurrently compute all values of the function f in terms of the first three values, which we denote by f(1) = a, f(2) = b and f(3) = c. Note that these numbers a, b, c are naturals, so we can divide by them (this is important, since we will often do this in future). Then,\r\n\r\n$f\\left(4\\right)=f\\left(2^2\\right)=f\\left(2-1\\right)f\\left(2+1\\right)=f\\left(1\\right)f\\left(3\\right)=ac$.\r\n\r\nNow,\r\n\r\n$af\\left(5\\right)=f\\left(1\\right)f\\left(5\\right)=f\\left(3-2\\right)f\\left(3+2\\right)=f\\left(3^2\\right)$\r\n$=f\\left(3-1\\right)f\\left(3+1\\right)=f\\left(2\\right)f\\left(4\\right)=b\\cdot ac$.\r\n\r\nDivision by a yields f(5) = bc. Furthermore,\r\n\r\n$bf\\left(6\\right)=f\\left(2\\right)f\\left(6\\right)=f\\left(4-2\\right)f\\left(4+2\\right)=f\\left(4^2\\right)$\r\n$=f\\left(4-1\\right)f\\left(4+1\\right)=f\\left(3\\right)f\\left(5\\right)=c\\cdot bc=bc^2$.\r\n\r\nDivision by b yields $f\\left(6\\right)=c^2$. Continuing the game,\r\n\r\n$cf\\left(7\\right)=f\\left(3\\right)f\\left(7\\right)=f\\left(5-2\\right)f\\left(5+2\\right)=f\\left(5^2\\right)$\r\n$=f\\left(5-1\\right)f\\left(5+1\\right)=f\\left(4\\right)f\\left(6\\right)=ac\\cdot c^2=ac^3$.\r\n\r\nDivision by c yields $f\\left(7\\right)=ac^2$. Hence,\r\n\r\n$a^2c^2=a\\cdot ac^2=f\\left(1\\right)f\\left(7\\right)=f\\left(4-3\\right)f\\left(4+3\\right)=f\\left(4^2\\right)$\r\n$=f\\left(4-1\\right)f\\left(4+1\\right)=f\\left(3\\right)f\\left(5\\right)=c\\cdot bc=bc^2$.\r\n\r\nDivision by $c^2$ yields $a^2=b$. Thus, f(2) = b becomes $f\\left(2\\right)=a^2$, and f(5) = bc becomes $f\\left(5\\right)=a^2c$.\r\n\r\nSo we have kicked out b from the parameters a, b, c - our first progress. Let's continue the computing spree:\r\n\r\n$ac\\cdot f\\left(8\\right)=f\\left(4\\right)f\\left(8\\right)=f\\left(6-2\\right)f\\left(6+2\\right)=f\\left(6^2\\right)$\r\n$=f\\left(6-1\\right)f\\left(6+1\\right)=f\\left(5\\right)f\\left(7\\right)=a^2c\\cdot ac^2=a^3c^3$.\r\n\r\nDivision by ac yields $f\\left(8\\right)=a^2c^2$. Thus,\r\n\r\n$a^4c^2=a^2\\cdot a^2c^2=f\\left(2\\right)f\\left(8\\right)=f\\left(5-3\\right)f\\left(5+3\\right)=f\\left(5^2\\right)$\r\n$=f\\left(5-1\\right)f\\left(5+1\\right)=f\\left(4\\right)f\\left(6\\right)=ac\\cdot c^2=ac^3$.\r\n\r\nDivision by $ac^2$ transforms this into $a^3=c$. Thus, the variable c is unnecessary and we are left with a only. Let's first combine:\r\n\r\n$f\\left(1\\right)=a$;\r\n$f\\left(2\\right)=a^2$;\r\n$f\\left(3\\right)=c=a^3$;\r\n$f\\left(4\\right)=ac=aa^3=a^4$;\r\n$f\\left(5\\right)=a^2c=a^2a^3=a^5$;\r\n$f\\left(6\\right)=c^2=\\left(a^3\\right)^2=a^6$;\r\n$f\\left(7\\right)=ac^2=a\\left(a^3\\right)^2=a^7$;\r\n$f\\left(8\\right)=a^2c^2=a^2\\left(a^3\\right)^2=a^8$.\r\n\r\nReminds you of something? Well, then you might be disappointed in a few moments. We note that\r\n\r\n$af\\left(9\\right)=f\\left(1\\right)f\\left(9\\right)=f\\left(5-4\\right)f\\left(5+4\\right)=f\\left(5^2\\right)$\r\n$=f\\left(5-1\\right)f\\left(5+1\\right)=f\\left(4\\right)f\\left(6\\right)=a^4\\cdot a^6=a^{10}$,\r\n\r\nso that, upon division by a, we get $f\\left(9\\right)=a^9$. But this leads to\r\n\r\n$a^9=f\\left(9\\right)=f\\left(3^2\\right)=f\\left(3-1\\right)f\\left(3+1\\right)=f\\left(2\\right)f\\left(4\\right)=a^2a^4=a^6$,\r\n\r\nand thus to $a^3=1$, so that a = 1. How boring! Anyway, the problem is now almost solved. We will show that f(n) = 1 holds for every natural n. This will be proven by induction over n. In fact, for n = 1, n = 2, n = 3, n = 4 and n = 5, this follows from the above formulas f(1) = a, $f\\left(2\\right)=a^2$, $f\\left(3\\right)=a^3$, $f\\left(4\\right)=a^4$ and $f\\left(5\\right)=a^5$ combined with the killer equation a = 1. Now, assume that N is a natural number $\\geq 5$, and the identity f(n) = 1 holds for all naturals n < N. Then, particularly, f(N - 1) = 1, f(N - 3) = 1 and f(N - 4) = 1. Thus,\r\n\r\n$f\\left(N\\right)=1\\cdot f\\left(N\\right)=f\\left(N-4\\right)f\\left(N\\right)=f\\left(\\left(N-2\\right)-2\\right)f\\left(\\left(N-2\\right)+2\\right)$\r\n$=f\\left(\\left(N-2\\right)^2\\right)=f\\left(\\left(N-2\\right)-1\\right)f\\left(\\left(N-2\\right)+1\\right)=f\\left(N-3\\right)f\\left(N-1\\right)$\r\n$=1\\cdot 1$,\r\n\r\nso that f(N) = 1, and the induction step is done. Thus, f(n) = 1 holds for every natural n.\r\n\r\nSo the only solution of our functional equation is the function f(x) = 1. It is indeed a solution, as a reader could easily verify.\r\n\r\n Darij", "Solution_2": "^^^^ Great solution.", "Solution_3": "what about f(x)=0", "Solution_4": "$0 \\not\\in \\mathbb{N}$ as long as nothing else is said.", "Solution_5": "why do you exlude 0\r\n\r\n it's a positive and negative integer", "Solution_6": "In most parts of the world $0$ is considered neither positive nor negative, but thats not important here (did anybody say anything about that\u00bf).\r\n$0 \\not\\in \\mathbb{N}$ for me and a lot of other peoples, and it's the problem of the problem-poster when it's not well defined what he wants (and there are good reasons here that $0$ is excluded, like that then also $f(0)$ is defined and the problem gets a at least not harder).", "Solution_7": "Yes 0 is not natural number in Bulgaria, too :D\r\n\r\nAnd I think that this problem is too easy to be in TST", "Solution_8": "Here is my approach:\r\nBy induction we obtain that :$f(m+2)=cf(m)$ for all $m\\geq 3$ where $c$ is constant.So :\r\n$F(2n)=c^{n-1}f(2)$ and $f(2n+1)=c^nf(1)$\r\nThis one is really classic !", "Solution_9": "[hide]Let $ g_p(n) \\equal{} v_p(f(n))$ for all primes $ p$, where $ v_p(n)$ is the largest integer $ k$ such that $ p^k | n$. Our recurrence relation becomes $ g_p(m\\plus{}n) \\plus{} g_p(m \\minus{} n) \\equal{} g_p(m^2)$, and $ g_p$ is a function from the naturals to the nonnegative integers. We claim that $ g_p(n) \\equal{} 0$ for all nonnegative $ n$. \n\nLet $ g_p(1) \\equal{} a_1$, $ g_p(2) \\equal{} a_2, g_p(3) \\equal{} a_3$. Note that if $ m > 2$, then $ g_p(m\\plus{}1) \\plus{} g_p(m\\minus{}1) \\equal{} g_p(m \\plus{} 2) \\plus{} g_p(m \\minus{} 2) \\equal{} g_p(m^2)$, so $ g_p(m\\plus{}2) \\equal{} g_p(m\\plus{}1) \\plus{} g_p(m\\minus{}1) \\minus{} g_p(m\\minus{}2)$, so for all $ m > 4$, $ g_p(m) \\equal{} g_p(m \\minus{} 1) \\plus{} g_p(m \\minus{} 3) \\minus{} g_p(m \\minus{} 4)$. \n\nNote that $ g_p(2^2) \\equal{} g_p(2 \\plus{} 1) \\plus{} g_p(2 \\minus{} 1) \\equal{} a_3 \\plus{} a_1$, so $ g_p(4) \\equal{} a_3 \\plus{} a_1$. With this recurrence, we can easily compute that $ g_p(5) \\equal{} a_2 \\plus{} a_3$, $ g_p(6) \\equal{} 2a_3$, $ g_p(7) \\equal{} a_1 \\plus{} 2a_3$, $ g_p(8) \\equal{} a_2 \\plus{} 2a_3$, and $ g_p(0) \\equal{} 3a_3$. But $ g_p(3^2) \\equal{} g_p(3 \\plus{} 2) \\plus{} g_p(3 \\minus{} 2) \\equal{} a_1 \\plus{} a_2 \\plus{} a_3$, so we see that $ 2a_3 \\equal{} a_1 \\plus{} a_2$. In addition, $ g_p(6^2) \\equal{} g_p(6 \\minus{} 1) \\plus{} g_p(6 \\plus{} 1) \\equal{} g_p(6 \\minus{} 3) \\plus{} g_p(6 \\plus{} 3)$, that is, $ (a_2 \\plus{} a_3) \\plus{} (a_1 \\plus{} 2a_3) \\equal{} a_3 \\plus{} 3a_3$. Rearrange this to get $ a_1 \\plus{} a_2 \\equal{} a_3$. But have just found that $ 2a_3 \\equal{} a_1 \\plus{} a_2$, so $ a_1 \\equal{} a_2 \\equal{} a_3 \\equal{} 0$. In other words, $ g_p(n) \\equal{} 0$ for all positive integers $ n$. \n\nIt follows trivially that $ f(n) \\equal{} 1$ for all nonnegative integers. [/hide]", "Solution_10": "[hide=\"Just do it\"]\nNote that:\n\n$f(1)f(3)=f(4)$\n$f(2)f(4)=f(1)f(5)=f(9)$\n\nSubstituting the former into the second gives $f(2)f(3)=f(5)$\n\nNow, we have $f(25)=f(8)f(2)=f(9)f(1)=[f(2)f(4)]f(1) \\implies f(1)f(4)=f(8)$\n\nThen, $f(8)f(2)=f(6)f(4) \\implies f(1)f(2)=f(6)$\n\nBut, $f(1)[f(2)]^2=f(2)f(6)=f(1)f(7) \\implies f(7)=f(2)^2$\n\nSince $f(9)f(3)=f(7)f(5)=[f(2)^2][f(2)f(3)]$, we have $f(9)=f(2)^3$\n\nAs $f(2)f(4)=f(9)$, $f(4)=f(2)^2=f(7)$.\n\nSince $f(7)f(1337)=f(4)f(1340)$, $f(1340)=f(1337)$ and $f(1340)f(6)=f(1337)f(9) \\implies f(6)=f(9)=f(2)^3$\n\nWe can similarly prove $f(2)f(1340)=f(5)f(1337) \\implies f(2)=f(5)$\n\nAs $f(6)=f(1)f(2)$, $f(1)=f(2)^2$\n\nNow, $f(7)f(1)=f(3)f(5) \\implies f(2)^4=f(2)^2 \\implies f(2)=1$ so $f(1)=f(2)=f(3)=f(4)=f(5)=f(6)=f(7)=f(8)=f(9)=1$, huzzah.\n\nThe result easily follows from strong induction; assume every integer less than or equal to $N \\geq 9$ is mapped to one, then:\n\n$f(N+1)=f(N+1)f(1)=f(N)f(2)=1$ for even $N$ and $f(N+1)=f(2)f(N+1)=f(3)f(N)=1$ for odd, so $\\boxed{f(x)=1}$ for all natural $x$.[/hide]\n\nEDIT: whoops did some stuff wrong, should be fixed now", "Solution_11": "[hide]$$[/hide]", "Solution_12": "If $n\\ge 3$, then $f(2n+2)f(2)=f(2n)f(4)=f((n+2)^2)$, so $\\frac{f(2n+2)}{f(2n)}=\\frac{f(4)}{f(2)}$.\n\nThus, for $n\\ge 3$, $f(2n)=AB^n$, for some $A,B \\ge 0$, and $f(4n)f(6)=f((2n+3)^2)$, or $A^2B^{4n+6}=AB^{(2n+3)^2}$. This is true as $n \\to \\infty$, so $A=B=1$, and $f(2n)=1$ for $n\\ge 3$ .\n\nThen for any $k \\ge 1$, take $N \\ge k+3$. $f(4N-k)f(k)=f(4N^2)=1, \\Rightarrow f(k)=1$.", "Solution_13": "It's easy to see $\\frac {f(m+2)} {f(m)}=\\frac {f((n+1)^2)} {f(n^2)}$\nSo $f(2n)=k^{n-1}f(2)$ and $f(2n+1)=k^nf(1)$\nnow it's obvious that $\\frac {f((m+2)^2)} {f(m^2)}=k^{2m+2}$, from another way it's also $k^2$\nso $k=1$,so $f(1)=f(2)=1$ so $f(n)=1$ for all $n$.", "Solution_14": "[hide=Flawed Solution]Let $P(m, n)$ be the assertion $f(m + n)f(m - n) = f(m^2).$ Let $a = f(1).$\n\n[b]Lemma.[/b] $f(4k + 1) = a$ and $f(4k + 3) = 1$ for all $k \\in \\mathbb{N}_0.$\n[b]Proof:[/b] We go by induction. For the base case, $k = 0$, note that $f(1) = a$ trivially and $P(2, 1) \\implies f(3) = 1.$ Now, for the inductive step, suppose that $f(4k - 3) = a$ and $f(4k - 1) = 1$ for $k \\ge 1.$ Then by the inductive hypothesis, $P(4k , 1) \\implies f(4k + 1) = a$ and $P(4k + 2, 1) \\implies f(4k + 3) = 1$, completing the induction. $\\blacksquare$\n______________________________________________________________________________________________________________________________________________________\nBack to the problem at hand, we will show that $a = 1.$ By applying the lemma, we make the following calculations:\n\\begin{align*}\nP(3, 2) &\\implies f(5)f(1) = f(4) \\implies a^2 = f(4) \\\\\nP(3, 1) &\\implies a^2f(2) = a \\implies af(2) = 1.\n\\end{align*}\nHence, $a \\mid 1 \\implies a = 1.$ From here, it follows quickly that $f(k) = 1$ for all $k \\in \\mathbb{N}.$ Indeed, choose an odd positive integer $m > k$, and let $n = m - k.$ Then as $m^2 \\equiv 1 \\pmod{4}$, the lemma implies that $f(m^2) = a = 1.$ Hence, $P(m, n) \\implies f(2m - k)f(k) = 1$, whence it follows that $f(k) \\mid 1.$ Therefore, $f(k) = 1$ for all $k \\in \\mathbb{N}$, as desired. It is immediate to check that this solution is indeed valid. $\\square$[/hide]\n\nEDIT: I mixed up $m$ and $n.$", "Solution_15": "Could someone check if my solution is correct? Thanks in advance! :D\n[hide = \"Solution\"]We start by proving the following lemma:\n[b]Lemma.[/b] $f(3) = 1$\n[hide = \"Proof of Lemma\"]From our relation, we have:\n1. $f(1)f(3)=f(4)$\n2. $f(2)f(4)=f(9)$\n3. $f(1)f(5)=f(9)$\n4. $f(6)f(2)=f(7)f(1)$\n5. $f(6)f(2)=f(5)f(3)$\n6. $f(7)f(3)=f(6)f(4)$\n7. $f(6)f(4)=f(1)f(9)$\n8. $f(8)f(4)=f(5)f(7)$\n9. $f(4)f(6)=f(2)f(8)$\n\nFrom 2 and 3, we have $f(2)f(4)=f(1)f(5)$ so $f(5)=\\frac{f(2)f(4)}{f(1)}=f(2)f(3)$\n\nFrom 5, we have $f(6)f(2)=f(5)f(3)=f(2)f^2(3)$ so $f(6)=f^2(3)$.\n\nFrom 6, $f(7)f(3)=f(6)f(4)$ so $\\frac{f(7)}{f(6)}=\\frac{f(4)}{f(3)}=f(1)$.\n\nFrom 4, $f(7)f(1)=f(6)f(2)$ so $\\frac{f(7)}{f(6)}=\\frac{f(2)}{f(1)}=f(1)$. Thus, $f(2)=f^2(1)$.\n\nFrom 7, $f(6)f(4)=f(1)f(9)=f(1)f(2)f(4)=f^3(1)f(4)$ so $f(6)=f^3(1)$.\n\nFrom 2 and 3, $f(1)f(5)=f(2)f(4)=f^2(1)f(4)$ so $f(5)=f(1)f(4)$.\n\nFrom 9, $f(6)f(4)=f(2)f(8)$ so $f^3(1)f(4)=f^2(1)f(8)$. Thus, $f(8)=f(1)f(4)=f(5)$.\n\nFrom 8, $f(8)f(4)=f(5)f(7)=f(8)f(7)$ so $f(4)=f(7)$.\n\nFrom 6, $f(7)f(3)=f(6)f(4)$ so $f(3)=f(6)=f^2(3)$. Thus, $f(3)=1. \\blacksquare$.[/hide]\n\nSince $f(6)=f^2(3)=f^3(1)$ and $f^2(1)=f(2)$, we have $f(1)=f(2)=1$. Now, we will show, by strong induction, $f(n)=1 \\forall n \\in \\mathbb{N}$ with $n \\geq 3$.\n\nFor the base case, consider $n=3$. By the above lemma, $f(3)=1$ so our base case satisfies what we are trying to show!\n\nFor the strong inductive step, suppose $\\exists k \\in \\mathbb{N}$ with $k \\geq 3$ such that $f(n)=1$ for $n=1,2,\\cdots,k$.\n\nFor the strong induction, consider $n=k+1$.\n[u]Case 1:[/u] $k$ is odd\nThen, we have $f(k)f(3)=f(k+1)f(2)$. However, we already know $f(2)=f(3)=f(k)=1$, implying $f(k+1)=1$.\n[u]Case 2:[/u] $k$ is even\nThen, we have $f(k)f(2)=f(k+1)f(1)$. However, we already know $f(1)=f(2)=f(k)=1$, implying $f(k+1)=1$.\n\nHence, regardless of the parity of $k$, we have $f(k+1)=1$, completing the induction.\n\nTherefore, $f(x)=1 \\forall x \\in \\mathbb{N}$ is the only solution.[/hide]", "Solution_16": "Note that for positive $x$ and $d$ such that all terms are positive, we have\n\\begin{align*}\nf(x-3d)f(x-d)f(x+d)f(x+3d)=(f(x-3d)f(x-d))(f(x+d)f(x+3d))&=(f(x-3d)f(x+d))(f(x-d)f(x+3d)) \\\\\n\\implies f((x-2d)^2)f((x+2d)^2)&=f((x-d)^2)f((x+d)^2) \\\\\n\\implies f(x^2+4d^2)&=f(x^2+d^2)\n\\end{align*}\nand \n\\begin{align*}\nf(x-5d)f(x-d)f(x+d)f(x+5d)=(f(x-5d)f(x-d))(f(x+d)f(x+5d))&=(f(x-5d)f(x+d))(f(x-d)f(x+5d)) \\\\\n\\implies f((x-3d)^2)f((x+3d)^2)&=f((x-2d)^2)f((x+2d)^2) \\\\\n\\implies f(x^2+9d^2)&=f(x^2+4d^2)\n\\end{align*}\n\nIn particular, this means that $f(17)=f(20)$ and $f(45)=f(40)$, by setting $(x,d)$ to be $(4,1)$ and $(6, 1)$, respectively. Now $f(17)=f(20)\\implies f(16)=f(19)$ because $f(18^2)=f(17)f(19)=f(20)f(16)$. \n\nThis actually implies for all $x$, we know $f(x)=f(x+3)$ because depending on whether $x$ is odd or even, we can multiply both sides by $f(16)$ or $f(17)$ to get an equality we know is true.\n\nSimilarly, from $f(45)=f(40)$, we prove that $\\forall x, f(x)=f(x+5)$. Now, since $3$ and $5$ are relatively prime, this implies that $\\forall x, y\\in\\mathbb{N}, f(x)=f(y)=k$, for some positive integer $k$. Then, the original equation becomes $k^2=k$, which implies $k=1$.\n\nThus, we've concluded that the only solution is $f(x)=1$." } { "Tag": [ "inequalities" ], "Problem": "If $a,b,c>0$ then prove that \r\n\r\n$\\frac{a^2}{b^2+c^2}+\\frac{b^2}{a^2+c^2}+\\frac{c^2}{a^2+b^2}\\geq\\frac{3}{2}$", "Solution_1": "Nesbit's inequality after substitution $x=a^2,y=b^2,z=c^2$ ;)", "Solution_2": "Say WLOG $a^2\\ge b^2 \\ge c^2$.\r\nChebyshev tells us that \\[ \\displaystyle \\frac{\\frac{a^2}{b^2+c^2}+\\frac{b^2}{a^2+c^2}+\\frac{c^2}{a^2+b^2}}{3} \\geq \\frac{a^2+b^2+c^2}{3}.\\frac{\\frac{1}{b^2+c^2}+\\frac{1}{a^2+c^2}+\\frac{1}{a^2+b^2}}{3} \\]\r\nNow from AM-HM: \\[ \\frac{\\frac{1}{b^2+c^2}+\\frac{1}{a^2+c^2}+\\frac{1}{a^2+b^2}}{3} \\geq \\frac{3}{2(a^2+b^2+c^2)} \\]\r\nSo replace and get \\[ \\displaystyle \\frac{\\frac{a^2}{b^2+c^2}+\\frac{b^2}{a^2+c^2}+\\frac{c^2}{a^2+b^2}}{3} \\geq \\frac{a^2+b^2+c^2}{3}.\\frac{3}{2(a^2+b^2+c^2)}=\\frac{1}{2} \\]\r\n\r\n$\\mathbb{QED}$! :D\r\n\r\nI think it is true for $a,b,c \\in \\mathbb{R}_0$, since the power makes them positive.", "Solution_3": "$f(ta, tb, tc) = f(a, b, c)$ so set $a^2 + b^2 + c^2 = 1$ and Jensen's gives $\\sum_{cyc} \\frac{a^2}{b^2 + c^2} \\ge \\frac{1}{2(a^2b^2 + a^2c^2 + b^2c^2)} = \\frac{1}{(a^2 + b^2 + c^2)^2 - a^4 - b^4 - c^4} = \\frac{1}{1 - a^4 - b^4 - c^4}$. \r\n\r\nNext, AM-QM gives us $\\sqrt{ \\frac{a^4 + b^4 + c^4}{3} } \\ge \\frac{a^2 + b^2 + c^2}{3} \\implies a^4 + b^4 + c^4 \\ge \\frac{1}{3}$.\r\n\r\nFinally, we have $\\sum_{cyc} \\frac{a^2}{b^2 + c^2} \\ge \\frac{1}{1 - (a^4 + b^4 + c^4)} \\ge \\frac{1}{1 - \\frac{1}{3}} = \\frac{3}{2}$. Q.E.D." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $a,b,c,d$ be four non-negative numbers such that $a^2+b^2+c^2+d^2=1$, find the minimum of\r\n\\[ f(a,b,c,d)= 5^{ab}+5^{bc}+5^{cd}+5^{da}. \\]", "Solution_1": "It seems very hard. I'm trying to solve your problem!", "Solution_2": "Thats trivial , because $\\forall x\\geq 0,\\ 5^x \\geq 1$. So $f(a,b,c,d) \\geq 4=f(1,0,0,0)$\r\n :cool:", "Solution_3": "I don't think so .because he said that $a,b,c,d > 0$ ;) .\r\n\r\n$A = f(a,b,c,d)$\r\n\r\n$A = 5^{ab+bc+ca+dc} ???$ :? \r\n\r\nfor me,i don't [u]think[/u] that it is possible to find the $MIN(f (a,b,c,d))$exception if $a,b,c,d \\geq 0$,in this case the $MIN(f (a,b,c,d))$ is 1 :huh: .", "Solution_4": "[quote=\"toetoe\"]I don't think so .because he said that $a,b,c,d > 0$ ;) [/quote]\nYou are wrong, he said non-negative numbers :D \n\n\n [quote=\"toetoe\"]... exception if $a,b,c,d \\geq 0$,in this case the $MIN(f (a,b,c,d))$ is 1 .[/quote] \r\nAre you sur ?\r\n\r\n\r\n :cool:", "Solution_5": "for the $MIN(f(a,b,c,d))$ i am sure it will be (in the second \r\n\r\ncase ):rotfl: ,it was just a mistake in pressing the '4'." } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "Let $ C$ be a point on a fixed circle whose diameter is $ AB \\equal{} 2R$ ($ C$ is different from $ A$ and $ B$). The incircle of $ ABC$ touches $ AB$ and $ AC$ at $ M$ and $ N$, respectively Find the maximum value of the length of $ MN$ when $ C$ moves on the given fixed circle.", "Solution_1": "This is a problem on M&Y in Vietnam. Pleases lock it!" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "The question goes as\r\nConsider a circle of radius $ 5cm$ with centre $ O$. Let a chord $ BC$ be drawn whose length is $ 6cm$.\r\nLet a point $ A$ be marked on the circumference of the circle nearer to $ B$ such that there exists [b]only one chord[/b] through $ A$ (say $ AD$) such that $ AD$ is bisected by $ BC$. \r\nLet $ \\angle{AOB}$ be $ \\theta$ then $ sin\\theta$ is $ \\minus{}\\minus{}\\minus{}\\minus{}\\minus{}\\minus{}\\minus{}\\minus{}$", "Solution_1": "Let's call $ { E } AD \\cap BC$. Since $ E$ is unique ( a single chord through A), it follows that $ BC$ is tangent at $ E$ to the circle of diameter $ OA$. But we know that $ D$ is the middle of the $ arc BC$; with $ E$ being the middle of $ AD$ we get $ OE \\perp AD$, hence $ E$ belongs to the circle diameter $ OD$. Further, it seems to be a piece of cake ( not for diabetics, of course).\r\n\r\nBest regards,\r\nsunken rock", "Solution_2": "[quote=\"sunken rock\"] But we know that $ D$ is the middle of the $ arc BC$; with $ E$ being the middle of $ AD$ we get $ OE \\perp AD$, hence $ E$ belongs to the circle diameter $ OD$.[/quote]\r\nHey\r\nHow do you tell that $ D$ is middle of $ arc{AB}$?? :maybe:\r\n\r\nI can't quite understand the question.\r\nI don't think unique $ AD$ exists" } { "Tag": [ "inequalities", "logarithms", "function", "inequalities unsolved" ], "Problem": "$a,b,c>0,\\ abc=1$. For $n\\geq1$ prove $\\sum\\frac{a^{3n}-1}{a^n}\\geq n\\sum\\frac{a^3-1}a.$", "Solution_1": "ok,a nice problem.\r\nI have draw the Graphics.\r\nit tells me jensen does work :) \r\nso I think maybe you will have a nice solution :)", "Solution_2": "Which graph did you draw? The one of $f(x)=\\frac{x^{3n}-1}{x^n}-n\\frac{x^3-1}x$?\r\nBecause for applying Jensen IMO we would need to first use logs to get an additive constraint,\r\nLike putting $x=\\ln a$ etc in the original ineq, so the condition becomes $x+y+z=0$. But then the function needed for Jensen would be $f(x)=\\frac{e^{3nx}-1}{e^{nx}}-n\\frac{e^{3x}-1}{e^x}$ and that is not convex, it behaves roughly like $x^3$. ???", "Solution_3": "I drawed $f(x)=\\frac{e^{3nx}-1}{e^{nx}}-n\\frac{e^{3x}-1}{e^x}$\r\nand didn't you notice what I said?\r\nI mean it is not convex,either", "Solution_4": "OK, finally I got it. If it is not all convex, there is still the Right-Convex Function Theorem ([url]http://www.mathlinks.ro/Forum/viewtopic.php?p=226013#p226013[/url]). And it works, as $f(x)=\\frac{e^{3nx}-1}{e^{nx}}-n\\frac{e^{3x}-1}{e^x}$ is convex for $x\\geq0$. We can use it with three variables and $s=0$ as soon as we can show $2f(x)+f(-2x)\\geq0$ for all $x\\geq0$. Writing again $a$ for $e^x$ (for convenience of notation), this is equivalent to\r\n$\\frac{(2a^n-1)(a^{3n}-1)}{a^{2n}}+n\\frac{a^3-1}{a^4}\\geq0$, which is obviously true as $a\\geq1$. :) :)", "Solution_5": "I found a result to this idea due to Titu Andreescu, also.\r\nLet $m,n$ positive integers. Prove that $\\frac{x^{mn}-1}{m}\\geq\\frac{x^n-1}{x}$\r\n\r\nfor any positive real number $x$ ." } { "Tag": [], "Problem": "Planet Ypsilon has a calendar similar to ours: A year consists of $365$ days, and every month has $28$, $30$ or $31$ days.\r\nProve that on Planet Ypsilon, a year must have $12$ months.", "Solution_1": "[hide]\n\ni ono if theres a more general way to do it, so i just took it in cases\n\nwhen there are 0 months with 28 days, we know there must be 5 months with 31 days (units digit of 365 is 5), and thus:\n\n$365 - 5(31) = 210 = 7(30)$\n\nso there will be 5+7 = 12 months\n\nwhen there is 1 month with 28 days, we know there must be 7 months with 31 days (again units digit of 365 is 5), and we have:\n\n$365 - 28 - 7(31) = 120 = 4(30)$\n\nso there will be 1+7+4 = 12 months\n\nsimilarly, when there are 2 months with 28 days, we will have 9 months with 31 days, giving us:\n\n$365 - 2(28) - 9(31) = 1(30)$\n\nso there will be 2+9+1 = 12 months\n\nfinally, we know the rest of the cases will be impossible, because:\n\n$3(28) + 9(31) = 363 < 365$, and thus we will never be able to reach 365 days again after this case\n\n[/hide]", "Solution_2": "[hide]If the year has at most 11 months, it has at most 31*11=341 days, not enough.\nIf the year has at least 13 months, the least number of days will be 28*13=364. Trying to increase the number of days by adding more months or making a month at least 30 days will create at least 366 days, too much.[/hide]" } { "Tag": [], "Problem": "In an attempt to find the concentration of iron, which is the main component of the mineral hematite ($Fe_{2}O_{3}$), a chemist takes $1.25$ g of the mineral, dissolve it in $2$ mol/L, $0.025$ L of $H_{2}SO_{4}$, and put in zinc to reduce all into $Fe^{2+}$. Then the leftover zinc is taken out, and the remaining solution is added $H_{2}SO_{4}$, and titrate with $0.1$ mol/L of $KMnO_{4}$. At the equivalence point, it is found that $0.02$ g of $KMnO_{4}$ has been used. Find the percentage of iron in hematite. Given the half-reactions involved:\r\n$MnO_{4}^{-}+8H^{+}+5e^{-}\\rightarrow Mn^{2+}+4H_{2}O$\r\n$Fe^{2+}\\rightarrow Fe^{3+}+e^{-}$", "Solution_1": "[quote=\"Cancer\"]At the equivalence point[/quote]\r\n\r\nYou mean, \"at the final point\".\r\n\r\nWell, let's start by writing the titration reaction:\r\n\r\n$5Fe^{2+}+MnO_{4}^{-}+8H^{+}\\longrightarrow 5Fe^{3+}+Mn^{2+}+4H_{2}O$.\r\n\r\nGiven the molar mass of potassium permanganate, 158.03 g/mol, then 0.02g of it correspond to $1.3 \\times 10^{-4}$ mol. Then, the number of moles of $Fe^{2+}$ that reacted must be five times that: $n(Fe^{2+}) = 6.5 \\times 10^{-4}$ mol, which corresponds to a mass of 0.036 g. But then the percentage of iron in hematite would be $\\frac{0.036g}{1.25 g}\\times 100 = 2.9per cent$, which is a too low percentage. So perhaps I am forgeting something, or some of your data is incorrect.", "Solution_2": "OK, it should be $0.02$ L instead of $0.02$ g\r\n\r\nBut what I don't understand is why is the equation $Fe^{2+}\\rightarrow Fe^{3+}+e^{-}$ used, when in the problem the $3+$ is converted to the $2+$...", "Solution_3": "In that case the number of moles of potassium permanganate that reacted were 0.002 mol, and so the number of moles of $Fe^{2+}$ that reacted were 0.01 mol, which correspond to a mass of 0.56 g and a percentage of iron of 45%, which still seems to be low but at least more reasonable.\r\n\r\n[quote=\"Cancer\"]But what I don't understand is why is the equation $Fe^{2+}\\longrightarrow Fe^{3+}+e^{-}$ used, when in the problem the $3+$ is converted to the $2+$...[/quote]\r\n\r\nThe $Fe^{3+}$ ions present in the sample (and possibly other higher oxidation states) are previously converted to a reduced state because you are going to use a redox titration using an oxidising agent (potassium permanganate). Potassium permanganate is commonly used in redox titrations for two reasons: is a good oxidising agent, and its not necessary to use an indicator because the pink $MnO_{4}^{-}$ ion have a very low minimum detectable concentration." } { "Tag": [ "linear algebra", "matrix", "vector", "induction", "combinatorics proposed", "combinatorics" ], "Problem": "In a regular $n$-gon each side is colored blue or yellow.From such a coloring a new coloring is obtained in one step as follow:if the two neighbors of a side have different colors,the new color of that side will be blue,otherwise it will be yellow(the colors are modified simultaneously).Show that after finitely many steps all side will be color yellow.", "Solution_1": "This isn't true in general, for all $n$, if you start with any coloring whatsoever. Think about an equilateral triangle ($n=3$) having two blue sides and one yellow side. \r\n\r\nI believe there is a proof on the forum that this is true (you end up with an all-yellow polygon no matter how you begin) iff $n$ is a power of $2$. Try searching in the 'Combinatorics' section (it won't be an easy task; the problem I'm referring to was posted quite a while ago, but I do remember the thread was started by orl).", "Solution_2": "Yes,grobber :) :) .I conjectured this problem after i solve for $n=2^{k}(k \\in n^{*}\\geq 2)$ and all $4 \\leq n \\leq 20$,so i think the general may be true.\r\nHave you already found a counter-example of $n \\geq 4$ for which the conclusion doesn't take place? :blush:", "Solution_3": "Take $n=5$, for instance. We'll call our pentagon $ABCDE$, and let's say we color the vertices instead of the edges. It's easier to work that way. Start with $B,E$ blue, and the rest of the vertices yellow. After three iterations you're back where you started.\r\n\r\nSo I guess you didn't really prove it for $4\\le n\\le 20$, did you? :)\r\n\r\n\r\nEdit:\r\n\r\nYes, Jose, it was a typo. I apologize.", "Solution_4": "[quote=\"grobber\"]We'll call our pentagon $ABCDEF$[/quote]\r\n\r\nTypo grobber", "Solution_5": "I tried something with Linear Algebra, but it didn't work out.\r\n\r\nLet $0 = \\textrm{yellow}, \\, 1 = \\textrm{blue}$ and $A$ be the matrix having: $\\left( 0, 1, 0, \\ldots, 0, 1 \\right)$ on the first line and cyclic permutations of that (one step to the right, then one more step to the right) on the next rows. Then $Ax$, where $x$ is the current state vector, is the state of the polygon after we apply our transformation.\r\n\r\n$A$ should look like this $\\left( \\begin{matrix}0 & 1 & 0 & \\ldots & 0 & 1 \\\\ 1 & 0 & 1 & \\ldots & 0 & 0 \\\\ 0 & 1 & 0 & \\ldots & 0 & 0 \\\\ \\ldots & \\ldots & \\ldots & \\ldots & \\ldots & \\ldots \\\\ 0 & 0 & 0 & \\ldots & 0 & 1 \\\\ 1 & 0 & 0 & \\ldots & 1 & 0 \\end{matrix}\\right)$.\r\n\r\nThe fact $A^{m}x = \\left( \\begin{matrix}0 \\\\ 0 \\\\ \\vdots \\\\ 0 \\end{matrix}\\right)$ for all $x \\in \\mathbb F_{2}^{n}$, where $m$ is dependent of $x$, is equivalent to $A$ being nilpotent.\r\n\r\nI computed $\\det \\left( A+\\lambda I_{n}\\right)$ with Mathematica for some small values. The results are as grobber predicted.\r\n\r\ne.g.:\r\n$n=7$: $P_{A}= \\lambda+\\lambda^{5}+\\lambda^{7}$ - bad, has roots other than $0$;\r\n$n=8$: $P_{A}= \\lambda^{8}$ - good;\r\n$n=9$: $P_{A}= \\lambda+\\lambda^{5}+\\lambda^{7}+\\lambda^{9}$ - bad, has roots other than $0$;\r\netc.\r\n\r\nBut I can't find a proof. I tried computing circulant determinants (in $\\mathbb C$, then reducing to $\\mathbb F_{2}$), but I kind of failed. :(\r\n\r\nIs the idea any good?\r\n\r\n\r\nPS:\r\n\r\n[quote=\"Jos\u00e9\"][quote=\"grobber\"]We'll call our pentagon $ABCDEF$[/quote]\n\nTypo grobber[/quote]\r\n\r\nNo s**t, Sherlock! Viva la spamolucion. All must bow before you :D", "Solution_6": "How about something like the following:?\r\n\r\nWorking modulo 2, we can label the sides as $(x_{1}, \\dots x_{n})$ where $x_{i}$ is 0 if side $i$ is yellow, 1 otherwise. The recurrence we are given is effectively \r\n\r\n$x_{i}(t+1)=x_{i-1}(t)+x_{i+1}(t)$, where addition in the subscripts is modulo $n$.\r\n\r\nInstead of following this recurrence directly, however, let us next shift the vector over by 1 unit to the left. This additional shift won't affect whether all sides eventually become yellow or not, but it allows us to think of the recurrence relation as \r\n\r\n$x_{i}(t+1)=x_{i}(t)+x_{i+2}(t)$.\r\n\r\nYou can now show via induction on $k$ that $x_{i}(t+2^{k-1})=x_{i}(t)+x_{i+2^{k}}(t)$. \r\n\r\nIf $n=2^{k}$, this immediately gives that all the sides are yellow at $t=2^{k-1}$.\r\n\r\nIf on the other hand $n=2^{k}*j$, $j$ odd, we can use this to reduce from an $n-$gon to a $j-$gon, so I guess it's enough to just give a counterexample for any odd $j$.", "Solution_7": "I found the thread I was referring to:\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1073[/url]\r\n\r\nAs you can see, it's not exactly a complete solution, just an outline.", "Solution_8": "Actually, I think the case $j$ odd isn't that bad either.\r\n\r\nThe only way we can reach all yellow is if in the prior step everything was the same. \r\n\r\nThe only way everything can be the same at time $t+1$ is if either\r\n\r\n-Everything was the same at time $t$\r\n\r\nor\r\n\r\n-Each side is a different color than the side two after it at time $t$. But this is impossible, since the polygon has an odd number of sides (We can \"go right two edges\" an odd number of times to get from an edge back to itself, so we can't have changed colors every time).\r\n\r\nThis implies that for an odd number of sides the only way we could ever have reached the all yellow state is if all sides started the same color in the first place." } { "Tag": [ "inequalities", "search", "function" ], "Problem": "Prove $\\frac{a}{b+c}+\\frac{b}{a+c}+\\frac{c}{a+b} \\geq \\frac{3}{2}$ for all $a,b,c$", "Solution_1": "This is known as Nesbitts inequality. If you want an good proof try a Google search on \"[b]Nesbitts inequality\" proof[/b]. Perhaps you will find a Ph. D's work.", "Solution_2": "There are two proofs of this in the Math Olympiad Treasures book, one by Cauchy-Schwarz and one by Jensen's. Both are pretty cool.", "Solution_3": "Heh, i got this from Engel's book. :) n I guess it's a pretty popular one.", "Solution_4": "This is one of the nicer applications of rearrangement.\r\n\r\nLet $a\\geq b\\geq c$, and so $\\frac {1}{b+c}\\geq \\frac {1}{a+c}\\geq \\frac {1}{a+b}$.\r\n\r\nNow,\r\n\r\n$\\sum \\frac {a}{b+c}\\geq \\frac {1}{2}\\sum\\left( \\frac {b}{b+c}+\\frac {c}{b+c}\\right)=\\frac {3}{2}$\r\n\r\nJensen's: (this is making up for my mistake in the other thread)\r\nIt's homogeneous, so let $a+b+c=1$. Then by the CONVEXITY of $\\frac {1}{x}$, we get that:\r\n\r\n$\\sum \\frac {a}{b+c}\\geq \\frac {1}{a(b+c)+b(c+a)+c(a+b)}$\r\n\r\nBut $2/3(a+b+c)^2\\geq 2(ab+bc+ca)$, and the result follows.", "Solution_5": "Engel gave a good 10,000,000 proofs of this inequality.", "Solution_6": "Another way to apply Jensen's (from MO Treasures):\r\n\r\nLet $s=a+b+c$. Consider the convexity of the function $f(x)=\\frac{x}{s-x}$ on [0,s).\r\n\r\nThen using equal weights, $f(a)+f(b)+f(c) \\ge 3f(\\frac{a+b+c}{3}) = 3f(\\frac{s}{3}) = 3\\cdot\\frac{\\frac{s}{3}}{s-\\frac{s}{3}} = \\frac{3}{2}$.", "Solution_7": "Well, I don't have the book yet, but here is the one I came up with:\r\n\r\n\\[\\frac{a}{b+c}+\\frac{b}{a+c}+\\frac{c}{a+b} \\geq \\frac{3}{2}\\]\r\n\r\nWe rewrite this as\r\n\r\n\\[\\frac{a+b+c}{b+c}+\\frac{a+b+c}{a+c}+\\frac{a+b+c}{a+b}-3 \\geq \\frac{3}{2}\\]\r\n\\[(a+b+c)(\\frac{1}{b+c}+\\frac{1}{a+c}+\\frac{1}{a+b})-3 \\geq \\frac{3}{2}\\]\r\n\\[\\frac{1}{2} ( (a+b)+(b+c)+(a+c) ) (\\frac{1}{b+c}+\\frac{1}{a+c}+\\frac{1}{a+b})-3 \\geq \\frac{3}{2}\\]\r\n\r\nNow let $x=a+b$, $y=b+c$, and $z=a+c$ to obtain\r\n\\[\\frac{1}{2}(x+y+z)(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z})-3 \\geq \\frac{3}{2}\\]\r\n\\[\\frac{(x+y+z)(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z})-6}{2} \\geq \\frac{3}{2}\\]\r\n\\[\\frac{x}{y}+\\frac{y}{x}+\\frac{x}{z}+\\frac{z}{x}+\\frac{y}{z}+\\frac{z}{y}+3-6 \\geq 3\\]\r\n\\[{\\frac{x}{y}+\\frac{y}{x}+\\frac{x}{z}+\\frac{z}{x}+\\frac{y}{z}+\\frac{z}{y}}-3 \\geq 3\\]\r\n\\[{\\frac{x}{y}+\\frac{y}{x}+\\frac{x}{z}+\\frac{z}{x}+\\frac{y}{z}+\\frac{z}{y}} \\geq 6\\]\r\n\r\nSince for all $m$ and $n$, $\\frac{m}{n}+\\frac{n}{m} \\geq 2$ $($Note: $m^2+n^2 \\geq 2mn )$ the last inequality is clearly true, thus we are done.", "Solution_8": "[quote=\"paladin8\"]There are two proofs of this in the Math Olympiad Treasures book, one by Cauchy-Schwarz and one by Jensen's. Both are pretty cool.[/quote]What is that Math Olympiad Treasures book you're talking about? :) I've been looking for a good treasure book for a while. What math level is it aimed at?", "Solution_9": "[quote=\"DPopov\"]\n\\[\\frac{1}{2}(x+y+z)(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z})-3 \\geq \\frac{3}{2}\\]\n[/quote]\r\n\r\nMore simply, $\\left(x+y+z\\right)\\left(\\frac {1}{x}+\\frac {1}{y}+\\frac {1}{z}\\right)\\geq 9$ holds by Cauchy and by AM-HM", "Solution_10": ":( I'm still working to learn to apply those", "Solution_11": "Is this AM-HM?\r\n\r\n\\[\\frac{x+y+z}{3} \\geq \\frac{3}{\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}}\\]\r\n\\[(x+y+z) \\left( \\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z} \\right) \\geq 9\\]", "Solution_12": "[quote=\"Peter VDD\"][quote=\"paladin8\"]There are two proofs of this in the Math Olympiad Treasures book, one by Cauchy-Schwarz and one by Jensen's. Both are pretty cool.[/quote]What is that Math Olympiad Treasures book you're talking about? :) I've been looking for a good treasure book for a while. What math level is it aimed at?[/quote]\r\n\r\nIt's called \"Math Olympiad Treasures\" by Titu Andreescu and Bogdan Enescu. It's level is probably like USAMO.\r\n\r\nYes, DPopov. The first statement is AM-HM. The second is also direct Cauchy-Schwarz.", "Solution_13": "LHS is homogeneous then you cam take $a+b+c=1$\r\nIt is easy to prove that \r\n$\\frac{a}{b+c}=\\frac{a}{1-a}\\geq \\frac{9a-1}{4}$ then you can easily can get the result", "Solution_14": "It's worth noting that that last inequality is just a statement about convex functions; that is, a convex function is essentially a function which lies above its tangent at any point (the RHS is just the linear approximation at x=1/3)", "Solution_15": "This message is for DPopov: there is an extra term in your proof-step in the last 3 lines of your reasoning in the post #8. There should be 6 terms in it, not 7 terms" } { "Tag": [ "geometry" ], "Problem": "What are all the awards? I know that bronze is something like 60-74, silver like 75-95, and gold 96+(which, btw, I'm NOT going to get :) ) But what about HM and the awards for getting 5's on certain problems?", "Solution_1": "I think the honorable mention range is from 40-59, but I could be wrong.", "Solution_2": "Bronze, silver, and gold all get key chains and I think gold and silver can choose two books or items (I'm not sure on how many so you have to check them out for sure). But I do know that gold choose items first then silver then bronze then HM. HM does NOT get any keychain. As for other people, I think you just get t-shirt. I'm not sure for that either. I just know that HM gets one shirt and one item because I was HM last year (could've done better but then time is something I do NOT have.. haha..).", "Solution_3": "http://www.usamts.org/About/U_AbPrizes.php The prize ranges are all correct. :D", "Solution_4": "So even Honorable Mention gets a book? That's cool :D . What kind of books are they?", "Solution_5": "Last year I chose 360 Problems for Mathematical Contests and Challenging Problems in Geometry as a Silver, but they also have more biographical or fun books if that's what you like. If I remember correctly, there were around 7 choices or so, not just books, either.", "Solution_6": "What kind of book is left for the honorable mentions? I think that I'm getting that this year.", "Solution_7": "Ugh...well I guess after my somewhat pitiful round 1, I'll be able to scrape out a bronze if I get around the same on the other rounds...", "Solution_8": "[quote=\"13375P34K43V312\"]Ugh...well I guess after my somewhat pitiful round 1, I'll be able to scrape out a bronze if I get around the same on the other rounds...[/quote]\r\n\r\nWait, did you already get your score?", "Solution_9": "[quote=\"SplashD\"][quote=\"13375P34K43V312\"]Ugh...well I guess after my somewhat pitiful round 1, I'll be able to scrape out a bronze if I get around the same on the other rounds...[/quote]\nWait, did you already get your score?[/quote]\r\nProbably an estimate. :D" } { "Tag": [ "quadratics", "number theory", "number theory unsolved" ], "Problem": "Could anyone help me on the following:\r\n\r\nSuppose we have an equation:\r\n\r\ny^2 = [ x^3 + 1 ] mod m\r\n\r\nNow I want to solve for x,\r\n[y^2 - 1]^1/3 = x mod m\r\n\r\nHere comes my question : Why is the above similar to:\r\n\r\n[y^2 - 1] ^ (2m-1)/3 = x mod m\r\n\r\nOr more explicitly: here is anothe equation where I want to find the quadratic residue\r\n\r\ny^2 = a mod m\r\nThis is equivalent to \r\n\r\n[y^2]^ (2m-1) = a mod m\r\n________________________\r\n\r\n\r\nI could not think of any relationship between the above and Fermal Theorem. Perhaps there isn't...\r\n\r\nThanks", "Solution_1": "Neither of those manipulations are valid unless $ m$ is prime; when they are valid, they follow from Fermat's Little Theorem, although they do not really tell you what the solutions are.", "Solution_2": "[quote=\"t0rajir0u\"]Neither of those manipulations are valid unless $ m$ is prime; when they are valid, they follow from Fermat's Little Theorem, although they do not really tell you what the solutions are.[/quote]\r\n\r\nThanks alot that will do" } { "Tag": [ "logarithms", "limit", "integration", "calculus", "LaTeX", "real analysis", "real analysis unsolved" ], "Problem": "Let $s_n = \\sum_{k=n}^{\\infty}\\frac{\\ln k}{k^3}$. For which values of $a$ does $\\sum_{n=1}^{\\infty}s_n^a$ converge?", "Solution_1": "I find $s_n = \\sum_{k=n}^{\\infty}\\frac{\\ln k}{k^3}=O(\\frac{\\ln n}{n^2})$\r\nis it useful?", "Solution_2": "Yes, from the comparison test you get that $a > 1/2$ immediately from that (if it is indeed true). You can compare $s_n^a$ with $\\left(\\frac{\\ln n}{n}\\right)^{2a}$ to get that $\\sum_{n=1}^{\\infty}s_n^a$ is convergent if $\\sum_{n=1}^{\\infty}\\left(\\frac{\\ln n}{n}\\right)^{2a}$ is convergent, which happens for $a > 1/2$. How did you find it?\r\n\r\nedit: I've edited this post many times now. Maybe it is not as simple as I thought it would be at first. The current idea doesn't show what happens for $a \\le 1/2$. But yeah, your findings must definitely help. Is it possible that you could say something directly about the limit of $\\frac{n^2}{\\ln n}\\sum_{k=n}^{\\infty}\\frac{\\ln k}{k^3}$ ?", "Solution_3": "Try the condensation test.\r\n\r\n$\\sum \\left(\\frac k {2^{3k}\\right)^}{\\alpha}$ is convergent iff $\\sum \\left(\\frac{k^{\\alpha} \\ln 2^{\\alpha}}{2^{(3\\alpha-1)k}}\\right)$ which is convergent for $\\alpha > \\frac 1 3$ use: $\\sum_k \\frac{k^t}{m^k}$ with $m>1$ is convergent; and diverget if $\\alpha \\le \\frac 1 3$ use :$\\sum_k k^t\\cdot m^k$ with $m>1$ is divergent.", "Solution_4": "[quote=\"Kalle\"] Is it possible that you could say something directly about the limit of $\\frac{n^2}{\\ln n}\\sum_{k=n}^{\\infty}\\frac{\\ln k}{k^3}$ ?[/quote]\r\n\r\nYes. it is $\\infty$ Maybe you want the limit: $\\frac{\\ln n}{n^2}\\sum_{k=n}^{\\infty}\\frac{\\ln k}{k^3}$. By Cesaro stolz method the limit is equal with:\r\n\r\n$\\lim_{n\\to \\infty } \\frac{\\frac{\\ln n+1 }{(n+1)^3}} {\\frac{(n+1)^2}{\\ln n+1}-\\frac{ n^2}{ \\ln n}}=$\r\n$\\lim_{n \\to \\infty } \\frac{\\frac{(\\ln n+1)^2 \\ln n} {(n+1)^3}}{ \\ln(n+1)n^2- \\ln n (n+1)^2}=$\r\n$\\lim_{n\\to \\infty} \\frac{\\frac{(\\ln n+1)^2 \\ln n} {(n+1)^3}}{\\ln n (2n+1)-ln(1+\\frac 1 n) n^2}$\r\nDividing bpth numerator an denominator by $n$ we get that limit is $\\frac 0 {\\infty-1}=0$.\r\n\r\nCorect me if I'm wrong.", "Solution_5": "I can't understand why this equation is true.\r\n$\\lim_{n \\to \\infty } \\frac{\\frac{(\\ln n+1)^2 \\ln n} {(n+1)^3}}{ \\ln(n+1)n^2- \\ln n (n+1)^2}=$\r\n$\\lim_{n\\to \\infty} \\frac{\\frac{(\\ln n+1)^2 \\ln n} {(n+1)^3}}{\\ln n (2n+1)-ln(1+\\frac 1 n) n^2}$?\r\n\r\nAny way here my solution for:$s_n = \\sum_{k=n}^{\\infty}\\frac{\\ln k}{k^3}=O(\\frac{\\ln n}{n^2})$\r\nnotice $\\frac{\\ln x}{x^3}$ was reduce when x large enough.\r\nSo we get:\r\n$\\int_{n}^{+\\infty}\\frac{\\ln x}{x^3} dx \\le \\sum_{k=n}^{\\infty}\\frac{\\ln k}{k^3} \\le \\int_{n-1}^{+\\infty}\\frac{\\ln x}{x^3} dx$\r\nand then :)\r\nmaybe I am wrong,I am not sure.", "Solution_6": "Complete and correct solution:\r\n\r\nFor large enough n:\r\n$\\int_{n}^{\\infty}\\frac{\\ln x}{x^3} dx \\le \\sum_{k=n}^{\\infty}\\frac{\\ln k}{k^3} \\le \\int_{n-1}^{\\infty}\\frac{\\ln x}{x^3} dx$\r\n\r\nFrom primitively calculating the integrals it follows that\r\n$\\frac{n^2}{\\ln n}\\cdot\\frac{2\\ln(n) + 1 }{4n^2} \\le \\frac{n^2}{\\ln n}\\sum_{k=n}^{\\infty}\\frac{\\ln k}{k^3} \\le \\frac{n^2}{\\ln n}\\cdot\\frac{2\\ln(n-1) + 1 }{4(n-1)^2}$\r\n\r\nTaking limits, we find by the squeeze theorem that $\\lim_{n\\to\\infty}\\frac{n^2}{\\ln n}\\sum_{k=n}^{\\infty}\\frac{\\ln k}{k^3} = \\frac12$\r\n\r\nNow, let $b_n = \\frac{\\ln n}{n^2}$.\r\n\r\n$b_n^a$ and $s_n^a$ are both positive sequences and we have that that $\\frac{s_n^a}{b_n^a} \\to \\frac{1}{2^a} > 0$ as $n\\to\\infty$. By the comparison test $\\sum s_n^a$ is convergent iff $\\sum b_n^a$ is convergent. The latter is convergent precisely when $a > \\frac12$.\r\n\r\nedit: You make so many mistakes when you type LaTeX on these forums :( I don't understand any of your posts, xirti :P", "Solution_7": "I forgot a \"-\" sign :D \r\n\r\n\\[ \\ln (n+1)n^2 -\\ln n(n+1)^2=[\\ln (n+1)-\\ln n]n^2 -(\\ln n)( 2n+1)= \\]\r\n\\[ \\ln (\\frac{n+1}{n})n^2-(\\ln n)(2 n+1)=\\ln (1+\\frac{1}{n})n^2-(\\ln n)(2 n+1) \\]\r\n\r\nthe limit remains $0$ however.\r\n\r\n\r\nyour method works even for $\\sum_{k=n}^{\\infty} \\left(\\frac{\\ln k}{k^3}\\right)^{\\alpha}$. I dare to state that that $s_n = \\sum_{k=n}^{\\infty}\\frac{\\ln k}{k^3}=o(\\frac{\\ln n}{n^2})$. That's what I proved." } { "Tag": [], "Problem": "the following incredible formula was discovered by Srinivasa Ramanujan:\r\n\\[ \\frac{1}{\\pi}\\equal{}\r\n\\sum_{n\\equal{}0}^{\\infty}{2n\\choose n}^3\\cdot\\frac{42n\\plus{}5}{2^{12n\\plus{}4}}\\]\r\n\r\n\r\n\r\n\r\nhave anyone seen the proof of this ??\r\n\r\nif yes, please tell me where can I find it :)", "Solution_1": "There's another formula similar to that one that also yields $ \\frac{1}{\\pi}$. It is somewhere on his wikipedia page.", "Solution_2": "You mean $ \\frac{1}{\\pi} \\equal{} \\frac{2\\sqrt{2}}{9801} \\sum^\\infty_{k\\equal{}0} \\frac{(4k)!(1103\\plus{}26390k)}{(k!)^4 396^{4k}}$?", "Solution_3": "Yes. Unless that is just the same thing written differently.", "Solution_4": "http://www.geocities.com/titus_piezas/Pi_formulas1.pdf", "Solution_5": "Do you Know that This is the fastest algorithm ever discovered to calculated the values of pie..", "Solution_6": "By fastest, do you mean fastest converging?", "Solution_7": "[quote=ghaprameya]Do you Know that This is the fastest algorithm ever discovered to calculated the values of pie..[/quote]\n\nThe Chudnovsky brothers formula $$\\frac{426880 \\sqrt{10005}}{\\pi} = \\sum_{k=0}^\\infty \\frac{(6k)! (13591409 + 545140134k)}{(3k)!(k!)^3 (-640320)^{3k}}$$ is actually faster but as you can see it is based on Ramanujan's formula." } { "Tag": [ "probability", "symmetry", "AMC", "AIME", "number theory", "relatively prime" ], "Problem": "Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let $ m$ and $ n$ be relatively prime positive integers such that $ \\frac{m}{n}$ is the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. Find $ m\\plus{}n$.", "Solution_1": "[hide=\"Solution\"]The probability one rolls one's die for $ n$ times until the first six appears is\n\\[ \\left(\\frac56\\right)^{n - 1} \\cdot \\frac16 = \\frac {5^{n - 1}}{6^n}.\n\\]\nThen the probaility that Linda will roll her die $ n$ times while Dave does for $ n - 1$, $ n$, or $ n + 1$ times (i.e. within one of the number of times Linda rolls her die) equals\n\\[ \\frac {5^{n - 1}}{6^n} \\left( \\frac {5^{n - 2}}{6^{n - 1}} + \\frac {5^{n - 1}}{6^n} + \\frac {5^{n}}{6^{n + 1}} \\right)\n\\]\nfor all $ n \\ge 2$. For $ n = 1$, we must exclude the case when Dave rolls 0 times to obtain a probability of\n\\[ \\frac16 \\left( \\frac16 + \\frac56 \\cdot \\frac16 \\right) = \\frac {11}{216}.\n\\]\nNow, summing for all natural numbers $ n$,\n\\begin{eqnarray*} & & \\frac {11}{216} + \\sum_{n = 2}^{\\infty} \\frac {5^{n - 1}}{6^n} \\left( \\frac {5^{n - 2}}{6^{n - 1}} + \\frac {5^{n - 1}}{6^n} + \\frac {5^{n}}{6^{n + 1}} \\right) \\\\\n& = & \\frac {11}{216} + \\sum_{n = 2}^{\\infty} \\frac {5^{n - 1}}{6^n} \\left( (36 + 30 + 25) \\cdot \\frac {5^{n - 2}}{6^{n + 1}} \\right) \\\\\n& = & \\frac {11}{216} + \\sum_{n = 2}^{\\infty} \\frac {91 \\cdot 5^{2n - 3}}{6^{2n + 1}} \\\\\n& = & \\frac {11}{216} + 91 \\cdot \\frac {5/6^5}{1 - 5^2/6^2} \\\\\n& = & \\frac {11}{216} + 91 \\cdot \\frac {5}{6^5} \\cdot \\frac {6^2}{11} \\\\\n& = & \\frac {11}{216} + \\frac {455}{216 \\cdot 11} \\\\\n& = & \\frac {121 + 455}{216 \\cdot 11} \\\\\n& = & \\frac {8}{33} \\\\\n& = & \\frac {m}{n} \\end{eqnarray*}\nand therefore $ m + n = 8 + 33 = \\boxed{41}$.[/hide]", "Solution_2": "[hide=\"Solution\"]\nLet $ p_n \\equal{} \\frac16\\left(\\frac56\\right)^{n \\minus{} 1}$ be the probability that it takes $ n$ rolls. Then\n$ P \\equal{} p_1(p_1 \\plus{} p_2) \\plus{} p_2(p_1 \\plus{} p_2 \\plus{} p_3) \\plus{} p_3(p_2 \\plus{} p_3 \\plus{} p_4) \\plus{} \\cdots$\n$ \\equal{} (p_1^2 \\plus{} p_2^2 \\plus{} \\cdots) \\plus{} 2(p_1p_2 \\plus{} p_2p_3 \\plus{} \\cdots)$\n$ \\equal{} \\left(\\left(\\frac16\\right)^2 \\plus{} \\left(\\frac16\\right)^2\\left(\\frac56\\right)^2 \\plus{} \\left(\\frac16\\right)^2\\left(\\frac56\\right)^4\\right) \\plus{} 2\\left(\\left(\\frac16\\right)^2\\left(\\frac56\\right) \\plus{} \\left(\\frac16\\right)^2\\left(\\frac56\\right)^3 \\plus{} \\left(\\frac16\\right)^2\\left(\\frac56\\right)^5 \\plus{} \\cdots\\right)$\n$ \\equal{} \\left(\\frac16\\right)^2\\left(\\frac1{1 \\minus{} \\left(\\frac56\\right)^2} \\plus{} 2\\cdot\\frac56\\cdot\\frac1{1 \\minus{} \\left(\\frac56\\right)^2}\\right) \\equal{} \\left(\\frac16\\right)^2\\left(1 \\plus{} \\frac53\\right)\\left(\\frac {6^2}{11}\\right)$\n$ \\equal{} \\frac83\\cdot\\frac1{11} \\equal{} \\frac8{33}\\implies8 \\plus{} 33 \\equal{} \\boxed{41}$.\n[/hide]", "Solution_3": "[hide=\"solution\"]\nLet $ p_0$ be the probability they end on same roll.\nThis is the chance they both roll 6, plus $ p_0$ times the chance they both roll non-6.\n$ p_0 \\equal{} \\left(\\frac {1}{6}\\right)^2 \\plus{} \\left(\\frac {5}{6}\\right)^2 p_0$\n$ p_0 \\equal{} \\frac {1}{11}$\nLet $ p_D$ be probability that Dave ends 1 roll before Linda.\nThis is the chance of Dave rolling 6, Linda rolling a non-6, then Linda rolling a 6 the roll after, added to $ p_D$ times the chance both of them roll non-6.\n$ p_D \\equal{} \\left(\\frac {1}{6}\\right)\\left(\\frac {5}{6}\\right)\\left(\\frac {1}{6}\\right) \\plus{} \\left(\\frac {5}{6}\\right)\\left(\\frac {5}{6}\\right)p_D$\n$ p_D \\equal{} \\frac {5}{66}$\nThe probability that Linda ends 1 roll before Dave is also equal to $ p_D$.\nNone of the three probabilities intersect (if one of them happens, the other two can't also happen). Thus the total probability is $ p_0 \\plus{} p_D \\plus{} p_D \\equal{} p_0 \\plus{} 2p_D \\equal{} \\frac {1}{11} \\plus{} \\frac {5}{33} \\equal{} \\frac {8}{33}$\n[/hide]", "Solution_4": "[hide]\nAll the times when 6's are rolled at the same time can be written as ${ (\\frac {5^{n-1}}{6^n}})^2=\\frac {5^{2n - 2}}{6^{2n}}$, all the times when they aren't rolled at the same time can be written as ${{ (\\frac {5^{n-1}}{6^n}})(\\frac {5^{n}}{6^n+1}})=\\frac {5^{2n - 1}}{6^{2n+1}}$, but either person may roll a 6 first, so it needs to be multiplied by two.\n\n$ \\sum_{i = 1}^{\\infty} \\frac {5^{2n - 2}}{6^{2n}} + 2 \\sum_{i = 2}^{\\infty} \\frac {5^{2n - 1}}{6^{2n+1}}$\n\n$ \\frac {\\frac{5^0}{6^2}}{1-\\frac{5^2}{6^2}}+2\\frac {\\frac{5^1}{6^3}}{1-\\frac{5^2}{6^2}}$\n$ \\frac{\\frac{1}{36}}{\\frac{11}{36}}+2\\frac {\\frac{5^1}{6^3}}{\\frac{11}{6^2}}$\n\n$ \\frac{1}{11}+2\\frac {5}{66}$\n\nWould have gotten it right if I had remembered that I needed to multiply by 2.\n\n$ \\frac{16}{66}=\\frac{8}{33}$[/hide]", "Solution_5": "Wouldn't it be more accurate to name this thread \"Rolling Sixes Simultaneous-ish-ly?\" :)", "Solution_6": "Surprisingly, I thought this was the easiest question on the test. I solved it in about 3-5 minutes (I failed #1 lol)", "Solution_7": "strange, i got to 11/216+455/216*11 and epic failed :(", "Solution_8": "Isn't there symmetry in this problem? 3 cases: 1) Dave has same # of rolls as Linda. 2) Linda has more rolls than Dave. 3) Dave has more rolls than Linda. \r\n\r\nNumber 1 has 1/11 chance of happening, and 2 and 3 have an equal chance of happening, since they both roll under the same conditions, so why can't you just divide 1-1/11 = 10/11 by 2? \r\nAs in, what's wrong with the logic there?", "Solution_9": "That's not what the question is asking for.", "Solution_10": "It's asking for same number of rolls or within. So, why couldn't you just add the chance of equal rolls (1/11), with the chance that Linda has more rolls?", "Solution_11": "Georgiamathdude, I initially interpreted the problem exactly the same way as you did--and I got exactly your answer.\r\n\r\nI kept reading the problem over and over again until I finally figured out my blind spot. I had overlooked the word \"one\" in the last line of the problem. \"Within [i]one [/i]of the number of times\" does not mean the same as \"Within the number of times.\"", "Solution_12": "Oh, i see. I saw it as within \"any one of the rolls\" Like, if Linda rolled a 6, it could be any ONE of 1, 2, 3, 4, 5, 6... alrite. thanks.", "Solution_13": "I agree that the wording of the problem lends itself to some ambiguity of interpretation. The English phrase \"one of the number of times\" is somewhat ambiguous, depending on the context. \r\n\r\nFor example, consider the use of the phrase in the sentence \"During [i]one of the number of times[/i] he visited me in the hospital, he read me poetry.\" In that context, the phrase \"one of the number of times\" means something entirely different than it does in the AIME problem. It means \"one of the numerous instances of times.\" Since \"during\" and \"within\" are somewhat similar, I can see how you and I might have been led to parse the problem statement incorrectly. \r\n\r\nIt really comes down to how your mind chops up the sentence into phrases. A careful reading indicates that the correct parsing should immediately connect together the word \"one\" inside the phrase \"within one of\" rather than inside the phrase \"one of the number of times.\" \r\n\r\nI certainly don't think there's any grounds for protest, as a careful reading makes the intended interpretation evident, but it would have been even clearer, in my opinion, to have worded the problem as follows:\r\n\r\nDave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Call the total number of times that Dave rolls $ D$, and call the total number of times that Linda rolls $ L$. Let $ m$ and $ n$ be relatively prime positive integers such that $ \\frac {m}{n}$ is the probability that $ |D \\minus{} L| \\le 1$. Find $ m \\plus{} n$.", "Solution_14": "Yea, I was going for 7 and i got the first 7 so when I \"got\" the eigth one, I was like whoa that too easy... nah don't go back I'm done.", "Solution_15": "[quote=worthawholebean]Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let $ m$ and $ n$ be relatively prime positive integers such that $ \\frac{m}{n}$ is the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. Find $ m\\plus{}n$.[/quote]\n\nOh man this problem was so confusing... I thought that if Linda rolled her dice $k$ then Dave could roll $1,2,3,\\cdots,k$ times :(\n\nRead the problem carefully.", "Solution_16": "Grinding [url=https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZS9iL2FjOWU0OTVjZDU0ZWMyM2I2MDMxYjU1OWEzM2RmNzg0NjljYjQyLnBkZg==&rn=QWltZUNvdW50aW5nUHJvYmxlbXMucGRm]@hurdler's AIME combinatorics handout[/url]\n\n[hide=solution]Consider three cases. Let the requested probability be $p$.\n\nIf both get a six on their first roll, game over. This occurs with probability $\\displaystyle\\frac{1}{36}$.\n\nIf one gets a six on the first roll and the other doesn't (which occurs with probability $2\\cdot\\displaystyle\\frac{5}{6}\\cdot\\displaystyle\\frac{1}{6}=\\displaystyle\\frac{5}{18}$), the one who didn't get a six initially will need to get a six on the second roll (otherwise the difference won't be $\\leq 1$). This occurs with probability $\\displaystyle\\frac{1}{6}$, so the entire situation has probability $\\displaystyle\\frac{5}{18}\\cdot\\displaystyle\\frac{1}{6}=\\displaystyle\\frac{5}{108}$.\n\nIf no one gets a six on the first roll (which occurs with probability $\\displaystyle\\frac{25}{36}$), then we're back to square one.\n\nTherefore $\\displaystyle\\frac{1}{36}+\\displaystyle\\frac{5}{108}+\\displaystyle\\frac{25}{36}p=p$, so $\\displaystyle\\frac{2}{27}=\\displaystyle\\frac{11}{36}p$ and $p=\\displaystyle\\frac{72}{297}=\\displaystyle\\frac{8}{33}$. Answer $8+33=\\boxed{041}$.[/hide]", "Solution_17": "We take three cases.\n\nCase 1: Dave = Linda\n\nThe probability is $\\frac{1}{6} \\cdot \\frac{1}{6} + \\frac{5}{6} \\cdot \\frac{5}{6} \\cdot \\frac{1}{6} + \\ldots$, a geometric sequence with common ratio $\\frac{25}{36}$ and first term $\\frac{1}{36}$. This gives a sum of $\\frac{\\frac{1}{36}}{1-\\frac{25}{36}} = \\frac{1}{11}$.\n\nCase 2: Dave = 1 + Linda\n\nThe probability is $\\frac{5}{6} \\cdot \\frac{1}{6} \\cdot \\frac{1}{6} + \\frac{5}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} \\cdot \\frac{1}{6} + \\ldots$, a geometric sequence with common ratio $\\frac{25}{36}$ and first term $\\frac{5}{216}$. This gives a sum of $\\frac{\\frac{5}{216}}{1-\\frac{25}{36}} = \\frac{5}{66}$.\n\nCase 3: Dave = -1 + Linda\n\nThis would be the same as the previous case by symmetry, so another $\\frac{5}{66}$.\n\nThe total is $\\frac{1}{11} + 2 \\cdot \\frac{5}{66} = \\frac{8}{33} \\implies \\boxed{41}$." } { "Tag": [ "ratio", "geometry", "3D geometry", "sphere" ], "Problem": "Three tennis balls stacked one on top of the other, fit snuggly in a cylindrical container. What is the ratio of the volume of the tennis balls to the volume of the space inside the container not occupied by the tennis balls?", "Solution_1": "[hide=\"Solution.\"]\nIf the radius of one tennis ball is $ r$, then clearly the radius of the base of the container is $ r$ and the height of the can is $ 6r$. Then the volume of the three balls is $ 3\\cdot \\left(\\frac {4\\pi r^3}{3}\\right) \\equal{} 4\\pi r^3$. Also, the volume of the can is $ (\\pi r^2)(6r) \\equal{} 6\\pi r^3$. Thus the desired ratio is:\n\n$ \\frac {4\\pi r^3}{6\\pi r^3 \\minus{} 4\\pi r^3} \\equal{} \\boxed{2: 1}$\n[/hide]", "Solution_2": "it is also useful to know that a sphere's volume is $ \\frac23$ of the cylinder with same radius and height equal to diameter.\r\nstacking 3 of these cylinders with spheres inside them, you get the structure described\r\nsince the occupied space is $ \\frac23$ and the unoccupied space is $ \\frac13$, the ratio is $ 2: 1$" } { "Tag": [ "geometry", "number theory proposed", "number theory" ], "Problem": "Given an equilateral triangle and a point in its interior. There is such a triangle with all 3 distances and the length of the triangle side being integers?", "Solution_1": "Which distances are you talking about?\r\n\r\nIf these are the distances to the sides of the triangle, the answer is no. By an area argument, the sum of the three distances is the altitude of the triangle, which is $\\sqrt 3/2$ times the side, so it's impossible the three distances and the side to be all integers.\r\n\r\nIf these are the distances to the vertices, I don't know the answer right now, but I can try it later.", "Solution_2": "I was rushing :blush: The question was....\r\n\r\nGiven an equilateral triangle and a point in its interior.\r\nThere is such a point with all 3 distances to the vertices and the length of the triangle side being integers?" } { "Tag": [ "quadratics", "algebra", "polynomial", "number theory", "modular arithmetic" ], "Problem": "Find all $ (a ; b)$ such that $ a ; b$ are [b]interger[/b] and:\r\n$ \\frac {a \\plus{} b}{a^2 \\minus{} ab \\plus{} b^2} \\equal{} \\frac {3}{7}$\r\nor can write $ 7a \\plus{} 7b \\equal{} 3a^2 \\plus{} 3b^2 \\minus{} 3ab$\r\nI found that $ a \\equal{} 4, b \\equal{} 5$ or $ a \\equal{} 5, b \\equal{} 4$ by an ugly proof \r\nHelp the new member please :(", "Solution_1": "if we work in $ \\mathbb{Z}/3\\mathbb{Z}$ and $ \\mathbb{Z}/7\\mathbb{Z}$ we find that:\r\n$ a \\equal{} 21k \\plus{} 4$ and $ b \\equal{} 21k \\plus{} 5$\r\nso we get:\r\n\r\n$ \\frac {a \\plus{} b}{a^{2} \\minus{} ab \\plus{} b^{2}} \\equal{} \\frac {3(14k \\plus{} 3)}{7(63k^{2} \\plus{} 27k \\plus{} 3)}$\r\n===> $ 14k \\plus{} 3 \\equal{} 63k^{2} \\plus{} 27k \\plus{} 3$\r\n===>$ k \\equal{} 0$ then:\r\n$ (a,b) \\equal{} (4,5) or (5,4)$", "Solution_2": "[quote=\"mathematikos\"]if we work in $ \\mathbb{Z}/3\\mathbb{Z}$ and $ \\mathbb{Z}/7\\mathbb{Z}$ we find that:\n$ a \\equal{} 21k \\plus{} 4$ and $ b \\equal{} 21k \\plus{} 5$[/quote] Working mod 3 and 7, the most you could possibly get is that $ a \\equal{} 21k \\plus{} 4$, $ b \\equal{} 21j \\plus{} 5$. There is no way modular arithmetic will tell you they differ by exactly 1, rather than by 1 modulo 21.\r\n\r\nThe equation you get is quadratic in $ a$, so the discriminant $ (3b \\plus{} 7)^2 \\minus{} 4\\cdot 3 \\cdot (3b^2 \\minus{} 7b)$ must be nonnegative. This implies $ 27b^2 \\minus{} 126b \\minus{} 49 \\leq 0$. This quadratic polynomial has reasonably nice roots, one in the interval $ ( \\minus{} 1, 0)$ and the other in the interval $ (5, 6)$. So we only need to check integer values of $ b$ from 0 to 5 and see if they have matching $ a$-values. This is only moderately ugly." } { "Tag": [ "inequalities", "function", "inequalities solved" ], "Problem": "a,b,c,d>0 ab+bc+cd+da=1\r\nProve that:\r\n\r\na^3/(b+c+d)+ b^3/(c+d+a)+ c^3/(a+b+d)+ d^3/(a+c+b) \\geq 1/3", "Solution_1": "by cauchy,\r\n[aaa/(b+c+d)+bbb/(c+d+a)+ccc/(d+a+b)+ddd/(a+b+c)][ab+ac+ad+bc+bd+ba+cd+ca+cb+da+db+dc]>=(aa+bb+cc+dd)^2\r\n2(ab+ac+ad+bc+bd+cd)<=3(aa+bb+cc+dd) (am-gm)\r\n\r\naaa/(b+c+d)+bbb/(c+d+a)+ccc/(d+a+b)+ddd/(a+b+c)\r\n>=aa+bb+cc+dd/3\r\n>=(ab+bc+cd+da)/3\r\n=1/3", "Solution_2": "Just an idea ... couldn't this be done using Jensen ?\r\n\r\nI mean, is the function f(x) = x/(a - x) convex on [0, a[ for a > 0 ?\r\n\r\nIf so, put a + b + c + d = S > 0. Put f(x) = x/(S - x).\r\nThen LHS = f(a) + f(b) + f(c) + f(d) \\geq 4 f(S/4).\r\nSo it suffices to prove that 1/3 \\leq 4 f(S/4) = S/12.\r\nSo it suffices to prove that S \\geq 4. Now\r\nS = a + b + c + d + 2(ab + bc + cd + da) + 2ac + 2bd\r\n= (a + c) + (b + d) + 2 \\geq ((a + c) + (b + d))/2 + 2 = S/2 + 2.\r\nSo S \\geq S/2 + 2, so indeed S \\geq 4.\r\n\r\nBut i'm not sure whether f(x) = x/(a - x) is convex.\r\nI'll calculate f\"(x). (Boring !!!!!!)\r\n\r\nTo be continued ...", "Solution_3": "I think f is indeed convex.\r\nCould someone verify my computations - e.g. with software ?\r\nYou know, I always make lots of mistakes in boring computations ...", "Solution_4": "A lemma Maverick posted before works fine here: for real ai and positive xi a1^2/x1+..+an^2/xn >= (a1+a2+..+an)^2/(x1+..+xn). We amplify each fraction by a, b, c, d respectively and we have to prove \r\n\r\n \\sum a^4/ab+ac+ad >=1/3. We apply the lemma and we are left with (a^2+b^2+c^2+d^2)^2/2(ab+ac+ad+bc+bd+cd) >= 1/3. (a^2+b^2+c^2+d^2)/2(ab+ac+ad+bc+bd+cd) >= 1/3 is well known and a^2+b^2+c^2+d^2 >= 1= ab+bc+cd+da by rearrangement inequality." } { "Tag": [ "function", "integration", "logarithms", "limit", "calculus", "calculus computations" ], "Problem": "Let $F: (1,\\infty) \\to \\R$ be the function defined by\r\n\r\n$F(x) = \\int^{x^{2}}_{x}\\frac{1}{\\ln{t}}\\,dt$\r\nfind $F(x)$ as $x\\to 1$", "Solution_1": "Why does the question read \"Find the minimum?\"", "Solution_2": "[quote=\"FieryHydra\"]Why does the question read \"Find the minimum?\"[/quote]\r\n\r\noh, because i found the function to be strictly increasing. hence, the minimum of it is as x approaches 1", "Solution_3": "[quote=\"violetcraze\"]Let $F: (1,\\infty) \\to \\R$ be the function defined by\n\n$F(x) = \\int^{x^{2}}_{x}\\frac{1}{\\ln{t}}\\,dt$\nfind $F(x)$ as $x\\to 1$[/quote]\r\nNice problem, and though I got hold of the correct technique the first time I attempted this problem, for some inexplicable reason I didn't finish it off from there. It was only after I confirmed with Mathematica there was a solution that I went back and solved the problem.\r\n\r\nAnyways, let me solve a more general problem: Find $\\lim_{x \\to 1}\\int_{x}^{x^{n}}\\frac{\\,dt}{\\ln t}.$\r\n\r\nSince the function $\\frac{1}{\\ln t}$ is not integrable in the elemenary sense, we must somehow determine if we can use Taylor series expansion and then integrate term by term. To achieve this, we make the substitution $y = \\ln t.$\r\n\r\nSo, we have $\\int_{x}^{x^{n}}\\frac{\\,dt}{\\ln t}$\r\n\r\n$= \\int_{\\ln x}^{n \\ln x}\\frac{e^{y}\\,dy}{y}$\r\n\r\n$= \\int_{\\ln x}^{n \\ln x}\\frac1{y}(1+\\frac{y}{1!}+\\frac{y^{2}}{2!}+\\frac{y^{3}}{3!}+\\ldots) \\,dy$\r\n\r\n$= \\int_{\\ln x}^{n \\ln x}\\frac{\\,dy}{y}+\\int_{\\ln x}^{n \\ln x}(1+\\frac{y}{2!}+\\frac{y^{2}}{3!}+\\ldots)\\,dy$\r\n\r\n$= \\ln y \\Big|_{\\ln x}^{n \\ln x}+(y+\\frac{y^{2}}{2\\cdot 2!}+\\frac{y^{3}}{3 \\cdot 3!}+\\ldots)\\Big|_{\\ln x}^{n \\ln x}$\r\n\r\n$\\Rightarrow \\boxed{\\lim_{x \\to 1}\\int_{x}^{x^{n}}\\frac{\\,dt}{\\ln t}= \\ln n}.$" } { "Tag": [ "abstract algebra", "number theory unsolved", "number theory" ], "Problem": "Let p be a prime. In Zp(the group with addition module p), let A and B be subsets of Zp and |A|+|B|-1<=p. Prove that\r\n |A+B|<=|A|+|B|-1, where A+B={a+b|a in A and b in B}\r\n\r\n(Note: I am not sure whether the problem is right, that is, maybe there are some counterexamples. But I strongly believe there aren't any and I've write a program to test the cases p<30. No counterexample is found)", "Solution_1": "It's called Cauchy-Davenport-Theorem:\r\n\r\n[url]http://planetmath.org/encyclopedia/CauchyDavenportTheorem.html[/url]\r\n[url]http://www.mathlinks.ro/Forum/topic-43784[/url]", "Solution_2": "Thank you.\r\n\r\nHow sad it has been proved before, I think it is the best problem I've ever found by myself ;) ." } { "Tag": [ "geometry" ], "Problem": "A wall to wall carpet cost $ 15 per square yard. Rita paid$330 to carpet a room 6 yards long. What is the width of the room?\r\n\r\nHow do I find the answer?", "Solution_1": "First, we find the number of square yards.\r\n\r\n=330/15\r\n=22 sq. yds.\r\n\r\nThen we divide 22 by 6 to get the width.\r\n\r\n=11/3 yards.", "Solution_2": "[hide=\"Solution\"]\nThe area of the carpet can be $ \\text{length} \\times \\text{width}$. Dividing $ 330$ by $ 15$ yields $ 22$, which is the area of the carpet. Plugging it into our equation, we get $ 6 \\times w \\equal{} 22$, or $ \\text{width} \\approx \\boxed{3. \\overline{6}}$.\n[/hide]\r\n\r\nEDIT: Wow someone beat me..." } { "Tag": [ "probability" ], "Problem": "Try this problem I made up:\r\n\r\nBob and Rob play toss an unfair coin 3 times. It has a $\\frac{2}{3}$ chance of landing heads and $\\frac{1}{3}$ tails. If exactly 2 out of 3 times the coin lands heads, Bob gets a point. If otherwise, Rob gets a point. They play until they have a total of 3 points. What is the probability that Bob has exactly 2 points?", "Solution_1": "[hide]After 3 coin flips, Bob has a $\\frac{2}{3}\\cdot \\frac{2}{3} \\cdot \\frac{1}{3} \\cdot {3} = \\frac{4}{9}$ chance of getting a point. So after 3 points, Bob has a $\\frac{4}{9} \\cdot \\frac{4}{9} \\cdot \\frac{5}{9} \\cdot {3} = \\frac{80}{243}$[/hide]", "Solution_2": "[hide]The probablility that there is exactly two heads out of three is $(\\frac{2}{3})^2\\cdot\\frac{1}{3}\\cdot3 = \\frac{4}{9}$. That means not getting 2/3 is 5/9. \n\n$(\\frac{4}{9})^2\\cdot\\frac{5}{9}\\cdot3=\\frac{80}{243}$. [/hide]", "Solution_3": "Yup that's correct! :D" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "if$ 0\\leq k<\\frac{1}{\\sqrt{2}}$,and $ a,b>0$,prove that\r\n\\[ (\\frac{a}{a\\plus{}kb})^2\\plus{}(\\frac{b}{b\\plus{}ka})^2\\geq\\frac{1\\minus{}2k^2}{(1\\minus{}k^2)^2}\\]", "Solution_1": "Let $ x\\equal{}\\frac{b}{a}$,\r\n$ LHS\\minus{}RHS\\equal{}(\\frac{1}{1\\plus{}kx})^{2}\\plus{}(\\frac{x}{x\\plus{}k})^{2}\\minus{}\\frac{1\\minus{}2k^{2}}{(1\\minus{}k^{2})^{2}}\\equal{}\\frac{(k^{3}x^{2}\\plus{}k^{3}\\plus{}3k^{2}x\\minus{}x)^{2}}{(1\\minus{}k^{2})^{2}(1\\plus{}kx)^{2}(x\\plus{}k)^{2}}\\geq 0$" } { "Tag": [ "algebra", "polynomial", "complex numbers", "algebra unsolved" ], "Problem": "Find all real polinomials $P(x)$ that $P(x)P(x+1)=P(x^2)$,for every $x\\in{R}$", "Solution_1": "$P(x)=0 \\forall x$or$P(x)=x^n(x-1)^n$for any natural number n", "Solution_2": "[quote=\"vietnamesegauss89\"]$P(x)=0 \\forall x$or$P(x)=x^n(x-1)^n$for any natural number n[/quote]\r\nPlease,then sent solution,and not only answer :mad: :) .", "Solution_3": "I just want to mention that this problem was proposed in 2000 in the last round of the Bulgarian Mathematical Olympiad as Problem 1, Day 2 :)", "Solution_4": "Let $\\omega$ be any root of $P$. Then, evaluating $P(x)P(x+1)=P(x^2)$ in $\\omega$, we see that the LHS is $0$, so the RHS is $0$, and then $\\omega^2$ is a root of $P$.\r\n\r\nSuppose that $P$ is not $0$; then it has finitely many roots. If it has a root $\\omega$ of modulus not $0$ or $1$, then $\\omega^2, \\omega^4, \\omega^8, \\ldots$ are all roots of $P$ with different modulus; so $P$ has infinitely many roots, contradiction. Then $P$ has roots of modulus only $0$ or $1$.\r\n\r\nHowever, let $\\omega$ be a root of $P$. Then the LHS is $0$ for $x=\\omega-1$ also. Then the RHS is $0$, so $(\\omega-1)^2$ is a root of $P$. So it has modulus $0$ or $1$, then $\\omega-1$ has modulus $0$ or $1$. This allows for two possibilities: either $\\omega$ is $0$ or $1$, or both $\\omega, \\omega-1$ are on the unit circle. This means that $\\omega$ has angle $60$ or $300$ in the complex plane. However, if $\\omega$ has angle $60$, then $\\omega^2$ is a root and has angle $120$ which is forbidden; if it has angle $60$, then $\\omega^2$ has angle $240$, which is forbidden. So all roots are $0$ or $1$.\r\n\r\nLet $P(x)=x^m(x-1)^n$. Then $P(x+1)=(x+1)^mx^n$ and $P(x^2)=x^{2m}(x^2-1)^n$. Since $P$ has $m$ zero roots, $P(x+1)$ has $n$ and $P(x^2)$ has $2m$, we have $m+n=2m$ so $m=n$. Also, $x^n(x-1)^n$ works.", "Solution_5": "Whoa that's hard...I worked like a functional one, studyng x(x+1) and subtraing x and using series I had to modify the polyomial to x^2-x=x(x-1) but it's a terrifing method ;) \r\n\r\nCan I reply? Find al polynomials for which $P(x)P((x-1))=P(x^2)$. And $P(x-1)P(x+1)=P(x^2)$", "Solution_6": "This also appeared on an Irish Olympiad.", "Solution_7": "Well, I had just discovered this technique and was trying to fit it into any problem I could... incidentally, it works in these ;) \r\n\r\n$P(x)P(x-1)=P(x^2)$ is similar to the first one: if $\\omega$ is a root, then so are $\\omega^2$ and $(\\omega+1)^2$. Because of the first, all roots have modulus $0$ or $1$. Besides, $\\omega+1$ must have modulus $0$ or $1$ If $1$ is a root, then $(1+1)^2$ is a root, contradiction. If $0$ is a root, then $(0+1)^2$ is a root, contradiction. If $-1$ is a root, then $(-1)^2=1$ is a root, contradiction. So, for $\\omega, \\omega+1$ to have both modulus $0$ or $1$, $\\omega$ must be a third root of unity. Since they come in conjugate pairs, either $P=0$ or $P=1$, or $P$ is divisible by $x^2+x+1=(x-\\omega)(x-\\omega^2)$. If the latter happens, we divide by $x^2+x+1$ and repeat the argument. Solutions are $P=0, 1, (x^2+x+1)^n$.\r\n[Edited because I had messed up the casework.]\r\n\r\nFor $P(x+1)P(x-1)=P(x^2)$, we see that if $\\omega$ is a root then $(w-1)^2, (w+1)^2$ are too. Now note that at least one of $\\omega+1, \\omega-1$ has modulus greater than $\\omega$ (+1 if it has positive real part, -1 otherwise). If $\\omega$ has modulus at least $1$, and $\\omega+1$ has greater modulus (the other case is analogous), then $(\\omega+1)^2$ has it even greater modulus; so we can construct infinitely many roots this way. So all roots are inside the unit circle. However, for any nonzero $\\omega$ in the unit circle, at least one of $\\omega+1, \\omega-1$ is outside, so its square is outside too, contradiction. Then only $\\omega=0$ is possible. But then $(\\omega+-1)^2=1$ which is nonzero, contradiction. There are no solutions except constants, $P=0, 1$.", "Solution_8": "[hide=\"another way\"]and what about divisibility? checking 0 and 1 in the polinomial you see that the known term is 0. So $x+1|P(x+1)$ and $x+1|P(x^2)$ and $x|P(x+1)$ (and $x(x-1)|P(x^2)-P(x)$[/hide]" } { "Tag": [ "probability", "AMC", "AIME", "USA(J)MO", "USAMO", "trigonometry", "AMC 10" ], "Problem": "Hmmm? \r\n\r\nBIlly", "Solution_1": "Or 8th.\r\n\r\n7. Though it should have been a 9...sigh.", "Solution_2": "9th grader. Got an 8. Meh.", "Solution_3": "[quote=\"bookaholic\"]Or 8th.\n\n7. Though it should have been a 9...sigh.[/quote]\r\n\r\nI'm sorry, but it clearly says 9th or 10th. You will be prosecuted to the fullest extent of the law. You have the right to remain silent; anything you say may be used against you in court.", "Solution_4": "But I'm a possible-floor-qualifier!\r\n\r\nActually, I'm merely a possible-floor-qualifier-providing-that-\r\nsolafidefarms's-and-mine-and-all-the-other-7-pointers'-wishful-\r\nthinking-somehow-pays-off. But that's okay.", "Solution_5": "Ah, probability, she never said she voted, did she now? :D (Not that I'd be offended if she did!)\r\n\r\nI think they'll make us have some sort of special qualifying process, since way too many of us got 7s. Or am I grasping at straws? :)", "Solution_6": "[quote=\"probability1.01\"]..anything you say may be used against you in court.[/quote]\r\n\r\nI believe it's \"in a court of law\". *pats*", "Solution_7": "Sophomore. 9. I'll probably make USAMO.", "Solution_8": "[quote]The lowest AIME score among those 160 first selected will determine a floor value. The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value.[/quote]\r\n\r\nIf they have too many qualifiers by the floor value then they'll look at your AMC score.", "Solution_9": "I'm fairly sure that 160 number is going to be more like 220-250 given the expansion.", "Solution_10": "I'm a seventh grade girl and I got an 8.. Do you think I'd qualify?", "Solution_11": "So there's a chance of a floor value of 6? :D", "Solution_12": "I'm a freshman, and I got an 8. I hope I qualify.", "Solution_13": "6. I really really really really really hope that the floor will be 6 :P. And good job to all of you :)", "Solution_14": "[quote=\"inyoung\"]I'm a seventh grade girl and I got an 8.. Do you think I'd qualify?[/quote]\r\nThey don't make a distinction between genders, but I think an 8 will probably qualify. Lucky duck.", "Solution_15": "Actually now I think of it the floor of 7 is very likely. \r\n\r\nHere is the 2005 AOPS AIME statistics:\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=2005+AIME&t=29581]www.artofproblemsolving.com/Forum/viewtopic.php?highlight=2005+AIME&t=29581[/url]\r\n\r\n41% of AoPSers last year scored the floor value(9) or above, \r\n\r\nNow look at this year's AIME statistics\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77873&start=160]www.artofproblemsolving.com/Forum/viewtopic.php?t=77873&start=160[/url]\r\n36% of AoPSers this year scored 9 or above, 47% scored 8 or above. Now we can expect AoPS to have gotten a little better since last year, so without expansion a floor value of 8 seems likely. Now we take account of the expansion, a floor of 7 seems likely. \r\n\r\nI hope this makes some sense :lol:", "Solution_16": "AOPS has expanded dramatically over the last year also, so we'll probably have more of the qualifiers.", "Solution_17": "[quote=\"filletwho\"]AOPS has expanded dramatically over the last year also, so we'll probably have more of the qualifiers.[/quote]\r\n\r\nYeah the poll for this year is probably not done yet(but we already have more voters than last year, meaning aops probably expanded).", "Solution_18": "Thank you beta for actually taking the time to analyze the stats! :D", "Solution_19": "[quote=\"K81o7\"]10th grade.\nGot a 7. Though it should have been 12...I should have been a bit careful and checked more thoroughly. Yes, I lost [b]5[/b] points out of utter carelessness :roll:[/quote] Yeah, me too. Kind of. I got a 10, but I thought I had a 13, and the two I didn't think I had gotten I got after the test in like 5 minutes each. My mistakes were as follows:\r\n\r\n8) Lack of geometric intuition...square is only max if side length is constant, off by factor of $\\sqrt2$.\r\n10) Just another stupid error, 4 and 5 are only one apart, not two, so we only need to shift 1/2 over...\r\n11) Never thought of recursion until after the test.\r\n12) Tried to overdo it and go the really long way about it (i.e. substituting trig identities for cos(ax) for a=3,4,5), drew a bad diagram and ended up being off by 9 degrees.\r\n14) Mixed up Stewart's Theorem, b is on the same side as n, m on the same side as c. Yeah, that was just stupid, because I got everything else in that problem.\r\n\r\nI also noticed that the three I thought I had but didn't (8, 10, and 14) were all geometry. Maybe that says something... :roll:\r\n\r\nSorry to anyone that feels envious of my 10, it was just really frustrating not to get the problems I thought I had gotten and end up with the same score as last year.", "Solution_20": "If the floor is 7, i might be screwed over because of my relatively low index (roughly 214). Any opinions, or speculations on the cutoff Index for 10th and under?", "Solution_21": "[quote=\"dhuang614\"]If the floor is 7, i might be screwed over because of my relatively low index (roughly 214). Any opinions, or speculations on the cutoff Index for 10th and under?[/quote]\r\n\r\nI don't think there's any such thing as an underclassman index. We qualify on the floor, and you'll qualify if the floor's 7.", "Solution_22": "sophomore: 7\r\nwhat do u think the floor will be?", "Solution_23": "It could be influenced by the AIME IIs, but right now it most likely will be an 8 (7's still possible though).", "Solution_24": "OKay since it's quite possible aops on general is decreasing in math ability, we'll take it from another approach.\r\nWe still refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=29581]this thread[/url]. \r\n\r\n In 2005, 76 people scored 9 or above in that poll, out of 378 people last year on AIME I scoring 9 or above, that's approximately 20%. \r\n\r\nBut the % of people scoring the floor value or above that are frequent AoPS visitors cannot possibly decrease, so this year we expect at least 20% of the people scoring the floor or above to be AoPSers. Since there's expansion we expect 570ish people scoring the floor or above. So at least 115ish people will come from AoPS frequent visitors(by that I mean people who vote in the poll). Now 92 people scored 9 or above this year on AoPS, 120 scored 8 or above. So floor will be 8(well, at most an 8)", "Solution_25": "Freshman\r\nAMC10:146.5\r\nAIMEI:7\r\n\r\nI'm also hoping for a floor of 7.", "Solution_26": "i predict an 8! i'm a 9th grader, got 11.", "Solution_27": "9th grade.\r\n130 AMC 12B.\r\n10 AIME.\r\n230 index.", "Solution_28": "[quote=\"mathclass\"]i predict an 8! i'm a 9th grader, got 11.[/quote]\r\n\r\nActually I don't the think the floor can possibly be 40320...", "Solution_29": "9th grade\r\n\r\nA.M.C. 12B : 132.5\r\nA.I.M.E. I : 10" } { "Tag": [ "function", "Euler", "algebra", "polynomial", "logarithms", "linear algebra", "calculus" ], "Problem": "The functions $y_{1}=x^{3}$ and $y_{2}=x^{4}$satisfies some linear homogeneous differential eq. of second order.\r\n\r\na)Show that the functions are linearly independent.\r\n\r\nb)Determine the equation.\r\n\r\nplease help me on this.", "Solution_1": "One way to determine whether two proposed solutions are linearly independent is to consider their quotient and if their quotient is a constant then they are linearly dependent while if it is not a constant then they are linearly independent.\r\n\r\nAnother way is to consider Wronskian's. If the Wronskian's is NEVER zero then they are linearly independent.\r\n\r\nOnce you have established linearly independence, then the general equation is given by $y = c_{1}y_{1}+c_{2}y_{2}$ ( for some constants $c_{1}$, $c_{2}$)\r\n\r\nI wasn't sure if your part b was asking to find the original differential equation which is satisfied by the given solutions.", "Solution_2": "[quote=\"akech\"]\n\nI wasn't sure if your part b was asking to find the original differential equation which is satisfied by the given solutions.[/quote]\r\n\r\n[color=red]Yes how do I find that one ? [/color](the original differential equation which is satisfied by the given solutions.)\r\n\r\nThanks a lot for your reply.", "Solution_3": "Note that the general differential equation in question looks like this:\r\n$y''+py'+qy = 0$\r\nNow the question reduces to finding $p$ and $q$\r\nPlugging in our two solutions we get two equations:\r\n$qx^{2}+3px+6 = 0$\r\n$qx^{2}+4px+12 = 0$\r\n\r\nBy elimination we get $px+6 = 0 \\Rightarrow p = \\frac{-6}{x}$ \r\n$q = \\frac{12}{x^{2}}$ by substitution?\r\n\r\nThe original equation reads:\r\n$y''-\\frac{6}{x}y'+\\frac{12}{x^{2}}y = 0$", "Solution_4": "also if we determine the wronskian $W$ of $y,x^{3},x^{4}$ and set $W=0$ we will have the eq.. Am I right?", "Solution_5": "That is right. \r\nThe method you are suggesting is somehow related to solving $AX = b$ in linear algebra (determinant of A and Wronskian).", "Solution_6": "In this case, the Wronskian is zero at $0$, but nowhere else. What does that mean? It means that our differential equation must be singular at zero, and we can't match arbitrary initial conditions there.\r\n\r\nIn general, equations with the Euler form $x^{n}y^{(n)}+a_{n-1}x^{n-1}y^{(n-1)}+\\cdots+a_{1}xy'+a_{0}y=0$ (each $a_{i}$ a constant) have solutions of the form $y=x^{r}$, where the $r$ are roots of an associated polynomial equation. If that equation has multiple roots, we also get solutions of the form $y=x^{r}\\ln^{k}x$.\r\nGiven that knowledge, we can reach the answer akech gave by simply matching coefficients." } { "Tag": [ "algebra", "polynomial", "search", "inequalities", "number theory", "number theory unsolved" ], "Problem": "Please help me with this one:\n\nLet $P\\left(x\\right)$ be a polynomial with complex coefficients, and of degree $n$. All its roots $z_1$, $z_2$, $z_3$, ..., $z_n$ are different and are such that all $i$ satisfy $\\left|z_i\\right|=1$.\n\nProve that there exists a complex number $z$ with $\\left|z\\right|=1$ such that $\\left|P(z)\\right| \\geq 2$.\n\nI couldn't solve it after a lot of thinking. :?", "Solution_1": "It's perfectly equivalent to this one:\r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=5774[/url]\r\n\r\nWe just represent the roots of $P$ on the unit circle. Did you notice that they were the same? Im asking because it seems strange that you're the author of both topics :).", "Solution_2": "Yeah ...I'm not so stupid :D There(in the algebra section), Myth said that it has been discussed before, but I couldnt find it in the forum, I wanted to put it in the number theory section, so i changed it a little bit, so that it looks more Number theoretical :D \r\n\r\nBy the way does any one know the solution, or knows where this problem has been discussed in the forum?\r\n\r\nPlease HELP :D", "Solution_3": "But didn't Myth (in the Algebra section) give a link to the topic where he solved the problem?", "Solution_4": "My internet connection is too poor. So I can't search forum for appropriate link. In particular, such searching is difficult due to \"crooked\" topic titles. But I suppose my post was last in desired topic, thus you can try to find all topic which end with my post.", "Solution_5": "If z is the n-th root of unity one easily shows\r\np(1)+p(z)+p(z^2)+..p(z^(n-1))=2n from here the desired inequality follows\r\n Note: Use that 1^i+z^i+(z^2)^i+..+(z^(n-1))^i=0 if n doesn't divide i, and n otherwise", "Solution_6": "Let's assume $P(x)=ax^2+bx+c$. In this case $P(1)+P(-1)=2a+2c$, which isn't necessarily equal to $2n=4$. I think that sum you wrote depends on the coefficients of the polynomial, iura. Am I misunderstanding something?", "Solution_7": "First it was not IMO selection test just an ordinary test.\r\nAlso there is a stronger problem from India 1998.\r\nYou can see it in book Problems and .... 1998-1999", "Solution_8": "is the polynomial monic? else we could multiply our polynomial by a constant and it would still satisfy the conditions, so in order to be always greater than 2 somewhere on the unit circle, there had to be a pole there, but polynomials don't like to have a pole...", "Solution_9": "[quote=\"Omid Hatami\"]First it was not IMO selection test just an ordinary test.\nAlso there is a stronger problem from India 1998.\nYou can see it in book Problems and .... 1998-1999[/quote]\r\nWhat is more stronger problem?", "Solution_10": "hi iura ,\r\ni don't understand your hint .. i can just prove that sup(|P(x)|) >= 1+1/n .. how do you achieve 2 with your hint ?", "Solution_11": "I have found my previous post on similar problem:\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=4532[/url]\r\nThough readability of this post is quite low :), I think you will be able extract its idea." } { "Tag": [ "calculus", "complex numbers", "imaginary numbers", "AMC" ], "Problem": "Let $ p(x) \\equal{} x^3 \\plus{} ax^2 \\plus{} bx \\plus{} c$, where $ a$, $ b$, and $ c$ are complex numbers. Suppose that\r\n\\[ p(2009 \\plus{} 9002\\pi i) \\equal{} p(2009) \\equal{} p(9002) \\equal{} 0\r\n\\]What is the number of nonreal zeros of $ x^{12} \\plus{} ax^8 \\plus{} bx^4 \\plus{} c$?\r\n\r\n$ \\textbf{(A)}\\ 4\\qquad \\textbf{(B)}\\ 6\\qquad \\textbf{(C)}\\ 8\\qquad \\textbf{(D)}\\ 10\\qquad \\textbf{(E)}\\ 12$", "Solution_1": "[hide=\"solution\"]Note that $ x^{12}\\plus{}ax^8\\plus{}bx^4\\plus{}c\\equal{}p(x^4)$. Therefore the roots are the fourth roots of $ p(x)$.\n\nThe square root of an imaginary number is an imaginary number, so we have that the fourth roots of $ 2009\\plus{}9002\\pi i$ are all imaginary.\n\nNote that the fourth roots of 2009 are $ \\sqrt[4]{2009}$, $ \\minus{}\\sqrt[4]{2009}$, $ i\\sqrt[4]{2009}$, and $ \\minus{}i\\sqrt[4]{2009}$. There are similar fourth roots of 9002. Therefore there are two imaginary roots for each of 2009 and 9002.\n\nAdding these all up we see that $ 4\\plus{}2\\plus{}2\\equal{}8$.[/hide]", "Solution_2": "To clarify:\r\n\r\nFrom the three zeroes, we have $ p(x) \\equal{} (x \\minus{} (2009 \\plus{} 9002\\pi i))(x \\minus{} 2009)(x \\minus{} 9002)$.\r\n\r\nThen $ p(x^4) \\equal{} (x^4 \\minus{} (2009 \\plus{} 9002\\pi i))(x^4 \\minus{} 2009)(x^4 \\minus{} 9002)$.\r\n\r\nLet's do each factor case by case:\r\n$ x^4 \\minus{} (2009 \\plus{} 9002\\pi i) \\equal{} 0$: Clearly, all the fourth roots are going to be complex.\r\n$ x^4 \\minus{} 2009 \\equal{} 0$: The real roots are $ \\pm \\sqrt [4]{2009}$, there are two complex roots.\r\n$ x^4 \\minus{} 9002 \\equal{} 0$: Same.\r\n\r\nSo the answer is $ 4 \\plus{} 2 \\plus{} 2 \\equal{} \\boxed 8$.", "Solution_3": "I feel like I must point out a major flaw in high school mathematics here. I had no clue how to do this problem 2 weeks ago before my precalc teacher taught my class how to take the complex roots of a number (purely for the intention of calculus nonetheless) but now it's trivial (other than realizing that $ x^{12} \\plus{} ax^8 \\plus{} bx^4 \\plus{} c \\equal{} p(x^4)$). Why is the purpose of high school math calculus rather than algebra or discrete mathematics?", "Solution_4": "Lol this is a major revival, but I would like to say that this question does not require any calculus at all, just realize that in polynomials, if $r$, $s$, and $t$ are roots to $x^3 + ax^2 + bx + c$, then it factors as $(x-r)(x-s)(x-t)$. Then plug in $x=x^4$ and find the nonreal roots..." } { "Tag": [ "geometry", "area of a triangle", "Heron\\u0027s formula", "algebra", "difference of squares", "special factorizations" ], "Problem": "find the area of a triangle whose sides have lengths a, b, and c if 4a^2*b^2-(a^2+b^2-c^2)^2=16. I got an answer, but im not sure if its right or if theres an easier way to solve it. Thanks for any help.", "Solution_1": "This was a fun problem, although rather easier than I thought it would be. \r\n\r\nThe original equation was 4a^2(b^2) - (a^2 + b^2 - c^2)^2 = 16. \r\nFor my hand's sake, I wrote a^2 + b^2 - c^2 = x. So now we have\r\n(2ab)^2 - x^2 = 16. Noting a difference of squares, we have \r\n(2ab - x)(2ab + x) = 16. Reintroducing x's equivalent, we have \r\n(2ab - a^2- b^2 + c^2)(2ab + a^2 + b^2 - c^2) = 16. Multiplying the first \r\nexpression by -1, we can then reduce down to \r\n((a-b)^2 - c^2)((a+b)^2 - c^2) = -16. Once again, we have a diff. of squares. Now we get ((a-b+c)(a-b-c))((a+b-c)(a+b+c))=-16. Multiplying the second term by -1, we get (a+c-b)(b+c-a)(a+b-c)(P) = 16, where P is the triangle's perimeter. We can clearly see an unrefined version of heron's formula here. Using that 2s=P and 2s-a=b+c (which we shall do for each side length), we can get 2s(2s-2b)(2s-2c)(2s-2a)=16. Taking a 2 from each expression, and then dividing by 16, we get that s(s-b)(s-c)(s-a)=1. Square rooting both sides yields the desired 1 due to the left side then beign heron's formula.", "Solution_2": "It is often useful to know the alternate form of Heron's Formula:\r\n\r\nA=1/4*sqrt((a+b+c)(a+b-c)(a-b+c)(-a+b+c)).", "Solution_3": "easier way\r\n\r\n2abcos C = a^2+b^2-c^2\r\n\r\nso the second term in the left hand side will be 4*a^2*b^2*(cos C)^2. Plugging in, you get 4*a^2*b^2*(sin C)^2=16. Solve that for absinC=2, K=1." } { "Tag": [ "function", "number theory", "least common multiple", "number theory theorems" ], "Problem": "What do you know about the function $ L(n) $ satisfying : \r\n Let $ n = p_1^{\\alpha_1}p_2^{\\alpha_2}...p_k^{\\alpha_k} $ \r\n$ L(n) = 1 $ for $ n=1 $\r\n$ L(n) = lcm( (p_1^{\\alpha_1-1}(p_1-1) , ..... , p_k^{\\alpha_k-1}(p_k-1) ) $ for $ n>1 $", "Solution_1": "So if a is prime to p,\r\n a^(L(n)) == 1 (mod n ) ? (and L(n) would be smaller than phi(n)) )", "Solution_2": "It looks like the Carmichael Function:\r\n[url=http://mathworld.wolfram.com/CarmichaelFunction.html]mathworld.wolfram.com/CarmichaelFunction.html[/url]" } { "Tag": [ "function", "geometry", "geometric transformation", "inequalities", "advanced fields", "advanced fields unsolved" ], "Problem": "Here is a piece of the proof in some paper I don't quite understand. \r\n\r\n[quote]Suppose we know that $ M$ is a subspace of the Hilbert space $ H$, and the mapping\n\\[ X: M \\rightarrow H \\oplus \\cdots \\oplus H\n\\]\ndefined by\n\\[ X(a) \\equal{} (X_1 (a) \\oplus \\cdots \\oplus X_1(a))\n\\]\nis an isometry. Since\n\\[ \\dim M \\equal{} \\dim (XM),\n\\]\nthe map $ X \\equal{} (X_1, \\ldots, X_n)$ can be extended to an isometry $ A \\equal{} (A_1, \\ldots, A_n),$ defined on $ H$ and with values in the direct sum $ H \\oplus \\cdots \\oplus H.$[/quote]\r\n\r\nI do not know why \r\n[hide]the map $ X$ can be extended to an isometry $ A,$ defined on $ H$.[/hide]\n\nThe professor wrote something like this:\n[hide]\\[ H \\equal{} M \\oplus (H \\ominus M)\n\\]\n\n\\[ H\\oplus \\cdots \\oplus H \\equal{} XM \\oplus (XM)^\\perp\n\\]\nSince\n\\[ \\dim (H \\ominus M) \\equal{} \\dim H \\minus{} \\dim M < n \\dim H \\minus{} \\dim M \\equal{} \\dim (XM)^\\perp,\n\\]\n\n\\[ \\dim (H \\ominus M) < \\dim (XM)^\\perp,\n\\]\nwe can construct an isometry from $ H\\ominus M$ to $ (XM)^\\perp.$\n[/hide]\r\nBut he does not explicitly define the extension. \r\n\r\nCan anyone provide the details for this argument showing how one can define the extension? \r\n\r\nAlso, in general, when can one say he can extend a linear operator from a subspace to the whole space?", "Solution_1": "In addition to this, I thought that the following might be helpful.\r\n\r\nIt is written in the paper that $ X\\equal{}(X_1, \\ldots, X_n)$ is an isometry on $ M \\subset H$ because we can show\r\n\\[ \\langle a,a \\rangle \\equal{} \\sum_{j\\equal{}1}^n \\langle X_j^T X_j a, a \\rangle, \\qquad \\text{for all } a \\in M\r\n\\]\r\n\r\n\r\nSo I'm assuming that the image of $ X$ is all the elements of the form\r\n\\[ \\sum_{j\\equal{}1}^n X_j (a), \\qquad a \\in M\r\n\\]\r\nand $ H \\oplus \\cdots \\oplus H$ has the induced inner product\r\n\\[ \\langle \\sum_{j\\equal{}1}^n a_j, \\sum_{j\\equal{}1}^n b_j \\rangle \\equal{} \\sum \\langle a_j, b_j \\rangle.\r\n\\]\r\nbecause then we can have\r\n\\[ \\langle \\sum_{j\\equal{}1}^n X_j(a), \\sum_{j\\equal{}1}^n X_j(a) \\rangle \\equal{} \r\n \\sum_{j\\equal{}1}^n \\langle X_j^T X_j a, a \\rangle.\r\n\\]\r\n\r\nIs this correct?", "Solution_2": "are $ H$ infinite-dimesional, $ M$ of dimension $ n$?\r\nit seems that these are reasonable hypotheses to make everything work.\r\nin this case (changing $ n$ if necessary) we can suppose and $ X_i \\equal{} \\phi_i f_i$, where $ \\phi_i$ is in $ H^*$ and $ f_i$ has norm 1.\r\nin this case, simply extend the dual basis $ e_1,\\dots,e_n$ relative to $ \\phi_1,\\dots,\\phi_n$ to a basis (in the hilbert sense) $ \\{e_\\alpha\\}_{\\alpha\\in A}$ of $ H$. hilbert spaces are classified up to isometry by cardinality of a basis, if we partition $ A$ in $ n$ sets of the same cardinality, with each of $ e_1,\\dots,e_n$ belonging to a different subset, we obtain an isometry $ H\\to H^{\\oplus n}$, which we can choose to extend the given one.\r\n\r\nanyway.. if $ M$ is not finite-dimensional, it doesn't seem to work: $ H$ is isometric to $ H_1\\oplus H_2$ (where $ H_1,H_2$ are copies of $ H$, just recalled for sake of clarity). now choose $ M \\equal{} H_1$ and an isometry $ H_1 \\to H \\equal{} H_1\\oplus H_2$. of course you can't extend it to an isometry of the full $ H$, since it's already bijective...\r\n\r\nhope this helps somehow..", "Solution_3": "Hi ma-go,\r\n\r\nThank you for replying. $ H$ should be finite-dimensional. [url=http://www.math.ucsd.edu/~helton/osiris/NONCOMMINEQ/sphereSubm.pdf]Here[/url] is the whole paper if you are interested. The part I'm mentioning is on page 5 to 7. What do you mean by $ H^*$??", "Solution_4": "$ H^*$ is the dual, the space of linear continuous functions $ H\\to R$.\r\n\r\nanyway, in the finite-dimensional case it's even easier: pick a (hilbert) basis $ B_i$ for $ H_i\\cap {\\rm Im}\\, M$, extend it to a (hilbert) basis $ E_i$ of $ H_i$ (the i-th factor in the sum), take the counterimages of $ B_i$'s, extend it to a (hilbert) base $ E$ of $ H$ (you can do it because the map is an isometry, so take bases to bases), and choose distinct elements from $ \\bigcup E_i\\setminus B_i$ to creat your map.\r\nyou have room enough, since $ \\dim H^{\\oplus n} \\minus{} \\dim M \\equal{} n\\dim H \\minus{} \\dim M \\ge \\dim H \\minus{} \\dim M$.", "Solution_5": "Hi ma_go,\r\n\r\nThanks again. I don't quite follow what you said. Could you explain more?\r\n\r\n1. It seems your are using the terms \"basis\" and \"base\" interchangeably. \r\n\r\n2. What do you mean by the counter-images of $ B_i$'s? I'm assuming you mean the pre-images.\r\n\r\n3. When you say \"choose distinct elements from $ \\bigcup E_i \\backslash B_i$ to create your map,\" do you mean \"choose the elements where $ E \\backslash \\bigcup \\operatorname{pre\\minus{}image}(B_i)$ can mapped to?\"\r\n\r\nSorry, I'm not quite used to this material so I'd appreciate if you can elaborate it for me.", "Solution_6": "I have more questions.\r\n\r\n1. Why do you pick a basis for $ H_i \\cap \\operatorname{Im} M$? Why don't you just pick a basis for $ \\operatorname{Im} M$ and take the pre-image of that? and then extend it to a basis of $ H$?\r\n\r\n2. When you say \"you can do it because the map is an isometry,\" which part exactly are you referring to?\r\n\r\n3. Are you assuming that the union of $ B_i$'s form a basis of $ \\operatorname{Im} M$? And then you take the pre-image of this union?\r\n\r\n4. Why is that the resulting map is also an isometry?", "Solution_7": "[quote=\"lenny\"]Hi ma_go,\n\nThanks again. I don't quite follow what you said. Could you explain more?\n\n1. It seems your are using the terms \"basis\" and \"base\" interchangeably. \n\n2. What do you mean by the counter-images of $ B_i$'s? I'm assuming you mean the pre-images.\n\n3. When you say \"choose distinct elements from $ \\bigcup E_i \\backslash B_i$ to create your map,\" do you mean \"choose the elements where $ E \\backslash \\bigcup \\operatorname{pre - image}(B_i)$ can mapped to?\"\n\nSorry, I'm not quite used to this material so I'd appreciate if you can elaborate it for me.[/quote]\n\n1. i thought \"bases\" stood for the plural of \"basis\" (i've been checking on wikipedia): please correct me if it's not like this (in italian \"basis\" is called \"base\", and i think \"base\" (en) could be a good translation for \"base\" (it), so i'm quite messed up :p\n\n2. pre-images, yes :) again, in italian you can use either \"preimmagine\" or \"controimmagine\" :)\n\n3. yes.\n[quote]\nI have more questions.\n\n1. Why do you pick a basis for $ H_i\\cap \\Im M$? Why don't you just pick a basis for $ \\Im M$ and take the pre-image of that? and then extend it to a basis of $ H$?\n\n2. When you say \"you can do it because the map is an isometry,\" which part exactly are you referring to?\n\n3. Are you assuming that the union of $ B_i$'s form a basis of $ \\Im M$? And then you take the pre-image of this union?\n\n4. Why is that the resulting map is also an isometry?[/quote]\r\n\r\n1. it's not the same, of course. it's wrong as i wrote it. you need to take a basis for $ \\Im M$: there are several \"transversal\" spaces that would made my proof totally wrong. your correction is ok.\r\n\r\n2. let's name it: call $ \\phi: M\\to H^{\\oplus n}$. since $ \\phi$ is an isometry, it takes (hilbert) basis to (hilbert) basis for $ \\Im \\phi$. so you can pull back a basis for $ \\Im \\phi$ to a basis of $ M$.\r\n\r\n3. yes. with \"new\" point 1. you don't need $ B_i$'s, but just $ B$.\r\n\r\n4. a map between hilbert spaces is an isometry if it takes a basis into an orthonormal set.\r\n\r\nby the way, this kind of problem could be easy generalized to a map $ M\\to H'$, with $ \\dim H'\\ge \\dim H$, even if $ H, H'$ are infinite-dimensional, and i think it works even for $ %Error. \"codim\" is a bad command.\n_H M \\le %Error. \"codim\" is a bad command.\n_{H'} M$, without any further hypothesis.", "Solution_8": "It seems like you can start by choosing some orthonormal basis for $ M$ as well instead of $ \\operatorname{Im} M.$ What do you think is the key in this proof? I'm still somehow not completely convinced by it.", "Solution_9": "the key is the relation on codimension: $ {\\rm codim}_H\\, M \\le {\\rm codim}_{H'}\\, M$, with the notation of my last post.", "Solution_10": "Yeah but also what role does isometry play in here? I think the key is that one can extend orthonormal sets and any subset of an orthonormal set is still orthonormal.\r\n\r\nAlso is it correct that you can start by choosing some orthonormal basis for M?\r\n\r\n[quote]and i think it works even for ....., without any further hypothesis.[/quote]\r\n\r\nHere I don't understand ur notation in ur second last post.", "Solution_11": "yes for both.\r\nsame setting, just instead of $ H^{\\oplus n}$ you put $ H'$, with the inequality on codimensions." } { "Tag": [], "Problem": "\u0388\u03c3\u03c4\u03c9 $ n\\in N$ \u03ba\u03b1\u03b9 $ a_{1},a_{2},...,a_{n}>0$ \u03bc\u03b5 $ \\sum^{n}_{i=1}a_{i}=1$. \u039d\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9 \r\n\r\n$ \\frac{a^{4}_{1}}{a^{2}_{1}+a^{2}_{2}}+\\frac{a^{4}_{2}}{a^{2}_{2}+a^{2}_{3}}+...+\\frac{a^{4}_{n}}{a^{2}_{n}+a^{2}_{1}}\\geq \\frac{1}{2n}$\r\n\r\n[hide]\u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03c0\u03bf\u03bb\u03cd \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03bc\u03b5 Andreescu[/hide]", "Solution_1": "\u0391\u03c5\u03c4\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03ac \u03b5\u03cd\u03ba\u03bf\u03bb\u03b7. :) \r\n\r\n\u038c\u03c0\u03c9\u03c2 \u03b5\u03af\u03c0\u03b5\u03c2:\r\n\r\nLHS>= $ [(a1)^2\\plus{}...\\plus{}(an)^2]^2/2[(a1)^2\\plus{}...\\plus{}(an)^2]>\\equal{}(a1\\plus{}...\\plus{}an)/(2n)\\equal{}1/(2n)$" } { "Tag": [ "search", "AMC", "AIME", "LaTeX", "inequalities", "function", "geometry" ], "Problem": "[quote=\"pbornsztein\"]This is a well-known result from Beatty (and the two involved sequences are so-called Beatty's sequences).\nNote that this kind of partition is no longer possible with three (or more) sequences.\n\nPierre.[/quote]\r\n\r\nI don't know if you realize it, but when someone asks for help on a problem, the answer \"this is a well-know result\" is the second-most unhelpful thing you could say, right after \"Oh, that's trivial!\"", "Solution_1": "[quote=\"JBL\"]I don't know if you realize it, but when someone asks for help on a problem, the answer \"this is a well-know result\" is the second-most unhelpful thing you could say, right after \"Oh, that's trivial!\"[/quote]Don't worry about it.. I already read the solution. I usually don't even attempt the problems in this book before I read the solutions.", "Solution_2": "That's probably not a good strategy for developing problem-solving skills. One thing you might do is go back to problems that you've looked at the solutions to several days or a week ago and see if you can solve them then -- in other words, check to make sure that the solutions you read are sticking.", "Solution_3": "[quote=\"JBL\"][quote=\"pbornsztein\"]This is a well-known result from Beatty (and the two involved sequences are so-called Beatty's sequences).\nNote that this kind of partition is no longer possible with three (or more) sequences.\n\nPierre.[/quote]\n\nI don't know if you realize it, but when someone asks for help on a problem, the answer \"this is a well-know result\" is the second-most unhelpful thing you could say, right after \"Oh, that's trivial!\"[/quote]I think Pierre was reffering to the fact that this problem has appeared at least 5 times on the forum before, the latest topic being in http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Beatty&t=22013", "Solution_4": "I don't really want to turn this into an argument, but to me it appears that he was making a reference to the literature without providing any way of reaching it or any idea of the proof. If he wished to indicate that this problem has appeared before on this site, it would be logical for him to make a reference to this problem on this site, and possibly even provide a link to one of the previous times it was discussed, but he did not. Thus, I do not think your conclusion is logical.", "Solution_5": "[quote=\"JBL\"]I don't really want to turn this into an argument, but to me it appears that he was making a reference to the literature without providing any way of reaching it or any idea of the proof. If he wished to indicate that this problem has appeared before on this site, it would be logical for him to make a reference to this problem on this site, and possibly even provide a link to one of the previous times it was discussed, but he did not. Thus, I do not think your conclusion is logical.[/quote]\r\n\r\nDon't worry, JBL. This is one of the most common problems I always encounter in this site. I have emphasized about this situation before but no one seemed to care about it. It's like Andrew Wiles saying that \"The proof of FLT is trivial, and QED; OK kiddos, you guys get it?\" If there is something to say, please post links/references/hints on how to do it. If not, please save a post. Thank you all for finally [b]comprehend[/b].", "Solution_6": "I think some of us must start using the search function. There are more than 5 topics on the same subject (Beatty sequences) here on the forum. \r\n\r\nThere were instances when people were asking for say shortlisted problems, and some members, moderators even, redirected them to other websites, instead of showing the [b]tons[/b] of problems presented here. \r\n\r\nAt this point, with 120,000 posts, I think that more than 75% of the problems being posted were already discussed on the forums, so a search for them appears natural. \r\n\r\nOk so Pierre didn't give a link, but he usually does, and he didn't do it this time probably because he was either in a hurry, or tired of the same and same problems showing up again and again :) Let's not jump to conclusions here.", "Solution_7": "In addition to what Valentin said, I'm sure that a fair number of us have the aforementioned USSR book, and as such know that it contains complete solutions.", "Solution_8": "Ooops... :blush: I didn't look at this thread for a while, and it seems that I've made something wrong (I do not say 'bad', but... :P ).\r\nI agree that it is more useful to give an appropriate link, but as Valentin said, this problem appeared many times on the forum, the latest before this one being less than one week ago :\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=beatty&t=22013\r\nfor which I gave a link after only writing 'Beatty' on Google, which seems to be the keyword here, and that's what I meant when I said that this problem is very well-known (so that any personal search about it should be very easy).\r\n\r\nPierre.", "Solution_9": "[quote=\"3X.lich\"]It's like Andrew Wiles saying that \"The proof of FLT is trivial, and QED; OK kiddos, you guys get it?\" If there is something to say, please post links/references/hints on how to do it. If not, please save a post. Thank you all for finally [b]comprehend[/b].[/quote]\r\n\r\n3X.lich is right. You may find it bothersome that the same problems appear multiple times: fine, but that's an immediate consequence of having a large and growing community, something I assume you wanted when you decided to join your forum. Not everyone will have seen everything on this website. Perhaps people should do a search before every single post, but to me that seems like a major imposition on people, especially newcomers to the site. And I find that a lot of \"old-timers\" simply don't do anything to make this community more friendly and inviting and are frequently counter-productive, especially in the intermediate and pre-olympiad forums, Pierre's post being just a single example. (Other popular methods include the belittling of the difficulties of problems and posting solutions below the level at which the poster really is, especially without hiding the solution.) I note that the link Pierre gives is to the advanced section of this site -- I would hardly expect someone with 26 posts to have thoroughly surfed through the large forum structure of that section.\r\n\r\nFinally, I'm not jumping to any conclusions whatsoever about Pierre -- I don't think he's a bad person, or even a bad member of this community. I do, however, think that this post is representative of a class of posts that makes this a (sometimes) unattractive community.", "Solution_10": "[quote=\"JBL\"]Finally, I'm not jumping to any conclusions whatsoever about Pierre -- I don't think he's a bad person, or even a bad member of this community. I do, however, think that this post is representative of a class of posts that makes this a (sometimes) unattractive community.[/quote]I personally do not think that there are more \"bad members\" than my fingers on my hands in this community. \r\n\r\nWhat we have to agree upon is the never-solved problem of difficulty. In my opinion this is an advanced subject, and should be treated there.\r\n\r\nFuthermore, all new students, especially those thinking about olympiad-like (that is more difficult than the average multiple-choice aime-type question) should search the forums because almost every problem they have has been previously solved before (especially if given in a ussr olympiad, or if it is a nice problem like this one). I must concur that when I join a very large board for the first time (happens every now and then) the first thing I usually do before posting anything is [b]search[/b] :) \r\n\r\nAs at this point we don't have the merge topics ability (will come in the future) it's hard to have tons of topics on the same subject, knowing that a simple search on the site (or google for that matter) will reaveal numerous solutions (also the book itself has a solution). \r\n\r\nWith all due respect the original author of the thread left the impression that he doesn't poses the solution, which in fact he did (either solved by himself or found in the book) so I woudln't call Pierre's answer as he didn't wanted to help him. \r\n\r\nTo make long things short, this topic was posted in the wrong place. :)", "Solution_11": ":mad: :mad: :mad: \r\nI am very angry.\r\nPeople on this site have different aims. Who granted rights to JBL to judge them?", "Solution_12": "Valentin, you didn't even respond to my points. But, if we're getting into this fundamentally unrelated issue, I shall defend myself: I didn't call anyone on this forum a bad member, nor did I even suggest that there are bad members -- only that there are bad posts, something that I think anyone can agree upon. (Myth apparently thinks there is at least one bad member, but that is beside the point.) I was very careful not to judge any individuals because I feel that the majority of the input of the majority of the people on this forum, especially those of you who were in mathlinks before the two forums joined, is wonderful. Nonetheless, I think that even some of the generally good contributors make the type of posts I complained about. Rather than addressing my complaint, you implied that I was attacking individuals and made the obvious statement that, if people did a search before they posted, it might reduce a little bit of clutter, without even mentioning my central complaint. You have yet to address, or even mention, it in 3 posts on this topic.", "Solution_13": "[quote=\"JBL\"]Valentin, you didn't even respond to my points. You have yet to address, or even mention, it in 3 posts on this topic.[/quote]I've addressed your points. In this case Pierre's posts as I said above, is not out of line, or improper, considering the background and difficulty level of the problem at hand. \r\n\r\nFor the \"making a reference to the literature without providing any way of reaching\" point, I've stated that Pierre usually (as other people do too) post links where such references can be obtained. Again he did not post any in this case because the simple mentioning of Beatty sequences give you [b]tons[/b] of ways to get the info (which was already avaialable in that book) as stated.\r\n\r\nFor the \"Other popular methods include the belittling of the difficulties of problems and posting solutions below the level at which the poster really is\" point, I think that if people whoose level is higher than the one of the problems posted in intermediate and pre-olympiad would stop posting then the people asking help would have no place to get it (as themselves could not solve the problems, and higher people would not be able to post). \r\n\r\nAs a final notice, this is in fact a forum of discussions, and discussions are welcomed in all forums. So what if Pierre didn't supply a solution? Is he by any means bound to supply one? I don't think so. Each member of this community contributes to it as he sees fit, and there are moderators (and admins :P) that must make sure that things are posted where they are supposed to, no vulgar language is used, and that's about it. \r\n\r\nIf he just wanted to give a hint, without actually saying the \"this is a hint: look for Beatty sequences\" and others would have posted a solution (which they did, btw) I do not see where the problem is. Again, no-one is forced to have his posts necesarily under the [hide ] tag or all of one's posts in a forum to represent solutions. \r\n\r\nPS Due to the fact that this discussion has really no connection with the actual problem (in a mathematical sense) I will split it.\r\n\r\nPPS [absolutely personal opinion - off the record] Besides the fact that the topic should have been posted in the Advanced Section >> Algebra >> Algebra Proposed Problems forum (which moderators should have moved it too), the author was not looking for a solution because he didn't solve it himself (as he proved by posting another solution) but rather for a discussion on the problem. In this case Pierre's post (which gives more information about the problem) is much helpfull than a solution.", "Solution_14": "I guess this topic should be sticky. :)", "Solution_15": "I think an important point is [b]STRUCTURE[/b]. And I guess JBL's argument is somehow misleading. It is obvious that you do not fully understand something new which is complex, i.e. theories, forum structures etc. Structure is also important though most people just will tend to stick to a small subset of the available forums. Some important points to make the best use of the search engine are as follows: \r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21455]\"please explicitely define all notations used instead of hoping that the reader will be able to guess their meaning\" (Darij)[/url]\r\n\r\n- proper naming of thread subjects/titles: not just \"easy and nice\"\r\n\r\n- [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15585]if from a certain source several problems are posted collect them in a thread, for example the IMO ShortList 1998 here[/url]\r\n\r\n- [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]please learn $\\LaTeX$ and post your problem in this style, this makes reading the problems much easier and people will be easier attracted by your problem[/url]\r\n\r\n- classification of problems as suggested by [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=285]Harazi[/url], for instance: \r\n\r\nDiscrete Mathematics, Graph Theory, Labeled Graphs, Labeled Tree", "Solution_16": "So we disagree about whether or not this single post of Pierre's was appropriate: fine. \r\n\r\n[quote]I think that if people whoose level is higher than the one of the problems posted in intermediate and pre-olympiad would stop posting then the people asking help would have no place to get it (as themselves could not solve the problems, and higher people would not be able to post). [/quote]\r\n\r\nSomehow, in the many months I spent on this forum prior to the union between AoPS and Mathlinks, I managed to make many posts in the Getting Started and Intermediate levels without either solving all the questions there or demeaning anyone else's questions as being \"too easy\" to appear on the forum. Both activities now seem quite popular. I can give you a large number of links to threads where such activities occur, if you wish. As someone who is most comfortable in the pre-olympiad forum, I note an extraordinary number of solutions that are written by people who probably should be solving in the advanced forums, frequently unhidden. This makes the forum, for me, a much less attractive place to turn to look for problems. \r\n\r\nI don't think Pierre should have supplied a solution, if for no other reason than he had seen the question solved within the past week. There were a number of things he could have done that I would have found more reasonable than what he did do, each of which would have been more helpful to someone (such as myself) who has not seen the problem before.", "Solution_17": "[quote=\"JBL\"]So we disagree about whether or not this single post of Pierre's was appropriate: fine. \n\n[quote]I think that if people whoose level is higher than the one of the problems posted in intermediate and pre-olympiad would stop posting then the people asking help would have no place to get it (as themselves could not solve the problems, and higher people would not be able to post). [/quote]\n\nSomehow, in the many months I spent on this forum prior to the union between AoPS and Mathlinks, I managed to make many posts in the Getting Started and Intermediate levels without either solving all the questions there or demeaning anyone else's questions as being \"too easy\" to appear on the forum. Both activities now seem quite popular. I can give you a large number of links to threads where such activities occur, if you wish. As someone who is most comfortable in the pre-olympiad forum, I note an extraordinary number of solutions that are written by people who probably should be solving in the advanced forums, frequently unhidden. This makes the forum, for me, a much less attractive place to turn to look for problems. \n\nI don't think Pierre should have supplied a solution, if for no other reason than he had seen the question solved within the past week. There were a number of things he could have done that I would have found more reasonable than what he did do, each of which would have been more helpful to someone (such as myself) who has not seen the problem before.[/quote]I think that Pierre's post could not have possibly shinder your chances of solving a problem. He mentioned that it is well-known (which it is!) and it's called Betty sequence because the [b]thread's name[/b] was \"I don't know how to name this thread\" (or something of the sort). In my opinion it's the best reply anyone could have given (aside from the fact that the a link was omitted). \r\n\r\nI am curious what exactly would have you expected him to post? Exactly, what do you think that was appropriate to post (that is post the \"post\" as it should have been posted :) - too many damn \"post\" words here). \r\n\r\nOn the other hand, I also do not agree with people saying that users' questions are \"too easy\", but it's exactly as when people say they are too hard (which also happen). These should be solely regarded as a person's personal opinion, and it is true that each of us defines his own \"simple\" problem and \"difficult\" problems, so one's mentioning them is really not that of a \"offence\" I think. It rather might demonstrate someone's *modesty* at most. \r\n\r\nI also think that you might be deliberately giving the Advanced Sections the etiquette \"too hard\" without even taking a closer look. Many of the Advanced Section problems seem to be that need to be placed in the intermediate or pre-olympiad and some of the intermediate or pre-olympiad need to be placed in the advanced sections. \r\n\r\nI have to go to have some dinner right now, but I'll be back to continue this conversation :)", "Solution_18": "I'm not sure we're communicating fully, probably my fault: I am expressing 2 different things. 1 of them is about Pierre's post, which I did not think incredibly greivous and which I have not been highly concerned with in my previous 2 posts. The other is a whole class of posts, and we could argue about whether or not Pierre's post belongs in that class and for what reasons. It seems to me that you address my more general comments as if they were all about Pierre's post -- this is why I said that I thought you were not addressing my major complaints. I do not think Pierre's post would have hindered anything -- I did think it was unhelpful, something you disagree about. I claimed it was a member of a class of \"counter-productive\" posts on this forum, particularly in the lower-level sections. As far as I can tell, your response was that people should do a search before they post something, which neglects to address my point entirely. I can imagine a number of things that Pierre could have said that would have been more helpful than what he did: among them, a link to one of the previous discussions of this problem on this site or to any other website discussing them or a suggestion for solving the problem (presumably coupled with the comment that this problem is difficult and has been tackled on the advanced fora).", "Solution_19": "[quote=\"JBL\"] As far as I can tell, your response was that people should do a search before they post something, which neglects to address my point entirely. I can imagine a number of things that Pierre could have said that would have been more helpful than what he did: among them, a link to one of the previous discussions of this problem on this site or to any other website discussing them or a suggestion for solving the problem (presumably coupled with the comment that this problem is difficult and has been tackled on the advanced fora).[/quote]\r\n\r\nJBL, you must understand that people like Pierre are here for their joy. They are NOT payed teachers. I think it is already very kind to say what he said and posting a link. Rewriting all the hints again each time is a very tiring job. I think it is not their job to do that just because people are too lazy to use the search engine. They are no service and convenience staff employed on ML/AoPS. If all the details are covered in other threads why should one do that ? Just that a certain user does not need to follow the links ? Personally I appreciate the efforts of Pierre and others. If users think hints are not sufficient they still may ask for more.", "Solution_20": "Yes that's exactly what I wanted to say orlando. JBL you still haven't made me the pleasure of seeing:\r\n\r\n1) what [b]exactly[/b] you would have wanted Pierre to post (literally I mean - put it in a quote)\r\n2) the links to the topics in which people are *offensed* by other people by calling their problems *not worth to be on the forum*. \r\n\r\nAs for the \"other whole class of posts\", I must repeat that this is an open forum and people are invited to discuss mathematics, especially problem solving, but that doesn't mean restricting one's expression about certain problems. \r\n\r\nOf course it would have absolutely no sense for me to do in the getting started and post in each of the problems there \"this is very easy for me\", and for any of us for that matter.\r\n\r\nWhat the issue is in fact is what some people consider difficult and what some people easy. Clearly there is no such thing as a general easy problem or a hard problem. It might be hard for you, easy for me, or vice-versa. Now if the problem that caused this topic would have been posted where it stands now in the first place, then this discussion would have not taken place. \r\n\r\nSimilarly there are many problems posted in the advanced topics which are too easy to be there (or so think some people): for example\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=146807&highlight=#p146807\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=22839\r\nand many more. This is more a matter of structure. \r\n\r\nWe are discussing, and we will at some point in the future implement post ratings, which will help in separating general hard from general easy problems. \r\n\r\nUntill then I wonder why you have posted in exactly 8 topics in the advanced sections (all of which ended up there after being moved from different parts of the site). Do you find it [b]that[/b] hard? Or is it more of a personal reason?", "Solution_21": "I unstickied this - while an interesting discussion among members of this community I highly respect, it doesn't need to be stuck on top of the forum as it isn't really a discussion about Other Problem Solving topics (or really anything besides 'How I Would Like People to Use the Forums'). \r\n\r\nI would like people to try to post in a constructive, clear, and educational way, but you can't force people to do what you want them to - people will use the board in different ways. Perhaps when we institute post rankings this will somewhat solve itself - maybe we can even implement a filter a la slashdot so you don't have to be bothered with low-rated posts. We'll see what we can do when we have time.", "Solution_22": "I don't see why there's all this ruckus going in this topic. If it is your style that \"Hey that's trivial!!!\" or \"Hey that's super-easy!!!\" or \"Hey that's well-known!!!\", maybe you should move on because these phrases [b]do not[/b] help at all to the readers. For example (and this is true), I saw a couple of posts about \"Popovicius Inequality\" and I do see many other posts where they say \"that's well known, just use Popovicius\". I remember the first time I saw that I was like, \"what is going on? Am I stupid to miss what Popovicius is?\" I think fairness has been worn out when such phrases like \"well-known\" or \"triviality\" comes up. Try writing \"Popovicius is well-known\" in an IMO paper, I'm sure you will get full marks (you will?).\r\nIf anyone disagrees to what I say, OK fine, I apologize to have stirred up your inner fire, and perhaps you can ignore me in the future.", "Solution_23": "[quote] Try writing \"Popovicius is well-known\" in an IMO paper, I'm sure you will get full marks (you will?). [/quote]\r\n\r\nIf you were to write something like - 'this inequality follows immediately from Popoviciu's inequality applied to the convex function $f(x)=x^2$, then yes, you would receive full marks.\r\n\r\nAnd I think that there's a few things that should be noted. First of all, if a problem comes up that is a named theorem, e.g, Nesbitt's Inequality, I think that it's helpful to say that. That means that you can use 1) the forum search, and 2) google, to look for a solution to the problem.\r\n\r\nSecond of all, if someone says, 'this inequality is an immediate consequence of $X$ theorem', I also think that that's helpful. I mean, again on inequalities, you can prove a lot of them using AM-GM. However, once you learn something like Muirhead, it TRIVIALIZES (oops) a lot of the traditional AM-GM inequalities. And there are similar results in the combinatorics, geometry - heck, all of the sections. So this should also be noted; people shouldn't be expected to manufacture an artificial solution using only the knowledge known to the proposer when there is a simple, clean solution using a more advanced theorem.", "Solution_24": "[quote=\"3X.lich\"]If it is your style that \"Hey that's trivial!!!\" or \"Hey that's super-easy!!!\" or \"Hey that's well-known!!!\", maybe you should move on because these phrases [b]do not[/b] help at all to the readers.[/quote]As me and Richard pointed out, helping other readers is [b]not[/b] mandatory. It is for Richard and me up to a level, and to the moderator of the respective forum, again up to a certain level. But the rest of the users are [b]not obligated[/b] to make each of their posts as helpfull as they can get. We cannot ban users from certain forum for these reasons (especially when the affirmations they make can also be interpreted by some people as a good helping post!). We introduced age ranges to create a certain feeling that \"hey, maybe you should post here if this is a forum where stuff that is usually discussed by 13 years olds\" but some of the visitators of those forums [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16644]weren't too happy[/url] about this (although it was somewhat in their winning, at least from JBL's point of view). \r\n\r\nAnd as I said, one cannot expect every post in a topic to be a \"here's how to solve this problem\" post. This is a forum and if someone expresses his opinion, and his opinion is \"this is too easy\" then you just skip to the next post. There are people on the site for which almost all problems are \"easy\", and some of them state that. If you disagree, just post your opinion .. \"it's hard\" :) \r\n\r\nWe understand that this must be for some users an annoying thing, but look on the bright side: if you further increase your problem solving skills, then you will like having also advanced people out there to help you, and the problems will become easy to you also (the ones that you at first said that are not easy). \r\n\r\nWe have plans to upgrade the forum in the future, which will further help you with these issues (introducing posting ranking, ability of ignoring users (so if you ignore a user, you will not see his posts, reveal only the first post some topics, etc.) while some of them are effective now (we have two accessing urls, collapsing forums to hide the unwanted stuff, favorite topics, local communities, descriptions systems) but you must try to fit everyone's shoes in this story :)", "Solution_25": "[quote=\"Valentin Vornicu\"][quote=\"3X.lich\"]If it is your style that \"Hey that's trivial!!!\" or \"Hey that's super-easy!!!\" or \"Hey that's well-known!!!\", maybe you should move on because these phrases [b]do not[/b] help at all to the readers.[/quote]As me and Richard pointed out, helping other readers is [b]not[/b] mandatory. It is for Richard and me up to a level, and to the moderator of the respective forum, again up to a certain level. But the rest of the users are [b]not obligated[/b] to make each of their posts as helpfull as they can get. We cannot ban users from certain forum for these reasons (especially when the affirmations they make can also be interpreted by some people as a good helping post!). We introduced age ranges to create a certain feeling that \"hey, maybe you should post here if this is a forum where stuff that is usually discussed by 13 years olds\" but some of the visitators of those forums [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16644]weren't too happy[/url] about this (although it was somewhat in their winning, at least from JBL's point of view). \n\nAnd as I said, one cannot expect every post in a topic to be a \"here's how to solve this problem\" post. This is a forum and if someone expresses his opinion, and his opinion is \"this is too easy\" then you just skip to the next post. There are people on the site for which almost all problems are \"easy\", and some of them state that. If you disagree, just post your opinion .. \"it's hard\" :) \n\nWe understand that this must be for some users an annoying thing, but look on the bright side: if you further increase your problem solving skills, then you will like having also advanced people out there to help you, and the problems will become easy to you also (the ones that you at first said that are not easy). \n\nWe have plans to upgrade the forum in the future, which will further help you with these issues (introducing posting ranking, ability of ignoring users (so if you ignore a user, you will not see his posts, reveal only the first post some topics, etc.) while some of them are effective now (we have two accessing urls, collapsing forums to hide the unwanted stuff, favorite topics, local communities, descriptions systems) but you must try to fit everyone's shoes in this story :)[/quote]\r\n\r\nOK, thx for clarifying me that no one is obliged to help others. I see your point, understood, and I apologize again to everyone who felt offended by my thoughts. My sole purpose was to help people in this forum but unfortunately the forum is not compromised to do that. I apologize to everyone for stirring up all these troubles reading through my posts and I will posts whenever I can help. If someone requests me to provide an answer, I am not obliged to provide any sort of information. It is up to me. Thank you for giving me the freedom of speech which I enjoy the most so I can express my opinions the way I want over the internet as long as it does not concern anything else besides it is related to the subject Mathematics. Thank you and I will not post in this topic anymore. Bye.", "Solution_26": "[quote=\"Valentin Vornicu\"]I think some of us must start using the search function...[/quote]\n[quote=\"Valentin Vornicu\"]But the rest of the users are not obligated to make each of their posts as helpfull as they can get. [/quote]\r\n\r\nI said, right at the beginning, that I didn't want to turn this into an argument, but it's happened anyway. Take your pick, Valentin: I just took 2 things you've said in this discussion and put them at the top of this post. In the first, you argue, more or less, that the blame for certain types of comments should be put upon those who post topics without first searching through the website. In the second, you say that you can't possibly force people to post in a certain way. That's logically inconsistent, and worse than that, it's a completely arbitrary distinction of demanding certain things of certain people and not of others for some unknowable personal reason. \r\n\r\n\r\n\r\nAside from that, which I think is just outrageous, are other smaller flaws in your argument that any mathematician should be apalled about. For example, you sent me a link to a post about people comlaining about the age guidelines added to the forum. (2 people in fact -- the same number complaining here. Apparently, you have again decided that certain people's complaints have more merit than others.) However, their complaints line up far more with mine than against them -- they are complaining on behalf of people who are older, but less skilled -- not those likely to be swooping down into lower-level forums, but those who the guidelines might push, unfairly, into higher-level forums. So that little sentence, which you seem to have thought was very clever, is completely bunk.\r\n\r\n\r\nBlahblahblah -- if I posted a question in the Getting Started forum, and you proved it with Muirhead and acted as if that was it, don't you think I would have a right to be annoyed? The post, by nature of where it lies, was obviously designed for solvers of a certain level who wouldn't have heard of Muirhead. Your using it isn't helpful at all to the hypothetical me, a middle schooler from Indiana. (Although, if you commented that it could be solved by a relatively advanced inequality named Muirhead and pointed me to some posts on the topic, I could perhaps learn something from it.)\r\n\r\n\r\n\r\nValentin, (in reference to your penultimate post), I don't see why you need me to give you specific example of bad things in order to admit that bad things exist -- you should recognize, again as a mathematician, that my refusal to produce an example does not prove that none exist. Instead, I feel that I am causing enough trouble between individual members of this forum and myself through my posts without needing to add the extra strain of labeling certain individuals posts negatively, since I already seem to be doing a number with you and Pierre. [color=red][censured by Admin - please do not mock other individuals][/color]", "Solution_27": "Taking two statements of mine, and putting them out of their context, and thus conlcuding that they are illogical is not a correct assumption. Please read carefully the context in which they were made. \r\n\r\nI am not discussing your ideas of my arguments being \"just outrageous, smaller flaws that any mathematician should be apalled about\", etc. and not the ones that I deleted in which you mock other people. \r\n\r\nAnyway, even if supposedly my arguments would be as you described them above, the problem which is discussed here is not a mathematical problem, and one should learn that mathematical arguments are not always the best and the most correct ones. \r\n\r\nAs I've said before, and you have reiterated it here, it seems that you have a personal problem with some members, including me probably, so henceforth this discussion will be taken on an e-mail or private message, to avoid further personal conflicts." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "Let $ \\langle F_{n}\\rangle$ be a sequence of nonempty closed sets of real numbers with $ F_{n\\plus{}1}\\subset F_{n}$. Show that if one of the sets $ F_{n}$ is bounded, then $ \\cap^{\\infty}_{i\\equal{}1}F_{i}\\neq\\emptyset$. Give an example to show that this conclusion may be false if we do not require one of the sets to be bouneded.", "Solution_1": "Try to come up with an unbounded closed subset of $ (\\minus{}\\infty,\\infty)$ that involves $ n$. :maybe:", "Solution_2": "There are tons of ways to prove this depending on what you are allowed to use. One way would be to use the Theorem: $ X$ is compact iff for every collection $ \\mathbb{C}$ of closed sets in $ X$ having the finite intersection property, the intersection $ \\cap_{C\\in\\mathbb{C}}C$ all of the elements of $ \\mathbb{C}$ is nonempty.\r\n\r\nSince these are sets of real numbers: compact iff closed and bounded. You have one of the sets $ F_{i}$ is bounded and all the sets are sitting inside of it after $ i$ (by nested property), so since $ \\cap_{n \\equal{} 1}^{\\infty}F_{n}\\subset\\cap_{n \\equal{} i}^{\\infty}F_{n}$ you can just look at your compact space $ X$ being $ F_{i}$ and apply the theorem in the forward direction. \r\n\r\nThere are ways to do it without theorems. You look at complements and use DeMorgan and things." } { "Tag": [ "abstract algebra", "vector", "superior algebra", "superior algebra unsolved" ], "Problem": "I need examples of groups , rings , and in fact for bands (boolean semigroups )?\r\n\r\nI know some standard examples :\r\n\r\n1) $ (\\mathsf(P(X),\\Delta,\\cap)$.\r\n\r\n2) $ \\mathbb Z_2 \\times \\mathbb Z_2 \\cdots \\times \\mathbb Z_2$\r\n\r\n3)$ \\mathbb Z_2 \\times \\mathbb Z_2 \\cdots\\cdots$\r\n\r\nPlease put down any example you know , may be it will helpful :)", "Solution_1": "I don't know what you mean by a \"Boolean group.\" Do you mean a group in which every element has order $ 2$? Any such group must be abelian, hence a vector space over $ \\mathbb{F}_2$. Vector spaces over a field are uniquely determined by their dimension.\r\n\r\nFor Boolean rings construct the corresponding Boolean algebra and use [url=http://en.wikipedia.org/wiki/Stone%27s_representation_theorem_for_Boolean_algebras]Stone's representation theorem[/url].", "Solution_2": "Boolean group means the square of any element gives itself.\r\n\r\nIn fact I tried to find a band (boolean semigroup ) of odd order ($ \\geq 3$)\r\n\r\nThx :)", "Solution_3": "The only such group is the trivial group. \r\n\r\nCould you define more thoroughly what a Boolean semigroup is? I can't find a good definition online.", "Solution_4": "[quote]\nCould you define more thoroughly what a Boolean semigroup is? I can't find a good definition online.[/quote]\r\n\r\nThe terminology \"Boolean\" means in algebra (as I know) if you take the element with itself in the operation it will gives itself .In other words, Boolean means every element is idempotent. \r\n\r\nThus, Boolean semigroup $ (S,\\bullet)$ , it means if $ s\\in S$ , then $ s^2 \\equal{} s$.\r\n\r\n :)", "Solution_5": "The semigroup $ \\{ e, a, b \\}$ with identity $ e$ and relations $ a^2 \\equal{} ab \\equal{} a, b^2 \\equal{} ba \\equal{} b$ is \"Boolean\" (I am not convinced this terminology is universal) of order $ 3$. Every word evaluates to its leftmost element.", "Solution_6": "Ok. that's nice one.\r\n\r\nbut , what about commutative Boolean semigroup of odd order? \r\n\r\n(I believe it should be universal terminology)", "Solution_7": "Take $ \\{ a, b, c \\}$ with multiplication as follows: any word evaluates to the letter in it which is lexicographically first. Hence $ a^2 \\equal{} a, b^2 \\equal{} b, c^2 \\equal{} c, ab \\equal{} ba \\equal{} ac \\equal{} ca \\equal{} a, bc \\equal{} cb \\equal{} b$. This works for any totally ordered set.", "Solution_8": "That's really very nice :) \r\n\r\n\r\nNow, could we find infinite Boolean semigroup ?infinite Commutative Boolean semigroup?\r\n\r\nThank you so much :)", "Solution_9": "The above works for [b]any[/b] totally ordered set. For example, take the integers with binary operation $ \\text{min}(x, y)$. The power set of any infinite set with either union or intersection is also still a commutative Boolean semigroup.\r\n\r\nCan you please just give the entire question at once?", "Solution_10": "Sorry, I just like to collect examples of several properties .\r\n\r\nLast question :\r\n\r\nI need two commutative Boolean semigroups , the first one is finite (odd order) and the other one is infinite . Such that both of them have a zero divisors (like the definition we know in the ring theory for zero divisors) and they have at least two elements are not a zero divisors .\r\n\r\nThank you in advanced :)", "Solution_11": "This is starting to sound like homework. At this point I've given you enough ideas that you should be able to do this yourself.", "Solution_12": "[quote=\"t0rajir0u\"]This is starting to sound like homework. At this point I've given you enough ideas that you should be able to do this yourself.[/quote]\r\n\r\n[b]This is not homework[/b] :!: .In fact,I'm trying to study the properties of zero graph of commutative bands , and I could not find a finite (odd order ) and infinite bands with some elements are zero divisors and some element are not zero divisors (at least two elements ) \r\n\r\nI can see we can add some elements in your finite commutative example and trying to make an example as I asked for , but I need to collect more of commutative bands :) \r\n\r\nThank you :wink:" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Given a,b,c > 0.Prove", "Solution_1": "We get :\r\n\r\n$ \\sum_{cyc} \\frac{1}{1\\plus{}\\frac{b}{a}\\plus{}(\\frac{b}{a})^2} \\ge 1$\r\n\r\nWe put $ \\frac{b}{a}\\equal{}x$ and so on.so we have $ xyz\\equal{}1$\r\n\r\nNow we have two solutions;we can either expand it and then we are done!\r\nOr we can put $ x\\equal{}\\frac{st}{r^2}$ and so on to get :\r\n\r\n$ \\sum \\frac{r^4}{r^4\\plus{}r^2st\\plus{}s^2t^2} \\ge 1$\r\n\r\nNow we can use Cauchy :\r\n\r\n$ \\sum \\frac{r^4}{r^4\\plus{}r^2st\\plus{}s^2t^2} \\ge \\frac{(\\sum r^2)^2}{\\sum r^4\\plus{} \\sum r^2st \\plus{} \\sum s^2t^2}$\r\n\r\nSo we have only to prove that $ \\sum s^2t^2 \\ge \\sum r^2st$\r\nWhich can be done by using AM-GM.", "Solution_2": "please learn to use TeX to write down your problem here...\r\n\r\n$ \\sum{\\frac{a^2}{a^2\\plus{}ab\\plus{}b^2}}\\ge1$\r\n\r\n$ \\iff a^4b^2\\plus{}b^4c^2\\plus{}c^4a^2 \\ge a^3bc^2\\plus{}a^2b^3c\\plus{}ab^2c^3,$\r\n\r\nby AM-GM, we have\r\n\r\n$ a^4b^2\\plus{}b^4c^2 \\ge 2a^2b^3c,$\r\n\r\n$ b^4c^2\\plus{}c^4a^2 \\ge 2ab^2c^3,$\r\n\r\n$ c^4a^2\\plus{}a^4b^2 \\ge 2a^3bc^2,$\r\n\r\nadding them together, done.", "Solution_3": "the solution of shoki is similar to me.I think it is an interesting way", "Solution_4": "Can you find another way?", "Solution_5": "[quote=\"duc 95\"]Can you find another way?[/quote]\r\nit is very old and well known inequality.\r\nAnother sulution:\r\nlet:\r\n$ {a}^{2} \\plus{} ab \\plus{} {b}^{2} \\equal{} x$\r\n$ {b}^{2} \\plus{} bc \\plus{} {c}^{2} \\equal{} y$\r\n$ {c}^{2} \\plus{} ca \\plus{} {a}^{2} \\equal{} z$\r\nwe have:\r\n$ \\left(\\frac {1}{x} \\plus{} \\frac {1}{y} \\plus{} \\frac {1} {z} \\right)\\left(\\frac {{a}^{2}}{x} \\plus{} \\frac {{b}^{2}}{y} \\plus{} \\frac {{c}^{2}} {z} \\minus{} 1 \\right) \\equal{} \\sum \\frac {{a}^{2}}{{x}^{2}} \\plus{} \\sum \\frac {{b}^{2} \\plus{} {c}^{2}}{yz} \\minus{} \\sum \\frac {1}{x}$\r\n$ \\equal{} \\sum \\left(\\frac {{a}^{2}}{{x}^{2}} \\minus{} \\frac {bc}{yz} \\right) \\plus{} \\sum \\frac {{b}^{2} \\plus{} bc \\plus{} {c}^{2}}{yz} \\minus{} \\sum \\frac {1}{x} \\equal{} \\sum \\left(\\frac {{a}^{2}}{{x}^{2} }\\minus{} \\frac {bc}{yz} \\right)$\r\n$ \\equal{} \\frac {1}{2}\\sum {\\left(\\frac {b}{y} \\minus{} \\frac {c}{z} \\right)}^{2}\\geq 0$\r\nSo $ \\sum{\\frac {a^2}{a^2 \\plus{} ab \\plus{} b^2}}\\ge1$\r\nEquality occurs for $ a \\equal{} b \\equal{} c$ :)" } { "Tag": [], "Problem": "You just post ways of making pictures using only your keyboard. You can put a word describing it.\r\n\r\n(-(-(-(-(-(-(-(-(-(- -)-)-)-)-)-)-)-)-)-) ninjas\r\n\r\n(\\__/) \r\n(+'.'+) bunny \r\n(\")_(\")", "Solution_1": "The Joker\r\n (.)(.)[color=red]\n \\____/[/color]", "Solution_2": 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\"]MWAHAHAHAHA You're computer has been invaded by an army of 1337 fishes MWAHAHAHAHAHA[/hide]", "Solution_3": "to ytrewq: what is that??? it looks sorta like a snakeskin if u scroll fast... is that it?", "Solution_4": "to ytrewq: what is that??? it looks sorta like a snakeskin if u scroll fast... is that it?", "Solution_5": "I think it's a lot of fish.", "Solution_6": "[code] ____\n /o \\ /\n< <\n \\____/ \\\n\n[/code]\r\nFEESH!", "Solution_7": "@o.o@\r\n\r\n<(^.^<)\r\n\r\n<(^.^)>\r\n\r\n(:::::[ ]:::::)", "Solution_8": "***** * ***** * * ***** * * * ***** * *\r\n * * * * * * * ** ** * * * * *\r\n * * * * * * *** * * * ***** * *****\r\n * * * * * * * * * * * * * *\r\n***** **** ***** * ***** * * * * * * *\r\n[code]\n***** * ***** * * ***** * * * ***** * *\n * * * * * * * ** ** * * * * *\n * * * * * * *** * * * ***** * *****\n * * * * * * * * * * * * * *\n***** **** ***** * ***** * * * * * * *\n[/code]", "Solution_9": "Whoa nice one 1=2!:coolspeak:\r\n\r\n(=^_^=) Kitty\r\n\r\n m m\r\n | |(o o)| |\r\n ||(~)|| Monkey\r\n\r\n~~~{~~@ Rose\r\n\r\nd-_-b Headphones\r\n\r\n'`--'`<'|) ,-, \n ,. |/ \\| ,-./ / \n _ | \\,-. ( ) | `-'`--. \n ( `' (_/|__ \\ (o / ,-' ,-' \n ; ) ,|`. - ,'|. `-. ) \\ \n | ( ,-' _/ `-.`\"\"',-' \\---. / ; \n | | ,-' \\ /\\ / \\ | |--/ | \n | |_,| / \\/ \\/ \\/\\ | | \n | ` \\ | \\ / ,' \n | \\ | | / _,' \n : \\ , `/------' \n `-.___,---') `. \n ,' \\ \n / \\ \n : : \n | _,| \n \\--.___ __,--' ; \n `. `\"\"\"\"\"\"\"' ,' \n | | \n | .____, | \n | | | \n | | | \n | | | \n | | | \n | | | \n -hrr- | | | \n | | | \n |-._____,-|-.____,-| \n |_ |_ | \n ,' `------'| `-----' \\ \n / _|_ \\ \n `--._____,-' `-.___,-' \n\n[/code]\r\n\r\nOK, I didn't make it... but it sure looks good :wink: Ascii art rules :lol: http://chris.com/ASCII/", "Solution_11": "<3-heart lol i noob\r\n\r\n:(|)-monkey!", "Solution_12": "[quote=\"winner2009\"]to ytrewq: what is that??? it looks sorta like a snakeskin if u scroll fast... is that it?[/quote]\n\n[quote=\"ytrewq\"]MWAHAHAHAHA You're computer has been invaded by an army of 1337 fishes MWAHAHAHAHAHA[/quote]\r\n\r\nThat's what it is.", "Solution_13": "_______\r\n:b:p:b:p:b:p:b:p:b:p:b :b:p:b:p:b:p:b:p:b:p:b\r\n:b:p:b:p:b:p:b:p:b:p:b \"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"''':b:p:b:p:b:p:b:p:b:p:b", "Solution_14": "What is it?", "Solution_15": "O-|--( STICK MAN!!!! (look from side and turn head to the left)", "Solution_16": "O-|--< Stick man\r\n\r\n___\r\n|||||-)\r\n|||||-) Four ninjas looking\r\n|||||-) around a corner. :ninja: \r\n|||||-)\r\n=-=", "Solution_17": "<'\\__ ~\r\n--//\\\\--\r\n--------\r\nhorse", "Solution_18": "---/\\---\r\n---\\/---\r\n---[]--- \r\n---[]--- \r\n-/\\[]/\\-\r\n-\\.[]. /-\r\n-/.....\\-\r\n-\\___/-\r\nIt's a GUITAR! xD" } { "Tag": [], "Problem": "\u03a4\u03bf \u03b5\u03c0\u03cc\u03bc\u03b5\u03bd\u03bf \u03b8\u03ad\u03bc\u03b1 \u03c4\u03bf \u03b1\u03c6\u03b9\u03b5\u03c1\u03ce\u03bd\u03c9 \u03c3\u03b5 \u03cc\u03bb\u03bf\u03c5\u03c2 \u03b5\u03c3\u03ac\u03c2 ... :) \r\n\r\n\u0394\u03af\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bf\u03b9 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03af $ x_i$, i = 1,2, ... n, \u03c0\u03bf\u03c5 \u03b1\u03bd\u03ae\u03ba\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf [0,1].\r\n\r\n\u0391\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd $ i, j \\in$ { 1, 2, ... n }, \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03ce\u03c3\u03c4\u03b5 $ |x_i \\minus{} x_j|> \\frac{1}{2}$\r\n\r\n\u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03b3\u03b9\u03b1 \u03ad\u03bd\u03b1\u03bd \u03c4\u03bf\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03b1\u03c0\u03cc \u03c4\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd\u03c2 $ x_i$, \u03ad\u03c3\u03c4\u03c9 $ X$, \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9\r\n\r\n$ ............$ $ \\frac{3}{4}\\le 4X^2\\minus{}6X\\plus{}3 \\le 1$", "Solution_1": "\u0397 \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03b1 \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7 \u03b4\u03af\u03bd\u03b5\u03b9 \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b1 \r\n\r\n$ \\frac{3}{4} \\leq \\left(2x\\minus{}\\frac{3}{2}\\right)^2\\plus{}\\frac{3}{4}\\leq 1 \\Leftrightarrow 0\\leq \\left(2x\\minus{}\\frac{3}{2}\\right)^2 \\leq \\frac{1}{4}$\r\n\r\n\u0391\u03c1\u03ba\u03b5\u03af \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd (\u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03b1) \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf $ x_i$ \u03b3\u03b9\u03b1 \u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \r\n\r\n$ \\minus{}\\frac{1}{2} \\leq 2x_i\\minus{}\\frac{3}{2}\\leq \\frac{1}{2} \\Leftrightarrow \\frac{1}{2} \\leq x_i \\leq 1$ \u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03bb\u03cc\u03b3\u03c9 \u03c4\u03b7\u03c2 \u03c5\u03c0\u03cc\u03b8\u03b5\u03c3\u03b7\u03c2.\r\n\r\n(\u0395\u03b1\u03bd \u03b3\u03b9\u03b1 \u03cc\u03bb\u03b1 \u03c4\u03b1 $ j \\in \\{1,\\ldots,n\\}$ \u03b5\u03af\u03c7\u03b1\u03bc\u03b5 $ 0\\leq x_j < \\frac{1}{2}$ \u03c4\u03cc\u03c4\u03b5 $ |x_k\\minus{}x_l|<\\frac{1}{2}$, \u03b3\u03b9\u03b1 \u03cc\u03bb\u03b1 \u03c4\u03b1 $ k,l \\in \\{1,\\ldots,n\\}$, \u03ac\u03c4\u03bf\u03c0\u03bf).\r\n\r\n\u0391\u03bb\u03ad\u03be\u03b1\u03bd\u03b4\u03c1\u03bf\u03c2", "Solution_2": "$ 3/4 < \\equal{} 4x^2 \\minus{} 6x \\plus{} 3 < \\equal{} 1$\r\n\u0393\u03b9\u03b1 \u03ba\u03b1\u03b8\u03b5 $ x$ \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9 \u03bf\u03c4\u03b9 $ 4x^2 \\minus{} 6x \\plus{} 3 > \\equal{} 3/4$ \r\n\u0393\u03b9\u03b1 \u03bd\u03b1 \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9 \u03bf\u03c4\u03b9 $ 4x^2 \\minus{} 6x \\plus{} 3 < \\equal{} 1$ \u03b8\u03b1 \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 $ 1/2 < \\equal{} x < \\equal{} 1$ \u03c0\u03bf\u03c5 \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9 \u03b3\u03b9\u03b1\u03c4\u03b9 \u03b1\u03bd $ x_i < \\equal{} 1/2$ \u03ba\u03b1\u03b9 $ x_j < 1/2$ \u03c4\u03bf\u03c4\u03b5 $ |x_i \\minus{} x_j| < 1/2$ \u03b1\u03c4\u03bf\u03c0\u03bf \u03b1\u03c6\u03bf\u03c5 \u03c5\u03c0\u03b1\u03c1\u03c7\u03bf\u03c5\u03bd $ i,j$ \u03c9\u03c3\u03c4\u03b5 $ |x_i \\minus{} x_j| > 1/2$ \u0391\u03bd \u03b5\u03c7\u03c9 \u03ba\u03b1\u03bd\u03b5\u03b9 \u03bb\u03b1\u03b8\u03bf\u03c2 \u03c0\u03b5\u03c3\u03c4\u03bf \u03bc\u03bf\u03c5. :)", "Solution_3": "\u039f\u039a \u03c0\u03b1\u03af\u03b4\u03b5\u03c2 ... :) \r\n\r\n\u0388\u03c7\u03c9 \u03c5\u03c0\u03cc\u03c8\u03b7 \u03bc\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ae \u03bb\u03cd\u03c3\u03b7 \u03b1\u03bd \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03b5\u03c3\u03c4\u03b5 \u03c0\u03b5\u03af\u03c4\u03b5 \u03bc\u03bf\u03c5 \r\n\r\n[i]\u039c\u03b1\u03ba\u03c1\u03ae\u03c2 \u0392\u03b1\u03b3\u03b3\u03ad\u03bb\u03b7\u03c2[/i]", "Solution_4": "\u03a6\u03c5\u03c3\u03b9\u03ba\u03b1 \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03b5\u03c1\u03bf\u03bc\u03b1\u03b9 \u0392\u03b1\u03b3\u03b3\u03b5\u03bb\u03b7...\u03a3\u03b5 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03c9 \u03bf\u03c0\u03bf\u03c4\u03b5 \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03b3\u03c1\u03b1\u03c8\u03c4\u03b7\u03bd. :)", "Solution_5": "\u03a0\u03b1\u03b9\u03b4\u03b9\u03ac \u03bd\u03b1 \u03bf \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03cc\u03c2 \u03c4\u03c1\u03cc\u03c0\u03bf\u03c2 ... :) ( \u03c3\u03b5 attachment )\r\n\r\n\u03a0\u03b5\u03af\u03c4\u03b5 \u03bc\u03bf\u03c5 \u03c4\u03b7 \u03b3\u03bd\u03ce\u03bc\u03b7 \u03c3\u03b1\u03c2 ...", "Solution_6": "\u0392\u03b1\u03b3\u03b3\u03b5\u03bb\u03b7 \u03c0\u03bf\u03bb\u03c5 \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03b5\u03c1\u03bf\u03c5\u03c3\u03b1 \u03bb\u03c5\u03c3\u03b7....\u03ba\u03b1\u03b9 \u03c3\u03b9\u03b3\u03bf\u03c5\u03c1\u03b1 \u03b7 \u03ba\u03b1\u03bb\u03c5\u03c4\u03b5\u03c1\u03b7. :wink: \u0391\u03c0\u03bf \u03c0\u03b5\u03c1\u03b9\u03b5\u03c1\u03b3\u03b9\u03b1 \u03c0\u03bf\u03c3\u03bf \u03c7\u03c1\u03bf\u03bd\u03c9\u03bd \u03b5\u03b9\u03c3\u03b1\u03b9? :)" } { "Tag": [], "Problem": "Solve in real numbers:\r\n\r\n$ x \\plus{} y \\plus{} z \\equal{} 2$\r\n\r\n$ (x \\plus{} y)(y \\plus{} z) \\plus{} (y \\plus{} z)(z \\plus{} x) \\plus{} (z \\plus{} x)(x \\plus{} y) \\equal{} 1$\r\n\r\n$ x^2(y \\plus{} z) \\plus{} y^2(z \\plus{} x) \\plus{} z^2(x \\plus{} y) \\equal{} \\minus{} 6$", "Solution_1": "[hide]Set $ \\sigma_1 = x + y + z,\\sigma_2 = xy + yz + zx, \\sigma_3 = xyz$ and get from the equations\n\\begin{align*} \\sigma_1 & = 2 & {} &{}\\\\\n\\sigma_1^2 + \\sigma_2 & = 1 & \\Rightarrow \\sigma_2 &= - 3 \\\\\n\\sigma_1 \\sigma_2 - 3\\sigma_3 & = - 6 & \\Rightarrow \\sigma_3 &= 0\n\\end{align*}\nSo $ (x,y,z)$ are (distinct) solutions of $ 0 = x^3 - 2x^2 - 3x = x(x + 1)(x - 3)$ (Viet\u00e0). This gives the solutions $ (0, - 1,3),(0,3, - 1),( - 1,0,3),( - 1,3,0),(3, - 1,0),(3,0, - 1)$ and we easily check that these are indeed solutions of the original system.[/hide]" } { "Tag": [ "geometry", "quadratics", "trigonometry", "Pythagorean Theorem", "algebra", "quadratic formula" ], "Problem": "Circle a is tangent to both the x and y axis. Circle b is tangent to circle a, and is tangent both the x and y axis also. Both circles rest completely in quadrant I and their point of tangency is the only point that they share. If the radius of circle a is 1, then what is the radius of circle b?\r\n\r\nplz help me solve this!", "Solution_1": "Smaller circle has radius $ 1$ and larger cicle has radius $ r$ (or vice versa).\r\n\r\nPythagoras gives $ (r\\plus{}2)^{2} \\equal{} 2r^{2}$\r\n\r\n$ r^{2} \\minus{}4r \\minus{}4\\equal{}0$\r\n\r\nr= 2 plus or minus the square root of 2", "Solution_2": "where did you get $ (r\\plus{}2)^{2}\\equal{} 2r^{2}$ from?", "Solution_3": "The equation should be $ (\\sqrt{2}\\plus{}1\\plus{}r)^2\\equal{}2r^2$ since the smaller circle doesn't intersect the origin. [geogebra]b423b85e232eb58d5ef25e02c645e4c926bbda9b[/geogebra] \r\n\r\nLet me know if you don't see where the equation comes from.", "Solution_4": "You're right, fishythe fish. I was a bit hasty.", "Solution_5": "It's okay. When I did the explanation, I almost didn't catch it myself until I drew it out.", "Solution_6": "got it. thx!", "Solution_7": "No problem. Using the Pythagorean Theorem on the larger triangle seems obvious, but the smaller right triangle you also have to solve is not.", "Solution_8": "How would you solve the equation $ (\\sqrt{2}\\plus{}1\\plus{}r)^{2}\\equal{}2r^{2}$ for r?\r\nI can't get r by itself!!! sry, I'm kind of a beginner...", "Solution_9": "Expand first:\r\n\r\n$ (\\sqrt{2}\\plus{}1\\plus{}r)^2\\equal{}2r^2$\r\n\r\n$ r^2\\plus{}2r(1\\plus{}\\sqrt{2})\\plus{}1\\plus{}2\\sqrt{2}\\plus{}2\\equal{}2r^2$\r\n\r\n$ 0\\equal{}r^2\\plus{}(2\\plus{}2\\sqrt{2})r\\plus{}(3\\plus{}2\\sqrt{2})$\r\n\r\nThis is kinda ugly, especially with the quadratic formula:\r\n\r\n$ x\\equal{}\\frac{\\minus{}(2\\plus{}2\\sqrt{2})\\pm\\sqrt{(2\\plus{}2\\sqrt{2})^2\\minus{}4(1)(3\\plus{}2\\sqrt{2})}}{2(1)}$\r\n\r\n$ x\\equal{}\\frac{\\minus{}2\\minus{}2\\sqrt{2}\\pm\\sqrt{4\\plus{}8\\sqrt{2}\\plus{}8\\minus{}12\\minus{}8\\sqrt{2}}}{2}$\r\n\r\n$ x\\equal{}\\frac{\\minus{}2\\minus{}2\\sqrt{2}\\pm\\sqrt{0}}{2}$\r\n\r\n$ x\\equal{}\\minus{}1\\minus{}\\sqrt{2}$\r\n\r\nSo it all works out in the end. Maybe it's not that ugly.", "Solution_10": "It's actually easier if you take the square root of both sides. We obtain\r\n\r\n$ \\sqrt{2} \\plus{} 1 \\plus{} r \\equal{} \\sqrt{2} r \\implies (\\sqrt{2} \\minus{} 1)r \\equal{} \\sqrt{2} \\plus{} 1\\implies r \\equal{} (\\sqrt{2} \\plus{} 1)^2 \\equal{} 3 \\plus{} 2\\sqrt{2}\r\n\\\\\r\n\\sqrt{2} \\plus{} 1 \\plus{} r \\equal{} \\minus{}\\sqrt{2} r \\implies \\minus{}(\\sqrt{2} \\plus{} 1)r \\equal{} \\sqrt{2} \\plus{} 1\\implies r \\equal{} \\minus{}1.$", "Solution_11": "Since both our answers are negative, I think we need to take absolute values. (Like in trigonometry, you can reverse the sign of the radius if you add $ \\pi$ radians to the angle, right?)\r\n\r\nAnd TachyonPulse, you can't go from\r\n\r\n$ (\\sqrt{2}\\minus{}1)r\\equal{}\\sqrt{2}\\plus{}1$\r\n\r\nto\r\n\r\n$ r\\equal{}(\\sqrt{2}\\plus{}1)^2$\r\n\r\nIf you divide both sides out, you get\r\n\r\n$ r\\equal{}\\frac{\\sqrt{2}\\plus{}1}{\\sqrt{2}\\minus{}1}\\equal{}\\frac{(\\sqrt{2}\\plus{}1)(\\sqrt{2}\\minus{}1)}{(\\sqrt{2}\\minus{}1)^2}\\equal{}\\frac{2\\minus{}1}{2\\minus{}2\\sqrt{2}\\plus{}1}\\equal{}\\frac{1}{3\\minus{}\\sqrt{2}}\\equal{}\\frac{3\\plus{}\\sqrt{2}}{7}$\r\n\r\nAnd something seems intuitively wrong with the second solution.\r\n\r\nI messed up in my earlier post. The equation should be\r\n\r\n$ 0\\equal{}r^2\\minus{}(2\\plus{}2\\sqrt{2})r\\minus{}(3\\plus{}2\\sqrt{2})$\r\n\r\n$ r\\equal{}\\frac{2\\plus{}2\\sqrt{2}\\pm\\sqrt{(2\\plus{}2\\sqrt{2})^2\\plus{}4(1)(3\\plus{}2\\sqrt{2})}}{2(1)}$\r\n\r\n$ r\\equal{}\\frac{2\\plus{}2\\sqrt{2}\\pm\\sqrt{4\\plus{}8\\sqrt{2}\\plus{}8\\plus{}12\\plus{}8\\sqrt{2}}}{2}$\r\n\r\n$ r\\equal{}\\frac{2\\plus{}2\\sqrt{2}\\pm\\sqrt{24\\plus{}16\\sqrt{2}}}{2}$\r\n\r\n$ r\\equal{}1\\plus{}\\sqrt{2}\\pm\\sqrt{6\\plus{}4\\sqrt{2}}$", "Solution_12": "but if you multiply TachyonPulse's answer in a different way...\r\n\r\n$ r\\equal{}\\frac{\\sqrt{2}\\plus{}1}{\\sqrt{2}\\minus{}1}\\equal{}\\frac{(\\sqrt{2}\\plus{}1)^{2}}{(\\sqrt{2}\\plus{}1)(\\sqrt{2}\\minus{}1)}\\equal{}\\frac{2\\plus{}2\\sqrt{2}\\plus{}1}{(\\sqrt{2})^{2}\\minus{}(1)^{2}}\\equal{}\\frac{3\\plus{}2\\sqrt{2}}{2\\minus{}1}\\equal{}3\\plus{}2\\sqrt{2}$", "Solution_13": "Yes and I am right because fishythefish dropped a $ 2$ in front of the $ \\sqrt {2}$ in the following step:\r\n\r\n$ \\frac {2 \\minus{} 1}{2 \\minus{} 2\\sqrt {2} \\plus{} 1} \\equal{} \\frac {1}{3 \\minus{} \\sqrt {2}}.$\r\n\r\nI also checked both the solutions I obtained through Wolfram Alpha and both worked.", "Solution_14": "Wow. I feel like an idiot. :blush: \r\n\r\n$ \\frac{1}{3\\minus{}2\\sqrt{2}}\\equal{}\\frac{3\\plus{}2\\sqrt{2}}{1}\\equal{}3\\plus{}2\\sqrt{2}$\r\n\r\nSo it turns out that if I do the math right, you don't need to multiply it a different way... :blush:" } { "Tag": [ "\\/closed" ], "Problem": "I have a question regarding WOOT: Is it possible for a company to sponsor you/pay part of the money for you to take WOOT? There are 4 or so people in my school who would really enjoy taking the class, except that we don't have that much money to pay for the class. Any help would be appreciated :)", "Solution_1": "[quote=\"Karth\"]I have a question regarding WOOT: Is it possible for a company to sponsor you/pay part of the money for you to take WOOT? There are 4 or so people in my school who would really enjoy taking the class, except that we don't have that much money to pay for the class. Any help would be appreciated :)[/quote]\r\nYes, but I think you'll have to figure out how to do that on your own. There was another topic similar to this one here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=100121" } { "Tag": [ "geometry", "circumcircle", "angle bisector", "geometry unsolved" ], "Problem": "Let $ ABCDEF$ be a convex hexagon, such that:\r\n\r\n$ \\angle CAB\\equal{}\\angle FBA, \\angle ECD\\equal{}\\angle BDC$\r\n\r\n$ \\angle AEF\\equal{}\\angle DFE, \\angle EAC\\equal{}\\angle FBD$\r\n\r\n$ \\angle ACE\\equal{}\\angle BDF, \\angle CEA\\equal{}\\angle DFB$\r\n\r\nProve that the \"central\" hexagon formed by the intersection of the lines $ AC,BD,CE,DF,EA$ and $ FB$ is circumscribable.", "Solution_1": "let $ A': AC \\cap FB$, $ B': BD \\cap AC$ and cyclicaly define C',D',E',F'. By angle the quads FF'DE', DD'B'C, BB'F'A are cyclic, so $ \\angle A'B'F' \\equal{} \\angle A'F'B' \\equal{} a$, $ \\angle C'B'D' \\equal{} \\angle C'D'B' \\equal{} b$, $ \\angle E'D'F' \\equal{} \\angle E'F'D' \\equal{} c$ and $ \\angle FF'E\\equal{} \\angle FD'E \\equal{} \\angle DD'C \\equal{} \\angle DB'C \\equal{} \\angle BB'A \\equal{} \\angle BF'A\\equal{}k$. Then the angle bisector of $ \\angle F'A'B'$, $ \\angle B'C'D'$, $ \\angle D'E'F'$ concur on the circumcenter O of $ \\triangle B'D'F'$. So we have to prove that the angle bisector of $ \\angle A'B'C'$ pass trought O and so that $ 2 \\left ( 90\\minus{}\\angle F'D'B' \\plus{} a \\right ) \\equal{} a\\plus{} b\\plus{} \\angle F'B'D' $$ \\Longleftrightarrow \\ (180 \\minus{} \\angle F'D'B' \\minus{} \\angle F'B'D') \\plus{}a \\equal{}\\angle F'D'B' \\plus{} b$ that is true because $ (180 \\minus{} \\angle F'D'B' \\minus{} \\angle F'B'D') \\plus{}a \\equal{} B'F'D' \\plus{} a \\equal{}$ $ 180 \\minus{}\\angle FF'D' \\equal{} 180 \\minus{} \\angle F'D'E \\equal{} \\angle F'D'B' \\plus{} b$." } { "Tag": [ "geometry", "ratio" ], "Problem": "Two concentric circles are drawn such that the inner circle covers 25% of the area of the outer circle. Given that the radius of the outer circle is 10 units, how many units are in the radius of the inner circle?", "Solution_1": "[hide]Since the small circle covers $25\\% = \\frac{1}{4}$, the ratio of the radii is $\\sqrt{\\frac{1}{4}}= \\frac{1}{2}$. So the radius of the small circle is $10\\cdot \\frac{1}{2}= 5$.[/hide]\r\n\r\nEDIT: thanks", "Solution_2": "[hide]The ratio of the areas of the two circles is $4: 1$. Thus the ratio of the radii is $sqrt{1}: sqrt{4}=1: 2$. Thus the radius of the smaller circle is $10/2=5$[/hide]", "Solution_3": "[quote=\"vishalarul\"][hide]Since the small circle covers $25\\% = \\frac{1}{4}$, the ratio of the radii is $\\sqrt{\\frac{1}{4}}= \\frac{1}{2}$. So the radius of the small circle is $50\\cdot \\frac{1}{2}= 25$.[/hide][/quote]\r\nWhere'd you get 50?", "Solution_4": "[hide]$10^{2}=100$\n$100*.25=25$\n$sqrt25=5$[/hide]" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let a,b,c>0 such that $ ab\\plus{}bc\\plus{}ca\\equal{}1$ and $ \\alpha$ be real number\r\nProve that\r\n $ \\frac{a^\\alpha}{b\\plus{}c}\\plus{}\\frac{b^\\alpha}{c\\plus{}a}\\plus{}\\frac{c^\\alpha}{a\\plus{}b} \\geq \\frac{3\\sqrt{3abc}}{2}(\\sqrt{a^{2\\alpha\\plus{}1}}\\plus{}\\sqrt{b^{2\\alpha\\plus{}1}}\\plus{}\\sqrt{c^{2\\alpha\\plus{}1}})$", "Solution_1": "Nobody's interested in my ineq" } { "Tag": [], "Problem": "Suppose that lists are formed in the following manner. Start with the list\r\n$1,1$\r\nThen, insert in between two consecutivey entries of the list $x$ and $y$, the number $x+y$. So the next list would be\r\n$1,2,1$\r\nand then\r\n$1,3,2,3,1$\r\n$1,4,3,5,2,5,3,4,1$\r\nand so on.\r\n\r\nProve that $a$ and $b$ can be consecutive in at most one list $L_{n}$.", "Solution_1": "A consecutive pair (a,b) can only come from exactly 1 of the pairs (a-b,b) or (a,b-a), more precisely the only in which both entries are positive(that is the smallest number of the 2 is preserved). Same story for that pair, and so on. This gives a unique path leading to (a,b) that must clearly terminate in a finite number of steps. Remark that (1,1) is the only possible pair of equal consecutive numbers. It must end with one of the entries being 1, and the issue is now settled because there is only one pair of the form (1,x) or (x,1)" } { "Tag": [ "superior algebra", "superior algebra solved" ], "Problem": "Let $ S$ be any commutative ring, $ x\\in S$, $ I\\subseteq S$ an ideal and $ P_1,\\ldots,P_k$ prime ideals of $ S$. Suppose that the coset $ x\\plus{}I$ is contained in $ \\cup_{i\\equal{}1}^kP_i$. Then there exists $ j$ such that $ xS\\plus{}I\\subseteq P_j$.", "Solution_1": "[url]http://www.math.lsa.umich.edu/~hochster/615W07/L03.19.pdf[/url]", "Solution_2": "[quote=\"hjbrasch\"][url]http://www.math.lsa.umich.edu/~hochster/615W07/L03.19.pdf[/url][/quote]\r\n\r\nWell, that is the course that I am following. I gave this problem for you to think at, not for you to show me where I can find it.\r\nAll the results that I have posted so far that have the subject description \"Tight closure course\" can be found there or close to there.", "Solution_3": "well, it is a slight modification of the normal prime avoidance lemma" } { "Tag": [ "geometry", "3D geometry", "sphere", "tetrahedron" ], "Problem": "Three mutually tangent spheres of radius 1 rest on a horizontal plane. A sphere of radius 2 rests on them. What is the distance from the plane to the top of the larger sphere?\r\n\r\nthe answer is : 3+root 69/3\r\n\r\nanyone can explain?[/img][/code]", "Solution_1": "find the height of the tetrahedron formed by the centers of the 4 spheres, then add 3 (=1+2)." } { "Tag": [ "function", "inequalities proposed", "inequalities" ], "Problem": "Let $a,b\\ge0$ and $n \\in \\mathcal N$:\r\n\r\n$\\sqrt[n+1]{a^{n+1}+b^{n+1}} \\le \\sqrt[n]{a^n+b^n}$", "Solution_1": "if ab=0 then equal.\r\n\r\nab>0\r\n\r\nWLOG b=1\r\n\r\nand define a function f(x)=(1+a^x)\r\n\r\ng(x)=f(x)^(1/x)\r\n\r\ng'(x)*x^2 *(1+a^x)(a^x(lna-ln1+a^x) -ln(1+a^x)))\r\n\r\ng'(x)<0 (for all x>=1)\r\n\r\ntherefore g(n)>g(n+1)\r\n\r\nQ.E.D." } { "Tag": [ "inequalities", "search", "inequalities proposed" ], "Problem": "Here is a joke:\r\n If $x_1 x_2...x_n=1$ and $ (x_1+x_2+x_3)(x_2+x_3+x_4)...(x_n+x_1+x_2)=(3.1)^n$ and $x_1,...,x_n>0$ then $(x_1+x_2)(x_2+x_3)...(x_n+x_1)<(2.1)^n$.", "Solution_1": "Sprry if I make a mistake, Harazi, but isn't this just consequence of Mongolian Inequality?", "Solution_2": "Iura, do not drop the atomic bomb! And the mongolian inequality was not refering to some order between the $x_i$? Anyway, even if it is a consequence of it(though I do not realize now), I want a solution in 3 lines.", "Solution_3": "Very strange...\r\nHere is my idea..\r\nI consider \r\n\\[\r\nx_i + x_{i + 1} + k\\frac{1}{k}x_{i + 2} \\ge (k + 1)\\left( {\\frac{1}{k}x_{i + 2} } \\right)^{\\frac{k}{{k + 1}}} \\left( {x_i + x_{i + 2} } \\right)^{\\frac{1}{{k + 1}}\r\n}\\]\r\nMultiply all of them together,\r\n\\[\r\n3.1^n \\ge (k + 1)^n \\left( {\\frac{1}{k}} \\right)^{\\frac{{nk}}{{k + 1}}} \\left( {\\prod {\\left( {x_i + x_{i + 2} } \\right)} } \\right)^{\\frac{1}{k}} \r\n\\]\r\nSuppose k satisfies\r\n\\[\r\n3.1^n \\ge (k + 1)^n \\left( {\\frac{1}{k}} \\right)^{\\frac{{nk}}{{k + 1}}} 2.1^{\\frac{n}{k+1}} \r\n\\]\r\nthen we are done.\r\nI used mathematica to solve the equation,\r\n\\[\r\n\\frac{{3.1k^{k + 1} }}{{k(k + 1)^{k + 1} }} = 2.1\r\n\\]\r\nit seems to have a root but i doubt if that's the mistake of mathematica. (the root is right between 0.4761904 and 0.4761905), because the graph is very weird...\r\ncan anyone verify if there is a root for this equation?", "Solution_4": "Maximum over $k$ is about $2.109362$. Computed by Maple.", "Solution_5": "[quote=\"Fedor Petrov\"]Maximum over $k$ is about $2.109362$. Computed by Maple.[/quote]\r\ndo you mean this value of k satisfies my equation?\r\nis my idea valid?", "Solution_6": "Sorry. Why did I search for a maximum? We are interested in minimum. Put $k=1/(2.1)$ and get desired result. \r\n\r\nAfter all, I do not consider this problem too easy. At least, it seems more interesting than usual inequalities which I see on olympiads. Or is there much easier solution?", "Solution_7": "I will try to remember wy idea, I know it was nice. I agree with Fedor about the inequalities in contest, they are all the same!", "Solution_8": "May I help you? We have $\\root n\\of{a_1a_2\\dots a_n}+\\root n\\of{b_1b_2\\dots b_n}\\le \\root n\\of{(a_1+b_1)(a_2+b_2)\\dots (a_n+b_n)}$. Proof: wlog, $a_i+b_i=1$ (else multiple $a_i$ and $b_i$ by the same factor $t$), then we have $\\root n\\of{a_1a_2\\dots a_n}+\\root n\\of{b_1b_2\\dots b_n}\\le \\sum a_i/n+\\sum b_i/n=1$.\r\n\r\nThen just put $a_i=x_i$ and $b_i=x_{i+1}+x_{i+2}$.", "Solution_9": "Oh, yes! Thanks, Fedor. Oh, so this is the reason for which you said that after all you do not consider the problem that easy: you always had this extremely easy solution? :)", "Solution_10": "I said it before I found this solution. But nevertheless, I consider a problem tricky enough after more than all. It looks stupid when you have a solution, but not before (at least for me).", "Solution_11": "very simple and nice solution!", "Solution_12": "[quote=\"iura\"]Sprry if I make a mistake, Harazi, but isn't this just consequence of Mongolian Inequality?[/quote]\r\n\r\nWhat is the Mongolian inequality?", "Solution_13": "If $x_1\\le x_2\\le x3\\le x_n$, then $\\prod_{i=1}^n \\frac{x_i+x_{i+1}+x_{i+2}}3\\ge \r\n\\prod_{i=1}^n \\frac{x_i+x_{i+1}}2$. (products are cyclic). I am sure it was posted on the forum." } { "Tag": [ "geometry", "analytic geometry", "graphing lines", "slope", "trigonometry", "articles", "vector" ], "Problem": "Hi,\r\n\r\nThis might sound like a stupid idea but here it is (I didn't know where to put it so I'll just place it here.)\r\n\r\nI just had a thought that occurred to me when it was raining: \r\n\r\nDepending on how heavy or how light the rainfall is, is there a certain speed at which you can run so that you can avoid a majority of the rainfalls that would have otherwise hit you? Can you avoid them all?\r\n\r\nAgain, this question might not make sense, but I'm just curious to see you guys' opinions/views on it. \r\n\r\nThank you very much :)", "Solution_1": "Unless you have a lightning cloud over you that you are trying to outrun, I don't really see any way to do it. The rain in front of the spot you just were at would hit you instead. In addition, running would provide a greater bird's eye view area than walking, so I think you would be hit by more, not less raindrops.", "Solution_2": "The rain that falls on your head varies with the speed you move. The rain you walk into does not. If you think of the rain as a continuous space of water that doesn't move, you move diagonally upward through it. The faster you move, the shallower the slope, and since you're going the same horizontal distance, you want to have as shallow a slope as possible.", "Solution_3": "If raindrops fell in a predictable array of some sort (for example, a falling 3D lattice array), then theoretically, you could probably get through without getting hit, if you were small enough compared to the distance between rain drops. However, that's not the case.\r\n\r\nI've thought about this, since I may have to decide between biking or walking in the rain.\r\n\r\nThere's two ways you get hit by rain -- rain that falls on you and drips down, and rain that you walk into. As it happens, if you're going faster, you tend to lean forward, so the extra surface area of your body as viewed from above will initially cancel out the reduced amount of travel time, and roughly the same amount of rain falls on you. But, as you speed up more and more, the reduced time has a larger impact than the increase in your surface area viewed from above, and it starts to become effective. In the limit case, your body will be horizontal and you'll be moving at velocity infinity (forget Einstein for a moment), and you won't be hit from above with any water at all.\r\n\r\nAs for the rain you walk into, it's about the same. The only difference is, if you're going really slowly, then the rain you walk into is more likely to hit your shoulders, and when you're, say, biking, it's more likely to get your whole body. The fact that you lean forward while going more quickly has little effect. Think about it -- at 90 degrees, a small change in angle will cause a negligible change in sine but a large change in cosine.\r\n\r\nUltimately, the only useful solution is to cover yourself by a means of your choice.", "Solution_4": "For some reason my English teacher of a few years ago was talking about his really smart physics buddy from Caltech.\r\n\r\nAnd apparently this buddy wrote a paper on how fast you would have to go through a rain storm to avoid getting hit.\r\n\r\nSo a solution exists, somewhere, I think.\r\n\r\nAnyways, if you go above mach one or something I'm sure the air pressure around you would cause any raindrops to slide off without technically hitting you, which is what I imagined this buddy's solution entailing.", "Solution_5": "I remember someone writing an article on this at least 20 years ago - probably the [i]American Mathematical Monthly,[/i] if someone wants to look for a reference. The author was using a continuous model - treating rain as having a certain (vector) mass flux. I vaguely remember that the conclusions were that in a headwind or crosswind, you intersect the least water by going as fast as possible, but in a fairly strong tailwind, there was another optimum where you roughly match the wind speed.\r\n\r\nI think the author he was treating the pedestrian as an essentially vertical rectangular box - the actual geometry may matter, and in particular, I would think that leaning forward while moving forward rapidly would help quite a bit. That's certainly an argument for riding the bicycle - at least as long as it's only the falling rain that matters; wheel-splash would be another issue.\r\n\r\n[quote]If raindrops fell in a predictable array of some sort (for example, a falling 3D lattice array),[/quote]\r\nNot a chance. I would assume that a good model for how many raindrops hit a specified area in a specified time would be a Poisson probability distribution.", "Solution_6": "Many references on the topic...\r\n\r\nhttp://www.flyingcircusofphysics.com/pdf/Chapter1_Ref_Com.pdf", "Solution_7": "Also, the rain isn't falling at the same rate all day long. Clearly if the rain is getting bigger you would do well to go fast, while if it's lightening up (arrgh... pun) you may as well take your time.", "Solution_8": "Interesting views. I was curious to know what you guys think about this. Thank you for your time :)", "Solution_9": "Yes, you can avoid them by bringing an umbrella! :lol:", "Solution_10": "[quote=\"Kent Merryfield\"]I vaguely remember that the conclusions were that in a headwind or crosswind, you intersect the least water by going as fast as possible, but in a fairly strong tailwind, there was another optimum where you roughly match the wind speed.[/quote]\r\nYes, that sounds about right. Roughly, if you match the wind speed no drops hit you horizontally, and the vertical drops are unavoidable." } { "Tag": [ "function", "algebra", "functional equation", "algebra proposed" ], "Problem": "I'd like to see your solutions to this functional equation:\r\n\r\nFind all functions $f: \\mathbb{Q}\\rightarrow \\mathbb{Q}$ such that:\r\n\r\n$f(x+y)=f(x)+f(y)+xy$", "Solution_1": "Denote $g(x)=f(x)-\\frac{x^{2}}{2}$, we have $g(x+y)=g(x)+g(y)\\forall x,y\\in\\mathbb{Q}$ and $g(x)=ax(a\\in\\mathbb{Q})$. Final $f(x)=\\frac{x^{2}}{2}+ax$." } { "Tag": [ "limit", "arithmetic sequence", "number theory unsolved", "number theory" ], "Problem": "Questions:\r\n\r\n (1) Is there 4 CONSECUTIVE primes such that they below an aritmetica progression ??\r\n\r\n (2) Is there 3 primes in aritmetica progression with arit razon: r=8??\r\n\r\n\r\nThanks in Advanced\r\n\r\nFaby :rotfl:", "Solution_1": "Look at the paper [url=http://www.dms.umontreal.ca/~andrew/PDF/PrimePatterns.pdf]Prime number patterns[/url] by Andrew Granville.", "Solution_2": "Ok.. thankz maxal...\r\n\r\nbt ..\r\nwhat about in these especific cases ..?\r\n\r\nFaby. ;)", "Solution_3": "[quote=\"fabycv\"] (2) Is there 3 primes in aritmetica progression with arit razon: r=8??\n[/quote]\r\nLet the primes be $p, p+8, p+16$. Since the three numbers are pairwise incongruent modulo 3, there is exacly one of them being a multiple of 3. Thus the only possibility is $p=3$. Hence the only sequence satisfying your condition is $3,11,19$.\r\nP.S. I hope \"arit razon\" means \"common difference\". :P", "Solution_4": "[quote=\"fabycv\"]\n (1) Is there 4 CONSECUTIVE primes such that they below an aritmetica progression ??\n[/quote]\r\nFor example: 251, 257, 263, 269", "Solution_5": "And of course for any $n$, there are $n$ primes in arithmetic progression.", "Solution_6": "Now, let us ask the more interesting question: for $n \\in \\mathbb{N}$ let $d(n)$ denote the smallest arithmetic difference that a sequence of $n$ primes in arithmetic progression can have. \r\n\r\nWhat is $\\lim_{n \\to \\infty} d(n)$? :)", "Solution_7": "[quote=\"maxal\"][quote=\"fabycv\"]\n (1) Is there 4 CONSECUTIVE primes such that they below an aritmetica progression ??\n[/quote]\nFor example: 251, 257, 263, 269[/quote]\r\n\r\nMaxal ....how could you find this arit progression??...\r\ni 'd like to know the way ... to try the t0rajir0u's QUESTION -\"Generalization\"....\r\n\r\nThank you for your answer!!!\r\n\r\n\r\nCarlos Bravo ;)\r\nLima - Peru", "Solution_8": "[quote=\"carlosbr\"]Maxal ....how could you find this arit progression??...[/quote]Simple [url=http://pari.math.u-bordeaux.fr/]PARI/GP[/url] program:\r\n\r\n[code]forprime(p=2,10^5,if(isprime(p+6)&&isprime(p+12)&&isprime(p+18)&&primepi(p+18)-primepi(p)==3,forprime(q=p,p+18,print1(\" \",q));print()))[/code]", "Solution_9": "[quote=\"t0rajir0u\"]Now, let us ask the more interesting question: for $n \\in \\mathbb{N}$ let $d(n)$ denote the smallest arithmetic difference that a sequence of $n$ primes in arithmetic progression can have. \n\nWhat is $\\lim_{n \\to \\infty} d(n)$? :)[/quote]\r\n\r\nI think it tends to $\\infty$\r\n\r\nFor an IrMO, we were asked to show that if we had 15 primes in aritmetic progression, then the common difference must be divisable by $2,3,5,7,11,13$ So continuing the pattern the result follows. Of couse, we'll need to prove it :lol:", "Solution_10": "[quote=\"maxal\"][quote=\"carlosbr\"]Maxal ....how could you find this arit progression??...[/quote]Simple [url=http://pari.math.u-bordeaux.fr/]PARI/GP[/url] program:\n\n[code]forprime(p=2,10^5,if(isprime(p+6)&&isprime(p+12)&&isprime(p+18)&&primepi(p+18)-primepi(p)==3,forprime(q=p,p+18,print1(\" \",q));print()))[/code][/quote]\r\n\r\nwell....\r\ni hoped ... a mathematical process....\r\n\r\nThanks...\r\n\r\nCarlos Bravo ;)\r\nLima Peru", "Solution_11": "[quote=\"seamusoboyle\"][quote=\"t0rajir0u\"]Now, let us ask the more interesting question: for $n \\in \\mathbb{N}$ let $d(n)$ denote the smallest arithmetic difference that a sequence of $n$ primes in arithmetic progression can have. \n\nWhat is $\\lim_{n \\to \\infty} d(n)$? :)[/quote]\n\nI think it tends to $\\infty$\n\nFor an IrMO, we were asked to show that if we had 15 primes in aritmetic progression, then the common difference must be divisable by $2,3,5,7,11,13$ So continuing the pattern the result follows. Of couse, we'll need to prove it :lol:[/quote]\r\n\r\nWell, it makes sense - if the common difference wasn't divisible by those primes then the remainder $\\bmod 2, 3, 5, 7, 11$ or $13$ from the first prime to the last would reach zero in $14$ steps or less... or something along those lines :)\r\n\r\nFor example, the progression maxal found had a common difference divisible by $2, 3$ for a $4$-length sequence. Yeah... that makes sense. So $d(n)$ would just be $\\prod_{p_k < n} p_k$?", "Solution_12": "No, $d(n)$ could be bigger, at least there are a lot of really hard problems about that.\r\nAlso take a look at http://www.mathlinks.ro/Forum/viewtopic.php?t=6075 .", "Solution_13": "When it says 'primes in amithmetic order', does it mean consecutively?", "Solution_14": "[quote=\"seamusoboyle\"]When it says 'primes in amithmetic order', does it mean consecutively?[/quote]\r\nYes, everything else is senseless.", "Solution_15": "As i realised just after i posted! Just thinking of $4a+1$ !!!" } { "Tag": [ "limit" ], "Problem": "\u0388\u03c7\u03c9 \u03bc\u03af\u03b1 \u03b1\u03c0\u03bf\u03c1\u03af\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03b7 \u03b3\u03b5\u03bd\u03af\u03ba\u03b5\u03c5\u03c3\u03b7 \u03bc\u03b9\u03b1\u03c2 \u03b2\u03b1\u03c3\u03b9\u03ba\u03ae\u03c2 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2. \u0388\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $ 2(a^{2}\\plus{}b^{2})\\geq (a\\plus{}b)^{2}$ \u03ba\u03b1\u03b9 $ 3(a^{2}\\plus{}b^{2}\\plus{}c^{2})\\geq (a\\plus{}b\\plus{}c)^{2}$. \u0386\u03c1\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b1\u03c5\u03c4\u03ae \u03b7 \u03bc\u03bf\u03c1\u03c6\u03ae \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c5\u03c2, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae $ n\\cdot \\sum^{n}_{i\\equal{}1}a^{2}_{i}\\geq (\\sum^{n}_{i\\equal{}1}a_{i})^{2}$???", "Solution_1": "\u039d\u03b1\u03b9. \u0395\u03af\u03bd\u03b1\u03b9 \u03b7 Cauchy-Schwarz \u03b3\u03b9\u03b1 $ x_i \\equal{} a_i$, $ y_i \\equal{} 1$.\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_2": "\u03a3\u03c5\u03bc\u03c6\u03c9\u03bd\u03ce \u03c0\u03c9\u03c2 \u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03bb\u03ad\u03b5\u03b9 \u03bf \u0393\u03b9\u03ce\u03c1\u03b3\u03bf\u03c2 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03bb\u03cc\u03b3\u03c9 \u03c4\u03b7\u03c2 bcs \u03b3\u03b9\u03b1 n \u03cc\u03c1\u03bf\u03c5\u03c2 \u03cc\u03c0\u03bf\u03c5 n \u03c6\u03c5\u03c3\u03b9\u03ba\u03cc\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2, \u03b4\u03b7\u03bb.\u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03cc\u03c1\u03c9\u03bd.\u0393\u03b9\u03b1 [u]\u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\u03c5\u03c2[/u] \u03cc\u03c1\u03bf\u03c5\u03c2 \u03c4\u03b9 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9?", "Solution_3": "$ s_n\\geq t_n \\implies \\lim s_n \\geq \\lim t_n$\r\n\r\n\u0397 LHS \u03c4\u03b7\u03c2 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2 \u03c0\u03ac\u03bd\u03c4\u03b1 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c3\u03c4\u03bf \u03cc\u03c1\u03b9\u03bf $ n\\to\\infty$. \u0391\u03bd \u03ba\u03b1\u03b9 \u03b7 RHS \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9, \u03b8\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1.\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_4": "\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce Durandal ,\u03cc\u03bc\u03c9\u03c2 \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03ac \u03c4\u03bf $ n\\rightarrow\\infty$ \u03b1\u03c0\u03cc \u03c4\u03bf $ n\\equal{}\\infty$?\r\nB\u03ad\u03b2\u03b1\u03b9\u03b1 \u03b1\u03c5\u03c4\u03cc \u03c3\u03c7\u03b5\u03c4\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03bc\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b7 \u03b1\u03c0\u03bf\u03c1\u03af\u03b1 \u03bc\u03bf\u03c5:\u03a4\u03af \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03ac \u03ad\u03c7\u03b5\u03b9 \u03cc\u03c4\u03b1\u03bd \u03bb\u03ad\u03bc\u03b5 \u03c4\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c4\u03c9\u03bd \u03cc\u03c1\u03c9\u03bd \u03bc\u03b9\u03b1\u03c2 \u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03c5\u03af\u03b1\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf \u03b1\u03c0\u03cc \u03c4\u03bf \u03bd\u03b1 \u03c4\u03b5\u03af\u03bd\u03b5\u03b9 \u03c3\u03c4\u03bf \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf :?:", "Solution_5": "\u0394\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03b1\u03bc\u03af\u03b1 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03ac \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 ... \u03c3\u03c4\u03b7\u03bd \u03c3\u03c4\u03b7\u03bd \u03c0\u03c1\u03cc\u03c4\u03b1\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03b3\u03c1\u03ac\u03c6\u03b5\u03b9\u03c2 ...\r\n\u0395\u03af\u03bd\u03b1\u03b9 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03b1 \u03bb\u03ac\u03b8\u03bf\u03c2 \u03cc\u03bc\u03c9\u03c2 \u03c4\u03bf $ n \\equal{} \\infty$\r\n\r\n[i]*\u0392\u03ad\u03b2\u03b1\u03af\u03c9\u03c2 \u03cc\u03bc\u03c9\u03c2 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03c3\u03bf\u03c5 \u03c0\u03b5\u03af \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03b1 \u03bf Panagiotis ![/i]", "Solution_6": "Heh, \u0392\u03b1\u03b3\u03b3\u03ad\u03bb\u03b7 \u03bc\u03b5 \u03c0\u03c1\u03cc\u03bb\u03b1\u03b2\u03b5\u03c2 :)\r\n\r\nGuys, \u03bc\u03b7\u03bd \u03bc\u03c0\u03b5\u03c1\u03b4\u03b5\u03cd\u03b5\u03c3\u03c4\u03b5...\r\n\r\n$ n \\equal{} \\infty$ \u03b4\u03b5 \u03b8\u03b1 \u03c4\u03bf \u03b4\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c4\u03ad \u03b3\u03b9\u03b1\u03c4\u03af \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9 \u03bd\u03cc\u03b7\u03bc\u03b1 (\u03b5\u03ba\u03c4\u03cc\u03c2 \u03ba\u03b1\u03c4\u03ac \u03c3\u03cd\u03bc\u03b2\u03b1\u03c3\u03b7), \u03bc\u03cc\u03bd\u03bf \u03c4\u03bf $ n\\to\\infty$ \u03ad\u03c7\u03b5\u03b9 \u03bd\u03cc\u03b7\u03bc\u03b1. \u0393\u03b9'\u03b1\u03c5\u03c4\u03cc \u03b1\u03c0\u03ac\u03bd\u03c4\u03b7\u03c3\u03b1 \u03ad\u03c4\u03c3\u03b9, \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03ce\u03bd\u03c4\u03b1\u03c2 \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03c9 \u03bc\u03b5 \u03c0\u03bf\u03b9\u03bf\u03bd \u03c4\u03c1\u03cc\u03c0\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c9\u03c3\u03c4\u03ac \u03b4\u03b9\u03b1\u03c4\u03c5\u03c0\u03c9\u03bc\u03ad\u03bd\u03b7 \u03b7 \u03b5\u03c1\u03ce\u03c4\u03b7\u03c3\u03b7.\r\n\r\n\u03a4\u03bf $ \\infty$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03bc\u03bf\u03bd\u03b1\u03b4\u03b9\u03ba\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c3\u03c5\u03c3\u03c3\u03ce\u03c1\u03b5\u03c5\u03c3\u03b7\u03c2 \u03c4\u03c9\u03bd \u03c6\u03c5\u03c3\u03b9\u03ba\u03ce\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03ce\u03bd, \u03ba\u03b1\u03b9 \u03b1\u03bd \u03c0\u03bf\u03c4\u03ad \u03b3\u03c1\u03ac\u03c8\u03bf\u03c5\u03bc\u03b5 \u03c0\u03c7 $ a_\\infty \\equal{} \\ldots$ \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03c3\u03c4\u03b7\u03bd \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b5\u03bd\u03bd\u03bf\u03bf\u03cd\u03bc\u03b5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 $ \\lim_{n\\to\\infty}a_n \\equal{} \\ldots$. \u0393\u03b9'\u03b1\u03c5\u03c4\u03cc \u03cc\u03c4\u03b1\u03bd \u03bb\u03ad\u03bc\u03b5 $ \\sum_{n \\equal{} 1}^\\infty$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03bb\u03ce\u03c2 \u03c3\u03c5\u03bd\u03c4\u03bf\u03bc\u03bf\u03b3\u03c1\u03b1\u03c6\u03af\u03b1 \u03c4\u03bf\u03c5 $ \\lim_{n\\to\\infty}\\sum_{k \\equal{} 1}^n$. \u0391\u03bd \u03b8\u03ad\u03bb\u03b5\u03b9\u03c2, \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \"\u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\" \u03b5\u03af\u03bd\u03b1\u03b9 \u03ad\u03bd\u03b1 \u03b1\u03c6\u03b7\u03c1\u03b7\u03bc\u03ad\u03bd\u03bf [i]\u03c3\u03b7\u03bc\u03b5\u03af\u03bf[/i] \u03b5\u03ba\u03c4\u03cc\u03c2 \u03c4\u03bf\u03c5 $ \\mathbb{N}$ \u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03c4\u03bf \u03c0\u03c1\u03bf\u03c3\u03b1\u03c1\u03c4\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03ba\u03ac \u03ce\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03ad\u03c7\u03b5\u03b9 \u03bd\u03cc\u03b7\u03bc\u03b1 \u03b7 \u03ad\u03ba\u03c6\u03c1\u03b1\u03c3\u03b7 $ n\\to\\infty$.\r\n\r\n\u03a4\u03bf \u03cc\u03c4\u03b9 \u03c4\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c4\u03c9\u03bd \u03cc\u03c1\u03c9\u03bd \u03bc\u03af\u03b1\u03c2 \u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03af\u03b1\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \"\u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\" \u03ad\u03c7\u03b5\u03b9 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03bc\u03b5 \u03c4\u03bf\u03bd \u03c0\u03bb\u03b7\u03b8\u03c5\u03ba\u03cc \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc $ \\aleph_0$ \u03c4\u03bf\u03c5 \u03c3\u03c5\u03bd\u03cc\u03bb\u03bf\u03c5 $ \\mathbb{N}$ \u03c4\u03c9\u03bd \u03c6\u03c5\u03c3\u03b9\u03ba\u03ce\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03ce\u03bd... \u03b5\u03bd\u03c4\u03b5\u03bb\u03ce\u03c2 \u03ac\u03bb\u03bb\u03b7 \u03b9\u03c3\u03c4\u03bf\u03c1\u03af\u03b1! \u0394\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf\u03c2, \u03ac\u03c1\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \"\u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\u03c2\", \u03cc\u03c0\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b1 \u03c3\u03c4\u03bf\u03b9\u03c7\u03b5\u03af\u03b1 \u03c4\u03c9\u03bd \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ce\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \"\u03ac\u03c0\u03b5\u03b9\u03c1\u03b1\" \u03b1\u03bb\u03bb\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03cc \"\u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\" \u03b1\u03c5\u03c4\u03cc \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03b7\u03c2 \u03c4\u03ac\u03be\u03b7\u03c2 \u03c4\u03bf\u03c5 \u03c3\u03c5\u03bd\u03b5\u03c7\u03bf\u03cd\u03c2 $ \\mathfrak{c}$ \u03ba\u03b1\u03b9.........\r\n\r\n\u038c\u03c0\u03c9\u03c2 \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03af\u03bd\u03b5\u03b9\u03c2, \u03c4\u03b1 \u03b4\u03cd\u03bf \"\u03ac\u03c0\u03b5\u03b9\u03c1\u03b1\" \u03c3\u03c4\u03b1 \u03bf\u03c0\u03bf\u03af\u03b1 \u03b1\u03bd\u03b1\u03c6\u03ad\u03c1\u03b5\u03c3\u03b1\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03cd\u03bf \u03b5\u03bd\u03c4\u03b5\u03bb\u03ce\u03c2 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ad\u03c2 \u03ad\u03bd\u03bd\u03bf\u03b9\u03b5\u03c2 (\u03c4\u03bf \u03ad\u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c3\u03c5\u03c3\u03c3\u03ce\u03c1\u03b5\u03c5\u03c3\u03b7\u03c2, \u03c4\u03bf \u03ac\u03bb\u03bb\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bb\u03b7\u03b8\u03c5\u03ba\u03cc\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2) \u03ba\u03b1\u03b9 \u03b1\u03bd \u03b4\u03b5\u03bd \u03b5\u03af\u03c3\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c3\u03b5\u03ba\u03c4\u03b9\u03ba\u03cc\u03c2 \u03c3\u03c4\u03b7 \u03c7\u03c1\u03ae\u03c3\u03b7 \u03c4\u03bf\u03c5\u03c2 \u03bf\u03b4\u03b7\u03b3\u03b5\u03af\u03c3\u03b1\u03b9 \u03c3\u03b5 \u03c0\u03bf\u03bd\u03bf\u03ba\u03b5\u03c6\u03ac\u03bb\u03bf\u03c5\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03b1\u03c1\u03b5\u03c1\u03bc\u03b7\u03bd\u03b5\u03af\u03b5\u03c2 ;)\r\n\r\nCheerio,\r\n\r\nDurandal 1707" } { "Tag": [ "MATHCOUNTS", "calculus", "integration" ], "Problem": "HOw many times does the digit \"4\" appear in the integer part of the quotient 10^100/7\r\n\r\n\r\ncalculators allowed", "Solution_1": "i got: [hide]16[/hide]", "Solution_2": "would that be (10^100)/7 or 10^(100/7) because one's a heck of a lot easier than the other...", "Solution_3": "(10^100)/7", "Solution_4": "my bad\n\nrevised answer: [hide]17[/hide]", "Solution_5": "show solution please. in spoiler", "Solution_6": "Ok. SOlution:\n\n[hide]10^0/7=0.142857142857....\n\n\n\nNow, since we have powers of 10, the decimal is shifted.\n\n\n\nSo in every 10^(6n), there are n 4's.\n\n\n\nSo 10^96 has 16 4's.\n\n\n\nBut notice that the integer part of 10^96 ends in a 7. This means that 10^98 has a 4 as its last digit.\n\n\n\nSO 10^100 has 16+1=17 fours[/hide]", "Solution_7": "I believe that is correct. Let's try something else. What is the sum of the digits in the integral part of that expression?", "Solution_8": "ceil(100/7) ... 15?" } { "Tag": [ "abstract algebra", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $K$ a finite extension of $Q$ and $A$ the set of algebraic integers in $K$. Prove that $A$ is a free abelian group of rank $[K:Q]$.", "Solution_1": "I know one proof of this is to show that the ring of integers of the Galois extension of K is finitely generated, which means that the ring of integers of K, which is a subset of that, is also finitely generated, and therefore forms a finitely generated torsion free z-module, meaning that it's additivie group is a free abelian group. Not sure if there is anything more basic.", "Solution_2": "it is classical because i don't want to copy the proof and being too lazy to type solution you can find it in this pdf (theorem 41 p.35)\r\nhttp://www.math.sc.edu/~filaseta/gradcourses/TheMath784Notes.pdf" } { "Tag": [ "inequalities", "quadratics", "analytic geometry", "inequalities proposed" ], "Problem": "The real numbers $a,b,c$ with $bc\\neq0$ satisfy $\\frac{1-c^2}{bc}\\geq0.$ Prove that $10(a^2+b^2+c^2-bc^3)\\geq2ab+5ac.$", "Solution_1": "[quote=\"nickolas\"]The real numbers $a,b,c$ with $bc\\neq0$ satisfy $\\frac{1-c^2}{bc}\\geq0.$ Prove that $10(a^2+b^2+c^2-bc^3)\\geq2ab+5ac.$[/quote]\r\nIf $bc<0$ then the result is obvious,else \r\n$c^2\\leq 1\\Rightarrow 10(a^2+b^2+c^2-bc^3)\\ge 10a^2+10b^2+10c^2-10bc\\ge 2ab+9a^2+9b^2+10c^2-10bc\\ge 2ab+9a^2+(10-\\frac{25}{9})c^2> 2ab+9a^2+7c^2> 2ab+5ac$", "Solution_2": "hehe,I think use $bc(1-c^2) \\geq 0$ will be a little easier. :)", "Solution_3": "[quote=\"nickolas\"]The real numbers $ a,b,c$ with $ bc\\neq0$ satisfy $ \\frac{1-c^{2}}{bc}\\geq0.$ Prove that $ 10(a^{2}+b^{2}+c^{2}-bc^{3})\\geq2ab+5ac.$[/quote]\r\nFrom $ \\frac{1-c^{2}}{bc}\\geq0.$,we have $ bc(1-c^{2}) \\geq 0$ or $ bc-bc^{3}\\geq 0$\r\nSo we have\r\n$ 10(a^{2}+b^{2}+c^{2}-bc^{3}) =10((\\frac{a^{2}}{2}+\\frac{b^{2}}{50})+(\\frac{a^{2}}{2}+\\frac{c^{2}}{8})+(\\frac{b^{2}}{2}+\\frac{c^{2}}{2}-bc^{3})+\\frac{24b^{2}}{50}+\\frac{3b^{2}}{8})\\geq 2ab+5ac$", "Solution_4": "Can it be made stronger?\r\nSubtract $ 10a^2$ from both sides, then the right side is a quadratic, since no restriction is plaed on a, we can assume a is the x-coordinate of the vertex, which is $ \\frac{2b\\plus{}5c}{20}$. Substituting in, it remains to show that\r\n$ 10(b^2\\plus{}c^2\\minus{}bc^3) \\ge \\frac{(2b\\plus{}5c)^2}{40}$, or\r\n$ 396b^2\\plus{}375c^2 \\ge 400 bc^3\\plus{}20bc$\r\nBut as others have shown, the given implies $ bc(1\\minus{}c^2) \\ge 0$, or $ bc \\ge bc^3$. So we have to show\r\n$ 396b^2\\plus{}375c^2 \\ge 420 bc$\r\nBut by AM-GM, $ 396b^2\\plus{}375c^2 \\ge 770.713954 bc$. Is this correct?", "Solution_5": "[quote=\"nickolas\"]The real numbers $ a,b,c$ with $ bc\\neq0$ satisfy $ \\frac {1 - c^2}{bc}\\geq0.$ Prove that $ 10(a^2 + b^2 + c^2 - bc^3)\\geq2ab + 5ac.$[/quote]\r\n\r\n$ \\frac{1-c^2}{bc}\\geq 0\\Longleftrightarrow -bc^3\\geq -bc$ \r\n\r\n$ 10( \\sum a^2-bc^3)\\geq 10( \\sum a^2-bc)$ So now we need to prove that $ 10( \\sum a^2-bc)\\geq 2ab+5ac\\Longleftrightarrow 10\\sum a^2\\geq 2ab+10bc+5ac\\Longleftrightarrow 30\\sum a^2\\geq 3( 2ab+10bc+5ac)$.\r\n\r\nBut from Cauchy we have that $ 30\\sum a^{2}\\geq ( a+2b+5c)^{2}$. Thus we need to prove that $ ( a+2b+5c)^{2}\\geq 6ab+30bc+15ac$ which is obviously true...", "Solution_6": "[quote=\"Nickolas\"]\nThe real numbers $ a,b,c$ with $ bc\\neq0$ satisfy $ \\frac {1 - c^2}{bc}\\geq0$. Prove that $ 10(a^2 + b^2 + c^2 - bc^3)\\geq2ab + 5ac.$\n[/quote]\r\n[hide=\"Click here to view solution\"]\nI used similar logic compared to Zhaobin.\n\n$ \\& 10\\left( {{a^2} + {b^2} + {c^2} - b{c^3}} \\right) \\\\\n\\& = 5\\left( {2{a^2} + 2{b^2} + 2{c^2} - 2b{c^3}} \\right) \\\\\n\\& = 5\\left[ {2{a^2} + {b^2} + {c^2} + \\left( {{b^2} + {c^2} - 2b{c^3}} \\right)} \\right]$\n\nUsing AM-GM we have $ {b^2} + {c^2} \\geq 2bc$\nsubstituting, we get -\n\n$ \\& 5\\left[ {2{a^2} + {b^2} + {c^2} + \\left( {{b^2} + {c^2} - 2b{c^3}} \\right)} \\right] \\\\\n\\& \\geq 5\\left[ {2{a^2} + {b^2} + {c^2} + \\left( {2bc - 2b{c^3}} \\right)} \\right] = 5\\left[ {2{a^2} + {b^2} + {c^2} + 2bc\\left( {1 - {c^2}} \\right)} \\right]$ ---------------------------------- $ \\left( i \\right)$\n\nbut we have $ \\frac {1 - {c^2}} {bc} \\geq 0$\n$ \\implies \\frac {\\left( {1 - {c^2}} \\right)} {bc}\\cdot{b^2}{c^2} \\geq 0$ -----[$ \\because b^2c^2 > 0$]\n\n$ \\implies bc\\left( {1 - {c^2}} \\right) \\geq 0 \\implies 2bc\\left( {1 - {c^2}} \\right) \\geq 0$\n\nre-substituting this in $ \\left( i \\right)$ , we have -\n\n$ \\& 5\\left[ {2{a^2} + {b^2} + {c^2} + \\left( {{b^2} + {c^2} - 2b{c^3}} \\right)} \\right] \\geqslant 5\\left[ {2{a^2} + {b^2} + {c^2} + 2bc\\left( {1 - {c^2}} \\right)} \\right] \\\\\n\\& \\geq 5\\left[ {2{a^2} + {b^2} + {c^2}} \\right] \\\\\n\\& = 5\\left[ {\\left( {{a^2} + {b^2}} \\right) + \\left( {{a^2} + {c^2}} \\right)} \\right] \\\\\n\\& \\geq 5\\left[ {2ab + 2ac} \\right] = 10ab + 10ac$\n\n$ \\implies 5\\left[ {2{a^2} + {b^2} + {c^2} + \\left( {{b^2} + {c^2} - 2b{c^3}} \\right)} \\right] \\geq 10ab + 10ac$\nand obviously,\n\n$ 10ab + 10ac \\geq 2ab + 5ac$\n[b]Note : the equality stands for the case when $ a = 0$[/b]\n\n$ \\therefore 10\\left( {{a^2} + {b^2} + {c^2} - b{c^3}} \\right) \\geq 2ab + 5ac$\n[b]\n[Q.E.D][/b]\n\n[/hide]" } { "Tag": [ "calculus" ], "Problem": "Is it true that lefties are better at math and logic and stuff than righties?\r\nI am a right hander I've happened to notice that each time someone beats me at a competition or a math exam or something they're always lefties...is there some sort of connection?\r\nIf yes, then what about ambidextrous people?? :lol:", "Solution_1": "Ambidextrous people are extremely lucky in the fact that if one hand is broken, they can use the other in a competition. (However, most ambidextrous people pretend they're not during a huge final exam :D when they can't write). \r\n\r\nI am left-handed and I believe it started when my father was teaching me how to write when I was young and he sat across from me. Naturally when he wrote with his right hand, my brain interpreted that I am supposed to write with my left. I guess it caught on over the years :wink: . My right hand-writing capabilities are okay but I need to work on speed and accuracy.", "Solution_2": "[quote=\"7h3.D3m0n.117\"]I am left-handed and I believe it started when my father was teaching me how to write when I was young and he sat across from me. Naturally when he wrote with his right hand, my brain interpreted that I am supposed to write with my left. I guess it caught on over the years :wink: . My right hand-writing capabilities are okay but I need to work on speed and accuracy.[/quote]\r\nHandedness is usually determined by birth, I thought. (I don't know if it's genetic or part of early brain development.) It's unlikely that you \"chose\" it by observing your father. It is possible to force someone to make the \"wrong\" hand dominant, but this will cause them to be relatively clumsy. It was not unusual for this to happen even as recently as within the past century. My grandfather was left-handed, but that was seen as bad so he was forced to do everything with his right hand.", "Solution_3": "[quote=\"Xevarion\"]Handedness is usually determined by birth, I thought. (I don't know if it's genetic or part of early brain development.) It's unlikely that you \"chose\" it by observing your father. [/quote]\r\n\r\nYeah I know this isn't true but for me its the earliest memory I have of myself and I remember observing my father writing with his \"left\" hand.", "Solution_4": "I'm ambidextrous but mainly use my right hand...and I know loadz of math whizzes who are lefties...more than the number whom I know are righties...I don't think there's any connection between doing well and hand...it doesn't matter which hand you use, but what you write with it :) \r\nI feel it's all in the brain... But if there's any bio-fan here then maybe you could just confirm...this is one subject I suck at so... :P", "Solution_5": "Now that you mention it, the topper in our class (an IMO bronzie) is a leftie, and he in the first two sems he has amassed around 980 out of 1000.\r\n\r\nBut no, if you seriously want to know it, I can assure you there is no connection. My intuition as a statistician tells me so!\r\n\r\nOh and that reminds me:\r\nQ: What do you call somebody who puts his right hadn into a monster?\r\nA: A leftie.", "Solution_6": "[quote=\"Xevarion\"][quote=\"7h3.D3m0n.117\"]I am left-handed and I believe it started when my father was teaching me how to write when I was young and he sat across from me. Naturally when he wrote with his right hand, my brain interpreted that I am supposed to write with my left. I guess it caught on over the years :wink: . My right hand-writing capabilities are okay but I need to work on speed and accuracy.[/quote]\nHandedness is usually determined by birth, I thought. (I don't know if it's genetic or part of early brain development.) It's unlikely that you \"chose\" it by observing your father. It is possible to force someone to make the \"wrong\" hand dominant, but this will cause them to be relatively clumsy. It was not unusual for this to happen even as recently as within the past century. My grandfather was left-handed, but that was seen as bad so he was forced to do everything with his right hand.[/quote]\r\n\r\nCenturies ago, they used to think left-handed people were witches. Hence the Latin \"sinistra\" for left, from which we derive \"sinister\".", "Solution_7": "I am left-handed when it comes to writing, eating, etc. but I am right-handed when it comes to playing sports, etc. However, I am not ambidextrous...\r\n\r\nI have a book about left-handed people... it does say that there are more neural connections between the two hemispheres of the brain.", "Solution_8": "I am right-handed. I do everything with it, writing, cooking, eating. I can do math with my left hand, but you can barely read it.\r\n\r\nI heard that which hand you are dominant with is on the same side as your dominant leg.", "Solution_9": "This year, there are three lefites in B&H IMO team :lol: \r\n\r\nThe theories of \"lefties' intelectual supremacy :D \" don't have scientific arguments on their side... But on the other hand, it is scientificaly proven that in a number of sports (tennis and ping-pong, for instance) lefties have an advantage (correct me if I'm wrong)... :) \r\n\r\nBTW, I'm a leftie.", "Solution_10": "I seem to recall an abnormally large number of lefties at MOP last year......not sure if it matters, but statistically....", "Solution_11": "[quote=\"hsiljak\"]But on the other hand, it is scientificaly proven that in a number of sports (tennis and ping-pong, for instance) lefties have an advantage (correct me if I'm wrong)... :) [/quote]\r\n\r\nI'm not sure what \"scientically proven\" means, but there's an obvious simple explanation: most people, and thus most players, are right-handed. Thus, most games any given player plays will be against right-handed people. In a game where handedness matters (of which tennis and ping-pong are surely examples, due to the distinction between forehands and backhands), a left-handed player playing a right-handed player will be used to that configuration while the right-handed player will be unused to it.\r\n\r\nBaseball (pitcher vs. hitter) is another famous example of this phenomenon, at least in America. However, left-handed players have other problems -- picking off players at first base requires a more difficult motion, and lefties can't play third base.\r\n\r\nOne would expect that other sports (e.g. soccer and American football) should not exhibit much leftie-rightie separation (although perhaps left-footed players are more likely to play on the left wing?).", "Solution_12": "Familiarity isn't that much of the difference between left-handers and right-handers in baseball; the \"platoon\" advantage is mostly about how typical pitches curve. Most right-handed pitchers throw pitches which tend to break toward the first base side, which is toward a left-handed hitter and away from a right-handed hitter. Left-handed pitchers throw pitches which break in the opposite direction. It is generally easier to hit pitches that are curving toward you than pitchers that curve the other way.\r\nThe size of the platoon advantage is very consistent across hitters. It's not for pitchers- some pitchers (such as Randy Johnson) have very large advantages against their own handedness, while some others have essentially none or even get the opposite handedness out better.\r\nThis also leads to the phenomenon of batters switch hitting, or even batting left-handed full time while throwing right-handed.\r\n\r\nAs for handedness on defense: being left-handed is a tremendous disadvantage at second base, shortstop, and third base, to the point where no left-handers play those positions at the major league level. At first base, being left-handed is generally an advantage- some of the hardest plays for a first baseman involve throws to second base, which are hard for a right-hander. Most teams choose a first baseman for his hitting, so there are still plenty of right-handers there. Being left-handed probably hurts a pitcher's fielding, but helps with holding runners on and picking them off. This is of course mostly ignored, since pitching left-handed is such a useful specialty. Handedness doesn't matter much in the outfield, so there are a lot of lefties chosen for their hitting.\r\n\r\nOther sports have more subtle effects. In basketball, being opposite-handed to your counterpart is good on defense and equally bad on offense. This leaves left-handers at about the same proportion as the general population, but likely to be defensive specialists.", "Solution_13": "You did not mention catchers: they are also right-handed, right?\r\n\r\nLefties should have an option of playing in a league where they run bases clockwise. :lol:", "Solution_14": "[quote=\"mlok\"]You did not mention catchers: they are also right-handed, right?\n\nLefties should have an option of playing in a league where they run bases clockwise. :lol:[/quote]\r\n\r\nYes, catchers are right-handed, i play catcher, and have been for the bast 8 years or so. :P \r\ni mean, you can catch and be left handed, but at the competitive level (high school and beyond...nearly all catchers are right handed, not all), along with first basemen being lefties.\r\n\r\nthe reason for this though, is not like because you can catch better if your a rightie, but it is because the majority of batters are right handed, and when someone steals or whatever, you want your throwing arm away from the batter; the same reason its hard to throw to third, because the batter is in the way.\r\n\r\nback on topic though--\r\ni only know one lefty who is not ambidexterous, and that is ragnarok23. so hes pretty smart.\r\n\r\non my baseball team though, there are many ambidexterous people, maybe 3, and all are...dumb. but i play on the southside/ghetto of tallahassee, so their lack of intellegence might be more of a cultural/childhood/upbringing thing, than a natural scientific thing.\r\n\r\n\r\nEDIT: someone wrote this a couple posts ago\r\n\r\n\"Baseball (pitcher vs. hitter) is another famous example of this phenomenon, at least in America. However, left-handed players have other problems -- picking off players at first base requires a more difficult motion, and lefties can't play third base. \"\r\n\r\nAre you crazy? lefties have the easiest pick-off move to first. its righties who cant do it for anything. but its all fair...righties have that same advantage when picking off to third, but no one does that much.", "Solution_15": "Running bases clockwise would change the patterns of defensive plays, making lefties preferable at third base, etc. This was the reason I brought it up -- not the baserunning part itself.", "Solution_16": "Actually, I don't think handedness makes much difference to a catcher's defensive responsibilities. Catchers are right-handed for other reasons; most importantly, lefties with the necessary arm strength and accuracy are pushed into pitching.", "Solution_17": "Hm about the math, one study showed that mathematically intellegent people were more likely to be male, left handed, and prone to allergies.", "Solution_18": "Baseball players have two types of handedness: throwing hand and batting hand. Throwing hand is right or left, batting hand can be right, left, or both. That makes for 6 possible combinations, and there have been historically great players in each of the 6 types. But here's a look at a few of those of the throws left, bats left type: Ty Cobb, Tris Speaker, Oscar Charleston, Babe Ruth, Lou Gehrig, Ted Williams, Stan Musial, Barry Bonds. The representation of that list among the greatest hitters of all time says something about the intrinsic advantages of left-handedness. (Note also that most of that list were outfielders, with the exceptions being Gehrig who played first base and Musial who divided time between first base and outfield.)", "Solution_19": "[quote=\"Kent Merryfield\"]Baseball players have two types of handedness: throwing hand and batting hand. Throwing hand is right or left, batting hand can be right, left, or both. That makes for 6 possible combinations, and there have been historically great players in each of the 6 types. But here's a look at a few of those of the throws left, bats left type: Ty Cobb, Tris Speaker, Oscar Charleston, Babe Ruth, Lou Gehrig, Ted Williams, Stan Musial, Barry Bonds. The representation of that list among the greatest hitters of all time says something about the intrinsic advantages of left-handedness. (Note also that most of that list were outfielders, with the exceptions being Gehrig who played first base and Musial who divided time between first base and outfield.)[/quote]\r\n\r\nthat could be misleading though, because it is easiest to hit your opposite handed pitcher, and since there are more right handed pitchers, it is an advantage to be a left-handed hitter in that matter.\r\n\r\nbut then again, a lefty-lefty matchup is much tougher than righty-righty.", "Solution_20": "[quote]because it is easiest to hit your opposite handed pitcher, and since there are more right handed pitchers, it is an advantage to be a left-handed hitter in that matter. [/quote]\r\nThat's exactly what I meant by \"intrinsic advantage of left-handedness.\"", "Solution_21": "When proctoring calculus exams, I take notice of who writes with which hand (there is not much else to do anyway). So far, I have not seen any clear correlation between handedness and test scores, except for the following: the students who hold their head with both hands for long periods of time (making it difficult to determine their handedness) usually do poorly.", "Solution_22": "[quote=\"Kent Merryfield\"][quote]because it is easiest to hit your opposite handed pitcher, and since there are more right handed pitchers, it is an advantage to be a left-handed hitter in that matter. [/quote]\nThat's exactly what I meant by \"intrinsic advantage of left-handedness.\"[/quote]\r\n\r\nLets remember this is a math forum, i get D's in English :D\r\n\r\ni thought you meant that they were perhaps better just because they were left handed, i.e. left handed people have better hand-eye or something of that nature.\r\n\r\nsorry about that... :oops:", "Solution_23": "I'm a leftie.\r\nI know people who have still lived under the ruthless oppressive regime by the righties. :mad: Back in the days when my grandparents grew up, a kid would be hit on his left hand if he dared to use it to write.\r\n\r\nBut the shameless discrimination still continues. One-person-tables are often only designed for righties. Classrooms always have the blackboards on that side such that the window is on the left hand side for the pupils.\r\n\r\n\r\nLefties around the world, unite! :!:", "Solution_24": "Does not it make sense to have blackboard on the side that is convenient for most pupils? You can't have both. \r\nFor better or for worse, I teach in windowless rooms. It would be nice to have some chairs designed for lefties, though.", "Solution_25": "Hmm... as I inventory in my head the classrooms I often teach in, it seems there are about as many with windows on the right as there are with windows on the left. And there are a couple of lefty desks in each room, although the left-handed students may or may not wind up sitting in them. Also, if you're in a room with left-side windows and stuck with a righty desk, don't you partially turn in your chair so that your back is to the window?", "Solution_26": "Considering that Daniel Li is a leftie and probably the smartest red MOPper this year, and I'm a rightie and perhaps the dumbest red MOPper this year, I'd say there's some correlation.", "Solution_27": "[quote=\"Treething\"]Considering that Daniel Li is a leftie and probably the smartest red MOPper this year, and I'm a rightie and perhaps the dumbest red MOPper this year, I'd say there's some correlation.[/quote]\r\n\r\nJust as a sidenote, I know Jacob Steinhardt is a lefty, but I don't remember much about the other MOPpers...", "Solution_28": "On the other hand, David Xiao was a lefty, and I'd rank him around my level at MOP.", "Solution_29": "[quote=\"Klebian\"][b]On the other hand[/b], David Xiao was a lefty, and I'd rank him around my level at MOP.[/quote]\r\n\r\nWorst pun ever.", "Solution_30": "Right-handed and I am useless at everything I do. I know the top student at my school is a leftie.", "Solution_31": "is ambidexterousness something one teaches themselves to do? or something you happen to have by birth?", "Solution_32": "[quote=\"sonny\"]is ambidexterousness something one teaches themselves to do? or something you happen to have by birth?[/quote]\r\nI don't think there's complete agreement on this in the scientific community, but I'd say it's leaning more towards birth than learning. At least I know that if you start too late, you can force yourself to learn everything with the off hand but it will still remain your off hand. So if it is learned, that happens very early (first couple of years probably).", "Solution_33": "I'm sort of ambidextrous, as I can do a lot of things with both hands, but my left hand is slightly weaker than my right.\r\n\r\nI have a friend does everything with his left hand, but he writes with his right.", "Solution_34": "I'm left-handed, but I've adapted to using right-handed mice. I think my grandma was born left-handed, but then was forced to turn into a right-handed person. Its really funny how my Chinese teachers (I don't take Chinese anymore, it was SO boring!) from a Chinese school called Yellow River Chinese school used to complain how I wrote the strokes in the wrong direction because I'm left handed. Some people think of being left-handed as having a disability. One parent was actually amazed that I could use chopsticks :furious: .", "Solution_35": "[quote=\"Nerd_of_the_Ages\"]One parent was actually amazed that I could use chopsticks :furious: .[/quote]\r\n\r\nChopsticks are an interesting challenge for lefties...I avoided them for a while because it was difficult to use (not to mention your family members staring at you when you mess up). \r\n\r\nHmph another thing against lefties: when writing gigantic essays, normally your left hand would smear across the paper as you write from left to right. Every time I take Cornell notes for AP bio it always smears :mad: and I can barely read a thing.", "Solution_36": "I'm a leftie, and I absolutely HATE having \"half-desks\" made in favor of right-handed people. I wish they'd just make \"full-size\" desks. When I took my SAT Subject Tests, there were 3 lefties, but only 2 leftie desks. I was the odd one out. :furious: \r\n\r\nBut, I've adapted somewhat to a right-handed world. Although, it is annoying when I have sit next to someone and eat at a restaurant... Oh, well.\r\n\r\nI read somewhere that before long-lasting tools were made (around the Bronze Age, when tools were made more durable), about half of the people were righties and about half were lefties. However, once tools could be made that would last several generations, people had to choose a handedness; apparently, right-handedness won out. I wonder why?\r\n\r\nMy dad is a leftie, but he has adjusted to a right-handed world, like me. We are the only two lefties in my immediate family. However, my younger brother, who is a rightie for most things, does many things in sports left-handed. Go figure. I wonder why that is? It's curious, to say the least.", "Solution_37": "[quote=\"Zakary\"]However, my younger brother, who is a rightie for most things, does many things in sports left-handed. Go figure. I wonder why that is? [/quote]\r\n\r\nWell, Rafael Nadal (number 2 seed tennis player) is right handed but plays a leftie because his forehand was ironically stronger on his left hand.", "Solution_38": "i dont think there is a very big relation between IQ and handedness.\r\npeople who are left handed arent automatically geniuses, and people who are right handed arent automatically dumb." } { "Tag": [ "puzzles" ], "Problem": "http://www.clickmazes.com its a great site and has a bunch of good games my favorite is 3-some maze in in the 2nd from the top on the right. btw to get more mazes for each one look at the top of the screen.", "Solution_1": "the tilt mazes are fun. also there is another site with similar puzzles\r\nhttp://www.logicmazes.com/theseus.html" } { "Tag": [ "logarithms", "function", "number theory unsolved", "number theory" ], "Problem": "lets say i have the following simple maths division\r\n\r\na = b / c \r\n\r\nnow is there a formula to test the if the result in \"a\" has \r\n\r\n1) a fix number of digits \r\n\r\ne.g. a = 2, 2 digit answer\r\n or a = 123, 3 digit answer\r\n or a = 12345, 5 digit answer\r\n\r\n2) a unlimited number of digits\r\n\r\ne.g. a = 1.12312312312312312312402389403289402349803985034985034589.......\r\n or a = 6.23409823409823409823049823048920398420398402398402384023894..... \r\n\r\n\r\nnow, if I have to write a C procedure called Checknum()\r\n\r\nb = Checknum( a );\r\n\r\nif a is of a fix digit answer, eg in example 1) above, Checknum() will return into b the number of digits in a\r\n\r\nif a is of an infinite number of digits, eg in example 2) above, Checknum() will return into b the number '0'\r\n\r\nnow the problem is how do i write the procedure Checknum()\r\n\r\n\r\nint Checknum( int number )\r\n{\r\n ???\r\n}\r\n\r\nThank you for your time.", "Solution_1": "One could do it as follows:\r\n\r\n(i) Loop until $2 \\not| c$, replace $c$ by $\\frac{c}{2}$ at each step.\r\n(ii) Loop until $5 \\not| c$, replace $c$ by $\\frac{c}{5}$ at each step.\r\n(iii) Now if $c \\not| b$, return $0$, otherwise return the floor of $\\log_{10} \\frac{b}{c} +1$.", "Solution_2": "[quote=\"Fiachra\"]One could do it as follows:\n\n(i) Loop until $2 \\not| c$, replace $c$ by $\\frac{c}{2}$ at each step.\n(ii) Loop until $5 \\not| c$, replace $c$ by $\\frac{c}{5}$ at each step.\n(iii) Now if $c \\not| b$, return $0$, otherwise return the floor of $\\log_{10} \\frac{b}{c} +1$.[/quote]\r\n\r\ncan someone please kindly explain this in plain simple english ???\r\n\r\nmy maths is very poor, but i am learning ...\r\n\r\nthank you so much for your time...", "Solution_3": "it is very simple \r\n\r\nThe formula is this >>> a = b / c \r\n\r\nExample (1) \r\n\r\na = 9999 / 13\r\n\r\nmeans 'a' = 769.153846153846840343495345345........ to infinity\r\n\r\na has an infinite digit answer\r\n\r\nExample (2) \r\n\r\na = 6 / 2\r\n\r\nmeans 'a' = 3\r\n\r\na has a 1 digit answer\r\n\r\nNow how do I test if a number after division has an infinite number of digits as in Example (1) and how do i test if a number after division has a set number of digits as in Example (2) \r\n\r\nThank you so much for your time\r\n\r\nAndy", "Solution_4": "[quote=\"vendetta\"]\ncan someone please kindly explain this in plain simple english ???\n\nmy maths is very poor, but i am learning ...\n\nthank you so much for your time...[/quote]\r\n\r\nThe loop is an ordinary programming loop (e.g. a for loop). $5|c$ means that 5 divides c, and $5 \\not| c$ means that it does not. Basically we eliminate the factors 2 and 5, and then check for divisibility. If $c$ divides in evenly, $a$ has a finite number of digits.\r\n\r\nBy the way, I assumed above that $b$ and $c$ were integers. If not, but they have a finite number of digits, then we could multiply by a sufficiently large power of ten. If they have infinitely many digits, one cannot store them directly on a computer.", "Solution_5": "[quote=\"Fiachra\"][quote=\"vendetta\"]\ncan someone please kindly explain this in plain simple english ???\n\nmy maths is very poor, but i am learning ...\n\nthank you so much for your time...[/quote]\n\nThe loop is an ordinary programming loop (e.g. a for loop). $5|c$ means that 5 divides c, and $5 \\not| c$ means that it does not. Basically we eliminate the factors 2 and 5, and then check for divisibility. If $c$ divides in evenly, $a$ has a finite number of digits.\n\nBy the way, I assumed above that $b$ and $c$ were integers. If not, but they have a finite number of digits, then we could multiply by a sufficiently large power of ten. If they have infinitely many digits, one cannot store them directly on a computer.[/quote]\r\n\r\nThank you so much for taking the time to explain everything to me.\r\n\r\nHowever, I still do not quite understand you.\r\n\r\nI was wondering if you would like to spend a few minutes to kindly write the C code for this function ???\r\n\r\nI really will appreciate it very much.\r\n\r\nThank you so much for your time and effort\r\n\r\nYour the BEST !\r\n\r\nAndy" } { "Tag": [ "geometry", "perimeter" ], "Problem": "What is the perimeter of a right-angled triangle if one of its sides is of length $ 47$ and the other two sides have integer lengths?", "Solution_1": "hello, and what else is given?\r\nSonnhard.", "Solution_2": "is 47 the lenght of the hypotenuse?", "Solution_3": "Sorry Sonnhard (and Tiger100: just seen you post), I should have said that the sides have integer lengths. I'll edit it in a moment or two.\r\n\r\nI used $ (m^{2} \\plus{} n^{2}), (2mn)$ and $ (m^{2} n^{2})$ to get $ m^{2} \\minus{} n^{2} \\equal{} (m \\plus{} n)(m \\minus{} n) \\equal{} 1\\times 47$.\r\n\r\nThen $ m \\equal{} 24, n \\equal{} 23$ and so the sides are $ 47, 1104$ and $ 1105$ and the perimeter is $ 47 \\plus{} 1104 \\plus{} 1105 \\equal{} 2256$.", "Solution_4": "[hide] Generally, suppose that $ k$ is an odd positive integer. Then, there is always a Pythagorean triple of the form $ k$, $ \\frac{k^2\\minus{}1}{2}$, $ \\frac{k^2\\plus{}1}{2}$. These obviously are all integers. Solving this general case for the perimeter, we have \n\n$ k\\plus{}\\frac{k^2\\minus{}1}{2}\\plus{}\\frac{k^2\\plus{}1}{2}$\n\n$ \\equal{}k\\plus{}\\frac{2k^2}{2}$\n\n$ \\equal{}k\\plus{}k^2$.\n\nPlugging in the specific case $ k\\equal{}47$, we have $ 47\\plus{}47^2\\equal{}47\\plus{}2209\\equal{}\\boxed{2256}$.[/hide]" } { "Tag": [ "trigonometry" ], "Problem": "Let $ x,y,z$ be pairwise distinct real numbers such that $ k\\equal{}\\dfrac{1\\plus{}xy}{x\\minus{}y},$ $ l\\equal{}\\dfrac{1\\plus{}yz}{y\\minus{}z}$ and $ m\\equal{}\\dfrac{1\\plus{}zx}{z\\minus{}x}$ are integers. Prove that $ k,l$ and $ m$ are pairwise relatively prime.", "Solution_1": "Nice problem!\r\n[hide]\nLetting $ x,y,z\\equal{}\\tan A,\\tan B,\\tan C$, respectively gives $ k,l,m\\equal{}\\cot(A\\minus{}B),\\cot(B\\minus{}C),\\cot(C\\minus{}A)$. We have $ A\\minus{}B\\plus{}B\\minus{}C\\plus{}C\\minus{}A\\equal{}0$ obviously. It is well known that for three angles which add up to a muliple of $ \\pi$, the sum of the pairwise products of their cotangents is equal to $ 1$. That is, we have $ kl\\plus{}lm\\plus{}mk\\equal{}1$. \n\nNow suppose that the claim in the problem was not true. Without loss of generality, assume $ t|k,t|l$ for $ t>1$. Then $ t|kl\\plus{}lm\\plus{}mk$. Contradiction![/hide]", "Solution_2": "[quote=cosinator]Nice problem!\n[hide]\nLetting $ x,y,z\\equal{}\\tan A,\\tan B,\\tan C$, respectively gives $ k,l,m\\equal{}\\cot(A\\minus{}B),\\cot(B\\minus{}C),\\cot(C\\minus{}A)$. We have $ A\\minus{}B\\plus{}B\\minus{}C\\plus{}C\\minus{}A\\equal{}0$ obviously. It is well known that for three angles which add up to a muliple of $ \\pi$, the sum of the pairwise products of their cotangents is equal to $ 1$. That is, we have $ kl\\plus{}lm\\plus{}mk\\equal{}1$. \n\nNow suppose that the claim in the problem was not true. Without loss of generality, assume $ t|k,t|l$ for $ t>1$. Then $ t|kl\\plus{}lm\\plus{}mk$. Contradiction![/hide][/quote]\n\nVery nice solution\n" } { "Tag": [ "trigonometry", "calculus", "integration", "function", "algebra", "polynomial", "vector" ], "Problem": "If I raise $a cis b$ to a power, should I reduce b (the theta) to be between 0 and 360? I am confused on this aspect, as if you raise $a cis b$ to a power, and then simplify the theta, wouldn't raising the new complex number to the recipricol power not yield the original theta?\r\n\r\nFurthermore, if I have say $\\sin\\frac{3}{8}\\theta = 1$, then I have $\\frac{3}{8}\\theta = 90^\\circ$ \r\n\r\nSo I get $\\theta = \\frac{8\\cdot 90^\\circ}{3}$, $\\frac{8\\cdot 90^\\circ}{3} + \\frac{8\\cdot360^\\circ}{3}$, $\\frac{8\\cdot 90}{3} + \\frac{8\\cdot2\\cdot 360^\\circ}{3}$\r\n\r\nI don't think it makes sense to reduce the angles to be an angle from between 0 and 360 because then if I multiply the reduced angles by $\\frac{3}{8}$, they will not be equal to $90^\\circ$?", "Solution_1": "just because the angles are not equal, does not mean that they have different sines", "Solution_2": "Okay, I gotcha.\r\n\r\nBut I am confused on this regard:\r\n\r\n$(r^{\\frac{2}{5}}$ $cis$ $\\frac{2\\theta}{5})^{\\frac{5}{2}}$ is not equal to $r$ $cis$ $\\theta$\r\n\r\nIt appears to be actually $5 \\cdot 2 = 10$ different values.\r\n\r\nDoes raising the multiple values of a power of a complex number to its recipricol power only yield the original complex number if it is an integral power and not a fractional power?", "Solution_3": "[quote=\"breez\"]Okay, I gotcha.\n\nBut I am confused on this regard:\n\n$(r^{\\frac{2}{5}}$ $cis$ $\\frac{2\\theta}{5})^{\\frac{5}{2}}$ is not equal to $r$ $cis$ $\\theta$\n\nIt appears to be actually $5 \\cdot 2 = 10$ different values.\n\nDoes raising the multiple values of a power of a complex number to its recipricol power only yield the original complex number if it is an integral power and not a fractional power?[/quote]\r\n\r\nThe square root function only takes the principal value by definition, so no, it doesn't take on ten values - it isn't $cis$ that takes on multiple values, it's $\\theta$, and this is only because every angle takes on the same $\\sin, \\cos$ angles as all of the angles that are congruent to it $\\bmod 2\\pi$ (as much as non-integer mods make sense).\r\n\r\n$\\sin \\theta = 1$ has one solution $0 \\le \\theta < 2\\pi$, but it has infinite solutions over the reals, as does $\\sin \\frac{3\\theta}{8} = 1$. So you get\r\n\r\n$\\frac{3\\theta}{8} = \\frac{\\pi}{2} + 2 \\pi k$\r\n$\\theta = \\frac{4\\pi}{3} + \\frac{16 \\pi k}{3}$,\r\n\r\nWhich in fact also only has one solution $0 \\le \\theta < 2\\pi$, given by $k = 0$.", "Solution_4": "Then it would be equal to 2 values correct? (My equation from above involving the recipricol powers)", "Solution_5": "I don't know where to post this, but does anyone have a clear explanation of the idea behind cis and the complex plane and powers $e$ etc.?\r\nI can perform all the calculations, but what does it really mean? How did we hatch such concepts?", "Solution_6": "[quote=\"me@home\"]I don't know where to post this, but does anyone have a clear explanation of the idea behind cis and the complex plane and powers $e$ etc.?\nI can perform all the calculations, but what does it really mean? How did we hatch such concepts?[/quote]\r\n\r\nDe Moivre came up with the form that didn't involve $e^{i\\theta}$, but Euler... I'm pretty sure he came up with it by messing around with Taylor series. (Short explanation: You can write $e^x, \\sin x, \\cos x$ as infinite polynomials, and then Euler's formula is true because the coefficients of the infinite polynomials of both sides are the same.)\r\n\r\nJust take for granted that mathematically, $e^{i \\theta} = \\cos \\theta + i \\sin \\theta$ is consistent, and ends up producing a vector of length $1$ and angle $\\theta$ in the complex plane. And everything else follows. :)\r\n\r\nbreez: $x^{ \\frac{1}{2} }$ only returns one value.", "Solution_7": "i hate using something I don't completely understand... oh well thanks! :(", "Solution_8": "But say your solving an equation, wouldn't you return all possible values? Then it would be 2 right?", "Solution_9": "[quote=\"breez\"]But say your solving an equation, wouldn't you return all possible values? Then it would be 2 right?[/quote]\r\n\r\nIf you're solving the equation $\\sqrt{9} = x$ then the answer is $x = 3$ because $f(x) = \\sqrt{x}$ is defined to return only the positive square root. If you're solving the equation $9 = x^2$ then the answer is $x = 3, -3$ because both are roots of the equation. \r\n\r\nIn your example, $\\left( r^{ \\frac{2}{5} } \\, cis \\, \\theta \\right)^{ \\frac{5}{2} }$ is a definite quantity with a definite value for any given $\\theta$ (one definite value), as is $r \\, cis \\, \\theta$." } { "Tag": [ "trigonometry", "inequalities" ], "Problem": "If A,B,C and D are angles of a quadilateral and sin A/2.sinB/2.sinC/2.sinD/2=1/4,prove that A=B=C=D=/2", "Solution_1": "A=B=C=D=/2 doesn't exactly make sense. \r\n\r\nDo you mean that ${ \\sin{\\frac{A}{2}}\\sin{\\frac{B}{2}}\\sin{\\frac{C}{2}}\\sin{\\frac{D}{2}} = \\frac{1}{4}}$?", "Solution_2": "Fix:\r\nIf $ A,B,C,D$ are angles of a quadrilateral and $ \\sin\\frac {A}2\\sin\\frac {B}2\\sin\\frac {C}2\\sin\\frac {D}2 \\equal{} \\frac14$, prove that $ A \\equal{} B \\equal{} C \\equal{} D \\equal{} \\frac {\\pi}2$.", "Solution_3": "Yes not /2 but pi/2", "Solution_4": "From Jensen's Inequality, \r\n\r\nsince $ \\frac{A}{2} \\plus{} \\frac{B}{2} \\plus{} \\frac{C}{2} \\plus{} \\frac{D}{2} \\equal{} \\pi$\r\n\r\n$ \\frac{\\sin \\frac{A}{2} \\plus{} \\sin \\frac{B}{2} \\plus{} \\sin \\frac{C}{2} \\plus{} \\sin \\frac{D}{2}}{4} \\le \\sin \\frac{\\pi}{4} \\equal{} \\frac{1}{\\sqrt 2}$\r\n\r\nHence, from AM-GM Inequality\r\n\r\n$ \\left (\\sin \\frac{A}{2} \\sin \\frac{B}{2} \\sin \\frac{C}{2} \\sin \\frac{D}{2} \\right )^\\frac{1}{4} \\le \\frac{1}{\\sqrt 2}$\r\n\r\nor $ \\sin \\frac{A}{2} \\sin \\frac{B}{2} \\sin \\frac{C}{2} \\sin \\frac{D}{2} \\le \\frac{1}{4}$\r\n\r\nSince, here equality holds, $ A \\equal{} B \\equal{} C \\equal{} D \\equal{} \\frac{\\pi}{2}$" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "$a, b, c \\in R$ and $a+b+c = 1$, prove\r\nthat $15(a^{2}+b^{2}+c^{2})+54abc-7$ is non-negative number.", "Solution_1": "[quote=\"Beat\"]$a, b, c \\in R$ and $a+b+c = 1$, prove\nthat $15(a^{2}+b^{2}+c^{2})+54abc-7$ is non-negative number.[/quote]\r\n\r\nI think the idea for this problem is \r\n1) prove that $F(a,b,c)\\ge{F(\\frac{a+b}{2},\\frac{a+b}{2},c)}$\r\n2)then we must prove that $F(x,x,1-2x)\\ge0$. that is all.\r\nBut maybe it needs some long computations. :) \r\nI will think on it.", "Solution_2": "15*(a+b+c)*(a^2+b^2+c^2)+54*a*b*c-7*(a+b+c)^3\r\n=8*(a*(a-b)*(a-c)+b*(b-c)*(b-a)+c*(c-a)*(c-b))\r\n+2*((b+c)*(a-b)*(a-c)+(c+a)*(b-c)*(b-a)+(a+b)*(c-a)*(c-b))>=0.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nhttp://www.actamath.com/actamath_new/cn/search_gkll.asp?page=1&pagesize=10&sel_zazhimc=\u6570\u5b66\u5b66\u62a5&sel_niandu=2006&sel_qihao=49(3)", "Solution_3": "a,b,c>=0\r\n =>\r\n 15*(a+b+c)*(a^2+b^2+c^2)+54*a*b*c-7*(a+b+c)^3 \r\n=8*(a*(a-b)*(a-c)+b*(b-c)*(b-a)+c*(c-a)*(c-b)) \r\n+2*((b+c)*(a-b)*(a-c)+(c+a)*(b-c)*(b-a)+(a+b)*(c-a)*(c-b))>=0. \r\n\r\na=1,b=1,c=-1\r\n=>\r\n15*(a^2+b^2+c^2)+54*a*b*c-7=-16<0." } { "Tag": [ "function", "floor function", "ceiling function", "combinatorics open", "combinatorics" ], "Problem": "Let $ n>k$ be positive integers. \r\nLet $ S\\equal{}\\{1,2,3,...,n\\}$. \r\nDenote by $ s(A)$ the sum of the elements of the set $ A$.\r\nHow many $ k$_element subsets $ A$ of $ S$ satisfying $ s(A) \\equiv 0 (mod 2)$\r\n\r\n[b]Can we solve this problem by gernerating function?[/b] :maybe: Thanks.", "Solution_1": "$ A$ should have an even number of odd elements. \r\n\r\nthere are $ m\\equal{}\\lfloor \\frac{n}{2} \\rfloor$ even numbers in $ S$.\r\n\r\nthere are $ n\\equal{}\\lceil \\frac{n}{2} \\rceil$ odd numbers in $ S$.\r\n\r\nthe number of $ k$ element subsets of $ A$ that have an even sum is $ a_k\\equal{}\\binom{n}{0} \\binom{m}{k}\\plus{}\\binom{n}{2}\\binom{m}{k\\minus{}2}\\plus{}\\binom{n}{4}\\binom{m}{k\\minus{}4}\\plus{}\\cdots\\plus{}\\binom{n}{2\\lfloor \\frac{k}{2} \\rfloor}\\binom{m}{k\\minus{}2\\lfloor \\frac{k}{2} \\rfloor}$\r\n\r\nnow note that $ a_k$ is the coefficient of $ x^k$ in $ \\left(\\binom{n}{0}\\plus{}\\binom{n}{2}x^2\\plus{}\\binom{n}{4}x^4\\plus{}\\cdots \\right)\\left(\\binom{m}{0}\\plus{}\\binom{m}{1}x\\plus{}\\binom{m}{2}x^2\\plus{}\\cdots\\right)$\r\n\r\nthe second factor is simply $ (1\\plus{}x)^m$. the first is $ \\frac{1}{2}[(1\\plus{}x)^n\\plus{}(1\\minus{}x)^n]$." } { "Tag": [ "algebra" ], "Problem": "Let $ P(n)$ and $ S(n)$ denote the product and the sum, respectively, of the digits of the integer $ n$. For example, $ P(23) \\equal{} 6$ and $ S(23) \\equal{} 5$. Suppose $ N$ is a two-digit number such that $ N \\equal{} P(N) \\plus{} S(N)$. What is the units digit of $ N$?\r\n\r\n$ \\textbf{(A) }2\\qquad\\textbf{(B) }3\\qquad\\textbf{(C) }6\\qquad\\textbf{(D) }8\\qquad\\textbf{(E) }9$", "Solution_1": "[quote=\"mysmartmouth\"]Let $P(n)$ and $S(n)$ denote the product and the sum, respectively, of the digits of the integer $n$. For example, $P(23) = 6$ and $S(23) = 5$. Suppose $N$ is a two-digit number such that $N = P(N)+S(N)$. What is the units digit of $N$?\n\n$\\text{(A) }2\\qquad\\text{(B) }3\\qquad\\text{(C) }6\\qquad\\text{(D) }8\\qquad\\text{(E) }9$[/quote]\r\n\r\n[hide]Let $N=10a+b$. $10a+b=ab+a+b\\Rightarrow ab-9a=0$ so $\\boxed{b=9}$.[/hide]", "Solution_2": "[hide=\"Solution\"]Let $a$ and $b$ represent the digits of the number $ab$. So the numerical value of the number $10a+b$. The sum of the digits is $a+b$ and the product is $ab$. Now combining this with the given,\n\n$10a+b = ab+a+b$\n$ab = 9a$\n\n$b=9$\n\nSo the last digit (units' digit) is $9$. The answer is $E$. [/hide]", "Solution_3": "b-9=0\nHence b=9" } { "Tag": [ "linear algebra", "matrix", "integration", "linear algebra unsolved" ], "Problem": "Let $A,B\\in M_n(C)$ having all their eigenvalues with strictly negative real part.Let $C\\in M_n(C)$.Show that there exists $M\\in M_n(C)$ such that $AM+MB=C$.", "Solution_1": "Similar stuff has been discussed before many times. The spectrum $\\lambda(\\mathcal T)$ of the operator $\\mathcal T$ on $\\mathcal M_n(\\mathbb C)$ sending $X$ to $AX+XB$ is contained in $\\Lambda(A)+\\Lambda(B)$, which does not contain $0$, so the operator is invertible, and, in particular, onto, which is what we wanted to prove.", "Solution_2": "killer have a look in RMS n\u00b04 (2004-2005) page 77\r\n\r\n$M=-\\int_{0}^{+\\infty}e^{tA}Ce^{tB}dt$" } { "Tag": [ "trigonometry", "conics", "complex numbers" ], "Problem": "Hi\r\nI am looking for an online, self-paced precalculus course\r\nThe two I'm considering so far are EPGY M013 and CTY Honors Precalculus with Trigonometry\r\nAnybody have opinions on those two or other suggestions?\r\nThanks", "Solution_1": "[quote=\"daermon\"]Hi\nI am looking for an online, self-paced precalculus course\nThe two I'm considering so far are EPGY M013 and CTY Honors Precalculus with Trigonometry\nAnybody have opinions on those two or other suggestions?\nThanks[/quote]\r\n\r\ni've taken M013 (well i never finished it, w/e)\r\nit's okay, i guess", "Solution_2": "Eh, don't try to go too high in terms of school type classes...", "Solution_3": "[quote=\"daermon\"]Hi\nI am looking for an online, self-paced precalculus course\nThe two I'm considering so far are EPGY M013 and CTY Honors Precalculus with Trigonometry\nAnybody have opinions on those two or other suggestions?\nThanks[/quote]\r\n\r\n\r\nDO NOT TO CTY! I went there and I hated it.\r\n\r\nDo EPGY.", "Solution_4": "He's not talking about going to CTY, he's talking about taking an online CTY class.\r\n\r\nCTY's \"ACS\" Algebra 1 and 2 classes were helpful, insightful, and impressive. Their precalc class, on the other hand, was not that impressive. However, that was in 2004... it might've changed by now.\r\n\r\nIMO, precalculus really isn't worth spending the big bucks on. Get a book off amazon and figure it out yourself. Probably the only thing in precalc that's actually new is trigonometry (possibly conic sections).\r\n\r\nFrom my experience, the best way to get a good feel of trigonometry is to go on wikipedia, get a list of various identities, and try to prove them geometrically and/or algebraically (i.e. not with complex numbers -- that approach is better to save for later).\r\n\r\n[b]Note:[/b] [i]I actually recently listed the book I used for CTY Precalc -- which is a good book -- on Amazon. It's in \"pristine condition\" and it's 32 bucks :P buy, buy, buy![/i]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Given $ a,b,c \\ge 0$\r\nprovethat\r\n$ (a^3b^3\\plus{}b^3c^3\\plus{}c^3a^3)[(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)\\minus{}8abc] \\ge abc(a\\minus{}b)^2(b\\minus{}c)^2(c\\minus{}a)^2$", "Solution_1": "[quote=\"babyloverain\"]Given $ a,b,c \\ge 0$\nprovethat\n$ (a^3b^3 \\plus{} b^3c^3 \\plus{} c^3a^3)[(a \\plus{} b)(b \\plus{} c)(c \\plus{} a) \\minus{} 8abc] \\ge abc(a \\minus{} b)^2(b \\minus{} c)^2(c \\minus{} a)^2$[/quote]\r\nLet $ a\\plus{}b\\plus{}c\\equal{}3u,$ $ ab\\plus{}ac\\plus{}bc\\equal{}3v^2,$ $ abc\\equal{}w^3$ and $ u^2\\equal{}tv^2.$\r\nHence, your inequality is equivalent to $ (27v^6\\minus{}27uv^2w^3\\plus{}3w^6)(9uv^2\\minus{}9w^3)\\geq$\r\n$ \\geq27w^3(3u^2v^4\\minus{}4v^6\\plus{}6uv^2w^3\\minus{}4u^3w^3\\minus{}w^6)\\Leftrightarrow$\r\n$ \\Leftrightarrow(4u^3\\plus{}4uv^2)w^6\\minus{}(12u^2v^4\\plus{}5v^6)w^3\\plus{}9uv^8\\geq0.$\r\n$ (12u^2v^4\\plus{}5v^6)^2\\minus{}4\\cdot4(u^3\\plus{}uv^2)\\cdot9uv^8\\equal{}(25v^2\\minus{}24u^2)v^{10}.$\r\nHence, for $ t\\geq\\frac{25}{24}$ our inequality is proven.\r\nBut for $ 1\\leq t\\leq\\frac{25}{24}$ we'll prove that $ w^3\\leq\\frac{12u^2v^4\\plus{}5v^6\\minus{}v^5\\sqrt{25v^2\\minus{}24u^2}}{8(u^3\\plus{}uv^2)}.$\r\n$ (a\\minus{}b)^2(a\\minus{}c)^2(b\\minus{}c)^2\\geq0$ gives $ w^3\\leq3uv^2\\minus{}2u^3\\plus{}2\\sqrt{(u^2\\minus{}v^2)^3}.$\r\nId est, it remains to prove that\r\n$ 3uv^2\\minus{}2u^3\\plus{}2\\sqrt{(u^2\\minus{}v^2)^3}\\leq\\frac{12u^2v^4\\plus{}5v^6\\minus{}v^5\\sqrt{25v^2\\minus{}24u^2}}{8(u^3\\plus{}uv^2)}.$\r\nBut $ 3uv^2\\minus{}2u^3\\plus{}2\\sqrt{(u^2\\minus{}v^2)^3}\\leq\\frac{12u^2v^4\\plus{}5v^6\\minus{}v^5\\sqrt{25v^2\\minus{}24u^2}}{8(u^3\\plus{}uv^2)}\\Leftrightarrow$\r\n$ \\Leftrightarrow8(t\\plus{}1)\\left(3t\\minus{}2t^2\\plus{}2\\sqrt{t(t\\minus{}1)^3}\\right)\\leq12t\\plus{}5\\minus{}\\sqrt{25\\minus{}24t}\\Leftrightarrow$\r\n$ \\Leftrightarrow16t^3\\minus{}8t^2\\minus{}12t\\plus{}5\\minus{}\\sqrt{25\\minus{}24t}\\geq16(t\\plus{}1)\\sqrt{t(t\\minus{}1)^3}\\Leftrightarrow$\r\n$ \\Leftrightarrow4t^2\\plus{}2t\\minus{}1\\plus{}\\frac{6}{1\\plus{}\\sqrt{25\\minus{}24t}}\\geq4(t\\plus{}1)\\sqrt{t^2\\minus{}t}\\Leftrightarrow$\r\n$ \\Leftrightarrow\\left(4t^2\\plus{}2t\\minus{}1\\plus{}\\frac{6}{1\\plus{}\\sqrt{25\\minus{}24t}}\\right)^2\\geq\\left(4(t\\plus{}1)\\sqrt{t^2\\minus{}t}\\right)^2\\Leftrightarrow$\r\n$ \\Leftrightarrow12t^2\\plus{}12t\\plus{}1\\plus{}\\frac{12(4t^2\\plus{}2t\\minus{}1)}{1\\plus{}\\sqrt{25\\minus{}24t}}\\plus{}\\frac{36}{\\left(1\\plus{}\\sqrt{25\\minus{}24t}\\right)^2}\\geq0,$ which is obvious.", "Solution_2": "\\[ (a^3b^3 \\plus{} b^3c^3 \\plus{} c^3a^3)[(a \\plus{} b)(b \\plus{} c)(c \\plus{} a) \\minus{} 8abc] \\ge k abc(a \\minus{} b)^2(b \\minus{} c)^2(c \\minus{} a)^2,\r\n\\]\r\n\r\n\\[ k_{max} \\equal{} 9.\r\n\\]", "Solution_3": "We can use cauchy for problem,try it :blush:", "Solution_4": "[quote=\"fjwxcsl\"]\n\\[ (a^3b^3 \\plus{} b^3c^3 \\plus{} c^3a^3)[(a \\plus{} b)(b \\plus{} c)(c \\plus{} a) \\minus{} 8abc] \\ge k abc(a \\minus{} b)^2(b \\minus{} c)^2(c \\minus{} a)^2,\n\\]\n\n\\[ k_{max} \\equal{} 9.\n\\]\n[/quote]\r\nYes, you are right! But I don't see an easy proof for your nice result. :(\r\nLet $ ab \\equal{} z,$ $ ac \\equal{} y$ and $ bc \\equal{} x.$\r\nHence, $ (a^3b^3 \\plus{} b^3c^3 \\plus{} c^3a^3)((a \\plus{} b)(b \\plus{} c)(c \\plus{} a) \\minus{} 8abc)\\ge k abc(a \\minus{} b)^2(b \\minus{} c)^2(c \\minus{} a)^2\\Leftrightarrow$\r\n$ \\Leftrightarrow(x^3 \\plus{} y^3 \\plus{} z^3)\\sum_{cyc}(x^2y \\plus{} x^2z \\minus{} 2xyz)\\geq k(x \\minus{} y)^2(x \\minus{} z)^2(y \\minus{} z)^2.$\r\nIf $ z \\equal{} 0$ and $ x^2 \\plus{} y^2 \\equal{} pxy$ then, should be $ (x^3 \\plus{} y^3)(x \\plus{} y)\\geq k(x \\minus{} y)^2xy$ and\r\n $ (x^3 \\plus{} y^3)(x \\plus{} y)\\geq k(x \\minus{} y)^2xy\\Leftrightarrow$\r\n$ \\Leftrightarrow(x^2 \\plus{} y^2 \\minus{} xy)(x^2 \\plus{} y^2 \\plus{} 2xy)\\geq k(x^2 \\plus{} y^2 \\minus{} 2xy)xy\\Leftrightarrow$\r\n$ \\Leftrightarrow(p \\minus{} 1)(p \\plus{} 2)\\geq k(p \\minus{} 2)\\Leftrightarrow p^2 \\minus{} (k \\minus{} 1)p \\plus{} 2k \\minus{} 2\\geq0,$ which gives $ k\\leq9.$\r\nThus, it remains to prove that \r\n$ (x^3 \\plus{} y^3 \\plus{} z^3)\\sum_{cyc}(x^2y \\plus{} x^2z \\minus{} 2xyz)\\geq9(x \\minus{} y)^2(x \\minus{} z)^2(y \\minus{} z)^2.$\r\nLet $ x \\plus{} y \\plus{} z \\equal{} 3u,$ $ xy \\plus{} xz \\plus{} yz \\equal{} 3v^3,$ $ xyz \\equal{} w^3$ and $ u^2 \\equal{} tv^2.$\r\nHence, $ (x^3 \\plus{} y^3 \\plus{} z^3)\\sum_{cyc}(x^2y \\plus{} x^2z \\minus{} 2xyz)\\geq9(x \\minus{} y)^2(x \\minus{} z)^2(y \\minus{} z)^2\\Leftrightarrow$\r\n$ \\Leftrightarrow(27u^3 \\minus{} 27uv^2 \\plus{} 3w^3)(9uv^2 \\minus{} 9w^3)\\geq$\r\n$ \\geq9\\cdot27(3u^2v^4 \\minus{} 4v^6 \\plus{} 6uv^2w^3 \\minus{} 4u^3w^3 \\minus{} w^6)\\Leftrightarrow$\r\n$ \\Leftrightarrow8w^6 \\plus{} (27u^2 \\minus{} 44v^2)uw^3 \\plus{} 9v^2(u^2 \\minus{} 2v^2)^2\\geq0.$\r\nId est, if $ t\\geq\\frac {44}{27}$ hence, our inequality is proven.\r\nIf $ 1\\leq t\\leq\\frac {44}{27}$ and $ (27u^2 \\minus{} 44v^2)^2u^2 \\minus{} 4\\cdot8\\cdot9v^2(u^2 \\minus{} 2v^2)^2\\leq0,$ which is equivalent to\r\n$ \\Leftrightarrow729t^3 \\minus{} 2664t^2 \\plus{} 3088t \\minus{} 1152\\leq0,$ we obtain $ 1.01648...\\leq t\\leq\\frac {44}{27}.$\r\nHence, it remains to prove our inequality for $ 1\\leq t\\leq1.01648...,$ for which we'll prove that\r\n$ w^3\\leq\\frac {(44v^2 \\minus{} 27u^2)u \\minus{} \\sqrt {729u^6 \\minus{} 2664u^4v^2 \\plus{} 3088u^2v^4 \\minus{} 1152v^6}}{16}.$\r\n$ (x \\minus{} y)^2(x \\minus{} z)^2(y \\minus{} z)^2\\geq0$ gives $ w^3\\leq3uv^2 \\minus{} 2u^3 \\plus{} 2\\sqrt {(u^2 \\minus{} v^2)^3}.$\r\nHence, enough to prove that\r\n$ 3uv^2 \\minus{} 2u^3 \\plus{} 2\\sqrt {(u^2 \\minus{} v^2)^3}\\leq\\frac {(44v^2 \\minus{} 27u^2)u \\minus{} \\sqrt {729u^6 \\minus{} 2664u^4v^2 \\plus{} 3088u^2v^4 \\minus{} 1152v^6}}{16},$\r\nwhich is equivalent to\r\n$ 5u^3 \\minus{} 4uv^2 \\minus{} \\sqrt {729u^6 \\minus{} 2664u^4v^2 \\plus{} 3088u^2v^4 \\minus{} 1152v^6}\\geq32\\sqrt {(u^2 \\minus{} v^2)^3}.$\r\nEasy to check that $ 5u^3 \\minus{} 4uv^2 \\minus{} \\sqrt {729u^6 \\minus{} 2664u^4v^2 \\plus{} 3088u^2v^4 \\minus{} 1152v^6}\\geq0$ is true for $ 1\\leq t\\leq1.01648....$\r\nBut $ 5u^3 \\minus{} 4uv^2 \\minus{} \\sqrt {729u^6 \\minus{} 2664u^4v^2 \\plus{} 3088u^2v^4 \\minus{} 1152v^6}\\geq32\\sqrt {(u^2 \\minus{} v^2)^3}\\Leftrightarrow$\r\n$ \\Leftrightarrow \\minus{} 135t^3 \\plus{} 184t^2 \\plus{} 16t \\minus{} 64\\geq(5t \\minus{} 4)\\sqrt {t(729t^3 \\minus{} 2664t^2 \\plus{} 3088t \\minus{} 1152)},$ where\r\n$ \\minus{} 135t^3 \\plus{} 184t^2 \\plus{} 16t \\minus{} 64\\geq0$ is still true for $ 1\\leq t\\leq1.01648....$\r\nIn the end we obtain \r\n$ \\minus{} 135t^3 \\plus{} 184t^2 \\plus{} 16t \\minus{} 64\\geq(5t \\minus{} 4)\\sqrt {t(729t^3 \\minus{} 2664t^2 \\plus{} 3088t \\minus{} 1152)}\\Leftrightarrow$\r\n$ \\Leftrightarrow45t^5 \\minus{} 162t^4 \\plus{} 213t^3 \\minus{} 116t^2 \\plus{} 16t \\plus{} 4\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow(t \\minus{} 1)^2(45t^3 \\minus{} 72t^2 \\plus{} 24t \\plus{} 4)\\geq0,$ which is obviously true.", "Solution_5": "With,k=1,we can use cauchy.\r\n$ [c(a\\minus{}b)^2\\plus{}b(a\\minus{}c)^2\\plus{}a(b\\minus{}c)^2][\\frac{1}{c^3}\\plus{} \\frac{1}{b^3}\\plus{} \\frac{1}{a^3} ] \\ge$\r\n$ ( \\frac{a\\minus{}b}{c}\\plus{} \\frac{b\\minus{}c}{a}\\plus{} \\frac{c\\minus{}a}{b} )^2 \\equal{} \\frac{(a\\minus{}b)^2(b\\minus{}c)^2(c\\minus{}a)^2}{a^2b^2c^2}$ \r\nand we get\r\n$ (a^3b^3\\plus{}b^3c^3\\plus{}c^3a^3)[(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)\\minus{}8abc] \\ge abc(a\\minus{}b)^2(b\\minus{}c)^2(c\\minus{}a)^2$", "Solution_6": "Comment: in arqady's proof, he multiplies by abc in order to do the substitution $ x\\equal{}bc$, etc. which has the problem if $ a\\equal{}0$ then he hasn't verified the inequality... But those cases are trivial. Anyway, the inequality is equivalent to the one:\r\n\r\n$ f(x,y,z)\\equal{}(x^3\\plus{}y^3\\plus{}z^3)((x\\plus{}y)(y\\plus{}z)(z\\plus{}x)\\minus{}8xyz)\\minus{} (x\\minus{}y)^2(y\\minus{}z)^2(z\\minus{}x)^2\\ge 0$ where $ x\\equal{}bc\\ge 0$, etc\r\n\r\nNow $ (x\\plus{}y)(y\\plus{}z)(z\\plus{}x)\\minus{}8xyz\\equal{}x(x\\minus{}y)^2\\plus{}y(y\\minus{}z)^2\\plus{}z(z\\minus{}x)^2$.\r\n\r\nSo it is easy to see that $ f(x,y,z)\\ge f(x\\minus{}d,y\\minus{}d,z\\minus{}d)$ where $ d\\ge 0$. So $ f(x,y,z)\\ge f(x\\minus{}y,x\\minus{}z,0)\\equal{}f(p,q,0)$ where WLOG $ z$ is the minimum of $ x,y,z$ and $ p,q\\ge 0$.\r\n\r\nBut $ f(p,q,0)\\equal{}p,q(p^2\\plus{}q^2)^2\\ge 0$.", "Solution_7": "[quote=\"Altheman\"]\n$ f(x,y,z) \\equal{} (x^3 \\plus{} y^3 \\plus{} z^3)((x \\plus{} y)(y \\plus{} z)(z \\plus{} x) \\minus{} 8xyz) \\minus{} (x \\minus{} y)^2(y \\minus{} z)^2(z \\minus{} x)^2\\ge 0$ \n\nSo it is easy to see that $ f(x,y,z) \\geq f(x \\minus{} d,y \\minus{} d,z \\minus{} d)$ where $ d\\ge 0$. [/quote]\r\nI have obtained $ f(x,y,z) \\minus{} f(x \\minus{} d,y \\minus{} d,z \\minus{} d) \\equal{} \\minus{} 6\\sum_{cyc}(x^2 \\minus{} xy)d^4 \\plus{}$\r\n$ \\plus{} 3\\sum_{cyc}(x \\plus{} y \\plus{} 2z)(x \\minus{} y)^2d^3 \\minus{} 3\\sum_{cyc}(x^2 \\plus{} y^2 \\plus{} 2z^2 \\plus{} xz \\plus{} yz)(x \\minus{} y)^2d^2 \\plus{}$\r\n$ \\plus{} \\sum_{cyc}(x^3 \\plus{} y^3 \\plus{} 4z^3 \\plus{} 3x^2z \\plus{} 3y^2z)(x \\minus{} y)^2d.$ :wink:", "Solution_8": "[quote=\"fjwxcsl\"]\n\\[ (a^3b^3 \\plus{} b^3c^3 \\plus{} c^3a^3)[(a \\plus{} b)(b \\plus{} c)(c \\plus{} a) \\minus{} 8abc] \\ge k abc(a \\minus{} b)^2(b \\minus{} c)^2(c \\minus{} a)^2,\n\\]\n\n\\[ k_{max} \\equal{} 9.\n\\]\n[/quote]\r\nReplacing $ a,b,c$ by $ \\frac{1}{a},\\frac{1}{b},\\frac{1}{c}$ respectively, then our inequality becomes\r\n$ (a^3\\plus{}b^3\\plus{}c^3)[(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)\\minus{}8abc] \\ge 9(a\\minus{}b)^2(b\\minus{}c)^2(c\\minus{}a)^2.$\r\nNow, assume that $ c\\equal{}\\min\\{a,b,c\\}$ then we have\r\n$ a^3\\plus{}b^3\\plus{}c^3 \\ge a^3\\plus{}b^3,$\r\n$ (a\\plus{}b)(b\\plus{}c)(c\\plus{}a)\\minus{}8abc\\equal{}2c(a\\minus{}b)^2\\plus{}(a\\plus{}b)(a\\minus{}c)(b\\minus{}c) \\ge (a\\plus{}b)(a\\minus{}c)(b\\minus{}c),$\r\nand\r\n$ (a\\minus{}b)^2(a\\minus{}c)^2(b\\minus{}c)^2 \\le ab(a\\minus{}b)^2(a\\minus{}c)(b\\minus{}c).$\r\nTherefore, we can reduce our inequality to\r\n$ (a^3\\plus{}b^3)(a\\plus{}b)(a\\minus{}c)(b\\minus{}c) \\ge 9ab(a\\minus{}b)^2(a\\minus{}c)(b\\minus{}c),$\r\nor\r\n$ (a^3\\plus{}b^3)(a\\plus{}b) \\ge 9ab(a\\minus{}b)^2,$\r\nwhich is equivalent to\r\n$ (a^2\\minus{}4ab\\plus{}b^2)^2 \\ge 0,$ which is trivial. :)", "Solution_9": "I should specify that min $ (x,y,z)\\ge d\\ge 0$.\r\n\r\n$ x^3\\ge (x\\minus{}d)^3$ so \r\n\r\n$ (x^3\\plus{}y^3\\plus{}z^3)\\ge (x\\minus{}d)^3\\plus{}(y\\minus{}d)^3\\plus{}(z\\minus{}d)^3\\ge 0$\r\n\r\nalso $ (x)(x\\minus{}y)^2\\ge (x\\minus{}d)(x\\minus{}y)^2\\equal{}(x\\minus{}d)((x\\minus{}d)\\minus{}(y\\minus{}d))^2\\ge 0$ so \r\n\r\n$ \\sum_{cyc} x(x\\minus{}y)^2\\ge \\sum_{cyc}(x\\minus{}d)((x\\minus{}d)\\minus{}(y\\minus{}d))^2\\ge 0$\r\n\r\nMultiplying these two relations and adding $ (x\\minus{}y)^2(y\\minus{}z)^2(z\\minus{}x)^2$ gives us the result\r\n\r\n\r\nObviously, I did not communicate my solution well (or at all for that matter), I apologize. Thanks for checking my post carefully.", "Solution_10": "Very nice and easy your proof, Can! :lol:" } { "Tag": [], "Problem": "In trigonometric form, is the complex number $z^{\\frac{x}{y}}$ equal to:\r\n\r\n$r^{\\frac{x}{y}} cis \\frac{x\\theta}{y} + \\frac{2\\pi n}{y}$ \r\n(equivalent of raising to xth power, then taking yth root)\r\n\r\n\r\nor \r\n\r\n$r^{\\frac{x}{y}} cis \\frac{x\\theta}{y} + \\frac{2\\pi xn}{y}$ \r\n(equivalent to taking yth power, then raising to xth power)", "Solution_1": "The key is to remember that $cis \\theta=cis(\\theta+2n\\pi)$, where n is an integer. Thus if $z=rcis\\theta$, then\r\n\r\n$z^{\\frac{x}{y}}=(rcis\\theta)^{\\frac{x}{y}}=(rcis(\\theta+2n\\pi))^{\\frac{x}{y}}=r^{\\frac{x}{y}}cis(\\frac{(\\theta+2n\\pi)x}{y})$.", "Solution_2": "Shouldn't the first expression be equivalent to the second one?", "Solution_3": "i believe so, but only if x is an integer." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "prove that doesnot exist the function f: R -> R satisfy f(f(x)) = x^2 - 2008", "Solution_1": "Prove that exist $ a,b$ different satisfy condition :\r\n$ a\\equal{}b^2\\minus{}2008$\r\n$ b\\equal{}a^2\\minus{}2008$\r\nThen consider $ f(a)$ and $ f(b)$" } { "Tag": [ "inequalities", "calculus", "algebra", "function", "domain", "inequalities unsolved" ], "Problem": "aaa", "Solution_1": "I thinkk it is well-known :)", "Solution_2": "You think it is well known.Can you tell me where have you seen it,please?\r\n By the way do you have I proof for this?", "Solution_3": "Cezar, this is everything except new and important. I suggest you to browse the forum and you will see it in different forms and generalizations. Anyway, it is direct consequence of Holder inequality. And a little sugestion: please use more appropriate names for your topics.", "Solution_4": "Sorry, I thought that this one was interesting. :oops:", "Solution_5": "The general inequal was posted a long time before:\r\n[tex]\\sum_{i=1}^n \\frac{a_i^u}{b_i^v}\\geq \\frac{(\\sum a_i)^u}{n^{u-v-1}(\\sum b_i)^v}[/tex] with [tex]u,v[/tex] be positive s.t:[tex]u\\geq v+1[/tex]", "Solution_6": "[quote=\"keira_khtn\"]The general inequal was posted a long time before:\n[tex]\\sum_{i=1}^n \\frac{a_i^u}{b_i^v}\\geq \\frac{(\\sum a_i)^u}{n^{u-v-1}(\\sum b_i)^v}[/tex] with [tex]u,v[/tex] be positive s.t:[tex]u\\geq v+1[/tex][/quote]\r\nI think this is not the most general one. This is stronger. http://www.mathlinks.ro/Forum/viewtopic.php?p=111563#p111563\r\njaloweic's proof is amazing. :lol:", "Solution_7": "cant we solve many inequalities in general with series expansions :idea: \r\n\r\nperhaps a silly idea :roll: \r\n\r\nbut i was thinking f(x;y;z;..)>g(x;y;z;...)\r\n\r\nturn them into series expansions like e.g. taylor and use some kind of calculus formula to decide if it is true or not ?? \r\n\r\nof course the taylor series must converge for all variables in the conjectured domain and integer conditions might make it \"diophantic-like\" but in general like algebraic inequalities with less than 5 variables just to say something...\r\n\r\njust \"dreaming\" about that \r\nguess someone is going to tell me nope it cant be done :D \r\nbut if you do plz tell me why not , im not that smart :oops: \r\n\r\ngreetings to all", "Solution_8": "do you mind telling me the relationship between your post and the inequality of this thread???", "Solution_9": "[quote=\"siuhochung\"]do you mind telling me the relationship between your post and the inequality of this thread???[/quote]\r\n\r\nyou have a point there :blush: \r\n\r\nbut the inequation lead to that ( silly ? ) theory.\r\nand i felt it was not worth starting a new topic for its since its pretty vague ... :|" } { "Tag": [], "Problem": "Anyone got any good ti-89 games to share. For a start, 68k mario on ticalc.org is extremely good (basically and abridged version of mario for the N.E.S.)", "Solution_1": "Block Man? :lol: \r\nSudoku might be fun on 89." } { "Tag": [ "analytic geometry", "graphing lines", "slope", "trigonometry", "conics", "parabola", "number theory" ], "Problem": "Source: 2005 AMC-12 B Problem 24\r\n\r\n[b]I am not asking for a solution, only an explanation.[/b]\r\n\r\nIn an AMC-12 solution packet, one solution uses trig to define the slopes of a triangle:\r\n\r\nThe problem states \"All three vertices of an equilateral triangle are on the parabola $y=x^{2}$, and one of its sides has a slope of $2$. The $x$-coordinates of the three vertices have a sum of $\\displaystyle\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is the value of $m+n$?\"\r\n\r\nPart of the solution states that \"The slope of one side is $2=\\tan\\theta$, for some angle $\\theta$, and the two remaining sides have slopes $\\tan(\\theta\\pm\\frac{\\pi}{3})$.\"\r\n\r\nTheir statement makes sense, and it simplifies the problem greatly, but I have never seen anything relating the slopes of the sides of a triangle with trig before, and I had a difficult time proving it to myself.\r\n\r\nIs there a simple relationship here? Does it apply to equilateral triangles only, or any triangle?", "Solution_1": "Well if you draw a line, you'll see the slope is $\\frac y x$ if you draw a right triangle to the x-axis, and letting y be the height and x the base of the right triangle. Then, if $\\theta$ is the angle the line makes with the x-axis, clearly $\\tan \\theta = \\frac y x \\implies \\tan \\theta = m$", "Solution_2": "I understand that part, but how do they get $(\\theta\\pm\\frac{\\pi}{3})$?", "Solution_3": "because you have the original slope defined by $\\tan{\\theta}$, then the next one is $60$ degrees more, so it's defined by $\\tan(\\theta+\\frac{\\pi}{3})$, and the next one is $120$ degrees more, or $60$ degree less (you can show this easily by drawing a parallel line line to that third side through the vertex opposite that side), so it's slope is defined by $\\tan(\\theta-\\frac{\\pi}{3})$.", "Solution_4": "Alright, thanks to both of you!" } { "Tag": [ "calculus", "integration", "algebra unsolved", "algebra" ], "Problem": "[b]1.[/b] Does there exists positive integral infinite sequences $\\{a_n\\}$ such that $a_{n+1}^2\\geq 2a_{n+2}a_{n}$ for any positive integer $n$?\r\n\r\n[b]2.[/b] Does there exists positive irrational infinite sequences $\\{a_n\\}$ such that $a_{n+1}^2\\geq 2a_{n+2}a_{n}$ for any positive integer $n$?", "Solution_1": "We get $\\frac{a_{n+1}}{a_n}\\le\\frac{a_2}{a_1}\\cdot\\frac 1{2^{n-1}}$. We may assume WLOG $a_2\\le a_1$, because otherwise, we start from $n>1$ s.t. $a_{n+1}\\le a_n$. Anyway, the sequence becomes strictly decreasing eventually, so there's no such sequence made up of positive integers.\r\n\r\nOf course there has to be one made up of irrationals. Just take a transcendental number $\\alpha$, put $a_1=\\alpha,a_2=\\alpha^2$, and then we can even construct the sequence inductively s.t. we have equality (not merely inequality). We get $a_n=r_n\\cdot \\alpha^n$, where $r_n$ are positive rationals." } { "Tag": [], "Problem": "Yes, T.O.'s a bad, baaaaaaad man, but the Patriots will pull this one out. They'll be ahead at the half by a touchdown, and will be up by 17 by the end of the game.\r\n\r\nIts been widely discussed that the advertising during the game will be \"toned down\", but there will certainly not be a lack of humor in the commercials. Watch for the Lays ad that features the unceremonious return of MC Hammer...\r\n\r\nWho do you think's going to win the big game??\r\n\r\n\r\nedit: i knew i thought i saw a previous thread about this...just now found it. oops. feel free to post here, though...can't have too many discussions about the super bowl!", "Solution_1": "Dynasty shall be continued\r\n\r\nGo Patroits :lol:", "Solution_2": "superbowls tomarrow!", "Solution_3": "Go Patriots! I agree. They are a DYNASTY!\r\n\r\nI'm not very good at debating so I practice my debating skills against people who think the Eagles are gonna win.... and I actually win these debates! That's how good the Patriots are! If I can convince others that they are good (and I can't convince anyone of anything), then there's no way they can lose!", "Solution_4": "16 year bumps" } { "Tag": [ "probability" ], "Problem": "there are 16 games you have to get them all correct you choose home or visitor win in each box how many different combinations are there to ensure you get perfect", "Solution_1": "hi Bobby,\r\n\r\nPlease do not post a topic twice. All math belongs in the math forums.\r\n\r\nThis one goes to Getting Started unless I misinterpreted the problem and it's actually much more difficult than I imagine.", "Solution_2": "I'm not quite sure what your asking.....\r\n\r\nAssuming that each game is unique:\r\n\r\n[hide=\"If your looking for number of combos\"]$2^16$=66536 possibilities[/hide]\n[hide=\"If your looking for probability that you will pick the correct combo\"]Let $x$ be the number of guesses you have, your probabiliby will be $x/(2^16)$ or $x/65536$[/hide]\n[hide=\"If you want to know how many picks you need to ensure you will get the correct combo\"]65536....[/hide]" } { "Tag": [ "function", "calculus", "derivative", "algebra", "polynomial", "limit", "calculus computations" ], "Problem": "how many critical points does the function provided have?\r\n\r\nf(x)=(x+2)^2 * (x-3)^4", "Solution_1": "What is its derivative? Find and factor that.", "Solution_2": "i got 2(x-3)^3 * (x+2) * (3x + 1)\r\n\r\ndo i then set that equal to 0 and see my solutions...basically do i, in essence, to find the critical points take the deriv. and set equal to 0?", "Solution_3": "Yes. This function has critical points at $-2,-\\frac13,$ and $3.$\r\n\r\nOur next task is to classify them. There are three basics families of information we can use to classify critical points.\r\n\r\n1. Information about the function values, or\r\n2. Inforamtion about the sign of the first derivitative in between the critical points, or\r\n3. Information about the sign of the second derivative at the critical points.\r\n\r\nAll three families of information have their uses. I'll show you #1 in this case.\r\n\r\nNote that $f(x)$ is a sixth degree polynomial with a positive coefficient of $x^{6}.$ Here's what we know:\r\n\r\n$\\lim_{x\\to\\infty}f(x)=\\infty$ and $\\lim_{x\\to-\\infty}f(x)=\\infty.$ (That is, both ends of the graph turn up.)\r\n$f(-2)=0$ and $f(3)=0.$\r\nBut with the function being a second power times a fourth power, we have that $f(x)\\ge0$ for all $x.$\r\n\r\nThat last item tells us what is happening at $-2$ and at $0:$ each point is local minimum - and these two points are tied for being the global mininum. Since the function is positive in between those two points, there must be a local maximum in that interval - and that tells us that the point at $-\\frac13$ is that local maximum. But it can't be the global maximum because of the behavior of this polynomial for large $x.$ There is no global maximum." } { "Tag": [ "geometry", "perimeter", "function", "calculus", "quadratics", "area of a triangle", "Heron\\u0027s formula" ], "Problem": "An isosceles triangle $ABC$ has a perimeter of $90 cm$. Find the side lenghts that will maximize the area of the triangle. (I know the answer... but the function is coming out ridiculous for the expected difficulty of the problem.)", "Solution_1": "Intuitively it is clearly the equilateral triangle, which is what you're thinking of. Without getting into sin/cosine, the equation to maximize is: $y\\sqrt{x+y}\\sqrt{|x-y|}$ given $2x+y=90$. Not so bad, but it is definitely not high school basics forum! (hint... move to preolympiad...)", "Solution_2": "You can use Heron's formula and $\\frac{a+b+c}{3}\\geq \\sqrt[3]{abc}$ for $a>0,\\ b>0,\\ c>0.$ This problem is beyond this forum, I think.", "Solution_3": "I don't see how this is too hard for HSB at all.\r\n\r\n[hide]The equilateral triangle is always the maximum area of a triangle with a certain perimeter. So you plug in 30 into $\\frac{s^{2}\\sqrt{3}}{4}$ to get $\\boxed{225\\sqrt{3}}$.[/hide]", "Solution_4": "You should prove your declare. :wink:", "Solution_5": "Ok guys... I'll be honest, this was a question a friend of mine had while I was tutoring him before his calculus test... I just figured that if you came up with a function for area expressed through one variable it would probably be quadratic and so the calculus part would just be a fancy way to find the vertex... It seems that the easiest way to do it however would be through Heron's and that function certainly isn't quadratic and so then for it to be easy it would require some calculus techniques. :blush: Could an admin move this wherever they see fit or delete it please?", "Solution_6": "One can prove strictly geometrically that the equilateral triangle is optimal (with the assumption that an optimal triangle exists). Note that the question is equivalent to showing that among triangles with a given area, the equilateral triangle has the least perimeter. (If you can shrink the perimeter with the same area, then there is a triangle similar to the new own with the original perimeter but larger area.) Rather than showing directly that the equilateral is optimal, we show that every other triangle is sub-optimal (which is why we need the assumption that an optimal triangle exists). For any non-equilateral triangle, choose two unequal sides, and draw a line through their shared vertex parallel to the third side. As we move the vertex along this line, the area of the triangle is unchanged. The perimeter is minimized when we choose the point which lies on the perpendicular bisector of the third side. (This is a particularly simple instance of the \"man needs to walk to the river and then back to his cottage\" problem.) This leaves us with those two sides equal, so we have a different triangle with smaller perimeter and the same area, done.\r\n\r\nAlso, if calculus student doesn't know Heron's, easier to do by dropping an altitude then by teaching Heron's. (It's isosceles, so that makes everything simpler.)", "Solution_7": "Yes, I showed him originally by dropping the altitude (he only new A= bh/2 so it's the formula I used) but the numbers get really dumb when you express A through the height or the base. Heron's makes it very much cleaner, but teaching it would be a pain." } { "Tag": [ "calculus", "integration", "logarithms", "trigonometry", "calculus computations" ], "Problem": "Can anyone help me to show me how to calculate this integral:\r\n$\\int_{0}^{\\frac{\\pi}{2}}{\\ln(\\sin x)}dx$", "Solution_1": "http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=sin+ln&t=135259", "Solution_2": "Or you can use the famous identity\r\n\r\n$\\prod_{k=1}^{n-1}\\sin \\left( \\frac{k \\pi}{n}\\right) = \\frac{n}{2^{n-1}}$\r\n\r\nto calculate the Euler's integral. Simple manipulation gives\r\n\r\n$\\sum_{k=1}^{n-1}\\ln \\sin \\left( \\frac{k \\pi}{n}\\right) \\frac{\\pi}{n}= \\frac{\\pi}{n}(\\ln n-(n-1) \\ln 2)$\r\n\r\nThe left side converges to $\\int_{0}^{\\pi}\\ln(\\sin x) \\, dx$ and the right side converges to $-\\pi \\ln 2$.", "Solution_3": "[quote=\"sos440\"]Or you can use the famous identity\n\n$\\prod_{k=1}^{n-1}\\sin \\left( \\frac{k \\pi}{n}\\right) = \\frac{n}{2^{n-1}}$\n\nto calculate the Euler's integral. Simple manipulation gives\n\n$\\sum_{k=1}^{n-1}\\ln \\sin \\left( \\frac{k \\pi}{n}\\right) \\frac{\\pi}{n}= \\frac{\\pi}{n}(\\ln n-(n-1) \\ln 2)$\n\nThe left side converges to $\\int_{0}^{\\pi}\\ln(\\sin x) \\, dx$ and the right side converges to $-\\pi \\ln 2$.[/quote]\r\nThank you very much!I thought about this but fogot how to calculate it." } { "Tag": [ "analytic geometry", "real analysis", "integration", "LaTeX", "geometry", "3D geometry", "sphere" ], "Problem": "Let $ S_{k\\minus{}1}$ be the unit circle in $ \\mathbb{R}^{k}$, i.e the set of all $ u\\in\\mathbb{R}^{k}$ whose distance from the origin is 1 . It is known that $ \\forall x\\in\\mathbb{R}^{k}$ then $ x\\equal{}ru$ for some $ u\\in S_{k\\minus{}1}$ and $ r\\geq 0$. Thus $ \\mathbb{R}^{k}\\minus{}\\{0\\}$ may be regarded as the Cartesian product of $ (0,\\infty) x S_{k\\minus{}1}$ \r\nLet $ m_{k}$ be the Lebesgue measure on $ \\mathbb{R}^{k}$ and define a measure $ \\sigma_{k\\minus{}1}$ on $ S_{k\\minus{}1}$ as follow : If $ A\\subset S_{k\\minus{}1}$ and $ A$ is a Borel set, let $ \\bar{A}$ be the set of all the points $ ru$ where $ 00.$\r\n\r\nThe cone $ C$ is then the convex hull of $ A$ and $ U\\times\\{0\\}$ - that is, the union of all the line segments from an arbitrary point on the base to the apex.\r\n\r\nNow take the section through this cone at height $ y,$ for some $ y\\in[0,h].$ That is, slice through the cone along a \"plane\" (OK, hyperplane) parallel to the base. The cross section is similar to $ U,$ with a coefficient of similarity equal to $ \\frac{h\\minus{}y}{h}.$ That is, the cross section has the same shape as $ U$ but with all linear dimensions multiplied by $ \\frac{h\\minus{}y}{h}$ (and possibly also translated, but the translation doesn't matter.) \r\n\r\nWe do need to know how $ (n\\minus{}1)$-dimensional measure behaves with respect to similarity. We need to know that the $ (n\\minus{}1)$-dimensional measure of that cross section is $ \\left(\\frac{h\\minus{}y}{h}\\right)^{n\\minus{}1}B.$\r\n\r\nWe then compute the $ n$-dimensional measure of the cone by integrating this in $ y.$ Call it Cavalieri's principle; call it Fubini's theorem. That $ n$-measure is:\r\n\r\n$ V\\equal{}\\int_0^h\\left(\\frac{h\\minus{}y}{h}\\right)^{n\\minus{}1}B\\,dy\\equal{}\\left.\\frac{B}{h^{n\\minus{}1}}\\cdot\\left(\\minus{}\\frac1n(h\\minus{}y)^n\\right)\\right|_0^h\\equal{}\\frac1n\\,Bh.$\r\n\r\nIs that what you were asking for, 1234567a?", "Solution_6": "Thanks prof. Kent, yes that is what I am asking for\r\n\r\nThanks a bunch \r\n\r\n\r\n--------------------------------------\r\n\r\n[b]For Fedja:[/b]\r\n[quote]\nBut we know the volume (k-dimensional Lebesgue measure), so to have everything consistent, we are forced to define the surface measure on the unit sphere by the cited formula at least for small sets. B \n\n[/quote]\r\n\r\n\r\nI still do not understand \" [b]the cited formula at least for small sets.[/b]\", would you mind explaining ?", "Solution_7": "Oh, that was just the reference to the formula $ \\sigma_{k\\minus{}1}(A)\\equal{}k\\cdot m_{k}(\\widetilde A)$ defining the measure. (\"cited\"=\"the one you rewrote from the book\") :) I was just saying that this formula should hold (at least, approximately) when the diameter of $ A$ is small. That's all." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "Let [$ a_n, b_n$], $ n \\equal{} 1,2,.....$ be closed intervals with [$ a_n, b_n$] $ \\cap$ [$ a_m, b_m$] $ \\neq \\emptyset$ for all $ n, m$. Prove that $ \\bigcap_{n\\equal{}1}^{\\infty}[a_n, b_n] \\neq \\emptyset$", "Solution_1": "one word: compactness...", "Solution_2": "Thanks for answer, pleurestique, but what do you mean by \"compactness\"? And aren't there any other approches to this problem (since the idea of \"compactness\", I guess, is pretty general)?", "Solution_3": "i meant use the definition of compact..." } { "Tag": [ "number theory", "relatively prime", "calculus", "calculus computations" ], "Problem": "here is another easy problem. :D \r\n\r\nthe value of the series\r\n\r\n$ \\sum _{n\\equal{}2}^\\infty \\; \\frac { x^4\\plus{} x^3 \\plus{}x^2 \\minus{} x \\plus{}1} {x^6\\minus{}1}$\r\n\r\ncan be expressed as $ \\frac{p}{q}$ where $ p$ and $ q$ are relatively prime $ \\plus{}ve$ numbers. \r\n\r\n[b]without[/b] explicitly solving for $ p$ and $ q$ , calculate $ p\\plus{}q$", "Solution_1": "I am confused :( , the index is n but it appears nowhere in the summation. Would it not blow up to infinity then?", "Solution_2": "sorry, that was a typo :( , the summation should read as $ \\sum_{x\\equal{}2}^\\infty$", "Solution_3": "Let's try to simplify our sum: \r\n\r\n$ \\textbf{S}\\equal{}\\sum _{x\\equal{}2}^\\infty \\; \\frac { x^4\\plus{} x^3 \\plus{}x^2 \\minus{} x \\plus{}1} {x^6\\minus{}1}\\equal{}\\sum _{x\\equal{}2}^\\infty \\; \\frac { x^4\\plus{} x^3 \\plus{}x^2 \\plus{} x \\plus{} 1 \\minus{} 2x} {x^6\\minus{}1}\\equal{}$\r\n\r\n$ \\equal{}\\sum _{x\\equal{}2}^\\infty \\; \\frac {\\frac{x^5\\minus{}1}{x\\minus{}1}\\minus{}2x} {x^6\\minus{}1}\\equal{}\\sum _{x\\equal{}2}^\\infty \\; \\frac {x^5\\minus{}1} {(x\\minus{}1)(x^6\\minus{}1)}\\minus{}2\\sum _{x\\equal{}2}^\\infty \\; \\frac {x} {x^6\\minus{}1}\\equal{}$\r\n\r\n$ \\equal{}\\sum _{x\\equal{}2}^\\infty \\; \\frac {x^6\\minus{}1\\minus{}x\\plus{}1} {x(x\\minus{}1)(x^6\\minus{}1)}\\minus{}2\\sum _{x\\equal{}2}^\\infty \\; \\frac {x} {x^6\\minus{}1}\\equal{}$\r\n\r\n$ \\equal{}\\sum _{x\\equal{}2}^\\infty \\; \\frac {1} {x(x\\minus{}1)}\\left(1\\minus{}\\frac{x\\minus{}1}{x^6\\minus{}1}\\right)\\minus{}2\\sum _{x\\equal{}2}^\\infty \\; \\frac {x} {x^6\\minus{}1}\\equal{}$\r\n\r\n$ \\equal{}\\sum _{x\\equal{}2}^\\infty \\; \\frac {1} {x(x\\minus{}1)}\\minus{}\\sum _{x\\equal{}2}^\\infty \\; \\frac {1} {x(x^6\\minus{}1)}\\minus{}2\\sum _{x\\equal{}2}^\\infty \\; \\frac {x} {x^6\\minus{}1}\\equal{}$\r\n\r\n$ \\equal{}1\\minus{}\\sum _{x\\equal{}2}^\\infty \\; \\frac {1} {(x^6\\minus{}1)}\\left(2x\\plus{}\\frac{1}{x}\\right)\\equal{}1\\minus{}\\textbf{G}$.\r\n\r\nWe can find that $ \\textbf{G}\\equal{}\\frac{1}{12}$. This gives $ \\textbf{S}\\equal{}\\frac{11}{12}$ and $ \\boxed{p\\plus{}q\\equal{}23}$. But how not to use calculation of $ \\textbf{G}$ find $ p\\plus{}q$ I don't know.", "Solution_4": "You're working too hard with the summand.\r\n\r\nI haven't figured out how to answer this question without explicitly evaluating the sum in question, though.\r\n\r\n[hide]\nWe factor the denominator:\n\n$ (x^6 \\minus{} 1) \\equal{} (x^3 \\minus{} 1)(x^3 \\plus{} 1) \\equal{} (x \\minus{} 1)(x^2 \\plus{} x \\plus{} 1)(x \\plus{} 1)(x^2 \\minus{} x \\plus{} 1)$.\n\nThen we wish to find coefficients $ A, B, C, D, E, F$ such that\n\n$ \\frac {A}{x \\minus{} 1} \\plus{} \\frac {B}{x \\plus{} 1} \\plus{} \\frac {Cx \\plus{} D}{x^2 \\minus{} x \\plus{} 1} \\plus{} \\frac {Ex \\plus{} F}{x^2 \\plus{} x \\plus{} 1} \\equal{} \\frac {x^4 \\plus{} x^3 \\plus{} x^2 \\minus{} x \\plus{} 1}{x^6 \\minus{} 1}$.\n\nMultiplying through and collecting like terms and solving the partial fraction decomposition gives\n\n$ A \\equal{} 1/2, B \\equal{} \\minus{} 1/2, C \\equal{} 0, D \\equal{} 1/2, E \\equal{} 0, F \\equal{} \\minus{} 1/2$,\n\ngiving\n\n$ \\sum_{x \\equal{} 2}^\\infty \\frac {x^4 \\plus{} x^3 \\plus{} x^2 \\minus{} x \\plus{} 1}{x^6 \\minus{} 1} \\equal{} \\frac {1}{2}\\sum_{x \\equal{} 2}^\\infty \\frac {1}{x \\minus{} 1} \\minus{} \\frac {1}{x \\plus{} 1} \\plus{} \\frac {1}{1 \\minus{} x \\plus{} x^2} \\minus{} \\frac {1}{1 \\plus{} x \\plus{} x^2}$.\n\nThe first two terms clearly telescope, but it also turns out that\n\n$ 1 \\plus{} x \\plus{} x^2 \\equal{} \\minus{} x \\plus{} (x \\plus{} 1)^2 \\equal{} 1 \\minus{} (x \\plus{} 1) \\plus{} (x \\plus{} 1)^2$,\n\nso the last two terms in the summand also telescope. Hence the sum is equal to\n\n$ \\frac {1}{2}\\left(\\frac {1}{2 \\minus{} 1} \\plus{} \\frac {1}{3 \\minus{} 1} \\plus{} \\frac {1}{1 \\minus{} 2 \\plus{} 2^2}\\right) \\equal{} \\frac {11}{12}$,\n\nand since 11 and 12 are relatively prime, their sum is 23.\n[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let a=[m^(m+1)+n^(n+1)]/(m^m+n^n) where m,n are from N*. Prove that a^m+a^n>=m^m+n^n.\r\n\r\ncheers!", "Solution_1": "Oh.. for a solution you could use the hint that if n>=m then m<=a<=n and the inequality to be proved is like so:\r\n\r\n(a-m)(a^(m-1)+a^(m-2)*m+..+am^(m-2)+m^(m-1))>=(n-a)(n^(n-1)+n^(n-2)*a+..+na^(n-2)+a^(n-1)).\r\n\r\nbut I am looking for another solution...cheers!", "Solution_2": "If I am not mistaking.. this problem was given for a test in USA :?" } { "Tag": [], "Problem": "general", "Solution_1": "", "Solution_2": "", "Solution_3": "144117", "Solution_4": "", "Solution_5": "", "Solution_6": "", "Solution_7": "", "Solution_8": "Q=65", "Solution_9": "Q=85", "Solution_10": "Q=125", "Solution_11": "Q=145 (left)", "Solution_12": "Q=145 (right)", "Solution_13": "57122", "Solution_14": "57122(2)", "Solution_15": "202654 - Property(Euler line)-2", "Solution_16": "204922 - B ta ha R (2nd case)", "Solution_17": "204922 - B ta ha R (1st case)", "Solution_18": "204922 - B ta ha R (discussion)", "Solution_19": "adjcacent numbers sum to a square", "Solution_20": "", "Solution_21": "", "Solution_22": "", "Solution_23": "", "Solution_24": "215035\\fig1", "Solution_25": "215035-fig2", "Solution_26": "215035-fig3", "Solution_27": "221092 - point P", "Solution_28": "221092 - point P (1)", "Solution_29": "221092 - point P (2)" } { "Tag": [ "induction", "algebra", "binomial theorem", "advanced fields", "advanced fields unsolved" ], "Problem": "( The better presentation of the question with correct symbols is here.\r\nhttp://www.geocities.com/johntyung/Q2.pdf )\r\n\r\n\r\n*** For this problem the best way to solve it by just using the identity and manipulating it algebraically. Without using induction. ***\r\n\r\nfor n even, prove that \r\n\r\n(n choose 0) + (n choose 2) + (n choose 4) +...+ (n choose n)=2^(n-1) \r\n\r\nand \r\n\r\n(n choose 1) + (n choose 3) +...+(n choose (n-1)) = 2^(n-1) \r\n\r\nusing mathematical induction and/or various identities involving binomial coefficients. \r\n(Hint: first establish the identity, \r\n\r\n((m+2) choose k) = (m choose k)+ 2(m choose (k-1)) +(m choose (k-2) ) \r\n\r\nThen, when doing the induction, do a simultaneous induction on both identities.", "Solution_1": "$\\sum_{k\\text{ is even}}2\\binom nk=\\sum_{k=0}^n(1+(-1)^k)\\binom nk=(1+1)^n+(1-1)^n$\r\n$\\sum_{k\\text{ is odd}}2\\binom nk=\\sum_{k=0}^n(1-(-1)^k)\\binom nk=(1+1)^n-(1-1)^n$\r\n\r\nThe binomial theorem is powerful.", "Solution_2": "ok smarty pants LOL\r\n\r\n\r\nhow about explaining your steps....\r\n\r\nthanks in advance any way.", "Solution_3": "From the binomial theorem, we know that\r\n\r\n$\\binom{n}{0}+\\binom{n}{1}+\\ldots+\\binom{n}{n}=2^{k}$\r\n\r\nWe also know that \r\n\r\n$\\binom{n}{0}-\\binom{n}{1}+\\ldots\\pm\\binom{n}{n}=0$\r\n\r\nNow add or subtract this to get what jmerry wrote (I'm assuming you are not comfortable with sigma notation)", "Solution_4": "OK. can u show the steps of what u subtract from what?\r\n\r\nU have to use the Identity to solve this.... Can u show how u use the identity?" } { "Tag": [ "inequalities", "function", "inequalities solved" ], "Problem": "Prove that :\n$\\frac{a}{a^2+1} + \\frac{b}{b^2+1} + \\frac{c}{c^2+1} \\leq \\frac{9}{10}$\nfor all real numbers $a$, $b$, $c$ satisfying $a+b+c=1$ and $a,b,c>-\\frac34$.", "Solution_1": "I'll try:\r\n\r\nLet there be a concave function f(x)=x/(x^2+1)\r\nFrom Jensen we get\r\n\r\nf(a)+f(b)+f(c)<=3f((a+b+c)/3)=3*1/3*9/10=9/10\r\n\r\nHope it's the corect solution!", "Solution_2": "Hey, yours soln is true if a,b,c>0. trie to solv :D", "Solution_3": "We'll prove that: a/(a2+1) <= (36a+3)/50, multiplying 50(a2+1), we get (4a+3)(3a-1)2>= 0, is true\r\n Similar, we have b/(b2+1) <= (36b+3)/50 ; c/(c2+1) <= (36c+3)/50.\r\n Then we get a/(a2+1) + b/(b2+1) + c/(c2+1) <= (9+36(a+b+c))/50 =9/10.\r\n Equality holds when a=b=c=1/3.", "Solution_4": "Comment:\r\n I found out (*), because I suppose a/(a2+1) <= xa +y with a>-3/4. multiplying (a2+1)and change the member of an inequality, we get :xa3 + ya2 +a(x-1) +y>=0.Put f(a)=xa3 + ya2 +a(x-1) +y\r\n We suppose f(a)=(3a-1)2(ra+t).By indentity, we get:\r\n 9r=x, 9t-6r=y, r-6t=x-1 and t=y. Solv this system, we have x=18/25, y=3/50.", "Solution_5": "[quote=\"hxtung\"]Comment:\n I found out (*), because I suppose a/(a2+1) <= xa +y with a>-3/4. multiplying (a2+1)and change the member of an inequality, we get :xa3 + ya2 +a(x-1) +y>=0.Put f(a)=xa3 + ya2 +a(x-1) +y\n We suppose f(a)=(3a-1)2(ra+t).By indentity, we get:\n 9r=x, 9t-6r=y, r-6t=x-1 and t=y. Solv this system, we have x=18/25, y=3/50.[/quote]\r\n\r\nNice solution! Since you are expecting f has a factor (3a-1) 2, we must have f(1/3)=f'(1/3) = 0. This gives us two linear equations in two variables x and y, then solve for them.\r\n\r\nIt's interesting to see how you applied the same technique to Japan1997. I found your technique can also be applied to #5 of USAMO2003.", "Solution_6": "We can solv this one with $a,b,c>-1;a+b+c=1$", "Solution_7": "[quote=hxtung]Prove that :\n$\\frac{a}{a^2+1} + \\frac{b}{b^2+1} + \\frac{c}{c^2+1} \\leq \\frac{9}{10}$\nfor all real numbers $a$, $b$, $c$ satisfying $a+b+c=1$ and $a,b,c>-\\frac34$.[/quote]\n\n$$\\text{RHS -LHS} = \\sum \\frac{(2a-b-c)^2 (4a+3)}{50(a^2+1)} \\geqq 0$$", "Solution_8": "[quote=hxtung]Prove that :\n$\\frac{a}{a^2+1} + \\frac{b}{b^2+1} + \\frac{c}{c^2+1} \\leq \\frac{9}{10}$\nfor all real numbers $a$, $b$, $c$ satisfying $a+b+c=1$ and $a,b,c>-\\frac34$.[/quote]\n$$\\dfrac{x}{x^2+1}\\le \\dfrac{18}{25}x+\\dfrac{3}{50}$$ \n[url=https://artofproblemsolving.com/community/c6h514132p3003386]for[/url] $x\\ge -\\dfrac{3}{4}$\n" } { "Tag": [], "Problem": "Find $ a,b \\in\\mathbb{Z}_ \\plus{} ^*$ such that $ (\\sqrt [3]{a} \\plus{} \\sqrt [3]{b} \\minus{} 1)^2 \\equal{} 49 \\plus{} 20\\sqrt [3]{6}$.", "Solution_1": "Its in PEN\r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=150693[/url]" } { "Tag": [ "calculus", "integration", "algebra", "polynomial", "real analysis", "real analysis unsolved" ], "Problem": "Evaluate $ \\int_{ \\minus{} 0.5}^{0.5}\\left( 2y \\plus{} 1\\right) ^{2009}\\left( \\sum_{k \\equal{} 0}^{2009}\\frac {\\left( y \\plus{} 0.5\\right) ^{k}}{k!}\\right) ^{ \\minus{} 1}dy$", "Solution_1": "Well as the integrand has no dependence on $ x$, you can pull it out, leaving you with $ C\\int_{\\minus{}0.5}^{0.5}\\,dx\\equal{}Cx\\Bigg|_{\\minus{}0.5}^{0.5}\\equal{}C$, where $ C$ is that whole big mess.", "Solution_2": "Have edited the typo.", "Solution_3": "First, make the obvious substitution $ t \\equal{} y \\plus{} \\frac12$. The integral becomes $ 2^N\\int_0^1 \\frac {y^N}{p_N(y)}\\,dy$, where $ p_N$ is the $ N$th Taylor polynomial of the exponential at the origin and $ N$ is $ 2009$.\r\nNote that $ p_N'(y) \\equal{} p_N(y) \\minus{} \\frac {y^N}{N!}$. Thus $ \\frac {y^N}{p_N(y)} \\equal{} N!\\frac {p_N(y) \\minus{} p_N'(y)}{p_N(y)} \\equal{} N! \\minus{} N!\\frac {p_N'(y)}{p_N(y)}$\r\nThis is easy to integrate, and we have\r\n$ 2^N\\int_0^1 \\frac {y^N}{p_N(y)}\\,dy \\equal{} 2^N\\left[N!y \\minus{} N!\\ln(p_N(y))\\right]_0^1$\r\n\r\n$ \\equal{} 2^N\\cdot N!\\cdot(1 \\minus{} \\ln(p_N(1)))$\r\n\r\nThat's an exact answer. If you want a reasonable approximation, the above calculation was basically a waste of time; the bad subtraction will kill you. To first order, it's approximately $ \\frac {2^N}{N\\cdot e}$." } { "Tag": [ "blogs" ], "Problem": "Hi guys. I just started my own blog site. I'm going to be putting up some cool, basic effects that you can make in Photoshop or Photoshop Elements. Take a look! [url=http://www.wangsblog.com/andrew/]My Site[/url]", "Solution_1": ":lol: it's cool and nice" } { "Tag": [ "search", "modular arithmetic", "number theory" ], "Problem": "Find remainder when $19^{92}$ is divided by 92.", "Solution_1": "[hide=\"Solution\"]\n$19^{92} \\equiv 361^{92} \\equiv (-7)^{92} \\equiv (-7)^{4} \\equiv 49^2 \\equiv 11^2 \\equiv 121 \\equiv \\boxed{7} \\bmod 92$\n[/hide]", "Solution_2": "note that $\\phi(92)=44$ and $(92,19)=1$ so you can reduce the exponent by multiples of $44$, and that is how t0rajirou got from $7^{92}$ to $7^4$, the theorem that he used is called euler's generalization of fermat's little theorem, if you don't know it, search first", "Solution_3": "How did you get $19^{92} \\equiv 361^{92}$?", "Solution_4": "Maybe because $19^2=361$", "Solution_5": "Then shouldnt that be $19^{92} \\equiv 361^{46}?$", "Solution_6": "If $7 \\equiv 19^{92}$, then $7^2 = 49 \\equiv 361^{92}$...", "Solution_7": "Err....no.\r\n\r\n[hide=\"Correct solution\"]\nBy Euler's totient theorem, $19^{\\phi(92)}\\equiv 1 \\pmod {92}$, since $92 = 2^2\\cdot 23$, $\\phi(92) = 92(1/2)(22/23) = 44$, so $19^{44}\\equiv 1 \\pmod {92}$. Squaring gives $19^{88}\\equiv 1 \\pmod {92}$. Thus $19^{92}\\equiv 19^4\\equiv (-7)^2\\equiv \\boxed {49}\\pmod {92}$.\n[/hide]", "Solution_8": "Oops. Did I write $361^{92}$? \r\n\r\n :blush:" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Theorem: All numbers are equal.\r\nProof: Choose arbitrary a and b, and let t = a + b. Then\r\na + b = t\r\n(a + b)(a - b) = t(a - b)\r\na^2 - b^2 = ta - tb\r\na^2 - ta = b^2 - tb\r\na^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4\r\n(a - t/2)^2 = (b - t/2)^2\r\na - t/2 = b - t/2\r\na = b", "Solution_1": "[hide]|a-t/2|=|b-t/2| :D[/hide]" } { "Tag": [ "algebra", "binomial theorem" ], "Problem": "Hi, Can any body help solve the following problems in Angel's book?\r\n\r\nWhat intergers m and n is valid for the equation \r\n(5 + 3 sqrt 2) ^ m = ( 3 + 5 sqrt 2 ) ^n.\r\n\r\nThe solution in the book gives the following hint:\r\n(5 + 3 sqrt 2) ^ m = ( 3 + 5 sqrt 2 ) ^m --> (5 - 3 sqrt 2) ^ m = ( 3 - 5 sqrt 2 ) ^n.\r\nI can not figure out the equivalence of the given equation to the second one.\r\n\r\nThank you,", "Solution_1": "[hide=\"One way to think about it\"] When the expressions are expanded via the Binomial Theorem, the terms such that $ 3\\sqrt{2}$ or $ 5\\sqrt{2}$ have an even exponent are integer, and are the same whether we are expanding $ (5\\minus{}3\\sqrt{2})$ or $ (5\\plus{}3\\sqrt{2})$, and the terms such that $ 3\\sqrt{2}$ or $ 5\\sqrt{2}$ have an odd exponent are not (and are positive when we expand the positive version and negative when we expand the negative version).\n\nIn other words, the \"integer part\" remains unchanged when we flip the sign on the radical, while the \"irrational part\" flips signs. Since the original statement implies that the \"integer\" and \"irrational\" parts are already equal, flipping the sign does not change equality. [/hide]", "Solution_2": "in other words\r\n$ (5\\plus{}3\\sqrt{2})^{m}\\equal{}A\\plus{}B\\sqrt{2}$\r\n$ (3\\plus{}5\\sqrt{2})^{n}\\equal{}C\\plus{}D\\sqrt{2}$\r\nwhere A;B;C;D are some integer numbers\r\nbecause $ (5\\plus{}3\\sqrt{2})^{m}\\equal{}(3\\plus{}5\\sqrt{2})^{n}$ so $ A\\equal{}C$ and $ B\\equal{}D$\r\nobviously $ A\\minus{}B\\sqrt{2}\\equal{}C\\minus{}D\\sqrt{2}$\r\nfinally we get $ (5\\minus{}3\\sqrt{2})^{m}\\equal{}(3\\minus{}5\\sqrt{2})^{n}$", "Solution_3": "If you mulitply those two expressions, you get $ 43^{m}\\equal{}53^{n}$...which has the only solution when $ m\\equal{}n\\equal{}0$...\r\n\r\nThis works in the original equation, hence it is a solution." } { "Tag": [ "geometry", "circumcircle", "perimeter", "inequalities", "inequalities proposed" ], "Problem": "For $m_a,m_b,m_c$ medians of a triangle of semiperimeter $s$ and circumradius $R$ prove\r\n\r\n\r\n$m_a+m_b+m_c < 2R +s $", "Solution_1": "First I solve for acute triangles .\r\n\r\nDenote p= semi perimeter\r\n[b]Lemma 1[/b] For acute triangles: $ p \\geq\\ 2R+r $\r\n\r\n[b]Lemma 2[/b]: For acute triangles: $ m_a + m_b + m_c \\leq\\ 4R+r $\r\n\r\nProof for lemma 2 \r\n Let A', B', C' be three midpoints of BC, CA, AB. \r\n We have $ m_a \\leq\\ R + OA' = R + RcosA $ ( ABC is an acute triangle)\r\n Similarly, $ m_b \\leq\\ R + OB' = R + RcosB $\r\n $ m_c \\leq\\ R + OC' = R + RcosC $\r\nAdd them together we will get $ m_a + m_b + m_c \\leq\\ R(3+cosA+cosB+cosC) $\r\nwhich is equivalent to lemma 2\r\n\r\nBy using two lemmas manlio's inequality is proved.\r\n\r\nI haven't proved for obstute triangles yet. I still try.", "Solution_2": "For obtuse triangle in angle A you must use $2m_ca when 0f(0)=0 \nthus, tanx - x>0 \ntanx>x[/hide]", "Solution_3": "[quote=\"Gabriel(fr)\"]We can prove it by the area of the $90^o$ arc[/quote]\r\ncan you elaborate?" } { "Tag": [ "geometry", "trigonometry", "AMC 10", "AMC" ], "Problem": "Hmmm, I'm going to, at least I hope I'm going to, take the AMC 10 or 12 in February for the first time, so I decided to read about it in the posts... I just heard about it this year... and I come across A and B versions of the tests... which have totally confuzzled me... what is the difference... what should I choose to take? I am relatively new at this math competition stuff... I was on a different competition track in elementary and middle school... and I am currently enrolled in Honors Alg. 2/Trignometry with a satisfactorily high grade... not super advanced, like some people I know are, but I'll just try to learn it well... By February, which is when I have the concept that the test is offered, I'll probably have finished most of Alg. 2 and delved into some trig.... perhaps I should self-study the whole book... Only problem is... I can do really well in school mathematics... but most things outside of that... well, I have difficulties. So, based on your knowledge of my mathematical ability, what do you think I should take? I really appreciate your help in advance ^^. thanks. Oh yes, and I don't know if this helps... but I also took this Mathematical Logic class at John Hopkins' CTY summer programs two years ago... load of interesting proofs... anyhow, Happy New Years everyone!", "Solution_1": "You definitely need to cover geomtry a lot. Geometry is strong at AMC-10 and if you're taking AMC-12, make sure you understand the trig and other logarithrms and stuffs very much. And it might be too late but do LOTS of problems. I think I did more problems this vacation than any other time in my lifetime for just this AMC.\r\n\r\nAMC A and B are basically same. They cover same topics, difficulty but some say A's easy, some say B's easy.\r\n\r\nHope this helps! :)" } { "Tag": [ "function", "number theory", "totient function" ], "Problem": "how do you write a program on a TI-84 that allows you to find the last $x$ digits of a number.\r\n\r\nfor example,\r\n\r\nwhat are the last 2 digits of $2^{100}$?", "Solution_1": "Use modular arithmetic.\r\n\r\nFirst, use Euler's totient theorem $\\left( a^{\\phi(n)}\\equiv 1 \\bmod n \\right)$, where $\\phi(n)$ represents the [url=http://mathworld.wolfram.com/TotientFunction.html]Totient function[/url], to reduce the exponent. In your case we would have $2^{\\phi(100)}=2^{40}\\implies 2^{100}\\equiv 2^{20}\\bmod 100$.\r\n\r\nFrom there, you can have the calculator multiply that out, or you can use the [url=http://mathworld.wolfram.com/ChineseRemainderTheorem.html]Chinese Remainder Theorem[/url] to (I think) simplify it further." } { "Tag": [ "group theory", "combinatorics proposed", "combinatorics" ], "Problem": "A polygon with 2n + 1 vertices is given. Show that it is possible to assign numbers 1, 2, \u2026, 4n + 2 to the vertices and midpoints of the sides of the polygon so that for each side the sum of the three numbers assigned to it is the same.", "Solution_1": "let the vertices of the polygon be(clockwise,say)$A_1,A_2,\\ldots,A_{2n+1}$.we label the vertices using the labels $1,2,\\ldots,2n+1$ as follows:\r\n Consider the cyclic group $\\mathbb{Z}_{2n+1}$.clearly $o(2)=2n+1$,so $2$ is a generator for the group.let the vertices of the polygon denote the group elements(i.e.$A_i\\equiv i$ in $\\mathbb{Z}_{2n+1}$).put the label $k$ on $A_{2k(mod\\ 2n+1)}$. it is easy to check that the edge sums now read $n+2,n+3,\\ldots,3n+2$.now insert the remaining numbers $2n+2,2n+3,\\ldots,4n+2$ along the midpoints of the edges such that the sums all read $5n+4$." } { "Tag": [], "Problem": "find all integers positives $x ,y ,z$ with $x^2+y^2-2z=18$", "Solution_1": "Isn't there an infinite amount of these? As long as $x$ and $y$ are both even or both odd, and $x^2+y^2>18$, there will always be a $z$ value to satisfy the equation. Am I missing something?", "Solution_2": "Yes that's right, the amount of solutions is infinite." } { "Tag": [ "Putnam", "Euler", "Gauss" ], "Problem": "If you were awarded a prize of maths, Which mathematican do you want to issue you the cup(or certificate,etc.) the most?", "Solution_1": "I don't reallyy know any famous mathematicians :oops: By the way, A Beautiful Mind was a great movie :D", "Solution_2": "A Beautiful Mind is about a great mathematician: John Forbes Nash,isn't it? ;)", "Solution_3": "[quote=\"cezar lupu\"]A Beautiful Mind is about a great mathematician: John Forbes Nash,isn't it? ;)[/quote]\r\n\r\nOh, I didn't know it was based on a real mathematician, haven't watched it in a while. Anyway, Cezar, you changed your avatar :(", "Solution_4": "Yes,I've changed it for a while. It is me there. ;)", "Solution_5": "Jan Siwanowicz, Putnam Fellow.", "Solution_6": "Well I like A Beautiful Mind mostly because I like J.F.Nash a lot.\r\nAs for me, he or P.Erd\u00f6s would be the best...", "Solution_7": "Andrew J. Wiles", "Solution_8": "[quote=\"Skyward_Sea\"]P.Erd\u00f6s would be the best...[/quote]\r\nif he was alive.", "Solution_9": "There r a ton of mathematicians that were simply great, just a few that could be names r:\r\n\r\nLeonard Euler\r\nRiemann\r\nGauss\r\nLagrange\r\nCauchy\r\nAbel\r\nGalois\r\nCayley\r\nPolya\r\nWiles\r\nVan Neumann\r\nKolmogorov\r\nChern\r\n\r\netc etc etc etc", "Solution_10": "does it have to be relatively recent? archimedes gets my vote then :surf:", "Solution_11": "Pythogoras :D", "Solution_12": "Fermat - maybe he could tell me his proof for FLT while he was giving it to me.", "Solution_13": "Gauss", "Solution_14": "i would have pythagoras if he was still alive :(", "Solution_15": "John Conway.", "Solution_16": "Archimedes", "Solution_17": "Alive: Andrew Wiles\r\n\r\nDead: Fermat." } { "Tag": [ "trigonometry", "symmetry", "calculus", "algebra unsolved", "algebra" ], "Problem": "solve in R\u00b2 the system:\r\nsin\u00b2x+cos\u00b2y=y\u00b2\r\nsin\u00b2y+cos\u00b2x=x\u00b2", "Solution_1": "Add the two equation to obtain $x^{2}+y^{2}= 2$. Substituting for y in the first equation gives $\\sin^{2}x+sin^{2}\\sqrt{2-x^{2}}= 1-x^{2}$.\r\n\r\nClearly $x=1$ is a solution. If we let $f(x) = \\sin^{2}x+sin^{2}\\sqrt{2-x^{2}}+x^{2}-1$, we see that $f'(x) = 2\\sin x \\cos x+\\frac{2x\\sin \\sqrt{2-x^{2}}\\cos \\sqrt{2-x^{2}}}{\\sqrt{2-x^{2}}}+2x > 0 \\ \\forall x \\in [0, \\sqrt{2})$. Thus $(1, 1)$ is the only solution in the 1st quadrant. By symmetry, the solutions are $(\\pm 1, \\pm 1)$.\r\n\r\n[b]Edit:[/b] Left out addition signs", "Solution_2": "An other one, whithout calculus ...\r\nFirst of all it's easy to see that $|x|\\leq \\sqrt 2< \\frac{\\pi}2 ,|y|\\leq \\sqrt 2< \\frac{\\pi}2$\r\nThere are 2 cases :\r\n$|x|\\leq |y|\\Longrightarrow y^{2}= sin^{2}x+cos^{2}y \\leq sin^{2}y+cos^{2}x=x^{2}$ , so $|x|=|y|$\r\n$|y|\\leq |x|\\Longrightarrow x^{2}= sin^{2}y+cos^{2}x \\leq sin^{2}x+cos^{2}y=y^{2}$ , so $|x|=|y|$\r\n\r\nFinally $|x|=|y|=1$\r\n :cool:", "Solution_3": "[quote=\"Diogene\"]\nThere are 2 cases :\n$|x|\\leq |y|\\Longrightarrow y^{2}= sin^{2}x+cos^{2}y \\leq sin^{2}y+cos^{2}x=x^{2}$ , so $|x|=|y|$\n$|y|\\leq |x|\\Longrightarrow x^{2}= sin^{2}y+cos^{2}x \\leq sin^{2}x+cos^{2}y=y^{2}$ , so $|x|=|y|$\n\nFinally $|x|=|y|=1$\n :cool:[/quote]\r\n\r\nI'm sorry, but could you explain this? I'm not sure how you got the inequalities...", "Solution_4": "[quote=\"mathisfun1\"][quote=\"Diogene\"]\nThere are 2 cases :\n$|x|\\leq |y|\\Longrightarrow y^{2}= sin^{2}x+cos^{2}y \\leq sin^{2}y+cos^{2}x=x^{2}$ , so $|x|=|y|$\n$|y|\\leq |x|\\Longrightarrow x^{2}= sin^{2}y+cos^{2}x \\leq sin^{2}x+cos^{2}y=y^{2}$ , so $|x|=|y|$\n\nFinally $|x|=|y|=1$\n :cool:[/quote]\n\nI'm sorry, but could you explain this? I'm not sure how you got the inequalities...[/quote]\r\n\r\nNo, It's wrong", "Solution_5": "[quote=\"hien\"][quote=\"mathisfun1\"][quote=\"Diogene\"]\nThere are 2 cases :\n$|x|\\leq |y|\\Longrightarrow y^{2}= sin^{2}x+cos^{2}y \\leq sin^{2}y+cos^{2}x=x^{2}$ , so $|x|=|y|$\n$|y|\\leq |x|\\Longrightarrow x^{2}= sin^{2}y+cos^{2}x \\leq sin^{2}x+cos^{2}y=y^{2}$ , so $|x|=|y|$\n\nFinally $|x|=|y|=1$\n :cool:[/quote]\n\nI'm sorry, but could you explain this? I'm not sure how you got the inequalities...[/quote]\n\nNo, It's wrong[/quote]\r\n??? I see nothing wrong.\r\n\r\nAssume $|x|\\leq |y|$ by symmetry.\r\nThen $y^{2}\\leq x^{2}+y^{2}=2\\Rightarrow |y|\\leq \\sqrt{2}<{\\pi\\over 2}$ and $0\\leq |x|\\leq |y|\\leq{\\pi\\over 2}$.\r\nThen $0\\leq \\sin |x|\\leq \\sin |y|$ and $\\cos |x|\\geq \\cos |y|\\geq 0$, \r\nas $\\sin$ is increasing positive on $[0,{\\pi\\over 2}]$\r\nand $\\cos$ is decreasing positive on $[0,{\\pi\\over 2}]$.\r\n\r\nSo \r\n$y^{2}= sin^{2}x+cos^{2}y= sin^{2}|x|+cos^{2}|y|\\\\ \\hspace*{0.85cm}\\leq sin^{2}|y|+cos^{2}|x|=sin^{2}y+cos^{2}x=x^{2}$\r\nand $|y|\\leq |x|$. ($\\sin$ is odd and $\\cos$ is even)\r\n\r\nTogether with the assumption this give $|x|=|y|$ and $2=x^{2}+y^{2}=|x|^{2}+|y|^{2}=2|y|^{2}$, \r\nand so $|x|=|y|=1$ and $(x,y)=(\\pm 1,\\pm 1)$.\r\n\r\nNice solution [b]Diogene[/b]!", "Solution_6": "Yes, i'm wrong. Sory of my stupid! :huh:", "Solution_7": "THanks Olorin. I wasn't thinking straight. :blush:", "Solution_8": "nice solution", "Solution_9": "it is czech- slovakia regional round 2006" } { "Tag": [ "geometry", "parallelogram", "geometry proposed" ], "Problem": "[color=darkred]In the acute triangle $ ABC$ denote the orthocenter $ H$ , the midpoint $ M$ of the side $ [BC]$ \n\nand the projections $ U$ , $ V$ of the point $ M$ on the lines $ AB$ , $ AC$ . Prove that $ \\widehat {AUH}\\equiv\\widehat {AVH}$ .[/color]", "Solution_1": "Let $ BB_1$ and $ CC_1$ are heights of the triangle. Then triangles $ HBC_1$ and $ HCB_1$ are similar and the points $ U,V$ are midpoints of segments $ BC_1, CB_1$.", "Solution_2": "[quote=\"Virgil Nicula\"][color=darkred]In the acute triangle $ ABC$ denote the orthocenter $ H$ , the midpoint $ M$ of the side $ [BC]$ \n\nand the projections $ U$ , $ V$ of the point $ M$ on the lines $ AB$ , $ AC$ . Prove that $ \\widehat {AUH}\\equiv\\widehat {AVH}$ .[/color][/quote]\r\nI think that this is similar to [b]baysa[/b]'s solution:\r\n[hide=\"Solution\"]\nLet $ CQ$ and $ BR$ be altitudes, let $ MU$ meet $ BR$ and $ X$, and let $ MV$ meet $ CQ$ at $ Y$. We have that $ MV\\parallel BR$ and $ MU\\parallel CQ$, so ${ \\frac {BX}{BH} = \\frac {BM}{MC} = \\frac {1}{2}\\implies BX = \\frac {BH}{2}\\implies XH} = \\frac {BH}{2}$. Similarly, $ HY = \\frac {CH}{2}$. Also, we have that $ \\frac {UX}{QH} = \\frac {BM}{MC} = \\frac {1}{2}\\implies UX = \\frac {QH}{2}$. Simiarly, we have that $ YV = \\frac {HR}{2}$. Now, notice that $ \\triangle QHB\\sim \\triangle RHC\\implies \\frac {QH}{BH} = \\frac {HR}{HC}$. Thus, $ \\frac {\\frac {QH}{2}}{\\frac {HB}{2}} = \\frac {\\frac {HR}{2}}{\\frac {CH}{2}}\\implies \\frac {UX}{XH} = \\frac {YV}{YH}$. Notice that $ XHYM$ is a parallelogram since $ XH\\parallel MY$ and $ HY\\parallel XM$. Thus, $ \\angle HXM = \\angle HYM\\implies \\angle UXH = \\angle HYV$. Therefore, by SAS similarity, we have that $ \\triangle HYV\\sim \\triangle HXU\\implies \\angle HUX = \\angle HVY\\implies \\angle AVH = \\angle AUH$, as desired. [/hide]" } { "Tag": [ "abstract algebra", "modular arithmetic", "function", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ G$ be a finite group of order $ pq$, where $ p,q$ are prime and $ p < q$. Suppose that $ q\\not\\equiv 1\\pmod{p}$. Prove that $ G$ is abelian and cyclic.", "Solution_1": "It's obvious with Sylow's Theorem, but can be done only with Cauchy:\r\n\r\nLet $ a$ be an element of order $ q$. Then for every element $ x$ of order $ q$, the cosets\r\n\\[ ,x,x^2,\\cdots ,x^p\\]\r\nare not all different since there are only $ p$ distinct cosets of $ $. If $ x^j\\ \\equal{}x^k$ for $ p\\geq j>k\\geq 0$, then $ x^{j\\minus{}k}\\in $ where $ q>p\\geq j\\minus{}k>0$. In particular, $ j\\minus{}k$ has a multiplicative inverse modulo $ q$, so $ x\\in $.\r\n\r\nNow that we know $ $ contains precisely the identity and every element of order $ q$, it's normal. Let $ b$ be an element of order $ p$, so conjugation by $ b$ (i.e. the function $ \\beta$ such that $ \\beta (g)\\equal{}bgb^{\\minus{}1}$) is an automorphism of $ $. The order of $ \\beta$ divides both $ p$ (as $ b^p\\equal{}1$) and $ q\\minus{}1$ (as an automorphism of $ $: $ $'s automorphism group has $ q\\minus{}1$ elements). But $ q\\not\\equiv 1\\pmod{p}$, so the order of $ \\beta$ is 1.\r\n\r\nThis means conjugation by $ b$ is the identity function on $ $. In particular, $ bab^{\\minus{}1}\\equal{}a$. We get $ G\\equal{}$ as a result." } { "Tag": [ "function", "real analysis", "real analysis solved" ], "Problem": "Find n so that this limit is finite and it is not 0:\r\n\r\nlim(x-->0) (x*e^(x^3)-sin(x)*e^(x^2)+sin(x)-x)/x^n.\r\n\r\nNot hard, but I like it...", "Solution_1": "n=3, the limit is -1" } { "Tag": [ "linear algebra", "research" ], "Problem": "Hey, I was just curious if any of you could share some independent study stratagies with me. thanks", "Solution_1": "That's a much, much, much too vague question to lead to anything useful. What aspect of independent study are you asking about? Finding a project? Actually carrying out mathematical research? Something else entirely?", "Solution_2": "Finding a project. For instance, learn about a specific topic in linear algebra or analysis with a alot of depth, rather than just a superficial skim." } { "Tag": [ "Olimpiada de matematicas" ], "Problem": "un problema de mis entrenamientos en yucat\u00e1n:\r\n\r\nsean a, b, c enteros tales que b distinto de c. si $ax^{2}+bc+c$ y $(c-b)x^{2}+(c-a)x+(a+b)$ tienen una raiz en comun, pruebe que a+b+2c es multiplo de 3.", "Solution_1": "[quote=\"Tony2006\"]un problema de mis entrenamientos en yucat\u00e1n:\n\nsean a, b, c enteros tales que b distinto de c. si $ax^{2}+bc+c$ y $(c-b)x^{2}+(c-a)x+(a+b)$ tienen una raiz en comun, pruebe que a+b+2c es multiplo de 3.[/quote]\r\ntengo dos preguntas\r\n1.- la raiz en comun es entera,racional, real?.\r\n2.-es asi $ax^{2}+bc+c$ o es asi $ax^{2}+bx+c$.\r\nchaufas espero ke respondas mis preguntas.", "Solution_2": "HOLA, CONSIDERANDO $ax^{2}+bx+c....(1)$. y \r\n$(c-b)(x_{0})^{2}+(c-a)(x_{0})+a+b..(2)$\r\nsea $x_{0}$ la solucion en comun.entonces\r\n$a(x_{0})^{2}+b(x_{0})+c = (c-b)(x_{0})^{2}+(c-a)(x_{0})+a+b$\r\n-> $(a+b-c)(x_{0})^{2}+(a+b-c)(x_{0})-(a+b-c)=0$\r\nfacorisando:\r\n$(a+b-c)[ (x_{0})^{2}+(x_{0})-1]=0$\r\nI CASO: \r\n$(x_{0})^{2}+(x_{0})-1]=0$-$x_{0}=(1+5^{1/2})/2$ \u00f3 $x_{0}=(1-5^{1/2})/2$\r\nremplasando en la ecuacion (1) y (2)\r\n$x=(1+5^{1/2})/2$\r\n en (1)\r\n$a[(1+5^{1/2})/2]^{2}+b(1+5^{1/2})/2+c] = 0$\r\n$a[(3+5^{1/2})/2]+b[(1+5^{1/2})/2]+c=0$\r\n$3a+b+2c+[5^{1/2})/2](a+b)=0...(3)$\r\n en (2)\r\n$(c-b)[(1+5^{1/2})/2]^{2}+(c-a)[(1+5^{1/2})/2]+a+b=0$\r\n$(c-b)[(3+5^{1/2})/2]+(c-a)[(1+5^{1/2})/2]+a+b=0$\r\n$3a-3b+c.5^{1/2}-b.5^{1/2}+c-a+c.5^{1/2}-a.5^{1/2}+2a+2b=0$\r\n$4a-b+c+5^{1/2}(2c-b-a)=0...(4)$\r\nde (3) y (4).sumamos\r\n$7a+3c+5^{1/2}(2c)=0 \\to c(2(5^{1/2})+3)=-7a$, de aca $c=0$, entonces $a=0 (\\to <-)$.\r\nanalogo para $x=(1-5^{1/2})/2$.\r\nentonces $a+b-c=0$, de aqui. $a+b=c..(5)$.\r\nahora nos piden demostrar que $a+b+2c$ es multiplo de 3, utilisando (5) tenemos que $a+b+2c=3c$, con lo cual se demuestra." } { "Tag": [], "Problem": "What did you all think.... I just finished a few minutes ago..\r\n\r\nI thought the Mech was pretty standard, the last response question was hard for me... I thought the E&M was exceptionally hard, I am a bit scared I might not get a 5...\r\n\r\nDon't forget, don't discuss any questions specifically.", "Solution_1": "How about don't discuss anything about it until tomorrow.", "Solution_2": "I am taking the Physics C next week, and I am wondering where I could find a simple explanation of some electricity concepts. I find it really difficult to attain an intuitive understanding of electricity.", "Solution_3": "I thought it was fairly difficult, but I did not take the class and did not study extensively. The last E&M FR was difficult - part of it, at least." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let a,b,c be positive numbers \r\nProve that\r\n $ (a\\plus{}b\\plus{}c)(\\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c}) \\geq \\frac{(a\\plus{}2b)^2}{a^2\\plus{}b(c\\plus{}a)}\\plus{}\\frac{(b\\plus{}2c)^2}{b^2\\plus{}c(a\\plus{}b)}\\plus{}\\frac{(c\\plus{}2a)^2}{c^2\\plus{}a(b\\plus{}c)}$", "Solution_1": "Nobody's interested in it?", "Solution_2": "[quote=\"8826\"]Nobody's interested in it?[/quote]\r\n\r\nNo :blush: , I'm trying.", "Solution_3": "$ LHS\\equal{}\\sum_{cyc}(1\\plus{}\\frac{b}{c}\\plus{}\\frac{b}{a})\\equal{}\\sum_{cyc}\\frac{a^2}{a^2}\\plus{}\\frac{4b^2}{bc\\plus{}ba}\\geq \\sum_{cyc}\\frac{{a\\plus{}2b}^2}{a^2\\plus{}bc\\plus{}ba}\\equal{}RHS$\r\nQ>E>D :)", "Solution_4": "[quote=\"tuandokim\"]$ LHS \\equal{} \\sum_{cyc}(1 \\plus{} \\frac {b}{c} \\plus{} \\frac {b}{a}) \\equal{} \\sum_{cyc}\\frac {a^2}{a^2} \\plus{} \\frac {4b^2}{bc \\plus{} ba}\\geq \\sum_{cyc}\\frac {{a \\plus{} 2b}^2}{a^2 \\plus{} bc \\plus{} ba} \\equal{} RHS$\nQ>E>D :)[/quote]\r\n\r\nThank you, very nice solution. You have some mistake, but no problem" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Each interior point of an equilateral triangle of side 1 lies in one of six congruent circles of radius [i]r.[/i] Prove that \r\n[i]r[/i] >= :sqrt: 3 /10", "Solution_1": ":blush: ...any help? ;) Hints are welcome, too :D \r\nArtiona", "Solution_2": "Suppose $r<\\dsp\\frac {\\sqrt 3}{10}$ and then consider the six centers $C_1,...,C_6$ of our circles. In order for our property to hold, any disk of radius $r$ centered in a point in the triangle has to contain at least one of the 6 centers. So construct 3 circles of radius $r$ centered at the vertices, and 6 more circles centered on the sides, at distance $2r$ from any of the vertices. Now, it's easy to see that these 9 circles do not intersect, except for the three pairs of circles centered on each of the sides, which intersect in a really small region (they intersect on a length of 0.04 of each side). In order for the 6 centers $C_1,...,C_6$ to lie in these 9 circles, three of them must lie in the circles that correspond to the vertices, and the other three in the common regions of the circles centered on the sides. But then it's easy to see that the centroid of the triangle is at a distance $>r$ from $C_1,...,C_6$ and thus we are done.\r\n\r\nActually, you can considerably improve the constant $\\dsp \\frac {\\sqrt 3}{10}$, but who wants to carry out the computation?", "Solution_3": "[quote=\"RHS\"]\n\nActually, you can considerably improve the constant $\\dsp \\frac {\\sqrt 3}{10}$, but who wants to carry out the computation?[/quote]\r\n\r\nThat's the most difficult part! :P \r\nCould you make an effort?", "Solution_4": "Stronger result: \\[ \\displaystyle r \\geq \\frac 1 {2(1+ \\sqrt 3)} \\]" } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "reflection", "incenter", "geometry proposed" ], "Problem": "Let ABC be a triangle with circumcentre O and Nagel point N.\r\nIf ON intersects the circumcircle(O) at P. \r\nProve that the simson line wrt. P is parallel to NH.", "Solution_1": "Take anticomplementary triangle $ \\triangle A_0B_0C_0,$ such that $ A, B, C$ are midpoints of $ B_0C_0, C_0A_0, A_0B_0.$ $ H$ is its circumcenter, $ O$ its 9-point circle center, $ N$ its incenter. Reflections of the diacentral line $ HN$ of $ \\triangle A_0B_0C_0$ in the sidelines $ BC, CA, AB$ concur at the orthopole $ P \\in (O)$ of $ HN$ WRT $ \\triangle A_0B_0C_0$ $ \\Longrightarrow$ $ HN$ is Steiner line of $ \\triangle ABC$ with the pole $ P,$ parallel to Simson line $ p$ of $ \\triangle ABC$ with the pole $ P.$ By 2nd Fontene theorem, circumcircle of pedal triangle (pedal circle), of $ \\triangle A_0B_0C_0$ WRT any point $ Q \\in HN$ goes through $ P \\in (O).$ When $ Q$ coincides with the incenter $ N,$ pedal circle coincides with the incircle, which is tangent to the 9-point circle $ (O)$ at the Feuerbach point $ \\Longrightarrow$ the orthopole $ P$ is identical with the Feuerbach point, which is identical with intersection of the ray $ \\overrightarrow{ON}$ with $ (O).$" } { "Tag": [ "logarithms" ], "Problem": "Let $\\log_{14}7=a,\\,\\log_{14}{5}=b$. Calculate: $\\log_{35}28$", "Solution_1": "[hide=\"Partial solution\"]\nWell, using the change of base formula, we find that:\n\n$\\log_{35}28 = \\frac{\\log_{14}28}{\\log_{14}35}$\n\n$= \\frac{\\log_{14}(4*7)}{\\log_{14}(7*5)}$\n\nFrom log properties, this equals\n\n$\\frac{\\log_{14}4+\\log_{14}7}{\\log_{14}7+\\log_{14}5}$\n\nFrom the values for $a$ and $b$, we see that this equals:\n\n$\\frac{\\log_{14}4+a}{a+b}$\n\nFrom here, though, I'm stuck on evaluating $\\log_{14}4$.[/hide]", "Solution_2": "[hide]Solution?\nThe beginning is similar to I Am Me's:\n\n$log_{35}28=\\frac{log_{14}28}{log_{14}35}$\n\n$=\\frac{log_{14}(14\\times2)}{log_{14}(7\\times5)}$\n\n$=\\frac{log_{14}14+log_{14}2}{log_{14}7+log_{14}5}$\n\nSo the only thing left to figure out is $log_{14}2$:\n\n$log_{14}7 = log_{14}(\\frac{14}{2}) = a$\n$log_{14}(\\frac{14}{2}) = log_{14}14-log_{14}2 = a$\nTherefore:\n$log_{14}2 = 1-a$ since $log_{14}14 = 1$\n\nBack to our original expression which becomes:\n$\\frac{1+(1-a)}{a+b}$\nor \n$\\frac{2-a}{a+b}$\n[/hide]", "Solution_3": "[hide]$\\log_{35}28=\\frac{\\log_{14}28}{\\log_{14}35}=\\frac{2\\log_{14}14-\\log_{14}7}{\\log_{14}7+\\log_{14}5}=\\frac{2-a}{a+b}$.[/hide]", "Solution_4": "Good solution, [b]pianoforte[/b] :10:" } { "Tag": [ "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "Solve Cauchy's problem: \\[ tx'-x-\\ln x'=0, \\] where $x(1)=1.$", "Solution_1": "notez x'=p. obtinem tp-x-lnp=0, care prin derivare conduce la $(t-\\frac{1}{p})\\frac{dp}{dt}=0$. se obtine $t=\\frac{1}{p}$ si $x=1-lnp$. solutia singulara este $x=1+lnt$, iar cea generala $x=tc+lnc$ care impreuna cu conditia initiala devine $x=t$" } { "Tag": [ "inequalities", "algebra", "polynomial", "binomial theorem", "inequalities proposed" ], "Problem": "Hi!\r\n\r\nLet $x_1,x_2,x_3$ be three real numbers such that $x_2x_3+x_3x_1\\geq 1+x_1x_2+x_3^2$. What is the minimal value of $c$ in the inequality:\r\n\\[\r\n(x_1-x_2)^{2n+1}+(x_2-x_3)^{2n+1}+(x_3-x_1)^{2n+1}\\geq\r\nc(x_1-x_2)(x_2-x_3)(x_3-x_1)\r\n\\]\r\nwhere $n\\in{\\Bbb N}$.", "Solution_1": "I assume $c$ has to satisfy the inequality for all real $x_1,x_2,x_3$, otherwise I've done a completely different question.\r\n\r\nJust one thing: suppose we have the inequality for some $x_1,x_2,x_3,c$. Then if we replace $x_i$ by $-x_i$, the requirement is preserved... but doesn't the inequality change direction?\r\n\r\nAnyway: let $a=x_3-x_2, b=x_1-x_3$. The requirement becomes $ab \\geq 1$, while the inequality is \r\n\\[(a+b)^{2n+1} - a^{2n+1} - b^{2n+1} \\geq cab(a+b)\\] Here we have the problem I outlined above: $a$ and $b$ must have the same sign, but the direction of the inequality depends on whether they are both positive or both negative. \r\n\r\nAssuming $a,b > 0$, we can let $a=b=1$ to get $2^{2n+1} - 2 \\geq 2c$. Hence $c \\leq 2^{2n} -1$. To prove the inequality for this value of c, expand $(a+b)^{2n+1}$ by the binomial theorem, cancel and divide across by ab. The LHS becomes a polynomial with $2^{2n+1} -2$ terms, the RHS simply $(2^{2n}-1)(a+b)$, and since $ab \\geq 1$, the inequality is trivial (just a case of repeated AM-GMs)." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ a,b,c,d>0$,prove that:\r\n $ \\frac{a\\minus{}b}{a\\plus{}2b\\plus{}c}\\plus{}\\frac{b\\minus{}c}{b\\plus{}2c\\plus{}d}\\plus{}\\frac{c\\minus{}d}{c\\plus{}2d\\plus{}a}\\plus{}\\frac{d\\minus{}a}{d\\plus{}2a\\plus{}b}\\geq 0$", "Solution_1": "where are you read this inequality?", "Solution_2": "[quote=\"super_math\"]where are you read this inequality?[/quote]\r\nMy friend told me about this inequality.why did you ask about it?", "Solution_3": "Can you write solution?", "Solution_4": "[quote=\"super_math\"]where are you read this inequality?[/quote]\r\nIf I don't wrong \r\nThis ineq is in Algebraic inequalities", "Solution_5": "Write solution if you know", "Solution_6": "[quote=\"super_math\"]Can you write solution?[/quote]\r\nIf i know solution,i would post it in \"Proposed and Own\" problems section,isn't it?", "Solution_7": "I think I've seen this before on the forum....anyway,\r\n\r\nThis is the type of problem for which it's quite difficult to find an elegant solution (olympiad type) if you don't know the idea and it's quite easy to if you do know it.\r\nThe idea is to \"get rid\" of the differences $ a\\minus{}b,\\ldots,d\\minus{}a$ as they may be negative and restrict our choices. We add $ \\frac{1}{2}$ to each term of the [i]LHS[/i].\r\nThe inequality rewrites:\r\n$ \\frac{3a\\plus{}c}{a\\plus{}2b\\plus{}c}\\plus{}\\frac{3b\\plus{}d}{b\\plus{}2c\\plus{}d}\\plus{}\\frac{3c\\plus{}a}{c\\plus{}2d\\plus{}a}\\plus{}\\frac{3d\\plus{}b}{d\\plus{}2a\\plus{}b}\\ge4$.\r\nAfter amplifying each by its numerator, and applying Cauchy we actually get equality, as $ (3a\\plus{}c\\plus{}3b\\plus{}d\\plus{}3c\\plus{}a\\plus{}3d\\plus{}b)^{2}\\equal{} 16(a\\plus{}b\\plus{}c\\plus{}d)^{2}$ and \r\n$ 4(a\\plus{}b\\plus{}c\\plus{}d)^{2}\\equal{} (3a\\plus{}c)(a\\plus{}2b\\plus{}c)\\plus{}$ $ (3b\\plus{}d)(b\\plus{}2c\\plus{}d)\\plus{}(3c\\plus{}a)(c\\plus{}2d\\plus{}a)\\plus{}(3d\\plus{}b)(d\\plus{}2a\\plus{}b)$.\r\n\r\nI've firstly seen this idea in a problem from [i]Old & New Inequalities (GIL Publishing House)[/i]. The problem was proposed by [i][b]Vasc[/b]:\nIf $ a,b,c,d$ are positive real numbers, then:\n$ \\frac{a\\minus{}b}{b\\plus{}c}\\plus{}\\frac{b\\minus{}c}{c\\plus{}d}\\plus{}\\frac{c\\minus{}d}{d\\plus{}a}\\plus{}\\frac{d\\minus{}a}{a\\plus{}b}\\ge0$\n[/i]", "Solution_8": "Very nice,thank you!", "Solution_9": "Yea, very nice", "Solution_10": "[quote=\"freemind\"]I've firstly seen this idea in a problem from [i]Old & New Inequalities (GIL Publishing House)[/i]. The problem was proposed by [i][b]Vasc[/b]:\nIf $ a,b,c,d$ are positive real numbers, then:\n$ \\frac {a \\minus{} b}{b \\plus{} c} \\plus{} \\frac {b \\minus{} c}{c \\plus{} d} \\plus{} \\frac {c \\minus{} d}{d \\plus{} a} \\plus{} \\frac {d \\minus{} a}{a \\plus{} b}\\ge0$\n[/i][/quote]\r\nWe have:\r\n$ \\frac{a\\minus{}b}{b\\plus{}c}\\plus{}\\frac{c\\minus{}d}{d\\plus{}a}\\equal{}\\frac{a\\plus{}c}{b\\plus{}c}\\plus{}\\frac{a\\plus{}c}{d\\plus{}a}\\minus{}2\\equal{}(a\\plus{}c)(\\frac{1}{b\\plus{}c}\\plus{}\\frac{1}{d\\plus{}c})\\minus{}2$\r\n$ \\geq \\frac{4(a\\plus{}c)}{a\\plus{}b\\plus{}c\\plus{}d}\\minus{}2$\r\nSimilar,we have:\r\n$ \\frac{b\\minus{}c}{c\\plus{}d}\\plus{}\\frac{d\\minus{}a}{a\\plus{}b} \\geq \\frac{4(b\\plus{}d)}{a\\plus{}b\\plus{}c\\plus{}d}\\minus{}2$\r\nDone!", "Solution_11": "[quote=\"freemind\"]The problem was proposed by [i][b]Vasc[/b]:\nIf $ a,b,c,d$ are positive real numbers, then:\n$ \\frac {a \\minus{} b}{b \\plus{} c} \\plus{} \\frac {b \\minus{} c}{c \\plus{} d} \\plus{} \\frac {c \\minus{} d}{d \\plus{} a} \\plus{} \\frac {d \\minus{} a}{a \\plus{} b}\\ge0$\n[/i][/quote]\r\nSee also here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=102924", "Solution_12": "In fact the stonger one holds (and it can be proved by Cauchy Swartz )\r\nLet $ a,b,c,d > 0$,prove that:\r\n$ \\frac {a}{a \\plus{} 2b \\plus{} c} \\plus{} \\frac {b }{b \\plus{} 2c \\plus{} d} \\plus{} \\frac {c }{c \\plus{} 2d \\plus{} a} \\plus{} \\frac {d }{d \\plus{} 2a \\plus{} b}\\geq 1\\geq\\frac {b}{a \\plus{} 2b \\plus{} c} \\plus{} \\frac {c }{b \\plus{} 2c \\plus{} d} \\plus{} \\frac {d }{c \\plus{} 2d \\plus{} a} \\plus{} \\frac {a}{d \\plus{} 2a \\plus{} b}$", "Solution_13": "[quote=\"silouan\"]In fact the stonger one holds (and it can be proved by Cauchy Swartz )\nLet $ a,b,c,d > 0$,prove that:\n$ \\frac {a}{a \\plus{} 2b \\plus{} c} \\plus{} \\frac {b }{b \\plus{} 2c \\plus{} d} \\plus{} \\frac {c }{c \\plus{} 2d \\plus{} a} \\plus{} \\frac {d }{d \\plus{} 2a \\plus{} b}\\geq 1\\geq\\frac {b}{a \\plus{} 2b \\plus{} c} \\plus{} \\frac {c }{b \\plus{} 2c \\plus{} d} \\plus{} \\frac {d }{c \\plus{} 2d \\plus{} a} \\plus{} \\frac {a}{d \\plus{} 2a \\plus{} b}$[/quote]\r\n\r\nWe have:\r\n$ \\frac {a}{a \\plus{} 2b \\plus{} c} \\plus{} \\frac {b }{b \\plus{} 2c \\plus{} d} \\plus{} \\frac {c }{c \\plus{} 2d \\plus{} a} \\plus{} \\frac {d }{d \\plus{} 2a \\plus{} b} \\geq$\r\n$ \\geq \\frac {(a \\plus{} b \\plus{} c \\plus{} d)^2}{\\sum a^2 \\plus{} 2(ab \\plus{} ac \\plus{} bc \\plus{} cd \\plus{} ad \\plus{} bd)} \\geq 1$\r\nThis inequality is true!\r\n$ \\frac {b}{a \\plus{} b \\plus{} b \\plus{} c} \\leq \\frac {b}{4(a \\plus{} b)} \\plus{} \\frac {b}{4(b \\plus{} c)}$\r\nSimilar,we have:\r\n$ \\frac {c}{b \\plus{} c \\plus{} c \\plus{} d} \\leq \\frac {c}{4(b \\plus{} c)} \\plus{} \\frac {c}{4(c \\plus{} d)}$\r\n$ \\frac {d}{c \\plus{} d \\plus{} d \\plus{} a} \\leq \\frac {d}{4(c \\plus{} d)} \\plus{} \\frac {d}{4(d \\plus{} a)}$\r\n$ \\frac {a}{d \\plus{} a \\plus{} a \\plus{} b} \\leq \\frac {a}{4(d \\plus{} a)} \\plus{} \\frac {a}{4(a \\plus{} b)}$\r\nSo we have $ 1\\geq\\frac {b}{a \\plus{} 2b \\plus{} c} \\plus{} \\frac {c }{b \\plus{} 2c \\plus{} d} \\plus{} \\frac {d }{c \\plus{} 2d \\plus{} a} \\plus{} \\frac {a}{d \\plus{} 2a \\plus{} b}$\r\nDone!", "Solution_14": "[quote=\"chien than\"][quote=\"silouan\"]In fact the stonger one holds (and it can be proved by Cauchy Swartz )\nLet $ a,b,c,d > 0$,prove that:\n$ \\frac {a}{a \\plus{} 2b \\plus{} c} \\plus{} \\frac {b }{b \\plus{} 2c \\plus{} d} \\plus{} \\frac {c }{c \\plus{} 2d \\plus{} a} \\plus{} \\frac {d }{d \\plus{} 2a \\plus{} b}\\geq 1\\geq\\frac {b}{a \\plus{} 2b \\plus{} c} \\plus{} \\frac {c }{b \\plus{} 2c \\plus{} d} \\plus{} \\frac {d }{c \\plus{} 2d \\plus{} a} \\plus{} \\frac {a}{d \\plus{} 2a \\plus{} b}$[/quote]\n\nWe have:\n$ \\frac {a}{a \\plus{} 2b \\plus{} c} \\plus{} \\frac {b }{b \\plus{} 2c \\plus{} d} \\plus{} \\frac {c }{c \\plus{} 2d \\plus{} a} \\plus{} \\frac {d }{d \\plus{} 2a \\plus{} b} \\geq$\n$ \\geq \\frac {(a \\plus{} b \\plus{} c \\plus{} d)^2}{\\sum a^2 \\plus{} 2(ab \\plus{} ac \\plus{} bc \\plus{} cd \\plus{} ad \\plus{} bd)} \\geq 1$\nThis inequality is true!\n$ \\frac {b}{a \\plus{} b \\plus{} b \\plus{} c} \\leq \\frac {b}{4(a \\plus{} b)} \\plus{} \\frac {b}{4(b \\plus{} c)}$\nSimilar,we have:\n$ \\frac {c}{b \\plus{} c \\plus{} c \\plus{} d} \\leq \\frac {c}{4(b \\plus{} c)} \\plus{} \\frac {c}{4(c \\plus{} d)}$\n$ \\frac {d}{c \\plus{} d \\plus{} d \\plus{} a} \\leq \\frac {d}{4(c \\plus{} d)} \\plus{} \\frac {d}{4(d \\plus{} a)}$\n$ \\frac {a}{d \\plus{} a \\plus{} a \\plus{} b} \\leq \\frac {a}{4(d \\plus{} a)} \\plus{} \\frac {a}{4(a \\plus{} b)}$\nSo we have $ 1\\geq\\frac {b}{a \\plus{} 2b \\plus{} c} \\plus{} \\frac {c }{b \\plus{} 2c \\plus{} d} \\plus{} \\frac {d }{c \\plus{} 2d \\plus{} a} \\plus{} \\frac {a}{d \\plus{} 2a \\plus{} b}$\nDone![/quote]\r\n\r\nVery nice :lol:", "Solution_15": "Similar problem:\r\n[b]Problem 1(Pham Kim Hung)[/b] Let $ a;b;c$ be positive real numbers.Prove that:\r\n$ \\frac {a \\minus{} b}{\\sqrt {b \\plus{} c}} \\plus{} \\frac {b \\minus{} c}{\\sqrt {c \\plus{} a}} \\plus{} \\frac {c \\minus{} a}{\\sqrt {a \\plus{} b}} \\geq 0$\r\n[b]Problem 2 [/b] Prove that:\r\n$ \\frac {a^2 \\minus{} b^2}{\\sqrt {b \\plus{} c}} \\plus{} \\frac {b^2 \\minus{} c^2}{\\sqrt {c \\plus{} a}} \\plus{} \\frac {c^2 \\minus{} a^2}{\\sqrt {a \\plus{} b}} \\geq 0$", "Solution_16": "[quote=\"chien than\"]Similar problem:\n[b]Problem 1(Pham Kim Hung)[/b] Let $ a;b;c$ be positive real numbers.Prove that:\n$ \\frac {a - b}{\\sqrt {b + c}} + \\frac {b - c}{\\sqrt {c + a}} + \\frac {c - a}{\\sqrt {a + b}} \\geq 0$\n [/quote]\r\nPham Kim Hung's first inequality is equivalent to this one:\r\n${{ \\frac{a+c}{\\sqrt{b+c}}+\\frac{b+a}{\\sqrt{c+a}}+\\frac{c+b}{\\sqrt{a+b}}\\geq \\sqrt{b+c}+\\sqrt{b+a}}+\\sqrt{a+c}}$,which is true \r\nfrom CBS inequality.", "Solution_17": "[quote=\"chien than\"]\n[b]Problem 2 [/b] Prove that:\n$ \\frac {a^2 \\minus{} b^2}{\\sqrt {b \\plus{} c}} \\plus{} \\frac {b^2 \\minus{} c^2}{\\sqrt {c \\plus{} a}} \\plus{} \\frac {c^2 \\minus{} a^2}{\\sqrt {a \\plus{} b}} \\geq 0$[/quote]\r\nVery nice too. :) \r\nProof:\r\nThis inequality is equivalent to this one:\r\n$ \\frac{(a\\plus{}b)(a\\plus{}c)}{\\sqrt{b\\plus{}c}}\\plus{}\\frac{(a\\plus{}b)(b\\plus{}c)}{\\sqrt{a\\plus{}c}}\\plus{}\\frac{(c\\plus{}b)(a\\plus{}c)}{\\sqrt{b\\plus{}a}}\\geq$ $ (a\\plus{}b)\\sqrt{b\\plus{}c}\\plus{}(c\\plus{}b)\\sqrt{a\\plus{}c}\\plus{}(a\\plus{}c)\\sqrt{b\\plus{}a}$,which is obviously true from CBS too.", "Solution_18": "[quote=\"Erken\"]which is true from CBS inequality.[/quote]\r\n\r\nWhat is this?", "Solution_19": "Cauchy-Buniakovsky-Schwarz", "Solution_20": "[quote=\"freemind\"]\nAfter amplifying each by its numerator, and applying Cauchy [/quote]\r\nDo you mean apply CS like this:\r\n$ (\\frac {3a \\plus{} c}{a \\plus{} 2b \\plus{} c} \\plus{} \\frac {3b \\plus{} d}{b \\plus{} 2c \\plus{} d} \\plus{} \\frac {3c \\plus{} a}{c \\plus{} 2d \\plus{} a} \\plus{} \\frac {3d \\plus{} b}{d \\plus{} 2a \\plus{} b})(\\sum (3a \\plus{} c)(a \\plus{} 2b \\plus{} c))\\ge...$??", "Solution_21": "[quote=\"FOURRIER\"][quote=\"freemind\"]\nAfter amplifying each by its numerator, and applying Cauchy [/quote]\nDo you mean apply CS like this:\n$ (\\frac {3a \\plus{} c}{a \\plus{} 2b \\plus{} c} \\plus{} \\frac {3b \\plus{} d}{b \\plus{} 2c \\plus{} d} \\plus{} \\frac {3c \\plus{} a}{c \\plus{} 2d \\plus{} a} \\plus{} \\frac {3d \\plus{} b}{d \\plus{} 2a \\plus{} b})(\\sum (3a \\plus{} c)(a \\plus{} 2b \\plus{} c))\\ge...$??[/quote]\r\nYes,he meant to apply CBS this way :wink:" } { "Tag": [ "geometry", "3D geometry" ], "Problem": "Find the volume in cubic inches of a cube whose surface area is $ 96$ square inches.", "Solution_1": "Each face has an area of 96/6=16, so each side is 4. Thus 4^3=64." } { "Tag": [ "algebra open", "algebra" ], "Problem": "Can anyone provide me with easy to understand links to \"rough set\" and \"Fuzzy set\" ?", "Solution_1": "Fuzzy set :\r\nIt's hard to explain in a few words here.\r\nI've got a good book about fuzzy sets, but it's in Polish :)\r\nMaybe google knows something..." } { "Tag": [ "geometry", "3D geometry", "sphere", "geometry unsolved" ], "Problem": "A minimum distance of a finite set points is defined as follows: Consider the length of all distances between each two points of this set. The shortest distance is denoted as minimum distance.\r\n\r\na.) Prove that on a sphere's surface K with radius R one can find 8 points whose minmum distance is greater than 1.15 R.\r\nb.) Prove that on a sphere's surface K with radius R one can find 8 points whose minmum distance is greater than 1.2 R.\r\nc.) Prove that the limit for the minimum distance for the situation described in a.) and b.) is 2/7* \\sqrt 7 * \\sqrt (4 - \\sqrt 2)", "Solution_1": "Are the distances mesured on the surface of the sphere or is it the usual euclidean distance?\r\n\r\nPierre.", "Solution_2": "It is the usual Euclidean distance.", "Solution_3": "To begin interrapted discussion...\r\n\r\nI have just managed to find construction with the minimal distance $d=\\frac{2R}{\\sqrt 3}\\approx 1.155$.\r\nIt looks like the following. Choose two opposite points $A$ and $B$. Construct now three edges from $A$ and $B$ or length $d$ which form equal angles. And the figure should be \"antisymmetrical\", it is if we watch at it from the top of the sphere, where $A$ is placed, then we should see six radial edges." } { "Tag": [ "inequalities", "geometry", "\\/closed" ], "Problem": "Can I ask where are all the other posts from the old forum? I hope we will have access to them.", "Solution_1": "look carefully: the inequalites are all in the inequalities. \r\n\r\nonly the topics that were in the proposed problems section (in advanced and math for college) I moved them all to algebra. from there the moderators are supposed to relocate them to their places (solved, unsolved, geometry etc.).", "Solution_2": "Upps, sorry. I noticed that earlier, but I forgot where I posted this. I hope I will acomodate with the new format (better). Interesting... Why exactly inequalities? I really like other things, too.", "Solution_3": "[quote]Why exactly inequalities? I really like other things, too.[/quote]\r\njust gave a quick example :)", "Solution_4": "there was a section includes many math resource links.\r\nwhere is it now?", "Solution_5": "Hm, will all the posts \"outside\" the subforums be moved into the subforums eventually? It's a bit strange to leave a pile of problems \"outside\" and just continue the rest \"inside\", as it might get messy if people alternately post \"outside\" and \"inside the subforums\".\r\n\r\nActually I think it might be better if the \"Post New Topic\" option inside the forum, but outside the subforums could be removed..." } { "Tag": [ "ratio", "function" ], "Problem": "If there is $ 3$ mouses eating $ 3$ cheeses in $ 3$ minutes. What is the time required if there is $ 100$ mouses eating $ 100$ cheeses?\r\n\r\nI think its $ 3$ minutes also? But I don't know a good mathematical method to do it. So its might be wrong :(", "Solution_1": "You are right.\r\n$ 1$ mouse eats $ 1$ cheese for $ 3$ minuts\r\nthen $ 100$ mouses eats$ 100$ cheeses for $ 3$ minuts.", "Solution_2": "Thanks a lot.\r\nBut how I can make it mathematically. For example if I face different numbers like:\r\n\r\nIf there is $ 5$ mouses eating $ 7$ cheeses in $ 9$ minutes. What is the time required if there is $ 91$ mouses eating $ 28$ cheeses?\r\n\r\nNow I can't solve this :(", "Solution_3": "You should just go step by step\r\n$ 5$ mouses eats $ 7$ cheeses in $ 9$ minutes , then \r\n$ 1$ mouse eats $ \\frac {7}{5}$ cheeses in $ 9$ minutes \r\n$ 1$ mouse eats $ \\frac {7}{5 \\cdot 9}$ cheeses in $ 1$ minute \r\n \r\n * * * \r\nGo on this way -step by step .I hope this is clear for you !", "Solution_4": "[quote=\"hasan4444\"]Thanks a lot.\nBut how I can make it mathematically. For example if I face different numbers like:\n\nIf there is $ 5$ mouses eating $ 7$ cheeses in $ 9$ minutes. What is the time required if there is $ 91$ mouses eating $ 28$ cheeses?\n\nNow I can't solve this :([/quote]\r\n\r\nThen in 9 minutes, 1 mouse eats $ \\frac{7}{5}$ cheeses. Thus in 9 minutes, 20 mice eat 28 cheeses.\r\n\r\nSince there are 91 mice, the time is $ \\frac{91}{20} \\cdot 9\\equal{}\\frac{819}{20}\\equal{}40.95$ minutes.\r\n\r\nThis can be generalized using quite a few variables.", "Solution_5": "Thanks a lot \"enndb0x\" and \"BOGTRO\".\r\nLOL I was mouses and its mice :rotfl:", "Solution_6": "We could also use a ratio: mice/cheeses/minutes, which is equivalent to (mice*minutes)/(cheese). Basically, this just makes the common sense (more mice=more cheese, more mice=less time) a bit more rigorous :lol:", "Solution_7": "[quote=\"aguoman\"]We could also use a ratio: mice/cheeses/minutes, which is equivalent to (mice*minutes)/(cheese). Basically, this just makes the common sense (more mice=more cheese, more mice=less time) a bit more rigorous :lol:[/quote]\r\nInteresting\r\nGood thinking anyway. :lol:", "Solution_8": "How about something like: 5 mice eat x slices of cheese in 9 minutes and x mice eat 10 slices of cheese in 2 minutes. Find x.", "Solution_9": "[quote=\"person1133\"]How about something like: 5 mice eat x slices of cheese in 9 minutes and x mice eat 10 slices of cheese in 2 minutes. Find x.[/quote]\r\nIs $ x$ is the same in both places?", "Solution_10": "Yes, both x's are the same.", "Solution_11": "[quote]Then in 9 minutes, 1 mouse eats 7/5 cheeses. Thus in 9 minutes, 20 mice eat 28 cheeses. \n\nSince there are 91 mice, the time is 91/20 * 9 = 819/20 = 40.95 minutes[/quote]\r\n\r\nThat doesn't make sense. If 20 mice eat 28 cheeses in 9 minutes, then 91 mice should be able to eat 28 cheeses in less than 9 minutes, not more.", "Solution_12": "Let there be $ m$ mice, $ c$ pieces of cheese that are eaten, and $ n$ minutes. \r\nAs $ n$ is kept the same, as $ m$ gets greater, so does $ c$. However,\r\nas $ c$ is kept the same, as $ m$ gets gets bigger, $ n$ gets smaller.\r\nTherefore, the relationship between $ c,m$ and $ n$ is $ \\frac{mn}{c}$, a\r\nfraction which is always constant.\r\nSo, $ \\frac{3*3}{3}\\equal{}3$ is our constant. Therefore, we have the equation:\r\n$ \\frac{100n}{100}\\equal{}3 \\rightharpoonup \\boxed{n\\equal{}3}$", "Solution_13": "[quote=\"hasan4444\"]But how I can make it mathematically. For example if I face different numbers like:\n\nIf there is $ 5$ mouses eating $ 7$ cheeses in $ 9$ minutes. What is the time required if there is $ 91$ mouses eating $ 28$ cheeses?\n\nNow I can't solve this :([/quote]You can look at that type of problem in terms of \"direct variation\" and \"inverse variation.\"\r\nIf doubling one variable makes the other double (for example, twice as much time, means twice as much cheese eaten), you have a direct variation. If doubling one variable makes the other half as much (for example, twice the number of workers will finish the job in half the time), you have an inverse variation.\r\nFor a direct variation, you'll make the dependent variable a factor in the function.\r\nFor an inverse variation, you make the inverse of the variable a factor. In other words, you put the variable in the denominator.\r\nIf the volume, $ V$ of a balloon varies directly with the temperature, $ T$, and inversely with the pressure, $ P$, you would write:\r\n$ V\\equal{} \\frac{kT}{P}$ $ (k$ is a constant to be determined$ )$ to show the \"joint variation\" of $ V$ with $ P$ and $ T$.\r\nThe number of cheeses eaten, $ c$, varies directly with the time in minutes, $ t$, and also directly with the number of mice, $ m$. You can write that as:\r\n$ c\\equal{}k t m$ where $ k$ is a constant to be determined.\r\nWith the data given, you can find $ k\\equal{} \\frac{7}{45}$, so $ c\\equal{} \\frac{7tm}{45}$ in your problem.\r\nNow you have $ c$ as a function of $ t$ and $ m$.\r\nYou can solve for $ t$ or for $ m$ if needed, to get\r\n$ t\\equal{} \\frac{45c}{7m}$ and $ m\\equal{} \\frac{45c}{7t}$\r\nIf they give you any two of the variables, you can find the other one.", "Solution_14": "All of these problems revolve around finding a constant $ k$, and then setting up an equation with the information you know." } { "Tag": [], "Problem": "[color=darkblue]Slove this equation: $ x^3\\minus{}3x^2\\plus{}3x\\minus{}16\\sqrt{x}\\minus{}9\\equal{}0$[/color]", "Solution_1": "Square it, and we get that as $ x$ divides every part of $ LHS$ except for $ 81$(we don't know) and it divides $ RHS$ than $ x\\mid 81$, so try all cases(I assumed that $ x\\in \\mathbb{Z}$).", "Solution_2": "YOU CAN JUST REPRESENT THE SQUARE ROOT OF THE UNKNOWN(X) BY A VARIABLE(P) SO THAT (X)=P^2. \r\nSO FOLLOWING THAT ARGUMENT,(X)^2=P^4.(X)^3=P^6.", "Solution_3": "Thank. You try again.", "Solution_4": "hello, with $ \\sqrt{x}\\equal{}t$ you have to solve\r\n$ t^6\\minus{}3t^4\\plus{}3t^2\\minus{}16t\\minus{}9\\equal{}0$.\r\nSonnhard.", "Solution_5": "[quote=\"Dr Sonnhard Graubner\"]hello, with $ \\sqrt {x} \\equal{} t$ you have to solve\n$ t^6 \\minus{} 3t^4 \\plus{} 3t^2 \\minus{} 16t \\minus{} 9 \\equal{} 0$.\nSonnhard.[/quote]\r\n\r\nYes it right. Equation $ t^6 \\minus{} 3t^4 \\plus{} 3t^2 \\minus{} 16t \\minus{} 9 \\equal{} 0$ has a root $ t\\equal{}3$?", "Solution_6": "[hide]and finally,(x)=9 :D [quote]all laws of nature are the same in all uniformly moving frames of refrence[/quote][/hide]" } { "Tag": [ "geometry", "Columbia" ], "Problem": "Hola,\r\n\r\nTal vez una forma de subir el n\u00famero de personas que participan en este foro sea si nos conocemos un poco. Pues por que no nos presentamos.\r\n\r\nBueno, all\u00ed voy yo. Mi nombre es David Jim\u00e9nez (el tipo de la foto de al lado), tengo 24 a\u00f1os, lo que me convierte en el mas viejo de los que participan en este foro con frecuencia. Lastimosamente, entre en el mundo de olimpiadas algo tarde como para hacer mucho como participante, pero Olimpiadas tuvo dos efectos positivos en m\u00ed: Me motiv\u00f3 a estudiar Matem\u00e1tica, y me hizo participar durante algunos a\u00f1os en el entrenamiento de los equipos que participan en la Centro y en la Ibero por Costa Rica. Actualmente vivo en Atlanta, Georgia, USA, donde curso mis estudios doctorales, de momento orientados hacia Teor\u00eda de C\u00f3digos. Pues al menos que alguien tenga una pregunta, creo que eso es todo lo que aplica decir por el momento. \u00bfQue tal ustedes?", "Solution_1": "Hola!, mi nombre es Leonardo Urbina y tengo 17 a\u00f1os (pronto 18 :)) . Me gradue de Bachiller en Julio de 2004, pero me quede un a\u00f1o fuera de la Universidad para poder ir por segunda vez a la IMO. Las Olimpiadas en las que he participado son la Centroamericana de 2003, la Ibero de 2003 y la IMO de 2004. Me gusta mucho la matematica, aunque no tengo un area preferida en particular, y ademas me gusta la musica, durante mi vida he tocado varios instrumentos: Violin, Piano, Cuatro y Guitarra :D. Quien mas se presenta??", "Solution_2": "Hola yo soy Pascual Restrepo de Colombia tengo 18 a\u00f1os y me gustan las matematicas mucho desde hace aproximadamente 3 a\u00f1os en los que he podido representar a colombia en varios eventos....Sin embargo quienes me conocen saben que soy un tipo tranquilo, alegre y amante de los placeres de la vida...jaja :D :D . Tambien acabo de terminar el colegio, espero participar en esta IMO (2005) y luego estudiar en Colombia que es para mi el paraiso! Suerte!!!", "Solution_3": "Yo soy Manuel Rivera de Puerto Rico (el mejor pais del mundo) y tengo 16 anos y pues llevo estudiando matematicas desde hace poco (1 ano) pero cada dia aprendo mas y me estoy fajando bastante en este tiempo. Me interesa mucho las matematicas en especial la combinatoria y grafos, y lo menos que me gusta es la geometria jaj pero necesito aprender y mejorar mas ahi. Fui a Atenas a la IMO2004 y espero ir a Mexico en el 2005 y a Eslovenia en el 2006. (y a las dos proximas iberos). Ademas de la mate me interesan los deportes, la musica (reggaeton puertorro, vdd pascual?jaja) y pasarla bien por ahi. Y pues nada espero aprender aki en mathlinks mas de mate cn ustedes que ya llevan mas tiempo!", "Solution_4": "Hola\r\nSoy David Torres de M\u00e9xico. Vivo en Le\u00f3n, Guanajuato y Me gustan mucho las m\u00e1tematicas. A la fecha he representado a M\u00e9xico en la Centro del 2004 y espero este a\u00f1o ir a la Internacional en M\u00e9rida (lej\u00edsimos por cierto (aunque puede ser que alguien de Puerto Rico o Cuba le quede m\u00e1s cerca que a m\u00ed)) y con suerte a la Ibero", "Solution_5": "Hola a todos, mi nombre es Jos\u00e9 Luis y soy de El Salvador, he participado en la centro 2003 y en la imo2005 y tambien soy parte del equipo de mi pa\u00eds para la ibero 2005, me gusta mucho la matem\u00e1tica y me perece \"chivo\" que existan sitios como este... :lol:", "Solution_6": "Bueno para extender la informacion de los miembros de este forum..\r\n\r\naqui pongo la direccion de un post \r\ncon la informacioin de los peruanos que estan participando en esta comunidad\r\nespero les interese conocer sobre este grupo !!!:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=56422\r\n\r\nCarlosBrav :)\r\nLima- PERU", "Solution_7": "me parece genial este foro, y mejor a\u00fan haber encontrado gente que habla espa\u00f1ol... :) \r\n\r\nmi nombre es Francisco Campos, tengo 16 a\u00f1os, y vivo en Santiago de Chile\r\n\r\nme encanta todo esto relacionado a olimp\u00edadas matem\u00e1ticas, fundamentalmente las internacionales\r\n\r\nencontr\u00e9 el foro por casualidad (qu\u00e9 raro haha), y me gust\u00f3 mucho saber que hay gente que se dedica a esto como una comunidad...\r\n\r\ncurr\u00edculum... mmm... s\u00f3lo una Olimp\u00edada del Cono Sur :blush: \r\npero espero ir este a\u00f1o a la Ibero... :lol: \r\n\r\nbien... eso ser\u00eda, amigos... saludos!!", "Solution_8": "Hola soy nuevo aqu\u00ed :blush: \r\n\r\nMe llamo Sebasti\u00e1n Illanes, tengo 14 a\u00f1os y soy de Chile, amigo de Gu\u00eda Rojo... encontr\u00e9 este foro gracias a otro amigo , y me pareci\u00f3 espectacular... este es mi primer a\u00f1o serio en las Olimpiadas y espero llegar a la Cono Sur el pr\u00f3ximo a\u00f1o :) . Bueno, que m\u00e1s decir, soy principiante en esto y espero subir mi nivel, que por lo cierto es bajo :wink: :blush: \r\n\r\nBueno, eso ser\u00eda\r\n\r\nSaludos :lol:", "Solution_9": "Bueno, nunca me present\u00e9, as\u00ed que ac\u00e1 vamos:\r\n\r\nMe llamo Jos\u00e9, tengo 17 a\u00f1os, y soy de Argentina. Vivo en una ciudad de 23.000 personas, que se dedica al turismo, por su patrimonio hist\u00f3rico. Participo en olimp\u00edadas desde los 10 u 11 a\u00f1os, todos los a\u00f1os. Juego al b\u00e1squet, me gusta estar en Internet, escuchar m\u00fasica, y estar con mis amigos y familiares. \u00bfLogros en olimp\u00edadas? Ninguno. Me gusta practicar matem\u00e1tica, y charlar (hablar de modo informal) con otras personas sobre matem\u00e1tica (de cualquier cosa, no solo de problemas y soluciones exactamente). Bueno, eso es todo, cualquier cosa me preguntan. Saludos a la Comunidad!", "Solution_10": "Hola...Me llamo C\u00e9sar N\u00fa\u00f1ez, tengo 19 a\u00f1os y soy de Santiago, Chile. Estudio Ingenier\u00eda en la Universidad de Chile. Desarfotunadamente, me enter\u00e9 muy viejo de estas competencias, por lo que apenas tengo un logro en una comptencia local; sin embargo, eso no implica que no quiera seguir aprendiendo cada vez m\u00e1s. Un saludo a toda la comunidad.", "Solution_11": "Un amigo m\u00edo tambi\u00e9n se llama as\u00ed, y tiene tambi\u00e9n 19 a\u00f1os! Linda casualidad", "Solution_12": "A great time to practice my Spanish!!\r\n\r\nHola. Me llamo Sean. Tengo 19 a\u00f1os. Vivo en Carolina del Sur en el EE.UU, pero yo soy Indian. Mi asignaturas favoritas son matematicas y ciencias. No me gusta musica y Ingles. Adios amigos! Corriges mis errores, por favor!!", "Solution_13": "[quote=\"mysmartmouth\"]A great time to practice my Spanish!!\n\nHola. Me llamo Sean. Tengo 19 a\u00f1os. Vivo en Carolina del Sur en el EE.UU, pero yo soy Indian. Mi asignaturas favoritas son matematicas y ciencias. No me gusta musica y Ingles. Adios amigos! Corriges mis errores, por favor!![/quote]\r\n\r\nSoy indio. \r\nNo me gustan ni la m\u00fasica y ni el ingl\u00e9s.\r\nCorrige mis errores...\r\n\r\nTen\u00e9s un buen espa\u00f1ol! (You have a good spanish!)", "Solution_14": "[quote=\"Jos\u00e9\"]\nCorrige mi errores...\n[/quote]\r\n\r\nSo I assume we are using the imperative tense of corrigar, which is corrige? And we don't use mis, even though we have a plural noun?", "Solution_15": "corrige mis errores / correg\u00ed mis errores\r\n\r\n(it was just a typo). Yes, you use the imperative (us\u00e1s el imperativo)", "Solution_16": "Hola a todos, no habia visto este topico de presentarse. A ver, yo soy Daniel Hernandez, competi por Cuba en las IMO, ahora estoy en Madrid haciendo el doctorado en f\u00edsica te\u00f3rica. Pero de vez en cuando entro y me quedo con algun problemita que me guste para no perder la costumbre. 24 a\u00f1os.\r\nLa cosa m\u00e1s divertida que hay es coger una casa de campa\u00f1a y una mochila e irse a la monta\u00f1a o a cualquier lugar que incite a tu curiosidad por descubrir. Por suerte eso todav\u00eda lo podemos hacer en los paises latinoamericanos, en Europa ya poco monte queda inexplorado.\r\nLa segunda cosa mas divertida es la fiesta con tus amigos. Bailar o tomar el te, da igual mientras estes compartiendo con la gente de tu bando\r\nLa tercera cosa mas divertida son los placeres intelectuales, leer, en el tema cientifico aparte de la fisica, me encantan desde luego las matematicas y disfruto programando.\r\nY ya! que vaya pedazo de presentacion!\r\n\r\nDaniel (La cosa blanca detras de mi Avatar es un rapido en el Nilo :))\r\n\r\n[color=red] Ah se me olvidaba preguntar, alguien de Cuba aqui?? alguien de Espa\u00f1a??[/color]", "Solution_17": "[color=green]Hola a todos!!! :) Mi nombre es Gustavo Esp\u00ednola, tengo 17 a\u00f1os, vivo en Capiat\u00e1, a 21 kil\u00f3metros de la capital, Asunci\u00f3n. Si bien estoy registrado desde hace un buen tiempo, estaba s\u00f3lo como oyente... Mas me gusta el grado de respeto con el que se trabaja, lo que hace ideal el ambiente para compartir mis experiencias. Es una satisfacci\u00f3n para m\u00ed estar con ustedes, y espero aprender un poco m\u00e1s y ayudar en lo que pueda...\n\nComo experiencia en olimpiadas matem\u00e1ticas, adem\u00e1s de la Nacional, s\u00f3lo Cono Sur en el 2005, mi objetivo ser\u00e1 la IBERO de Portugal... ya que me voy a la universidad desde el pr\u00f3ximo a\u00f1o (chau IMO...). \n\nCreo que es todo... Saludos!!![/color]", "Solution_18": "[color=blue]Soy Rafael Arias. Hab\u00eda creado, sin saber que exist\u00eda este t\u00f3pico, uno similar, que es el siguiente:\n\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=110616[/url]\n\nDisc\u00falpenme por mi error. :blush: A\u00fan era nuevo en el foro cuando sucedi\u00f3 eso.[/color]", "Solution_19": "Soy Jorge Castillo. No soy olimpico, sino que universitario :blush: . Estudio ingenieria civil matem\u00e1tica en la universidad de concepcion, en 2 \u00b0 a\u00f1o, tgo 19 a\u00f1os, y eso..saludos", "Solution_20": "Hola, soy Roland Hablutzel, venezolano, y particip\u00e9 en la OMCC y en la OIM en el 2004, y en el 2005 en la IMO y en la OIM, y con premios en todas ellas (aunque nada importante :blush: ). A partir de ah\u00ed empez\u00e9 la universidad en f\u00edsica, ya que conociendo tan profundamente la matem\u00e1tica no quise estudiarla, me volvi\u00f3 loco, jajaja, sin embargo me ha sido \u00fatil todo lo que he aprendido en ellas. En mathlinks entr\u00e9 un poco tarde... pero me sirve para no perder el ritmo de resolver problemas. Me da gusto esta reuni\u00f3n de hispanohablantes ya que hay pocos sitios as\u00ed.", "Solution_21": "Hola!\r\n\r\nSoy Esteban Arreaga, tengo 17 a\u00f1os, vivo en la ciudad de Guatemala, Guatemala. Hasta ahora solo he participado en una centro y una ibero, espero estar al menos en otra ibero. Me encanta la matem\u00e1tica, en especial las cosas algebraicas, soy malo en geometr\u00eda aunque tambien la disfruto. Entro a la universidad en el 2009, as\u00ed que talvez todav\u00eda tengo chance.\r\n\r\n\r\nSaludos a todos!!\r\n\r\n\r\n :)" } { "Tag": [ "function", "geometry", "pigeonhole principle", "combinatorics proposed", "combinatorics" ], "Problem": "We have a closed path on a vertices of a $ n$\u00d7$ n$ square which pass from each vertice exactly once . prove that we have two adjacent vertices such that if we cut the path from these points then length of each pieces is not less than quarter of total path .", "Solution_1": "A really nice one, it is hard to explain without pictures but I will try to give a sketch of the proof. \r\nInitially take the adjacent squares with the maximum lenght of the smaller path between them and assume that this small path's lenght is lesser than one quarter of the total path. Now, divide into two parts, first if at least one of these two squares is not on an edge and second both of them are on an edge. First case is really simple, a casework using the maximality of these two squares leads directly to a contradiction. (Yet, it's hard to explain it without too many pictures so i'm leaving this part now, if a problem occurs, tell me and i will answer it.) The main part is the second part. Let's look at the picture, the squares we have selected are the ones with a red and black edge connected to them. (since there is no edge between them, the edges that pass from them must be like red and black edges) Start from one of the black edges and go along with the path without passing through our selected squares. We can't reach a red edge before we reach the other black edge. So divide the path into two parts, the black edges and the path connects them without passing through our selected squares; and the red edges and the path connects them without passing through our selected squares. Say the first part black path and the second part red path. We can say from the maximality that the red path is the smaller one. Otherwise we could find a bigger smaller part by taking the squares that are on the just right of our selected squares.The squares that are on the black path are black squares and the others are red squares, two selected squares are also black. \r\nClaim 1: Red squares are a connected set and there is no black squares inside red squares. Moreover there is no red square that is adjacent to the edge of the big square. \r\nProof: Easy properties that can be proved by using the black squares are also connected. So I'm again leaving the proof. \r\nTherefore, the red squares has a shape like the grey colored squares in the picture. Say P to the polygon that formed of the edges of the shape of red squares. Now let's define a f function for the edges of lenght 1 on P (like the yellow edge on picture) A be an edge like that, one side of A is a red square and one side is a black square from the claim 1. Start from the edge next to the upside selected square say this edge A_1, go along P from the direction that doesn't pass through the edge of downside selected square and tell the i th reached edge lenght 1 A_i. (For example in the picture the yellow edge is A_7, while the downside selected squares' edge is A_24) \r\nClaim 2: If i>j, the black path between the upside black edge (the one that drawn on picture) and the black square that contains A_i is longer than the one with the square that contains A_j.\r\nProof: Again start from the upside black edge and go along the black path, assume that we first reached A_i. Since A_i and A_j are on the edge of P and the black path can't go inside P, it is clear that it's impossible to connect A_j and the downside black edge. Contradiction. So claim 2 is true. Moreover the same for the red path is also true, I mean If i>j, the red path between the upside black edge (the one that drawn on picture) and the red square that contains A_i is longer than the one with the square that contains A_j. \r\nf(A_i) is equal to the lenght of the path between the two squares that contain A_i as an edge and passing through the upside red and black edges (on the picture again). From claim 2, f(A_i) is a non decreasing function. Because of the maximality of the first selected squares, f(A_i) must be either lesser than one quarter or more than 3 quarter of the total path.\r\nClaim 3: There is an i such that f(A_{i+1})-f(A_i) is bigger than the half of the total path.\r\nProof: f(A_1)=1 and f(the downside selected squares' edge)= Total path-1. So f(A_i) starts to be more than 3 quarter of the total path at some i, take the smallest such i. Then, f(A_{i-1}) is lesser than 1 quarter of the total path. Result follows.\r\nNow we have a little to do remaining. Assume that for i, f(A_{i+1})-f(A_i) is bigger than the half of the total path. There are 3 cases\r\ni) A_i and A_{i+1} are not perpendicular to each other. The red path between the red squares that contains A_i and A_{i+1} is lesser than one quarter of the total path, so the black path between the black squares that contain A_i and A_{i+1} must be bigger than one quarter of the total path. Moreover the other path between the black squares that contain A_i and A_{i+1} includes the whole red path, so it can't be the smaller path between the black squares that contain A_i and A_{i+1} because of the maximality of the first selected squares. So it's again a contradiction!\r\nii) A_i and A_{i+1} are perpendicular and has an angle 90 degrees in P. The path between the black squares that contain A_i and A_{i+1} as an edge is longer than the half of the total path. Let's look at the black square that is a neighbour to the both black squares that contain A_i and A_{i+1} as an edge. Say this square X, an argument like in claim 2 shows us that the black path between the black squares that contain A_i and A_{i+1} as an edge passes through X. So the black path between X and one of the black squares that contain A_i and A_{i+1} as an edge must be bigger than one quarter of the total path. So it's again a contradiction as in i)\r\niii) A_i and A_{i+1} are perpendicular and has an angle 270 degrees in P. since the red path is lesser than one half, this is absurd.\r\nFinally we are done, if it is unclear, I am sorry. I can answer your questions if you ask.", "Solution_2": "very very nice problem :). here's my solution:\n\nit's easy to see that $n$ must be even for such a path to exist, so from now for comfort we write it as $2n$.\nthis closed path divides the area of the square into two parts, inside and outside. if we consider each square in the inside part as a vertice of a graph and connect two vertices if their corresponding squares are adjacent, we get a tree. (one can easily show that this graph is connected and does not have any cycles). by pick's theorem we get that this tree has $2n^2-1$ vertices. the maximum degree of this tree is $4$, since each square can be adjacent to at most $4$ squares.\n\n[i]Lemma[/i]: if the maximum degree of a tree is $4$, then there is an edge in it that divides the graph into two parts, each having at least $\\frac{n-1}{4}$ vertices.\n\n[i]Proof[/i]:if an edge has the above property, we call it good. let $v$ be an arbitrary vertex and make it the root of the tree. let it's neighbors be $v_1$ (possibly $v_2,v_3,v_4$). let the subgraph of our tree with the root $v_i$ be $V_i$. now, by pigeonhole principle, one of the $V_i$'s, has at least $\\frac{n-1}{4}$ vertices, namely $V_1$. if the edge $vv_1$ is not good, we conclude that $V_1$ must have at least $\\frac{3(n-1)}{4}$ vertices except $v_1$. now do the same, again by pigeonhole principle, we get that in one of the subgraphs with roots neighbours of $v_1$, there is at least $\\frac{n-1}{4}$ vertices. if the edge $v_1v_2$ is not good, again we conclude that this subgraph must have at least $\\frac{3(n-1)}{4}$ vertices except $v_2$ and so on. this process must stop somewhere and we must get a good edge, since if not, we get a graph with infinite vertices. so the proof is complete.\n\nnow back to the main problem, there is an edge $ab$ dividing the graph into two parts each having at least $\\frac{n^2-1}{2}$ vertices. this edge cuts the lattice edge between two vertices, namely $t$ and $s$. we claim that $t$ and $s$ have the desired property. delete the edge $ab$ of the graph. now we have two trees, each having at least $\\frac{n^2-1}{2}$ vertices. again using pick's theorem, we easily find that each of the two paths connecting $t$ and $s$ have at least $n^2$ edges, so the proof is complete. :)", "Solution_3": "This beautiful problem has a solution using this problem as its main idea : https://artofproblemsolving.com/community/c6h85771p499536" } { "Tag": [ "geometry", "parallelogram", "vector", "geometric transformation", "rotation", "complex numbers" ], "Problem": "Four squares are constructed externally on the sides of a parallelogram. Show that the centers of these squares form a square. \r\n\r\nI know there's a simple vector solution, but what about a purely Euclidean solution?", "Solution_1": "Let the intersection of the diagonals of the parallelogram be $O$, the origin of the complex plane $\\Pi$. Let the vertices of the parallelogram be $a,b,-a,-b$. Also, let the centers of the squares be $s_{1},s_{2},s_{3},s_{4}$. Then note that the diagonals are orthogonal since upon division of the complex numbers $s_{i}-s_{j}$ and $s_{m}-s_{n}$, you obtain $\\pm i$. The conclusion follows.\r\n\r\nI haven't thought of the Classical Geometry proof, yet.\r\n\r\nMasoud Zargar", "Solution_2": "hmm... seems like there should be a solution involving rotation. Maybe you could show that if $X$ is the intersection of the diagonals of the parallelogram, then a $90^\\circ$ rotation around $X$ would result in each center being sent to another center. While I can't figure out how I could prove this, I know that a $180^\\circ$ rotation clearly does this (because it leaves the parallelogram unchanged), which at least provides an easy proof that the centers form a parallelogram." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "This problem is generalization of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5918]this one[/url].\r\nSuppose $G$ is a graph and $S\\subset V(G)$. Suppose we have arbitrarily assign real numbers to each element of $S$. Prove that we can assign numbers to each vertex in $G\\backslash S$ that for each $v\\in G\\backslash S$ number assigned to $v$ is average of its neighbors.", "Solution_1": "I wanted to post this generalization after seeing these problems:\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=468314#p468314[/url]\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=468314#p468314[/url]\r\n\r\nIf $S$ does not meet every connected component of $G$, then it's easy to take care of the components disjoint from $S$: just assign the value $0$ to each point in such a component. This means that it suffices to restrict our attention to the case when $S$ does meet every component, and in this case the proof in that first link works, with only the obvious modifications (that last argument in that topic basically says that in that particular case, $S$ meets every component; since here we're already assuming this, no such argument is needed).", "Solution_2": "Let $T = V(G) / S.$ We interpret the desired labelling as a system of equations. Consider the adjacency matrix $A$ of the vertices in $T$, with the one difference that we write $-deg(v_i)$ along the main diagonal. In this way, the problem's condition can be interpreted as:\n\n$$A \\cdot v = v',$$\n\nwhere $v$ is the vector of the labellings of the elements of $T$, and $v'$ is some fixed vector. We need to show that this has a solution.\n\nFirst of all, as [b] grobber [/b] did above, we can WLOG assume that there is no connected component in which $S$ contains no elements. \n\nWith this assumption, we're ready to show that the rows of $A$ are linearly independent. Indeed, suppose that they were not for contradiction. Then, if we let $r_i$ be the vector corresponding to the $i$th row of $A$, then there exist real numbers $x_i$ so that\n$$\\sum x_ir_i = 0.$$\nTo see that this cannot occur, consider the largest $x_i.$ Then since $-x_ideg(v_i)$ must be summed with other terms to make the $i$th coordinate zero, we know that all neighbors $v_j$ of $v_i$ must be in $T$, and have $x_j = x_i.$ If we continue this logic, we get that the connected component of the subgraph of $G$ spanned by $T$ which contains $x_i$ is adjacent to no vertices of $S$, i.e. it should've been removed by the above assumption, contradiction.\n\n$\\square$ \n", "Solution_3": "Another approach. The idea is at first to assign some number to each vertex of $V(G)\\setminus S$ and then iterate assigning to each vertex of $V(G)\\setminus S$ the average of its neighbours. If the corresponding real sequences, obtained for each vertex, converge then their limits will do the job. \n\tIt's not certain that each initial assigning will bring convergence, it's even not true, but we show it holds for a specific initial assignment.\n\t\n\tTo begin with, it's enough to prove the claim when all numbers assigned to $S$ are non negative. Indeed, in general case, we can represent each number in $S$ as difference of non negative numbers, then get missing numbers to both configurations and finally subtract vertex-wise corresponding values. So, we may assume the assigned numbers to $S$ are non negative.\n\n\tLet $M$ be the maximal number assigned to $S$. By $v^{k}_j$ we denote the number assigned to the vertex $v_j\\in V$ at step $k$. We set the initial assigning as $v^{(0)}_j:=M\\,,\\,\\forall j,v_j\\notin S $. We construct inductively the sequence $v^{k}_j, k=1,2,\\dots$ \n\t$$ (*)\\,\\,\\,\\,\\,\\,\\,\\, v^{(k+1)}_j:=\n\t\\begin{cases} \n\t\\displaystyle \\frac{1}{|N(v_j)|} \\sum_{v_i\\in N(v_j)} v^{(k)}_i\\,,& v_j\\notin S\\\\\n\tv^{(k)}_j\\,,& v_j\\in S \n\t\\end{cases}\n\t$$\n\twhere $N(v_j)$ denotes the set of verticies adjacent to $v_j$ . Clearly $v^{(1)}_j\\leq v^{(0)}_j\\,,\\,\\forall v_j\\in V$. \n\tNow, by induction, applying $(*)$, we conclude $v^{(k+1)}_j\\leq v^{(k)}_j\\,,\\,\\forall v_j\\in V;k=0,1,\\dots$ . It means the sequence $v^{(k)}_j$ for fixed $j$ and $k=0,1,\\dots$ is monotonically decreasing and non negative, hence it converges to $v^*_j$. \n\tFinally, taking the limit in $(*)$, we conclude the assignment $v^*_j$ satisfies the requirement. " } { "Tag": [ "trigonometry", "function", "trig identities", "Law of Cosines" ], "Problem": "Find the length of a side of a regular polygon of $24$ sides, incribed in a circle of unit radius.", "Solution_1": "[quote=\"Hikaru79\"]Find the length of a side of a regular polygon of $24$ sides, incribed in a circle of unit radius.[/quote]\r\n[hide] Draw 2 radii which intercept one side of the 24-gon. This forms an isoceles triangle with vertex angle $\\frac{360}{24}=15^\\circ$. With the Law of Cosines, $x^{2}=1^{2}+1^{2}-2\\cdot1\\cdot1\\cdot\\cos15=2-2\\cos15=2-2(\\cos45\\cos30+\\sin45\\sin30)=2-2(\\frac{\\sqrt6+\\sqrt2}{4})=\\boxed{2-\\frac{\\sqrt6+\\sqrt2}{2}}$ [/hide]", "Solution_2": "You've found $x^{2}$. Still have to take a root of that.", "Solution_3": "I think there's an easier way.\r\n[hide]\nWe know that the angle from vertice $v_{1}$ to vertice $v_{2}$ is $15^\\circ$ (because $\\frac{360}{24}=15$) Lucky for us, two of the well known trig functions are $15$ apart -- $30$ and $45$! So we know that there is a vertice at $(\\cos{30},\\sin{30})$ and one at $(\\cos{45},\\sin{45})$. The length of the vertice is just the distance between those two. Plug those two points into a distance formula and you're done. The choice of the points is important because we know exact values for 30 and 45.\n[/hide]" } { "Tag": [ "topology", "induction", "geometry", "3D geometry", "sphere", "function", "combinatorial geometry" ], "Problem": "It is a well-known old Kvant problem that any $n$ red and $n$ blue points in the plane, no three of which are collinear, can be grouped into $n$ pairs in such a way that every pair consists of points of different colors, and no two of the $n$ segments formed by the pairs intersect (to show this, just consider the grouping that minimizes the sum of the lengths of these segments). Here is an analogous problem for three dimensions:\r\n\r\n[b]Problem. [/b] Consider any set of $3n$ points in $\\mathbb{R}^3$, no four of which are coplanar. Color the points in three colors (red, green, blue) arbitrarily, with each color used exactly $n$ times. Is it always possible to partition the points into triples $(A_i,B_i,C_i)$ with $A_i$ red, $B_i$ green, $C_i$ blue, in such a way that the $n$ triangles $A_iB_iC_i$ are pairwise disjoint?\r\n\r\n[Edited. ]\r\n\r\n--Vesselin", "Solution_1": "I think this problem is also rather well known. Take a line not parallel to any of the lines determined by the triangles \"far away\" from the set of points, and then start moving it (keep its direction fixed) towards the set of points. It will meet our points one by one. We form our triangles from the first three points it meets, then the next three, then the next three, and so on.", "Solution_2": "I didn't say that the problem is difficult :) . (Only you have to take a [i]plane[/i] instead of a line: the points are in $\\mathbb{R}^3$).", "Solution_3": "Oh, sorry about that. I missed that $\\mathbb R^3$ there :).", "Solution_4": "Oh, silly me, I knew that I had missed something :blush: :blush: :blush:. Now, this will make the problem non-trivial; look back at my edited post above :) .", "Solution_5": "I have seen a proof that uses topology, namely the Ham Sandwich theorem.", "Solution_6": "another general result:Let $A_1,\\dots,A_d$ be d pairwise disjoint subsets of $\\mathbb{R}^d$,each containing n points ,and suppose that no hyperplane contains $d+1$ of the points in the union of all the sets $A_j$,then there is a partition of $\\cup A_j$ into disjoint sets $S_1,\\dots S_n$ each containg pricisely one point from each $A_i$,such that the n simplices $conv(S_1),\\dots ,conv(S_n)$ are pairwise disjoint.", "Solution_7": "Can someone prove Vess's edited problem. :( \r\nplease and thanks ;)", "Solution_8": "Man, is this going to be long.. :)\r\n\r\nAssume we prove the following result:\r\n\r\n[b]Theorem 1[/b]\r\n\r\nGiven $3n$ points in space together with another point $P$, no four coplanar, s.t. among the first $3n$ points there are $n$ of each color (red, green, blue), we can find a plane $\\pi$ through $P$ s.t. the complement of $\\pi$ in $\\mathbb R^3$ consists of two disjoint open half-spaces, each one containing equal numbers of points of the three colors. \r\n\r\nThen the positive answer to the problem in post #1 would follow by induction on $n$:\r\n\r\nTake the point $P$ inside the convex hull of the $3n$ points, not lying on any of the planes determined by the $3n$ colored points. Take a plane $\\pi$ as in the statement of Theorem 1. One of the open half-spaces it determined contains $3u$ points, $u$ of each color, while the other one contains $3v$ points, $v$ of each color, with $u+v=n,\\ u,v>0$. $u$ and $v$ must be positive because $P$ is inside the convex hull of the $3n$ points, so there are no half-planes through $P$ which have all the $3n$ points on the same side. By the induction hypothesis, we can construct $3$-colored disjoint triangles on the $3u$ points and also on the $3v$ points, since $u,v$ enclosed in the interval $\\left( \\frac{9-\\sqrt{81-4e^{2}}}{2}, \\frac{9+\\sqrt{81-4e^{2}}}{2}\\right)$. It is equivalent to two inequalities:\r\n1. $\\frac{9-\\sqrt{81-4e^{2}}}{2}<1$\r\n2. $e^{2}<\\frac{9+\\sqrt{81-4e^{2}}}{2}$.\r\nProof 1. The inequality is equivalent to $\\sqrt{81-4e^{2}}>7\\iff e^{2}<8$ which is obvious. \r\nProof 2. It is equivalent to $\\sqrt{81-4e^{2}}>2e^{2}-9$. Both sides are positive, so squaring we get $81-e^{2}>4e^{4}-36e^{2}+81 \\iff$ $4e^{4}-35e^{2}<0\\iff 4e^{2}-35<0\\iff e^{2}<\\frac{35}{4}$ which is obvious because $e^{2}<8$ and $8<\\frac{35}{4}$. The proof of $xe^{x^{2}}+ye^{y^{2}}<3$ is ended. Of course $xe^{x^{2}}+ye^{y^{2}}<3$ implies $xe^{x^{2}}+ye^{y^{2}}\\ne 3$ so we are done." } { "Tag": [ "limit", "trigonometry", "irrational number", "continued fraction", "real analysis", "real analysis unsolved" ], "Problem": "Do you know any problems or references on the sequence $\\sin(n)^{n}$ ?\r\nthank's", "Solution_1": "I made the following problem.\r\n\r\nEvaluate $\\lim_{n\\to\\infty} \\frac{1}{(n+1)^n} \\sum_{k=1}^n \\sin ^{-1} (\\sin k^k).$", "Solution_2": "kunny, is that $\\sin\\left(k^{k}\\right)$ or $\\sin\\left(k\\right)^{k}$?", "Solution_3": "[quote=\"amcavoy\"]kunny, is that $\\sin\\left(k^{k}\\right)$ or $\\sin\\left(k\\right)^{k}$?[/quote]\r\n\r\nIt's $\\sin\\left(k^k\\right).$", "Solution_4": "I wonder if someone can explain how we might be sure that $\\lim_{n\\to\\infty}\\sin^n(n) = 0$. I mean, obviously the terms will be very small for most $n$, but for some $n$, we will have $\\sin(n)$ very close to 1. Can we prove that we will never have some $n$ so close to a $k\\pi$ that $\\sin(n)^n$ still is close to 1? It seems to me that this should never happen ($n$ will be too large) but I don't know how to make a formal argument.", "Solution_5": "It doesn't go to zero. There exists a $c$ and infinitely many pairs $m,n$ such that $|n-m\\pi-\\frac\\pi2|<\\frac cn$, whence $|\\sin n|>1-\\frac{c^2}{2n^2}$ and $|\\sin n|^n>1-\\frac{c^2}{2n}$.\r\n\r\nThis would be much easier to prove for $\\cos n$, but I'm sure it's true for $\\sin n$ as well.", "Solution_6": "Oh, right! Now I recall something, is it Hurwitz' irrational number theorem?\r\nIf $x$ is irrational, then there are infinitely many $p, q$ so that \r\n$|x - \\frac{p}{q}| < \\frac{c}{q^2}$ for some $c$ (which I think is even less than 1). \r\nWill that give what we want? Maybe it doesn't, or at least not immediately. But it's the same idea, I guess.", "Solution_7": "You can take $c=1$ in that theorem- every convergent of the continued fraction is that good. The problem we're looking at is harder, since it's shifted by $\\frac12$.", "Solution_8": "[quote=\"jmerry\"] The problem we're looking at is harder, since it's shifted by $\\frac12$.[/quote]\r\nThis is to formally justify jmerry's claim. The statement we need is the following: if $x$ is irrational, then there exist infinitely many [b]odd[/b] $n$ such that $\\text{dist}(nx,\\mathbb Z)\\le \\frac Cn$ where $C$ is some absolute constant. By the theorem mentioned by Xevarion, we can find infinitely many irreducible fractions $\\frac pq$ such that $\\left|x-\\frac pq\\right|<\\frac 1{q^2}$. If $q$ is odd, take $n=q$. If $q$ is even, find $n,m\\in \\mathbb Z$ such that $0\r\nBlogger \r\nGet your own blog Flag Blog\r\nNext blog\r\n\r\nBlogThis! \r\n\r\n\r\nThe Daily Demarche \r\nThursday, July 21, 2005\r\nAl Jazeera supporters are right.\r\nAs London recovers from the latest act of terror to strike the city a\r\nterrorist and mass murderer in Iraq begins his march to justice. Saddam\r\nhas begun his day in court, unfortunately media coverage is minimal due\r\nthe continued threat to the west posed by al Qaeda and its sympathizers.\r\nLuckily al Jazeera is covering the trial\r\n:\r\n\r\n/Long-bearded Saddam, who was wearing a white shirt, a suit jacket and\r\nglasses, appeared to be defiant and stared directly into the judge's\r\neyes, who in turn seemed uncomfortable and avoided eye contact with the\r\nousted president./\r\n\r\nI'd be uncomfortable too, looking into the eyes of a man who ordered a\r\nprovince of his own country to be gassed, although I assume al Jazeera\r\nfinds the judge's revulsion to be a victory of will for Hussein. Not\r\nsurprisingly, some of their readers feel the same way (from the comments\r\non the AJ site):\r\n\r\n/he lived like a lion and even living like lion. these pupets don't dare\r\nto look into his eyes. they know he can not do a thing but their guilty\r\nand fear don't let them to look into his eyes. salute to a lion heart./\r\n\r\nand\r\n//\r\n/saddam husseins trial should be open so that the whole world can\r\nunderstand the history of these past crimes and that there are many\r\nothers deeply implicated, particularly the usa. /\r\n\r\nLuckily, there are other sources of news to refer to- the Internet\r\nallows anyone interested in learning more to do so, easily. Take this\r\nview of\r\nSaddam, for example:\r\n\r\n/A man describes how Saddam Hussein's secret police shoved a dissident's\r\nbaby into a sack with a vicious cat that scratches it. /\r\n\r\n/Undercover agents throw a man to his death from the roof of a building./\r\n\r\n/Iraqiya state television is reviving images of life under Saddam as a\r\ncourt prepares to announce his trial date./\r\n\r\n/``I wish they were here to see the day when Saddam is finished,'' a\r\ntearful woman who lost her relatives under Saddam tells viewers of\r\nIraqiya, which broadcast footage of abuses filmed by members of Saddam\r\nsecurity forces as they committed them./\r\n//\r\n/Grainy footage of senior officials, including Ali Hassan al-Majid -\r\nnicknamed Chemical Ali because his men allegedly gassed 5,000 Kurds in\r\n1988 - shows them questioning Shiites after a failed rebellion in 1991./\r\n\r\n/One official calmly smokes a cigarette and then kicks one of them in\r\nthe face. The bound men were later executed. Other reminders include a\r\nblindfolded man with his hands bound behind his back being pushed off\r\nthe roof of a building./\r\n\r\n/Another scene captured on video shows a man being held on the ground\r\nwith an arm extended. The arm is beaten with a club until the bone breaks./\r\n//\r\nSo, for perhaps the first and last time, I agree, at least in part, with\r\nsentiments expressed on the al Jazeera site- the trial of Saddam Hussein\r\nshould be open and transparent, it should be broadcast on the satellite\r\nchannels and the Internet, the entire world should have the\r\nopportunityto view the process of the rule of law in action. The lion of\r\nthe desert will quickly be exposed as the hyena he is, or was- with full\r\napologies to hyenas.\r\n\r\n(End of post.)\r\n\r\n\r\n 7/21/2005 09:52:00 PM posted by Dr. Demarche\r\n \r\n \r\n \r\n \r\n\r\n\r\nComments (4)\r\n |\r\nTrackback (0) \r\n\r\n<< Home \r\n\r\n\r\nd\u00c3\u00a9\u00c2\u00b7marche 1) A course of action; a maneuver. 2) A diplomatic\r\nrepresentation or protest 3) A statement or protest addressed by\r\ncitizens to public authorities.\r\n\r\n\r\n Contributors\r\n\r\n * Dr. Demarche \r\n * Smiley \r\n\r\n\r\n A blog by members of the State Department Republican Underground-\r\n conservative Foreign Service Officers serving overseas commenting\r\n on foreign policy and global reactions to America.\r\n\r\n Send us mail: Dr.Demarche (or) Smiley.George AT gmail.com\r\n\r\n\r\n Recent Posts\r\n\r\nAl Jazeera supporters are right. <#112199712354356739>\r\n\r\n\r\n Blogroll\r\n\r\nA Guy in Pajamas \r\nAmerican Future \r\nAustin Bay Blog \r\nBarcepundit \r\nBarcepundit (English version) \r\nBlogs of War \r\nCacciaguida \r\nCheese & Crackers \r\nChicago Boyz \r\nCold Fury \r\nConsul At Arms \r\nDaniel Drezner \r\nDhimmi Watch \r\nDouble Cannister at Ten Yards \r\nEagle Speak \r\nExpat Yank \r\nFlight Pundit \r\nFresh Bilge \r\nGeoPolitical Review \r\nInstapundit \r\nJihad Watch \r\nMad Minerva \r\nMarcvs The Bard's Tales \r\nMover Mike \r\nMy Blog is Your Blog \r\nNew Sisyphus \r\nNo Jihad \r\nNo Pundit Intended \r\nOliver Kamm \r\nOpinionated Jerks \r\nOraculations \r\nPower and Control \r\nRight Intention \r\nRoger L. Simon \r\nSmadeNek \r\nSolomon's House \r\nThe Atlantic Review \r\nThe Big Cat Chronicles \r\nThe Diplomad \r\nThe Glittering Eye \r\nThe Seeker Blog \r\nThe Sundries Shack \r\nThe Truth Laid Bear \r\nThe Word Unheard \r\nTotal Information Awareness \r\nUrthshu \r\nView from Tonka \r\nZen Pundit \r\n\r\nby BlogRolling \r\n\r\n\r\n Non-Blog Links\r\n\r\n10 Myths About Islam \r\nAmerican Future Resources\r\n\r\nFaithFreedom.Org \r\nAsk Imam \r\nMEMRI \r\nSecularizing Islam \r\nWomen's Forum Against Fundamentalism in Iran \r\n\r\nArchives\r\nNovember 2004\r\n\r\n/ December 2004\r\n\r\n/ January 2005\r\n\r\n/ February 2005\r\n\r\n/ March 2005\r\n\r\n/ April 2005\r\n\r\n/ May 2005\r\n\r\n/ June 2005\r\n\r\n/ July 2005\r\n\r\n/ August 2005\r\n\r\n/ September 2005\r\n /\r\n\r\n\r\n Link to us:\r\n\r\nBlogroll Me!\r\n\r\n\r\nListed on Blogwise Site Meter\r\n Weblog\r\nCommenting and Trackback by HaloScan.com \r\nPowered by Blogger \r\n\r\n\r\nUnder Politics \r\n\r\n\r\n\r\nI'm a\r\nLarge Mammal\r\n\r\nin the\r\nTTLB Ecosystem ", "Solution_1": "So basically you have to read from files and reprint them in form you want.\r\n\r\nDeleting tags is simple. When you encouter opening of tag '<' then don't print things untile you will encounter closing bracket.\r\n\r\nIn that way you can eliminate whether you want.\r\n\r\nHope that helped.", "Solution_2": "I don't know the syntax to do this.\r\nI have a basic algorithm:\r\n\r\n-read in file character by character\r\n-print it to an output file character by character.\r\n-if '<' is encountered, don't print it\r\n-don't print anything else until '>' is encountered.\r\n-keep printing characters to the output file.\r\n\r\nBut I don't know how to tell Java not to print stuff.", "Solution_3": "bool inside = false;\r\nchar c;\r\n\r\nfor(i = 0; i < textsize; i++){\r\n fscanf(fin, \"%c\", c);\r\n if(isalpha(c) && !inside)fprintf(fout, \"%c\", c);\r\n else if(c=='<')inside = true;\r\n else if(c=='>')inside = false;\r\n}\r\n\r\nJust translate it to java and it's functions. (I could write it in java but it looks like you need some practice ;-) )", "Solution_4": "Thank you a lot." } { "Tag": [ "geometry" ], "Problem": "Determine Whether or Not There are exist a set of infinity integer points S such that for every pair(A,B,C) lies on S then the orthocenter of triangle ABC lies on S,too.", "Solution_1": "Wikipedia has the following theorem: [i]An orthocentric system is a set of four points in the plane one of which is the orthocenter of the triangle formed by the other three. If four points form an orthocentric system, then each of the four points is the orthocenter of the other three.[/i] This may help.", "Solution_2": "uhm.May be.But It's My Home Work, My teacher give me last week.U Just trying to help me^^^" } { "Tag": [ "modular arithmetic", "number theory", "prime numbers" ], "Problem": "show that any number raised to the 4th power is either 0 or 1 mod 16", "Solution_1": "[hide=\"i guess you could do it this way\"]do a lot of casework!\nassume that some number $ n\\equiv0,1,2...15\\pmod{16}$ take the 4th power of each of them and show that each of those is $ \\equiv 1/0\\pmod{16}$[/hide]\r\n\r\nis there a faster/more creative way?", "Solution_2": "Maybe you could find what $ x^{4}$ is congruent to when $ x$ is congruent to 0, 1, and prime numbers that are less than 16, which are 2, 3, 5, 7, 11, 13. If $ x^{4}$ is congruent to $ 1$ or $ 0$ when $ x \\equal{} 0, 1, 2, 3, 5, 7, 11, 13$, then you know that all other numbers that x is congruent to that are below 16 can either be congruent to 0 or 1 using modular arithmetic rules. This isn't that clever, but it saves you from checking 8 cases. :wink:", "Solution_3": "[quote=\"mdk\"]show that any number raised to the 4th power is either 0 or 1 mod 16[/quote]\r\n[hide]$ n\\equiv0,1\\bmod2\\,\\forall\\,n\\in\\mathbb{N}\\Longrightarrow n^{4}\\equiv0,1\\bmod2^{4}\\,\\forall\\,n\\in\\mathbb{N}$[/hide]", "Solution_4": "That works... basically an odd number to the fourth is 1 mod 16 (An odd number squared is 1 mod 8, which is a far more useful conclusion).\r\n\r\nDo I even need to explain for an even number?", "Solution_5": "[quote=\"i_like_pie\"][quote=\"mdk\"]show that any number raised to the 4th power is either 0 or 1 mod 16[/quote]\n[hide]$ n\\equiv0,1\\bmod2\\,\\forall\\,n\\in\\mathbb{N}\\Longrightarrow n^{4}\\equiv0,1\\bmod2^{4}\\,\\forall\\,n\\in\\mathbb{N}$[/hide][/quote]\n\nThis is a restatement of the question; the lemma you are implicitly trying to assert is not, in general, true. $ \\bmod 2^{k}$ is a special case, and similar reasoning does not hold in any other cases. For example,\n\n$ n\\equiv 0, 1, 2\\bmod 3$\n\nbut\n\n$ n^{2}\\equiv 0, 1, 4, 7\\bmod 9$\n\n[hide=\"Straightforward solution\"] $ (2k\\plus{}1)^{4}\\equal{} 16k^{4}\\plus{}32k^{3}\\plus{}24k^{2}\\plus{}8k\\plus{}1\\equiv 16\\frac{k(3k\\plus{}1)}{2}\\plus{}1\\equiv 1\\bmod 16$ [/hide]\n[hide=\"How this fact relates to general theory\"] Let $ \\lambda(m)$ denote the smallest exponent $ e$ such that $ a^{e}\\equiv 1\\bmod m$ for all $ gcd(a, m) \\equal{} 1$. For an odd prime $ p$ and for $ m \\equal{} 1, 2, 4, p^{k}, 2p^{k}$ we have $ \\lambda(m) \\equal{}\\varphi(m)$.\n\n[b]Lemma:[/b] $ \\lambda(2^{k\\plus{}2}) \\equal{} 2^{k}, k\\in\\mathbb{N}$ [/hide]", "Solution_6": "[quote=\"i_like_pie\"][quote=\"mdk\"]show that any number raised to the 4th power is either 0 or 1 mod 16[/quote]\n[hide]$ n\\equiv0,1\\bmod2\\,\\forall\\,n\\in\\mathbb{N}\\Longrightarrow n^{4}\\equiv0,1\\bmod2^{4}\\,\\forall\\,n\\in\\mathbb{N}$[/hide][/quote]\r\n\r\nYou are implicitly assuming that\r\n\r\n$ a\\equiv b\\pmod{p}$\r\n\r\nimplies\r\n\r\n$ a^{k}\\equiv b\\pmod{p^{k}}$,\r\n\r\nwhich is false. For example,\r\n\r\n$ 3\\equiv 1\\pmod{2}$\r\n\r\n$ 3^{3}\\not\\equiv 1\\pmod{2^{3}}$", "Solution_7": "[quote=\"vishalarul\"]$ 3^{3}\\not\\equiv 1\\pmod{2^{3}}$[/quote]\r\nBut $ 3^{4}\\equiv 1\\pmod{2^{4}}$, and the problem involves $ 2^{4}$, not $ 2^{3}$.", "Solution_8": "[quote=\"i_like_pie\"][quote=\"vishalarul\"]$ 3^{3}\\not\\equiv 1\\pmod{2^{3}}$[/quote]\nBut $ 3^{4}\\equiv 1\\pmod{2^{4}}$, and the problem involves $ 2^{4}$, not $ 2^{3}$.[/quote]\r\n\r\n$ 10\\equiv 1\\pmod{9}$\r\n\r\n$ 10^{4}\\not\\equiv 1\\pmod{9^{4}}$\r\n\r\nEither way, it doesn't work. And don't tell me that the problem involves $ \\pmod{2}$, because the problem is equivalent to proving:\r\n\r\n$ x\\equiv y\\pmod{2}\\rightarrow x^{4}\\equiv y\\pmod{2^{4}}$, which you implicitly assumed." } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "reflection", "perpendicular bisector", "geometry proposed" ], "Problem": "In an acute triangle ABC, AC > BC and M is the\r\nmidpoint of AB. Let altitudes AP and BQ meet at H, and let lines\r\nAB and PQ meet at R. Prove that the two lines RH and CM are\r\nperpendicular.", "Solution_1": "Note that $ CPHQ$ and $ BPQA$ are cyclic quadrilaterals. Let $ C_1,C_2,C_3$ be the circumcircles of $ CPHQ,BPQA,BHA$ respectively. Let $ C_1$ and $ C_3$ intersect at $ H$ and $ T$. Let $ O_1$ and $ O_3$ be the centers of $ C_1$ and $ C_3$ respectively. Let $ O$ be the circumcenter of triangle $ ABC$. Since $ \\angle CPH\\equal{}90^{\\circ}$, then $ CH$ is the diameter of $ C_1$, so $ O_1$ is the midpoint of $ CH$. It is well-known that $ CH\\equal{}2OM$, therefore $ CO_1\\equal{}OM$. Note that $ \\angle BOA\\equal{}2\\angle BCA\\equal{}2(180^{\\circ}\\minus{}\\angle PHQ)\\equal{}2(180^{\\circ}\\minus{}\\angle BHA)$. But $ \\angle BO_3A\\equal{}2(180^{\\circ}\\minus{}\\angle BHA)$, so $ \\angle BOA\\equal{}\\angle BO_3A$. Since $ O_3A\\equal{}O_3B$ and $ OA\\equal{}OB$, $ O_3$ and $ O$ lie on the perpendicular bisector of segment $ AB$. Thus $ O$ is the reflection of $ O_3$ with respect to $ AB$. Hence $ O_3M\\equal{}OM\\equal{}CO_1$. Consider $ CO_1O_3M$. We have $ CO_1\\equal{}O_3M$. Also $ CH\\perp BC,BC\\perp O_3M$, so $ CH\\parallel O_3M$. So $ CO_1O_3M$ is a parallelogram. Since $ HT\\perp O_1O_3$ and $ HT\\perp CT$, so $ O_1O_3\\parallel CT$. Therefore $ CT\\parallel CM$, and hence $ CTM$ is a straight line. Consider $ C_1,C_2,C_3$. The three radical axes (for each pair of circles) are concurrent. So they concur at $ R$. We now have $ RH\\parallel RT\\perp CM$, as desired." } { "Tag": [ "geometry", "rectangle", "search", "combinatorics proposed", "combinatorics" ], "Problem": "In how many ways can you tile a $ 4\\times n$ by $ 3\\times 1$ dominoes ? Justify fully your answer", "Solution_1": "I guess we can proceed similarly as in http://www.mathlinks.ro/viewtopic.php?search_id=1699599031&t=54081", "Solution_2": "Yes: a quick examination revealed five possible variables: $ a_n$ for a $ 4 \\times 3n$ block, $ b_n$ for such a $ 4 \\times 3(n \\minus{} 1)$ block with a $ 3\\times 3$ block glued on one end, $ c_n$ for a $ 4 \\times 3(n \\minus{} 1)$ block with a $ 3 \\times 2$ block glued on, $ d_n$ for a $ 4 \\times 3(n \\minus{} 1)$ block with a $ 3 \\times 1$ block glued on, and $ e_n$ for a $ 4 \\times (3n \\minus{} 1)$ block with a $ 1 \\times 1$ block glued on the end in the top- or bottom-most row. Then we have equations like\r\n$ a_n \\equal{} b_n \\plus{} e_n$\r\n$ b_n \\equal{} c_n \\plus{} a_{n \\minus{} 1}$\r\n$ c_n \\equal{} d_n \\plus{} e_{n \\minus{} 1}$\r\n$ d_n \\equal{} a_{n \\minus{} 1} \\plus{} d_{n \\minus{} 1}$\r\nand $ e_n \\equal{} c_n$, or something close to this. There's lots of simplification that can be done, computation of the actual answer, etc. -- if I have time later, I may come and complete this argument." } { "Tag": [ "calculus", "integration", "logarithms", "trigonometry", "function", "geometric series", "calculus computations" ], "Problem": "Evaluae $ \\int_{0}^{1} \\left( \\frac{1}{1\\minus{}x} \\plus{} \\frac{1}{\\ln x} \\minus{} \\frac{1}{2}\\right) \\frac{1}{\\ln x} \\, dx$\r\n\r\nOf course it would be easy for some people here...", "Solution_1": "somewhat of a long solution... :| :( \r\n\r\nwe are given,\r\n$ I \\equal{} \\int_{0}^{1} \\left( \\frac {1}{1 \\minus{} x} \\plus{} \\frac {1}{\\ln x} \\minus{} \\frac {1}{2}\\right) \\frac {1}{\\ln x} \\, \\textbf dx$\r\n\r\nlet us transform it using $ x \\equal{} e^{ \\minus{} y}$ to get\r\n\r\n$ I \\equal{} \\int_0^\\infty\\left(\\frac {1}{2} \\plus{} \\frac {1}{y} \\minus{} \\frac {1}{1 \\minus{} e^{ \\minus{} y}}\\right)\\;\\frac {e^{ \\minus{} y}}{y}\\;\\textbf dy$\r\n\r\nnoting that we want $ I(1)$, let us define\r\n$ I(z) \\equal{} \\int_0^\\infty\\left(\\frac {1}{2} \\plus{} \\frac {1}{y} \\minus{} \\frac {1}{1 \\minus{} e^{ \\minus{} y}}\\right)\\;\\frac {e^{ \\minus{} yz}}{y}\\;\\textbf dy$\r\n\r\nthen,\r\n$ \\frac {d}{dz}\\;I(z)\\; \\equal{} \\; \\minus{} \\int_0^\\infty\\left(\\frac {1}{2} \\plus{} \\frac {1}{y} \\minus{} \\frac {1}{1 \\minus{} e^{ \\minus{} y}}\\right)\\;e^{ \\minus{} yz}\\;\\textbf dy$\r\n\r\nusing mittag-leffler, we get\r\n$ \\frac {1}{2} \\plus{} \\frac {1}{y} \\minus{} \\frac {1}{1 \\minus{} e^{ \\minus{} y}}\\; \\equal{} \\; \\minus{} 2\\sum_{k \\equal{} 1}^\\infty\\frac {y}{y^2 \\plus{} 4\\pi^2\\,k^2}$\r\n\r\nso, $ \\frac {d}{dz}\\;I(z)\\; \\equal{} \\;2\\int_0^\\infty\\sum_{k \\equal{} 1}^\\infty\\frac {y}{y^2 \\plus{} 4\\pi^2\\,k^2}\\; e^{ \\minus{} yz}\\;\\textbf dy$\r\n\r\nnow using the fact\r\n$ \\int_0^\\infty e^{ \\minus{} au}\\;\\sin\\,(bu)\\;\\textbf du\\; \\equal{} \\;\\frac {b}{b^2 \\plus{} a^2}\\qquad \\ldots (1)$\r\n\r\nwe have\r\n $ \\frac {d}{dz}\\;I(z)\\; \\equal{} \\;2\\int_0^\\infty\\sum_{k \\equal{} 1}^\\infty\\int_0^\\infty e^{ \\minus{} 2\\pi\\,k\\,u}\\;\\sin\\,(yu)\\;\\textbf du\\; e^{ \\minus{} yz}\\;\\textbf dy$\r\n\r\nwe know, since it is a geometric series,\r\n $ \\sum_{k \\equal{} 1}^\\infty\\;e^{ \\minus{} 2\\pi\\,k\\,u}\\; \\equal{} \\;\\frac {1}{e^{2\\pi\\,u} \\minus{} 1}$\r\n\r\nso, $ \\frac {d}{dz}\\;I(z)\\; \\equal{} \\;2\\int_0^\\infty\\int_0^\\infty\\frac {\\sin\\,(yu)}{e^{2\\pi\\,u} \\minus{} 1}\\;\\textbf du\\; e^{ \\minus{} yz}\\;\\textbf dy$\r\n\r\nagain, using the identity in $ (1)$, this time w.r.t. $ \\textbf dy$,\r\n\r\n$ \\frac {d}{dz}\\;I(z)\\; \\equal{} \\;2\\int_0^\\infty\\frac {u}{(u^2 \\plus{} z^2)\\; (e^{2\\pi\\,u} \\minus{} 1) }\\;\\textbf du$\r\n\r\nintegrating w.r.t. $ z$ between limits $ 1$ and $ \\infty$ gives\r\n\r\n$ \\Rightarrow 0 \\minus{} I(1)\\; \\equal{} \\;2\\int_0^\\infty\\int_1^\\infty \\frac {u}{(u^2 \\plus{} z^2)\\; (e^{2\\pi\\,u} \\minus{} 1) }\\;\\textbf dz\\;\\textbf du$\r\n\r\nsince we know $ \\int\\;\\frac {u}{u^2 \\plus{} z^2}\\;\\textbf dz \\equal{} \\tan^{ \\minus{} 1}\\,\\frac {z}{u}\\qquad \\text{and}\\qquad \\frac {\\pi}{2} \\minus{} \\tan^{ \\minus{} 1}\\;\\frac {1}{u} \\equal{} \\tan^{ \\minus{} 1}\\;u$, \r\n\r\nwe have\r\n$ I(1)\\; \\equal{} \\; \\minus{} 2\\int_0^\\infty \\frac {\\tan^{ \\minus{} 1}\\,u}{e^{2\\pi\\,u} \\minus{} 1}\\;\\textbf du$\r\n\r\nthe integral appearing on the right hand side of this equation was asked by me on this forum before, \r\ndon't know if anyone showed it, but i can show it equals $ \\frac {1}{2}\\left(1 \\minus{} \\ln\\,\\sqrt {2\\pi}\\right)$\r\n\r\nhence, $ I(1)\\; \\equal{} \\;\\boxed{\\ln\\,\\sqrt {2\\pi} \\minus{} 1}$\r\n\r\nsos, i would like to ask this question on the next exam in my class, hope that is OK with you. i am sure most of the students will like this question. :)", "Solution_2": "[quote=\"misan\"]sos, i would like to ask this question on the next exam in my class, hope that is OK with you. i am sure most of the students will like this question. :)[/quote]\r\n\r\nSure. -o-;;\r\n\r\nYour method is very impressive to me, since it involves some neat techniques which are new to me. Anyway, here is my solution:\r\n\r\n\r\n[b]Step 1) Preliminary[/b]\r\n\r\nOne can show the following equalities. You may use the infinite product formula for Gamma function to prove these.\r\n\r\n$ (1) \\quad \\ln \\Gamma(1 + x) = - \\gamma x + \\sum_{n = 2}^{\\infty} \\frac {( - 1)^{n}}{n}\\zeta(n) \\, x^{n}\\quad \\quad (|x| < 1),$\r\n\r\n$ (2) \\quad \\int_{0}^{1} \\frac {1 - t^{x}}{1 - t} \\, dt = \\gamma + \\frac {d}{dx} \\ln \\Gamma(1 + x).$\r\n\r\n\r\n[b]Step 2) Evaluation of $ \\int_{0}^{1} \\ln \\Gamma(1 + x) \\, dx$[/b]\r\n\r\nOne can confirm the equality $ \\left( 1 + \\frac {1}{n} \\right)^{n} = \\frac {n!}{n^n} \\prod_{k = 1}^{n} \\left( 1 + \\frac {1}{k}\\right)^{k}.$ Taking $ \\ln$ to both side,\r\n\r\n\\begin{eqnarray*} \\ln \\left( 1 + \\frac {1}{n} \\right)^{n} & = & \\ln \\left( \\frac {n!}{n^n} \\right) + \\sum_{k = 1}^{n} k \\ln \\left( 1 + \\frac {1}{k}\\right) \\\\\r\n& = & \\ln \\left( \\frac {n!}{n^n} \\right) + \\sum_{k = 1}^{n} \\sum_{j = 0}^{\\infty} \\frac {( - 1)^{j}}{j + 1} \\frac {1}{k^{j}} \\\\\r\n& = & \\ln \\left( \\frac {n!}{n^n} \\right) + n - \\frac {1}{2} \\mathrm{H}_n + \\sum_{j = 2}^{\\infty} \\sum_{k = 1}^{n} \\frac {( - 1)^{j}}{j + 1} \\frac {1}{k^{j}}. \\end{eqnarray*}\r\n\r\nTaking $ n \\to \\infty$ to both side, Stirling formula yields $ 1 = \\ln \\sqrt {2\\pi} - \\frac {\\gamma}{2} + \\sum_{j = 2}^{\\infty} \\frac {( - 1)^{j}}{j + 1} \\zeta(j).$ Together (1), this confirms\r\n\r\n$ (3) \\quad \\int_{0}^{1} \\ln \\Gamma(1 + x) \\, dx = \\ln \\sqrt {2\\pi} - 1.$\r\n\r\n\r\n[b]Step 3) An integral formula for $ \\ln \\Gamma (1 + x)$[/b]\r\n\r\nIntegrating both side of (2) on $ [0, x]$ and changing the order of integration immediately yields \r\n\r\n$ (4) \\quad \\int_{0}^{1} \\left( \\frac {x}{1 - t} + \\frac {1 - t^x}{1 - t}\\frac {1}{\\ln t} \\right) \\, dt = \\gamma x + \\ln \\Gamma (1 + x).$\r\n\r\nEspecially, \r\n\r\n$ (5) \\quad \\gamma = \\int_{0}^{1} \\left( \\frac {1}{1 - t} + \\frac {1}{\\ln t} \\right) \\, dt.$\r\n\r\nCombining (4) and (5), we obtain an integral formula\r\n\r\n$ (6) \\quad \\ln \\Gamma (1 + x) = \\int_{0}^{1} \\left( \\frac {1 - t^x}{1 - t} - x \\right) \\frac {dt}{\\ln t}.$\r\n\r\n\r\n[b]Step 4) Final Result[/b]\r\n\r\nNow integrate both side of (6) on $ [0, x]$ again, then we have\r\n\r\n\r\n$ (7) \\quad \\int_{0}^{x} \\ln \\Gamma (1 + t) \\, dt = \\int_{0}^{1} \\left( \\frac {x}{1 - t} + \\frac {1 - t^x}{1 - t}\\frac {1}{\\ln t} - \\frac {x^2}{2} \\right) \\frac {dt}{\\ln t}.$\r\n\r\nThis completes the evaluation by plugging $ x = 1$.", "Solution_3": "Wow. Those are some amazing solutions.\r\n\r\nUnfortunately, I'm not familiar with the gamma and zeta functions because I've never used them in classes.", "Solution_4": "about 2 step:\r\n$ G: \\equal{} \\int_{0}^{1}\\ln \\Gamma(1 \\plus{} z)dz \\equal{} \\int_{0}^{1}\\ln z dz \\plus{} \\int_{0}^{1}\\ln \\Gamma(z)dz \\equal{} \\minus{} 1 \\plus{} \\int_{0}^{1}\\ln \\Gamma(z)dz$ since $ \\Gamma(1\\plus{}z)\\equal{}z\\Gamma(z)$\r\n$ I \\equal{} \\int_{0}^{1}\\ln \\Gamma(z)dz$ say.\r\nput $ 1 \\minus{} z\\equiv y$ then \r\n$ I \\equal{} \\int_{0}^{1}\\ln \\Gamma(1 \\minus{} y)dy$\r\nthen\r\n$ 2I \\equal{} \\int_{0}^{1}\\ln \\Gamma(1 \\minus{} z)dz \\plus{} \\int_{0}^{1}\\ln \\Gamma(z)dz \\equal{} \\int_{0}^{1}\\ln \\Gamma(z)\\Gamma(1 \\minus{} z)dz \\equal{} \\int_{0}^{1}\\ln \\frac {\\pi}{\\sin\\pi z}dz$ since \r\n $ \\Gamma(z)\\Gamma(1 \\minus{} z) \\equal{} \\frac {\\pi}{\\sin\\pi z}$\r\nthus:\r\n$ 2I \\equal{} \\ln \\pi \\minus{} \\int_{0}^{1}\\ln \\sin \\pi z \\equal{} \\ln \\pi \\plus{} \\ln 2$\r\nhence:\r\n$ I \\equal{} \\ln \\sqrt {2\\pi}$ and $ G \\equal{} \\minus{} 1 \\plus{} \\ln \\sqrt {2\\pi}$", "Solution_5": "Oh, that's much easier than mine!", "Solution_6": "Look also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148672]here[/url].", "Solution_7": "[quote=\"Carcul\"]Look also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148672]here[/url].[/quote]\r\n\r\nI forgot that post... :blush: It's surprising that there are lots of (seemingly) different ways to show something. This is amusing." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Find the greatest constant k such that the following inequality holds for any positive real numbers $ a, b, c$ \r\n$ \\frac{a^2\\plus{}b^2\\plus{}c^2}{ab\\plus{}bc\\plus{}ca}\\le k$", "Solution_1": "[quote=\"vo thanh van\"]Find the greatest constant k such that the following inequality holds for any positive real numbers $ a, b, c$ \n$ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}\\le k$[/quote]\r\nI think U want find the Minimum of k .I sure $ k\\equal{}2$. And I have seen this problem on forum :)", "Solution_2": "[quote=\"onlylove_math\"][quote=\"vo thanh van\"]Find the greatest constant k such that the following inequality holds for any positive real numbers $ a, b, c$ \n$ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}\\le k$[/quote]\nI think U want find the Minimum of k .I sure $ k \\equal{} 2$. And I have seen this problem on forum :)[/quote]\r\nif $ a,b,c$ are side of a triangle,$ k\\equal{}2$ is true :)but in here $ a,b,c$ be positive real numbers", "Solution_3": "Dear,anh Van! Can you post your solution,I want to see it. :lol:", "Solution_4": "[quote=\"vo thanh van\"][quote=\"onlylove_math\"][quote=\"vo thanh van\"]Find the greatest constant k such that the following inequality holds for any positive real numbers $ a, b, c$ \n$ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}\\le k$[/quote]\nI think U want find the Minimum of k .I sure $ k \\equal{} 2$. And I have seen this problem on forum :)[/quote]\nif $ a,b,c$ are side of a triangle,$ k \\equal{} 2$ is true :)but in here $ a,b,c$ be positive real numbers[/quote]\r\nOh no I think with $ a,b,c$ be positive real numbers . Also $ k\\equal{}2$", "Solution_5": "[quote=\"vo thanh van\"]Find the greatest constant k such that the following inequality holds for any positive real numbers $ a, b, c$ \n$ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}\\le k$[/quote]\r\n\r\nLet $ c \\equal{} 0,b \\equal{} 1$ and $ a \\to \\plus{} \\infty$\r\n\r\nClearly we have the expression $ \\frac {{a^2 \\plus{} b^2 \\plus{} c^2 }}{{ab \\plus{} bc \\plus{} ca}} \\to \\plus{} \\infty$ and hence there is no $ k$ which satisfy that inequality. :|\r\n\r\nFor onlylove_math, we know $ \\frac{{a^2 \\plus{} b^2 \\plus{} c^2 }}{{ab \\plus{} bc \\plus{} ca}} \\ge 1$ is true. With the $ \\frac{{a^2 \\plus{} b^2 \\plus{} c^2 }}{{ab \\plus{} bc \\plus{} ca}} \\ge 2$ you can try with $ b\\equal{}c\\equal{}1$ and $ a \\le 2$. :)" } { "Tag": [ "probability", "probability and stats" ], "Problem": "Assume that $ P(A) \\equal{} 0.6$ and that $ P(B) \\equal{} 0.3$ and A and B are independent.\r\nFind $ P(A\\cup \\overline{B})$.\r\n\r\n\r\nI've tried with the following $ P(A) \\equal{} P(A\\cup E) \\equal{} P(A\\cap (B\\cup \\overline{B})) \\equal{} P(A\\cap B) \\plus{} P(A\\cap \\overline{B})$\r\nNow from $ P(A\\cup B) \\equal{} P(A)P(B) \\equal{} 0.18$ because A and B are independent we find the probability of\r\nintersection of A and B's complementary so $ P(A\\cap \\overline{B}) \\equal{} 0.42$\r\nUsing this and the relation above we get $ P(A\\cup \\overline{B}) \\equal{} P(A) \\plus{} P(\\overline{B}) \\minus{} P(A\\cap\\overline{B}) \\equal{} 0.6 \\plus{} 0.9 \\minus{} 0.42 > 1$ so there is a problem with this solution.\r\nAny comments and suggestions are much appreciated.\r\n\r\nEDIT:corrected a typo", "Solution_1": "$ P(\\overline{B})\\equal{}0.7,$ not $ 0.9.$ Fix that and your difficulties will go away.\r\n\r\nAs a check or as an alternative computation, try finding $ P(\\overline{A}\\cap B)$ and then subtracting that from $ 1.$", "Solution_2": "thanks Kent Merryfield,now it makes sense and the solution coincides with the one in the book." } { "Tag": [ "quadratics", "number theory unsolved", "number theory" ], "Problem": "Decide det irrational integer that has the continued fractions [9; 9,18]!\r\n\r\nI know that the answer is sqrt(83), but how do i get there if i don't that is a quadratic irrational?\r\n\r\n\r\nbig thanks for any solutions!", "Solution_1": "You have (neglect that $9$ at the beginning):\r\nLet $x=[0;9,18,9,18,...]$\r\n$x=\\frac{1}{9+\\frac{1}{18 + x}}$ (since $x$ stands just for the infinite chain of fractions, thus we can 'get out' the first period of $x$ without changing $x$).\r\nBut this equation gives $x=\\frac{18 + x}{163 + 9x} \\iff 9x^2 + 163x = 18+x \\iff x^2+18x-2=0$.\r\nSo solving this gives $x= -9 +\\sqrt{83}$ (and another root we can exclude to be correct since its $<0$, or just by testing) as it also should be, since than the desired number is $x+9= \\sqrt{83}$." } { "Tag": [], "Problem": "Hello,\r\nI'm reading a book about Number Theory. And its presenting an example to work through it.\r\n\r\nLet $ n$ be an integer greater than $ 1$. Prove that $ 2^n$ is the sum of two odd consecutive integers.\r\n\r\nFor me I just tried it in a piece of paper but how can I prove it? I never proof anything before. When I see the solution I get :huh: . \r\n\r\nThis was the solution: the relation $ 2^{n} \\equal{} (2k \\minus{} 1) \\plus{} (2k \\plus{} 1)$ implies $ k \\equal{} 2^{n \\minus{} 2}$ and we obtain $ 2^n \\equal{} (2^{n \\minus{} 1} \\minus{} 1) \\plus{} (2^{n \\minus{} 1} \\plus{} 1)$.\r\n\r\nbut from where he bring all that relations. How he get $ 2^n \\equal{} (2k \\minus{} 1) \\plus{} (2k \\plus{} 1)$ and so on :( .", "Solution_1": "What is a sum of two odd consecutive integers? How would you write down an odd integer? How would you write down the next odd integer?", "Solution_2": "OK first of all how he come up with $ 2^n\\equal{}(2k\\minus{}1)\\plus{}(2k\\plus{}1)$ :?: and as I understand that after this he implied that $ k\\equal{}2^{n\\minus{}2}$ OK before that how he know what is value of $ k$. There is something I didn't get it maybe if I do everything will be OK with me.", "Solution_3": "I think the solution assumes that it exists and then finds it. (If it didn't exist, some contradiction would occur.) Is this book by any chance 104 NT Problems?", "Solution_4": "Yes it is 104 NT problems but I'm still in the beginning. So do you mean that you should try it in small numbers and draw a proof :huh:", "Solution_5": "\"$ 2^n$ is the sum of two odd consecutive integers\" is equivalent to \"there exists an integer $ k$ such that $ 2^n \\equal{} (2k \\minus{} 1) \\plus{} (2k \\plus{} 1)$.\" Do you understand this? That's all the solution is.", "Solution_6": "OK I prefer to work with an example:\r\nlets say that $ 2^5$ . I know that the numbers are $ 15$ and $ 17$ but the problem with the proof thing. Is the proof need you to come to the solution or a method to apply with every case?", "Solution_7": "And how did you know that the numbers are $ 15$ and $ 17$?", "Solution_8": "By: $ 2^5\\equal{}32$ so if we say $ \\frac{32}{2}\\equal{}16$ then $ 16\\plus{}1\\equal{}17$ and $ 16\\minus{}1\\equal{}15$\r\n\r\nSo I can call it: $ 2^n\\equal{}S$, so the answer is $ \\frac{S}{2} \\pm 1$ from my view.", "Solution_9": "Two consecutive odd integers always add up to what?", "Solution_10": "I see: I understand now. $ 2k \\plus{} 1$ and $ 2k \\minus{} 1$ are all ways to make odd numbers. Which makes an even number when added.\r\nThanks to \"t0rajir0u\" and \"sunehra\" and also for \"Ihatepie\" for his post. :)" } { "Tag": [ "geometry", "geometric transformation", "rotation" ], "Problem": "In how many ways can we seat 6 people around a table if Fred and Gwen insist on sitting opposite each other? (Two seatings are considered equivalent if one is a rotation of the other.)", "Solution_1": "Isn't there 6 ways they can sit opposite from each other?", "Solution_2": "Yes, but there are $ 6$ rotations which we consider the same, so you over count and need to end up dividing it by $ 6$ anyways. \r\n[hide=\"Solution\"]\nFirst, we place Fred. Fred has $ 6$ different places to sit. Gwen has no choice but the seat directly opposite Fred. The rest of the people can be seated in $ 4!$ ways. So we have $ 6\\cdot4! \\equal{} 144$. However, there are $ 6$ different rotations, so to correct for our over counting, we do $ \\frac {144}6 \\equal{} \\boxed{24}$\n[/hide]", "Solution_3": "Or, we can just say that Fred is somewhere, then Gwen is directly opposite, and we have $ 4!\\equal{}24$ ways to arrange the rest.", "Solution_4": "6 ways to seat Fred and only 1 way to seat Gwen. Then the rest is just 4 factorial. $\\frac{6 \\cdot 1 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}{6}=24$" } { "Tag": [ "\\/closed" ], "Problem": "What do you do if someone locked a topic just to shut you up? :mad: And especially gave no explanation!!", "Solution_1": "[quote=\"jtw\"]What do you do if someone locked a topic just to shut you up?[/quote]There's a reason someone wants you to shut up.", "Solution_2": "Because I'm right and he/she can't prove me wrong.", "Solution_3": "First of all you're not right. Secondly, there is already a thread discussing (the science of) evolution versus (the myth of) creationism. All discussions about that will take place there not in new threads." } { "Tag": [ "function", "induction", "geometry", "geometric transformation", "abstract algebra", "combinatorics proposed", "combinatorics" ], "Problem": "Let $C = \\{1, 2, 3, \\cdots, p\\}$ where $p$ is a prime number.\r\nLet $r : C \\to C$ be a function defined by\r\n$\\forall i \\in C, ~ r(i) \\equiv i+1 \\mod{p}$\r\nLet $h : C \\to C$ be a function that is not onto.\r\nProve that there exists a constant function $f : C \\to C$ that can be written as $f = h_{1}\\circ h_{2}\\cdots \\circ h_{k}, \\quad h_{i}\\in \\{r,h\\}$.", "Solution_1": "For a function $f$, we denote it's range by $f(C)$. We will apply induction on the size of $h(C)$. \r\n\r\nIf the range has one element then there's nothing to prove. Let $|h(C) | = k < p$. Suppose that the statement is true for all functions $f$ such that $|f(C)| < k$. Let $h(C) = \\{h_{1}, \\ldots,h_{k}\\}$ and $S_{i}= h^{-1}(h_{i})$. We will show that there exists some $c \\in C, n \\in [k]$, such that \\[|\\,(h(C)+c) \\cap S_{i}\\,| \\geq 2 .\\] Let us first see that the existence of $c, n$ as above is enough to solve the problem. Let $f_{1}= h\\circ r^{c}\\circ h$ ($r^{0}$ is the identity function). Note that $(r^{c}\\circ h) (C) = \\{c+h_{1},\\ldots, c+h_{k}\\}$. So there are $k$ elements in $(r^{c}\\circ h)(C)$ and two of them belong to $S_{i}$ for some $i$. Therefore the range of $f_{1}= h \\circ r^{c}\\circ h$ has size at most $k-1$. So that the statement of the problem holds for $f_{1}$ by induction, which in turn implies that it holds for $h$.\r\n\r\n[b]Proof of existence of $c, n$:[/b] Suppose that there exists no cuch $c, n$. We will show that in this case $k | p$, which is a contradiction (as $1 < k < p$). We have that for all $c, i$, $|(h(C)+c ) \\cap S_{i}| < 2$. This implies that for all $i, c$, we have $|(h(C)+c) \\cap S_{i}| = 1$. So each translation of $h(C)$ has one element in each of the $S_{i}$'s. Hence, for a fixed $i$, (call this equation (1)) \\[\\sum_{c \\in C}|(h(C)+c)\\cap S_{i}| = p .\\] Further, each element $a \\in S_{i}$ lies in $h(C)+c$ for exactly $k$ values of $c$ (these are $a-h_{1}, a-h_{2}, \\ldots$). Therefore (call this equation (2)) \\[k \\cdot |S_{i}| = \\sum_{c \\in C}|(h(C)+c) \\cap S_{i}|.\\] From (1) and (2) we have $k \\cdot |S_{i}| = p$, a contradiction as $1 < k < p$.", "Solution_2": "Nice solution. \r\n\r\nin fact, this is a lemma that G. L. O'Brien used in a proof of partial result of Road Coloring Conjecture. (G. L. O'Brien. The road coloring problem, Israel J. Math., 39 (1981) )\r\n\r\nO'Brien also tried to generalize the lemma (G. L. O'Brien. Zero-inducing functions on finite abelian groups. PACIFIC JOURNAL OF MATHEMATICS, 1982 )", "Solution_3": "\"n fact, this is a lemma that G. L. O'Brien used in a proof of partial result of Road Coloring Conjecture. (G. L. O'Brien. The road coloring problem, Israel J. Math., 39 (1981) )\r\n\r\nO'Brien also tried to generalize the lemma (G. L. O'Brien. Zero-inducing functions on finite abelian groups. PACIFIC JOURNAL OF MATHEMATICS, 1982 )\"\r\n\r\nCould you give some details of these?", "Solution_4": "[url=http://en.wikipedia.org/wiki/Road_coloring_conjecture]road coloring conjecture[/url]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let be given a real number $k$ in the interval $(-1;2)$ and three pairwise distinct real numbers $a;b;c$.Prove that:\r\n$[a^{2}+b^{2}+c^{2}+k(ab+bc+ca)].[\\frac{1}{(a-b)^{2}}+\\frac{1}{(b-c)^{2}}+\\frac{1}{(c-a)^{2}}] \\geq \\frac{9}{4}(2-k)$", "Solution_1": "The problem was proposed by Nam Dung in [b]Mathematics and Youth Mgazine\"[/b] in Viet Nam. It's hae still in period dicussing. I think we hadn't better post discussion here" } { "Tag": [ "probability", "counting", "derangement", "FTW" ], "Problem": "The numbers 1, 2, 3 and 4 are written on four cards and randomly placed in the boxes below. What is the probability, expressed as a common fraction, that no card number appears in the box of the same number?\n[asy]draw((0,0)--(2,0)--(2,1)--(0,1)--cycle);\ndraw((3,0)--(5,0)--(5,1)--(3,1)--cycle);\ndraw((6,0)--(8,0)--(8,1)--(6,1)--cycle);\ndraw((9,0)--(11,0)--(11,1)--(9,1)--cycle);\nlabel(\"1\",(1,.5));\nlabel(\"2\",(4,.5));\nlabel(\"3\",(7,.5));\nlabel(\"4\",(10,.5));[/asy]", "Solution_1": "Even before we start, we can determine there are $ 4!$ or $ 24$ possibilities.\r\n\r\n1234, 1243, 1324, 1342, 1423, 1432, 2134, [color=green]2143[/color], 2314, [color=green]2341[/color], [color=green]2413[/color], 2431, 3124, [color=green]3142[/color], 3214, 3241, [color=green]3412[/color], [color=green]3421[/color], [color=green]4123[/color], [color=green]4132[/color], 4213, 4231, [color=green]4312[/color], [color=green]4321[/color]\r\n\r\nThe answer is $ \\frac{10}{24}$ or $ \\boxed{\\frac{5}{12}}$.", "Solution_2": "The answer 3/8 is correct. The answer 5/12 was suggested, but 4231 was mistakenly counted as a possibility.\r\n\r\nThere are 24 possible ways to fill all four boxes, if we ignore the restrictions, and there are 9 ways to fill the boxes with non-matching card numbers.\r\n-- 3 options to fill the first box, then\r\n-- 3 options for card number 1 to go into one of the remaining boxes, then\r\n-- 1 option for the remaining two cards, since one of the two remaining cards must match the number of an unfilled box, leaving only one option for placement of the two cards\r\n\r\n3*3*1/24 = 9/24 = 3/8", "Solution_3": "You can use the dearrangement theorem, without having to make cases.\nIt says that there are n![1/0!-1/1!+1/2!-1/3!+1/4!......1/n!] ways to arrange objects such that no object goes in the right place.\n\nSo 4![1/0!-1/1!+1/2!-1/3!+1/4!] = 12-4+1=9", "Solution_4": "Solution: Let's start with the card with #1 on it.\nWe can put it in 2,3, or 4. Let's call the box we put it in $n$ [b]3 possibilities[/b]\n\nnow we find a place for the card $n$. that goes in 3 places as well. assume we don't pick the first. there's 2 possibilities. And that leaves the last card (not the one that's same as the box just picked by $n$) one spot. That's [b]2 ways[/b] And if we put $n$ in box 1, there's 1 possibility for last 2 cards as well. So that's [b]1 way[/b] \n\nSo $3\\cdot(2+1)=9$\n\nbasically using whatever the derangement theorem is\n\nAnd then $\\frac9{24}=\\frac38$", "Solution_5": "You could also use casework:\n\nWe can either have $2$ cards being flipped (like $1$ with $3$ and $2$ with $4$) or all $4$ cards being flipped ($1$ with $2$, $2$ with $3$, $3$ with $4$ and $4$ with $1$ for example). (Odd number cards don't work because if $3$ cards are switched then one card remains the same, and if we have \"$1$ card switched\" that just means no cards are switched, since you can't switch a card with itself.)\n\nSo if we have $2$ cards flipped, we have $\\binom{4}{2}$ to choose a pair of cards to be flipped, and the other two will be flipped no matter what. Hence there are $6$ ways for this case.\n\nIf we have $4$ cards flipped, then there are just $3$ ways since we have $\\binom{4}{1} = 4$ ways to arrange this but of course we can't have $1,2,3,4$. \n\nSo $6 + 3 = 9$, and out of $4! = 24$ ways the probability is $\\boxed{\\frac{3}{8}}$.\n\nPhew longest FTW solution I ever typed :)", "Solution_6": "Whoever does not understand this should look at the first solution, but don't count 4132, because 3 is in the box number 3.", "Solution_7": "ya 4!=24\n9 possiblilities so 9/24 or 3/8\n", "Solution_8": "Just a quick note: the number of ways you can place the numbers into a different position is called a [b]derangement[/b]. it's represented with a $!$ before the number.\n\n$!1=0$\n$!2=1$\n$!3=2$\n$!4=9$\n$!5=44$\n\njust to name the first five ($!0=1$, but that's confusing)\n\nknowing this, it's easy to see that the answer is $\\frac{!4}{4!} \\implies \\boxed{\\frac{3}{8}}$" } { "Tag": [ "MATHCOUNTS", "analytic geometry", "FTW", "email", "quadratics", "probability", "percent" ], "Problem": "Hello!\r\n\r\nMany of you are aware that I recently hosted a contest for 5th-6th graders. This 10 Round Tournament is open to all grade levels. Each Round will have 10 questions, each one worth 10 points. So the highest possible score is 100 for each round, and the highest for the whole Tournament is 1000. See if you can reach 1000!\r\n\r\nThis Tournament is similar the Wickedestjr's tournament in that people who move on to the next round are a few short. (In other words, for each round, there will be a few people who don't make it. The number depends on how many sign up.)\r\n\r\nI will give you all until Friday, June 13th, midnight forum time to sign up. Once everyone who wishes to do so signs up, I will give more information.\r\n\r\nSo please, sign up! :)", "Solution_1": "I'm in :D", "Solution_2": "[b][u]The Sign-Up List[/u][/b]\r\n\r\n1. 12markkram34\r\n\r\n2. mewto55555\r\n\r\n3. BOGTRO\r\n\r\n4. dynamo729\r\n\r\n5. D3m0n Shad0w\r\n\r\n6. shentang\r\n\r\n7. Urc\r\n\r\n8. hunter34\r\n\r\n9. Canton\r\n\r\n10. Bacteria\r\n\r\n11. scast\r\n\r\n12. AIMEman\r\n\r\n13. moogra\r\n\r\n14. 7h3.D3m0n.117\r\n\r\nI will update this list as more people join.", "Solution_3": "I'll join.", "Solution_4": "I will join.", "Solution_5": "I'll join..... I have never done MOEMS so what is required??", "Solution_6": "I will join except I am attending a sleep-over on Friday, June 13th to June 14th at 10:00 am.", "Solution_7": "@D3m0n Shad0w\r\n\r\nI'll go into more detail later, but that shouldn't be a problem. :)", "Solution_8": "I guess I will join...but I don't know whats in my plans.", "Solution_9": "By the way, conflicts don't really ruin this, because you have 3 days for each round.", "Solution_10": "Can I be in please?", "Solution_11": "Me but I may be a bit late.", "Solution_12": "I'll sign up....", "Solution_13": "I will join.", "Solution_14": "Sign me in.", "Solution_15": "Sure. Advice: It's beastly hard. Good luck!", "Solution_16": "Can you post all the questions and answers somewhere?", "Solution_17": "[quote=\"isabella2296\"]Sure. Advice: It's beastly hard. Good luck![/quote] I'll try my best...here goes...", "Solution_18": "izzy-can you post full solutions to each of the problems?\r\n\r\ni understna dif you cant :P", "Solution_19": "To fishy the fish:what kind of excuse is that???\r\nto isabella:when do the answers come out, and theres no round ten???", "Solution_20": "crazy4math, try actually reading some of the posts. :P", "Solution_21": ":o I DO\r\ndo u know HOW hard it is to read 15 pages of posts i have been here since july and have trying to read all the posts and thats way to much, because i don't know abut u but i cant't spend my whole day at the computer reading posts of which some have unimportant answers. i read the last few pages so, i found out.\r\n1)there's no round ten because too many people dropped out. so that answers one of my questions but as hard as i look i cannot find where the answers are posted, \r\nu have offended me :mad: \r\ni really am trying hard to keep up with all of this.", "Solution_22": "crazy-this tournament is OVER. Don't post here anymore. You're just adding to the spam (i guess this post is also...)\r\n\r\nMods-lock this please", "Solution_23": "sorry. :blush: \r\nfine, but i'm blaming it on fishy", "Solution_24": "and i'm blaming it on you, all the :spam:\r\n\r\nnot relaly lol :rotfl: \r\n\r\nwhoops\r\nmore :spam:", "Solution_25": "This tourneys over, now we just have 1000000000 pages of spam, please let this die.", "Solution_26": "anyway, can you post the answers to the problems?", "Solution_27": "That's 90 problems....anyway my answer key disappeared, sorry. Ask Yongyi, he got almost all perfects.", "Solution_28": "I'll be posting answers to my problems! and thats' more than 90 for sure...", "Solution_29": "For the final time:\r\n\r\nDO NOT SPAM!" } { "Tag": [], "Problem": "(page 32, number 35)\r\n\r\n\"Mike and Joey bought identical loaves of bread and packages of bologna. Mike made sandwiches with 5 slices of bologna and had 4 slices of bread left when he ran out of meat. Joey made sandwiches with 4 slices of bologna and had 4 slices of meat when he ran out of bread. How many slices of bread were in each loaf?\"\r\n\r\nThis is in the linear equations section. Don't know if that helps...", "Solution_1": "I never liked this problem either. In fact, most people don't... Anyway, it was discussed before: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47721", "Solution_2": "I did the problem on a long car trip, checked the solution manual months later, and found that I got the wrong answer. I hated that problem.\r\n\r\nSpeaking of solution manuals, do you not have a solution manual, imber.matutinis? Because you can always peek at that and get the hints you want.", "Solution_3": "[quote=\"mathnerd314\"]I did the problem on a long car trip, checked the solution manual months later, and found that I got the wrong answer. I hated that problem.\n\nSpeaking of solution manuals, do you not have a solution manual, imber.matutinis? Because you can always peek at that and get the hints you want.[/quote]\r\n\r\nI do not have a solutions manual...because someone told me that I shouldn't get one. o.O I thought they had the answers at least in the book, but apparently they don't. >.<", "Solution_4": "Who told you that you shouldn't get a solutions manual? You should get one. It's not too costly anyway." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "1)$\\int_0^\\infty \\frac 1 {(1+x^a)(1+x^2)}dx,a>0$\r\n2)$\\int_1^\\infty sin(x^2)dx$", "Solution_1": "[quote=\"xirti\"]\n2)$\\int_1^\\infty sin(x^2)dx$[/quote]\r\n\r\nCould the check the 1 in the integral\r\n\r\nThis one is known as Fresnel integral $\\int_0^\\infty sin(x^2)dx$", "Solution_2": "Actually, this was the form in my book.. I looked up frensel integrals and found a proof. Thank you! What about the first?", "Solution_3": "[quote=\"xirti\"]1)$\\int_{0}^\\infty \\frac 1{(1+x^{a})(1+x^{2})}dx,a>0$\n2)$\\int_{1}^\\infty sin(x^{2})dx$[/quote]\r\nThis had been lying around for quite sometime.\r\n\r\nSolution to [b]problem (1)[/b]:\r\n\r\nLet $x = \\tan\\theta$. \r\n\r\nThen, $I = \\int_{0}^{\\frac{\\pi}{2}}\\frac{\\,d\\theta}{1+{\\tan}^{a}\\theta}= \\int_{0}^{\\frac{\\pi}{2}}\\frac{{\\cos}^{a}\\theta \\,d\\theta}{{\\cos}^{a}\\theta+{\\sin}^{a}\\theta}$\r\n$\\Rightarrow I = \\int_{0}^{\\frac{\\pi}{2}}\\frac{{\\sin}^{a}\\theta \\,d\\theta}{{\\cos}^{a}\\theta+{\\sin}^{a}\\theta}$ ... [using the identity $\\int_{0}^{a}f(x)\\,dx = \\int_{0}^{a}f(a-x)\\,dx$]\r\n$\\Rightarrow 2I = \\int_{0}^{\\frac{\\pi}{2}}\\,d\\theta$\r\n$\\Rightarrow I = \\boxed{\\frac{\\pi}{4}}$ ... [Not so hard after all!]" } { "Tag": [ "algebra", "polynomial", "group theory", "abstract algebra", "invariant", "Ring Theory", "Galois Theory" ], "Problem": "Hi, I am interested in the folowing theorem but I do not know where to find information on it. \r\n\r\nLet $ F$ be a polinomial with integer coeficients that and $ k$ irreducible components. Let $ s(p)$ be the number of solutions to the congruence $ F\\equiv 0\\mod p$, for each prime $ p$. Prove that the average of $ s(p)$ over all primes is $ k$. \r\n\r\nBy average it seems that one the folowing may be understood. \r\n-- $ \\displaystyle\\frac {1}{\\pi(x)}\\sum_{p\\le x}{s(p)}$ is asymptotically equal to $ k$\r\n\r\n-- $ \\displaystyle\\sum_{p\\le x}{\\frac {s(p)}{p}\\log(p)}$ is asymptotically equal to $ k\\log(x)$\r\n\r\nand so on\r\n.\r\n.\r\n.\r\n\r\nDoes the theorem have a name? Has anybody a literature suggestion? I am interested in the proof or related developments which might have inspirede the theorem. \r\n\r\nThanks in advance.", "Solution_1": "You should alsy assume that the irreducible components are pairwise different. This means, that for almost all $ p$, they have no common root $ \\mod p$. Thus the problem reduces to the case when $ F$ is irreducible.\r\n\r\nNow if $ F$ is irreducible:\r\nLet $ K$ be splitting field of $ F$ over $ \\mathbb Q$ and let $ G$ be the corresponding Galois group. If $ \\mathcal O_K$ is $ K$'s number ring and $ \\mathfrak p$ is a prime ideal over $ p$, then the map $ \\mathcal O_K / \\mathfrak p \\to \\mathcal O_K / \\mathfrak p$ with $ x \\mapsto x^p$ is called the Frobenius; it permutes the roots of $ F \\mod \\mathfrak p$, and a fixed point corresponds to a solution of $ F(x) \\equiv 0 \\mod p$, so we want to count those. Note that a $ \\sigma \\in G$ with $ \\sigma \\mathfrak p \\equal{} \\mathfrak p$ (these $ \\sigma$ build a subgroup $ G_{\\mathfrak p}$) induces some automorphism $ \\mod \\mathfrak p$, which by elementary field theory is a power of the Frobenius. For unramified primes $ p$ (those that are \"nice\"), this induces an isomorphism between $ G_{\\mathfrak p}$ and the Galois group of $ \\mathcal O_K / \\mathfrak p$ over $ \\mathbb Z /(p)$.\r\n\r\nNow we can write $ G$ as a union of it's conjugacy classes $ G_i$ and apply Chebotarev's theorem, which states that for $ \\sigma \\in G$, the density of the primes $ p$ such that the Frobenius of $ p$ is of type $ \\sigma$ (as a permutation of the roots of $ F$) is $ \\frac{|G_\\sigma|}{|G|}$, where $ G_\\sigma$ is $ \\sigma$'s conjugacy class. Note that the number of fixed points of $ \\sigma$ is invariant under conjugation, so $ |G_\\sigma| \\cdot \\mathrm{Fix}(\\sigma) \\equal{} \\sum_{\\tau \\in G_\\sigma} \\mathrm{Fix}(\\tau)$. This gives us that the average of $ s(p)$ over $ p$ is equal to $ \\frac{1}{|G|} \\sum_{\\sigma \\in G} \\mathrm{Fix}(\\sigma)$, which equals $ 1$ by elementary group theory. This is what we wanted to prove.\r\n\r\n\r\nAs a literature suggestion, most texts on Chebotarevs theorem should do (but most will not mention this application). I can't give a good reference, but most good books on algebraic number theory should have a reference or a proof for Chebotarev's theorem.", "Solution_2": "Thank you very much ZetaX." } { "Tag": [], "Problem": "192. 1\r\n193. a.6.24\r\n b.6.24\r\n[hide=\".\"] Btw, hi strangers from antarctica and germany\n:D :rotfl: :P [/hide]", "Solution_1": "let's wait for a few more people to reply, which i very much doubt will happen :D", "Solution_2": "mr antarctica, cant you say whether its right or wrong atleast? :huuh: :D", "Solution_3": "i too think so :roll: \r\n\r\n192. 1\r\n193. a.6.24\r\n b.6.24 :)", "Solution_4": "1.infinity\r\n2.--6.24(both)\r\n\r\ni m pretty sure 1 is infinity :D", "Solution_5": "No way man! :huuh: :D :rotfl:", "Solution_6": "it has 2 b infinity\r\nno. of eggs is fixed--u can't write as 20.00000000000000000000000eggs", "Solution_7": "What eggs? :D :D Well i dont think that might be the case :maybe:", "Solution_8": "that's the case :w00t: :whistling: :sleeping:", "Solution_9": "thats what shrids says is written in the 11th chem text book, but the book in which i read abt it (some guide i dont remember) says that its only 1 :maybe:", "Solution_10": "the guide guided u wrong!!!!!!!!!!! :rotfl: :rotfl: :D", "Solution_11": "i dont know which one man but ive seriously read that somewhere :D", "Solution_12": "don't say books can't b wrong!!!!!", "Solution_13": "ok i made a mistake bcos i didnt see that he was counting a definite quantity like egs or sth :D :D :o", "Solution_14": "ritu, you're spot on :thumbup: \r\nlooks like you've nv'd chem text book very well :D", "Solution_15": "[quote=\"hash_include\"]looks like you've nv'd chem text book very well \n[/quote]\r\nthat was a wild guess :rotfl:" } { "Tag": [ "function", "trigonometry", "algebra proposed", "algebra" ], "Problem": "Determine all monotonous functions $ f: \\mathbb{R} \\rightarrow \\mathbb{R}$ which satisfy \\[ f^3(x)\\minus{}3f^2(x)\\plus{}6f(x)\\equal{}4x\\plus{}3,\\] for all $x \\in \\mathbb{R}$, where $ f^3\\equal{}f \\circ f \\circ f$ and $ f^2\\equal{}f \\circ f$.", "Solution_1": "[quote=\"moldovan\"]Determine all monotonous functions $ f: \\mathbb{R} \\rightarrow \\mathbb{R}$ which satisfy: $ f^3(x) - 3f^2(x) + 6f(x) = 4x + 3, (\\forall) x \\in \\mathbb{R}$, where $ f^3 = f \\circ f \\circ f$ and $ f^2 = f \\circ f.$[/quote]\r\n\r\nLet $ u\\in\\mathbb R$. Consider the sequence $ a_n=f^{[n]}(u)$\r\n\r\nWe get $ a_n=3a_{n-1}-6a_{n-2}+4a_{n-3}+3$ with $ a_0=u,a_1=f(u)$, and $ a_2=f(f(u))$\r\n\r\nThe equation $ x^3-3x^2+6x-4$ has 3 roots $ 1,2e^{\\frac{i\\pi}3}$ and $ 2e^{-\\frac{i\\pi}3}$ and so we get :\r\n\r\n$ a_n=u+n+r2^n\\cos(\\frac{n\\pi}3+\\theta)-r\\cos(\\theta)$ for some $ r>0,\\theta$ which may be be computed from $ a_1$ and $ a_2$\r\n\r\nSince $ f(x)$ is monotonous, we need to have $ u_n=\\frac{a_{n+2}-a_{n+1}}{a_{n+1}-a_n}$ $ =\\frac{f(a_{n+1})-f(a_n)}{a_{n+1}-a_n}$ with a constant sign.\r\n\r\nBut $ u_n=\\frac{1-r\\sqrt 32^{n+1}\\sin(\\frac{(n+1)\\pi}3+\\theta)}{1-r\\sqrt 32^{n}\\sin(\\frac{n\\pi}3+\\theta)}$ $ =\\frac{1-r\\sqrt 32^{n+1}v_{n+1}}{1-r\\sqrt 32^{n}v_n}$ where $ v_n=\\sin(\\frac{n\\pi}3+\\theta)$\r\n\r\nAnd, obviously, when $ n\\to +\\infty$, we always have some $ n$ where $ v_{n+1}$ and $ v_n$ have the same sign and some $ n$ where $ v_{n+1}$ and $ v_n$ have opposite sign.\r\n\r\nSo $ u_n=\\frac{1-r\\sqrt 32^{n+1}v_{n+1}}{1-r\\sqrt 32^{n}v_n}$ can have a constant sign when $ n\\to +\\infty$ only if $ r=0$\r\n\r\nSo $ a_n=u+n$ and $ f(u)=u+1$ which indeed is a solution.\r\n\r\nSo the unique monotonous solution to this equation is $ \\boxed{f(x)=x+1}$" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "n boys are sitting in a circle. Each has an even number of sweets. Each then gives half his sweets to the boy on his left. The teacher then gives any boy who ends up with an odd number of sweets an extra one. Show that after a sufficient number of repetitions every boy has the same number of sweets.", "Solution_1": "I'm not sure if this is correct. I just tried with the following example. Here there are 3 boys having 2, 4 and 6 sweets and the boy who has 6 sweets starts giving his sweets to the boy in left with 4 sweets and so on.\r\n\r\n2\t4\t6\r\n2\t7\t3\r\n2\t8\t4\r\n6\t4\t4\r\n3\t4\t7\r\n4\t4\t8\r\n4\t8\t4\r\n8\t4\t4\r\n\r\nafter this the same 8, 4 and 4 is going to repeat.\r\n\r\nSorry if I have understood the problem wrongly. Correct me if I'm wrong.", "Solution_2": "i think all of them pass on the sweets instantly :D :lol:", "Solution_3": "Yes. I think gaurav's correct.\r\n[code] 2 4 6\n1+2 2+3 3+1\n 3 5 4\n 4 6 4\n2+3 3+2 2+2\n 5 5 4\n 6 6 4\n3+3 3+2 2+3\n 6 5 5\n 6 6 6[/code]", "Solution_4": "[hide=\"Solution\"]Suppose at some point define $ D \\equal{} \\max \\minus{} \\min$\nClearly the $ \\max$ can never increase(even if the teacher gives the extra), nor the $ \\min$ can decrease. If, after some turn, $ D$ remains same, then The $ \\min$ must have remained same. But in that case the number of minimum values must have decreased unless $ \\min \\equal{} \\max$. Thus after finitely many turns everyone will have same number of sweets.[/hide]", "Solution_5": "[url]http://www.cut-the-knot.org/Curriculum/Algebra/IntergerIterationsOnACircle.shtml[/url]", "Solution_6": "Why is everybody assuming that $ n\\equal{}3$? Consider the general case please.", "Solution_7": "[quote=\"Johan Gunardi\"]Why is everybody assuming that $ n \\equal{} 3$? Consider the general case please.[/quote]\r\n\r\nI had the same idea that is in the link that dbgreen posted. Let the largest number of sweets any one person has be l, and the smallest be m. Then after each step i.e. passing on half and giving an extra one to the odd ones if any, l cannot increase and m cannot decrease. The number of people with m sweets must decrease because there must be someone with m sweets to the left of someone with > m sweets, otherwise we are done. So the number of people with m sweets goes to 0, and then m to goes to at least m+2 etc.", "Solution_8": "[quote=\"Johan Gunardi\"]Why is everybody assuming that $ n \\equal{} 3$? Consider the general case please.[/quote]\r\nI have been so stupid.\r\nNow I have edited." } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "Let circle O1 and circle O2 be tangent internally to circle O at A.B, respectively. Line g1 and line g2 are internal tangent of circle O1 and O2. g1 intersects g2 at C. Circle O1, cirlce O2 touches line g3 at D,E, respectively. M is midpoint of the line segment DE. N is foot of perpendicular from C to g3. AM intersects circle O again at P. BN intersects circle O again at Q. Prove that PQ is parallel to g3.", "Solution_1": "where is the line g3......?", "Solution_2": "g3 is external tangent of circle O1 and O2", "Solution_3": "please send your solution Fang_jh(preferably hidden) i wanna know how did you solve it.....\r\nhow can you call such a problem EASY :maybe: ???????????!!!!!!!!!!!!", "Solution_4": "Let line g3 intersect center line of circle O1,O2 at K. \r\nThe three points K,A,B are collinear (by Monge's theorem). \r\nLet the circle O1 tangent to g1 at H, the circle O2 tangent to g2 at I.\r\n DH intersects EI at T. The three points T,C,N are collinear (?). \r\nLet DH, EI intersect center line of the circle O1,O2 at S,R, respectively.\r\nES is perpendicular to DT(?)\r\nDR is perpendicular to TE(?)\r\nM,N,S,R are concyclic (NPC wrt triangle TDE)\r\nKN*KM=KS*KR\r\nE,D,S,R are concyclic\r\nKS*KR=KD*KE\r\nE,D,A,B are concyclic(?)\r\nKD*KE=KA*KB\r\nyielding KN*KM=KA*KB\r\nas a result M,N,A,B are concyclic\r\n\u2220AMN=\u2220ABQ=\u2220APQ \r\nyielding PQ \u2225g3", "Solution_5": "My solution may be similar to your solution, but it is simpler\r\nLet line g3 intersect center line of circle O1,O2 at K.\r\nThe three points K,A,B are collinear (by Monge's theorem). \r\nAll we have to do is prove M,N,A,B are concyclic \r\nCall $r$ and $R$ are the radius of (O1) and (O2) ($r < R$)\r\n\r\nWe have \r\n$\\frac{r}{R}= \\frac{{ND}}{{NE}}= \\frac{{KD}}{{KE}}= \\frac{{KN}}{{KE+NE}}$ (Because K,C are the external and internal center of similitude of the circumcircle (O1) and incircle (O2).And $CN//O1D//O2E$)\r\n$\\begin{array}{l}\\Rightarrow \\frac{{KN}}{{KD}}= \\frac{{KE+NE}}{{KE}}= 1+\\frac{{NE}}{{KE}}\\\\ \\Leftrightarrow \\frac{{KN}}{{KD}}+\\frac{{KN}}{{KE}}= 1+\\frac{{NE}}{{KE}}+\\frac{{KN}}{{KE}}= 2 \\\\ \\Leftrightarrow KD \\times KE = KN \\times \\left({\\frac{{KD+KE}}{2}}\\right) = KN \\times KM \\\\ \\end{array}$\r\nWe also have $KD \\times KE = KA \\times KB$\r\nSo $KA \\times KB = KN \\times KM$\r\n\r\nSo that M,N,A,B are concyclic", "Solution_6": "[img]6824[/img]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that $\\sqrt{\\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}}\\ge{\\sqrt[3]{\\frac{abc+abd+acd+bcd}{4}}}$\r\nwhere $a,b,c,d>0.$", "Solution_1": "this is weaker than maclaurin's symmetric mean ineq\r\n\\[\r\n\\sqrt {\\frac{{\\sum {a^2 } }}\r\n{4}} \\geqslant \\sqrt {\\frac{{ab + bc + cd + da + ac + bd}}\r\n{6}} \\geqslant \\sqrt[3]{{\\frac{{\\sum {abc} }}\r\n{4}}}\r\n\\]", "Solution_2": "Yeah, I knew it , I wanted to post it after this one. :D", "Solution_3": "It also (of course) has an easy Muirhead solution. Raising to the sixth power, we get\r\n\r\n$\\frac {1}{4^3}\\left(\\sum a^6+3\\sum a^4b^2+6\\sum a^2b^2c^2\\right) \\geq \\frac {1}{4^2}\\left(\\sum a^2b^2c^2+2\\sum a^2b^2cd\\right)$\r\n\r\nwhich is pretty obvious since all of the sums on the LHS majorize the sums on the RHS." } { "Tag": [ "LaTeX" ], "Problem": "Like, say I want the margin to be different on one page than the other ones, what do I do?\r\nAnd say I want to end a line, but I don't want to have the huge space that appears when you have an empty line between them, what do I do?\r\nAnd uhh, say I want to have a vertical space, but not a line skipping space, why doesn't it work when i do something like this?: \r\nblah blah blah\r\n\\vspace{10mm}\r\nblah blah blah\r\n\\vspace{10mm}\r\nblah blah blah\r\nIt still either all end up on one line for some reason, or the space only happens to one of the lines, and the space created becomes too big.", "Solution_1": "I'm not sure for your first question, but I can answer your second and third.\r\n\r\nFor your second question, use \\\\ at the end of a line and then hit Enter once, then begin typing your next line. (Do not skip lines by hitting Enter twice.)\r\n\r\nFor your third question, you should set \\vspace off in a separate line; skip lines, basically.", "Solution_2": "[quote=\"ZhangPeijin\"]Like, say I want the margin to be different on one page than the other ones, what do I do?[/quote]\nTry the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=chngpage]chngpage package[/url]. The documentation is inside the file itself. It will do a lot of things but for changing margins use the adjustwidth environment:\n[quote] [b]adjustwidth[/b]\n\n Within an adjustwidth environment the left and right margins can be\n adjusted. The environment takes one optional argument and two required\n length arguments: \n\n \\begin{adjustwidth}[]{leftmargin}{rightmargin}\n\n A positive length value will increase the relevant margin (shortening\n the text lines) while a negative length value will decrease the margin\n (lengthening text lines). An empty length argument means no change\n to the margin. At the end of the environment the margins revert to\n their original values.\n\n For example, to extend the text into the right margin:\n \\begin{adjustwidth}{}{-8em}[/quote]" } { "Tag": [], "Problem": "$ abc\\equal{}216$\r\n$ a \\plus{} b \\plus{} c \\equal{}19$\r\n\r\n$ a,b,c \\in \\mathbb{Z}$", "Solution_1": "[hide=\"Hint\"]$ 216\\equal{}2^3 \\times 3^3$\n\n[hide=\"Answer\"]$ (a, b, c)\\equal{}(4, 6, 9)$ (INPO)[/hide][/hide]", "Solution_2": "[quote=\"mathwizarddude\"]$ abc \\equal{} 216$\n$ a \\plus{} b \\plus{} c \\equal{} 19$\n\n$ a,b,c \\in \\mathbb{Z}$[/quote]\r\nNote that $ 216 \\equal{} 6^3$..\r\nCase $ a,b,c>0$:\r\nWe can't have $ 3 \\mid a,b,c$ and on the other hand we can't have $ 27 | a$ or $ 27 | b$ or $ 27 | c$. So assume $ 9 \\mid a$, since $ a \\equal{} 18$ is not a solution: $ a\\equal{}9$. Let $ b \\equal{} 3k$ So $ abc \\equal{} 9bc \\equal{} 27kc\\equal{}27k(10\\minus{}k)$ $ \\iff$ $ k(10\\minus{}3k) \\equal{} 8 \\iff k \\equal{} 2$. So $ (a,b,c) \\equal{} (9,6,4)$ or any permutation of that is a solution.\r\nCase $ a>0$ and $ b,c<0$.\r\nObviously $ a>19$ and $ a \\mid 2^3 \\cdot 3^3$. If $ a > 27$ we have that $ abc \\ge 28(\\minus{}8)(\\minus{}1) \\equal{} 224 > 216$. So $ a \\in [20;27]$.\r\nSo we can split into cases:\r\ncase $ a \\equal{} 24$: $ b\\plus{}c \\equal{} \\minus{}5$ and $ bc \\equal{} 9$ but $ |bc| \\le (\\frac{b\\plus{}c}{2})^2 < 9$\r\ncase $ a \\equal{} 27$: $ b\\plus{}c \\equal{} \\minus{}8$ and $ bc \\equal{} 8$, but this also impossible.\r\n\r\nSo the only solution is $ (9,6,4)$ and it's permutations.." } { "Tag": [ "trigonometry" ], "Problem": "Let a+b+c = pi/2 Write cos(a)cos(b)cos(c) as a sum of sines.", "Solution_1": "Let x=a+b, then:\r\ncos(a)cos(b)cos(c)=\r\ncos(a)cos(x-a)cos(pi/2-x)=\r\ncos(a)*sin(x)*[cos(x)cos(a)+sin(x)sin(a)]=\r\ncos(a)cos(x)sin(a)sin(x)+cos(a)^2*sin(a)^2=\r\ncos(a)cos(x)sin(a)sin(x)+sin(a)^2-sin(a)^4=\r\ncos(a)cos(x)sin(a)sin(x)-(sin(4a)-1)/8=\r\n???\r\n...I don't know where to go from here.", "Solution_2": "@lokito: I don't think that's the best way to go.\r\n\r\n[hide=\"Solution\"]\\begin{eqnarray*}\\cos a\\cos b\\cos c &=& \\cos a\\cos b\\cos (\\pi/2-a-b)\\\\ &=& \\cos a \\cos b \\sin(a+b)\\\\ &=& \\left[ \\frac{\\cos(a+b)+\\cos(a-b)}{2}\\right]\\sin(a+b)\\\\ &=& \\frac{\\cos(a+b)\\sin(a+b)}{2}+\\frac{\\cos(a-b)\\sin(a+b)}{2}\\\\ &=& \\frac{\\sin(2(a+b))}{4}+\\frac{\\sin(2a)+\\sin(2b)}{4}\\\\ &=& \\frac{\\sin(2a+2b)+\\sin(2a)+\\sin(2b)}{4}\\\\&=& \\frac{\\sin(2c)+\\sin(2a)+\\sin(2b)}{4}\\end{eqnarray*}\n\nI think that's good enough. I have repeatedly made use of the product-sum identities.[/hide]" } { "Tag": [ "algebra", "polynomial", "calculus", "integration" ], "Problem": "Find $ m$ such that $ x^2\\plus{}3xy\\plus{}x\\plus{}my\\minus{}m$ be the product of two factors of degree $ 1$ and with integral coefficients.", "Solution_1": "[hide][b]Form:[/b] $ (x \\plus{} ay \\plus{} b)(x \\plus{} cy \\plus{} d)$\n\n$ \\blacktriangleright$ There is no $ y^2$ term, so $ c \\equal{} 0$.\n$ \\blacktriangleright \\; xy$ coeff. is $ 3$, so $ a \\plus{} c \\equal{} a \\equal{} 3$\n$ \\blacktriangleright \\; x$ coeff. is $ 1$, so $ b \\plus{} d \\equal{} 1 \\implies b \\equal{} 1 \\minus{} d$\n\n$ (x \\plus{} 3y \\plus{} 1 \\minus{} d)(x \\plus{} d) \\equal{} (x^2 \\plus{} 3xy \\plus{} x) \\plus{} d \\minus{} d^2 \\plus{} 3dy$\n$ \\therefore my \\minus{} m \\equal{} d \\minus{} d^2 \\plus{} 3dy$\n\nPlugging in $ y \\equal{} 1$ gives $ d(1 \\minus{} d) \\plus{} 3d \\equal{} 0 \\implies d \\equal{} 0,4$\n\n[b]Answer:[/b]\n$ (x \\plus{} 4)(x \\plus{} 3y \\minus{} 3) \\implies m \\equal{} 12$\n$ x(x \\plus{} 3y \\plus{} 1) \\implies m \\equal{} 0$[/hide]" } { "Tag": [ "geometry", "function", "trapezoid", "algebra", "system of equations", "national olympiad" ], "Problem": "1. Solve the system of equations:\r\n$(1+\\frac{12}{3x+y})\\sqrt{x}=2\\\\(1-\\frac{12}{3x+y})\\sqrt{y}=6$\r\n\r\n2. Given $x,y$ are integer number and different to -1, satisfying:\r\n$\\large \\frac{x^{4}-1}{y+1}+\\frac{y^{4}-1}{x+1}\\in Z$ . Prove that $x^{4}y^{44}-1 \\vdots (x+1)$.\r\n\r\n3. A triangle ABC with BC are fixed. H and G are the orthocenter and centroid of triangle ABC, respectively. A is moving such that the midpoint of HG belongs to the line BC. Find the locus of A.\r\n\r\n4. Given a regular 2007-gon. Find the minimal number $k$ such that: Among every $k$ vertexes of the polygon, there always exists 4 vertexes forming a convex quadrilateral such that 3 sides of the quadrilateral are also sides of the polygon.\r\n\r\n5. Given $b$ is a positive integer. Find all the function $f : R \\to R$ such that:\r\n\r\n$f(x+y)=f(x)3^{b^{y}+f(y)-1}+b^{x}(3^{b^{y}+f(y)-1}-b^{y})$\r\n\r\n6. Given a trapezoid ABCD, BC is the bigger base, is incribed by (O). . The point P varies on the line BC and lies outside the segment BC such that: PA is not tangent to BC. The circle having PD as diameter cuts (O) at E ( $E\\neq D$). M is the interection point of BC and DE. N ($N \\neq A$)is the intersection point of PA and (O). Prove that the line MN goes through a fixed point.\r\n\r\n7. Given a number $a > 2$. Denote $f_{n}(x)=a^{10}x^{n+10}+x^{n}+...+x+1$, $n$ is positive integer number.\r\n\r\nProve that $f_{n}(x)=a$ has only one solution $x_{n}\\in (0,+\\infty)$. Prove that the sequence $(x_{n})$ has finite limit", "Solution_1": "Very long! :D\r\nWhat about on students of you? :P", "Solution_2": "Our students solved 5 to 6 problems. They waste time on the analytical geometry problem.", "Solution_3": "So if, Our students only sloved at most four problem :D I think :P", "Solution_4": "Problem 1 http://www.mathlinks.ro/Forum/viewtopic.php?t=132535\r\nProblem 2 http://www.mathlinks.ro/Forum/viewtopic.php?t=132536\r\nProblem 3 http://www.mathlinks.ro/Forum/viewtopic.php?t=132547\r\nProblem 4 http://www.mathlinks.ro/Forum/viewtopic.php?p=750364#750364\r\nProblem 5 http://www.mathlinks.ro/Forum/viewtopic.php?t=132556\r\nProblem 6 http://www.mathlinks.ro/Forum/viewtopic.php?t=132551\r\nProblem 7 http://www.mathlinks.ro/Forum/viewtopic.php?t=132554", "Solution_5": "This is version pdf", "Solution_6": "This is result \r\n\r\nTotal: 41 ( Have all 399 candidate) :( \r\n\r\nRank 1: 2\r\nRank 2: 7\r\nRank 3: 7\r\nRank 4: 25" } { "Tag": [ "trigonometry" ], "Problem": "When Babe Ruth hit a homer over the 12-m-high right-field fence 100 m from home plate, roughly what was the minimum speed of the ball when it left the bat? assume the ball was hit 1.0 m above the ground and its path initially made a 40 degree angle with the ground.\r\n\r\nWell, I used the following formula, where $ R$ is range, $ v_0$ is initial velocity, and $ \\theta$ is the angle. I'm assuming that if everything started on the ground instead of one meter up, then the minimum height the ball would have to go to clear the fence is 11 meters. Since the fence is 100 meters away from the home plate, and it is the maximum height, then the ball will land 100 meters away (neglecting air resistance). So the $ R$ is 200 m.\r\n\r\n$ R \\equal{} \\frac {(v_0^2)(\\sin{2\\theta})}{g}$.\r\n\r\nPlugging in values gives:\r\n\r\n$ 200 \\equal{} \\frac {(v_0^2)(\\sin{80})}{9.8}\\Leftrightarrow$\r\n$ v_0 \\equal{} 44.6 \\frac {m}{s}$\r\n\r\nThe book says the answer is roughly 34, so there is some flaw in my logic. I tried various ways of solving this, and gotten a variety of answers, but never 34. How did the book get this answer?\r\n\r\nAlso, in real life, why isn't the ball at its maximum height at $ \\frac {t}{2}$, where $ t$ is the time its in the air? Wouldn't the air resistance affect it the same way on the journey up and down?\r\n\r\nEDIT: Another problem...\r\n\r\nAgent Tim, flying a constant 200km/hr horizontally in a low-flying helicopter, wans to drop an explosive onto a master criminal's automobile traveling 150 km/hr on a lvel highway 88 m below. At what angle (with the horizontal) should the car be in his sights when the bomb is released?\r\n\r\nWell, the first think I did was calculate the time it would take for the bomb to fall 88 m:\r\n$ 88 \\equal{} \\frac {1}{2}(9.8)(t^2)\\Leftrightarrow$\r\n$ t \\equal{} 4.24 \\sec$\r\n\r\nI'm not sure if this is allowed, but from the frame of reference of the car, the plane is moving 50 km/hr. Since the plane is dropping a bomb, not shooting it, that means the bomb's initial horizontal velocity is going to be 50 km/hr as well. So I converted that to m/s:\r\n\r\n$ \\frac {50 km}{hr}\\cdot\\frac {1hr}{60min}\\cdot\\frac{1min}{60sec}\\cdot\\frac{1000m}{1km} \\equal{} 13.9 \\frac {m}{s}$\r\n\r\nSince the bomb will be traveling 4.24 seconds, I multiplied the rate I just found with the time to find the distance the bomb travels, giving me 58.9 meters. So...\r\n$ \\tan^{ \\minus{} 1}\\theta \\equal{} \\frac {88}{58.9}\\Leftrightarrow$\r\n$ \\theta \\equal{} 56.2$\r\n\r\nIs this feasible?", "Solution_1": "Start by writing the parametric equations for the ball motion:\r\n\r\n$ x \\equal{} v_o \\cos \\theta t$\r\n$ y \\equal{} v_o \\sin \\theta t \\minus{} \\frac{1}{2}gt^2$.\r\n\r\nAccording to the data of the problem, we want that, when x = 100 m we must have y = 11 m. Substituting in the above two equations (with $ \\theta \\equal{} 40^o$) we get $ v_o \\equal{} 33.8\\, m/s$. Formulas should be used with care." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "The roots of $ x^3$ $ \\plus{}$ $ ax^2$ $ \\plus{}$ $ bx$ $ \\plus{}$ $ c$ $ \\equal{}$ $ 0$ are $ \\alpha$, $ \\beta$ and $ \\gamma$.Find the cubic whose roots are $ \\alpha$^3, $ \\beta$^3 and $ \\gamma$^3", "Solution_1": "Hello\r\n[hide=\"hint\"]\nLet assume that the required eq. be $ x^3 \\plus{} px^2 \\plus{} qx \\plus{} r \\equal{} 0$ whose roots are $ \\alpha^3,\\beta^3,\\gamma^3$.\nnow \nfrom the given eq. we have\n$ \\alpha \\plus{} \\beta \\plus{} \\gamma \\equal{} \\minus{} a$\n$ \\alpha\\beta \\plus{} \\gamma\\beta \\plus{} \\alpha\\gamma \\equal{} b$\n$ \\alpha\\beta\\gamma \\equal{} \\minus{} c$\n\nIn our eq.\n$ p \\equal{} \\alpha^3 \\plus{} \\beta^3 \\plus{} \\gamma^3 \\equal{} (\\alpha \\plus{} \\beta \\plus{} \\gamma)((\\alpha \\plus{} \\beta \\plus{} \\gamma)^2 \\minus{} 3(\\alpha\\beta \\plus{} \\alpha\\gamma \\plus{} \\gamma\\beta)) \\minus{} 3\\alpha\\beta\\gamma$\n where r=$ (\\alpha\\beta\\gamma)^3$=$ \\minus{} c^3$\nfind q by using same method that we have used for p.\n[/hide]\r\n\r\nThank u", "Solution_2": "so we have:\r\n[hide=\"answer\"]$ x^3\\plus{}(a^3\\plus{}3c\\minus{}3ab)x^2\\plus{}(b^3\\minus{}2abc\\plus{}3c^2)x\\plus{}c^3\\equal{}0$\n[/hide]" } { "Tag": [ "combinatorics open", "combinatorics" ], "Problem": "Accordingly,there are 2,279,184 solutions to any N-Queen problems for n=15.Can anyone show me how such calculations happened?How about 14,772,512 solutions to N-Queens problem for n=16;And 95,815,104 solutions to n-queens problem for n=17? I'm just intrigue about how such calculating possibilities happened.\r\n\r\n Respectfully,\r\n A.Eddington\r\n\r\n [/b]", "Solution_1": "See http://oeis.org/A000170 and references there." } { "Tag": [ "geometry", "geometric transformation", "reflection", "circumcircle", "Euler", "power of a point", "radical axis" ], "Problem": "Let $ ABC$ be a triangle and let $ M, N, P$ be the midpoints of sides $ BC, CA, AB$ respectively. Denote by $ X, Y, Z$ the midpoints of the altitudes emerging from vertices $ A, B, C$ respectively. Prove that the radical centre of the circles $ AMX, BNY, CPZ$ is the centre of the nine-point circle of triangle $ ABC$.\r\n\r\nThis is from Math Reflections 2008 Issue 1, and since deadline is over for those problems I think I can post them here.", "Solution_1": "Let $ G$ be the centroid and $ O$ the circumcenter of the $ \\triangle ABC.$ $ G$ is also the centroid of the medial $ \\triangle MNP$ and $ O$ is its orthocenter. Let $ U, V, W$ be midpoints of $ NP, PM, MN.$ Let $ \\mathcal Q_m, \\mathcal Q_n, \\mathcal Q_p$ be circles internally tangent to the circles $ \\odot (AMX), \\odot (BNY), \\odot (CPZ)$ at $ M, N, P$ and with similarity coefficients $ 1/3.$ \r\n\r\nLet $ D, E, F$ be the altitude feet of the medial $ \\triangle MNP.$ Since the circles $ \\odot (AMX), \\odot (BNY), \\odot (CPZ)$ pass through the reflections $ A', B', C'$ of $ M, N, P$ in the perpendicular bisectors of $ AX, BY, CZ$ and $ \\overline {MA'} \\equal{} 3\\ \\overline {MD},$ $ \\overline {NB'} \\equal{} 3\\ \\overline {NE},$ $ \\overline {PC'} \\equal{} 3\\ \\overline {PF},$ the circles $ \\mathcal Q_m, \\mathcal Q_n, \\mathcal Q_p$ pass through the altitude feet $ D, E, F$ of the medial $ \\triangle MNP$ and they are centered on the sidelines $ VW, WU, UV$ of its medial $ \\triangle UVW.$ Thus the powers of the orthocenter $ O$ of the medial $ \\triangle MNP$ to the circles $ \\mathcal Q_m, \\mathcal Q_n, \\mathcal Q_p$ are all equal to half the power of orthocenter $ O$ to the circumcircle $ (K)$ of this medial triangle, identical with the 9-point circle of the $ \\triangle ABC$:\r\n\r\n$ \\overline{OD} \\cdot \\overline {OM} \\equal{} \\overline{OE} \\cdot \\overline {ON} \\equal{} \\overline {OF} \\cdot \\overline {OP} \\equal{} \\frac {_1}{^2}\\ p(O, (K)).$ \r\n\r\nSince the circles $ \\odot (AMX), \\odot (BNY), \\odot (CPZ)$ pass through the reflections $ A, B, C$ of $ M, N, P$ in $ U, V, W$ and $ \\overline {MA} \\equal{} 3\\ \\overline {MG},$ $ \\overline {NB} \\equal{} 3\\ \\overline {NG},$ $ \\overline {PC} \\equal{} 3\\ \\overline {PG},$ the circles $ \\mathcal Q_m, \\mathcal Q_n, \\mathcal Q_p$ all pass through the centroid $ G.$ Thus the powers of the centroid $ G$ of the medial $ \\triangle MNP$ to the circles $ \\mathcal Q_m, \\mathcal Q_n, \\mathcal Q_p$ are all equal to zero.\r\n\r\nConsequently, the circles $ \\mathcal Q_m, \\mathcal Q_n, \\mathcal Q_p$ are coaxal and the common Euler line $ GO$ of the $ \\triangle MNP, \\triangle ABC$ is their common radical axis. The circumcenter $ K$ of the medial $ \\triangle MNP,$ identical with the 9-point center of the $ \\triangle ABC,$ then lies on their common radical axis. Let the lines $ MK, NK$ cut the circles $ \\mathcal Q_m, \\mathcal Q_n$ again at $ M_q, N_q.$ Since $ \\overline{KM} \\equal{} \\overline{KN}$ and $ \\overline{KM} \\cdot \\overline{KM_q} \\equal{} \\overline{KN} \\cdot \\overline {KN_q},$ it follows that $ \\overline {KM_q} \\equal{} \\overline {KN_q}$ and $ \\overline {MM_q} \\equal{} \\overline {NN_q}.$\r\n\r\nLet the lines $ KM, KN$ cut the circles $ \\odot (AMX), \\odot (BNY)$ again at $ M_0, N_0.$ Since $ M, N$ are similarity centers of $ \\mathcal Q_m \\sim \\odot (AMX), \\mathcal Q_n \\sim \\odot (BNY),$ we get $ \\overline {MM_0} \\equal{} 3\\ \\overline {MM_q} \\equal{} 3\\ \\overline {NN_q} \\equal{} \\overline {NN_0}.$ Using $ \\overline{KM} \\equal{} \\overline{KN}$ again, this yields $ \\overline{KM_0} \\equal{} \\overline{KN_0}$ and $ \\overline{KM_0} \\cdot \\overline{KM} \\equal{} \\overline{KN_0} \\cdot \\overline{KM}.$ In conclusion, $ K$ is on the radical axis of the circles $ \\odot (AMX), \\odot (BNY)$ and similarly, $ K$ is on the pairwise radical axes of $ \\odot (BNY), \\odot (CPZ)$ and $ \\odot (CPZ), \\odot (AMX).$ As a result, $ K$ is (at least) the radical center of these 3 circles.", "Solution_2": "Let $D$ be the foot of the A-altitude and $ H,N$ the orthocenter and the 9-point center of $ \\triangle ABC.$ Let $ L$ denote the midpoint of $AH$ (orthocenter of APN) and let the ray $ ML$ cut the circle $ \\mathcal{G}_a \\equiv \\odot(AMX)$ at $ T.$ Since $ LM$ is a diameter of the 9-point circle $ (N),$ we have that $ p(N,\\mathcal{G}_a) \\equal{} NM \\cdot NT$ and from the power of $ L$ WRT $ \\mathcal{G}_a$ we get $ R \\cdot TL \\equal{} AL \\cdot LX$ since $ \\triangle ABC \\sim \\triangle APN$ with similarity coefficient $ 2.$ Then the power $ AL \\cdot LX$ equals a quarter of the power $ AH \\cdot HD.$\n\n$ p(N,\\mathcal{G}_a) \\equal{} \\frac {R}{2} \\left (\\frac {R}{2} \\plus{} TL \\right ) \\equal{} \\frac {R}{2} \\left (\\frac {R}{2} \\plus{} \\frac {AH \\cdot HD}{4R} \\right )$ \n\n$ p(N,\\mathcal{G}_a) \\equal{} \\frac {R^2}{4} \\plus{} \\frac {AH \\cdot HD}{8} \\equal{} \\frac {R^2}{4} \\plus{} \\frac {k^2}{8}$\n\n$ AH \\cdot HD$ equals the power of inversion $ k^2$ that takes the circumcircle $ (O)$ of $ \\triangle ABC$ into its 9-point circle $(N),$ thus we conclude that $ p(N,\\mathcal{G}_a) \\equal{} p(N,\\mathcal{G}_b) \\equal{} p(N,\\mathcal{G}_c)$ $ \\Longrightarrow$ $ N$ is the radical center of $ \\mathcal{G}_a,\\mathcal{G}_b,\\mathcal{G}_c.$" } { "Tag": [ "trigonometry" ], "Problem": "While going about trying to figure out a problem about the static equilibrium of a rigid beam, I found the following equations after setting the sum of the forces equal to zero and setting them to zero and setting the sum of the torques equal to zero:\r\n\r\n$ T\\cos{\\theta} \\minus{} F\\cos{\\phi} \\equal{} 0$\r\n$ T\\sin{\\theta} \\plus{} F\\sin{\\phi} \\minus{} mg \\equal{} 0$\r\n$ \\minus{}\\frac{1}{2}lmg \\plus{} lT\\sin{\\theta} \\equal{} 0$\r\n\r\nwhere $ m$, $ l$, and $ \\theta$ are knowns. (Of course, $ g$ is known as well.) The $ l$ can be divided out of the last equation. This leaves three equations in three unkowns: $ T$, $ F$, and $ \\phi$. Now how do I get $ F$ in terms of the knowns? In particular, I'm having trouble with the $ \\phi$ and not getting something nasty......\r\n\r\nAny help would be greatly appreciated.", "Solution_1": "Making $ (T\\cos{\\theta} \\minus{} F\\cos{\\phi})^2 \\plus{} (T\\sin \\theta \\plus{} F\\sin{\\phi})^2$ gives cosine rule." } { "Tag": [ "trigonometry" ], "Problem": "For $x^5=1,$ find the value of $2x+\\frac{1}{1+x}+\\frac{x}{1+x^2}+\\frac{x^2}{1+x^3}+\\frac{x^3}{1+x^4}.$", "Solution_1": "WLOG x=1 :lol:", "Solution_2": "$x^5=1$ should have 5 roots? ;)", "Solution_3": "Well the way i did it was by very ugly algebra and mass calculation, then using the fact tha $x^5=1$ into reducing powers of remaining $x$ and then finally using $x^5-1=0$ as a way to chomp down a lot of factored terms. I will show the work at a later time..... but here is my answer:\r\n\r\n[hide]1[/hide]", "Solution_4": "Regrettably your answer is incorrect. :(", "Solution_5": "is it $4$?", "Solution_6": "Yes, this is one of them.", "Solution_7": "You should edit your question to find the value\"s\" i think....", "Solution_8": "\"Politics/Economics\"? Whoa, these Japanese University entrance exams are hard!", "Solution_9": "[hide]perhaps something like $4\\cos 72k^o$ where k=0,1,2 \n[hide]Of course, $4 cos 72^o$ is somethng like $\\sqrt 5 - 1$ so you could also say the other two values are $-1 \\pm \\sqrt 5$ \n[/hide]\n[/hide]", "Solution_10": "That's right.", "Solution_11": "Well, one answer is obviously $4$...", "Solution_12": "[quote=\"xxreddevilzxx\"]Well the way i did it was by very ugly algebra and mass calculation, then using the fact tha $x^5=1$ into reducing powers of remaining $x$ and then finally using $x^5-1=0$ as a way to chomp down a lot of factored terms. I will show the work at a later time..... but here is my answer:\n\n[hide]1[/hide][/quote]\nFWIW There may be (and is) elegant methods but even \"ugly algebra\" and brute force is not too hard, if one remembers that:\n[hide]$x^5= 1$ and either $x=1$ or $1+x+x^2+x^3+x^4=0$ and the second equation is same as $y^2+y-1=0$ (where y=x+1/x) and final answer gets to be 2y :) [/hide]", "Solution_13": "Hello, guys! Thank you for your replies. :) \r\n\r\nI will give a hint. Remark that $1+4=2+3=5.$\r\n\r\nGood luck!\r\n\r\nkunny", "Solution_14": "[quote=\"Gyan\"]perhaps something like $4\\cos 72k^o$ where k=0,1,2[/quote] \r\n\r\nHow?", "Solution_15": "[quote]How?[/quote] Just a fancy way to say $2(x+1/x)$ :) \r\n\r\n[hide]Doesn't matter, if you use [hide=\"trig or algebra\"]and write $x=cos a + i sin a$ where a=k 72^o [/hide] , its not too hard, and if you are lucky you may even notice:\n[hide=\"this\"]$\\frac 1 {1+x} + \\frac{x^3}{1+x^4} = \\frac x {1+x^2} + \\frac{x^2}{1+x^3} = \\frac 1 x = x^4$ [/hide] right away .\n[hide]$\\frac 1 {1+x} + \\frac{x^3}{1+x^4} = \\frac{x^5} {1+x} + \\frac{x^4}{x+x^5} = \\frac{x^5+x^4}{1+x} = x^4$ [/hide]\n\n[/hide]" } { "Tag": [ "algebra", "polynomial", "algebra solved" ], "Problem": "Find all polynomials [tex]f(x) = x^n + a_{n-1}x^{n-1}+...+a_{1}x+a_{0} , a_{0} \\neq 0[/tex] with integer coefficients such that \r\n[tex]f(a_i)=0, i=0,1,...,n-1[/tex]", "Solution_1": "using the fact that if $ \\frac {p}{q} $ is a root of an eqn, then $p$ divides $a_0$ and $q$ divides $a_n$ Using this we get that all of $ a_i \\ i=0,1,..... n-1 $ divide $a_0$\r\n\r\nAlso the factorisation of the eqn is $ \\prod _{i=0} ^n ( x - a_i) $", "Solution_2": "Also we have $ a_0 a_1 a_2 .... a_{n-1} = (-1)^n a_0 $\r\n\r\nNow if $a_0 $ is zero , it is trivial for all nos. will divide zero (obviously)\r\n\r\nthus we get that if $ a_0 \\not= 0 $,\r\n\r\n$ (a_1 a_2 .... a_{n-1} - (-1)^n ) = 0 $\r\nThus if n is even, we have that an even no. of $ a_i $ can be $-1$ and the rest should be $1$ , since $a_i$ are integers\r\n\r\nAlso, if n is odd, the situation is the opposite of the case for n = even\r\n\r\nIs this right????? I believe that's the best we can get!!!", "Solution_3": "we know that a[0] is not 0, so for all i in{1,..,n-1} a[i] is 1 or -1,now if a[0] is 1 or -1 ,let p the number of 1 and q is the number of -1 between the root of our polynom, with p>q(because a [n-1] is not 0) we get that p-q=1 and p(p-1)/2 + q(q-1)/2 -pq=-1 ( by the formuls between roots and coefficients) we get (p,q)=(2,1) the symetric case p1 then using 0=|P(a[0])|> |(a[0])^n|-....-|a[1]|-|-|a[0]|-1>-2 we get \r\n|a[2]|=2 and that a[i]=(-1)^(i+1) for i in {1,...,n-1} so a[2]=-2(using the rull of signe), if n>2 then a[2]=-1 but then \r\nP(-1) will not 0 so n is less than 2 , we get the solution P=(X-1)(X+2)" } { "Tag": [ "algebra", "polynomial" ], "Problem": "If $a_{1}, a_{2}\\cdots , a_{n}$ are distinct integers, prove that the polynomial $(x-a_{1})(x-a_{2})\\cdots (x-a_{n})-1$ is irreducible over the integers.", "Solution_1": "[hide]Suppose it can be factored as , where . Then for all . Since and are factorizations over the integers, then and are both integers, and thus they must be either 1 or -1. In fact, for all we have . Since is zero at different points, it must have at least roots. However, it is well known that if , then , but then , but is at least degree , so we have a contradiction, and the polynomial cannot be reduced over the integers.[/hide]" } { "Tag": [ "geometry" ], "Problem": "I need help solving this problem..\r\n\r\nthe colored part and the combined circle ( the middle part ) part have the same areas....\r\n\r\nwhat's the length between the point A nad point B?\r\n\r\n( it was from a korean worksheet, so I did my best to translate it.. :D )", "Solution_1": "Wait, is it the sum of both colored parts equals the middle part, or one of the colored part equals the middle part?", "Solution_2": "I'm pretty sure it's both, but the book doesn't say it clearly....\r\n\r\nthat's one of the reasons y I asked AoPS", "Solution_3": "Ok if its both combined, then I got 2 :pi: as the answer. Ok here's how I did it-\r\n\r\nLets assume AB as x. The area of the rectangular part in the middle is 8x. The area of half of each circle is \r\n8:pi: . So we can write the quation-\r\n8x=8:pi: +8:pi: (We dont need to add the colored part because the area of the comined part equals the colored part)\r\n\r\nTherefore x=2:pi:", "Solution_4": "[quote=\"GoBraves\"]Ok if its both combined, then I got 2 :pi: as the answer. Ok here's how I did it-\n\nLets assume AB as x. The area of the rectangular part in the middle is 8x. The area of half of each circle is \n8:pi: . So we can write the quation-\n8x=8:pi: +8:pi: (We dont need to add the colored part because the area of the comined part equals the colored part)\n\nTherefore x=2:pi:[/quote]\r\n\r\nWait, but if the area of the rectangular part is 8x, and you're saying 8x=16pi, doesn't 16pi only represent the area of the two half circles? And don't they overlap? And what about the shaded portion?\r\n\r\nOh wait. I think I get it. The overlap is the same as the shaded portion! Right? :blush:" } { "Tag": [ "algebra", "polynomial" ], "Problem": "Prove that the polynomial $ P(x)\\equal{}x^{n\\plus{}1}\\minus{}(x^{n\\minus{}1}\\plus{}x^{n\\minus{}2}\\plus{}...\\plus{}x\\plus{}1)$ has only one positive root, that we call it $ a_n$ Prove that the sequence has the limit $ \\frac{1\\plus{}\\sqrt{5}}{2}$", "Solution_1": "[hide=\"Hint\"]$ (x \\minus{} 1) P$[/hide]", "Solution_2": "could you expand your hint please?", "Solution_3": "[quote=\"TZF\"][hide=\"Hint\"]$ (x \\minus{} 1) P$[/hide][/quote]\r\n\r\nAm-Gm of bracketed term: =a(r^n-1)/r-1=(x^n-1)/x-1\r\n\r\nTherefore, P(x)=x^n+1-(x^n-1)/x-1\r\n\r\n(x-1)P(x)=x^n+1(x-1)-(x^n-1)\r\n\r\n(x-1)P(x)=x^n+2-x^n+1-x^n+1\r\n\r\n\r\nThats all i got... can someone else work off that maybe?" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Show that every positive integer can be written as a sum of no more than $ 53$ fourth powers of integers.", "Solution_1": "Before someone starts to post the proof with densities ;)\r\nNote that $6\\cdot (x_1^2+x_2^2+x_3^2+x_4^2)^2 = \\sum_{1 \\leq m < n \\leq 4} (x_m+x_n)^4 + (x_m-x_n)^4$.\r\nLet $n$ be our integer. Write $n=6q+r$ ($0 \\leq r <6$) and then $q=a_1^2+a_2^2+a_3^2+a_4^2$ (the last one by the Four-Square-Theorem).\r\nEvery term of type $6a_i^2$ can be written as $6a_i^2=6(b_{i,1}^2+b_{i,2}^2+b_{i,3}^2+b_{i,4}^2)^2$ after writing $a_i$ as sum of four squares.\r\nThus, by the above, $6a_i^2$ can be written as sum of $12$ fourth powers. Now $6q=6a_1^2+6a_2^2+6a_3^2+6a_4^2$ is a sum of $48$ fourth powers, and $r$ is the sum of at most $5$ ones, thus in total $n$ is a sum of $53$ fourth powers.\r\n\r\nNote that by such identities one gets a nice proof that when every integer can be written as sum of enough $n$-th powers, then every integer can be written as sum of enough $2n$-th powers." } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra solved" ], "Problem": "find all [b]idean[/b] of rings $\\mathbb{Z}*\\mathbb{Z}$.\r\n :(", "Solution_1": "Do you mean ideals of the ring $\\mathbb{Z}* \\mathbb{Z}$? How is this ring defined?", "Solution_2": "I'm sorry.I'm wrong that is [b]ideals[/b] not [b]ideans[/b] :lol:", "Solution_3": "Is it $\\mathbb{Z}\\times \\mathbb{Z}$\u00bf", "Solution_4": "yes, that's right", "Solution_5": "Let $I$ be an ideal of $\\mathbb Z^{2}$.\r\nLet $n\\mathbb Z$ and $m\\mathbb Z$ be the projections of $I$ on the two components.\r\nThere exists in $I$ an element of the form $(n,u)$ and an element $(v,m)$ for some $u,v\\in\\mathbb Z$.\r\nSince $I$ is an ideal, $I$ contains $(n,u)\\times(1,0)=(n,0)$ and similarly $I$ contains $(0,m)$.\r\nIt is now obvious that $I=n\\mathbb Z\\times m\\mathbb Z$.", "Solution_6": "Ideals in product rings are very easily described!\r\n\r\nNotice that AxB = A+B (direct product = direct sum)\r\n\r\nLet s1 : A -> A+B and s2 : B -> A+B\r\n p1 : AxB -> A and p2 : AxB -> B\r\n\r\nThen we have id(A) = p1*s1 and id(B) = p2*s2\r\n\r\nAlso id(A+B) = s1*p1 + s2*p2\r\n\r\nLet I be any ideal in A+B and a = p1(I) and b = p2(I) the ideals in A and B respectively. We compute I = id(I) = (s1*p1 + s2*p2)(I) = (s1*p1)(I) + (s2*p2)(I) = s1(a) + s2(b) = a+b (direct) = axb (direct).\r\n\r\nWe conclude that all ideals in A+B are of the form a+b with a and b ideals in A and B respectively.\r\n\r\nSorry, I don't know latex.", "Solution_7": "That's right :). Of course, it can be generalized to arbitrary finite products.\r\n\r\nA counterexample for infinite products:\r\n\r\n$R : = \\prod_{n \\in \\mathbb{N}}\\mathbb{Z}/2 ,~ I : = \\{r \\in R : \\{n \\in \\mathbb{N}: r_{n}= 1\\}\\text{~ is finite~}\\}$\r\n \r\nThis comes from the correspondence between ideals in products of fields and the filters of the index set; here one uses the frechet-filter, which is not generated by any subset.", "Solution_8": "Is the conclusion really correct? There is something wrong with the proof, no? \r\n\r\nLet f and g be homomorphisms A -> B of commutative rings with 1. When is it true that (f+g)(I) = f(I) + g(I) for any ideal I?", "Solution_9": "Yes the proof has the same error as discussed in the other thread. But you can easily correct the proof.", "Solution_10": "Here's a corrected proof to clean up this mess:\r\n\r\nWe want to prove that the only ideals in AxB are axb where a and b are ideals in A and B respectively.\r\n\r\nLet I be any ideal in AxB and put a = p1(I), b = p2(I). Suppose (x,y) is in I. Then so is (x,y)(1,0) = (x,0). Similarly, if (x',y') is in I, so is (0,y'). Therefore (x,0) + (0,y') = (x,y') is in I. But then axb < I. The converse is obvious so I = axb.\r\n\r\nThis proof also shows why the result doesn't carry over to abelian groups. :) \r\n\r\nIn the infinite case direct product and direct sum do not coincide and ideals in product(Ai) taken over an infinite index set are not determined by the projection ideals in the components. If I is an ideal with components ai, there are at least two ideals with these components: product(ai) and sum(ai). These are different.", "Solution_11": "Yes, that's right :-). One also has to prove, that a,b are ideals (images of ideals are not always ideals), but this is no problem. Besides, the characterization also works for left- or rightideals.", "Solution_12": "The projections are surjective so ideals are taken to ideals.\r\n\r\nYou are right, commutativity is not required. But in the proof given it is essential to have 1." } { "Tag": [ "MATHCOUNTS", "geometry" ], "Problem": "Are there any home school MathCounts teams in the NorthEast/ MetroWest area?\r\n\r\nWe are looking to join one or form one if there is enough interest in the Lexington/ Bedford area.\r\n\r\nThanks.", "Solution_1": "I know Voyagers has one, but I don't really know much about it.", "Solution_2": "I would like to mention that if one lives in Lexington then we would be extremely supportive of any decision to abandon the (relative) freedom, enjoyment, and overall excellent education of home-schooling and attend an overcrowded public school with ridiculous class sizes, stupid classes, and my personal favorite, the smoking corner. Just because we have an awesome math team.\r\n\r\n :) \r\n\r\nOn a slightly more relevant note, Voyagers is actually based in Acton, not Lexington/Bedford; however they do in fact send a team which competes in the Metrowest chapter. However you probably have to join them. Mr. J.S. Trabucco (perfect628) and I, way back in 1974 (ok maybe it was 2006) went to nationals with someone who competed with them...shoot me a PM and I'll see if I can find her e-mail address.\r\n\r\nalso I noticed that Mr. J. S. Trabucco has the same first two initials as Mr. J. S. Bach" } { "Tag": [ "geometry", "3D geometry", "octahedron", "probability", "symmetry", "rotation", "counting" ], "Problem": "Six ants simultaneously stand on the six vertices of a regular octahedron, with each ant at a different vertex. Simultaneously and independently, each ant moves from its vertex to one of the four adjacent vertices, each with equal probability. What is the probability that no two ants arrive at the same vertex?\r\n\r\n$ \\textbf{(A)}\\ \\frac {5}{256} \\qquad\r\n\\textbf{(B)}\\ \\frac {21}{1024} \\qquad\r\n\\textbf{(C)}\\ \\frac {11}{512} \\qquad\r\n\\textbf{(D)}\\ \\frac {23}{1024} \\qquad\r\n\\textbf{(E)}\\ \\frac {3}{128}$", "Solution_1": "We approach this problem by counting the number of ways ants can do their desired migration, and then multiple this number by the probability that each case occurs.\r\n\r\nLet the octahedron be ABCDEF, with points B,C,D,E coplanar. Then the ant from A and the ant from F must move to plane BCDE. Suppose, without loss of generality, that the ant from A moved to point B. Then, we must consider three cases.\r\n\r\nCase 1: Ant from point F moved to point C\r\nOn the plane, points B and C are taken. The ant that moves to D can come from either E or C. The ant that moves to C can come from either B or D. Once these two ants are fixed, the other two ants must migrate to the \"poles\" of the octahedron, points A and F. Thus, there are two degrees of freedom in deciding which ant moves to D, two degrees of freedom in deciding which ant moves to C, and two degrees of freedom in deciding which ant moves to A. Hence, there are 2*2*2=8 ways the ants can move to different points.\r\n\r\nCase 2: Ant from point F moved to point D\r\nOn the plane, points B and D are taken. The ant that moves to C must be from B or D, but the ant that moves to E must also be from B or D. The other two ants, originating from points C and E, must move to the poles. Therefore, there are two degrees of freedom in deciding which ant moves to C and two degrees of freedom in choosing which ant moves to A. Hence, there are 2*2=4 ways the ants can move to different points.\r\n\r\nCase 3: Ant from point F moved to point E\r\nBy symmetry to Case 1, there are 8 ways the ants can move to different points.\r\n\r\nGiven a point B, there is a total of 8+4+8=20 ways the ants can move to different points. We oriented the square so that point B was defined as the point to which the ant from point A moved. Since the ant from point A can actually move to four different points, there is a total of 4*20=80 ways the ants can move to different points.\r\n\r\nEach ant acts independently, having four different points to choose from. Hence, each ant has probability 1/4 of moving to the desired location. Since there are six ants, the probability of each case occuring is (1/4)(1/4)(1/4)(1/4)(1/4)(1/4)=1/4096. Thus, the desired answer is 80(1/4096)=5/256.\r\n\r\nA.", "Solution_2": "[hide=\"simple solution\"]There are of course, $4^6$ possible movements.\n\nEach movement in which ants from point move to adjacent point can be generalized with a pattern.\n\nEach movement will be like a continuous line of arrows. Starting from a tip, going down to the $4$ point area, there are $4$ options. Then $2$, choosing the direction of flow. Then finally, $2$ more for each of these variations as it can be from bottom to top or top to bottom. Therefore, each possible movement has $4*2*2 = 16$ variations.\n\n$\\frac{16n}{4096}$ is the form that probability must be in, where n is number of possible movements.\n\nThus $P$ of the event must be in the form: $\\frac{n}{256} = \\boxed{A}$[/hide]", "Solution_3": "Breez, I don't think your solution ensures that the answer is A. If $n$ was even, then the fraction could be further reduced, so E could also be a possible answer.", "Solution_4": "Let the points $ C,D,E,F$ form a square, and point $ A$ above the square and point $ B$ underneath the square. So the six points form a octahedron. \r\n\r\nAfter some observations, we see that the ants can be seprate into two groups with each group forming a triangle opposite the other triangle (ex. triangle $ ACD$ vs triangle $ BEF)$ \r\nThere are $ 4$ such pairs of triangles, and since on each triangle the ants can either rotates clockwise or anticlockwise there are a total of $ 4*2*2 \\equal{} 4^2$ possible ways.\r\n\r\nAnother way is to have four points forming a diamond and the rest two points forming a line segment (ex. $ ADBC$ vs $ EF$). There are $ 4$ such pairs and again the ants can either choose to rotate clockwise or anticlockwise in the diamond and can either choose to moves from left to right or right to left on the line segment, there are a total of $ 4*2*2 \\equal{} 4^2$ ways.\r\n\r\nThe last way is to have a single path starting from any point and ending at that point again while traveling through all the vertices only once. An example of such path is $ A \\minus{} D \\minus{} B \\minus{} C \\minus{} F \\minus{} E \\minus{} A$. There are $ 4*6$ such paths since they can start at any of the six vertices. But again since this is symmetric each path can either be traveling in right or left directions which gives a total of $ 4*6*2 \\equal{} 3*4^2$ ways.\r\n\r\nThis is out of $ 4^6$ so the probability is $ \\frac {5*4^2}{4^6} \\equal{} \\boxed{\\frac5{256}}.$", "Solution_5": "I think you can biject the total number of possible paths to the number of ways the ants can be on the six faces of a cube and each travel to a different face. There are two different ways to unfold a cube in a straight path, one is an \"s\", the other is a zigzag shape, so these are the ways the ants can move in one path without returning to a previous vertice. Can anyone finish this up?", "Solution_6": "First note that there are $ 4^6$ possible ways the ants can move\r\n\r\nNow we consider their final destination.\r\nConsider an octahedron $ ABCDEF$ with opposite vetecies; $ AB, CD, EF$\r\nAn ant on $ A$ or $ B$ can move to $ C,D,E,F$,\r\nSimilarly an ant of $ C$ or $ D$ can move to $ A,B,E,F$\r\nAnts on $ E$ or $ F$ can move to $ A,B,C,D$\r\n\r\nSo all we need to ask is;\r\nhow many ways can we pick two elements from each of $ \\{A,B,C,D\\},\\{C,D,E,F\\},\\{A,B,E,F\\}$ in order to make $ \\{A,B,C,D,E,F\\}$\r\n\r\nWell every elements can come from one of two sets so we have $ 2^6$ ways, but we cannot have $ 0,1, 3$ or $ 4$ elements comming from the same set so,(from symetry) we have over counted by $ 3\\cdot 4 \\plus{} 4\\cdot 3 \\equal{} 24$\r\n\r\nSo there are a total of $ 2^6 \\minus{} 24$ ways to pick $ A,B,C,D,E,F$, but each one could have come from either of two opposite vertecies so there are $ 2(2^6 \\minus{} 24)$ ways for the ants to move\r\n\r\n$ \\Rightarrow \\frac {2^7 \\minus{} 48}{4^6} \\equal{} \\frac {2^4(8 \\minus{} 3)}{2^{12}} \\equal{} \\frac {5}{256} \\longrightarrow \\boxed{A}$", "Solution_7": "Very nice solution ocha!", "Solution_8": "I don't understand this part of ocha's solution: \"but we cannot have 0,1, 3 or 4 elements comming from the same set so,(from symetry) we have over counted by $3\\cdot 4 + 4\\cdot 3 = 24$.\" Can someone help clarify? Thanks.", "Solution_9": "@cheesyicecream\nPlease do not revive old threads, start new ones with a link to the old one.", "Solution_10": "Best to think of this problem in terms of closed loops in the octahedron, ensuring that whenever an ant moves to a new vertex, that vertex's occupant has also left.\n\nCase 1: There is 1 closed loop. For a point you choose, the path starts out and ends at the same point without passing vertices twice. If you count them out, there are $4\\times(2+2\\times3)=32$ of them.\n\nCase 2: There are 2 closed loops. These can either be along a single edge and a quadrilateral, or on two triangles. There are 12 edges on which to create a closed path and they all go 1 way, while the resulting quadrilateral can be either \"clockwise\" or \"counterclockwise,\" Making the total $12\\times2=24$. For two triangles, there must be one on one of the top faces of the octahedron and one on one of the bottom faces, making 4 pairs of triangles that can all go clockwise or counterclockwise: So $4\\times2\\times2=16$ paths. For 2 closed loops, there are $24+16=40$ paths.\n\nCase 3: There are 3 closed loops. These can only occur in 3 edges of the octahedron. If you pick one point, there are 4 edges coming out of that point, and the other 2 edges from remaining unused vertices can be arranged 2 ways. For 3 closed loops, there are $4\\times2=8$ Paths.\n\nIn Total there are $32+40+8=80$ different combinations; $\\frac{80}{4096}=\\frac{5}{256}$", "Solution_11": "Would it be possible to solve this problem with some graph theory? Because if we can represent the octahedron as a graph, then looops and stuff would be much easier. I'm not very experienced in this area, anyone have an idea?", "Solution_12": "ocha could you explain the part on how to remove the overcounted part? I don't understand how you arrived at 3x4 + 4x3 = 24.\nthanks!\n\nalso, its ok to revive old threads as long as it isnt spam.", "Solution_13": "[quote=\"dgreenb801\"]I think you can biject the total number of possible paths to the number of ways the ants can be on the six faces of a cube and each travel to a different face. There are two different ways to unfold a cube in a straight path, one is an \"s\", the other is a zigzag shape, so these are the ways the ants can move in one path without returning to a previous vertice. Can anyone finish this up?[/quote]\nis this related to duality?", "Solution_14": "Sorry for the uber revive but for the solution of considering closed loops: is the case of 3 \"mutual swaps\" applicable? For example, A -> B, and B -> A; C -> D, and D -> C; E -> F, and F -> E?", "Solution_15": "I think I have a new solution:\nBecause the ants must end up at a unique position different from their start point, we can consider using derangement.\nBut one ant can only take one move, so we have to use PIE to account for the impossible configuration:\n!6-(3*!4-3*1+1) = 240\nWe did not account for orientation when using derangement, so we have to divide by the number of rotations (which is 3)\nSo we have (240/3)/4^6 = 5/256", "Solution_16": "One can also use Rook polynomials. If $A, B, C$ are the vertices with the opposite vertices $A',B', C'$ respectively, then consider the $6 \\times 6$ grid with $A, A', B, B', C, C'$ and we need to count the number of placements of 6 rooks in the grid such that the rook corresponding to any vertex is not placed in its column or its opposite vertex column.", "Solution_17": "Hello,\n\n\n[hide]Casework is very simple for this problem. Notice that the 4 vertices on the base, 2 of them must move into the top/bottom vertices.\n\nWe have:\nAdjacent ants move: that means there are 4 ways to choose the ants. There are also 2 ways to choose which one goes up and down. Finally, when we chose those, the 2 ants left at the base, there are 3 ways they can go. This is easy to count out. Finally, wherever those move, we have 2 vertices at the base where the ants that were previously on the top and bottom can go.\nDiagonal ants move: There are 2 ways to chose these ants, and 2 ways to chose which ones go up and down. Also, there are 4 ways to choose where the ones at the base go after that, and 2 ways left for the ants previously on the top and bottom.\n\nAdding these up you get 4*2*3*2+2*2*4*2=48+32=80, which means that the answer is A[/hide]" } { "Tag": [ "vector", "analytic geometry", "parameterization", "function", "geometry", "3D geometry", "combinatorial geometry" ], "Problem": "Suppose that $ K_i$ ($ i\\equal{}1,\\dots,n$) are compact connected sets in $ \\mathbb R^d$ such that each set is contained in the ball of radius $ \\delta$ centered at the origin and the sum $ K_1\\plus{}\\dots\\plus{}K_n\\equal{}\\{y_1\\plus{}\\dots\\plus{}y_n: y_i\\in K_i\\}$ is not contained in any half-space $ \\{y: (y,e)\\le 2\\}$ where $ e$ is any unit vector and $ (\\cdot,\\cdot)$ is the usual scalar product. Show that if $ \\delta>0$ is small enough, then this sum must contain the unit ball centered at the origin.", "Solution_1": "Well, if $ \\delta<\\frac2n$, there are no such sets satisfying the condition and it's trivially true. This probably isn't really what you were thinking of; I'll assume that $ \\delta$ is chosen before $ n$, to make small $ \\delta$ more meaningful.\r\n\r\nIt seems the core here is that the sum of many small connected pieces is almost a convex set. Easy to see intuitively, hard to quantify.", "Solution_2": "[quote=\"jmerry\"] I'll assume that $ \\delta$ is chosen before $ n$, to make small $ \\delta$ more meaningful.[/quote]\r\nYes, of course.", "Solution_3": "I also tried to post it on [url=http://mathoverflow.net/]Math Overflow[/url]. Here are two observations that may be useful:\r\n\r\n1) If $ a\\equal{}\\sum_ i x_ i$ and $ b\\equal{}\\sum_ i y_ i$ are in the sum, then the vectors $ v_ i\\equal{}y_ i\\minus{}x_ i\\minus{}(b\\minus{}a)/n$ are small and add up to $ 0$. There is a cute result that then you can rearrange them in such order that all partial sums are small (just constant times larger than the vectors themselves). If you switch from $ x_ i$ to $ y_ i$ along $ K_i$ in this order, you get a curve that travels from $ a$ to $ b$ within a small neighborhood of the segment $ [a,b]$.\r\n\r\n2) The sum is $ 2d\\delta$ dense in its convex hull. Indeed, if $ a\\equal{}\\sum_ i v_ i$ and $ v_ i$ are in the convex hull of $ K_ i$, then we can start moving $ v_ i$ until their representations as convex combinations of points in $ K_ i$ get shorter and we can do it as long as there are at least $ d\\plus{}1$ vectors $ v_ i$ that do not belong to $ K_ i$ themselves (any $ d\\plus{}1$ vectors in $ \\mathbb R^d$ are linearly dependent, so once you are given directions in which you can move $ v_i$, you can always choose speeds that keep the sum in place). Thus, in the representation of every point in the convex hull as a sum, we need only $ d$ vectors from convex hulls and the rest may be taken in $ K_ i$. Replacing these $ d$ vectors by arbitrary points in the corresponding $ K_i$, we get a close point in the sum itself.\r\n\r\nI feel that these two observations put together should be enough and I just do not see how to add 2 and 2 here. :wallbash:", "Solution_4": "All right, I found the missing piece (though not entirely by myself). We shall prove that if the diameters of $ K_j$ are less than $ \\delta$ and the convex hull of their sum contains the unit ball, then the sum contains the origin. \r\n\r\n[b]Step 1[/b] Without loss of generality we may assume that $ K_j$ are contained in the $ \\delta$-neighborhood of the origin. Indeed, since $ 0$ is in the convex hull of the sum, we can represent it as $ \\sum_j v_j$ with $ v_j\\in\\text{conv\\,}K_j$. Now just replace $ K_j$ with $ K_j\\minus{}v_j$. \r\n\r\n[b]Step 2[/b] Since the coordinate vectors $ \\pm e_k$ are in the convex hull, we can find close vectors $ f_k^{\\pm}$ in the sum. Now write their representations $ f_k^\\pm\\equal{}\\sum_j v_k^{\\pm}(j)$, put $ w_k^{\\pm}(j)\\equal{}v_k^{\\pm}(j)\\minus{}\\frac 1n f_k^{\\pm}$ and change the order of summation if necessary so that all partial sums of $ w_k^{\\pm}(j)$ are small. Now split the sum of $ K_j$ into $ d$ equal pieces (from $ 1$ to $ \\frac nd$, from $ \\frac nd$ to $ 2\\frac nd$ and so on). The corresponding sums of $ v_k^{\\pm}(j)$ will be about $ \\pm \\frac 1d e_k$.\r\n\r\n[b]Step 3[/b] In the $ k$-th sum of $ K_j$, choose a continuous path that goes from approximately $ \\minus{}\\frac 1d e_k$ to $ \\frac 1d e_k$ staying close to the straight segment (to be honest, this requires path connectedness rather than connectedness but we can always switch to small open neighborhoods and then use the routine limit procedure to take care of this minor issue). Call this path $ F_k(t)$, $ t\\in[\\minus{}1,1]$ (I do not care about the exact parameterization as long as $ F_k(\\pm 1)$ is close to $ \\pm\\frac 1d e_k$). \r\n\r\n[b]Step 4[/b] Let $ G_k(t)\\equal{}\\frac td e_k$ be the straight path. Now let $ \\psi: [\\minus{}1,1]^d\\to [0,1]$ be any continuous function such that $ \\psi\\equal{}0$ on the boundary of the cube $ [\\minus{}1,1]^d$ and $ 1$ on the slightly smaller cube $ [\\minus{}1\\plus{}\\varepsilon,1\\minus{}\\varepsilon]^d$. For $ \\bar t\\equal{}(t_1,\\dots,t_d)\\in[\\minus{}1,1]^d$, define \r\n\\[ F(\\bar t)\\equal{}\\sum_{k\\equal{}1}^d [(1\\minus{}\\psi(\\bar t))G_k(t_k)\\plus{}\\psi(\\bar t)F_k(t_k)]\\,.\\] \r\nThen $ F$ maps the cube $ [\\minus{}1,1]^d$ to $ \\mathbb R^d$ continuously so that the boundary is mapped to the boundary of the cube $ [\\minus{}1/d,1/d]^d$ in the canonical way. But then the image of $ F$ must contain the origin.\r\n\r\n[b]Step 5[/b] It remains to note that the origin cannot be obtained when one of the coordinates $ t_k$ of $ \\bar t$ is close to $ \\pm 1$ because then $ \\langle F(\\bar t),e_k\\rangle$ is close to $ \\pm\\frac 1d$, so the origin is attained deep inside the cube $ [\\minus{}1,1]^d$ where $ \\psi(\\bar t)\\equal{}1$. Thus, $ \\sum_k F_k(t_k)\\equal{}0$ but this sum is contained in the sum of $ K_j$.\r\n\r\nSo, jmerry was right: the sum of small connected compact pieces is almost convex, indeed. Fortunately, it turned out not so \"hard to quantify\" after all! :jump:." } { "Tag": [ "analytic geometry", "graphing lines", "slope", "exterior angle" ], "Problem": "How can you solve this... \r\nTwo circles intersect at points P and Q. A line l that intersects PQ intersect the two circles at the points A, B,C, and D( in that order along the line). It is known that $\\prod_{k=1}^na_k$ and when is a ratio. Thak you very much .", "Solution_9": "Add \\limits to get \\prod\\limits_{k=1}^na_k \n\n$ E_n=\\frac{2^{n-1}\\cdot\\left(1+\\prod\\limits_{k=1}^na_k\\right)}{\\prod\\limits_{k=1}^n\\left(1+a_k\\right)}\\ ,\\ n\\in\\mathbb{N}^* $", "Solution_10": "WOW ... it is very simply and v.v. nicely ==> $E_n=\\frac {2^{n-1}\\cdot\\left(1+\\prod\\limits_{k=1}^na_k\\right)}{\\prod\\limits_{k=1}^n\\left(1+a_k\\right)}\\ ,\\ n\\in\\mathbb N^*$ . Thank you v.v. much." } { "Tag": [ "ARML", "analytic geometry", "graphing lines", "slope" ], "Problem": "For $x$ and $y$ in radians, compute the number of solutions in ordered pairs $(x,y)$ to the following system:\r\n\r\n$\\sin(x+y)=\\cos(x+y)$\r\n\r\n$x^{2}+y^{2}=\\left(\\frac{1995\\pi}{4}\\right)^{2}$", "Solution_1": "[hide=\"Method\"]\nWell, from the first equation we have that $x+y = \\frac{\\pi}{4}+k\\pi$ for some integer $k$.\n\nNow, we can take the (geometric) tangent, with slope $-1$, of the circle determined by the second equation, and find where it hits the y-axis. Call that value $n$. We will then have that $n \\geq \\frac{\\pi}{4}+k\\pi \\geq-n$. Now, just calculate the number of solutions in that interval.\n\n[b]EDIT[/b]: It's easy to find that $n = \\pi \\cdot \\frac{1995 \\sqrt{2}}{4}\\approx \\left( 705.3 \\right) \\pi$, which means that\n\\[705.3-.25\\geq k \\geq-705.3-.25 \\implies 2 \\left( 2 \\cdot 706-1 \\right)= \\boxed{ 2822 \\text{ solutions}}\\]\n[/hide]" } { "Tag": [], "Problem": "In a calm city, the physics make a party every night.\r\n\r\nexperimentally, they establish that the variation of the number of persons in a day is proportional to the numbers of people that had gone in another pary in some day that they made and the present day. or matematically.\r\n\r\n${\\frac{\\partial N(t)}{\\partial t}=\\alpha N(t-\\epsilon ) N(t)}$\r\n\r\n\r\nHint: Need lagrange inversion", "Solution_1": "Not to say I have any idea how to solve this but what's the question asking? It seem to just stopped...." } { "Tag": [ "USAMTS", "algorithm", "calculus", "integration", "ARML", "probability", "blogs" ], "Problem": "It should be possible theoretically to prove, given some source code, that pending the outcome of a certain result upon running the program, a certain result may have been proven. It seems that at some level the source code is \"accessible\" to human thought processes and hence may justifiably constitute a part of a valid proof. The compilation and implementation of the source code concerns me, though, as this part seems less accessible. You don't have any way to check that all the bits are flowing the way they're supposed to, so even if the program does output a certain result upon completion, how do you know your computer wasn't just screwing up? Granted, there's a small chance of that, since computers don't usually randomly screw up, but it's been known to happen, and I don't think people usually accept proofs which run along the lines of \"well I can't check it completely but y'know it's got a pretty good chance of being true\". On the other hand, it would be disappointing to rid oneself of a powerful potential proof technique for handling a plethora of special cases. So?", "Solution_1": "If you do math for the sake of discovering new truths, then the only question is---do you find the result of the computer program convincing? I would be eager to use whatever tools are available to discover new knowledge. Maybe there is a point in math beyond which we can't go without computer proofs.\r\n\r\nThat said, computer programs usually have bugs in them, and we will always have to worry about that. (Then again, a written proof can have a \"bug\" as easily as a computer program can.)", "Solution_2": "The more interesting question to me is whether the nature of certain problems require computer proof. Casework never seems particularly elegant to me - in Erdos' words, I don't think any of the proofs in The Book are by computer. \r\n\r\nConsider also that proofs like Wiles' proof of Fermat's Last Theorem are not important because they are of an important result but because the ideas in the proof were new and exciting and shed light on things. Does a proof by computer have the same effect?", "Solution_3": "@tor well, arguably so - it may demonstrate a new class of programs which could be implemented in various proofs which could then be used to show various other results and, in the limiting case, open entire theories. I'm still not sure whether we should accept [i]any[/i] of these proofs by computer, though.\r\n\r\n@sing I tried to argue in the opening post that I'm not concerned about bugs since this is something visible on the source code level, which is accessible to human checking at much the same level as a standard mathematical proof, as you touch on. I'm more concerned with how one accepts that the conversion from source code to result is reliable.", "Solution_4": "Here's a thought I had. If you're worried about errors in the compilation process, you could write the program in lots of different languages and use lots of different compilers on lots of different kinds of computers, and make sure you get the same result. You may still worry that there is some subtle error that all the computers or compilers are consistently making, all in the same way. But then again, when it comes to checking a traditional written proof, you may worry that there is some subtle mistake in all of our brains that causes us to all get confused in the same way when faced with a certain mathematical argument. Which is more likely? I don't know, both seem very unlikely.", "Solution_5": "I think it is implicit in the nature of proof that one accepts the inherent limitations of the human brain, but there is no [i]a priori[/i] reason to extend this to computer compilers.\r\n\r\n\r\n\r\nby the way, note that this question isn't purely philosophical in nature as it crops up in my work as a grader, both for classes (if students use various programs such as mathematica) and for USAMTS", "Solution_6": "Yay! What a good topic. I've had this argument with my professors when we discuss whether the Four Color Theorem has been proven. \r\n\r\n\"Here's a thought I had. If you're worried about errors in the compilation process, you could write the program in lots of different languages and use lots of different compilers on lots of different kinds of computers, and make sure you get the same result. You may still worry that there is some subtle error that all the computers or compilers are consistently making, all in the same way.\" -Singularitarian\r\n\r\nThis is the argument I make that computer proof will always introduce the idea of empirically validating a claim. You can check it in as many ways and on as many systems as you want, but all you are still only gaining empirical evidence to the truth of the claim. Most mathematicians will say empiricism is not a valid way to do math.\r\n\r\nHere is the kicker, though. Most mathematicians accept the 4CT as proven. What they reject is that a computer proof done in as rigorous a way as that introduces empiricism. So we always both agree that it is proven, we just disagree on whether or not empirical methods were used.\r\n\r\nI agree with Singularitarian. All math has the possibility of error (well, at least at the level of Wile's proof). I actually find it more convincing if a computer tells me something than a human.", "Solution_7": "Er. If I understand you correctly, as your amenable to accepting large amounts empirical evidence as valid proof? \r\n\r\nAnd I acknowledge that mathematical proofs may have entirely human flaws, but these things are all open to human thought -- we can pore over what's been written down on the page and the definitions that have been made and so forth. We can't pore over the inner workings of the computer.", "Solution_8": "I think it depend on cases... it might be hard to explain, consider following example:\r\n given a 4x4 table without 2 corners, prove that the table cannot be covered by 6 1x2 dominoes.\r\n this problem is very similar as the classical problem that can be solved by coloring(the one with a chesstable without 2 corners), and because it is so small people can eventually prove it by evaluating each case.... and i believe most people will consider it valid. some case happenned with the problem 4 of the iberoamerican math olimpiad, with is one about a cominatorics problem about a 19x19 table I remember there is people who evaluate all the draganian path to every single square(by filling a 19x19 table by numbers) and they get 7. with means, if the cases is finite, the checking every case is a proof, even not elegant.\r\n if about problem is about a much bigger table, such that it is to big for human to check every case, then if a computer just check it for human should be a fine.\r\n\r\nbut however many conjectures or problems are consisted to have infinite specific cases, so in that case we cannot simply accept computer proof.", "Solution_9": "I'll admit, I've actually wondered before if computer proofs are acceptable because of questions like are \"all the bits are flowing the way they're supposed to\"? But my biggest question is, how do you know there is no bug in your program? You could check small cases, I suppose, but what if something changes as it gets bigger? What about, say, floating point inaccuracies? But, I suppose one would say a program can be checked as rigorously as a proof (as mentioned above).\r\n\r\nBut here's something weird. If we're worried about that, why can't we be worried about weird things in non-computer proofs? For instance, do we consider the fact that other mathematicians have confirmed a proof to be \"empirical evidence\" of its correctness? Would we need our own mind to confirm something to be correct? How does one know he himself isn't insane, and that every logical conclusion he come to isn't completely wrong?\r\n\r\nBut after all those worries, I would say we are safe to assume that our proofs are correct, even ones by computers. Again, the chances of the computer messing up are ridiculously small, and if we have it checked in many languages and many different computers, I'm pretty sure we are safe. Of course, whether or not the proof by computer is \"nice\" is a different question.", "Solution_10": "@chichi I'm not making an argument against casework here. Casework derived and proven by human thought is still perfectly ok because everything is right there, written down, and is accessible to other human thought.\r\n\r\n@tj Good point in your first paragraph -- there are different kinds of bugs than simply the \"logic bugs\" (i.e. the program not doing what it's supposed to) which I primarily focused on in my previous argument. I'm still not worried about non-computer proofs -- I acknowledge we have no way to say they are \"entirely correct in the utmost sense as if god had come down and handed it to us or whatever\", but I think it is implicitly accepted that \"correct up to the point of not conflicting with basic human logic\" is a reasonable standard of acceptance for proof. I don't understand how or why this extends to computer proofs.", "Solution_11": "Your example of grading for USAMTS is suggestive - it is true that programs like Mathematica do occasionally give wrong answers, but the only examples I can think of are abstruse series problems. As far as I know, the rest of Mathematica is reliable, but I can see where you're coming from.", "Solution_12": "[quote=\"t0rajir0u\"]Your example of grading for USAMTS is suggestive - it is true that programs like Mathematica do occasionally give wrong answers, but the only examples I can think of are abstruse series problems. As far as I know, the rest of Mathematica is reliable, but I can see where you're coming from.[/quote]\r\n\r\nYAH SO GUYS IF ANYBODY WANTS ME TO GIVE THEM POINTS FOR COMPUTER PROOFS ON USAMTS THIS IS TOTALLY THE PLACE TO CONVINCE ME TO DO SO", "Solution_13": "[quote=\"MysticTerminator\"]I don't understand how or why this extends to computer proofs.[/quote]\r\n\r\nI don't understand what you don't understand. Many proofs (especially in combinatorial fields) are algorithmic: here's a process that finds (or counts, or whatever) an object of interest, and here's a proof that it really does this. If this algorithm happens to be in C, what exactly is the problem?\r\n\r\nI'm not sure I understand the objection that computer proofs are \"uncheckable,\" either. After all, I'm well on my way to being a professional mathematician but 95% or more of what the people around me are proving is totally inaccessible to me. Even many papers in my own field are \"uncheckable\" as far as I am concerned, but I believe their results because, in principle, someone with enough knowledge and time could check them (and maybe already has). A computer proof is checkable in exactly the same sense: any results the computer claims could be checked one by one by sufficiently many people with sufficiently much time. (This is true of case-checking algorithms like 4CT and also of proof-writing algorithms (which should always be able to present a proof, not just the claim that the result is provable).)\r\n\r\n\r\nSome interesting personalities worth reading about on this subject are Doron Zeilberger (a mathematician at Rutgers who sometimes lists his computer as a coauthor on papers) and Gregory Chaitin* (who believes that incompleteness means that mathematicians shouldn't be so timid about adding new axioms given sufficient empirical support).\r\n\r\n* The book of his I read was called Meta-Math, and I really hated the first half of it, but it did have some interesting ideas.", "Solution_14": "I'm drawing a distinction between the part of the proof that is actually accessible to human intelligence and the part that is not. A computer proof, in my opinions, consists of three things: some source code that describe a program, a computer that runs the program and outputs some result, and an analysis of this result. The first and the third are essentially what you describe in your post above and are what I've been describing as accessible to human thought. I have no problem with this. The second part is not accessible to human thought. You have no way of confirming from pure logic that the bits in the computer are flowing the way they're supposed to be. Note that none of the examples you cite (algorithmic proofs in combinatorics, proofs in general which you have insufficient knowledge to understand, and so forth) suffer this flaw.", "Solution_15": "I don't see why it matters: if the program outputs a proof of some result, it doesn't matter what the bits were doing if the proof is right. Similarly, if the program tells me \"there are no examples of X among the 100000 input graphs,\" this result is entirely susceptible to human thought. Just because no one [i]wants[/i] to spend their energy checking it doesn't mean it's inaccessible to checking [i]in principle[/i].", "Solution_16": "Yes, those types of proofs seem valid. It's a bit annoying since you know nobody'll ever check them, but then again I know there are a lot of perfectly \"handwritten\" proofs that nobody wants to bother checking too. :wink: \r\n\r\nSo I'm referring to something more like what you might see as USAMTS grader. Problem says \"HEY FIND ALL THE INTEGERS THAT DO SUCH AND SUCH\" and contestant is all like \"hey here's my source code that should tests each integer and here is what the computer outputted\". Do I accept this or not?\r\n\r\nSimilarly as a grader for a course at college -- someone \"proves\" or \"solves\" something through the use of mathematica to evaluate an integral or some such -- do I accept this? (if anyone's curious, I did, but in much the same way as one \"accepts\" calculus on olympiads)\r\n\r\nHAH I USED A LOT OF QUOTATION MARKS IN THIS POST", "Solution_17": "[quote=\"JBL\"]I don't see why it matters... Just because no one [i]wants[/i] to spend their energy checking it doesn't mean it's inaccessible to checking [i]in principle[/i].[/quote]\r\n\r\nIsn't this quite a fine line to be treading? You consider a proof correct just because it is possible to check in principle. I think there is a distinction between a proof having been checked by people capable (Fermat's Last Theorem) and is not possible for me to check. And to no one having checked FLT, but saying that it is correct because it is possible to check in principle.\r\n\r\nI'm of the opinion that computer proofs can be accepted as proof. My goal is to get others to accept them, but under a proper title. Empiricism isn't bad. But let's not pretend that its not there when we accept one of these proofs.", "Solution_18": "what sort of a \"proper title\" would you formulate? since it seems you are drawing a distinction between empirically supported evidence and the standard notion of proof.", "Solution_19": "[quote=\"HilbertThm90\"]You consider a proof correct just because it is possible to check in principle.[/quote] This sentence misses a major piece of the idea: if a proof is produced [i]by some generally reliable means[/i] and is in principle checkable, I am perfectly happy to conclude that it's true. This is, after all, exactly what I do with all the theorems whose proofs I've never read. The major objection raised in this thread seems to be that computer proofs might be suspect because we can't, in principle, check that they are correct. I think that this is, in general, wrong (although it is true of certain \"proofs\" I've seen, e.g. on this year's ARML power round: \"We enumerated all cases by brute force, and the answer is [such and such]\"). \r\n\r\n(I couldn't parse the rest of the associated paragraph, unfortunately, so I'm not sure exactly of the distinction you drew there involving FLT.)\r\n\r\nOn problem sets I don't see what the problem is: a grader has discretion to grade not strictly on grounds of correctness -- \"being lame and using mathematica to do your homework\" seems like a perfectly legitimate reason to dock a point or two, just as \"having the right idea but making stupid computational errors\" or \"handing in work that reaches the right answer but is otherwise illegible\" are. I don't remember what grading standards USAMTS claims -- one possible resolution is to demand a demonstration that the algorithm described, if carried out correctly, would yield the correct answer.", "Solution_20": "well, I grade based on whether or not I think things are correct.", "Solution_21": "I admit that paragraph was weirdly worded.\r\n\r\nClear your mind of all history of knowing how FLT played out. Say Andrew Wiles said that he proved FLT. I'd say that him being a professional mathematician is a pretty \"reliable\" means. Since he wrote it down, it is also in principle checkable. But I don't think a single person would say, \"Ah, you proved it and it is correct!\" without someone actually doing the checking.\r\n\r\nI don't see how this is different from the computer...unless you are saying that the computer provides a reliable means and a human never does. In that case, I would buy your stance as sound, but I must say that it is still an empirical discovery (due to the slight chance of error and not checking for it), and anyone that says that empiricism is not a valid proof method cannot hold the stance you are holding (and thus reject 4CT as being proven).", "Solution_22": "If someone did present to me a proof of a theorem by computer, then I would want to see the source code and a proof that the code and/or its output actually proves the theorem.\r\n\r\nAs mentioned before, there is always a possibility of floating point error, unless you use an environment that can maintain exact values of numbers.", "Solution_23": "I think the problem that Mystic is trying to say (maybe?) is that even if you verify that the program you wrote would definitely produce the correct output based on the semantics of the language, there are more issues.\r\n\r\nHow can you guarantee that the binary you compile executes your code?\r\nHow can you guarantee that your computer's memory is not corrupted?\r\n\r\nI think the concerns are at compile time (maybe link time too) and run time. Furthermore, it's almost impossible to verify that all of the bits in your computer are what you expected them to be. It becomes a physical problem rather than a mathematical one.\r\n\r\nThat said, I do trust computer programs being a computer scientist myself :). The probability of such failures is small enough to convince me that things are true/false.", "Solution_24": "well, sure, I trust them too. as we all know from our olympiad days, though, proofs are tricksy because you have to maintain this kind-of annoying level of rigor. I'm still unconvinced that the problems I describe (and which you picked up on, yay) aren't enough to bring this down.", "Solution_25": "[quote=\"paladin8\"]How can you guarantee that the binary you compile executes your code?\nHow can you guarantee that your computer's memory is not corrupted?[/quote]\r\n\r\nSee my post #16 on this thread: a computer proof isn't simply a black-box into which outputs theorems: it has to, along the way, do something equivalent to actually proving whatever it is you want it to prove. And if it does this, it can output data that allows you to check a proof. Proof-checking is in principle trivial, so proof-writing is, in some sense, NP, so whether the physical computer did exactly what it was supposed to is irrelevant. (Of course, the reliability of a computer is far greater than the reliability of human mathematicians, so all this worrying seems rather misplaced to me.)", "Solution_26": "What if you need the computer to brute force check a million - or a billion - cases? You can have it output all the data that it uses, but how can you check it all? If you actually have the time or ability to check what the computer outputs, then this entire question becomes useless.", "Solution_27": "so joel, I'm acknowledging that if the computer outputs a proof, that's fine. I don't care where the proof came from -- from human creation, from a computer, or from strange arrangements of rocks. Once you have a proof that is written down (or printed or whatever) somewhere, it's there and that's all that matters. I'm objecting to the use of computers IN proofs. Not in CREATING the proofs but [i]in the actual proofs themselves[/i].\r\n\r\n@hance joel also raised this point and I agree with him here -- if it's in principle checkable, I'm going to be ok with it. honestly, there are really a surprising number of mathematical results (generated \"by hand\", as it were) that nobody would ever want to check. ok, some people probably do check them, but most people are just like ehhhh I'm going to do something else now instead.", "Solution_28": "I've thought about the validity of computer-assisted proofs before, and my own personal conclusion is that we tend to ask the wrong question: Is a computer-assisted proof a valid proof? I consider myself to be a mathematician, not a proof-ician. In other words, we write proofs in order to further the study of mathematics, not vice versa. For me, the right question is whether computer-assisted mathematics is good mathematics? This question can only be answered subjectively on a case-by-case basis.\r\n\r\nThe study of mathematics is a more complicated and beautiful enterprise than the mere cataloguing of formal statements as proved, disproved, and not-yet-proved-or-disproved. Instead of focusing on whether something is or is not a proof, we should think about the *reasons why* we write down proofs to begin with, and then evaluate computer-assisted mathematical work based on how well it achieves those ends. I wrote a little blog post one day on this subject if you're interested:\r\n\r\nhttp://yenlee.blogspot.com/2007/08/why-do-we-prove-things-or-what-is-good.html\r\n\r\nAs for the practical issue of grading, I think these issues are irrelevant. You're trying to teach people something, and you can set the agenda for what you want them to learn. You can decide to allow computers, not allow computers, or require the use of computers. If there is no official policy, then all parties involved should use a degree of common sense. For example, the \"common sense\" standard is used all the time in determining what constitutes a rigorous proof.", "Solution_29": "hah. just leave terry tao alone, huh? :lol: \r\n\r\ninteresting blog post, but I'm afraid I depart with you right from the start -- in my view, proofs are all about criterion A that you set down. B is good, necessary, and probably far more important than A, and usually the idea does come through in the proof, but the proof need not have anything to do with B if it doesn't want to. I'm also confused as to how you draw this distinction between \"being convinced by something\" and \"rigour\". If the proof's rigorous and I can understand it, I'm convinced. If the proof's not rigorous and I can understand it, I'm convinced if and only if I can see how to make it rigorous by certain standard arguments myself (and this is, in my understanding, usually the case in texts aimed at graduate-level education and beyond). If the proof seems pretty believable but I have no idea how to actually make it rigorous, I'll just be kinda like ehhhh.", "Solution_30": "[quote=\"MysticTerminator\"]It should be possible theoretically to prove, given some source code, that pending the outcome of a certain result upon running the program, a certain result may have been proven. It seems that at some level the source code is \"accessible\" to human thought processes and hence may justifiably constitute a part of a valid proof. The compilation and implementation of the source code concerns me, though, as this part seems less accessible. You don't have any way to check that all the bits are flowing the way they're supposed to, so even if the program does output a certain result upon completion, how do you know your computer wasn't just screwing up? Granted, there's a small chance of that, since computers don't usually randomly screw up, but it's been known to happen, and I don't think people usually accept proofs which run along the lines of \"well I can't check it completely but y'know it's got a pretty good chance of being true\". On the other hand, it would be disappointing to rid oneself of a powerful potential proof technique for handling a plethora of special cases. So?[/quote]\r\n\r\nYou could run the exact same program on many computers (probably 50-100 to be the most accurate) and use the most common result. Since computers rarely mess up and when they do they rarely do it in the same way, there should only be one of each invalid result.", "Solution_31": "in other words, you're happy with accepting probabilistic proof. I'm not.", "Solution_32": "Here is a tangentially related discussion that occurred a few years ago: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9991", "Solution_33": "thanks, gauss. I don't feel like necroing right now, so I'll respond here.\r\n\r\nI'm very much a left-winger, since I accept that there may well be some weird thing in everybody's brain that makes everyone ignore a certain logical flaw or whatever the case may be. and I'm ok with that. as long as it convinces me, I'm ok with putting up with the inherent limitation my neurological system imposes upon me. I simply happen to have high standards of what does convince me.\r\n\r\nthe prevalence of sociology in math is also interesting -- the whole hilbert-brouwer acrimony for instance and so forth, but I hardly know enough history of math to offer anything to that discussion (and this thread isn't really the place, I suppose).", "Solution_34": "Mystic: Perhaps you are confused, because I did NOT attempt to draw a distinction between \"being convinced by something\" and \"rigour.\" I essentially said that they are the same thing. I drew a distinction between *convincing* someone *that* something is true versus *explaining* to someone *why* something is (or is likely to be) true. You are correct that the \"convincing\" part is what generally qualifies a putative proof as a legitimate proof, but what I'm concerned about is not what makes a putative proof a legimate proof but rather what makes a putative proof a good piece of mathematics.\r\n\r\nWe indoctrinate the importance of the rigorous mathematical proof in undergraduate mathematical education because it's useful as a pedagogical tool: it teaches students how to think clearly and communicate clearly. But in the end, the proof is not the goal. Understanding mathematics is the goal. This is why partial results can be good math. This is why examples and counterexamples can be good math. This is why computer experiments can be good math. And most importantly, this is why there is value in giving two different proofs of the same result. Sometimes the less rigorous proof may be the more valuable one.\r\n\r\nRegarding the Devlin article: Mildly interesting, but nothing newsworthy or controversial. Nothing at all happened in 2003 that made mathematicians think differently about proofs. Obviously, a putative proof is a legitimate proof when the mathematical community says it is. What other reasonable standard could there possibly be? It's a lot like a physics experiment. The results of a physics experiment are supposed to be undeniable *facts* (just like theorems are undeniable truths), but the validity of the experiment can only be determined by the physics community as a whole. Perhaps this viewpoint is slightly \"postmodern,\" but it's essentially common sense, isn't it?", "Solution_35": "[quote=\"MysticTerminator\"]Not in CREATING the proofs but [i]in the actual proofs themselves[/i].[/quote] Okay, so obviously some of my disagreement with you is based on the fact that I'm not sure what you mean by this :). Since you seem to agree that a finite case-bash by a computer as part of a proof is acceptable, can you give an example or two of something that you find questionable? (Really, I guess I mean, \"can you give an example not in the context of USAMTS solutions?\")", "Solution_36": "I see nothing wrong with using a computer for proving theorems. As long as I can\r\na) Verify the validity of the algorithm;\r\nb) Check the code;\r\nc) Check the general ability of the computer to function,\r\nI believe the output. The human verification ability is by no means superior to the computer ability to function without an error. I had at least 5 occasions where the proofs verified by humans (i.e., published in refereed journals, cited, etc.) were wrong (in 2 cases the statements were wrong, in other cases there was a gap that was quite hard (but, fortunately, possible) to fill) and I have never seen a code that repeatedly malfunctioned because of the computer errors, not because the program was flawed (the famous story with Pentium multiplication error may be the only exclusion). \r\n\r\nIn general, in most of the situations, the task delegated to the computers is just some routine and straightforward verification of a large amount cases. There is nothing there to understand really. Once you know exactly what to do in each particular case and why you should do it, no extra understanding is achieved by actually doing them (or verifying that they are done correctly).\r\n\r\nA completely different question is whether the computer-assisted proofs are \"elegant\" or satisfy other beauty standards. I hold the point of view that such proofs may be beautiful if they contain a nice idea of how exactly you should reduce the problem to checking just finitely many things (it is not always obvious) and if the algorithm/computer code satisfy my \"programming beauty standards\", but it is just a personal view and I do not see any point in arguing about such things. \r\n\r\nAlso, my embracing computer assisted proofs doesn't mean that I won't appreciate an alternative \"conventional\" proof when it appears (provided that the latter is based on a different idea, not just reduces the casework somehow). On the other hand, a long and tangled \"conventional\" proof may, in my opinion, be inferior to an ideologically short and clear proof using the computer to check a few hundred similar numerical inequalities (or whatever is there to check).\r\n\r\nThe last relevant thing is the question of whether seeing that something happens in all the 1000000 cases one has to consider and not seeing the common underlying reason for that means the lack of true understanding of what is really going on. I believe that sometimes it does and sometimes it doesn't. Not all things, even in mathematics, have an ultimate simple reason behind them. Sometimes they are true just because they are (like $ x^2\\plus{}x\\plus{}41$ produces a long list of primes for no particular reason as far as I can tell; this may be not the best example, but, I hope you see the point), so if we can find that particular reason, great, if not, we can live without it and go further." } { "Tag": [ "search", "inequalities", "inequalities unsolved" ], "Problem": "$x$,$y$,$z$>0 \r\n$\\frac{x}{\\sqrt{y+z}}$+$\\frac{y}{\\sqrt{x+z}}$+$\\frac{z}{\\sqrt{y+x}}$<=$\\frac{5}{4}$$ \\sqrt{ x+y+z}$\r\nand tank you", "Solution_1": "[quote=\"bellahsayn\"]$x$,$y$,$z$>0 \n$\\frac{x}{\\sqrt{y+z}}$+$\\frac{y}{\\sqrt{x+z}}$+$\\frac{z}{\\sqrt{y+x}}$<=$\\frac{5}{4}$$ \\sqrt{ x+y+z}$\nand tank you[/quote]\r\n\r\nTake x->0 and y=z then it's false?", "Solution_2": "I remember this one! Boy, what I trouble I had with this one. :D . For the correct statement search in the solved or unsolved problems inequalities the post of Fedor Bekharev \"a lovely inequality\".", "Solution_3": "harazi, I think his inequality is false, because it's different from that in \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=1459\r\n(note the denominator)", "Solution_4": "I know, but the presence of $5/4$ makes me think that he made a typo and in fact he wants to prove that monster inequality. :?", "Solution_5": "for this problem I think there is no k,\r\nbecause it is a little from which siuhochung mentioned." } { "Tag": [ "algorithm" ], "Problem": "OK Guys. Listen up. I am a 4th year high school student (graduating in 2006). Till Class 10( I'm in class 12 now) , I was an avid computer enthusiast - learning HTML, XML, Javascript, a bit of C and Networking... but all this ended in Class 11. If you know, in India, after class 10, almost everyone interested in engineering attends tuitions and prepare for IIT-JEE-- and I was no exception. So, for the past year, I was totally cut off from computing. But now, it seems my interest is returning. As I was seeing the different competitons like Informatics Olympiad and APs etc. whcih might be useful in US admissions( just keeping all otions open!) I got the first volume of Knuth's \"Art of Computer Programming\" recently and am searching for knowledege! So I need your help on a no. of things(the previous para just established the context!)\r\n\r\n1) Is it realistic for me to consider taking part in these programming competitions an AP exams etc. now? I'm willing to spend at most 2 hours $\\pm$ 1 hour per day!\r\n\r\n2) Which are the different languages that I can learn now. Which are the best? Please recommend online pages where I can get the best compilers for the different languages asa well as documentation for learning these langugages(besides mathlinks.ro of course!!)\r\n\r\n3) I believe the competitons focus more on finding algorithms etc. and analysing them rather than just purely writing long , tedious programs. Am I right? If yes, please recommend the best books/online pages on ALgorithms etc. and general computing that will be helpful. Does Knuth's Volume I cover the requisite material for IOI?? Which other books are there? I f somebody who lives in india can suggest locally available books, that ill be the best!\r\n\r\n4) What else do I need to do if the answer to (1) is yes??", "Solution_1": "[size=150][b]Does nobody have any views on my pitiful situation??????[/b][/size]", "Solution_2": "i dont know too much since im not much of a programmer myself, but here are my answers:\r\n1) The AP tests have a reputation of being relatively straightforward, so that is definitely doable in a year. Try the USACO contests at ace.delos.com/usacogate. I've learned a lot from doing the excercises and problems on that site.\r\n2)This is a matter of taste. For AP you need to know java. For USACO you need java, c, or pascal. I use c++. the best free compiler as far as i can tell is the bloodshed devc++ (or something like that...google it).\r\n3) USACO is a great introduction as far as I can tell." } { "Tag": [ "invariant", "quadratics", "algebra", "polynomial", "linear algebra", "matrix", "vector" ], "Problem": "Suppose $ A\\in M_{n}(\\mathbb{R}).$ Suppose that the quadratic polynomial $ x^{2}\\plus{}\\alpha x\\plus{}\\beta$ for real $ \\alpha,\\beta$ is a factor of the characteristic polynomial of $ A.$ (For the sake of nontriviality, you might as well assume this quadratic is irreducible over the reals.)\r\n\r\nProve that the subspace $ E\\equal{}\\{v\\in\\mathbb{R}^{n}: (A^{2}\\plus{}\\alpha A\\plus{}\\beta I)v\\equal{}0\\}$ is at least two-dimensional.\r\n\r\nNow suppose $ Q$ is an orthogonal real matrix whose first two columns lie in $ E.$ (Given two independent vectors in $ E,$ we can always construct such a $ Q$ by the Gram-Schmidt process.) What can you say about the structure of $ Q^{T}AQ?$", "Solution_1": "Let's replace $ \\mathbb R$ with an arbitrary field $ \\mathbb F$. Suppose that $ p_{A}$ factors as $ gh$, where $ g$ is an irreducible polynomial of degree $ m$.\r\nThe conclusion should be that $ \\text{dim}\\ker\\left( g(A)\\right)\\geq m$.\r\n\r\nNow assume that we can find $ v\\in\\ker\\left( g(A)\\right),\\, v\\neq 0$.\r\nThe idea is to prove that $ v, Av, A^{2}v,\\ldots, A^{m\\minus{}1}v$ are linearly independent.\r\nIf not, then there is $ q\\in\\mathbb F[X]$ such that $ q(A) v \\equal{} 0$ and $ \\deg q < m$.\r\nBut $ g(A) v \\equal{} 0$ and $ g,q$ are coprime, so $ I_{n}v \\equal{} 0$, contradiction.\r\nHowever, $ v, Av,\\ldots, A^{m\\minus{}1}v$ are all in $ \\ker\\left( g(A)\\right)$, so this is enough to prove the claim.\r\n\r\nMy question is: How do we know that $ v$ exists? Of course, we could consider the algebraic closure of $ \\mathbb F$, but I want to avoid such complications. I just can't move on to the second part 'til I figure this out.", "Solution_2": "By Cayley-Hamilton, $ g(A)h(A)\\equal{}0.$ If $ g(A)$ were invertible, then $ h(A)\\equal{}0.$ Can you do anything with that?", "Solution_3": "Let's try this again. Forget Cayley-Hamilton for now, and stop talking about the characteristic polynomial.\r\n\r\n$ A$ has a minimal polynomial, $ M_{A}$ such that $ M_{A}(A)\\equal{}0.$ We get this just from noting that $ \\{I,A,A^{2},\\dots,A^{r}\\}$ must be linearly dependent for some $ r$ and finding the smallest $ r$ for which that is true. [Cayley-Hamilton tells us that $ m(A)$ divides the characteristic polynomial of $ A,$ but I said we're staying away from that for now.]\r\n\r\nNow let's change the second sentence of perfect_radio's post to \"Suppose that $ M_{A}$ factors as $ gh,$ where $ g$ is an irreducible polynomial of degree $ m.$\" Change the second sentence of my original post to say \"minimal polynomial\" instead of \"characteristic polynomial.\"\r\n\r\nWe know that $ g(A)h(A)\\equal{}0.$ If $ g(A)$ were invertible, then $ h(A)\\equal{}0.$ But that contradicts that $ m_{A}$ was the minimal polynomial for $ A.$ Therefore we conclude that $ g(A)$ is singular and there exists nontrivial $ v\\in\\ker(g(A)).$\r\n\r\nOK, perfect_radio - can you take it from here?", "Solution_4": "Actually, I can't say too much about $ Q^{T}A Q$. If $ \\dim E$ is [i]precisely[/i] $ m$, then it's pretty obvious that the resulting matrix has a $ (n\\minus{}m)\\times m$ zero sub-matrix in the left-down corner. Other than that, I didn't find nothing :dry:", "Solution_5": "Can we prove the existence of v this way?:\r\n\r\nsuppose that only v=0 belongs to Ker(g(A)), so dim(Ker(g(A))) = 0. Because n = dim(Ker(g(A))) + dim(Im(g(A))) we conclude that dim(Im(g(A))) = n => rank(g(A)) = n => det(g(A)) \\= 0. This contradicts the hipothesis that g(x) is a factor of P_A" } { "Tag": [ "function", "inequalities", "induction", "Cauchy Inequality", "algebra unsolved", "algebra" ], "Problem": "Let there be $ f: (0; \\infty) \\to \\mathbb{R}$ with the propriety that :\r\n$ \\frac{f(x)\\plus{}f(y)}{2} \\equal{} f(\\sqrt{xy})$ , $ \\forall x , y \\in (0, \\infty)$ .\r\n\r\nProve that :\r\n\r\n$ \\frac{f(x_1) \\plus{} f(x_2) \\plus{} \\dots \\plus{} f(x_n)}{n}\\equal{} f(\\sqrt[n]{x_1x_2 \\dots x_n})$ , $ \\forall x_i \\in (0; \\infty) \\text{ with } i \\equal{} \\overline{1,n}$ .", "Solution_1": "[hide=\"Hint\"] Similarly with ways of prove Cauchy inequality: by Cauchy induction\nProve equality is true with $ k$ then true with $ 2k$, and true with $ k$ then true with $ k\\minus{}1$! Then OK![/hide]", "Solution_2": "[quote=\"tdl\"], and true with $ k$ then true with $ k \\minus{} 1$! Then OK![/hide][/quote]\r\n\r\nWould you please post your solution for this step ? Thanks", "Solution_3": "$ f(\\sqrt[n\\minus{}1]{x_1\\cdots x_{n\\minus{}1}})\\equal{}f(\\sqrt[n]{x_1\\cdots x_{n\\minus{}1}\\sqrt[n\\minus{}1]{x_1\\cdots x_{n\\minus{}1}}})\\\\\r\n\\hspace*{1.1in}\\equal{}{f(x_1)\\plus{}\\cdots \\plus{}f(x_{n\\minus{}1})\\plus{}f(\\sqrt[n\\minus{}1]{x_1\\cdots x_{n\\minus{}1}})\\over n}$\r\n\r\n$ \\Rightarrow f(\\sqrt[n\\minus{}1]{x_1\\cdots x_{n\\minus{}1}})\\equal{}{f(x_1)\\plus{}\\cdots \\plus{}f(x_{n\\minus{}1})\\over n\\minus{}1}$", "Solution_4": "Let $ x=e^u,y=e^v$ then \r\n$ \\frac{f(e^u)+f(e^v)}{2}=f(e^{\\frac{u+v}{2})}$\r\nConsider $ g(x)=f(e^x)$ then\r\n$ g(x)+g(y)=2g(\\frac{x+y}{2})$\r\nIt is the Jensen's equation!" } { "Tag": [ "MIT", "college", "Harvard", "Duke" ], "Problem": "I took both the SAT and ACT in the fall of my sophomore year, and I am satisfied with the scores I got on both, so I don't want to retake either exam. I am looking to go to college after my junior year, btw. So i've been looking at a lot of admission applications, and I'm starting to get the feeling that I'll have to retake the tests anyway because they're going to be \"out of date\" or something by the time colleges will want to look at them, since everyone says I should take the tests spring of my junior year or fall of my senior year. Should I retake the tests? I'd appreciate any help/advice anyone could give me on this. Thanks :)", "Solution_1": "Don't value my advice that much, because I have none, zippo, zilch nonya, zero, etc. experience on that. I just got into high school, but if it were me, I'd just do whatever I felt like doing. If I was going to think that if I didn't take it, I'd be screwed for college, then I'd take it. But if I was pretty sure it wouldn't make a difference, I wouldn't retake it. If I was in the middle, I might take it just to be sure but I dunno.", "Solution_2": "The important thing to check on the SAT I is which version (the \"old,\" used through January 2005, or the \"new,\" used beginning March 2005) is required by your favorite college(s) for people enrolling in the year you want to enroll. This shouldn't be an issue for the ACT in quite the same way. \r\n\r\nOne of the books I mention in the \"[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16376]Books on College Admission[/url]\" thread suggests never taking the SAT I more than three times--taking it four times or more looks obsessive. But taking it twice is very normal, and it will be especially normal for people who have one \"old\" SAT I score to take the \"new\" SAT I at least once too. \r\n\r\nGood luck in your applications.", "Solution_3": "You should really check with the school that you want to go to. I just went to an MIT info session and they said that they are flexible and will take pretty much anything (i.e. old sat, new sat, or act). The day after the MIT session I went to one that had Harvard, Penn, Georgetown, and Duke and they were pretty much the same. So I get the impression that for most schools it wouldn't really matter." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "We know about the numbers $ a_0, a_1\\ldots, a_{n\\minus{}1},a_n$ that $ a_0\\equal{}a_n\\equal{}0$ and that $ a_{k\\minus{}1}\\minus{}2a_k\\plus{}a_{k\\plus{}1}\\ge 0: \\,\\forall 1\\le k\\le n\\minus{}1$. Prove that $ \\forall a_k\\le 0$.", "Solution_1": "assume that $ a_k>0$ for some $ k$. then $ (a_n\\minus{}a_{n\\minus{}1})\\plus{}\\cdots \\plus{}(a_{k\\plus{}1}\\minus{}a_k)\\equal{}\\minus{}a_k<0$, while $ (a_k\\minus{}a_{k\\minus{}1})\\plus{}\\cdots \\plus{}(a_1\\minus{}a_0)\\equal{}a_k>0$. so there exists $ i>j$ for which $ a_i\\minus{}a_{i\\minus{}1}0$.Then $a_{n}-a_{n-1} \\ge a_{n-1}-a_{n-2} \\ge \\cdots \\ge a_{j}-a_{j-1}>0 \\implies a_n>a_{n-1} \\cdots>a_j>0$,a contradiction." } { "Tag": [], "Problem": "Solve this ecuation: \\[ \\sqrt{1+a*x}=x+\\sqrt{1-a*x} , a \\in R \\]", "Solution_1": "The answer is as follows.\r\n\r\nFor $|a|=1$ or $a=0,\\ x=0.$\r\nFor $|a|<1$ and $a\\neq 0,\\ x=\\pm 2\\sqrt{1-a^2}.$\r\nFor $|a|>1,$ no solutions.", "Solution_2": "[hide]Rewrite equation as $\\sqrt{1+ax}-\\sqrt{1-ax}=x$ $(1)$, and multiply both sides by $\\sqrt{1+ax}+\\sqrt{1-ax}$ to obtain $2ax=x(\\sqrt{1+ax}+\\sqrt{1-ax})$. Clearly $x=0$ is a solution gor all $a$. Dividing by $x$ we have $2a=\\sqrt{1+ax}+\\sqrt{1-ax}$ $(2)$ so if $a \\leq 0$ there are no more solutions.\nBy adding $(1)$ and $(2)$ we get $x+2a=2\\sqrt{1+ax}$ so $x+2a \\geq 0$. From this we obtain $x^2=4(1-a^2)$ so when $a \\geq 1$ no more solutions.\nElse $x_1=-2\\sqrt{1-a^2}$ and $x_2=2\\sqrt{1-a^2}$.\nObviously $x_2+2a \\geq 0$ and checking when $x_1+2a \\geq 0$ leads to $a \\geq \\frac{1}{\\sqrt2}$.\nFinally $x=0$ is a solution for all $a$\nif $a \\leq 0$ or ${a \\geq 1}$ no more solutions\nfor $01 and without use Bertrand's theorem...... :huh:", "Solution_1": "Let p prime and $\\frac{n}{2}1. Obviosly $ord_{p}(n!)=1$ odd, therefore $n1\\not =m^{2}$.", "Solution_2": "$ord_{p}(n!)$ is an order so if you say that $ord_{p}(n!)=1$ this means that $n!^{1}\\equiv_{p}1$, you probably wanted $v_{p}(n!)$ as multiplicity of $p$ in factorization of $n!$ :D\r\n\r\nbye", "Solution_3": "[quote=\"Rust\"]Let p prime and $\\frac{n}{2}1. [...][/quote]\r\nIn fact, I'm almost sure that \u0391\u03c1\u03c7\u03b9\u03bc\u03ae\u03b4\u03b7\u03c2 6 has been quite clear: he wrote \"without appealing to Bertrand's postulate\"... :maybe:", "Solution_4": "\u03a5es without use this theorem.If sb can prove that the product of n successive integers isn't a square of integer(accept of 1!)tthen it's obviously that the problem have been solved.But hou you prove this without Bertrand's theorem? :wink:", "Solution_5": "He didn't use any known theorem, he said that there is some prime number between $\\frac{n}2$ and $n$, nad that's OK, now if you look at $1\\cdot 2 \\cdot 3 \\cdots (n-1) \\cdot n$ then $p|n!$ but $2p>n$ so $p^{2}\\not | n!$, and that's all, none theorem is used in this...", "Solution_6": "[quote=\"SpongeBob\"]He didn't use any known theorem, he said that there is some prime number between $\\frac{n}2$ and $n$, nad that's OK, now if you look at $1\\cdot 2 \\cdot 3 \\cdots (n-1) \\cdot n$ then $p|n!$ but $2p>n$ so $p^{2}\\not | n!$, and that's all, none theorem is used in this...[/quote]But we cannot deduce there is a prime between $\\frac{n}{2},n$ without using any theorem.", "Solution_7": "It's obviously that we can't suppose that betwen n/2 and n there is a prime without Bertrand's theorem! :maybe:", "Solution_8": "Yah, I'm having a discussion about it, too, with my friends... (although with different questions, but using this theorem)\r\nIs this Bertrand's thing a postulate or a theorem ?? 'Cause I've read in two books. They say Conjecture and Postulate.... :huh: \r\nThx for any helpful replies :)", "Solution_9": "[quote=\"RDeepMath91\"]\nIs this Bertrand's thing a postulate or a theorem ?? [/quote]\r\nIt's a theorem, it was demonstrated first by Pafnuti Chebyshev. There's an absolutely simple proof for this involving binomials. As far as I know, it's due to Erdos.", "Solution_10": "http://www.mathlinks.ro/Forum/viewtopic.php?t=102730", "Solution_11": "\u0399 don't believe that without Bertrand's the problem is unsolved. :D", "Solution_12": "An old theorem of Erdos states that the product of consecutive integers is never a perfect square. So this would yield an easy solution to our problem except Erdos uses Sylvester's theorem:\r\n\r\nThe product of $k$ consecutive integers is divisible by a prime greater than $k$, which is a generalization of Chebychev's theorem.\r\n\r\nHowever Erdos' proof of Chebychev's theorem simplifies drastically since if $(2n)!$ is a perfect square, then so is $\\binom{2n}{n}.$\r\nAs a consequence all prime factors of $\\binom{2n}{n}$ must be less than or equal to $\\sqrt{2n}.$\r\n\r\nCan anyone show:\r\nThe product of $k$ consecutive integers with $n, n+1, \\dots n+k-1$ is never a perfect square given $n \\le k.$\r\n\r\nThe following estimate might be useful (or not sharp enough)\r\n\r\nThe product of all primes less or equal than $n$ is less that $4^{n}.$ \r\n\r\nReference:\r\n\r\nMR0000022 (1,4e)\r\nErd\u00f6s, P.\r\nNote on products of consecutive integers.\r\nJ. London Math. Soc. 14, (1939). 194--198." } { "Tag": [ "floor function", "inequalities", "algebra" ], "Problem": "Find all real solutions to the equation $4x^2 - 40 \\lfloor x \\rfloor + 51 = 0$.", "Solution_1": "Denote $[x]=n\\in\\mathbb{Z},\\{x\\}=\\alpha\\in[0,1)$. Then the equation becomes\r\n\r\n$4(n+\\alpha)^2-40n+51=0\\iff \\alpha=-n\\pm\\sqrt{10n-{51\\over 4}}$\r\n\r\nWe must have $10n-{51\\over 4}\\geqslant 0\\iff n\\geqslant{51\\over 40}\\implies n\\geqslant 2$ and also $0\\leqslant\\alpha <1$, therefore we take only the plus sign and have two inequalities:\r\n\r\n1. $-n+\\sqrt{10n-{51\\over 4}}\\geqslant 0\\iff 10n-{51\\over 4}\\geqslant n^2\\iff n^2-10n+{51\\over 4}\\leqslant 0$\r\n\r\nThe roots are $1.5$ and $8.5$, hence the inequality is satisfied for $2\\leqslant n\\leqslant 8$\r\n\r\n2. $-n+\\sqrt{10n-{51\\over 4}}<1\\iff 10n-{51\\over 4}<(n+1)^2\\iff n^2-8n+{55\\over 4}>0$\r\n\r\nThe roots are $2.5$ and $5.5$, hence the inequality is satisfied for $n\\leqslant 2$ or $n\\geqslant 6$\r\n\r\nNow $(n\\geqslant 2)\\land(2\\leqslant n\\leqslant 8)\\land (n\\leqslant 2\\lor n\\geqslant 6)$ gives $n\\in\\{2,6,7,8\\}$\r\n\r\nFrom $x=n+\\alpha=\\sqrt{10n-{51\\over 4}}=\\frac{\\sqrt{40n-51}}{2}$ we find the corresponding values for $x$:\r\n\r\n$x\\in\\{\\frac{\\sqrt{29}}{2},\\frac{3\\sqrt{21}}{2},\\frac{\\sqrt{229}}{2},\\frac{\\sqrt{269}}{2}\\}$", "Solution_2": "[hide=\"Alternate Method\"]Let $a$ be an integer such that $a=\\lfloor x \\rfloor \\le x.$ Substituting, our equation becomes \\[ 4a^2-40a+51=(2a-3)(2a-17)\\le 0. \\]$(2a-3)(2a-17)$ is only less than $0$ when $2a-3$ is positive and when $2a-17$ is negative. This means $\\frac32 0$,I will use the trivial inequalities $x-1<[x] \\le x$\n$0=4x^2-40[x]+51 \\ge 4x^2-40x+51=(2x-3)(2x-17)$,so we obtain $1.5 \\le x \\le 8.5$\n\n$0=4x^2-40[x]+51 < 4x^2-40(x-1)+51=4x^2-40x+91=(2x-7)(2x-13) \\Rightarrow x>6.5$ or$x<3.5$.\n\nCombining these bounds for $x$ we obtain the following possible values for $[x]:3,2,1,6,7,8$\n\nThus the required solutions are:$\\frac{\\sqrt{189}}{2},\\frac{\\sqrt{229}}{2},\\frac{\\sqrt{269}}{2},\\frac{\\sqrt{29}}{2}$(Note that for $[x]=3$ we obtain $x=\\frac{\\sqrt{69}}{2}>\\frac{\\sqrt{49}}{2}=3.5$ which is not a solution)", "Solution_4": "Note that we have $4x^2+51 = 40 \\lfloor x \\rfloor.$ Since $40 \\lfloor x \\rfloor \\leq 40x$, we know that $4x^2+52 \\leq 40x$. After subtracting $40x$ and factoring, this inequality becomes $$(2x-3)(2x-17) \\leq 0 \\implies \\frac{3}{2} \\leq x \\leq \\frac{17}{2}.$$ Since $\\lfloor x \\rfloor$ must be an integer, it must equal one of the integers from $1$ to $8$, inclusive. We take cases on what $\\lfloor x \\rfloor$ equals. We can plug the value of $\\lfloor x \\rfloor$ into the original equation and find the resulting value of $x$. \n\n[list]\n[*]$\\lfloor x \\rfloor=1:$ We get $4x^2=-11$; not possible\n[*]$\\lfloor x \\rfloor =2: $ We get $4x^2=29 \\implies x=\\frac{\\sqrt{29}}{2}$; works\n[*]$\\lfloor x \\rfloor =3:$ We get $4x^2=69 \\implies x=\\frac{\\sqrt{69}}{2}$; $x > 4$ so not possible\n[*]$\\lfloor x \\rfloor =4:$ We get $4x^2=109 \\implies x=\\frac{\\sqrt{109}}{2}$; $x> 5$ so not possible\n[*]$\\lfloor x \\rfloor =5:$ We get $4x^2=149 \\implies x=\\frac{\\sqrt{149}}{2}$; $x> 6$ so not possible\n[*]$\\lfloor x \\rfloor =6:$ We get $4x^2=189 \\implies x=\\frac{3\\sqrt{21}}{2}$; works\n[*]$\\lfloor x \\rfloor =7:$ We get $4x^2=229 \\implies x=\\frac{\\sqrt{229}}{2}$; works\n[*]$\\lfloor x \\rfloor =8:$ We get $4x^2=269 \\implies x=\\frac{\\sqrt{269}}{2}$; works\n[/list]\nHence, the solutions are $x=\\frac{\\sqrt{29}}{2}, \\frac{3\\sqrt{21}}{2}, \\frac{\\sqrt{229}}{2}, \\frac{\\sqrt{269}}{2}$.", "Solution_5": "Alternative, by solving for $x=\\frac{\\pm \\sqrt{40[x]-51}-4[x]}{2}$ and bounding\n$0 \\leq x-[x] <1$ and use increasing/decreasing functions. ", "Solution_6": "If anyone could help me to line up two inequalities in an align* environment that would be helpful.\n\n[hide=Solution]The idea is to use bounding to find the possible values of $\\lfloor x\\rfloor$ and then just plug them in.\n\nTo do this, we bound:\n\n\\begin{align*}\n\\lfloor x\\rfloor^2 &\\leq x^2 < (\\lfloor x\\rfloor +1)^2\\\\\n4\\lfloor x\\rfloor^2 &\\leq 4x^2 < 4\\lfloor x\\rfloor^2 + 8\\lfloor x\\rfloor +4\\\\\n4\\lfloor x\\rfloor^2 &\\leq 40\\lfloor x\\rfloor-51 < 4\\lfloor x\\rfloor^2 + 8\\lfloor x\\rfloor +4\\\\\n4\\lfloor x\\rfloor^2 -40\\lfloor x\\rfloor +51 &\\leq 0 < 4\\lfloor x\\rfloor^2 -32\\lfloor x\\rfloor+55\\\\\n(2\\lfloor x\\rfloor -3)(2\\lfloor x\\rfloor -17) &\\leq 0 < (2\\lfloor x\\rfloor -5)(2\\lfloor x\\rfloor -11)\\\\\n\\end{align*}\n\nThis implies that $\\lfloor x\\rfloor\\in\\{2,6,7,8\\}$.\n\nPlugging each of these in gives: \\[\\boxed{\\left\\{\\frac{\\sqrt{29}}{2}, \\frac{3\\sqrt{21}}{2}, \\frac{\\sqrt{229}}{2}, \\frac{\\sqrt{269}}{2}\\right\\}}\\]\n\nrespectively.\n[/hide]", "Solution_7": "This problem is a familiar one since lhs is greater than $4x^2-40x+51$ now factoring we get a bound which is as follows 3/2<=x<=17/2 now we do some casework and get the answer", "Solution_8": "[quote=4everwise][hide=\"Alternate Method\"]Let $a$ be an integer such that $a=\\lfloor x \\rfloor \\le x.$ Substituting, our equation becomes \\[ 4a^2-40a+51=(2a-3)(2a-17)\\le 0. \\]$(2a-3)(2a-17)$ is only less than $0$ when $2a-3$ is positive and when $2a-17$ is negative. This means $\\frac32 0$.\r\nThen, just use that $1+4x^2 \\geq 4x$ so that $y \\leq x$.\r\nSimilarly, we have $z \\leq y$ and $x \\leq z$.\r\nIt follows that $x=y=z$.\r\nReturning to the initial equations we deduce that $x=y=z = \\frac 1 2$, which is indeed a solution.\r\n\r\nPierre." } { "Tag": [ "function", "induction", "floor function", "Putnam", "modular arithmetic", "search" ], "Problem": "I have a solution by induction, but I'm curious to see if there's a non-inductive solution. \r\n\r\nProve that \r\n\\[\\left\\lfloor \\frac{n+2^{0}}{2}\\right\\rfloor+\\left\\lfloor \\frac{n+2^{1}}{2^{2}}\\right\\rfloor+\\left\\lfloor \\frac{n+2^{2}}{2^{3}}\\right\\rfloor+...+\\left\\lfloor \\frac{n+2^{n-1}}{2^{n}}\\right\\rfloor = n \\]\r\nfor any positive integer $n$.", "Solution_1": "I got a nice solution using binary, but im in too much of a rush right now...", "Solution_2": "[quote=\"seamusoboyle\"]I got a nice solution using binary, but im in too much of a rush right now...[/quote]\r\n\r\ni think theres exists even a nicer solution than that!\r\n\r\n[hide=\"hint\"]\nrecall the identity:\n$\\lfloor x\\rfloor+\\lfloor x+\\frac{1}{n}\\rfloor+\\lfloor x+\\frac{2}{n}\\rfloor+...+\\lfloor x+\\frac{n-1}{n}\\rfloor=\\lfloor nx \\rfloor$\n[/hide]\n\n[hide=\"solution\"]\nby the identity mentioned in the hint we have:\n$\\left\\lfloor x+\\frac12\\right\\rfloor= \\left\\lfloor 2x\\right\\rfloor-\\left\\lfloor x\\right\\rfloor$.\n\n$\\sum_{i=0}^{\\infty}\\lfloor \\frac{n+2^{i}}{2^{i+1}}\\rfloor= \\sum_{i=0}^{\\infty}\\lfloor \\frac{n}{2^{i+1}}+\\frac{1}{2}\\rfloor$\nby identity above:\n$= \\sum_{i=0}^{\\infty}\\lfloor \\frac{2n}{2^{i+1}}\\rfloor-\\lfloor \\frac{n}{2^{i+1}}\\rfloor=$ $\\sum_{i=0}^{\\infty}\\lfloor \\frac{n}{2^{i}}\\rfloor-\\sum_{i=0}^{\\infty}\\lfloor \\frac{n}{2^{i+1}}\\rfloor= \\lfloor \\frac{n}{2^{0}}\\rfloor= n$[/hide]\r\n\r\nit can also be shown via induction, show that if it holds for n, then it will hold for 2n and 2n+1.\r\nedit: this problem appeared in University of Washington Putnam team selection test.", "Solution_3": "It can also be shown using induction by noting that each term increases by 1 iff \r\n$n \\equiv 1 \\pmod{2}$\r\n$n \\equiv 2 \\pmod{4}$\r\n...\r\n$n \\equiv 2^{k}\\pmod{2^{k+1}}$\r\n...\r\nand it's clear that exactly one of these statements will be true.", "Solution_4": "[quote=\"amirhtlusa\"][quote=\"seamusoboyle\"]I got a nice solution using binary, but im in too much of a rush right now...[/quote]\n\ni think theres exists even a nicer solution than that!\n\n[hide=\"hint\"]\nrecall the identity:\n$\\lfloor x\\rfloor+\\lfloor x+\\frac{1}{n}\\rfloor+\\lfloor x+\\frac{2}{n}\\rfloor+...+\\lfloor x+\\frac{n-1}{n}\\rfloor=\\lfloor nx \\rfloor$\n[/hide]\n\n[hide=\"solution\"]\nby the identity mentioned in the hint we have:\n$\\left\\lfloor x+\\frac12\\right\\rfloor= \\left\\lfloor 2x\\right\\rfloor-\\left\\lfloor x\\right\\rfloor$.\n\n$\\sum_{i=0}^{\\infty}\\lfloor \\frac{n+2^{i}}{2^{i+1}}\\rfloor= \\sum_{i=0}^{\\infty}\\lfloor \\frac{n}{2^{i+1}}+\\frac{1}{2}\\rfloor$\nby identity above:\n$= \\sum_{i=0}^{\\infty}\\lfloor \\frac{2n}{2^{i+1}}\\rfloor-\\lfloor \\frac{n}{2^{i+1}}\\rfloor=$ $\\sum_{i=0}^{\\infty}\\lfloor \\frac{n}{2^{i}}\\rfloor-\\sum_{i=0}^{\\infty}\\lfloor \\frac{n}{2^{i+1}}\\rfloor= \\lfloor \\frac{n}{2^{0}}\\rfloor= n$[/hide]\n\nit can also be shown via induction, show that if it holds for n, then it will hold for 2n and 2n+1.\nedit: this problem appeared in University of Washington Putnam team selection test.[/quote]\r\n\r\nI dont assume any of those inequalities. mine looks at how you write the number n out in binary. ie divide out the highest power of two continousally...", "Solution_5": "[quote=\"amirhtlusa\"][quote=\"seamusoboyle\"]I got a nice solution using binary, but im in too much of a rush right now...[/quote]\n\ni think theres exists even a nicer solution than that!\n\n[hide=\"hint\"]\nrecall the identity:\n$\\lfloor x\\rfloor+\\lfloor x+\\frac{1}{n}\\rfloor+\\lfloor x+\\frac{2}{n}\\rfloor+...+\\lfloor x+\\frac{n-1}{n}\\rfloor=\\lfloor nx \\rfloor$\n[/hide]\n\n[hide=\"solution\"]\nby the identity mentioned in the hint we have:\n$\\left\\lfloor x+\\frac12\\right\\rfloor= \\left\\lfloor 2x\\right\\rfloor-\\left\\lfloor x\\right\\rfloor$.\n\n$\\sum_{i=0}^{\\infty}\\lfloor \\frac{n+2^{i}}{2^{i+1}}\\rfloor= \\sum_{i=0}^{\\infty}\\lfloor \\frac{n}{2^{i+1}}+\\frac{1}{2}\\rfloor$\nby identity above:\n$= \\sum_{i=0}^{\\infty}\\lfloor \\frac{2n}{2^{i+1}}\\rfloor-\\lfloor \\frac{n}{2^{i+1}}\\rfloor=$ $\\sum_{i=0}^{\\infty}\\lfloor \\frac{n}{2^{i}}\\rfloor-\\sum_{i=0}^{\\infty}\\lfloor \\frac{n}{2^{i+1}}\\rfloor= \\lfloor \\frac{n}{2^{0}}\\rfloor= n$[/hide]\n\nit can also be shown via induction, show that if it holds for n, then it will hold for 2n and 2n+1.\nedit: this problem appeared in University of Washington Putnam team selection test.[/quote]\r\n\r\nI dont assume any of those inequalities. mine looks at how you write the number n out in binary. ie divide out the highest power of two continousally...", "Solution_6": "The solution using Hermites identity is quite nice. Id be interested to see yours seamusboyle. (I apologize for the lack of apostraphes; every time I hit the apostraphe button it brings up the Mozilla search, no idea why).", "Solution_7": "seamusoboyle, i know exactly what you are talking about.\r\nin fact the binary expansion solution, was my solution, when i first wrote the solution for our test. :)", "Solution_8": "Woah woah woah! Thats two names! Seamus O' Boyle! And does anyone know how to delete posts? My double post is annoying me, and i cant find the option :(", "Solution_9": "there's an icon on the bottom right of ur post .. i believe..umm do u mind posting ur binary solution", "Solution_10": "This is IMO 1968/6. This problem is pretty straight forward when considering binary numbers. Also, the Hermite identity is a nice exercise in periodicity. :) Seamus O' Boyle, you cannot delete posts when someone has already posted after you." } { "Tag": [ "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Let $f$ be a bounded continuous function on $(a,\\infty)$. For any $T>0$ show that there exists $x_{n}\\to\\infty$ such that $\\lim_{n\\to\\infty}(f(x_{n}+T)-f(x_{n})) = 0$", "Solution_1": "$g(x)=f(x+T)-f(x)$ is bounded and continuous. If it changes sign infinitely often near $\\infty$, then by the Mean Value Theorem we're done. If not, WLOG suppose that it's positive. Then either $\\liminf_{x \\to \\infty}g(x) > 0$, which implies $f$ unbounded, contradiction, or $\\liminf_{x \\to \\infty}g(x) = 0$, which gives a possible sequence." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "$ b_1\\equal{}1,b_n\\equal{} n^{f(n)}b_{n\\minus{}1},(n\\geq2),f(n)\\equal{}2^{1\\minus{}n},\\Longrightarrow b_n<3.$", "Solution_1": "here is a solution............. :)", "Solution_2": "$ b_1 = 1,b_n = n^{2^{1 - n}}b_{n - 1}(n\\geq2)\\Longrightarrow$\r\n\r\n${{ b_n = \\prod_{k = 2}^{n}k^{2^{1 - k}}} < \\prod_{k = 2}^{n}\\left[\\frac {(k + 1)^2}{k + 2}\\right]^{2^{1 - k}}} = \\frac {3}{(n + 2)^{2^{1 - n}}} < 3.$\r\n\r\n[hide=\"Remark.\"]$ \\frac {(143k^2 + 221k - 6)^2[11(k + 1) + 6]}{13(11k + 6)^2[143(k + 1)^2 + 221(k + 1) - 6]} - k$\n\n$ = \\frac {(k - 2)(92807k^2 + 105963k - 306)}{13(11k + 6)^2(143k^2 + 507k + 358)}\\geq0(k\\geq2)\\Longrightarrow$\n\n${{ b_n = \\prod_{k = 2}^{n}k^{2^{1 - k}}} < \\prod_{k = 2}^{ + \\infty}\\left\\{\\frac {(143k^2 + 221k - 6)^2[11(k + 1) + 6]}{13(11k + 6)^2[143(k + 1)^2 + 221(k + 1) - 6]}\\right\\}^{2^{1 - k}}}$\n\n$ = \\frac {143\\times 2^2 + 221\\times 2 - 6}{13(11\\times 2 + 6)} = \\frac {36}{13} = 2.76\\cdots.$\n\n${{ b_n = \\prod_{k = 1}^{n}k^{2^{1 - k}}} < \\prod_{k = 1}^{ + \\infty}k^{2^{1 - k}}} = 2.7612068419574\\cdots.$[/hide]" } { "Tag": [ "blogs", "search", "\\/closed" ], "Problem": "Everytime I try to remove a contributor, I get this error message:\r\n\r\nSQL requests not achieved\r\n\r\nDEBUG MODE\r\n\r\nSQL Error: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ')' at line 3\r\n\r\nSQL Request: DELETE FROM phpbb_weblog_contributors WHERE weblog_id = 717 AND user_id IN ( )\r\nLine : 261\r\nFile : class_blog_options.php\r\n\r\n\r\nWhat do I do?", "Solution_1": "[quote=\"xxazurewrathxx\"]What do I do?[/quote]\r\nObviously not using the search function.", "Solution_2": "I have searched, I don't think anyone has given a conclusive answer to the question yet, besides, \"we're working on it\". I was hoping that there's been an update to the situation since then.", "Solution_3": "Nope, no update.", "Solution_4": "[quote=\"xxazurewrathxx\"]I have searched, I don't think anyone has given a conclusive answer to the question yet, besides, \"we're working on it\". I was hoping that there's been an update to the situation since then.[/quote]\r\nRemember me that I should always start new threads everytime I've something to say or want soemthing older than five minutes. Oh sorry for not starting a thread concerning this post. :roll:", "Solution_5": "Yes it's probably better to reply to one of the existing threads and say \"This problem is still occurring, here's what I was doing when it happened, any updates?\" instead of posting it again (and this really is like the 4th or 5th time, it happened to me the other day and I searched for it and saw so many so I decided not to post anything....).", "Solution_6": "Problem is now fixed. Small glitch, so small that it was hard to find :)" } { "Tag": [ "USAMTS" ], "Problem": "Are you allowed to do USAMTS if you are in 6th grade?", "Solution_1": "Yes, there is no lower grade limit. As long as you haven't graduated high school, you may participate." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let p be a prime and n is a positive integer.\r\nFind the number of $p$-Sylow groups in a symmetric group $S_n$ .", "Solution_1": "It was posted before - but not solved:\r\n[url=http://www.mathlinks.ro/Forum/topic-56995]www.mathlinks.ro/Forum/topic-56995[/url]", "Solution_2": "Sorry,can anyone show me a formular in a general case of this problem?I think it is a very nice problem.I mean the problem not to describe all the $p$ Sylow subgroups ,COUNT them !\r\nFor most simple cases , i have some answers:\r\n1/$n$<$2p$ then the answer is $n!$/$(n-p)!p(p-1)$\r\n2/$n=2p$ then the answer is $(2p)!$/$(2(p(p-1))^2)$" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "suppose you have the generally functional A Who can you find B in the case that f is independent of x?\r\n[img]http://img413.imageshack.us/img413/930/functionaalbisoi4.jpg[/img]\r\n\r\nThanks.", "Solution_1": "This isn't the place for calculus. I'm sure your post will be moved to the calculus forum where someone can help.", "Solution_2": "Sorry can someone replace it?" } { "Tag": [ "limit" ], "Problem": "Vom nota cu $f(n)$ numarul de 1-uri din reprezentarea binara a lui $n$.Demonstrati ca exista un sir $(a_{i})_{i\\geq 1}$ astfel incat $\\lim_{i\\rightarrow \\infty}\\frac{f(a_{i}^{2})}{f(a_{i})}$=0", "Solution_1": "Nu s-a gandit nimeni??? Mie mi se pare interesanta...da nu prea am idei...cum ar treb sa arate sirul ala :( Sper sa posteze pana la urma cineva o solutie:) Intre timp sper sa ma mai gandesc si eu (intre 2 teze si doua aplicatii :( )", "Solution_2": "Fie $a_{2}= (1101111)_{2}= (111)_{10}$ si $a_{n+1}= 2^{2^{n+1}}a_{n}-1, \\, \\forall n \\geq 2$. $a_{n}$ arata asa in baza $2$: $11$, apoi $0$, apoi $111$ ($2^{2}-1$), apoi $0$, apoi $1111111$ ($2^{3}-1$), apoi $0$, $\\ldots$, apoi $0$, apoi $11 \\ldots 1$ ($2^{n-1}-1$), apoi $0$, apoi $11 \\ldots 1$ ($2^{n}$).\r\n\r\n$1^\\circ$ $f \\left( a_{n}\\right) = 2^{n+1}-n$\r\n\r\n$2^\\circ$ $f \\left( a_{n}^{2}\\right) = \\frac{n^{2}-n+6}2$\r\n\r\nSe poate demonstra destul de usor $1^\\circ$, dar $2^\\circ$ are deocamdata statutul de conjectura (merge pentru [b]c\u00e2teva[/b] [i]n[/i]-uri). Chiar daca $2^\\circ$ nu se va dovedi a fi adevarata, exista sanse bune ca $\\lim_{n \\to \\infty}\\frac{f \\left( a_{n}^{2}\\right)}{f \\left( a_{n}\\right)}= 0$.\r\n\r\nCum m-am gandit monstrul de mai sus? Trisand un pic, bineinteles. Am facut un program care imi calcula raportul $\\frac{f(x^{2})}{f(x)}$ pentru $x$-urile intre $2$ si $5 \\cdot 10^{4}$.\r\n\r\nPentru numarul $x_{3}= (28415)_{10}= (110111011111111)_{2}$ am obtinut cea mai mica valoare, dupa care mi-a venit ideea de mai sus.", "Solution_3": "Ok :) Imi pare bine ca s-a gandit cineva. In lipsa totala a unor cunostinte de informatica, nu prea am avut rabdare sa ma joc cu calculatorul de buzunar (ajunsesem pe la 65 :blush: ). Ideea data de perfect_radio e frumoasa si poate in curand apare cineva cu o solutie completa :)", "Solution_4": "abcdefghijkl" } { "Tag": [ "geometry", "perimeter" ], "Problem": "How many scalene triangles with perimeters less than $ 16$ have sides whose measures are integers?\r\n\r\n[hide=\"Answer\"]\n9\n[/hide]\r\n\r\nI suppose we could use casework, but that would take a REALLY long time, anyone have a fast solution?\r\n\r\nEDIT: Thanks! By the way, this was #10 on Target :D", "Solution_1": "Brute force is the best way. For example, none of the sides can be $ 1$. After that, note that if $ 2$ is one of the sides, then the other sides must be $ n$ and $ n\\plus{}1$, where $ 3\\le n\\le6$ (the second half comes from the problem restrictions). If $ 3$ is the smallest side, then quick casework gives $ \\{(3,4,5),(3,4,6),(3,5,6),(3,5,7)\\}$, and if $ 4$ is the smallest side, we similarly get $ \\{(4,5,6)\\}$. $ 5\\plus{}6\\plus{}7\\equal{}18\\ge16$, so $ 5$ can't be the smallest side. Adding up, this is $ (6\\minus{}3\\plus{}1)\\plus{}4\\plus{}1\\equal{}\\boxed9$ triangles." } { "Tag": [ "trigonometry", "integration", "calculus", "logarithms", "function", "geometry" ], "Problem": "A ball is thrown from the ground with initial velocity $ v_o$ making an angle $ \\alpha$ with the horizontal. Find the total distance traveled by the ball (neglect air resistance).", "Solution_1": "Here's my opinion\r\n\r\n[hide]$ dx \\equal{} v_0\\cos\\alpha dt$\n\n$ dy \\equal{} v_0\\sin\\alpha dt \\plus{} gtdt$\n\n$ ds \\equal{} \\sqrt {dx^2 \\plus{} dy^2}$\n\nThe time that the ball flies is $ \\Delta t \\equal{} \\frac {2v_0\\sin\\alpha}{g}$\n\nThe distance that the ball travels\n\n$ s \\equal{} \\int ds \\equal{} \\int_0^{\\Delta t} \\sqrt {dx^2 \\plus{} dy^2}$[/hide]", "Solution_2": "There is something wrong there. Also, I want to see the final answer.", "Solution_3": "[hide=\"yet another opinion\"]\nclearly the distance travelled is \n$ \\int \\sqrt {1 + f'(x)^{2}}dx$ where $ y = f(x)$ is the equation of the motion of the body \nclearly $ f'(x) = \\tan \\alpha - \\frac {gt}{2v_{o}\\cos \\alpha}$\nhence the problem reduces nothing but integration an integral of the form \n$ \\int x\\sqrt {1 + {b(x - c)}^{2}dx \\bigstar}$\nafter substituting $ \\frac {dx}{dt} = v_{0}\\cos \\alpha$\nthe integral of $ \\bigstar$ can be solved by substituting\n$ b(x - c) = u$\nthus $ \\bigstar \\equiv \\int u \\sqrt {1 + u^{2}}du + \\int \\sqrt {1 + u^{2}} du$\nleaving out the constants obviuosly.\nthe first one can be done by substituting $ u^{2} = x$ and second one is in by itself a standard form\nP.S. i would like to see a tricky proof of this :) \n[/hide]", "Solution_4": "I still don't see the final answer.", "Solution_5": "yes of course that was the method :wink:", "Solution_6": "Assuming the same level for shooting point and target\r\n\r\n[hide=\"we get:\"]\n\\[ {v_0^2 \\over g}(sin\\alpha \\plus{} cos^2\\alpha.arc \\,sinh \\,tg\\alpha)\n\\]\n[/hide]", "Solution_7": "That is very close to [hide=\"my answer\"]$ L \\equal{} \\frac{v_o^2}{2g}\\left[2 \\sin \\alpha \\plus{} \\cos^2 \\alpha \\ln\\left(\\frac{1\\plus{} \\sin \\alpha}{1 \\minus{} \\sin \\alpha}\\right)\\right]$.[/hide]", "Solution_8": "Sorry, I forgot the factor $ \\,cos^2\\alpha\\,$.\r\n\r\nNow, $ \\,arc\\,sinh \\,tg \\alpha\\,$ is \"short' to $ {1\\over 2}ln\\,{{1 \\plus{} sin\\alpha}\\over{1 \\minus{} sin\\alpha}}\\,$", "Solution_9": "Yes, that's right. But its \"$ argsinh(\\tan \\alpha)$\", not \"$ \\arcsin(\\tan \\alpha)$\".", "Solution_10": "http://en.wikipedia.org/wiki/Inverse_hyperbolic_function\r\n\r\nhttp://mathworld.wolfram.com/InverseHyperbolicSine.html :The inverse hyperbolic sine sinh^(-1)z (Beyer 1987, p. 181; Zwillinger 1995, p. 481), sometimes called the area hyperbolic sine (Harris and Stocker 1998, p. 264) and sometimes denoted arcsinhz (Jeffrey 2000, p. 124), is the multivalued function that is the inverse function of the hyperbolic sine. The variants Arcsinhz or Arsinhz (Harris and Stocker 1998, p. 263) are sometimes used to refer to explicit principal values", "Solution_11": "Thanks for the links. However, in my country we write \"argsinh\" which stands for \"argument of the hyperbolic sine\", and that is the notation I use. \"Arcsin\" means (in my country) \"(the) arc (whose) sine (is)\"." } { "Tag": [ "inequalities", "trigonometry", "incenter", "geometric inequality", "inequalities proposed" ], "Problem": "Let $ x,y,z>0$ such that $xy+yz+zx+xyz=4$ . Prove that $ 1+x+y+z\\le xyz+\\frac 1x+\\frac 1y+\\frac 1z\\ .$", "Solution_1": "I think it can be solved by using trigonometry!Put [tex]cosA=\\frac{\\sqrt{yz}}{2},cosB=\\frac{\\sqrt{xz}}{2},cosC=\\frac{\\sqrt{xy}}{2}[/tex].Since hypothesis we see:[tex]A,B,C[/tex] are 3 angles of a triangle...But the rest is not easy!", "Solution_2": "Replacing $x,y,z$ by $1/x,1/y,1/z$, the statement becomes as follows:\r\nIf $x,y,z>0$ such that $x+y+z+1=4xyz$, then\r\n$1+x+y+z \\ge xyz+\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}$.\r\nFor $x+y+z=constant$ and $xyz=constant$, by n-1EV Principle it follows that the expression $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}$ is maximal when $x\\le y=z$. Thus, it suffices to prove the inequality for $y=z$, when the hypothesis condition $x+y+z+1=4xyz$ becomes $y=\\frac{x+1}{2x}$.\r\nHave you a nice solution to this nice problem, Levi ?", "Solution_3": "This algebraic inequality is obtained from the following geometric inequality in a triangle $ABC\\ ,$\n$\\ 1+\\frac{b+c-a}{a}+\\frac{c+a-b}{b}+\\frac{a+b-c}{c}\\le \\frac{2r}{R}+\\frac{a}{b+c-a}+\\frac{b}{c+a-b}+\\frac{c}{a+b-c}$\nby the system of substitusions $b+c=a(x+1), c+a=b(y+1), a+b=c(z+1)\\ \\ (*)$. The condition of the compatibility of the system $(*)$ is the relation $xy+yz+zx+xyz=4$ and his solution $(a,\\ b,\\ c)$ is\n$\\frac{a}{(y+2)(z+2)}=\\frac{b}{(z+2)(x+2)}=$ $\\frac{c}{(x+2)(y+2)}=\\frac{a+b+c}{(x+y)(y+z)(z+x)}t\\ge 0$\nand we can to suppose that $t=1$.\nWe show easy that $a,\\ b,\\ c$ are the lenghts of the sides a triangle iff $x,\\ y,\\ z>0$ a.s.o.\nRemark: From the distance between the orthocentre $H$ and the centre $I$ of the incircle,\n$HI=4r(R+r)+3r^2-s^2$ results the inequality $s^2 \\leq 4R(R+r)+3r^2$ which is equivalently with the up geometric inequality.", "Solution_4": "The last relation is writing thus $0 < k = \\frac {a}{(y+2)(z+2)} = \\frac {b}{(z+2)(x+2)} = $ $\\frac {c}{(x+2)(y+2)} = \\frac {a+b+c}{(x+2)(y+2)(z+2)}$", "Solution_5": "I think I got the most simple proof of this problem,still using Trigonometry:After some conclusions and the initial relationship ,it becomes:\r\n$(2-xy)(2-yz)(2-zx)\\leq x^2y^2z^2(*)$\r\nSubstituting $\\sqrt{yz}=2sin\\frac{A}{2}$,... then we have $A+B+C=\\pi$.And (*) becomes:\r\n$cosAcosBcosC\\leq (1-cosA)(1-cosB)(1-cosC)$\r\nwhich is well-known.", "Solution_6": "[quote=\"Virgil Nicula\"][color=darkred]Prove that in any triangle $ABC$ there is the following inequality :[/color]\n\n$\\boxed{\\ \\ 1+\\frac{b+c-a}{a}+\\frac{c+a-b}{b}+\\frac{a+b-c}{c}\\le \\frac{2r}{R}+\\frac{a}{b+c-a}+\\frac{b}{c+a-b}+\\frac{c}{a+b-c}\\ }\\ \\ (*)$[/quote][color=darkblue][b][u]Proof[/u].[/b] Prove easily that $\\left\\|\\begin{array}{c}\\sum\\frac{b+c-a}{a}=\\frac{2}{abc}\\cdot\\sum bc(p-a)=\\frac{p^{2}+r^{2}-8Rr}{2Rr}\\\\\\\\ \\sum \\frac{a}{b+c-a}=\\frac{1}{2(p-a)(p-b)(p-c)}\\cdot \\sum a(p-b)(p-c)=\\frac{2R-r}{r}\\end{array}\\right\\|\\ .$ Thus ,\n\n$(*)$ $\\Longleftrightarrow$ $\\frac{p^{2}+r^{2}-8Rr}{2Rr}+\\frac{R-2r}{R}-\\frac{2R-r}{r}\\le 0$ $\\Longleftrightarrow$ $p^{2}\\le 4R(R+r)+3r^{2}\\ \\ (**)$\n\nwhat is truly because the distance between the incenter $I$ and the orthocenter $H$ is given by\n\n$\\boxed{\\ HI^{2}=4R(R+r)+3r^{2}-p^{2}\\ }\\ge 0\\ .$ In conclusion , $(*)$ $\\Longleftrightarrow$ $(**)\\ .$[/color]", "Solution_7": "Using transformations $ x=\\frac{bc}{a},y=\\frac{ca}{b},z=\\frac{ab}{c} $ inequality is shown to:\n$ a, b, c >0,a^2+b^2+c^2+abc=4\\Rightarrow a^2b^2c^2+2(a^2+b^2+c^2)\\geq 4+b^2c^2+c^2a^2+a^2b^2\\Leftrightarrow 4a^2b^2c^2+8(a^2+b^2+c^2)\\geq 16+4(b^2c^2+c^2a^2+a^2b^2) $\nWithhold inequality shown:\n1) $ 4(a^2b^2+b^2c^2+c^2a^2)\\leq 4a^2b^2c^2+8(a^2+b^2+c^2)-16 $\nWe have by Schur:\n2) $ 4(a^2b^2+b^2c^2+c^2a^2)\\leq \\frac{(a^2+b^2+c^2)^3+9a^2b^2c^2}{a^2+b^2+c^2} $\nIt is sufficient to show that:\n3) $ \\frac{(a^2+b^2+c^2)^3+9a^2b^2c^2}{a^2+b^2+c^2}\\leq 4a^2b^2c^2+8(a^2+b^2+c^2)-16 $\nor\n $ (a^2+b^2+c^2)^2+\\frac{9a^2b^2c^2}{a^2+b^2+c^2}\\leq 4a^2b^2c^2+8(a^2+b^2+c^2)-16\\Leftrightarrow $ \n$ \\Leftrightarrow (a^2+b^2+c^2-4)^2+\\frac{9a^2b^2c^2}{a^2+b^2+c^2}\\leq 4a^2b^2c^2 $\nThe last relationship, considering that $a^2+b^2+c^2+abc=4 $, we get:\n$ (a^2+b^2+c^2-4)^2+\\frac{9a^2b^2c^2}{a^2+b^2+c^2}\\leq 4a^2b^2c^2\\Leftrightarrow a^2b^2c^2+\\frac{9a^2b^2c^2}{a^2+b^2+c^2}\\leq 4a^2b^2c^2\\Leftrightarrow $\n$ \\Leftrightarrow\\frac{9a^2b^2c^2}{a^2+b^2+c^2}\\leq 3a^2b^2c^2\\Leftrightarrow 3\\leq a^2+b^2+c^2 $\nBecause, by GM-AM we have $ abc\\leq 1\\Rightarrow a^2+b^2+c^2\\geq 3 $ proof is finished!\n__________________\nSandu Marin", "Solution_8": "Using transformations inequality is shown to:\n$ a, b, c >0,a^2+b^2+c^2+abc=4\\Rightarrow a^2+b^2+c^2+a^2b^2c^2 \\geq abc+b^2c^2+c^2a^2+a^2b^2 $", "Solution_9": "[quote=\"younesmath2012maroc\"]Using transformations inequality is shown to:\n$ a, b, c >0,a^2+b^2+c^2+abc=4\\Rightarrow a^2+b^2+c^2+a^2b^2c^2 \\geq abc+b^2c^2+c^2a^2+a^2b^2 $[/quote]\n\nFurther, considering that $ abc=4-a^2-b^2-c^2 $ obtain:\n$ a^2+b^2+c^2+a^2b^2c^2 \\geq abc+b^2c^2+c^2a^2+a^2b^2\\Leftrightarrow a^2+b^2+c^2+a^2b^2c^2 \\geq 4-a^2-b^2-c^2+b^2c^2+c^2a^2+a^2b^2 $\n$ \\Leftrightarrow a^2b^2c^2+2(a^2+b^2+c^2)\\geq 4+b^2c^2+c^2a^2+a^2b^2 $ Q.E.D", "Solution_10": "[quote=\"Virgil Nicula\"]Let $ x,y,z>0$ such that $xy+yz+zx+xyz=4$ . Prove that $ 1+x+y+z\\le xyz+\\frac 1x+\\frac 1y+\\frac 1z\\ .$[/quote]\n\nBecause\n $ 1+x+y+z\\le xyz+\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\ . $ $\\Leftrightarrow (2-xy)(2-yz)(2-zx)\\leq x^2y^2z^2 $\ndemonstrated inequality can be written as:\n$ x,y,z>0,xy+yz+zx+xyz=4\\Rightarrow (2-xy)(2-yz)(2-zx)\\leq x^2y^2z^2 $", "Solution_11": "Very nice problem! Congratulations to Virgil Nicula!\nMy current solution:\nProblem is equivalent to:\n$ x,y,z>0, x+y+z+1=4xyz\\Rightarrow 1+\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\le x+y+z+\\frac{1}{xyz} $\nWith $ p+x+y+z,q+xy+yz+zx,r=xyz $ it return to:\n$ q\\ge p\\ge 3r\\ge 3,p+1=4r\\Rightarrow 1+\\frac{q}{r} \\le\\frac{1}{r}+p\\Leftrightarrow $\n$ 2+\\frac{q}{r} \\le\\frac{1}{r}+p+1\\Leftrightarrow 2+\\frac{q}{r} \\le\\frac{1}{r}+4r $\n$ \\Leftrightarrow 2r+q\\le 1+4r^2\\Leftrightarrow q\\le 4r^2-2r+1\\Leftrightarrow 4q\\le p^2+3 $\n(1) $ \\Leftrightarrow q\\le\\frac{p^2+3}{4} $\nBy Schur of three degree we have:\n$ (x+y+z)^3+9xyz\\ge 4(x+y+z)(xy+yz+zx)\\Leftrightarrow p^3+9r\\ge 4pg\\Leftrightarrow $\n(2) $ q\\le\\frac{4p^3+9p+9}{16p} $\nIt is sufficient that:\n$ \\frac{4p^3+9p+9}{16p}\\le\\frac{p^2+3}{4}\\Leftrightarrow p\\ge 3 $\nWhich is true!!\n", "Solution_12": "[quote=sandumarin]Very nice problem! Congratulations to Virgil Nicula!\nMy current solution:\nProblem is equivalent to:\n$ x,y,z>0, x+y+z+1=4xyz\\Rightarrow 1+\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\le x+y+z+\\frac{1}{xyz} $\nWith $ p=x+y+z,q=xy+yz+zx,r=xyz $ it return to:\n$ q\\ge p\\ge 3r\\ge 3,p+1=4r\\Rightarrow 1+\\frac{q}{r} \\le\\frac{1}{r}+p\\Leftrightarrow $\n$ 2+\\frac{q}{r} \\le\\frac{1}{r}+p+1\\Leftrightarrow 2+\\frac{q}{r} \\le\\frac{1}{r}+4r $\n$ \\Leftrightarrow 2r+q\\le 1+4r^2\\Leftrightarrow q\\le 4r^2-2r+1\\Leftrightarrow 4q\\le p^2+3 $\n(1) $ \\Leftrightarrow q\\le\\frac{p^2+3}{4} $\nBy Schur of three degree we have:\n$ (x+y+z)^3+9xyz\\ge 4(x+y+z)(xy+yz+zx)\\Leftrightarrow p^3+9r\\ge 4pg\\Leftrightarrow $\n(2) $ q\\le\\frac{4p^3+9p+9}{16p} $\nIt is sufficient that:\n$ \\frac{4p^3+9p+9}{16p}\\le\\frac{p^2+3}{4}\\Leftrightarrow p\\ge 3 $\nWhich is true!![/quote]\n\n" } { "Tag": [ "LaTeX", "inequalities", "inequalities proposed" ], "Problem": "prove that if abc=1 then we have a^3 +b^3 +c^3 +2ab\\ a^2 +b^2 2ac\\a^2+c^2 +2bc\\b^2+c^2 >= 6 it can be solved by sos but i want another solution\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n :ninja:", "Solution_1": "[quote=\"fermat3\"]prove that if abc=1 then we have a^3 +b^3 +c^3 +2ab\\a^2+b^2+2ac\\a^2+c^2+2bc\\b^2+c^2 >= 6\nI can be solved by sos but i want another solution.[/quote]\nI think you meant:\n\n[quote]prove that if $abc=1$ then we have:\n\\[ a^3+b^3+c^3+\\frac{2ab}{a^2+b^2}+\\frac{2ac}{a^2+c^2}+\\frac{2bc}{b^2+c^2} \\ge 6 \\][/quote]\r\n\r\nYou should learn $\\LaTeX$, it's easy!", "Solution_2": "[quote=\"fermat3\"]...I can be solved by sos but i want another solution.[/quote]\r\n\r\nYou got it!\r\n\r\n$a^3+b^3+c^3+\\frac{2ab}{a^2+b^2}+\\frac{2ac}{a^2+c^2}+\\frac{2bc}{b^2+c^2}$\r\n\r\n$= (a^3+b^3+c^3)+\\frac{2}{(a^2+b^2)c}+\\frac{2}{(a^2+c^2)b}+\\frac{2}{(b^2+c^2)a}$\r\n\r\n$\\ge (a^3+b^3+c^3)+ \\frac{18}{\\sum_{sym}a^2b}$\r\n\r\n$\\ge 2\\sqrt{(a^3+b^3+c^3)\\cdot \\frac{18}{\\sum_{sym}a^2b}} \\ge 6$, since $2(a^3+b^3+c^3) \\ge{\\sum_{sym}a^2b}$.", "Solution_3": "Here's my solution.\r\n\r\nWe knew the well-known inequality:\r\n\\[ \\frac{a^3+b^3}{2} \\ge \\left(\\frac{a^2+b^2}{2}\\right)^{\\frac{3}{2}} \\]\r\nThen:\r\n$LHS \\ge \\sum_{cyc} \\left( \\frac{2ab}{a^2+b^2}+\\left(\\frac{a^2+b^2}{2}\\right)^{\\frac{3}{2}} \\right) \\ge \\sum_{cyc} \\left( 2\\sqrt{(x^2+y^2)^{\\frac12} \\cdot \\frac{2ab}{\\sqrt{2^3}}} \\right) \\ge \\sum_{cyc} \\left( 2 \\sqrt{\\sqrt{2xy}\\cdot \\frac{2}{2\\sqrt{2}} ab}\\right) = 2\\sum_{cyc} (ab)^{\\frac34}$\r\n$\\implies LHS \\ge 6$ by AM-GM and $abc=1$\r\nQ.E.D." } { "Tag": [], "Problem": "Hello,\r\nDoes anybody know how class rank is determined?\r\nthanks \r\n:)", "Solution_1": "Class rank is determined by two factors: your grades and the difficulty of the course your are taking. \r\n\r\nFor AP/GT courses, an A earns you 6 points, a B earns you 5 points, and so on. \r\n\r\nFor honors courses, an A earns you 5 points, a B earns you 4 points, and so on.\r\n\r\nAnd for standard courses, an A earns you 4 points. \r\n\r\nYour class rank is found by averaging your final(?) grades of each year and comparing that with your peers. \r\n\r\n-----------\r\n\r\nThat's how my high school does class rankings, but yours might be different, as the difficulty of the courses might have different points.", "Solution_2": "Some schools do GPAs with 100 being the most possible in the lower leveled classes and higher leveled classes able to earn above that. Aside from scale differences, it is based on grades earned in classes taken.", "Solution_3": "Each high school sets its own rules. Some high schools do not rank students." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Are there any two 8-digit numbers (which have the same digits but in different order), whose sum is equal to 20062007?", "Solution_1": "$ 10031089,10030918$" } { "Tag": [ "geometry", "algebra", "polynomial", "induction", "analytic geometry", "invariant", "advanced fields" ], "Problem": "Show that $ d\\plus{}1$ distinct points $ p_1,...,p_{d\\plus{}1} \\in \\mathbb{A}^n(\\mathbb{Q})$ always impose independent conditions on polynomials in $ P_{n,d}$\r\n\r\n[quote]Definition:\nGiven $ S$ $ \\subset \\mathbb{A}^n(k)$, the number of conditions imposed by $ S$ on polynomials of degree $ \\le d$ is defined $ C_d (S): \\equal{} dim P_{n,d} \\minus{} dim I_d (S)$, where $ I_d (S)$ is the vector space of polynomials of degree $ \\le d$ vanishing at each of the points.\n$ S$ is said to imposed independent conditions on $ P_{n,d}$ if $ C_d (S) \\equal{} |S|$[/quote]", "Solution_1": "Try induction on $ n$: For $ n \\equal{} 1$ you get $ I_1(S) \\equal{} 0$, because every polynomial of degree $ \\le d$ vanishing at $ d \\plus{} 1$ points vanishes identically. $ \\dim P_{1,d} \\equal{} d \\plus{} 1$ is easy.\r\n\r\nAssume that the claim is true for $ n \\minus{} 1$. Given $ q_1,\\ldots,q_{d \\plus{} 1}\\in\\mathbb Q^n$ we need to show that the linear map $ \\Phi_{n,d}^{(q_i)}: P_{n,d}\\rightarrow \\mathbb Q^{d \\plus{} 1}, f\\mapsto (f(q_i))_{1\\le i\\le d \\plus{} 1}$ is surjective, because $ I_{n,d} \\equal{} \\ker(\\Phi_{n,d}^{(q_i)})$ and $ C_d((q_i)) \\equal{} \\dim P_{n,d} \\minus{} \\dim I_{n,d} \\equal{} rk(\\Phi_{n,d}^{(q_i)})$.\r\n\r\nWlog we can assume that the $ q_i$'s are in sufficiently general position, i.e. that the projection $ \\pi_n: \\mathbb Q^{n}\\to\\mathbb Q^{n \\minus{} 1}$ dropping the last coordinate produces $ d \\plus{} 1$ pairwise different points $ q'_i \\equal{} \\pi_n(q_i)$. If this is not the case, then decompose $ \\mathbb Q^n$ in a hyperplane and a one-dimensional subspace such that the induced projection on the hyperplane produces pairwise different points. After a linear base change we can assume that the $ q_i$'s are in sufficiently general position. Note that the assertion to be proven is invariant under linear base changes such that everything is fine after our little manipulation.\r\n\r\nNow the induction step is easy: Consider the restriction of $ \\Phi_{n,d}^{(q_i)}$ to $ P_{n \\minus{} 1,d}$ which coincides with $ \\Phi_{n \\minus{} 1,d}^{(q_i')}$, but this map is surjective by the induction hypothesis." } { "Tag": [ "\\/closed" ], "Problem": "Hi,\r\n\r\nIf I check the option log in automatically, it loges in, but next time I enter the site:\r\n\r\n[u]First:[/u] It doesnt log in automatically, \r\n[u]Also[/u]: It even doesn't want to log in(when I enter username/password I see the first page http://www.mathlinks.ro/ without the \"Welcome FOURRIER!\") , so I get obliged to use Internet Explorer for some time...\r\n\r\nThank you for helping me.\r\n\r\n :)", "Solution_1": "Do you have cookies enabled? What version of Opera are you using?", "Solution_2": "Yes, cookies are enabled,\r\nMy opera version is 9.23.8808.0 .", "Solution_3": "I can't recreate the bug. It must be something in your browsers' settings I can only assume. \r\nAs a suggestion: try downloading Opera 9.24 (which is the latest stable version) and reinstalling it. That might fix the problem." } { "Tag": [ "function", "algebra", "domain", "trigonometry", "symmetry" ], "Problem": "For each function state the:\r\n-domain\r\n-range \r\n-is it one to one \r\n-what symettry it has (y-axis, x-axis, origin, or none)\r\n-is it even, odd or neither?\r\n\r\n1. $ y \\equal{} x^3\\minus{}5x$\r\n2. $ y \\equal{} \\sqrt{x^2\\minus{}4}$\r\n3. $ y \\equal{} \\frac{2}{x\\plus{}3}$\r\n4. $ y \\equal{} 2 \\cos{x}$", "Solution_1": "[hide=\"#1\"]\nDomain: All real numbers.\nRange: (-Infinity, Infinity)\nNot one-to-one\nOrigin symmetry\nOdd\n[/hide]\n\n[hide=\"#2\"]\nDomain: All real numbers >= 2\nRange: [0, Infinity)\nOne-to-one\nY-axis symmetry\nEven\n[/hide]\n\n[hide=\"#3\"]\nDomain: All real numbers except for -3\nRange: (-Infinity, 0) or (0, Infinity)\nOne-to-one\nOrigin symmetry\nOdd\n[/hide]\n\n[hide=\"#4\"]\nDomain: All real numbers.\nRange: [-2, 2]\nNot one-to-one\nY-axis symmetry\nEven\n[/hide]" } { "Tag": [ "inequalities", "algebra", "polynomial", "trigonometry", "induction", "inequalities proposed" ], "Problem": "this is again a wonderful problem.\r\n\r\nlet $-1\\leq p_i\\leq 1$, $i=1,...,n$ be real numbers. define $T_k=\\prod_{i=1,i\\neq k}^{n} |p_i-p_k|$. prove that\r\n\r\n\\[\\sum_{k=1}^{n} \\frac 1{T_k}\\geq 2^{n-2}\\]\r\n\r\nwhen do we have equality(i ask this question since once you've answered this question you know where to go in order to prove it)?", "Solution_1": "I have already posted a solution for this problem, using polynomials. I agree, great problem!", "Solution_2": "Would you mind posting it one more time?\r\nI don't find it, but I would love to see the solution to this beautiful problem", "Solution_3": "i guess harazi's solution is the same as mine, so here it goes:\r\n\r\nfirst assume that $1\\geq p_1>...>p_n\\geq -1$. define\r\n\r\n$P(x)=\\sum_{k=1}^{n} \\prod_{i=1,i\\neq k}^{n} \\frac{x-p_i}{|p_i-p_k|}$.\r\n\r\nthe polynomial $P(x)$ has degree $n-1$, leading coefficient $\\sum_{k=1}^{n} \\frac 1{T_k}$ and $P(p_i)=(-1)^i$. this last condition is the key to proving that the leading coefficient has to be quite large. now, we need some trick to show that it is actually at least $2^{n-2}$. define\r\n\r\n$Q(x)=\\cos( (n-1)\\arccos x )$ for $-1\\leq x\\leq 1$ (the n-1'th chebyshev polynomial).\r\n\r\nit is quite easy to see by induction that this is actually a polynomial of degree $n-1$ in $x$ with leading coefficient $2^{n-2}$. it obviously has the wonderful property that $|Q(x)|\\leq 1$ for $-1\\leq x\\leq 1$. now it's quite clear what we have to do: define\r\n\r\n$R(x)=P(x)-Q(x)$.\r\n\r\nwe want to show that it has a nonnegative leading coefficient. we know that $R(x)$ has degree $n-1$ and since $P(p_i)=(-1)^i$ and $|Q(p_i)|<1$ after slightly changing the variables we know that there are zeroes between $p_1$ and $p_2$, $p_2$ and $p_3$, ..., $p_{n-1}$ and $p_n$, in total $n-1$ real zeroes in $[-1; 1]$. therefore there are no other zeroes and since $P(p_1)=1>Q(p_1)$ this inequality holds for all $x\\geq p_1$. this gives the estimate of the leading coefficients, QED." } { "Tag": [ "analytic geometry", "probability", "expected value", "probability and stats" ], "Problem": "Assume that in the beginning we are at the origin of $\\mathbb{Z}^{n}$. At every step we move randomly one unit in the coordinate system, i.e. we move from the point $(z_{1},z_{2},\\cdots,z_{i},\\cdots,z_{n})$ to the point $(z_{1},z_{2},\\cdots,z_{i}\\pm 1,\\cdots,z_{n})$ for some $1 \\leq i\\leq n$ and the probability that we choose a specific index $i$ is $1/n$ for every $1\\leq i\\leq n$. Also with probability $1/2$ we decrease the index and with the same probability we increase it. Calculate the probability $P(n,m,q)$ such that after $m$ steps in $\\mathbb{Z}^{n}$ we are $q$ steps away from the origin.\r\n\r\n(I don't know even a solution for the case $q=0$)", "Solution_1": "The exact formula is a mess, but the central limit theorem can be used to do approximations for large $m.$", "Solution_2": "If you know $P(1, m, q)$, you can get $P(n, m, q)$ by summing $P(1, a_{1}, b_{1})\\cdot P(1, a_{2}, b_{2}) \\cdot\\cdots\\cdot P(1, a_{n}, b_{n})$ over all solutions in non-negative integers to $a_{1}+a_{2}+\\ldots+a_{n}= m$ and $b_{1}+b_{2}+\\ldots+b_{n}= q$. For fixed small values of any of the three variables, this is something that maybe you could evaluate and not find horrific.", "Solution_3": "There are some other very interesting questions to ask. One such question: what is the probability of the walk eventually returning to the origin? Or (closely related), what is the expected number of returns to the origin over the lifetime of the walk? For those questions, it very much matters what $n,$ the dimension, is." } { "Tag": [], "Problem": "[b][i][/i][/b]", "Solution_1": "Okay... sure....\r\n\r\nI'm from washington.", "Solution_2": "i was borni n washington!!!", "Solution_3": "I was born in washington and I'm in there right now!!! i'm doing aops during math class...heehee...my teacher doesn't care... :lol: :P", "Solution_4": "This forum... is like, dead. I'm from Washington... lived here... for like my whole life.\r\n:P" } { "Tag": [ "factorial", "rational numbers" ], "Problem": "Prove that numbers of the form \\[\\frac{a_{1}}{1!}+\\frac{a_{2}}{2!}+\\frac{a_{3}}{3!}+\\cdots,\\] where $0 \\le a_{i}\\le i-1 \\;(i=2, 3, 4, \\cdots)$ are rational if and only if starting from some $i$ on all the $a_{i}$'s are either equal to $0$ ( in which case the sum is finite) or all are equal to $i-1$.", "Solution_1": "I just wonder how this can be correct... if all are equal to $ i\\minus{}1$ gives rational number $ q\\in\\mathbb{Q}$, then putting all equal to $ \\frac{i\\minus{}1}{2}$ instead simply gives you $ \\frac{1}2 q\\in\\mathbb{Q}$...", "Solution_2": "[quote=\"Peter\"]I just wonder how this can be correct... if all are equal to $ i\\minus{}1$ gives rational number $ q\\in\\mathbb{Q}$, then putting all equal to $ \\frac{i\\minus{}1}{2}$ instead simply gives you $ \\frac{1}{2} q\\in\\mathbb{Q}$...[/quote]\r\n\r\nI don't have the book right now, I'm gonna check it later!", "Solution_3": "[quote=\"ideahitme\"][quote=\"Peter\"]I just wonder how this can be correct... if all are equal to $ i \\minus{} 1$ gives rational number $ q\\in\\mathbb{Q}$, then putting all equal to $ \\frac {i \\minus{} 1}{2}$ instead simply gives you $ \\frac {1}{2} q\\in\\mathbb{Q}$...[/quote]\n\nI don't have the book right now, I'm gonna check it later![/quote]Have you had time to check already? :)", "Solution_4": "The $ a_i$ have to be integers (everything else cannot make sense!). And then this can be proved by mimicing the proof that $ e$ is irrational.", "Solution_5": "[quote=\"Peter\"]I just wonder how this can be correct... if all are equal to $ i \\minus{} 1$ gives rational number $ q\\in\\mathbb{Q}$, then putting all equal to $ \\frac {i \\minus{} 1}{2}$ instead simply gives you $ \\frac {1}2 q\\in\\mathbb{Q}$...[/quote]\r\n$ \\frac {i \\minus{} 1}{2}$ is not integer for even $ i$ therefore we can not put all $ a_{i} \\equal{} \\frac {i \\minus{} 1}{2}$\r\nsorry :oops: when I begin writing this post,zetaX dosen't post...", "Solution_6": "[quote=\"ZetaX\"]The $ a_i$ have to be integers (everything else cannot make sense!). [/quote]That makes sense, thanks!", "Solution_7": "Actually, what you probably want to do first is to prove that every rational number has a unique \"factorial base\" expansion. The rest of the proof should then proceed analogously to the proof for rational numbers in base $ b$." } { "Tag": [ "FTW" ], "Problem": "How many distinct ordered pairs of positive integers $ (m,n)$ are there so that the sum of the reciprocals of $ m$ and $ n$ is $ \\frac14$?", "Solution_1": "In other words, $ \\frac {1}{m} \\plus{} \\frac {1}{n} \\equal{} \\frac {1}{4}$\r\n$ \\frac {1}{m} \\plus{} \\frac {1}{n} \\equal{} \\frac {1}{4} \\equal{} \\frac {m \\plus{} n}{mn}$\r\nThus $ mn \\minus{} 4m \\minus{} 4n \\equal{} 0$.\r\n$ mn \\minus{} 4m \\minus{} 4n \\plus{} 16 \\equal{} (m \\minus{} 4)(n \\minus{} 4) \\equal{} 16$\r\n\r\nThere are 3 unique ways of expressing 16 as a product of two positive integers: 1*16, 2*8, and 4*4.\r\n\r\nTherefore, [b]3[/b].\r\n\r\n(in case you were wondering they correspond to (5,20),(6,12),(8,8))", "Solution_2": "I believe the answer is $ 5$--minor technicality.", "Solution_3": "He is referring to the fact that (5,20) and (20,5) both work.", "Solution_4": "i was doing this problem on FTW and it said that the answer is 5", "Solution_5": "The answer is supposed to be $5$ because we can have $m - 4 = 1,2,4,8,16$; there are $5$ factors of $16$ and $m - 4$ has a one-to-one correspondence with $m$ so $\\boxed{5}$ should be the answer." } { "Tag": [ "AMC 12 B", "AMC 10", "AMC 10 B", "AMC" ], "Problem": "I'm trying to get my school to register for the AMC 12 B, and I have a few questions about it.\r\n\r\n1. Who can proctor the test? Can it be any adult, like a parent, or does it have to be someone associated with the school?\r\n\r\n2. Is there a time period that the test has to be taken within? Or can it be taken any time on the official date?\r\n\r\nThanks. Any suggestions for convincing reluctant schools (we tried to take the B test last year, and failed...) would also be really appreciated. (Taking the AMC at another school won't be possible.)", "Solution_1": "[quote=\"RenaudianOntology\"]1. Who can proctor the test? Can it be any adult, like a parent, or does it have to be someone associated with the school?[/quote]\n\nAny responsible adult can proctor the test, except someone who is related to a test-taker. IT MUST be given in a public building; if you can't do it in your school, you must do it in a library or other public building.\n\n[quote=\"RenaudianOntology\"]2. Is there a time period that the test has to be taken within? Or can it be taken any time on the official date?[/quote]\r\n\r\nThere is no set time; any time on Feb 16 is fine (but it must be on that day).\r\n\r\n\r\nKeep in mind that the latest possible registration date is Feb 7. For more information, see the [url=http://www.unl.edu/amc/index.html]AMC website[/url].", "Solution_2": "I suggest you go to the AMC Website \r\nhttp://www.unl.edu/amc\r\nthen on the right side, about mid-way under What's New, choose the link for the 2005 AMC 10 A/AMC 12 A Teachers' Manual. The rules, instructions and policy statements are all there, starting on about page 5.\r\n\r\nThe 2005 AMC 10 B/ AMC 12 B Teachers' Manual will be identical except for date changes and contest name changes.\r\n\r\nThe direct link to jump there now is:\r\nhttp://www.unl.edu/amc/d-publication/d1-pubarchive/2004-5pub/05TM1012/TM05hs.html\r\n\r\nSteve Dunbar\r\nAMC Director", "Solution_3": "Can someone explain instruction #1? \"Take the unopened contest package(s) and Certification Form to your Principal (or equivalent) to certify that the package was opened within an hour before the contest. Have the Principal sign the Certification Form at that time.\"\r\n\r\nDoes it really have to be the principal, or could it be a math teacher or math department head? The school principal isn't here very often, and I'm not sure what to do if we didn't take the exam during the school day.\r\n\r\nThanks.", "Solution_4": "The person who kept custody of the test forms might not in all cases be the school's principal, but it should be SOME identifiable person who signs the form to indicate that no one else had access to the test materials until the test was administered. The person to sign the form is the person who would take responsibility for making that statement truthfully if there were any doubts about the security of the test.", "Solution_5": "I haven't yet found a proctor for the B exam, but went to the online registration form anyways just to see what was on it, and all it shows me is that my school has already registered for the A exam. It doesn't show any form for a B exam. Does this mean I can't register online, or is there some way to get the B form?\r\nAlso, if I can somehow manage to register online, is the latest time to register 11:59 pm on February 7th? (The deadline is February 7th.)\r\n\r\nThanks", "Solution_6": "If you try registering online for the B test, after already having registered for the A test, it won't let you. You have to fax in your registration. And either put credit card number (I think) or if you're paying by check, that the check is in the mail (That's what the AMC people told me to do.) I'd fax it in today...The latest it would probably be okay would be 5:00PM Central time on Monday, if you really want to do it at the latest possible moment..." } { "Tag": [ "geometry", "probability", "FTW", "Alcumus" ], "Problem": "In AoPS City, there is only one area of trees, so if there is a forest fire, all the trees will catch on fire. The probability of a tree in AoPS City surviving in the year 2009 is 13/14. The probability of there being a forest fire is 1/52. The probability of a tree dying if there is a forest fire is 1/1. The probability of a tree in AoPS City dying by some cause other than a forest fire is 3/52. Trees A and B are both in AoPS City. What is the probability that tree A will die, but tree B will not die in 2009?", "Solution_1": "All of the rest of the stuff is filler information. We don't care how or why a tree dies, just whether or not it does. We quickly pick up on the key sentence:\r\n\r\nThe probability of a tree in AoPS City surviving in the year 2009 is 13/14.\r\n\r\nThus the probability of A dying is $ 1\\minus{}\\frac{13}{14}\\equal{}\\frac{1}{14}$\r\n\r\nThe probability of B living is $ \\frac{13}{14}$\r\n\r\nTherefore the probability of A dying and B living is $ \\frac{1}{14}\\cdot\\frac{13}{14}\\equal{}\\frac{1\\cdot13}{14\\cdot14}\\equal{}\\boxed{\\frac{13}{196}}$", "Solution_2": "That is correct. I tried to include alot of irrevelant information to throw you guys off.", "Solution_3": "Wait, so why did you post this?", "Solution_4": "I believe it was an exercise for other people to solve.", "Solution_5": "Yeah, that's why it is called warm-up. Otherwise I would have titled it Help!!!.", "Solution_6": "Quick Q. Im new here and i have a couple questions. \r\n1. How do you make a post?\r\n2. What are some of the basic controls on this site?\r\n3. Which links should i click on in the beginning?\r\n\r\nTY 2 all the ppl that answer. And sorry if this thing bothers u on ur post :maybe:", "Solution_7": "1. To make a new thread, click New Topic, in any one of the forums. (there are many!)\r\n\r\n2. Some basic places you should go to are FTW (a multiplayer game), Alcumus (single player), and maybe some AoPS classes.\r\n\r\n3. All the main stuff is on the left side above the graphic.", "Solution_8": "[quote=\"Cheese12345\"]Quick Q. Im new here and i have a couple questions. \n1. How do you make a post?\n2. What are some of the basic controls on this site?\n3. Which links should i click on in the beginning?\n\nTY 2 all the ppl that answer. And sorry if this thing bothers u on ur post :maybe:[/quote]\r\nIt won't bother anyone =]\r\nNext time, (goes for everyone too) if you have a question about the Forum in general, please post it in the Questions, Suggestions, and Announcements part of the Forum. (This will avoid lots of spam in the future.)\r\nThanks, and enjoy AoPS =]", "Solution_9": "Your problem has a problem. Like you said, the probability that a tree dies is 1/14. However, in the rest of the problem you say that the probability that it dies by fire is 1/52 and by any other cause is 3/52. However, 1/52 + 3/52 = 4/52, 4/52 = 1/13, and 1/13 is not equal to 1/14. Its fine to put filler info, but it should be correct.", "Solution_10": "[quote=\"myyellowducky82\"][quote=\"Cheese12345\"]Quick Q. Im new here and i have a couple questions. \n1. How do you make a post?\n2. What are some of the basic controls on this site?\n3. Which links should i click on in the beginning?\n\nTY 2 all the ppl that answer. And sorry if this thing bothers u on ur post :maybe:[/quote]\nIt won't bother anyone =]\nNext time, (goes for everyone too) if you have a question about the Forum in general, please post it in the Questions, Suggestions, and Announcements part of the Forum. (This will avoid lots of spam in the future.)\nThanks, and enjoy AoPS =][/quote]\r\n\r\nHe didn't know how to make a new topic.", "Solution_11": "[quote=\"12markkram34\"]Your problem has a problem. Like you said, the probability that a tree dies is 1/14. However, in the rest of the problem you say that the probability that it dies by fire is 1/52 and by any other cause is 3/52. However, 1/52 + 3/52 = 4/52, 4/52 = 1/13, and 1/13 is not equal to 1/14. Its fine to put filler info, but it should be correct.[/quote]\r\n\r\nOh, actually I was concentrating on it. \r\n\r\n:fool: I keep forgetting that 4(14) is 56. Sorry. Now please stop posting on this thread." } { "Tag": [ "ratio", "geometry", "3D geometry", "sphere" ], "Problem": "A cylinder is inside a cube, which is inside a sphere, which is inside a cube, which is inside a sphere. If the smaller sphere has radius $x,$ what is the ratio of the volumes of the bigger sphere and the cylinder?", "Solution_1": "[quote=\"i_like_pie\"]A cylinder is inside a cube, which is inside a sphere, which is inside a cube, which is inside a sphere. If the smaller sphere has radius $x,$ what is the ratio of the volumes of the bigger sphere and the cylinder?[/quote]\r\n\r\n[hide]Becuase the radius of the sphere is x, the distance from the center of the smaller cube to one of its corners is also x. Therefore, the distance from the center of the cube to the center one of its faces is x/(3^(1/2)). This is also the radius of the cylinder. Its hight is twice this or 2x/(3^(1/2)). This means that its volume is 2pi*x^3*3^(1/2)/9.\n\nThe radius of the sphere is also the distance from the center of the larger cube to one of its faces. Therefore, the distance from the center of this cube to one of its corners is x*3^(1/2). This is also the radius of the larger sphere. Thus, the volume of the larger sphere is 4pi*x^3*3^(1/2).\n\nTherefore the ratio of the volume of the larger sphere to that of the cylinder is 2:1/9 or 18:1.[/hide]", "Solution_2": "[hide]We write all the important lengths and volumes in terms of x:\nSide length of larger cube: 2x\nSpace diagonal of larger cube:$2\\sqrt{3}x$\nRadius of larger sphere: $\\sqrt{3}x$\nVolume of the larger sphere: $\\frac{4}{3}\\pi(x\\sqrt{3})^{3}=4\\pi\\sqrt{3}x^{3}$\n\nSpace Diagonal of smaller cube: 2x\nSide length of smaller cube: $\\frac{2x}{\\sqrt{3}}$\nHeight of cylinder: $\\frac{2x}{\\sqrt{3}}$\nRadius of cylinder: $\\frac{X}{\\sqrt{3}}$\nVolume of cylinder: $(\\frac{x}{\\sqrt{3}})^{2}\\pi\\times (\\frac{2x}{\\sqrt{3}})=\\frac{2x^{3}\\pi}{3\\sqrt{3}}$\n\nThen the ratio of the volumes is $4\\pi\\sqrt{3}x^{3}: \\frac{2x^{3}\\pi}{3\\sqrt{3}}$, or $18: 1$.[/hide]", "Solution_3": "You're both correct. :)\r\n\r\n@ samath - You forgot the last } of \\frac for volume of cylinder." } { "Tag": [ "search", "function", "combinatorics unsolved", "combinatorics" ], "Problem": "How many ways are there to write either $ 0$ or $ 1$ in each cell of a $ 4$x$ 4$ board so that the product of numbers in any two cells sharing an edge is always $ 0$?", "Solution_1": "Posted recently (and with a source listed, even!). Search functions are wonderful things.\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=444709738&t=169651", "Solution_2": "I solved it but I can't write it because I used very many diagrams \r\nthe answer is 1234" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "If $ a,b,c$ are positive real numbers such that $ abc\\equal{}1$, then\r\n\r\n$ \\frac {12a\\plus{}7}{2a^2\\plus{}1}\\plus{}\\frac {12b\\plus{}7}{2b^2\\plus{}1}\\plus{}\\frac {12c\\plus{}7}{2c^2\\plus{}1} \\le 19$.", "Solution_1": "[quote=\"Vasc\"]If $ a,b,c$ are positive real numbers such that $ abc \\equal{} 1$, then\n\n$ \\frac {12a \\plus{} 7}{2a^2 \\plus{} 1} \\plus{} \\frac {12b \\plus{} 7}{2b^2 \\plus{} 1} \\plus{} \\frac {12c \\plus{} 7}{2c^2 \\plus{} 1} \\le 19$.[/quote]\r\nIt is just equivalent to your previous inequality, VasC:\r\nIf $ x,y,z>0,xyz\\equal{}1$, then\r\n\\[ \\sum \\frac{7\\minus{}6x}{x^2\\plus{}2} \\ge 1\\]\r\nIn here $ x\\equal{}\\frac{1}{a},y\\equal{}\\frac{1}{b},z\\equal{}\\frac{1}{c}$ ;)", "Solution_2": "You are right. :oops:", "Solution_3": "[quote=\"can_hang2007\"][quote=\"Vasc\"]If $ a,b,c$ are positive real numbers such that $ abc \\equal{} 1$, then\n\n$ \\frac {12a \\plus{} 7}{2a^2 \\plus{} 1} \\plus{} \\frac {12b \\plus{} 7}{2b^2 \\plus{} 1} \\plus{} \\frac {12c \\plus{} 7}{2c^2 \\plus{} 1} \\le 19$.[/quote]\nIt is just [b]equivalent to your previous inequality[/b], VasC:\nIf $ x,y,z > 0,xyz \\equal{} 1$, then\n\\[ \\sum \\frac {7 \\minus{} 6x}{x^2 \\plus{} 2} \\ge 1\n\\]\nIn here $ x \\equal{} \\frac {1}{a},y \\equal{} \\frac {1}{b},z \\equal{} \\frac {1}{c}$ ;)[/quote]\r\nCan you show me the previous inequality of VasC.", "Solution_4": "Is my proof right? :lol: \r\n$ \\frac {12a\\plus{}7}{2a^2\\plus{}1}\\plus{}\\frac {12b\\plus{}7}{2b^2\\plus{}1}\\plus{}\\frac {12c\\plus{}7}{2c^2\\plus{}1} \\le 19$\r\n$ \\sum\\limits_{cyc} {9 \\minus{} \\frac{{12a \\plus{} 7}}{{2a^2 \\plus{} 1}}} \\ge 8$\r\n$ \\sum\\limits_{cyc} {\\frac{{2(3a \\minus{} 1)^2 }}{{2a^2 \\plus{} 1}}} \\ge 8$\r\n$ \\sum\\limits_{cyc} {\\frac{{(3a \\minus{} 1)^2 }}{{2a^2 \\plus{} 1}}} \\ge 4$\r\nBy Holder\r\n$ LHS \\ge \\frac{{(3a \\plus{} 3b \\plus{} 3c \\minus{} 3)^2 }}{{2a^2 \\plus{} 2b^2 \\plus{} 2c^2 \\plus{} 3}}$\r\nSo we need\r\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} 18(a \\plus{} b \\plus{} c) \\plus{} 18(ab \\plus{} bc \\plus{} ac) \\minus{} 3 \\ge 0$ which can be shown by Mixing-variable :lol:", "Solution_5": "Very nice and correct your solution, T.T. Hung. :lol:", "Solution_6": "[quote=\"TRAN THAI HUNG\"]Is my proof right? :lol: \n$ \\frac {12a \\plus{} 7}{2a^2 \\plus{} 1} \\plus{} \\frac {12b \\plus{} 7}{2b^2 \\plus{} 1} \\plus{} \\frac {12c \\plus{} 7}{2c^2 \\plus{} 1} \\le 19$\n$ \\sum\\limits_{cyc} {9 \\minus{} \\frac {{12a \\plus{} 7}}{{2a^2 \\plus{} 1}}} \\ge 8$\n$ \\sum\\limits_{cyc} {\\frac {{2(3a \\minus{} 1)^2 }}{{2a^2 \\plus{} 1}}} \\ge 8$\n$ \\sum\\limits_{cyc} {\\frac {{(3a \\minus{} 1)^2 }}{{2a^2 \\plus{} 1}}} \\ge 4$\nBy Holder\n$ LHS \\ge \\frac {{(3a \\plus{} 3b \\plus{} 3c \\minus{} 3)^2 }}{{2a^2 \\plus{} 2b^2 \\plus{} 2c^2 \\plus{} 3}}$\nSo we need\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} 18(a \\plus{} b \\plus{} c) \\plus{} 18(ab \\plus{} bc \\plus{} ac) \\minus{} 3 \\ge 0$ which can be shown by Mixing-variable :lol:[/quote]\r\nVery nice your solution. :lol:" } { "Tag": [ "search" ], "Problem": "Prove that if $ p$ is a prime number,then $ 7p \\plus{}3^{p} \\minus{}4$ is not a perfect square.(JBMO 2007)", "Solution_1": "When posting a problem from a national, regional or international competition, please check whether that problem has already been posted [see the Resources/Contests section!]\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?p=874754", "Solution_2": "Hi,Thank you hsiljak,I've searched but I couldn't find it .I think the name of the topic should presents information about the problem inside.\r\nHow could I find the problem from JBMO ,when the name of the topic you've posted was \"prime number implies non-square number\"", "Solution_3": "As a matter of fact, the name of that topic I linked to could be used as a typical example of [b]good titling[/b].\r\n\r\n[quote=\"pohoatza\"]prime number implies non-square number\nJBMO 2007, Bulgaria, problem 4\n[/quote]\r\n\r\nThe title of the topic tells about the type of the problem inside, and the subtitle informs us about the source. You could have simply used the '[url=http://www.mathlinks.ro/search.php]Search[/url]' button to find it (keyword could have been [i]jbmo 2007[/i]), or even better:\r\n\r\n[url=http://www.mathlinks.ro/resources.php]Contests/Resources[/url] -> [url=http://www.mathlinks.ro/resources.php?c=1&cid=21]Junior Balkan MO[/url] -> [url=http://www.mathlinks.ro/resources.php?c=1&cid=21&year=2007]2007[/url]->[url=http://www.mathlinks.ro/viewtopic.php?p=874754]Problem 4[/url]", "Solution_4": "Good day,\r\nsorry for asking this question here, bt i could not find an appropriate place.\r\nI need the contact information for 13th junior balkan mathematics olympiad. I mean, the contact information of the organisators of the competition.\r\ni will be very happy if you can help me.\r\nThank you.", "Solution_5": "Bosnia and Herzegovina is the host, maybe hsiljak knows more.\r\nI have also one question: will there be a web site for JBMO 09? Last year there wasn't one...", "Solution_6": "Samanyolu was PMed with something that looks like contact info (people who know how organized Bosnian Mathematical Society is, know why it's hard to give any kind of information about them :lol: ). And as far as I can expect, there could and should be some sort of website- i don't expect it to be something extremely special, but it should list the problems and results." } { "Tag": [ "geometry", "modular arithmetic", "geometric transformation", "reflection", "homothety", "circumcircle", "geometry unsolved" ], "Problem": "Let $ D. E, F$ be the feet of the angle bisectors of angles $ A, B, C$, respectively, of triangle $ ABC$, and let $ K_a, K_b, K_c$ be the points of contact of the tangents to the incircle of $ ABC$ through $ D, E, F$ (that is, the tangent lines not containing sides of the triangle). \r\n\r\nProve that the lines joining $ K_a, K_b, Kc_$ to the midpoints of $ BC, CA. AB$ respectively, pass through a single point on the incircle of $ ABC$.", "Solution_1": "[hide=\"Hint\"]The currency point is also the [b]Feurebach point[/b] wrt $ \\triangle ABC$ due to the fact that $ \\triangle K_aK_bK_c\\sim \\triangle M_aM_bM_c$, where $ M_a,M_b,M_c$ respectively are the midpoints of $ BC,CA,AB$. I will be back with more detailed solution.[/hide]\r\n\r\nBack with a full solution.\r\n\r\nLet $ A_0,B_0,C_0$ be the tangency points of $ (I)$ with the triangle sides $ BC,CA,AB$ respectively, where $ (I)$ is the incircle of $ \\triangle ABC$. Denote $ M_a,M_b,M_c$ respectively by the midpoints of $ BC,CA,AB$. We have $ (IB_0,IK_a)\\equiv (IB_0,IA_0) \\plus{} (IA_0,IK_a)$ $ \\equiv (CB_0,CA_0) \\plus{} (DA_0,DK_a)$ $ \\equiv (CB_0,CA_0) \\plus{} 2(DA_0,DA)$ $ \\equiv (CA,CB) \\minus{} 2(DA,AC) \\minus{} 2(CA,CD)$ $ \\equiv (CA,CB) \\minus{} (AB,AC) \\minus{} 2(CA,CB)\\equiv (BC,BA) \\pmod \\pi$. Hence $ (IB_0,IK_a)\\equiv (BC,BA)$. With the same argument, we also have $ (IB_0,IK_c)\\equiv (BA,BC)\\Longrightarrow (IB_0,IK_c)\\equiv \\minus{} (IB_0,IK_a) \\pmod \\pi$, which implies that $ K_c\\mapsto K_a$ through the reflection by $ IB_0$. Therefore $ K_aK_c\\bot IB_0\\Longrightarrow K_aK_c\\parallel{}M_aM_c$. Argue the same we also have $ M_aM_b\\parallel{}K_aK_b$, $ M_bM_c\\parallel{}K_bK_c$. Hence $ \\triangle M_aM_bM_c \\sim \\triangle K_aK_bK_c$. Moreover, $ K_aM_a,K_bM_b,K_cM_c$ are concurrent at the point $ F$ such that $ \\triangle M_aM_bM_c\\mapsto \\triangle K_aK_bK_c$ through a homothety with center $ F$, let us denote this homothety by $ \\mathcal {H}_{F}$. Hence $ \\mathcal {H}_{F}: (M_aM_bM_c)\\mapsto (K_aK_bK_c)$, or in other word, $ (\\mathcal {E})\\mapsto (I)$, where $ (\\mathcal {E})$ is the $ 9$- point circle wrt $ \\triangle ABC$. But we have already know that $ (I)$ externally tangents to $ (\\mathcal {E})$, which implies that $ F\\equiv (I)\\cap (\\mathcal {E})$. Therefore, $ K_cM_c,K_aM_a,K_bM_b$ are concurrent at a point $ F\\in (I)$. This $ F$ is also known as the [b]Feurebach point[/b] wrt $ \\triangle ABC$.\r\nOur proof is completed then $ \\square$", "Solution_2": "[quote=\"mathVNpro\"] $ \\triangle K_aK_bK_c\\cong \\triangle M_aM_bM_c$, where $ M_a,M_b,M_c$ respectively are the midpoints of $ BC,CA,AB$. [/quote] I believe that $ \\triangle K_aK_bK_c, \\triangle M_aM_bM_c$ are centrally similar, not congruent. :|", "Solution_3": "Actually,KcC,KaA,KbB,OI are concurrent", "Solution_4": "Denote KcF\u2229BC=L , KbE\u2229BC=K\r\nL is the reflection of A with respect to CF K is the reflection of A with respect to BE\r\nThus JL=JK => KcKb//LK => KcKb//BC\r\nAnalogously, KcKa//CA KaKb//AB\r\n\u22bfKcKaKb\u223d\u22bfCAB \r\nKcC,KaA,KbB,OI are concurrent at their inner similitude center\r\n\r\nBack to our problem:\r\nIt is obvious,that \u22bfMaMbMc\u223d\u22bfKcKaKb\u223d\u22bfCAB with corresponding sides parallel \r\nThus MaKa MbKb McKc are concurrent at Fe \r\nTherefore, IO pass through Fe\r\nNotice the circumcenter of \u22bfMaMbMb is the nine-point circle of \u22bfABC whose circumcenter is O\u2019 => Fe is the Feurebach point", "Solution_5": "[quote]Let $ D. E, F$ be the feet of the angle bisectors of angles $ A, B, C$, respectively, of triangle $ ABC$, and let $ K_a, K_b, K_c$ be the points of contact of the tangents to the incircle of $ ABC$ through $ D, E, F$ (that is, the tangent lines not containing sides of the triangle). \nProve that the lines joining $ K_a, K_b, Kc_$ to the midpoints of $ BC, CA. AB$ respectively, pass through a single point on the incircle of $ ABC$.[/quote]\r\nHere goes my solution. \r\nIt is enough to prove that $ \\triangle K_aK_bK_c$ and $ \\triangle PQR$ are centrally similar. ($ PQR$ is the medial triangle of $ ABC$). Because the center of homothety $ T$ of incircle and nine point circle is the Feuerbach point of $ \\triangle ABC$. So, If $ \\triangle K_aK_bK_c$ and $ \\triangle PQR$ are centrally similar, $ T$ will be their center of homothety and the problem will follow.\r\n\r\n$ X,Y,Z$ be the points where the incircle touch the sides of the triangle $ ABC$.\r\nNow we have\r\n\\begin{align*} \\angle K_cK_bK_a & = & \\angle K_cK_bX + \\angle XK_bY + \\angle YK_bK_a \\\\\r\n& = & \\left (\\frac \\pi2 - \\frac C2 - A\\right ) + \\left ( \\frac \\pi2 - \\frac B2\\right ) + \\left ( \\frac \\pi2 - \\frac A2 - C \\right ) & = & B \\end{align*}\r\nAnalogously chasing other angles we can say that $ \\triangle K_aK_bK_c \\sim \\triangle PQR$.\r\n\r\nNow we shall prove that, $ K_cK_a \\parallel RP$\r\nWe have $ \\measuredangle(IY,K_cK_a) = \\left (\\pi - A - 2C\\right ) + \\left (\\frac \\pi2 - B\\right ) = \\frac \\pi2 - B$ also $ \\measuredangle (IY,RP) = \\frac \\pi2 - B$.\r\nSo, $ K_cK_a \\parallel RP$ and the proof is complete. :)", "Solution_6": "Dear Mathlinkers,\r\nsee also for example :\r\nhttp://perso.orange.fr/jl.ayme vol. 4 Sym\u00e9triques de (OI) par rapport aux c\u00f4t\u00e9s des triangles de contact et m\u00e9dian p. 10\r\nSincerely\r\nJean-Louis", "Solution_7": "[b]REMARK-[/b] As [b]plane geometry[/b] have said that $ CK_c,AK_a,BK_b$ are concurrent is a special case of this problem:\r\n\r\n[b][color=darkblue]\"Let $ \\triangle ABC$ with $ (I)$ is its incircle. $ A_0,B_0,C_0$ are the tangency point of $ (I)$ with $ BC,CA,AB$ respsectively. Let $ A_1,B_1,C_1$ be the points on $ BC,CA,AB$ such that $ AA_1,BB_1,CC_1$ are concurrent. From $ A_1,B_1,C_1$, let $ A_2,B_2,C_2$ be the points on $ (I)$ such that $ AA_2,BB_2,CC_2$ are the second tangents from $ A_1,B_1,C_1$ to $ (I)$. Prove that: $ AA_2,BB_2,CC_2$ are concurrent.\"[/color][/b].", "Solution_8": "Let $ X,Y,Z$ be the tangency points of $(I)$ with $ BC,CA,AB.$ It is clear that $ YK_b \\parallel ZX$ and $ ZK_c \\parallel YX.$ Therefore, $ \\angle K_cZX \\equal{} \\angle ZXY \\equal{} \\angle K_bYX$ $ \\Longrightarrow$ $ K_cK_b$ is parallel to the tangent $ BC$ to $ (I)$ at $ X.$ By similar reasoning, this means that $ \\triangle K_aK_bK_c$ and $ \\triangle ABC$ are homothetic $\\Longrightarrow$ $ \\triangle K_aK_bK_c$ and the medial triangle $ \\triangle M_aM_bM_c$ are homothetic. Therefore, $ M_aK_a,$ $M_bK_b,$ $M_cK_c$ concur at their homothetic center, which is the exsimilicenter of their circumcircles $ (I)$ and the nine-point circle $(N),$ i.e. the Feuerbach point of $\\triangle ABC.$\n\n[hide=\"Another approach\"]Let $ X (0: p \\minus{} c: p \\minus{} b).$ Parallel through $ X$ to $ YZ$ meets the incircle $ (I)$ at \n\n$ K_a (( \\minus{} a \\plus{} b \\plus{} c)(b \\minus{} c)^2: (a \\minus{} b \\plus{} c)b^2: (a \\plus{} b \\minus{} c)c^2)$\n\nThen the line joining $ M_a(0: 1: 1)$ and $ K_a$ has equation\n\n$ (b^2 \\plus{} c^2 \\minus{} a(b \\plus{} c))x \\plus{} (b \\minus{} c)( \\minus{} a \\plus{} b \\minus{} c)y \\plus{} (a \\minus{} b \\minus{} c)(b \\minus{} c)z \\equal{} 0$ \n\nWhich clearly contains the Feuerbach point \n\n$ F_e (( \\minus{} a \\plus{} b \\plus{} c)(b \\minus{} c)^2: (a \\plus{} c \\minus{} b)(a \\minus{} c)^2: (a \\plus{} b \\minus{} c)(a \\minus{} b)^2)$ \n\nBy similar reasoning, we conclude that $ M_aK_a,M_bK_b,M_cK_c$ meet at $ F_e.$[/hide]", "Solution_9": "Let (O) be the circumcircle of triangle ABC. Ay is the tangency of (O). Mx is the tangency of the nine-point circle (E) of triangle ABC. G is the centroid of triangle ABC.\r\nBy angle charsing, it's easy to prove that $ DK_a//Ay$\r\nThe dilatation $ D^{\\frac { \\minus{} 1}{2}}_G$: $ A\\rightarrow M, B\\rightarrow N, C\\rightarrow P \\Rightarrow \\Delta ABC\\rightarrow \\Delta MNP$\r\nThen $ Ay//Mx$\r\nSo $ DK_a//Mx$\r\nSimilarly for $ EK_b, FK_c$ and we get $ K_aK_b//MN, K_bK_c//NP, K_aK_c//MP$\r\nAnd it's easy to prove that $ DK_a, EK_b, FK_c$ are concurrent at J.\r\nThe Dilatation $ D^{\\frac{r}{r_E}}_J$: $ M\\rightarrow K_a, N\\rightarrow K_b, P\\rightarrow K_c$ then J lies on IE (J is the tangent of (I) and (E))\r\n$ \\Rightarrow$ QED" } { "Tag": [ "email" ], "Problem": "Are we allowed to send in applications on May 1 or must it be before May 1?\r\n(online application)", "Solution_1": "I'd say get it in as soon as possible.", "Solution_2": "I think that you can send it in on May 1st as well, but I'm not sure.", "Solution_3": "Quote from an email from Administrative Coordinator John Woodman:\r\n\r\n[quote]You can proof it on May 1, but try to get it in asap.[/quote]\r\n\r\nSo they'll take it on May 1st. Hope this helps!", "Solution_4": "Ah. This is what I get for not reading the forum much during the end of application season.\r\n\r\nYes, May 1 is fine. :)", "Solution_5": "Nevermind, I got an answer, thanks." } { "Tag": [ "calculus", "integration", "function", "logarithms", "calculus computations" ], "Problem": "I need some help with this type of exercise. There are several of them, and I can't really solve a single one. Here's one that looks simple:\r\n\r\nFor what values of $a$ does the generalized integral $\\int_0^{\\infty}\\frac{\\ln(1+x^2)}{x^a}dx$ converge?\r\n\r\nYou need to find a good function $g$ to compare with, but no matter what I choose I get a limit which tends to 0, and isn't all that useful. How should I think?\r\n\r\nOther similar ones are: $\\int^{\\infty}_0\\frac{\\arctan x}{(1+x)^a(1+x^2)x^a}$, $\\int_0^{\\infty}\\frac{\\ln(1+x)}{x^a + x^{a+2}}$. I'm looking for the general idea of what to do. I am lost.", "Solution_1": "Look at your first example. That integral is improper for two reasons - it's potentially improper at zero, and it's improper at infinity. Break that into two pieces and consider each piece separately.\r\n\r\nAs $x\\to0,\\,\\ln(1+x^2)$ looks like $x^2.$ Hence at that end, we're comparing to\r\n\r\n$\\int_0^bx^{2-a}\\,dx.$ This converges when $2-a>-1,$ or $a<3.$\r\n\r\nAs $x\\to\\infty,\\,\\ln(1+x^2)$ looks like $2\\ln x.$ Hence at that end, we're comparing to\r\n\r\n$\\int_c^{\\infty}x^{-a}\\ln x\\,dx.$\r\n\r\nThe logarithm is of very little significance to that; it converges when $a>1.$ Thus the whole integral converges when $1b. :P", "Solution_7": "[quote=\"iqtimo\"]it is :\nlet $ a,b,c$ is nonegative.not two of which by zero.find best $ k \\ge 0$ such that:\n$ \\frac {a \\plus{} kb}{c \\plus{} kb} \\plus{} \\frac {b \\plus{} kc}{a \\plus{} kc} \\plus{} \\frac {c \\plus{} ka}{b \\plus{} ka} \\ge 3$\n\nthis is one problem nice and hard ! you can post your solution :wink:[/quote]\r\n\r\nIt is not hard,you can try. :P", "Solution_8": "[quote=\"shfdfzhjj\"][quote=\"Guest\"][quote=\"tuantam1lan\"]let $ a,b,c$ is nonegative.not two of which by zero.find best $ k \\ge 0$ such that:\n$ \\frac {a \\plus{} kb}{c \\plus{} kb} \\plus{} \\frac {b \\plus{} kc}{a \\plus{} kc} \\plus{} \\frac {c \\plus{} ka}{b \\plus{} ka} \\ge 3$\n i found $ k\\sim 2,876121406$[/quote]\nExactly speaking, the best $ k$ is the unique real root of the following equation:\n\n$ k^7 \\minus{} 4k^6 \\plus{} 8k^5 \\minus{} 16k^4 \\plus{} 16k^3 \\minus{} 28k^2 \\plus{} 4k \\minus{} 4 \\equal{} 0.$[/quote]\n\nI think you are wrong, counter-example: a=c>b. :P[/quote]\r\nWhat do you mean? What does your so-called \"counter-example\" mean?", "Solution_9": "[quote=\"Guest\"][/quote][quote=\"shfdfzhjj\"][quote=\"Guest\"][quote=\"tuantam1lan\"]let $ a,b,c$ is nonegative.not two of which by zero.find best $ k \\ge 0$ such that:\n$ \\frac {a + kb}{c + kb} + \\frac {b + kc}{a + kc} + \\frac {c + ka}{b + ka} \\ge 3$\n i found $ k\\sim 2,876121406$[/quote]\nExactly speaking, the best $ k$ is the unique real root of the following equation:\n\n$ k^7 - 4k^6 + 8k^5 - 16k^4 + 16k^3 - 28k^2 + 4k - 4 = 0.$[/quote]\n\nI think you are wrong, counter-example: a=c>b. :P[/quote][quote=\"Guest\"]\nWhat do you mean? What does your so-called \"counter-example\" mean?[/quote]\r\n\r\nBut you have edited your post,professor Guest. :roll:" } { "Tag": [ "pigeonhole principle", "modular arithmetic" ], "Problem": "Can anyone help me with the below problems?\r\n\r\n1. Show that for any list of ten distinct positive integers, there is a consecutive sequence of some of them such that the sum is divisible by 10. \r\n\r\n2. Find the number of positive integers p=<80 that are divisible by at least one of the numbers 2,3,5 and 7.", "Solution_1": "[hide=\"1.\"]\n\nLet the integers be $a_1, a_2, \\ldots, a_{10}$. Consider the sums\n\n$S_1 = a_1$,\n$S_2 = a_1+a_2$\n...\n$S_{10} = a_1+a_2+\\cdots+a_{10}$.\n\nIf any of them is divisible by $10$, we are done. So suppose none of them are. Then they are all $1,2,\\ldots,9 \\pmod{10}$. But since there are ten of them by Pigeonhole we have some two congruent modulo $10$. Let these be $S_i$ and $S_j$ with $i < j$.\n\nWe have $S_j-S_i$ divisible by $10$ but that is just $a_{i+1}+a_{i+2}+\\cdots+a_j$, which is a consecutive sequence, so we are done.[/hide]", "Solution_2": "Thanks! I have also solved #2. :)", "Solution_3": "And, just a little remark, one can easily generalize the first one (i.e. : n integers instead of 10), and it is solved exactly as [b]paladin8[/b] did for the case n=10. :)" } { "Tag": [ "geometry", "perimeter", "percent" ], "Problem": "Engineers place a sidewalk along the perimeter of a circular park so that the radius of the park is increased by $ 20\\%$. By what percent is the park's area increased?", "Solution_1": "The area increases with the square of the radius, so the park increases by $ (\\frac {6}{5})^2 \\equal{} \\frac {36}{25} \\equal{} \\fbox{44}\\%$. We use $ \\frac{6}{5}$ since that fraction represents an increase of $ 20\\%$." } { "Tag": [ "Support", "LaTeX", "\\/closed" ], "Problem": "I'm not sure if this is a bug or just something that went wrong on that post. Anyways, take a look at:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=205272#p205272", "Solution_1": "[quote=\"MithsApprentice\"]I'm not sure if this is a bug or just something that went wrong on that post. Anyways, take a look at:\n\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=205272#p205272[/quote]\r\n\r\nEverything I get is a window telling me\r\n\r\nSorry, but only users granted special access can read topics in this forum. \r\n\r\nWhat forum is that on? I mean, many of the national communities fora are closed to the general public (At least Georgia's Forum is :D ).", "Solution_2": "Quite so. This is our state forum. I was aiming the link at the admin (mostly Valentin since he apparently does most of the tech support around here...). :D", "Solution_3": "It's the old emoticons in the $\\LaTeX$ code bug :) I've fixed Mathew's post." } { "Tag": [ "function", "algorithm", "algebra", "domain" ], "Problem": "Hey all,\r\n\r\nI have a sorting algorithm that works only on integers, and I need to expand it to give the same order for double precision numbers. To do this, I need to map my doubles, which are on (-inf,inf) to a range where the \"exponent bits\" of the double are 0. Which means the range from (I think) [2^-52,1]. However, while the doubles could be anywhere on the real number line, most of them will be clustered very near 0, at r < 2^-52. So I need a non-linear map that will take miniscule numbers to larger ones, and large numbers to small ones, while still preserving their order. \r\n\r\nIf it makes things any easier, you can consider only the domain +/- (10^-19,10^4), which is where 99.99% of the numbers will be. \r\n\r\nOnce I have that I can easily map them to long integers with a simple multiplication and apply the sorting algorithm, but I am having trouble working out an appropriate map. Any suggestions are appreciated. \r\n\r\nOnce caveat is that this map can't be computationally intensive. For example, a shift of arctan(x) or some variation of log(x) would achieve what I want, but it takes too long to compute to make this sorting algorithm worthwhile.\r\n\r\nI have no idea if there is such a map in existence, but if anyone has ideas it would be greatly appreciated.\r\n\r\nBy the way, if this is in the wrong place feel free to move it.", "Solution_1": "Why not do something like take an 8th or 16th root? Should be relatively computationally gentle while still having the same sort of shape as arctan near 0, no? (I guess really I mean $ \\text{sign}(x) \\cdot \\sqrt[8]{|x|}$ to account for negative numbers.)", "Solution_2": "It's a possibility, but roots are still pretty intensive computationally when you do a lot of them. So not really as fast as I would like if there are other alternatives. Roots do have the nice property of mapping numbers < 1 to larger numbers < 1, though. Hmm, wonder if I can find an algorithm that approximates roots quickly while preserving order.", "Solution_3": "exactly what is this \"sorting function\"?\r\nyou should try to make it applicable to doubles, because if you shift the numbers to longs, numbers that are too close together will become equal, which would result in loss of precision", "Solution_4": "The sorting function is z-order sorting\r\n\r\n[url]http://en.wikipedia.org/wiki/Z-order_(curve)[/url]\r\n\r\nAnd yeah, that'w why I was hoping to map my numbers to somewhere between 2^-52 and 1, which I would then multiply by a constant factor before shifting to long integers in order to keep enough precision.", "Solution_5": "Sorry, the URL didn't work for some reason\r\n\r\ntry [url]http://en.wikipedia.org/wiki/Z-order_(curve)[/url]\r\n\r\nArg - just copy paste into the browser and be sure to include the closing bracket, I can't wrestle with this forum software ^_^\r\n\r\nAlso, could someone explain to me why simple bit interleaving works for integers but fails for doubles? Since the exponent is stored before the mantissa in a double (ignoring the sign bit for the moment) should you not be able to treat the exponent bits and mantissa bits as an 11 bit and 53 bit integer and order them with simple bit interleaving? This seems logical to me, but it doesn't work.\r\n\r\n\r\nAnother question:\r\n\r\nIf I used Newton's method with only one or two iterations in order to approxiate roots, will that preserve the order? I suspect the answer would be dependent on the initial guess somehow though.", "Solution_6": "this is probably because the exponent commonly does not have a \"sign\" bit.\r\n-1 is commonly (at least on my comp) stored as 111...11\r\n-2 would be stored as 111...10\r\n-3 : 111...101\r\netc.\r\nwhile positive numbers would be stored as\r\n000...01, 000...10 etc.\r\nhowever, any implementation that utilizes this would be unportable", "Solution_7": "[quote=\"KBriggs\"]I can't wrestle with this forum software[/quote] It was a bug that's now been fixed -- you can also always go back and edit your posts (for up to a day) using the \"edit\" button in the lower-right corner of your post.", "Solution_8": "Thanks JBL\r\n\r\n@lyn:\r\n\r\nexponents in double precision numbers are actually stored as unsigned 11 bit integers and a 'bias' of -1023(or maybe -1024?) is applied to them to give negative exponents. So I am not quite sure how it having a sign bit or not would change anything. Perhaps you could clarify what you meant?", "Solution_9": "yea that's basically what i meant (in terms of bits).\r\nwhen you take -1, it underflows and is equivalent to -1 + 1024\r\nbut if you compare the bits, 1 would be less than -1..., which i don't think you want", "Solution_10": "To my understanding, the exponent is stored as an unisgned 11 bit integer, and only biased when the number is interpreted as a double. Since the number is never interpreted during the bit-interleaving, the bias is irrelevent and the comparison will be between positive numbers between 1 and 2046... so negatives should never come up at all. \r\n\r\nAm I misunderstanding how the bias is applied?" } { "Tag": [ "search", "MATHCOUNTS", "AMC", "AMC 8", "FTW", "geometry" ], "Problem": "I've finally decided, to start posting on this AoPS thing. As to who I am, you can go on http://www.xtremesystems.org/forums look for user serialk11r and search through my 4000 posts to find the one where I revealed it. Nah here's a shortcut: I failed at mathcounts last year, 36th in the state. Yes I'm the lone JLS person in the top 25% at state out of the 3 individuals who made state (which wasn't celebrated to my disappointment :mad: jk, I'm failing at math so I don't deserve it). I just moved to Palo Alto last year, so I don't really have JLS pride anyways :P\r\nSorry to Lynnelle for not joining up 1 year ago, I was sorta busy with that other forum as you can see :oops: \r\nGood to see some familiar people :)\r\n\r\nPS dang at 3 pm I managed to make a ton of errors when typing that simple message.", "Solution_1": "Ahaha. I fail more than you.\r\n\r\nAt JLS I am known as a self-proclaimed failure... by some people.\r\n\r\nDid this make sense? :huh:", "Solution_2": "Are you another of Ubemaya's accounts?\r\n\r\n(It does if you are Ubemaya)\r\n :D", "Solution_3": "[quote=\"bowei\"]Are you another of Ubemaya's accounts?\n\n(It does if you are Ubemaya)\n :D[/quote]\r\nNo, I'm not Ubemaya(albert), I know him though sorta...I wish I could \"suck\" as much as he does at math :) I go to Gunn, I'm a freshman. The palo alto people know who I am. I joined cuz I finally decided to take a look here, one year after Lynnelle tried to force me and the rest of the JLS math club (which is dead) to, and I was like wow this forum is dead, and I love forums!!! so I decided I'm gonna try to revive it, get some more math pro people around here to join, get older people to post more.", "Solution_4": "Yay, hi! JLS MathCounts isn't dead, it's just its online presence that I've given up on. JLS MathCounts is pretty awesome right now--it won't start dying until around March/April. So John joined too? What's his username?", "Solution_5": "[quote=\"bookaholic\"]Yay, hi! JLS MathCounts isn't dead, it's just its online presence that I've given up on. JLS MathCounts is pretty awesome right now--it won't start dying until around March/April. So John joined too? What's his username?[/quote]\r\nJohn didn't join yet, he'll be making a little intro post too :) He forgot his username you see...\r\nI should go visit JLS sometime, see how math club is going. I should also visit jordan math club, only problem is I don't know when it is. If it was the same time as last year, then Gunn math club will be in the way.", "Solution_6": "I think Jordan math club is Friday after school. If you ever visit JLS math club, we're now Thursday after school in room 350.", "Solution_7": "[quote=\"bookaholic\"]I think Jordan math club is Friday after school. If you ever visit JLS math club, we're now Thursday after school in room 350.[/quote]\r\nHmmm that means I'll have to miss chess to go to JLS and I'll have to miss math club to visit jordan :(", "Solution_8": "Yoyo, everyone.\r\n\r\nThe Jordan math club just took the test to arrange Mathcounts teams. Compared to us, it was epic fail. I believe the highest scores on the sprints were three 21's and a no-name 19 (gaaaaaaaah!). Oh well... we've got some promising 6th grade noobs, though, and I think we can train them to own Mathcounts in a couple of years.", "Solution_9": "[quote=\"John the Scavenger\"]Yoyo, everyone.\n\nThe Jordan math club just took the test to arrange Mathcounts teams. Compared to us, it was epic fail. I believe the highest scores on the sprints were three 21's and a no-name 19 (gaaaaaaaah!). Oh well... we've got some promising 6th grade noobs, though, and I think we can train them to own Mathcounts in a couple of years.[/quote]\r\nBoylie, nice to see you!\r\nHahahahaha so it turns out JLS is STILL sucking worse than Jordan.", "Solution_10": "[quote=\"serialk11r\"][quote=\"John the Scavenger\"]Yoyo, everyone.\n\nThe Jordan math club just took the test to arrange Mathcounts teams. Compared to us, it was epic fail. I believe the highest scores on the sprints were three 21's and a no-name 19 (gaaaaaaaah!). Oh well... we've got some promising 6th grade noobs, though, and I think we can train them to own Mathcounts in a couple of years.[/quote]\nBoylie, nice to see you!\nHahahahaha so it turns out JLS is STILL sucking worse than Jordan.[/quote]\r\n1. Hi John!\r\n2. O rly? Explain.", "Solution_11": "well JLS rocked in '06 so it doesn't matter and appararently 1992", "Solution_12": "Apparently Steven SUDDENLY turned super beast at mathcounts...so I guess JLS has a chance. But on thursday I went to JLS and to my horror, they were taking LAST YEAR's school round.", "Solution_13": "W...T...F? Are you serious? Okay, I'm emailing Mr. Peterson.", "Solution_14": "[quote=\"bookaholic\"]W...T...F? Are you serious? Okay, I'm emailing Mr. Peterson.[/quote]\r\nOf course I'm serious, you can ask Steven too.\r\nIf JLS isn't able to participate that would really suck...seeing that Steven has improved so much.", "Solution_15": "Oh, that's not a problem--MathCounts doesn't care how schools choose team members. I'm more annoyed about 1) the unfair advantage people who remember problems better got and 2) the sheer silliness of it.", "Solution_16": "[quote=\"serialk11r\"]Apparently Steven SUDDENLY turned super beast at mathcounts...[/quote]\r\n\r\nNo. Definitely not. \r\n\r\nI still suck. You can't come to a conclusion just based on how I did on AoPS FTW. I'm doing pretty horribly now though. My rating died in the last week.", "Solution_17": "[quote=\"CheeseIsGood\"][quote=\"serialk11r\"]Apparently Steven SUDDENLY turned super beast at mathcounts...[/quote]\n\nNo. Definitely not. \n\nI still suck. You can't come to a conclusion just based on how I did on AoPS FTW. I'm doing pretty horribly now though. My rating died in the last week.[/quote]\r\nuh, I think I can.\r\nAlso, please explain how you got a perfect AMC8 score, when you got like 17 last year or something like that.\r\nI think you'll totally destroy everyone ;) maybe not the redwood and harker kids but whatever.", "Solution_18": "I did not take the AMC 8 last year. \r\n\r\nWhere did you get the number 17 from?", "Solution_19": "[quote=\"CheeseIsGood\"]I did not take the AMC 8 last year. \n\nWhere did you get the number 17 from?[/quote]\r\nLOL okay okay okay sorry I thought you took it and failed miserably...\r\nSeriously though, on FTW you had INSANE speed and last year you were no where near that fast.", "Solution_20": "Welcome to the (100% 1337) bay area forums. Enjoy your stay, reception is to your right.", "Solution_21": "Which way is right on a forum?", "Solution_22": "[quote=\"xscapezaer\"]Which way is right on a forum?[/quote]\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=48329", "Solution_23": "[quote=\"serialk11r\"][quote=\"CheeseIsGood\"][quote=\"serialk11r\"]Apparently Steven SUDDENLY turned super beast at mathcounts...[/quote]\n\nNo. Definitely not. \n\nI still suck. You can't come to a conclusion just based on how I did on AoPS FTW. I'm doing pretty horribly now though. My rating died in the last week.[/quote]\nuh, I think I can.\nAlso, please explain how you got a perfect AMC8 score, when you got like 17 last year or something like that.\nI think you'll totally destroy everyone ;) maybe not the redwood and harker kids but whatever.[/quote]\r\n\r\nRedwood sucks this year. Which is surprising, considering that I'm not on the team anymore.", "Solution_24": "[quote=\"Ubemaya\"][/quote][quote=\"serialk11r\"][quote=\"CheeseIsGood\"][quote=\"serialk11r\"]Apparently Steven SUDDENLY turned super beast at mathcounts...[/quote]\n\nNo. Definitely not. \n\nI still suck. You can't come to a conclusion just based on how I did on AoPS FTW. I'm doing pretty horribly now though. My rating died in the last week.[/quote]\nuh, I think I can.\nAlso, please explain how you got a perfect AMC8 score, when you got like 17 last year or something like that.\nI think you'll totally destroy everyone ;) maybe not the redwood and harker kids but whatever.[/quote][quote=\"Ubemaya\"]\n\nRedwood sucks this year. Which is surprising, considering that I'm not on the team anymore.[/quote]\r\n\r\nIf you think that you suck, then you saying Redwood sucks means they will kill everyone, am I right?", "Solution_25": "[quote=\"Jack Sparrow\"]\nYou're not making any sense.\n[/quote]", "Solution_26": "Um. Do you guys know what JLS's AMC-8 team score ended up being? Like, what were the top few scores, other than Steven's? (Congratulations on the perfect, by the way!)", "Solution_27": "[quote=\"bookaholic\"]Um. Do you guys know what JLS's AMC-8 team score ended up being? Like, what were the top few scores, other than Steven's? (Congratulations on the perfect, by the way!)[/quote]\r\nNo... Lemme guess though, 17, 17, 16, 15.....? It doesn't matter if everyone else at JLS sucks (that's the case if I remember correctly...no offense to some of the other kids if they ever see this), if Steven beasts mathcounts like I expect him to, then he can go to state as an individual easily, and I'm hoping top 16 in state :D you'll be pretty good at countdown. This year chapter will be really easy to beast since Jordan has no one left, Terman has no one left, Bowditch's 2 good people left, eh...no other schools are worth mentioning.\r\nLast year I was a total failure. I managed to get a 20...I only finished 20 problems though...my time management on tests is pretty bad I must say, but usually tests are easy enough for that not to matter. Argh another example of where school is destroying my math...School math is messing up my thinking processes because they teach the most retarded ways to do problems.", "Solution_28": "I know that there was a 19. The scores haven't been announced yet.... I got my score by asking Mr. Peterson.", "Solution_29": "[quote=\"CheeseIsGood\"]I know that there was a 19. The scores haven't been announced yet.... I got my score by asking Mr. Peterson.[/quote]\r\ndannggg that's pretty low...\r\nlast year was 22, 21, 20, then a whole bunch of crappy scores. :(\r\nIf I wasn't being stupid JLS could've made the honor whatever list thingy." } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Let $x_1, x_2,\\cdots,x_n$ be real numbers such that $0\\le x_i\\le1$ for $i=1,2,\\cdots,n.$ Prove that\r\n\\[\r\n\\frac 1 n\\sum_{i=1}^n x_i^2-(\\frac 1 n\\sum_{i=1}^n x_i)^2\\le\\{\\begin{array}{ll}\r\n\\frac 1 4 & \\textrm{if n is even }\\\\\r\n\\frac 1 4-\\frac 1{4n^2}& \\textrm{if n is odd }\r\n\\end{array}\r\n\\]\r\n\r\n[i]Edited by Myth[/i]", "Solution_1": "Indeed, LHS is a convex function on each $x_i$, so it attains maximum for $x_i\\in\\{0,1\\}$. The result follows." } { "Tag": [ "LaTeX", "algebra", "polynomial", "Vieta" ], "Problem": "\u0395\u03bc\u03ad\u03bd\u03b1 \u03bc\u03bf\u03c5 \u03ac\u03c1\u03b5\u03c3\u03b1\u03bd \u03c4\u03bf\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd.... :D :D \r\n\r\n\r\n1) \u039d.\u03b4.\u03bf \u03c4\u03bf \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03c4\u03c9\u03bd \u03c3\u03c5\u03bd\u03c4\u03b5\u03bb\u03b5\u03c3\u03c4\u03ce\u03bd \u03c4\u03bf\u03c5 \u03c0\u03bf\u03bby\u03c9\u03bd\u03cd\u03bc\u03bf\u03c5 \r\n\r\n[b]P(x) = Q(x)*(x-1) + 1 [/b]\u03b1\u03c0\u03bf\u03c4\u03b5\u03bb\u03b5\u03af \u03bb\u03cd\u03c3\u03b7 \u03c4\u03b7\u03c2 \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7\u03c2 [b]P(P(x))=x[/b]\r\n\r\n\r\n2) \u039d\u03b1 \u03b2\u03c1\u03b5\u03b8\u03b5\u03af \u03c4\u03bf \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03c4\u03c9\u03bd \u03c3\u03c5\u03bd\u03c4\u03b5\u03bb\u03b5\u03c3\u03c4\u03ce\u03bd \u03c4\u03bf\u03c5 \u03c0\u03bf\u03bb\u03c5\u03c9\u03bd\u03cd\u03bc\u03bf\u03c5 P(x) \u03b1\u03bd \u03c4\u03bf P(x) \u03ad\u03c7\u03b5\u03b9 \u03c4\u03b7\u03bd \u03b5\u03be\u03ae\u03c2 \u03b9\u03b4\u03b9\u03cc\u03c4\u03b7\u03c4\u03b1\r\n\r\n\r\n[b]P(x+1) = P(x) +2007 - a[/b]\r\n\r\n\r\n\u03cc\u03c0\u03bf\u03c5 a \u03bf \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03cc\u03c2 \u03cc\u03c1\u03bf\u03c2 \u03c4\u03bf\u03c5 P(x)", "Solution_1": "[img]http://img347.imageshack.us/img347/6774/226200700sc.png[/img]", "Solution_2": "\u03a0\u03bf\u03bb\u03cd \u03c9\u03c1\u03b1\u03af\u03b1 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 (\u03ba\u03b1\u03b9 \u03bb\u03cd\u03c3\u03b7). \r\n\u039c\u03c0\u03c1\u03ac\u03b2\u03bf! :wink:", "Solution_3": "\u03a9\u03c1\u03b1\u03af\u03bf\u03c2 \u0393\u03b9\u03ac\u03bd\u03bd\u03b7...\u039a\u03b1\u03b9 \u03b7 \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7 \u03ba\u03ac\u03c0\u03c9\u03c2 \u03ad\u03c4\u03c3\u03b9 \u03bb\u03cd\u03bd\u03b5\u03c4\u03b1\u03b9....", "Solution_4": "\u0395\u03c6\u03c4\u03b9\u03b1\u03be\u03b1 \u03ba\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b7 \u03c0\u03bf\u03c5 \u03bc\u03bf\u03c5 \u03ac\u03c1\u03b5\u03c3\u03b5....\u0395\u03af\u03bd\u03b1\u03b9 \u03bb\u03af\u03b3\u03bf \u03c0\u03b9\u03bf \u03b4\u03cd\u03c3\u03ba\u03bf\u03bb\u03b7 \u03b1\u03c0\u03cc \u03c4\u03b9\u03c2 \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03b5\u03c2.\r\n\r\n\r\n\u0395\u03c3\u03c4\u03c9 \u03c0\u03bf\u03bb\u03c5\u03ce\u03bd\u03c5\u03bc\u03bf [b]P(x)[/b] \u03bc\u03b5 (\u03bc\u03cc\u03bd\u03bf) \u03b8\u03b5\u03c4\u03b9\u03ba\u03ad\u03c2 \u03c1\u03af\u03b6\u03b5\u03c2 [b]X1,X2,....Xn[/b]\r\n\r\n\u03c4\u03ad\u03c4\u03bf\u03b9\u03b5\u03c2 \u03ce\u03c3\u03c4\u03b5 [b]X1*X2*X3*.....Xn = 1.[/b]\r\n\r\n\u039d.\u03b4.\u03bf [b]P(1)<=0[/b]\r\n\r\n :D", "Solution_5": "Fotis \u03b5\u03af\u03c3\u03b1\u03b9 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03b1 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7;\r\n\u03a4\u03bf \u03c0\u03bf\u03bb\u03c5\u03ce\u03bd\u03c5\u03bc\u03bf [b]P(x)=(x-1/6)(x-2)(x-3)[/b] \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03b5\u03af \u03c4\u03b1 \u03b4\u03b5\u03b4\u03bf\u03bc\u03ad\u03bd\u03b1 \u03c4\u03bf\u03c5 \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03bf\u03c2 ,\u03b1\u03bb\u03bb\u03ac [b]P(1)=5/3 > 0[/b] .\r\n\r\n\u0393\u03b9\u03b1 \u03c4\u03b7\u03bd 2 \u03b8\u03ad\u03c4\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c0\u03bf\u03c5 [b]x[/b] \u03c4\u03bf [b]0[/b] \u03ba\u03b1\u03b9 \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 [b]P(1)=2007[/b].\r\n\r\n\u03a3\u03ba\u03ad\u03c6\u03c4\u03b7\u03ba\u03b1 \u03ba\u03b9 \u03b5\u03b3\u03ce \u03bc\u03b9\u03b1 \u03c3\u03c7\u03b5\u03c4\u03b9\u03ba\u03ae(?) \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03c4\u03b7 \u03b2\u03ac\u03b6\u03c9.\r\n'\u0395\u03c3\u03c4\u03c9 [b]P(x)[/b] \u03ad\u03bd\u03b1 \u03c0\u03bf\u03bb\u03c5\u03ce\u03bd\u03c5\u03bc\u03bf \u03bc\u03b5 \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03bf\u03c5\u03c2 \u03c3\u03c5\u03bd\u03c4\u03b5\u03bb\u03b5\u03c3\u03c4\u03ad\u03c2. \u0391\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 [b]P(P(P(1)))=1[/b] \u03bd\u03b1 \u03b2\u03c1\u03b5\u03b8\u03b5\u03af \u03c4\u03bf \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03c4\u03c9\u03bd \u03c3\u03c5\u03bd\u03c4\u03b5\u03bb\u03b5\u03c3\u03c4\u03ce\u03bd \u03c4\u03bf\u03c5 [b]P[/b].", "Solution_6": "\u0394\u03af\u03ba\u03b9\u03bf \u03ad\u03c7\u03b5\u03b9\u03c2 Stedes...\u03a4\u03ce\u03c1\u03b1 \u03c0\u03bf\u03c5 \u03be\u03b1\u03bd\u03b1\u03ba\u03bf\u03b9\u03c4\u03ac\u03c9 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5 \u03b2\u03c1\u03af\u03c3\u03ba\u03c9 \u03bb\u03ac\u03b8\u03bf\u03c2.... :maybe: :blush:", "Solution_7": "[quote=\"stedes\"]\n'\u0395\u03c3\u03c4\u03c9 [b]P(x)[/b] \u03ad\u03bd\u03b1 \u03c0\u03bf\u03bb\u03c5\u03ce\u03bd\u03c5\u03bc\u03bf \u03bc\u03b5 \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03bf\u03c5\u03c2 \u03c3\u03c5\u03bd\u03c4\u03b5\u03bb\u03b5\u03c3\u03c4\u03ad\u03c2. \u0391\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 [b]P(P(P(1)))=1[/b] \u03bd\u03b1 \u03b2\u03c1\u03b5\u03b8\u03b5\u03af \u03c4\u03bf \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03c4\u03c9\u03bd \u03c3\u03c5\u03bd\u03c4\u03b5\u03bb\u03b5\u03c3\u03c4\u03ce\u03bd \u03c4\u03bf\u03c5 [b]P[/b].[/quote]\r\n\r\n=1..... :wink: \r\n\r\n :D", "Solution_8": "Ok, \u03b3\u03b9\u03b1 \u03c4\u03b7 2 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5:\r\n\r\n\u0386\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03c3\u03c5\u03bd\u03c4\u03b5\u03bb\u03b5\u03c3\u03c4\u03ce\u03bd \u03b4\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b3\u03b9\u03b1 $P(1)$. \u0395\u03c0\u03af\u03c3\u03b7\u03c2, \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03cc\u03c4\u03b9 $x=0$. \u038c\u03bc\u03c9\u03c2 $P(x)=a$ \u03b1\u03c0\u03cc \u03c5\u03c0\u03cc\u03b8\u03b5\u03c3\u03b7. \u0386\u03c1\u03b1 $P(1)=2007\\ldots$", "Solution_9": "\u039c\u03b5 \u03c4\u03b7\u03bd \u03af\u03b4\u03b9\u03b1 \u03bb\u03bf\u03b3\u03b9\u03ba\u03ae \u03bc\u03b5 \u03c4\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03b3\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03bf \u03c4\u03c9\u03bd \u03c1\u03b9\u03b6\u03ce\u03bd \u03b5\u03bd\u03cc\u03c2 \u03c0\u03bf\u03bb\u03c5\u03c9\u03bd\u03cd\u03bc\u03bf\u03c5 \u03bd \u03b2\u03b1\u03b8\u03bc\u03bf\u03cd \u03b1\u03ba\u03cc\u03bc\u03b7 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03ad\u03c7\u03b5\u03b9 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03ad\u03c2 \u03c1\u03af\u03b6\u03b5\u03c2 \u03b8\u03b1 \u03b9\u03c3\u03bf\u03cd\u03c4\u03b1\u03b9 \u03bc\u03b5\r\n\\[a_{0}\\]\r\n\u03b1\u03bd \u03bd=2\u03ba \u03ba\u03b1\u03b9 \u03bc\u03b5\r\n\\[-a_{0}\\]\r\n\u03b1\u03bd \u03bd=2\u03ba+1.\r\n\u03a0\u03c9\u03c2 \u03b2\u03ac\u03b6\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c5\u03c4\u03ac \u03c0\u03bf\u03c5 \u03b3\u03c1\u03ac\u03c6\u03bf\u03c5\u03bc\u03b5 \u03c3\u03b5 Latex \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03ad\u03c7\u03b5\u03b9\u03b1 \u03c4\u03bf\u03c5 \u03ba\u03b5\u03b9\u03bc\u03ad\u03bd\u03bf\u03c5??\r\n\u0394\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03c9 \u03bd\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ac\u03b2\u03c9...... :wallbash: :wink:", "Solution_10": "[quote=\"bouzoukman\"]\u039c\u03b5 \u03c4\u03b7\u03bd \u03af\u03b4\u03b9\u03b1 \u03bb\u03bf\u03b3\u03b9\u03ba\u03ae \u03bc\u03b5 \u03c4\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03b3\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03bf \u03c4\u03c9\u03bd \u03c1\u03b9\u03b6\u03ce\u03bd \u03b5\u03bd\u03cc\u03c2 \u03c0\u03bf\u03bb\u03c5\u03c9\u03bd\u03cd\u03bc\u03bf\u03c5 \u03bd \u03b2\u03b1\u03b8\u03bc\u03bf\u03cd \u03b1\u03ba\u03cc\u03bc\u03b7 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03ad\u03c7\u03b5\u03b9 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03ad\u03c2 \u03c1\u03af\u03b6\u03b5\u03c2 \u03b8\u03b1 \u03b9\u03c3\u03bf\u03cd\u03c4\u03b1\u03b9 \u03bc\u03b5\n\\[a_{0}\\]\n\u03b1\u03bd \u03bd=2\u03ba \u03ba\u03b1\u03b9 \u03bc\u03b5\n\\[-a_{0}\\]\n\u03b1\u03bd \u03bd=2\u03ba+1.\n\u03a0\u03c9\u03c2 \u03b2\u03ac\u03b6\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c5\u03c4\u03ac \u03c0\u03bf\u03c5 \u03b3\u03c1\u03ac\u03c6\u03bf\u03c5\u03bc\u03b5 \u03c3\u03b5 Latex \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03ad\u03c7\u03b5\u03b9\u03b1 \u03c4\u03bf\u03c5 \u03ba\u03b5\u03b9\u03bc\u03ad\u03bd\u03bf\u03c5??\n\u0394\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03c9 \u03bd\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ac\u03b2\u03c9...... :wallbash: :wink:[/quote]\r\nan katalava kala bouzoukman legontas sinexia tu kimenu malon enois ayto...\r\n\r\nisute me $a_{0}$ an n=2k i me $-a_{0}$ an n=peritos..\r\n\r\nan enois ayto,tote kane quote ayto to post pu diavazis tora gia na dis pos to exo grayi... :wink: \r\n\r\n :D", "Solution_11": "Thanks Rakar!!!!!!!\r\n\u039a\u03b1\u03bb\u03ac \u03ba\u03b1\u03c4\u03ac\u03bb\u03b1\u03b2\u03b5\u03c2,\u03c4\u03bf \u03b2\u03c1\u03ae\u03ba\u03b1 \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03ae\u03b8\u03b5\u03bb\u03b1..\r\n\u0399\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03b5\u03af\u03c0\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03b3\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03bf \u03c4\u03c9\u03bd \u03c1\u03b9\u03b6\u03ce\u03bd \u03b5\u03bd\u03cc\u03c2 \u03c0\u03bf\u03bb\u03c5\u03c9\u03bd\u03cd\u03bc\u03bf\u03c5??\r\n\u0394\u03b9\u03cc\u03c4\u03b9 [b]P(\u03c7)=(\u03c7-$x_{1}$)..(x-$x_{n}$)[/b] \u03bf\u03c0\u03cc\u03c4\u03b5 [b]P(0)=(-$x_{1}$)..(-$x_{n}$)[/b].\r\n\u0386\u03c1\u03b1 \u03b3\u03b9\u03b1 n=2\u03ba \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \r\n[b]P(0)=$x_{1}$..$x_{n}$[/b]\r\n\r\n\u03b5\u03bd\u03ce \u03b3\u03b9\u03b1 n=2\u03ba+1\r\n[b]P(0)=-($x_{1}$..$x_{n}$)[/b].\r\n\u038c\u03bc\u03c9\u03c2 P(0)=$a_{0}$ \u03bf\u03c0\u03cc\u03c4\u03b5 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03b2\u03b3\u03b1\u03af\u03bd\u03bf\u03c5\u03bd \u03b1\u03c5\u03c4\u03ac \u03c0\u03bf\u03c5 \u03ad\u03bb\u03b5\u03b3\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9...", "Solution_12": "@bouzoukman:\r\nontos etsi ine...ala ayto mporis na to pis +amesos ap tus tipus tu vieta afu\r\n$r_{1}r_{2}...r_{n}=(-1)^{n}\\frac{a_{0}}{a_{n}}$\r\n\r\n@stedes:\r\ni lisi pu ixa vri =1....ida oti kolai kapu,ala parolayta pali nomizo oti =1...8a to ksanayakso...\r\n\r\n@thatin:\r\ndes+ta pm su...\r\n\r\n :D \r\n\r\nedit:iparxi mipos kanenas pu na exi enan [b]olokliromeno[/b] odigo xrisis tu mathematica?an kapios mpori na voi8isi, na mu stili pm...\r\nthanks.." } { "Tag": [ "algebra", "polynomial", "number theory", "prime numbers", "algebra unsolved" ], "Problem": "Prove that if $a^p+b^p$ is integer for infinitely many primes $p$, then $a$ and $b$ are integers. \r\n\r\n\r\nI can only prove that $a$ and $b$ are algebraic numbers.", "Solution_1": "What if $a=2+\\sqrt{3}$ and $b=2-\\sqrt{3}$? :)", "Solution_2": "Also, check problem S5 [url=http://reflections.awesomemath.org/2006_1/2006_1_problems.pdf]here[/url]", "Solution_3": "Oh.. I see there were stonger conditions. I think I have a proof.\r\n \r\nI can get now, that there are two polynomials of degree 4 which vanishes: one in a, and the other in b, and which have the coeff for $X$ zero. \r\n [hide=\"hint\"]$b^6=(k+a^2)^3=(p+a^2)^3$ ,with k,p integers[/hide]\r\nhence $a^4-b^4=(a^2-b^2)(a^2+b^2) \\in \\mathbb{Z}$. And since $a^2-b^2\\in\\mathbb{Z}$ we get that $a^2+b^2\\in \\mathbb{Q}$ then $2b^2\\in \\mathbb{Q}\\Rightarrow b^2,b^4\\in\\mathbb{Q}\\Rightarrow$ since there is a polynomial of degree 4 with coeff for X 0 whih vanishes in b $b^3\\in \\mathbb{Q}\\Rightarrow b\\in \\mathbb{Q}$.\r\n\r\nHence $a,b$ are rational numbers. Write them as $\\frac m n$ and $\\frac k n$ .Assume that $n\\neq 1$ and fix an $a$. We have now that $m^q\\equiv k^q (\\mod n^a)$ for all primes $q>a$. Since prime numbers take all residues mod $\\phi(n^a)$ we get that $m^q\\equiv k^q (\\mod n^a)\\forall q\\in \\{1,\\dots, \\phi(n)\\}\\Rightarrow m\\equiv k (\\mod n^a)$. since a was chosen arbitrarly we have that $n^a\\cdot m-n \\forall a$, hence $m=n$ which leads us to $a=b$. Contradiction. Hence $n=1$, and so, a,b are integers. \r\n\r\nAm I wrong somewhere?", "Solution_4": "[quote=\"xirti\"]Oh.. I see there were stronger conditions. [/quote]\r\nAnd also, $a^p-b^p$ instead of $a^p+b^p$.\r\nPlease, don't post any solutions until March 1st." } { "Tag": [], "Problem": "Chuck Norris can roundhouse kick someone in 15 nanoseconds, exactly. Him working together with his best friend, Norris Chuck, can roundhouse kick someone in 10 nanoseconds. How long would it take Norris Chuck to roundhouse kick someone?\r\n\r\nYay for roundhouse kicks! And somehow I have this feeling that someone will post some kind of Chuck Norris joke here... :huh:", "Solution_1": "How can two people land a roundhouse kick faster if both of them kick slower than that rate?\r\nUnless they like help them or something with the technique lol.", "Solution_2": "Let's convert this to:\r\n\r\nChuck Norris can mow his lawn in 15 nanoseconds. Working with Norris Chuck, they can mow it in 10 nanoseconds. How long does it take for Norris Chuck to mow the lawn by himself?\r\n\r\n$ \\frac{1}{15}\\plus{}\\frac{1}{x}\\equal{}\\frac{1}{10} \\implies \\frac{1}{x}\\equal{}\\frac{1}{30} \\implies x\\equal{}\\boxed{30}$ nanoseconds.", "Solution_3": "Gah I made this problem earlier, but mixed up the numbers! Nooooo! :mad:\r\n\r\nEDIT: AIME15, that was the intended answer. :roll:", "Solution_4": "chuck norris's kick is so fast u cant time it lol :P", "Solution_5": "[quote=\"Cheese12345\"]chuck norris's kick is so fast u cant time it lol :P[/quote]\r\n1) This is spam\r\n2) Bruce Lee pwns Chuck Norris", "Solution_6": "1. Stop spamming, especially if it's off-topic.\r\n2. This is also spam, but all the posts on this page have been rated.\r\n3. In general, for these problems, if person A does an action in $ n$ hours, and person B does an action in $ k$ hours, and the amount of time it takes for them to do it together takes $ h$ hours, we can say $ \\frac{1}{n}\\plus{}\\frac{1}{k}\\equal{}\\frac{1}{h}$. This post is no longer spam." } { "Tag": [ "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Can we show that $\\mathcal{C}[0,1]$ is complete with the norm $\\|f\\|=\\max |f(x)|$ by using this argument.\r\n$C[0,1]\\subset L^\\infty[0,1]$ by using the norm $\\|f\\|_\\infty$. It is known that $L^\\infty$ is complete. So we just need to show that $C[0,1]$ is closed. But convergence in sup norm (max norm) in this space means convergent uniformly, so every convergent sequence converges to continuous function. So $C[0,1]$ is closed and we are done.\r\n\r\n\r\n\r\nthanks in advance", "Solution_1": "Yes, it's ok :)", "Solution_2": "An argument that's simpler ($L^\\infty$ is a pretty ugly object):\r\nA Cauchy sequence in the uniform norm is pointwise Cauchy, so it converges pointwise. This convergence is clearly uniform, and the uniform limit of continuous functions is continuous.", "Solution_3": "It's not wrong, but it is very odd. We have to ask, how did we prove the completeness of $L^{\\infty}$ in the first place? If we were to look at the proof of that, we would find arguments that, much simplified, could be used in the case of $\\mathcal{C}.$\r\n\r\nThe first problem in any completeness proof is to produce a candidate: an object that might be the sought-for limit.\r\n\r\nIn this case: assume that $(f_n)$ is a uniformly Cauchy seqence of functions. Then it follows that for each $x,\\,(f_n(x))$ is a Cauchy sequence of real (or maybe complex) numbers. By the completeness of $\\mathbb{R}$ (or $\\mathbb{C}$), $\\lim f_n(x)$ exists. Define $f(x)$ as that limit.\r\n\r\nWe then have our candidate. The two remaining steps are to show that $f_n$ converges uniformly to $f$ and that $f$ (as the uniform limit of continuous functions) is itself continuous and hence a member of $\\mathcal{C}.$\r\n\r\n(OK, jmerry just said the same thing in many fewer words. I'll finish the post anyway.)\r\n\r\nThe proof that $L^{\\infty}$ is complete is harder that that; in particular, we have to work harder to produce the candidate, because we can't claim that $(f_n(x))$ is a Cauchy sequence of numbers for every $x.$", "Solution_4": "[quote=\"Kent Merryfield\"]It's not wrong, but it is very odd. We have to ask, how did we prove the completeness of $L^{\\infty}$ in the first place? If we were to look at the proof of that, we would find arguments that, much simplified, could be used in the case of $\\mathcal{C}.$\n\nThe first problem in any completeness proof is to produce a candidate: an object that might be the sought-for limit.\n\nIn this case: assume that $(f_n)$ is a uniformly Cauchy seqence of functions. Then it follows that for each $x,\\,(f_n(x))$ is a Cauchy sequence of real (or maybe complex) numbers. By the completeness of $\\mathbb{R}$ (or $\\mathbb{C}$), $\\lim f_n(x)$ exists. Define $f(x)$ as that limit.\n\nWe then have our candidate. The two remaining steps are to show that $f_n$ converges uniformly to $f$ and that $f$ (as the uniform limit of continuous functions) is itself continuous and hence a member of $\\mathcal{C}.$\n\n(OK, jmerry just said the same thing in many fewer words. I'll finish the post anyway.)\n\nThe proof that $L^{\\infty}$ is complete is harder that that; in particular, we have to work harder to produce the candidate, because we can't claim that $(f_n(x))$ is a Cauchy sequence of numbers for every $x.$[/quote]\r\n\r\nI know that the usual procedure to show completeness of a space. I used the above argument because the problem is appear in a section about $L^p$ space (Royden book). So I just suspect that he expect to use properties of $L^p$ space. Thanks a lot for all commentts (yassin, jmerry, and Dr. Kent)", "Solution_5": "but if we dont use completeness of L\u00b0\u00b0 , we will use the same argument that prove that L\u00b0\u00b0 is a Banach to show the other space is complet ( i say almost the same technic are used ...)", "Solution_6": "[quote=\"yassinus\"]but if we dont use completeness of L\u00b0\u00b0 , we will use the same argument that prove that L\u00b0\u00b0 is a Banach to show the other space is complet ( i say almost the same technic are used ...)[/quote]\r\n\r\n\r\nRoyden claim that completeness of $L^\\infty$ is easy, so he left it as exercises\r\nOne of the exercises, there is a problem asking to prove that convergence in $L^\\infty$ is equivalent to uniform convergence outside a set of measure zero.\r\n\r\n\r\nUsing this exercise to prove completeness of $L^\\infty$ I proceed as the following:\r\n$(f_n)$ is Cauchy in $L^\\infty$ implies $f_n$ Cauchy (uniformly) outside a set of measure zero $A$. So by redefining $f_n$ and set $f_n(x)=0$ in $A$, we have a new $f_n$ that Cauchy in $\\mathbb{R}$. By completeness of $\\mathbb{R}$ we have the candidate $f$ for our limit. MOreover $f_n\\to f$ uniformly on $A$, so $f_n\\to f$ in $L^\\infty$ by the above exercise.\r\nHopefully its ok" } { "Tag": [ "inequalities", "trigonometry", "inequalities proposed" ], "Problem": "Let $ x_i>0, y_i\\in \\Bbb{R}$ such that $ \\sum_{i\\equal{}1}^n x_i\\equal{}\\sum_{i\\equal{}1}^n y_i\\equal{}\\pi$ prove that\r\n\\[ \\sum_{i\\equal{}1}^n\\frac{\\cos y_i}{\\sin x_i}\\le \\sum_{i\\equal{}1}^n\\cot x_i\\]", "Solution_1": "Any idea? :wink: actually I don't have solution? Can anyone move it to unsolve problems?" } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial" ], "Problem": "$ c$ is a constant real and $ \\{a_n\\}$ is a sequence defined as\r\n\r\n$ a_1\\equal{}a\\in \\mathbb R$,\u3000$ a_{n\\plus{}1}\\equal{}\\frac{ca_n\\minus{}1}{a_n} (n\\equal{}1,2,\\cdots )$.\r\n\r\nIf the sequence $ \\{a_n\\}$ has just three distinct values,then determine the possible value of $ c$ ,and find the range of $ a$ for each $ c$.", "Solution_1": "[hide]Let $ f(x) \\equal{} \\frac {cx \\minus{} 1}{x}$. Our sequence can be defined as $ a_{n \\plus{} 1} \\equal{} f(a_n)$\n\nObserve that the sequence must be periodic -- therefore, $ a_{n \\plus{} 3} \\equal{} a_n \\implies f(f(f(a_n))) \\equal{} a_n$\n\nWe solve the equation $ f(f(f(a))) \\equal{} a$ for $ c$. The LHS simplifies to $ \\minus{} \\frac {1 \\minus{} c^2 \\minus{} 2 c a \\plus{} c^3 a}{c \\plus{} a \\minus{} c^2 a} \\equal{} a$.\n\nClearing the denominators and factoring, we get that $ c \\equal{} 1$ or $ c \\equal{} \\minus{} 1$ or $ c \\equal{} \\frac {1 \\plus{} a^2}{a} \\equal{} a \\plus{} \\frac {1}{a}$\n\nWe know this gives at most $ 3$ distinct values, but we need to have at least $ 3$ distinct values. We need to eliminate the cases in which $ f(a) \\equal{} a$. Solving this equation for $ c$ gives $ c \\equal{} a \\plus{} \\frac {1}{a}$, so we have to strike out that entire class of solutions.\n\nFinally, we have to make sure that we're never dividing by $ 0$. For $ c\\equal{}1$ and $ c\\equal{}\\minus{}1$, we have to make sure none of $ a$, $ f(a)$, $ f(f(a))$ are zero.\n\n$ \\blacktriangleright$ For $ c\\equal{}1$, $ a \\neq 0$, $ f(a) \\neq 0 \\implies a \\neq 1$, and $ f(f(a)) \\neq 0$, which is always satisfied.\n$ \\blacktriangleright$ For $ c\\equal{}\\minus{}1$, $ a \\neq 0$, $ f(a) \\neq 0 \\implies a \\neq \\minus{}1$, and $ f(f(a)) \\neq 0$, which is always satisfied.\n\nThe final solution set is therefore:\n\n$ c \\in \\{ \\minus{}1, \\; 1 \\} \\; \\& \\; a \\not \\in \\{ 0, c \\}$\n\n[/hide]", "Solution_2": "[hide=\"Alternately\"] The third power of $ \\mathbf{M} \\equal{} \\left[ \\begin{array}{cc} c & \\minus{}1 \\\\ 1 & 0 \\end{array} \\right]$ must be a multiple of the identity matrix, so its satisfies $ \\mathbf{M}^3 \\equal{} k \\mathbf{I}_2$ (but no smaller power should be the identity matrix) and its characteristic polynomial should share roots with it. That polynomial is\n\n$ \\lambda(\\lambda \\minus{} c) \\plus{} 1 \\equal{} \\lambda^2 \\minus{} c \\lambda \\plus{} 1$.\n\nThis is clearly possible when $ c \\equal{} \\pm 1$. Otherwise, there are either repeated roots when $ c \\equal{} \\pm 2$ (which would make the period $ 1$ and not $ 3$) or there are two roots with different absolute values. [/hide]" } { "Tag": [ "summer program", "Mathcamp", "AwesomeMath", "PROMYS", "HCSSiM" ], "Problem": "Does anyone know any GREAT math summer camps that they would recommend? :)", "Solution_1": "I know of [url=http://cty.jhu.edu/]CTY[/url], [url=http://www.mathcamp.org]MathCamp[/url], and [url=http://www.awesomemath.org]AwesomeMath[/url]. But there may be more I haven't heard of (as well as the MOP, but qualification for that is extremely, extremely difficult).", "Solution_2": "[quote=\"worthawholebean\"]I know of [url=http://cty.jhu.edu/]CTY[/url], [url=http://www.mathcamp.org]MathCamp[/url], and [url=http://www.awesomemath.org]AwesomeMath[/url]. But there may be more I haven't heard of (as well as the MOP, but qualification for that is extremely, extremely difficult).[/quote]\r\nI think mathcamp is the favored of those three.\r\nTry promys as well.", "Solution_3": "There is quite a comprehensive list in the resources section of this site, and it had links too, i would look there to start.", "Solution_4": "HCSSiM!\r\n :lol:", "Solution_5": "Ah. thank you. i will check all of those out.", "Solution_6": "It also depends on what state you live in. If you live in Texas, I would recommend the math camp at Texas State University." } { "Tag": [ "probability", "AMC", "AIME", "geometry", "function", "quadratics", "algebra", "\\/closed" ], "Problem": "Hi,\r\n\r\nI am debating on whether to take the Intermediate Algebra or the Intermediate Counting and Probability online course (I am only going to take one of them). I am not familiar with the concepts that are going to taught in both those classes. So, I am basically aiming to take the course that proved to be comparatively difficult between Intermediate Algebra, and Intermediate Counting and Probability for the majority (it is true that everybody has a different perspective on what is difficult, but I am just looking for a general opinion).\r\n\r\nMy level right now: I am almost done with Introduction to Counting and Probability (2 more chapters to go.) I am also almost done with Introduction to Algebra (4 more chapters to go.) I have only done like 2 chapters from Introduction to Number Theory (I'm planning to do more of it when I am done with Intro. to Algebra and Intro. to Counting and Probability), and I am taking the online course on Introduction to Geometry. \r\n\r\nThank you very much for the help.", "Solution_1": "I would recommend you to take Intermediate Algebra. \r\n\r\nI myself have gotten Intermediate Algebra a few days back and even Chapter 1 challenge problems proved to be difficult for me (though you're probably way smarter than me). I would assume the course would be around the same level as the book.\r\n\r\nOn the other hand, I was able to do about half of Chapter 11 problems from Intermediate Counting and Probability (I know this isn't much) without any lessons whatsoever. \r\n\r\nLooking at it from a competition perspective, I see that you're still in middle school (I'm not a stalker). For that age, I would recommend Intermediate Algebra as the AMCs, AIME, etc. focus much more on Algebra than on probability. Surely, taking the Algebra course would steeply improve your score on such competitions. \r\n\r\nNext, we look at the size of the book (though it is the course you are talking about). As Intermediate Algebra is a much bigger book, definitely you will do more problems and learn more material by taking that course.\r\n\r\nFinally, it comes to your preference. No matter what others say, you should always take whichever branch of mathematics that interests you more regardless of the pros and cons.\r\n\r\n(I hade the same dilemma when it came to the books)", "Solution_2": "I think definitely Intermediate Algebra. You will need the material for Intermediate Algebra a lot more and earlier than Intermediate C & P in my opinion.", "Solution_3": "You should work through some/most of Intro NT and work through Intro to Geometry with the book as a first priority. The fundamentals taught in the Intro series are probably more important for you (in terms of succeeding in contests) right now.\r\n\r\nAs for your question, the Intermediate Algebra class probably suits you better (this is the same advice I gave to pytheagle).", "Solution_4": "Yes, the Intermediate Algebra book/Algebra 3 class is the right next step for you.", "Solution_5": "Thank you very much for the replies. Here is the course outline for Algebra 3: \r\n\r\nTentative Class Outline\r\n\r\nWeek 1: Equations+Functions Review\r\nWeek 2: Equations+Functions Review\r\nWeek 3: Complex Numbers\r\nWeek 4: Quadratics\r\nWeek 5: Advanced Quadratics and Generalizations\r\nWeek 6: Higher-Degree Polynomials 1\r\nWeek 7: Higher-Degree Polynomials 2\r\nWeek 8: Higher-Degree Polynomials 3\r\nWeek 9: Multi-Variable Polynomials\r\nWeek 10: Polynomial Problem Solving\r\nWeek 11: Factorizations\r\nWeek 12: Sequences & Series\r\nWeek 13: Advanced Sequences & Series\r\nWeek 14: Identities & Induction\r\nWeek 15: Exponents & Logarithms\r\nWeek 16: Special Functions\r\nWeek 17: More Special Functions\r\nWeek 18: Functional Equations\r\nWeek 19: Binomial Expansion\r\nWeek 20: Inequalities\r\nWeek 21: Advanced Inequalities\r\nWeek 22: Symmetry in Algebra\r\nWeek 23: Substitution & Systems of Equations\r\nWeek 24: Advanced Problems\r\n\r\nWill this cover a majority of the algebra that AIME has before next year's AIME which I think will be sometime in March 2010? The first day of the course is 7th Oct, 2009, and the last day is April 7th 2010.\r\n\r\nThank you.", "Solution_6": "It will cover a vast majority of the non-trig algebra you'll see on the AIME.", "Solution_7": "Got it! Thank you very much for the replies, I really appreciate it! :)", "Solution_8": "limac, just curious what grade are you going to?", "Solution_9": "I am going to 8th grade next year." } { "Tag": [ "vector", "geometry", "parallelogram", "linear algebra", "matrix", "complex numbers" ], "Problem": "Ok. Could somone please answer this, or give me a link to a web site that can help me...\r\n\r\nif a + b = 4j - 1 and a - b = i + 3j find a and b algebraically.\r\n\r\nThen, i need to be able to draw a diagram to show how to find them geometrically. Any ideas?\r\n\r\nThanks", "Solution_1": "Algebraically, write $a=\\langle a_{1},a_{2}\\rangle$ and similarly for $b$. Each vector equation (in 2 dimensions) can be written as 2 scalar equations. This gives four equations with four unknowns (this is easier than it sounds :) ). $-1$ is probably $-i$.\r\n\r\n Geometrically: if $a$ and $b$ are sides of a parallelogram, then $a+b$ and $a-b$ are the diagonals.", "Solution_2": "[quote]Each vector equation (in 2 dimensions) can be written as 2 scalar equations. This gives four equations with four unknowns [/quote]\r\nWell, that $4\\times 4$ matrix can be effeciently block-structured as a $2\\times 2$ with diagonal blocks.\r\n\r\nJust write:\r\n\r\n$\\vec{a}+\\vec{b}=\\vec{u}$\r\n$\\vec{a}-\\vec{b}=\\vec{v}$\r\n\r\nAdd these to get $2\\vec{a}=\\vec{u}+\\vec{v}$ or $\\vec{a}=\\frac12\\left(\\vec{u}+\\vec{v}\\right).$\r\n\r\nSubtract to get $2\\vec{b}=\\vec{u}-\\vec{v}$ or $\\vec{b}=\\frac12\\left(\\vec{u}-\\vec{v}\\right).$", "Solution_3": "[quote=\"rockin_woody\"]Ok. Could somone please answer this, or give me a link to a web site that can help me...\n\nif a + b = 4j - 1 and a - b = i + 3j find a and b algebraically.\n\nThen, i need to be able to draw a diagram to show how to find them geometrically. Any ideas?\n\nThanks[/quote]\r\n\r\nu find answer on complex plane? j looks like sqrt(-1), i it's 1 (hard to believe in four variables :lol: )???? If so, really sum equations algebraically: shortly a=[(4j-1)+(1+3j)]/2=7j/2.\r\n\r\nTo sum the same geometrically: draw (4j-1) and (1+3j) as vectors on complex plane with basis (1,j), build parallelogram on them, take 2a as diagonal. Find center of diagonal known in geometry methods. The same ideas with b (bring - to coefficients). If i is unknown constant, it changes nothing especially, it will be presented at formulaes (if i and j it's unknown constants u can consider basis (i,j) with the same geometrical pictures).\r\n\r\nVector notations could be used, but u'llve no 3D? \r\nTo vectorize a and b like complex numbers (and corresponding vectors) u have to decide, what is it i,j?? Can be written also simply v=(a,b) then $A$=[1 1; 1 -1], so $A$v=d is usual system, d consists right part. In any case it'll be 2D. Why u call it 3D vector algebra question? Where is the reson? Of course, complex plane could be observed and in 3D space but it looks like too complicated approuch to this sample." } { "Tag": [ "Support", "\\/closed" ], "Problem": "WAP comes from Wireless Access Protocol. It's a sort of WWW for mobile phones.\r\n\r\nHere are some FAQs\r\n\r\n[b]1) How can I access this WAP version of the site? [/b]\r\nYou must first posses a mobile device such as a mobile phone (recommended), pda or i-pod. If you have a mobile phone, please contact your mobile operator to obtain details about connection to WAP, settings and pricings. \r\n\r\n[b]2) What is GRPS? What is the difference between GPRS and CSD? [/b]\r\nGPRS is a techology that allows instant connections to the WAP network. Most recent phones poses this tech. It can reach high speeds. Usually charged by the amount of traffic (kB) made.\r\n\r\nThe connection on CSD support is a dial-up connection type made by dialling a number supplied by your mobile network. Generally slow. Usually charged by the amount of time (s) the phone is connected.\r\n\r\nThe difference between the two are most similar with the difference between cable internet and dial-up. \r\n\r\n[b]3) What is the purpose of this WAP access method?[/b]\r\nWhen you are away from your computer, or unable to reach internet, you can still view posts, reply, quote, edit your posts, send and receive private messages, and much more. Surfing unlimited.\r\n\r\n[b]4) What other WAP sites are there in the world?[/b]\r\nThe best WAP site I know, besides AoPS - MathLinks ;), is http://tagtag.com \r\n\r\n[b]5) I cannot see anything in viewforum list. I receive a network error. [/b]\r\nPlease check if you have the appropriate settings correct with your mobile operator. Or you can try accessing the tagtag site to see if your WAP phone works correctly. \r\n\r\n[b]6) At some people their mobile device from which they posted appears written in the post. At mine it just says \"mobile device\". Why?[/b]\r\nBecause your phone is not (yet) in our database, so the WAPgate doesn't know \"who\" your phone is. But it learns quickly, so just have patience! :) \r\n\r\n[b]7) Does this WAP mod make coffee? [/b]\r\nI am sorry, but this version of the AoPS - MathLinks WAPgate does not support coffee machines. It will however be a feature that might be implemented in the next versions. \r\n\r\nIf it something does not work correctly, please report this in the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16274]bug / support thread.[/url] \r\n\r\n[to be completed in the future as questions arrise].", "Solution_1": "[quote=\"Valentin Vornicu\"][b]1) How can I access this WAP version of the site? [/b]\nYou must first posses a mobile device such as a mobile phone (recommended), pda or i-pod. If you have a mobile phone, please contact your mobile operator to obtain details about connection to WAP, settings and pricings. \n[/quote]\r\n\r\niPods don't have internet... yet ;)", "Solution_2": "[quote=\"nr1337\"] iPods don't have internet... yet ;) [/quote] I was looking into the future there ;-)

[i]Posted from a Sony Ericsson K700i [/i]" } { "Tag": [], "Problem": "Farz konid $ a,b,c\\ge 0$ va $ a^2\\plus{}b^2\\plus{}c^2\\equal{}3$ sabet konid\r\n\r\n\\[ a^2b\\plus{}b^2c\\plus{}c^2a\\le 3\\]", "Solution_1": "9=$ \\sqrt3 (a^2 \\plus{} b^2 \\plus{} c^2)^{\\frac {3}{2}} \\geq (a^2 \\plus{} b^2 \\plus{} c^2)(a \\plus{} b \\plus{} c) \\geq 3(a^2b \\plus{} b^2c \\plus{} c^2a)$\r\nso $ (a^2b \\plus{} b^2c \\plus{} c^2a) \\leq 3$", "Solution_2": "Hale khobi bod sumita :wink: \r\nva yek soale jadid\r\n\r\nFarz konid $ a,b,c,d\\ge 0$ va $ a^2\\plus{}b^2\\plus{}c^2\\plus{}d^2\\equal{}4$ sabet konid\r\n\r\n\r\n\\[ a^2b\\plus{}b^2c\\plus{}c^2d\\plus{}d^2a\\le 4\\]", "Solution_3": "Inam asoone . avval tavvajoh konin ke :\r\n$ a^{2}b^{2} \\plus{} b^{2}c^{2} \\plus{} c^{2}d^{2} \\plus{} d^{2}a^{2} \\equal{} (a^{2} \\plus{} c^{2})(b^{2} \\plus{} d^{2}) \\leq 4$\r\nhala tebghe cauchy darim :\r\n$ (a^{2}b^{2} \\plus{} b^{2}c^{2} \\plus{} c^{2}d^{2} \\plus{} d^{2}a^{2})(a^{2} \\plus{} b^{2} \\plus{} c^{2} \\plus{} d^{2}) \\geq (a^{2}b \\plus{} b^{2}c \\plus{} c^{2}d \\plus{} d^{2}a)^{2}$\r\npas hokm natije mishe .", "Solution_4": "Khob bod :) \r\n\r\nFarz konid $ a,b,c,d,e\\ge 0$ va $ a^2\\plus{}b^2\\plus{}c^2\\plus{}d^2\\plus{}e^2\\equal{}5$ maximome ebarate zir ra bedast avarid\r\n\r\n\\[ a^2b\\plus{}b^2c\\plus{}c^2d\\plus{}d^2e\\plus{}e^2a\\]", "Solution_5": "Aghaye M.A age soalaye shortliste $ 2007$ ro dide bashin soale $ 6$ e jabr daghighan yechizi too hamin mayehast .\r\nba tavvajoh be oon chizi ke oon too neveshte shode mishe sabet kard ke maximum ye adadi bozorgtar az $ 5.04568$ hast va kamtar az $ 5.13065$ hast . \r\noon meghdare $ 5.04568$ ke goftam be ezaye adade zir be dast miad :\r\n$ a\\equal{}1.31324$ , $ b\\equal{}1.51404$ , $ c\\equal{}0.9445$ , $ d\\equal{}0.3005$ , $ e\\equal{}0.02973$" } { "Tag": [ "LaTeX", "\\/closed" ], "Problem": "Is it possible to configure a webserver like Apache to handle latex for testing locally? If so can someone please give me the directions or a link to the directions?\r\n\r\nI am using Apache 2.2 and have configured PHP 5.26 on it\r\nOS = Windows", "Solution_1": "I'm not sure what the problem would be. Install MiKTeX or TeXLive to a directory without spaces (usually c:\\texmf), make sure that the path environment includes the bin directory and that's it.\r\n[quote=\"mathemagician1729\"]Is it possible to configure a webserver like Apache to handle latex for testing locally?[/quote]Don't you mean remotely since otherwise what would Apache have to do with it?\r\nTo access LaTeX from your browser you write pages using PHP's exec command to produce something like the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_TeXer.php]TeXer[/url] for small images or a complete docment compiler like [url=http://nirvana.informatik.uni-halle.de/~thuering/php/latex-online/latex.php?sprachauswahl=2&aufruf=22103]LaTeX-Online-Compiler[/url]." } { "Tag": [ "MATHCOUNTS", "email" ], "Problem": "Due to suggestions for a Mathcounts type competition in the Getting Started forum I am thinking about hosting a Mock Mathcounts Math Chat. Basically what would happen is that I will post the problem in the chat room and then the first 3 or so ppl to correctly answer it get a point. It will be somewhere along those lines. Anyone who is interested please post here(or in the getting started forum where they are talking about this). It should run around an hour. The time will probably be on Friday or the weekends of one of the next two weeks(not this weekend but the two after it). \r\n\r\nAlso, if there happens to be too many people I might try to open two chat rooms to keep the lag down... does anyone know if this would work?", "Solution_1": "I'd like to try it", "Solution_2": "I'd like to sign up also.", "Solution_3": "I would like to sign up for this.", "Solution_4": "I'd like to do it. You could have a chatroom with easy problems like state level, and one with national level problems.", "Solution_5": "I don't think we should do that untill more people want to participate.", "Solution_6": "[quote=\"h_s_potter2002\"]I don't think we should do that untill more people want to participate.[/quote]\r\n\r\nI don't think we'll need to worry about lag if this many people participate.", "Solution_7": "[quote=\"RC-7th\"][quote=\"h_s_potter2002\"]I don't think we should do that untill more people want to participate.[/quote]\n\nI don't think we'll need to worry about lag if this many people participate.[/quote]\r\n\r\nI think what h_s_potter2002 means is that if there are only 5 or 6 people participating in the chats in total, it wouldn't be very convenient to have multiple chats up at once.\r\n\r\nGreat idea Joml! [color=green][b]stickificated[/b][/color]", "Solution_8": "4 people doesn't seem a like a lot to me.", "Solution_9": "I would be interested. I understand what the others were saying about how it would be silly to have 2 with so few people, but the idea of two at different levels would be convenient. It would give people a variety and would allow people who are past states to do something valuable with their time.", "Solution_10": "For those attending, please post if you would like this to be on a certain day or at a certain time. I would like to have it so that as many people that want to attend can. How is Friday? If not Friday it would have to be either Saturday or Sunday due to school. For the time I can start it anytime from 6 at the earliest to 7:30 at the latest. It should run for about an hour. I think that 7 PM EST will probably suit me best. I can arrange it at another time if there are complaints, however. As for the setup, I am going to have only one chatroom(unless a massive amount of ppl just decide to do it in the near future ;) ). I am thinking that the first 30 minutes will be about Local/State level and the last 30 minutes will be National level. I will clear the scoreboard after the Local/State level is done. I encourage ppl participating to do both levels.", "Solution_11": "I can make it on Friday and Sunday, but I probably won't be able to make it on Saturday.", "Solution_12": "Friday evening at like 6 or 6:30 would be fine with me. Saturday and Sunday is ok too, but I prefer Friday.", "Solution_13": "Friday at 6 is good for me", "Solution_14": "I am going to make it Friday at 6:15 EST since I have cross country and I am not sure when that gets over. O yeah, could those who are participating please notify me of you AIM sn. You can either post it here, IM me(BaseballJoe2002), pm me, or email me at jlaurendi@gmail.com\r\n\r\nAnyone can still come. If you know how to invite yourself in a chatroom, we will be using Math Chat. Those who aren't in middle school can still come(But if your at an extremely high level I would ask that you don't steal all the fun from everyone). Also, I am thinking that the best way to conduct this is for me to post the problem in the chat room and then for you to IM me your answer. This way we can give points to the first three to correctly answer the question.", "Solution_15": "Joml88 might not make it the first few minutes, so he gave me the problems, and I can host it. If people want to start coming to the chat room.", "Solution_16": "If anyone else wants to do this we are starting right now. IM me if you want in BaseballJoe2002", "Solution_17": "Can we have the scores from the Mathchat now? I couldn't be there for the whole thing.", "Solution_18": "Ok here are the scores that I have. I am putting the AIM sn's on here because I don't know all of their AoPS names.\r\n\r\nAoPS Mathlinks 20\r\n\r\nlonelytraveler8 6\r\n\r\nhspotter2002 7\r\n\r\nspillsdaughter 1\r\n\r\n[edit]koopakingz0 4\r\n\r\nKillinginSleep 0", "Solution_19": "The screename u forgot is koopakingz0. And it's me, dinoman.", "Solution_20": "May I join?\r\n\r\nAIM: Gupti Aga", "Solution_21": "this kinda happened a long time ago and i dont think there has been one since\r\nit would be great if we could do another one", "Solution_22": "Maybe... I could do one... on Friday... but then again... im often busy.", "Solution_23": "I could host one...\r\n\r\nOf course, I would have to cut through tons of parental red tape... 10 PM bedtime, 1 hour for piano, 30 minutes for cello, tons of schoolwork to study, independent study, etc.\r\n\r\nI think I could host one on Monday at 4:30 PM, right after MathCounts practice...", "Solution_24": "Hmm... sounds good. I'll comoderate then... if I can make it. I'll notice people in the beginner's forum too.", "Solution_25": "[quote=\"white_horse_king88\"]Hmm... sounds good. I'll comoderate then... if I can make it. I'll notice people in the beginner's forum too.[/quote]\r\n\r\nIf you're one of the people who care, that's notify, not notice.", "Solution_26": "Well, EXCUSE ME. Alright????", "Solution_27": "[quote=\"Treething\"]I could host one...\n\nOf course, I would have to cut through tons of parental red tape... 10 PM bedtime, 1 hour for piano, 30 minutes for cello, tons of schoolwork to study, independent study, etc.\n\nI think I could host one on Monday at 4:30 PM, right after MathCounts practice...[/quote]\r\n\r\n4:30 pm in which time zone?", "Solution_28": "I meant 7:30 east coast. Sorry.\r\n\r\nAIM SN: Gupti Aga", "Solution_29": "I'm going to un-sticky this for now, since there's a lack of interest for the moment. If a new math-chat is set up or scheduled, I will consider stickying it again.", "Solution_30": "my team practices every Saturday from 9:30 - 2:30 if you ever wanted to run one of these on a Saturday during those times, I could have 8-12 kids all do it online together as long as the school's server doesn't crash :lol: \r\n\r\nLet me know, \r\n\r\nJosh Frost", "Solution_31": "I'm thinking of holding another math chat this weekend.\r\n\r\nIf interested, please post.\r\n\r\nP.S. If you're not sure I'm going to be the good math chat leader, trust me, I started this originally.", "Solution_32": "if you can give a firm time, then I'll tell you if I(we) can make it, \r\n\r\nJosh Frost", "Solution_33": "i'm interested\r\nif its gonna be on saturday it would be after like 12:30 to work for me", "Solution_34": "I can't do on saturday. I would be back really late.\r\n\r\nI'm thinking of like 2:00 EST (dunno that in other period) and like for 30 mins (for ones that I'm going to be on)\r\n\r\nWhat we might do is since it's only 30 mins, we can like gather some problems and talk about it.\r\n\r\nI'll bring some from AHSME.", "Solution_35": "I can't do it on Friday between 3:45 and 6:30.\r\n\r\nSaturday is fine, though. All I need to do is convince my parents to go online for an educational event, which shouldn't be too hard." } { "Tag": [ "algebra", "polynomial", "limit", "algebra open" ], "Problem": "Does there exist a sign to present the first coeffients of the polynomail $P(x)$?(My english is horrible, I hope you can understand me)", "Solution_1": "Sorry, but I don't understand what you ask for.", "Solution_2": "Do you mean $\\lim_{x \\rightarrow \\infty}\\frac{P(x)}{x^{n}}$ where $n$ is the degree of $P(x)$?", "Solution_3": "[quote=\"modeler\"]Do you mean $\\lim_{x \\rightarrow \\infty}\\frac{P(x)}{x^{n}}$ where $n$ is the degree of $P(x)$?[/quote]\r\nYes,this is what I want,But do there anther sign for it (like $deg P(x)$ means the degree.", "Solution_4": "My teacher used to write things like $coef_{0}\\{P(x)\\}$ or $coef_{n}\\{P(x)\\}$ and everybody just assumed it was standard notation. It seems international too.", "Solution_5": "I've never seen that one before...", "Solution_6": "me neither, I think it's just called leading coefficient. :P", "Solution_7": "Thank you all." } { "Tag": [ "geometry", "3D geometry", "trigonometry", "trig identities", "Law of Cosines" ], "Problem": "Let $ ABC$ be the axial section of a cone with diametre $ AB\\equal{}3 cm$ and generatrix $ AC\\equal{}6 cm$. Find the shortest path to reach the midpoint of $ BC$ from point $ A$.", "Solution_1": "[hide]The generatrix is the slant height, so we have that the diameter is $ 3$ and the slant height is $ 6$.\nUnfolding the cone, we have a sector of a circle with a radius of the slant height, which is $ 6$. Then, the arc intercepted is the circumference of the base of the cone, which is $ 3\\pi$. Since the circumference of the circle with radius $ 6$ is $ 2\\cdot6\\pi\\equal{}12\\pi$, the sector is a quarter of a circle. Drawing a diagram, and using the Law of Cosines, we have:\n\\[ x^2 \\equal{} 6^2\\plus{}3^2\\minus{}2\\cdot 3\\cdot 6\\cdot \\cos \\frac{\\pi}{4} \\\\\n\\equal{} 45\\minus{}36\\cdot \\frac{\\sqrt{2}}{2} \\\\\n\\equal{}\\boxed{\\sqrt{45\\minus{}18\\sqrt{2}}}\\][/hide]", "Solution_2": "[quote=\"123456789\"][hide]The generatrix is the slant height, so we have that the diameter is $ 3$ and the slant height is $ 6$.\nUnfolding the cone, we have a sector of a circle with a radius of the slant height, which is $ 6$. Then, the arc intercepted is the circumference of the base of the cone, which is $ 3\\pi$. Since the circumference of the circle with radius $ 6$ is $ 2\\cdot6\\pi \\equal{} 12\\pi$, the sector is a quarter of a circle. Drawing a diagram, and using the Law of Cosines, we have:\n\\[ x^2 \\equal{} 6^2 \\plus{} 3^2 \\minus{} 2\\cdot 3\\cdot 6\\cdot \\cos \\frac {\\pi}{4} \\\\\n\\equal{} 45 \\minus{} 36\\cdot \\frac {\\sqrt {2}}{2} \\\\\n\\equal{} \\boxed{\\sqrt {45 \\minus{} 18\\sqrt {2}}}\n\\]\n[/hide][/quote]\n\nAlmost, except $ \\frac {\\pi}{4}$ is an eighth of a circle. Well, since he's not online, I'll correct it right now.\n\n[hide]$ x^2 \\equal{} 6^2 \\plus{} 3^2 \\minus{} 2\\cdot 3\\cdot 6\\cdot \\cos \\frac {\\pi}{2} \\equal{} 36 \\plus{} 9 \\minus{} 0 \\equal{} 45$, so $ x \\equal{} \\sqrt{45}$.[/hide]", "Solution_3": "[quote=\"syckls\"][quote=\"123456789\"][hide]The generatrix is the slant height, so we have that the diameter is $ 3$ and the slant height is $ 6$.\nUnfolding the cone, we have a sector of a circle with a radius of the slant height, which is $ 6$. Then, the arc intercepted is the circumference of the base of the cone, which is $ 3\\pi$. Since the circumference of the circle with radius $ 6$ is $ 2\\cdot6\\pi \\equal{} 12\\pi$, the sector is a quarter of a circle. Drawing a diagram, and using the Law of Cosines, we have:\n\\[ x^2 \\equal{} 6^2 \\plus{} 3^2 \\minus{} 2\\cdot 3\\cdot 6\\cdot \\cos \\frac {\\pi}{4} \\\\\n\\equal{} 45 \\minus{} 36\\cdot \\frac {\\sqrt {2}}{2} \\\\\n\\equal{} \\boxed{\\sqrt {45 \\minus{} 18\\sqrt {2}}}\n\\]\n[/hide][/quote]\n\nAlmost, except $ \\frac {\\pi}{4}$ is an eighth of a circle. Well, since he's not online, I'll correct it right now.\n\n[hide]$ x^2 \\equal{} 6^2 \\plus{} 3^2 \\minus{} 2\\cdot 3\\cdot 6\\cdot \\cos \\frac {\\pi}{2} \\equal{} 36 \\plus{} 9 \\minus{} 0 \\equal{} 45$, so $ x \\equal{} \\sqrt {45}$.[/hide][/quote]But doesn't $ BC$ bisect the sector?", "Solution_4": "[quote=\"123456789\"]But doesn't $ BC$ bisect the sector?[/quote]\r\n\r\nDoh, you're right. I had it right the first time I worked out this problem, but then I forgot that part.", "Solution_5": "[quote=\"syckls\"][quote=\"123456789\"]But doesn't $ BC$ bisect the sector?[/quote]\n\nDoh, you're right. I had it right the first time I worked out this problem, but then I forgot that part.[/quote]I almost made that mistake at first too.", "Solution_6": "But if the circumference is $ 6\\pi$ and $ BC$ bisects it , doesn't that mean that there is indeed a quarter of circle and the answer is $ \\sqrt{45}$? :maybe:", "Solution_7": "That was the way i solved the problem too but my book says the answer is $ \\sqrt{23}$, that's way i posted it...", "Solution_8": "[quote=\"sunnykim\"]That was the way i solved the problem too but my book says the answer is $ \\sqrt {23}$, that's way i posted it...[/quote]I'm pretty sure that's wrong, unless we're all misinterpreting the question...", "Solution_9": "General principle of forum posting: there is [i]never[/i] a good reason to block-quote the entire post preceding yours.\r\n\r\nI could read the problem in two ways: the first is that the path is supposed to lie on the surface of the cone, in which case 123456789 got it. The second is that the path is supposed to lie in space. Neither gives you $ \\sqrt{23}$ -- textbooks have been known to contain the occasional error :wink:", "Solution_10": "[quote=\"mon\"]But if the circumference is $ 6\\pi$ and $ BC$ bisects it , doesn't that mean that there is indeed a quarter of circle and the answer is $ \\sqrt {45}$? :maybe:[/quote]\r\n\r\nOk..but finally the answer is $ \\sqrt {45 \\minus{} 18\\sqrt {2}}$ or $ \\sqrt{45}$? :?:" } { "Tag": [ "inequalities", "geometry", "trapezoid", "geometry proposed" ], "Problem": "We have the trapezoid ABCD with m(=2AB*CD.", "Solution_1": "What do you mean with m( y \\Longrightarrow 3^x> 3^y$\n$3^x=4 \\Longrightarrow x=\\frac{log 4}{log 3}$\n$\\sqrt{2}> \\frac{log4}{log 3} \\Longrightarrow 3^{\\sqrt{2}}> 4$\n\n$x>y \\Longrightarrow 2^x>2^y$\n$2^x=4 \\Longrightarrow x=2$\n$2>\\sqrt{3} \\Longrightarrow 4>2^{\\sqrt{3}}$\n\nThen, $3^{\\sqrt{2}}>2^{\\sqrt{3}}$ [/hide]", "Solution_3": "[quote=\"Jos\u00e9\"][quote=\"Centy\"]Determine without a calculator which is bigger: $2^{\\sqrt{3}} \\text{ or } 3^{\\sqrt{2}}$.[/quote]\n\n[hide=\"is this correct?\"]$x> y \\Longrightarrow 3^x> 3^y$\n$3^x=4 \\Longrightarrow x=\\frac{log 4}{log 3}$\n$\\sqrt{2}> \\frac{log4}{log 3} \\Longrightarrow 3^{\\sqrt{2}}> 4$\n\n$x>y \\Longrightarrow 2^x>2^y$\n$2^x=4 \\Longrightarrow x=2$\n$2>\\sqrt{3} \\Longrightarrow 4>2^{\\sqrt{3}}$\n\nThen, $3^{\\sqrt{2}}>2^{\\sqrt{3}}$ [/hide][/quote]Jos\u00e9, why is this true: $\\sqrt{2}> \\frac{log4}{log 3}$? Remember no calculators...", "Solution_4": "$\\sqrt{2}>\\frac{7}{5}$\r\n$3^{\\frac{7}{5}}>4$", "Solution_5": "[quote=\"Jos\u00e9\"]$\\sqrt{2}>\\frac{7}{5}$\n$3^{\\frac{7}{5}}>4$[/quote]\r\nOK, and how do you know $\\frac75\\approx\\frac{\\log 4}{\\log 3}$? Show me something that you have not used a calculator please.", "Solution_6": "[hide=\"A slightly different solution\"]\nIf $3^{\\sqrt{2}}>2^{\\sqrt{3}}$, then $3^{\\sqrt{2}\\cdot \\sqrt{3}} >2^{\\sqrt{3}\\cdot \\sqrt{3}}=8$\n\nSince $3^{\\sqrt{6}}>3^{\\sqrt{4}}=9>8$, the assumption was true, so $3^{\\sqrt{2}}>2^{\\sqrt{3}}$\n[/hide]", "Solution_7": "[hide]$(2^{\\sqrt{3}})^{\\sqrt{3}}=8$\n$(3^{\\sqrt{2}})^{\\sqrt{3}}=3^{\\sqrt{6}}$\n$9=3^{\\sqrt{4}}<3^{\\sqrt{6}}<3^{\\sqrt{9}}=27$\nso $\\boxed{3^{\\sqrt{2}}}$ is bigger.[/hide]", "Solution_8": "Jose, your solution seems to use calculator work over the exact values of $\\log 4$ and $\\log 3$.\r\n\r\nThe solutions which simply use $x < y \\Rightarrow x^k < y^k$ if $x, y, k > 1$ are far nicer.", "Solution_9": "Which is bigger,\r\n\r\n$\\frac{1}{e}$ or $\\frac{\\ln 3}{3}$ ?", "Solution_10": "[quote=\"Kalle\"]Which is bigger,\n\n$\\frac{1}{e}$ or $\\frac{\\ln 3}{3}$ ?[/quote]I did it 'backwards' and keeping the binary relation in question:\r\n\r\n[hide=\"a not-so-neat solution\"]$\\frac1e\\stackrel{?}{<<}\\frac{\\ln3}3$\n\n$e\\stackrel{?}{>>}\\log_3e^3$\n\n$3^e\\stackrel{?}{>>}e^3$\n\n$3^\\frac13\\stackrel{?}{>>}e^\\frac1e$\n\nUsing the fact that the maximum of $f(x)=x^\\frac1x$ occurs when $x=e$, then $\\stackrel{?}{>>}\\ =\\ <$ and so $\\frac1e$ is bigger.[/hide]\r\n\r\nPS: this doesn't seem like a Getting Started level type of problem", "Solution_11": "Squaring each side, we see that $8<9$. Thus, $2^{\\sqrt{3}}<3^{\\sqrt{2}}$.", "Solution_12": "[quote=\"LordoftheMorons\"]Squaring each side, we see that $8<9$. Thus, $2^{\\sqrt{3}}<3^{\\sqrt{2}}$.[/quote]\r\n\r\nYou cant manipulate them like that.\r\n\r\n$(2^{\\sqrt{3}})^2 = 2^{2\\sqrt{3}}$", "Solution_13": "About this problem, which is bigger\r\n$\\ 1000*1000...*1000(1000 therms)$ or $\\ 1*3*5*...*1999$?", "Solution_14": "[quote=\"stancioiu sorin\"]About this problem, which is bigger\n$\\ 1000*1000...*1000(1000 \\text{ terms}$ or $\\ 1*3*5*...*1999$?[/quote]\r\n\r\n[hide]Notice, that $S=1000+1000+...+1000 (1000 \\text{ terms})=1+3+5+...+1999$. By AM-GM we know that the product of $n$ values assumes it's biggest value when $a_1=a_2=...=a_n$. Thus, $1000*1000...*1000 (1000 \\text{ terms}) > 1*3*5*...*1999$.[/hide]", "Solution_15": "[hide]$\\sqrt{1^3+2^3+...+n^3}=\\sqrt{\\frac{n^2(n+1)^2} 4}=\\frac{n(n+1)} 2.$\n\n$\\frac{(n+\\frac{1}{2})^2}{2}=\\frac{n^2+n+\\frac{1}{4}}{2}$\n\n$n^2+n+\\frac1 4 = n^2+n$\n$\\frac 1 4>0$\n\nSo we can conclude that $\\sqrt{1^3+2^3+3^3+4^3+\\cdots+n^3}>\\frac \\frac{(n+\\frac{1}{2})^2}{2}$[/hide]", "Solution_16": "Yes, the culprit is me. I did it again, posting hard problems in Getting Started. So yeah, sorry guys that this got moved. I removed all off-difficulty problems (all the previous bonus problems are removed so please don't ask me for the solutions), and I will not post anymore off-difficulty problems. Better yet, this is not a marathon for me, so I'm out. See ya.", "Solution_17": "Well, let's continue. Anybodycan post the next one.......", "Solution_18": "I'll post a relatively simple one to determine who posts next.\r\n\r\n$\\sqrt{1+\\sqrt{1+\\sqrt{1+\\cdots}}}-\\frac{1}{1+\\frac1{1+\\frac1{1+\\frac1{\\vdots}}}}$ or $\\ln{\\frac{e}{\\pi}}$\r\n\r\nIt's actually very simple ;)", "Solution_19": "[hide]$\\sqrt{1+\\sqrt{1+\\sqrt{1+\\cdots}}}$\n$x=\\sqrt{1+x}$\n$x= \\phi=\\frac{1 + \\sqrt 5} 2$\n\n$\\frac{1}{1+\\frac1{1+\\frac1{1+\\frac1{\\vdots}}}}$\n$y=\\frac{1}{1+y}$\n$0=y^2+y-1$\n$y=\\frac{-1+\\sqrt 5}{2}$\n\nBy adding them we get $\\sqrt 5$\n\n$e<\\pi$, $\\frac e {\\pi}<1,\\ln{\\frac{e}{\\pi}}<1$\n$\\sqrt 5>2$, so\n$\\sqrt{1+\\sqrt{1+\\sqrt{1+\\cdots}}}-\\frac{1}{1+\\frac1{1+\\frac1{1+\\frac1{\\vdots}}}}> \\ln {\\frac{e}{\\pi}}$[/hide]\r\n\r\nAnyone else can post the next problem....", "Solution_20": "[quote=\"Hamster1800\"]I'll post a relatively simple one to determine who posts next.\n\n$\\sqrt{1+\\sqrt{1+\\sqrt{1+\\cdots}}}-\\frac{1}{1+\\frac1{1+\\frac1{1+\\frac1{\\vdots}}}}$ or $\\ln{\\frac{e}{\\pi}}$\n\nIt's actually very simple ;)[/quote]\r\n\r\nYou could just say $\\sqrt{1+\\sqrt{1+\\sqrt{1+\\cdots}}}$ is greater than 1 while $\\frac{1}{1+\\frac1{1+\\frac1{1+\\frac1{\\vdots}}}}$ is smaller than 1.\r\n\r\nSo the LHS is positive.\r\n\r\n$\\ln{\\frac{e}{\\pi}}$ is obviously negative.\r\n\r\nThus the LHS is larger.", "Solution_21": "$500!500!$ or $700!200!100!$", "Solution_22": "[hide]\n$\\frac{700!200!100!}{500!500!}$\n\n$=\\frac{500!200!100!(700\\cdot 699\\cdot ...\\cdot 501)}{500!200!(500\\cdot 499\\cdot ...\\cdot 201)}$\n\n$=\\frac{100!(700\\cdot 699\\cdot ...\\cdot 501)}{500\\cdot 499\\cdot ...\\cdot 201}$\n\n$=\\frac{100\\cdot 99\\cdot ...\\cdot 1}{500\\cdot 499\\cdot ...\\cdot 401}\\cdot \\frac{700\\cdot 699\\cdot ...\\cdot 501}{400\\cdot 399\\cdot ...\\cdot 201}$\n\n$<\\left(\\frac{1}{5}\\right)^{100}\\cdot \\left(\\frac{5}{2}\\right)^{100}$\n\n$=\\left(\\frac{1}{2}\\right)^{100}<1$\n\n$\\therefore 500!500!>700!200!100!$\n[/hide]\r\n\r\nSimilar problem: $400!400!200!$ or $600!300!100!$", "Solution_23": "is this right?\r\n\r\n[hide]\n\n$\\frac{600!300!100!}{400!400!200!} = \\frac{401 \\cdot 402 \\cdot \\dots \\cdot 599 \\cdot 600}{301 \\cdot 302 \\dots 399 \\cdot 400 \\cdot 101 \\cdot 102 \\dots 199 \\cdot 200}$\n\nand both the numerator and the denominator are the poduct of 200 integers. Furthermore, all those in the numerator are greater than those in the denominator, so the numerator ( $600!300!100!$ ) is greater. [/hide]", "Solution_24": "Yes- you can post the next problem.", "Solution_25": "I'll just post another one.\r\n\r\n$\\left(e^{100}\\right)\\left(100!\\right)$ or $100^{100}$", "Solution_26": "Using Sterling's...\r\n\r\n$\\sqrt{2 \\cdot 100\\pi} \\cdot (\\frac{100}{e})^{100} \\cdot e^{100}=\\sqrt{200\\pi} \\cdot 100^{100}$\r\n\r\n$\\sqrt{200\\pi} \\cdot 100^{100} > 100^{100}$\r\n\r\n$100! \\cdot e^{100} > 100^{100}$", "Solution_27": "Although I don't quite follow you, you got the right answer, so I'll post another one.\r\n\r\nLet $a$ be the sum of the absolute values of the solutions of\r\n\r\n$3x^4 + 24x^3 + 72x^2+96x+\\frac{671}{27} = 0$\r\n\r\nLet $b$ be the sum of the absolute values of the solutions of\r\n\r\n$\\left(25^{25}\\right)\\left( x^{23}\\right) - \\frac{10^{3^3}}{2^{2^2}} = 0$\r\n\r\nWhich is greater, $a$ or $b$?", "Solution_28": "Only real solutions or are complex solutions included also?", "Solution_29": "Complex solutions included in both cases." } { "Tag": [ "geometry" ], "Problem": "A regular hexagon is inscribed in a circle of radius $ 10$ inches. Its area is:\r\n\r\n$ \\textbf{(A)}\\ 150\\sqrt{3} \\text{ sq. in.} \\qquad\r\n\\textbf{(B)}\\ \\text{150 sq. in.} \\qquad\r\n\\textbf{(C)}\\ 25\\sqrt{3} \\text{ sq. in.} \\qquad\r\n\\textbf{(D)}\\ \\text{600 sq. in.} \\qquad\r\n\\textbf{(E)}\\ 300\\sqrt{3} \\text{ sq. in.}$", "Solution_1": "[hide=\"Click for solution\"]\n$ A\\equal{}\\frac{1}{2}ap\\equal{}\\frac{1}{2} \\cdot 5\\sqrt{3} \\cdot 60\\equal{}150\\sqrt{3}$, or $ \\boxed{\\textbf{(A)}}$.\n[/hide]" } { "Tag": [ "function", "quadratics", "calculus", "derivative", "linear algebra", "matrix", "algebra" ], "Problem": "If $ f \\equal{} f(x_1,x_2,...,x_n)$, $ \\frac {{\\partial}^2 f}{\\partial x_i^2} \\geq 0$ for $ i \\equal{} 1,2,...,n$ and $ \\sum_{i \\equal{} 1}^{n}{x_i} \\equal{} 1$, than prove or disprove that $ f_{max} \\equal{} f(1,0,0,...,0)$ or any permutation of set $ (1,0,...,0)$ -f reaches maximum for an permutation of a set (1,0,...,0). \r\n\r\n* $ f_{max}$ means maximum value of a function f.", "Solution_1": "Obviously false. Take $ n\\equal{}2$ and $ f(x)\\equal{}x_1x_2$. You need the second differential to be a non-negative quadratic form for this, not just pure partial derivatives being non-negative.", "Solution_2": "[quote=\"fedja\"]Obviously false. Take $ n \\equal{} 2$ and $ f(x) \\equal{} x_1x_2$. You [b]need the second differential to be a non-negative quadratic form[/b] for this, not just pure partial derivatives being non-negative.[/quote]\r\n\r\nThat means if this bold is true, than the question I've asked is true?\r\n\r\nThanks!", "Solution_3": "I don't think so.\r\n\r\nUnless I'm seriously mistaken (and note that it's highly probable, as multivariate calculus still doesn't sit quite right with me :)), take $ n \\equal{} 2$ and $ f(x,y) \\equal{} x^2 \\plus{} y^2$. Obviously, this has no maximum, even taking $ x \\plus{} y \\equal{} 1$ into account. However, the Hessian of $ f$ (if that's what fedja was referring to) is $ H_f \\equal{} \\left( \\begin{array}{cc} 2 & 0 \\\\\r\n0 & 2 \\end{array} \\right)$, which is obviously a positive-definite matrix (meaning the second differential is a positive quadratic form) and it also satisfies your original condition.\r\n\r\n(By the way, I fail to see the reasoning behind the idea that $ (1,0,0,\\ldots,0)$ is a maximum - besides being a member of canonical basis for $ \\mathbb{R}^n$, I don't see what's so special about it.)\r\n\r\nHaving a positive-definite Hessian does indeed, if I'm not mistaken, guarantee that there are no local maximums in the interior of the domain, which is kind of close to what you seem to be looking for. However, I guess the domain here would be $ \\mathbb{R}^n$, as the set defined with $ \\sum_{i \\equal{} 1}^n x_i \\equal{} 1$ isn't an open set (nor it even has an interior, if I'm interpreting this correctly) in $ \\mathbb{R}^n$, so differentiating the function on that set might be problematic. \r\n\r\nSo, if $ \\mathbb{R}^n$ is indeed the domain of $ f$, it contains no maximums whatsoever. Please note that doesn't mean we can say anything about the existence or locations of possible maximums of $ f$ on its restriction defined by $ \\sum_{i \\equal{} 1}^n x_i \\equal{} 1$ - again, since this is not an open set, differentiation might not be a way to go here.\r\n\r\nIf you were to modify the condition into $ \\sum_{i \\equal{} 1}^n |x_i|\\leq 1$, the argument which I described would, I believe, guarantee that the maximum is attained somewhere on the edge (where $ \\sum_{i \\equal{} 1}^n |x_i| \\equal{} 1$), but that still probably isn't what you were looking for.\r\n\r\nUrgh, I wrote way too much than I wanted to :D (And much of it is probably wrong :blush:... my luck these days :))", "Solution_4": "You are formally right, of course, but I'm ready to bet \\$1000 against 1c that OP just forgot to write the condition $ x_j\\ge 0$ in his definition of the domain over which the maximum should be taken ;).", "Solution_5": ":lol: Yeah, you're probably right.\r\n\r\nIn that case, then, the short answer is yes - I think it follows easily from the convexity of the function, which is guaranteed if $ H_f\\geq 0$ (in this context, that means $ H_f$ is positive-semidefinite).", "Solution_6": "That's true that I forgot to write that $ x_1,...,x_n$ are positive numbers. \r\nThank you! :lol:" } { "Tag": [ "probability", "percent" ], "Problem": "Given three standard six-sided dice, what is the probability that the sum of the numbers on each dice is an odd number?", "Solution_1": "[quote=\"drunner2007\"]Given three standard six-sided dice, what is the probability that the sum of the numbers on each dice is an odd number?[/quote]\r\n[hide]There is odd-odd-odd and odd-even-even. \n1) a) If no odds are the same, there are 6 ways to arrange, $6(1/2)(1/3)(1/6)=1/6$\nb) If two odds are the same, there are 3 ways to arrange, $3(1/2)(1/2)(1/3)=1/4$\nc) If all odds are the same, there is one way to arrange, $(1/2)(1/6)(1/6)=1/72$\n2) a) If the evens are different, there are 6 ways to arrange, $6(1/2)(1/2)(1/3)=1/2$\nb) If the evens are the same, there are 3 ways to arrange, $3(1/2)(1/2)(1/6)=1/8$ \nBut then the probability is more than 1...can someone tell me what I did wrong?[/hide]", "Solution_2": "[hide] is it 1/2?\n\n27 ways to get o-o-o\n81 ways to get o-e-e or permutation.[/hide]\nI'm probably missing something though.[/hide]", "Solution_3": "[quote=\"drunner2007\"]Given three standard six-sided dice, what is the probability that the sum of the numbers on each dice is an odd number?[/quote]\r\n\r\nWhat do you mean by sum of the numbers on each dice? Is it like, the sum of the top faces or the sum of all of the faces? It is not clear to me.", "Solution_4": "i mean sum on all of the [i]top[/i] faces. i'm not 100 percent sure of the answer, but i got jclarke's answer using the exact same reasoning after looking at this problem for a while.\r\n\r\ni'm not sure about ch1n353ch3s54a1l 's answer\r\n\r\nif i can make sure that my answer correct, I'll probably use this on the tournament that I'm writing.", "Solution_5": "[hide=\"Possibility\"]Well since the low sum is 3 and the high is 18, there is an even number of sums. The median sum is 10.5 and there is a correspondence between the terms in the lower half and upper half, so the 10 and 11 have the same probability, 9 and 12, etc. down to 3 and 18 so the probability is ...1/2...?\n[/hide]", "Solution_6": "Think of it as three coins. One side of a coin is \"odd,\" and the other is \"even.\" When you toss these coins you must end up with an odd number of odds. There is an equal probability, however, of getting an odd number of evens and thus an even number of odds. So the probablity of getting an odd number of odds is 1/2.", "Solution_7": "[hide]$\\frac{1}{2}$[/hide]", "Solution_8": "[hide=\"question\"]I also thing it's a half, but does the calculate is right: the first throwing doesn't matter, and then $2\\cdot(\\frac{1}{2})^2=\\frac{1}{2}$?[/hide]", "Solution_9": "yeah, that seem's right.\r\n\r\nwow, that's the best solution yet.", "Solution_10": "[quote=\"drunner2007\"]wow, that's the best solution yet.[/quote]\r\n\r\nHow about this one:\r\n\r\nThe probability distribution doesn't change with each added die. :)", "Solution_11": "that's good to :)", "Solution_12": "[hide]there are 6^3=216 ways to get a number with combinations, there are 27 ways to get odd-odd-odd and 81 ways to get odd-even-even, this is just permutations, add them and divide by 216, you get 108/216=[b]1/2[/b][/hide]" } { "Tag": [ "geometry", "cyclic quadrilateral", "perpendicular bisector", "geometric transformation", "geometry solved", "USA TST", "steiner line" ], "Problem": "Let $ ABCD$ be a cyclic quadrilateral and let $ E$ and $ F$ be the feet of perpendiculars from the intersection of diagonals $ AC$ and $ BD$ to $ AB$ and $ CD$, respectively. Prove that $ EF$ is perpendicular to the line through the midpoints of $ AD$ and $ BC$.", "Solution_1": "Posted before, even quite recently (few months ago only!), if I am not wrong.\r\n\r\nBest regards,\r\nsunken rock", "Solution_2": "This problem is missing from the [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=35&year=2000&]resource page[/url].\n\n[hide=\"solution\"]Let $AC$ and $BD$ meet at $P$, the midpoints of $AD,BC,DP,AP$ are $M,N,R,S$, respectively. We have $FR=\\frac12DP=MS$, $ES=\\frac12AP=MR$, \\[\\angle MRF=\\angle MRD+\\angle DRF=\\angle ASM+2\\angle DPF=\\angle ASM+2(90^{\\circ}-\\angle PDF)=\\angle ASM+2(90^{\\circ}-\\angle PAE)=\\angle ASM+2\\angle APE=\\angle ASM+\\angle ASE=\\angle MSE.\\] Hence $\\triangle MRF\\cong\\triangle MSE$, which implies $ME=MF$. Similarly, we get $NE=NF$. Therefore $MENF$ is a kite, whence $EF\\perp MN$.[/hide]", "Solution_3": "It has now been corrected :lol: \n\nFor reference, also posted here:\n\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=296040\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=12098", "Solution_4": "See and [url=http://www.artofproblemsolving.com/blog/52010][color=red][b]here[/b][/color][/url].", "Solution_5": "Another way is possible also.\n\nYou can use just a bit of complex numbers. Let $P$ be the origin, and $E$ be at $e$, $F$ at $f$. We have that for some $k$, $A=e+ike$ and $D=f-ikf$ because of the perpendicularity and $PFD$ is similar to $PEA$.\n\nLet the midpoint of $AD$ be $M$, which is half of $e+f+ike-ikf$. Now, $ME$ is half $-e+f+ike-ikf$ and $MF$ is half $e-f+ike-ikf$. These are $(e-f)(-1+ik)$ and $(e-f)(1+ik)$ respectively, and they have the same magnitude because $-1+ik$ and $1+ik$ have the same magnitude.\n\nTherefore, $ME=MF$, and the midpoint of $AD$ is on the perpendicular bisector of $EF$. Similar holds for the midpoint of $BC$ so we are done.", "Solution_6": "[quote=\"jgnr\"]This problem is missing from the [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=35&year=2000&]resource page[/url].\n\n[hide=\"solution\"]Let $AC$ and $BD$ meet at $P$, the midpoints of $AD,BC,DP,AP$ are $M,N,R,S$, respectively. We have $FR=\\frac12DP=MS$, $ES=\\frac12AP=MR$, \\[\\angle MRF=\\angle MRD+\\angle DRF=\\angle ASM+2\\angle DPF=\\angle ASM+2(90^{\\circ}-\\angle PDF)=\\angle ASM+2(90^{\\circ}-\\angle PAE)=\\angle ASM+2\\angle APE=\\angle ASM+\\angle ASE=\\angle MSE.\\] Hence $\\triangle MRF\\cong\\triangle MSE$, which implies $ME=MF$. Similarly, we get $NE=NF$. Therefore $MENF$ is a kite, whence $EF\\perp MN$.[/hide][/quote]\nDayummm this is a nice solution. Although, what is the motivation for constructing the midpoints of S and R besides the right angles ad PF and PE?", "Solution_7": "Does anyone have a proof using spiral similarity?", "Solution_8": "bump :D $\\text{ }$", "Solution_9": "^^^Here is a solution that is absolutely effortless. So, note that the line adjoining the midpoints is antiparallel to the line $XY$ where $X, Y = AC \\cap BD, AB \\cap CD$. But clearly, $XY$ is anti-parallel to the line perpendicular to $EF$, so the lines adjoining the midpoints must be perpendicular to $EF$ so done.", "Solution_10": "@mathisfun7 yes I found a spiral similarity solution last night! Let M be the midpoint of AD, N of BC, R of DP, and S of CP. I will prove that F is the center of the spiral smiilarity bringing quad. FRPS to FMEN. Since clearly RS||CD perpendicular to FP then this will solve the proof. \nDenote by X 0$ \u03c4\u03bf \u03c0\u03bf\u03bb\u03cd \u03b3\u03b9\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 $ \\alpha\\in A$.\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_1": "Cute.\r\n\r\n\u03a3\u03b1\u03c4\u03b1\u03bd\u03b9\u03ba\u03ad\u03c2 \u03c3\u03c5\u03bc\u03c0\u03c4\u03ce\u03c3\u03b5\u03b9\u03c2? :diablo: :roll: (pun intended) \u0391\u03bd \u03b7 \u03c6\u03c1\u03b1\u03b3\u03bc\u03ad\u03bd\u03b7 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 $ f : R \\to \\mathbb{R}$, $ R \\equal{} [a_1,b_1] \\times \\cdots \\times [a_n,b_n]$, $ n \\geq 1$, \u03b5\u03af\u03bd\u03b1\u03b9 Riemann-\u03bf\u03bb\u03bf\u03ba\u03bb\u03b7\u03c1\u03ce\u03c3\u03b9\u03bc\u03b7 \u03c3\u03c4\u03bf $ R$, \u03c4\u03cc\u03c4\u03b5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c7\u03ae\u03c2 \u03c3\u03c7\u03b5\u03b4\u03cc\u03bd \u03c0\u03b1\u03bd\u03c4\u03bf\u03cd (\u03ba\u03b1\u03b9 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03c1\u03bf\u03c6\u03b1, \u03ba\u03b1\u03b9 \u03b2\u03b3\u03ac\u03b6\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03b9 Lebesgue \u03bf\u03bb\u03bf\u03ba\u03bb\u03b7\u03c1\u03c9\u03c3\u03b9\u03bc\u03cc\u03c4\u03b7\u03c4\u03b1)." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "$E$ is a normed vectorial space of dimension $n$.\r\n$C$ is a connex subset of $E$. Is it true that $Conv(C)$, the minimal convex set that contains $C$, is the set of all the points that are barycentre of at most $n$ points of $C$.", "Solution_1": "If $C$ is the union of two sides of a triangle (endpoints included) in the plane, how can you get the third side?", "Solution_2": "I don't understand what you mean :?", "Solution_3": "Looks like [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=caratheodory&t=24464]it's been discussed before[/url].\r\n\r\nI'm not sure about what I wrote there, though :).", "Solution_4": "Clarifying Xamog's counterexample:\r\n\r\nLet $C$ be the union of the segments $OA$ and $OB$, where $O=(0,0),A=(1,0),B=(0,1)$. Then the set of barycenters (midpoints) between points of $C$ is the union of the square $0\\le x,y\\le\\frac12$ and the two segments $OA$ and $OB$, which is not a convex set.", "Solution_5": "I'm sure that by \"barycenter\" alekk did not mean merely midpoints, but convex combinations in general.", "Solution_6": "yes :)" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra open" ], "Problem": "So given a finite group G, a subgroup H that contains NG(P) and P a p-sylow subgroup demonstrate the following:\r\n\r\n(a) H=NG(H) <--I was able to prove this\r\n(b) [G:M] \\equiv 1 mod p where p is a prime[/code]", "Solution_1": "What is M in your second question?" } { "Tag": [], "Problem": "Because I'm receiving a \"Forbidden\r\nYou don't have permission to access / on this server.\"", "Solution_1": "Works for me-- [url]http://www.naclo.cs.cmu.edu/[/url] Which site/page were you trying to access?" } { "Tag": [ "function" ], "Problem": "The function [i]f[/i] satisfies [i]f(ax)[/i] = [i]af(x) [/i]for all [i]a [/i]and [i]x[/i], and f(2) = 3. What is the value of f(19)?\r\n\r\nPlease show all of you work. :lol:\r\n\r\nThank you :)", "Solution_1": "[quote=\"dogseatcheese\"]The function [i]f[/i] satisfies [i]f(ax)[/i] = [i]af(x) [/i]for all [i]a [/i]and [i]x[/i], and f(2) = 3. What is the value of f(19)?\n\nPlease show all of you work. :lol:\n\nThank you :)[/quote]\r\n\r\n[hide]f(19)=28.5? I have an extremely dumb solution so i'll only embarrass myself if i'm right [/hide] :)", "Solution_2": "That IS the right answer..... yet I would be happy to know any solution to this problem so......do you mind telling me? Unlike you, I have NO solution :oops: :D", "Solution_3": "[quote=\"dogseatcheese\"]That IS the right answer..... yet I would be happy to know any solution to this problem so......do you mind telling me? Unlike you, I have NO solution :oops: :D[/quote]\r\n?!?! whoa :P \r\n\r\nwell i assumed the function to be linear, i don't know why. the equation can't have a y intercept because then the values of f(ax) and af(x) would be different since a is multipled by x in the first and the whole function is multiplied by a in the second and the y intercept would make them different. So if the form is y=mx, then the equation is y=1.5x and $1.5\\times19=28.5$", "Solution_4": "It has to be in the form $f(x)=mx$; you don't need to assume it. It can't have a constant on the end. Say it was $f(x)=mx+b$. Then, you have $f(ax)=m(ax)+b$ and $af(x)=a(mx+b)=m(ax)+ab$, which obviously does not satisfy $f(ax)=af(x)$ for all values. Therefore, it cannot have a constant on the end. If it is $f(x)=x^n$, then you have $f(ax)=a^nx^n$ and $af(x)=a(x^n)$, which is again not true for every value. Therefore, it must simply be $f(x)=mx$, and you go from there. ;)", "Solution_5": "[quote=\"JesusFreak197\"]It has to be in the form $f(x)=mx$; you don't need to assume it. It can't have a constant on the end. Say it was $f(x)=mx+b$. Then, you have $f(ax)=m(ax)+b$ and $af(x)=a(mx+b)=m(ax)+ab$, which obviously does not satisfy $f(ax)=af(x)$ for all values. Therefore, it cannot have a constant on the end. If it is $f(x)=x^n$, then you have $f(ax)=a^nx^n$ and $af(x)=a(x^n)$, which is again not true for every value. Therefore, it must simply be $f(x)=mx$, and you go from there. ;)[/quote]\r\n\r\nWait, but can you automatically assume that the function is linear? other functions might not have constants at the end either", "Solution_6": "I already addressed the other possibilities of $x$ to some power $n$. ;)", "Solution_7": "[quote=\"JesusFreak197\"]I already addressed the other possibilities of $x$ to some power $n$. ;)[/quote]\r\n\r\noh oops. sorry about that.", "Solution_8": "Or in a more strict manner:\r\n\r\nIf you put $x=1$ in $f(ax)=af(x)$, you'll get $f(a)=af(1)$. And now here's the trick - the letter [b]a[/b] can be replaced by any other letter we choose, the function will not change. Hence let's put [b]x[/b] instead of [b]a[/b]. We'll get $f(x)=xf(1)$. So $f(x)$ is indeed linear, of the form $f(x)=mx$ where $m=f(1)$.", "Solution_9": "[hide=\"different solution\"]\nSince $f(2)=3$, $f(2 \\times 1)=2f(1)$ so $f(1)=1.5$\n$f(19)=19f(1)=19\\times1.5=28.5$[/hide]", "Solution_10": "YAY :!: Eureka :!: I understand. Thank you very much" } { "Tag": [ "number theory", "least common multiple", "number theory proposed" ], "Problem": "Prove that for any $ n$, there is a subset $ \\{a_1,\\dots,a_n\\}$ of $ \\mathbb N$ such that for each subset $ S$ of $ \\{1,\\dots,n\\}$, $ \\sum_{i\\in S}a_i$ has the same set of prime divisors.", "Solution_1": "Let $ b_i$, $i = 1,2,\\ldots,n$, be any distinct positive integers, and $ P = \\operatorname{rad}\\left (\\operatorname{lcm}\\limits_{S \\in \\{1,2,\\ldots,n\\}}\\left [\\sum_{i\\in S}b_i\\right ]\\right )$. Take $ a_i = Pb_i$.", "Solution_2": "Simpler, let $ b_i$, $i = 1,2,\\ldots,n$, be any distinct positive integers, and $ P = \\prod_{S \\in \\{1,2,\\ldots,n\\}}\\sum_{i\\in S}b_i$. Take $ a_i = Pb_i$." } { "Tag": [], "Problem": "Prove that an infinite number of triangles each having a given interior point as centroid can be inscribed in a given circle.", "Solution_1": "[hide=\"Method 1 (constructive)\"]The naive approach is to guess that any point on the circle can be a vertex of a triangle with its centroid at any given interior point. It turns out this doesn't work (see Method 2), but it comes suprisingly close: the second-most naive approach is to just take any chord passing through the given internal point $G$. Say it has endpoints $A$ and $P$ on the circle, and without loss of generality $AG < GP$. (If $G$ happens to be the midpoint of segment $AP$, either $G$ is the center of the circle, for which the problem is trivial (all equilateral triangles work) or $G$ is not the center, in which case we happened to pick the one chord of which $G$ is the midpoint, and we can pick any other chord instead.) Let $M$ be the point on $GP$ such that $2GM = AG$, and let $BC$ be a chord(actually \"the chord,\" unless $M$ happens to be the center of the circle) passing through $M$ such that $BM = MC$. Then $G$ is the centroid of triangle $ABC$, and since the original chord $AP$ was arbitrary, we can in fact repeat this infinitely many times to get infinitely many such triangles, Q.E.D.[/hide]\n\n[hide=\"Method 2 (non-constructive)\"]Take any point $A$ on the circle. The set of points which are the centroid of some triangle inscribed in the circle with a vertex at $A$ is a disk whose boundary is internally tangent to the given circle at $A$, with diameter two thirds that of the diameter of the given circle. Any interior point of the given circle is covered by infinitely many of these disks. The fact that each triangle is counted 3 times obviously doesn't matter, and we're done.[/hide]" } { "Tag": [ "conics", "ellipse", "geometry", "circumcircle", "ratio", "incenter", "power of a point" ], "Problem": "An ellipse-shaped billiard table doesn't have holes. When a ball hits the table border in a point $P$, it follows the symmetric line in respect to the normal line to the ellipse in $P$. Prove that is a ball starts from a point $A$ of the ellipse and, after hitting the table at $B$ and $C$, returns to $A$, then it will return to $B$.", "Solution_1": "Consider a different triangle $\\triangle ABC$ with circumcircle (O). Let the tangent to (O) at A meet BC at L. A circle (P) with diameter OL meets BC at the midpoint M of BC, because $\\angle OML$ is right. Inversion in (O) carries (P) passing through $A \\in (O)$ and through the inversion center O into the line AU, where U is the intersection of tangents to (O) at B, C so that AU is the A-symmedian of $\\triangle ABC.$ Let $K \\in BC$ be the foot of the A- symmedian AU. Since the line AU is also the radical axis of (O), (P), $AK \\equiv AU \\perp OP \\equiv OD.$ It is a well known fact that the points K, L divide the segment BC harmonically, $\\frac{KB}{KC}\\cdot \\frac{LC}{LB}=-1.$\r\n\r\n[hide=\"proof\"]K is on the radical axis of (O), (P), hence $KB \\cdot KC = KM \\cdot KL.$ Denote a = BC, b = CA, c = AB. Assuming c > b,\n\n$KB = \\frac{ac^{2}}{c^{2}+b^{2}},\\ \\ KC = \\frac{ab^{2}}{c^{2}+b^{2}}$\n\n$KM = KB-MB =\\frac{ac^{2}}{c^{2}+b^{2}}-\\frac{a}{2}= \\frac{a(c^{2}-b^{2})}{2(c^{2}+b^{2})}$\n\n$KL = \\frac{KB \\cdot KC}{KM}= \\frac{2ac^{2}b^{2}}{(c^{2}+b^{2})(c^{2}-b^{2})}$\n\n$LM = KM+KL = \\frac{a(c^{2}+b^{2})}{2(c^{2}-b^{2})}$\n\n$LC = LM-CM = \\frac{a(c^{2}+b^{2})}{2(c^{2}-b^{2})}-\\frac{a}{2}= \\frac{ab^{2}}{c^{2}-b^{2}}$\n\n$LB = LM+MB = \\frac{a(c^{2}+b^{2})}{2(c^{2}-b^{2})}+\\frac{a}{2}= \\frac{ac^{2}}{c^{2}-b^{2}}$\n\nFor oriented segments KB, KC, LB, LC we then get $\\frac{KB}{KC}\\cdot \\frac{LC}{LB}=-1$ as expected.[/hide]\r\nUsing a parallel projection, we can project the $\\triangle ABC$ inscribed in the circle (O) into a $\\triangle A'B'C'$ so that the line OD stays in place, the line $AK \\perp OD$ is carried into itself, and the angle $\\angle DA'K'$ becomes right. The circumcircle (O) is carried into an ellipse $\\mathcal E$ circumscribed around the triangle $\\triangle A'B'C'$ and with the main axes lines OD, AK. Since the cross ratio $\\frac{K'B'}{K'C'}\\cdot \\frac{L'C'}{L'B'}= \\frac{KB}{KC}\\cdot \\frac{LC}{LB}=-1$ is preserved and the angle $\\angle DA'K'$ is right, the lines $A'K' \\equiv AK$ and $A'D$ are the internal and external bisectors of $\\angle C'A'B',$ their feet K', D dividing the segment B'C' harmonically. Conversely, if we project the $\\triangle ABC$ into the $\\triangle A'B'C'$ so that the line $A'K' \\equiv AK$ bisects the angle $\\angle C'A'B',$ then $A'D$ is the external bisector of this angle, which makes the angle $\\angle DA'K'$ right. In either case, the tangents BU, CU of the circle (O) at B, C are carried into tangents B'U', C'U' of the ellipse $\\mathcal E$ at B', C' and the angle bisector A'K' passes through their intersection U'. Thus we proved the following lemma:\r\n\r\n[color=darkblue][b]Lemma:[/b] Let $A, B, C$ be 3 points on an ellipse. The normal $l_{a}$ to the ellipse tangent $t_{a}$ at $A$ bisects the angle $\\angle CAB,$ iff it passes through the intersection $U$ of the ellipse tangents $t_{b},\\ t_{c}$ at $B,\\ C.$[/color]\r\n\r\nBy the problem condition, the normals $l_{b},\\ l_{c}$ to ellipse tangents $t_{b},\\ t_{c}$ at B, C bisect the angles $\\angle ABC,\\ \\angle BCA,$ hence they meet at the triangle incenter I and $l_{a}\\equiv AI$ is the remaining angle bisector. By the above lemma, $l_{b}\\equiv BI,\\ l_{c}\\equiv CI$ pass through the intersections $V \\equiv t_{c}\\cap t_{a},$ $W \\equiv t_{a}\\cap t_{b}$ of the ellipse tangents $t_{a},\\ t_{b},\\ t_{c}$ at A, B, C. It is therefore possible to project the $\\triangle ABC$ inscribed in the given ellipse with center O into a $\\triangle A'B'C'$ inscribed in a circumcircle (O), so that the incenter I of $\\triangle ABC$ is carried into the symmedian point I' of the $\\triangle A'B'C'.$ The angle bisectors AI, BI, CI are then carried into the symmedians A'I', B'I', C'I' and the remaining symmedian A'I' passes through the intersection $U' \\equiv t_{b}' \\cap t_{c}'$ of the circumcircle tangents at B', C'. Consequently, the original angle bisector AI passes through the intersection $U \\equiv t_{b}\\cap t_{c}$ of the ellipse tangents at B, C, hence by the above lemma, it is perpendicular to the ellipse tangent $t_{a}$ at A.", "Solution_2": "Hmm... I used the well-known fact that the tangent to an ellipse in $P$ is the external bisector of $\\angle FPF'$, where $F$ and $F'$ are the foci of the ellipse. From this fact it's not hard to see that $F$ and $F'$ are isogonal conjugates respect to the triangle $ABC$, and the result follows.", "Solution_3": "\"The billiard sharp who any one catches,\r\nHis doom's extremely hard \u2014\r\nHe's made to dwell \u2014\r\nIn a dungeon cell\r\nOn a spot that's always barred.\r\nAnd there he plays extravagant matches\r\nIn fitless finger-stalls\r\nOn a cloth untrue\r\nWith a twisted cue\r\nAnd elliptical billiard balls!\"\r\n\r\n(from [i]The Mikado[/i]. Apologies for irrelevance -- I couldn't resist.)" } { "Tag": [ "function", "calculus", "integration", "probability", "geometry", "search", "algebra" ], "Problem": "Here's the function:\r\n\r\n(x^n)*[(1-x)^(d-n)]\r\n\r\nI'm trying to evaluate an indefinite integral with respect to x (d and n are constants). It's been several years since I've done any calculus, so I've been looking at integration tables online, but they haven't helped me. I tried the Wolfram Mathematica Online Integrator, but it failed (even when I had it formatted properly). That suggests to me that this function might not be integratable...\r\n\r\nIf it helps, I am only interested in integrals on the interval 0 <= x <= 1.", "Solution_1": "This is the integral that defines the (incomplete) [url=http://en.wikipedia.org/wiki/Beta_function]beta function[/url]. If $ n, d \\minus{} n$ are non-negative integers, the answer over $ [0, 1]$ is particularly simple: it is $ \\frac {n!(d \\minus{} n)!}{(d \\plus{} 1)!}$. There is also a pretty neat probabilistic proof in this case.", "Solution_2": "Thanks for the quick response! And thanks for the link; I've never heard of beta functions before.\r\n\r\nEDIT: I re-read your response more closely and I see that it only supplies part of my answer. Let me be more specific about what I am trying to do here. Yes, n and d are positive integers. I am using the function as a probability histogram over the interval [0,1], so I need to divide the function by its indefinite integral over [0,1] to ensure that the area is 1. I then wish to (for a given d and n) find a mean and a confidence interval, which is what I would need the indefinite integral for. Is there a nifty way to get an indefinite integral too?\r\n\r\nPS. The full function has this coefficient:\r\n\r\nd!/[(d-n)!n!]\r\n\r\nbut I omitted it because it doesn't vary. It looks like with the coefficient, the integral over [0,1] would become d!/(d+1)!, which simplifies to 1/(d+1)", "Solution_3": "$ \\frac{d!}{(d\\plus{}1)!} \\equal{} \\frac1{d\\plus{}1}$ :roll:", "Solution_4": "Here is yet more background on what I'm trying to do here. I found a formula in my stats textbook that gave error estimates for the true chance of a binomial event, given an experiment. However, I thought it would be neat to find a formula for the EXACT confidence interval, but I couldn't find that formula anywhere (perhaps I'm just not looking hard enough).\r\n\r\nI started with the binomial probability formula:\r\n\r\nP(n) = d!/(d-n!)n!*p^n*(1-p)^(d-n)\r\n\r\n, where d is the number of trials and n is the number of successes. I've seen this formula derived and it makes sense to me.\r\n\r\nSo here was my reasoning after that: we have an experiment with d trials and n successes, and we're trying to find a probability distribution for p, which can be any number from 0 to 1. For any given p, there we can use this formula to find the chance that that p would have resulted in this n for this d. I would simply graph P(n) as a function of p, normalize it over the interval [0,1] (by dividing by the definite integral over [0,1]), and I would have my probability distribution for p.\r\n\r\nI could then use an indefinite integral to establish my confidence interval. For example, if I wanted a 90% confidence interval, I would find my left error bound (a) such that the integral from a to the peak was 0.45. The right error bound would be found in a similar fashion.\r\n\r\nFirst of all, does this reasoning make sense? Secondly, how do I find the integral :)\r\n\r\n[quote=\"Tyl\"]$ \\frac {d!}{(d \\plus{} 1)!} \\equal{} \\frac1{d \\plus{} 1}$ :roll:[/quote]Yeah, that. :blush: I've edited my post above.", "Solution_5": "[quote=\"hypehuman\"]I could then use an indefinite integral to establish my confidence interval. [/quote]\r\nYes, that's pretty much what I thought you were doing. Scroll down further in the Wikipedia article.", "Solution_6": "Okay, I assume you're hinting that I look at the \"Incomplete beta function\" subheading. I don't really understand how they worked out the integral, but I'll figure that out later if I feel like it. So basically, I'm supposed to use this formula: (a+b-1)!/[j!(a+b-1-j)!]*x^j*(1-x)^(a+b-1-j), substituting for j all integer values from a to a+b-1, and adding all the results. Since, in my case, a=n and b=(d-n), I would have (d-1)!/[j!(d-1-j)!]*x^j*(1-x)^(d-1-j), with j taking values from n to (d-1). And to normalize it, I would divide the whole thing by (d+1).\r\n\r\nI would then find the value of the integral at the peak (x=n/d), and find a such that I(n/d)-I(a) = 0.45. Trouble is... I'm not sure how I would do that except by trial and error, since that sigma is scary to me.\r\n\r\nEDIT: hold up... they're saying that I(0) = 0 and I=(1) = 1. Doesn't that mean that I don't have to use the normalization factor of 1/(d+1)?", "Solution_7": "[quote=\"hypehuman\"]I don't really understand how they worked out the integral, but I'll figure that out later if I feel like it.[/quote]\nBinomial theorem and term-by-term integration.\n\n[quote=\"hypehuman\"]I would then find the value of the integral at the peak (x=n/d), and find a such that I(n/d)-I(a) = 0.45. Trouble is... I'm not sure how I would do that except by trial and error, since that sigma is scary to me.[/quote]\r\nNumerically. I'm pretty doubtful that there's a \"neat\" way to do this.", "Solution_8": "I did some monkeying around with a grapher, and I found that this function:\r\n\r\n(d+1)*(dCn)*p^n*(1-p)^(d-n)\r\n\r\ngives the appropriately normalized function every time (that is, the integral over [0,1] is 1). If I'm doing this on a computer, should I just give up and use Euler's method? :(", "Solution_9": "Okay, I've decided to take a different approach to finding the error limits. Instead of going 0.45 to the left and right, I think it would make more sense to cut off from the bottom. Imagine a bell curve with a horizontal line cut through it. Draw vertical lines where the line intersects the curve, and cut off the tails. I think that would make the most sense because you're eliminating the least likely values.\r\n\r\nIn the computer program, all I would have to do is find the height such that it leaves an area of 90%. I would start by setting it to half the height of the peak, finding where it intersects the probability curve, and plugging those x values into the integral equation and subtracting the two values. I would do a binary search until I come as close to 90% as my accuracy requires.\r\n\r\nDoes that look like it would come up with a correct answer? Even if it does, is there a more efficient way of getting it?" } { "Tag": [ "geometry", "congruent triangles" ], "Problem": "Given:\r\n\r\n(1) an angle $ ABC$\r\n(2) a point $ M$ on ray $ BA$, and a point $ P$ on ray $ BC$, so that $ |MB| \\equal{} |PB|$\r\n(3) a point $ N$ distinct from $ M$ on ray $ BA$, and a point $ Q$ distinct from $ P$ on ray $ BC$, so that $ |NB| \\equal{} |QB|$\r\n(4) a point $ E$ a the intersection of $ MQ$ and $ NP$\r\n\r\nProve that $ BE$ bisects angle $ ABC$.[/img]", "Solution_1": "In $ \\triangle BNP$ apply Menelaus theorem $ \\implies \\frac {NM}{MB}\\cdot\\frac {BQ}{QP}\\cdot\\frac {PE}{EN} \\equal{} 1$\r\n\r\n$ \\Longleftrightarrow \\frac {PE}{EN} \\equal{} \\frac {MB}{BQ}$, because $ PQ \\equal{} MN$ .\r\n\r\nBut $ MB \\equal{} BP$ and $ BQ \\equal{} BN$, so $ \\frac {PE}{EN} \\equal{} \\frac {BP}{BN}$ which means that $ BE$ is the bisector of angle $ \\angle ABC$ .", "Solution_2": "Thank you, Constantin.\r\n\r\nI haven't yet learned Menelaus's Theorem. Is there a solution that involves congruent triangles, CPCTC, and perhaps some isosceles-triangle theorems? I believe there must be a solution that doesn't rely on Menelaus.", "Solution_3": "You can prove it using three pairs of congruent triangles, in this order: $ \\triangle NBP\\cong\\triangle MBQ,\\ \\triangle NEM\\cong\\triangle QEP,\\ \\triangle NEB\\cong\\triangle QEB$.", "Solution_4": "[quote=\"Bictor717\"]You can prove it using three pairs of congruent triangles, in this order: $ \\triangle NBP\\cong\\triangle MBQ,\\ \\triangle NEM\\cong\\triangle QEP,\\ \\triangle NEB\\cong\\triangle QEB$.[/quote]\r\n\r\nExcellent, thanks." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "A number $ B$ is obtained from a positive integer number $ A$ by permuting its decimal digits. The number $ A\\minus{}B\\equal{}11...1$ ($ n$ of $ 1's$). Find the smallest possible positive value of $ n$.", "Solution_1": "Obviosly $ 9|n$. n=9 for example $ A\\equal{}123456790,B\\equal{}012345679$." } { "Tag": [], "Problem": "Walter gets up at 6:30 am, catches the school bus at\n7:30 am, has 6 classes that last 50 minutes each, has 30\nminutes for lunch, and has 2 hours additional time at school. He\ntakes the bus that arrives at 4:00 pm. How many minutes\nhas he spent on the bus?", "Solution_1": "Hello\r\n$ 6: 30\\minus{}7: 30$ Bus\r\n$ 7: 30\\minus{}12: 30$-Class\r\n$ 12: 30\\minus{}13: 00$-Lunch\r\n$ 13: 00\\minus{}15: 00$ -Additional time\r\n$ 15: 00\\minus{}1600$-Bus\r\nThus the ans-$ 60$min.\r\nThank u." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "let a,b,c be positive real number,prove that:\r\n$ \\frac{a}{b} \\plus{} \\frac{b}{c} \\plus{} \\frac{c}{a} \\plus{} 1 \\ge 4\\sqrt {\\frac{{{a^2} \\plus{} {b^2} \\plus{} {c^2}}}{{ab \\plus{} bc \\plus{} ca}}}$", "Solution_1": "[quote=\"Toan_VN_LC\"]let a,b,c be positive real number,prove that:\n$ \\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{a} \\plus{} 1 \\ge 4\\sqrt {\\frac {{{a^2} \\plus{} {b^2} \\plus{} {c^2}}}{{ab \\plus{} bc \\plus{} ca}}}$[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=217810" } { "Tag": [], "Problem": "There are 10 1s and 10 0s, how many ways to distribute them so that they sequence does not have 3 consecutive 0s?", "Solution_1": "[hide=\"this is really a combinatorics problem...\"]\n\nanyway, there are $ 20!/(10!10!) \\equal{} 184756$ arrangements possible, but some of these will contain 3 consecutive zeros.\n\nSo lets line up the 1s, and consider sticking either one 0 or 2 zeros in each gaps or on the ends:\n1 1 1 1 1 1 1 1 1 1\n\nSo what we have is 11 spaces to start with, which we will stick pairs of zeros in, and then 11-(# of pairs) spaces left to stick in 10 - 2(# of pairs) 0's.\nFor no pairs of consecutive zeros: 11 take 10 = 11\nFor 1 pair of consecutive zeros: 11 take 1 * 10 take 8 = 11*45=495\nfor 2 pairs of consecutive zeros: 11 take 2 * 9 take 6 = 55**84=4620\nfor 3 pairs of consecutive zeros: 11 take 3 * 8 take 4 = 165*70=11550\nfor 4 pairs of consecutive zeros: 11 take 4 * 7 take 2 = 330*21=6930\nfor 5 pairs of consecutive zeros: 11 take 5 = 462\n\nor equivalently $ \\sum_{n \\equal{} 0}^5 (_{11}C_n)( _{11 \\minus{} n}C_{10 \\minus{} 2n}) \\equal{} 24068$\n\nfor a total of 24068 possibilities :D \n[/hide]", "Solution_2": "Sorry, I don't understand why they have to be in pairs...", "Solution_3": "If we have no 3 consecutive 0's, then we either have single zeros or pairs of zeros, as all other combinations are not allowed." } { "Tag": [ "inequalities" ], "Problem": "Find the minimum value of $ a\\plus{} \\frac{1}{(a\\minus{}b)(b)}$ if $ a>b>0$", "Solution_1": "[quote=\"Zellex\"]Find the minimum value of $ a \\plus{} \\frac {1}{(a \\minus{} b)(b)}$ if $ a > b > 0$[/quote]\r\n\r\n[hide=\"possible start\"]\n$ (a \\minus{} b)(b)$ reaches its maximum when $ b \\equal{} \\frac {a}{2}$\nthis is easy to see when you compare $ (a \\minus{} b)(b)$ ? $ \\frac {a^2}{4}$\nonce you move the $ (a \\minus{} b)(b)$ to the RHS, the equation becomes $ 0$ ? $ \\frac {a^2 \\plus{} 4ab \\plus{} 4b^2}{4}$\nobviously $ 0 \\le \\frac {(a \\plus{} 2b)^2}{2}$\nso, we put b as $ \\frac {a}{2}$ and solve\n$ a \\plus{} \\frac {4}{a^2}$\n\nnot sure where to go from here\n\nEDIT: maybe try setting it to the am-gm minimum?\n\n$ a \\plus{} \\frac {4}{a^2} \\equal{} 2\\sqrt {\\frac {4}{a}}$\n$ \\frac {a}{2} \\plus{} \\frac {2}{a^2} \\equal{} \\sqrt {\\frac {4}{a}}$\n$ \\frac {a^2}{4} \\plus{} \\frac {4}{a^4} \\plus{} \\frac {2}{a} \\equal{} \\frac {4}{a}$\n$ \\frac {a^6}{4a^4} \\plus{} \\frac {16}{4a^4} \\minus{} \\frac {8a^3}{4a^4} \\equal{} 0$\n$ a^6 \\minus{} 8a^3 \\plus{} 16 \\equal{} 0$\n$ a^3 \\equal{} 4$\n\nhmm\nso if that's right the minimum is $ 2\\cdot (4)^\\frac {1}{3}$\n\n[/hide]", "Solution_2": "Write $ a \\equal{} c \\plus{} b$ for a positive real $ c$. Then we want the minimum of $ c \\plus{} b \\plus{} \\frac {1}{bc}$. By AM-GM, $ c \\plus{} b \\plus{} \\frac {1}{bc} \\geq 3 \\sqrt [3]{c \\cdot b \\cdot \\frac {1}{bc}} \\geq 3$, and $ a \\equal{} 2, b \\equal{} 1$ achieves this lower bound.\r\n\r\n(to above poster: it's easy to show $ 3 < 2 \\cdot \\sqrt [3]{4}$, so that's not the minimum)", "Solution_3": "[quote=\"MellowMelon\"]Write $ a \\equal{} c \\plus{} b$ for a positive real $ c$. Then we want the minimum of $ c \\plus{} b \\plus{} \\frac {1}{bc}$. By AM-GM, $ c \\plus{} b \\plus{} \\frac {1}{bc} \\geq 3 \\sqrt [3]{c \\cdot b \\cdot \\frac {1}{bc}} \\geq 3$, and $ a \\equal{} 2, b \\equal{} 1$ achieves this lower bound.\n\n(to above poster: it's easy to show $ 3 < 2 \\cdot \\sqrt [3]{4}$, so that's not the minimum)[/quote]\r\n\r\naha, you're right (and i'm wrong as always)\r\ncan someone explain why my way produces the wrong answer? (maybe i multiplied wrong?)\r\n\r\n\r\n\r\nnote: fixed latex.", "Solution_4": "So you have $ a \\plus{} \\frac {4}{a^2}$, and you want to use AM-GM.\r\n\r\nThe problem with your application of AM-GM is that you have not set a concrete lower bound. You may have that at one point AM = GM for $ a$ and $ \\frac {4}{a^2}$, but that does not mean that anything is minimized because your lower \"bound\" itself varies with $ a$.\r\n\r\nYou can actually get a concrete lower bound with AM-GM if you rewrite this as:\r\n\r\n$ a \\plus{} \\frac {4}{a^2} \\equal{} \\frac {a}{2} \\plus{} \\frac {a}{2} \\plus{} \\left( \\frac {2}{a} \\right) ^2$\r\n\r\nNote that this is equal to $ 3 \\cdot \\frac {\\frac {a}{2} \\plus{} \\frac {a}{2} \\plus{} \\left( \\frac {4}{a^2} \\right)^2 }{3} \\geq 3 \\cdot \\sqrt [3]{\\frac {a}{2} \\cdot \\frac {a}{2} \\cdot \\left( \\frac {4}{a^2} \\right)^2 } \\equal{} 3$, which is the minimum.\r\n\r\nNow you only have to check that you can have $ \\frac {a}{2} \\equal{} \\left( \\frac {2}{a} \\right)^2$, which works just fine when $ a \\equal{} 2$.\r\n\r\n[b]Edit:[/b] [i]Haha, looks like kunny was after the same idea [/i] :P", "Solution_5": "In your solution, you can proceed such like: $ a\\plus{}\\frac{4}{a^2}\\equal{}\\frac{a}{2}\\plus{}\\frac{a}{2}\\plus{}\\frac{4}{a^2}\\geq 3\\sqrt[3]{\\frac{a}{2}\\cdot \\frac{a}{2}\\cdot \\frac{4}{a^2}}\\equal{}3$. The equality holds when $ \\frac{a}{2}\\equal{}\\frac{4}{a^2}\\Longleftrightarrow a\\equal{}2$." } { "Tag": [ "geometry", "ratio", "geometry proposed" ], "Problem": "$ABCD$ is a quadrilateral,\r\n$E,F,G,H$ are midpoints of $AB,BC,CD,DA$.\r\nFind the point P such that\r\n$area (PHAE) = area (PEBF) = area (PFCG) = area (PGDH)$.", "Solution_1": "Let $(S)$ denote the area of the figure $S$\r\n\r\n\r\nOf course $(PEA)=(PEB)$, because $E$ is the midpoint of $AB$\r\nSimilarly for the other pairs of triangles\r\n\r\nFor any point $P$ in the interior of $ABCD$ we have:\r\n\r\n\r\n$(PHAE) = (PHA)+(PEA)$\r\n$(PEBF) = (PFB)+(PEB)$\r\n\r\n$(PHAE) = (PEBF) \\Rightarrow (PHA)=(PFB)$ (the 'opposite' triangles have equal areas)\r\n\r\nSo it is $(PHA) = (PHD) = (PFB) = (PFC) \\Rightarrow $\r\n\r\n$(PAD) = (PBC)$\r\n\r\nand similarly $(PAB) = (PCD) \\ \\ (1)$\r\n\r\nLet $h_{AB}, h_{CD}$ be the distances from $P$ to $AB,CD$ respetively.\r\n\r\nFrom $(1)$ we get $\\frac{h_{AB}}{h_{CD}}= \\frac{CD}{AB}= \\lambda$\r\n\r\n$a)$ If $AB \\parallel CD$, then the locus of $P$ such that $(1)$ holds is a line $d$ parallel to $AB,CD,$ at the proper ratio $\\lambda$\r\n\r\n$b)$ If $AB \\not \\parallel CD$\r\n\r\nLet $O \\in AB \\cap CD$. We will see that the locus of $P$ such that $(1)$ holds is a ray $Ox$, in the interior of $\\angle AOD$\r\n\r\n\r\n\r\n\r\nWe draw two lines $m,n$ parallel to $AB,CD$ respectively, such that $\\frac{d(m,AB)}{d(n,CD)}= \\lambda.$\r\nWe choose these lines, such that they have common points with the angle $\\angle AOD$\r\n\r\nLet $T\\in m \\cap n$. Then $T$ is in the interior of $\\angle AOD$\r\n\r\n\r\n\r\nIt's easy to prove (by similarity) that for every point $P$ on the ray $OT$ has the property\r\n\r\n$\\frac{d(P,AB)}{d(P,CD)}=\\lambda$, hence $(1)$ holds\r\n\r\n\r\n\r\nSimilalry, we draw one more ray $OT'$ between the lines $BC,AD.$\r\n\r\nThe point $P$ can be found on the intersection of $OT$ and $OT'$, and it is clearly in the interior of $ABCD$\r\n\r\n\r\n\r\n[u]Note[/u]\r\nIf $AB \\parallel CD$, then, of course, instead of $OT$ we use the line $d \\parallel AB$, which defined above. \r\nSimilarly if $BC \\parallel DA$", "Solution_2": "[quote=\"Johann Peter Dirichlet\"]$ABCD$ is a quadrilateral,\n$E,F,G,H$ are midpoints of $AB,BC,CD,DA$.\nFind the point P such that\n$area (PHAE) = area (PEBF) = area (PFCG) = area (PGDH)$.[/quote][color=darkblue][b]Lemma.[/b] Denote $\\delta_{d}(X)$- the distance of the point $X$ to the line $d$. Given are two lines $d_{1}$, $d_{2}$ with the common point $A$ and a pozitive real number $k$.\nThen the geometrical locus of the point $L$ for which $\\frac{\\delta_{d_{1}}(L)}{\\delta_{d_{2}}(L)}=k$ is the reunion of two lines with its intersection in the point $A$.\n\n[b]Proof of the proposed probem.[/b] Suppose w.l.o.g. that $X\\in AB\\cap CD$ and $Y\\in BC\\cap AD\\ .$ Denote the area $S=[ABCD]\\ .$\n$[APE]=BPE]=[CPG]=$ $DPG]=x\\ ,\\ [APY]=[DPH]=[CPF]=[BPF]=y\\ ;\\ x+y=\\frac{1}{4}\\cdot S\\ .$\n$[PAD]=[PBC]=2y$ $\\Longrightarrow$ $\\frac{\\delta_{BC}(P)}{\\delta_{AD}(P)}=\\frac{AD}{BC}\\ \\mathrm{(constant)}\\ ;\\ [PAB]=[PCD]=2x$ $\\Longrightarrow$ $\\frac{\\delta_{AB}(P)}{\\delta_{CD}(P)}=\\frac{CD}{AB}\\ \\mathrm{(constant)}\\ .$\n[b]Construction of the point[/b] $P$. Draw the points : $U\\in (YC)\\ ,\\ YU=BC$ ; $V\\in (YD)\\ ,\\ YV=AD$ ; $R\\in (XA)\\ ,\\ XR=AB$ ; $S\\in (XD)\\ ,\\ XS=CD\\ .$\nConstruct the middlepoints $E$, $F$ of the segments $[UV]$, $[RS]$ respectively. Then the point $P$ is the intersection of the lines $YE$ and $XF\\ .$[/color]", "Solution_3": "What needs to be proved is \"Area phae = area pebf = area pfcg = area pgdh =Area of ABCD/4\"\r\n HE//DB//GF and EF//AC//HG by mid-point theorem\r\nLet m, n be the mid-point of diagonals AC and BD respectively.\r\nClearly, Area of PHAE = Area of ABCD/4 when P=M as MHAE is similar to CDAB.\r\n So Area of PHE= Area of MHE . As they have common base HE, their heights to HE are the same.We can therefore construct a line L1 through M and parallel to HE (DB) and P must be on the line. Similarly P is on L1 to make Area of PFCG = Area of ABCD/4. \r\n Construct another line L2 through N and parallel to AC. Then ,P is also on L2 to make Area of PEBF=Area of PGDH=Area of ABCD/4.\r\n So P is the intersection point of L1 and L2", "Solution_4": "just construct the point : join $ {EF,FG,HG,HE}$; draw $ {l1}\\parallel{CD}$;$ {l2}\\parallel{AD}$ meet at point $ {x}$;draw $ {q}\\parallel{GH}$; THEN THE LAST STEP: draw a line $ {p}$ // $ {HE}$ such that the distance from $ {p}$ to $ {HE}$ is equal to the altitude of $ {triangle FGC}$ ;Let $ {p}\\cap{q}\\equal{}{P}$ $ {P}$ is the point we want .Easy to prove the area of triangles with the same color are same with S(HEP)=S(GFC) ;we get : $ {Area PHAE \\equal{} area PEBF \\equal{} area PFCH \\equal{} area PGDH\\equal{}1/4S(ABCD)}$. :)", "Solution_5": "$ \\mbox{Let M and N midpoints of } [AC]\\mbox{ and } [BD] \\mbox{, respectively. If } \\\\\r\nPM \\parallel BD \\mbox{ and }PN \\parallel AC \\mbox{, then: } S_{PHAE}\\equal{}S_{PEBF}\\equal{}S_{PFCH}\\equal{}S_{PEDH}.$" } { "Tag": [ "symmetry" ], "Problem": "Solve the following simultaneous equations on the set of integers:\r\n ab + cd = -1\r\n ac + bd = -1\r\n ad + bc = -1", "Solution_1": "If you call them (1), (2), (3) in order as you write then\r\n\r\n(1)-(2)=(b-c)(a-d)=0\r\n(1)-(3)=(a-c)(b-d)=0\r\n(2)-(3)=(c-d)(a-b)=0\r\n\r\nIt means that at least three of a,b,c,d are equal. By symmetry let a=b=c.\r\nThen it turns to:\r\n$a^2+ad=-1$\r\n$a(a+d)=-1$\r\nSo solutions are obvious: a=b=c=1, d=-2 and a=b=c=-1, d=2. Also all cyclic with this are solutions. It's all solutions.", "Solution_2": "[quote=\"ondrob\"]If you call them (1), (2), (3) in order as you write then\n\n(1)-(2)=(b-c)(a-d)=0\n(1)-(3)=(a-c)(b-d)=0\n(2)-(3)=(c-d)(a-b)=0\n\nIt means that at least three of a,b,c,d are equal. By symmetry let a=b=c.\nThen it turns to:\n$a^2+ad=-1$\n$a(a+d)=-1$\nSo solutions are obvious: a=b=c=1, d=-1 and a=b=c=-1, d=2. Also all cyclic with this are solutions. It's all solutions.[/quote]\r\n\r\nI think you meant [b]a=b=c=1, d =-2[/b] and a=b=c=-1, d=2.\r\n\r\n[/b]", "Solution_3": "Yes, Nukular, you're right. I find it very hard to write solutions without some small mistakes in typing :( ." } { "Tag": [ "geometry" ], "Problem": "Find the area of the square inscribed in the quarter circle of radius $ r$.", "Solution_1": "[hide]\nNote that the square's diagonal is the circle's radius, so the answer is $ \\boxed{r^2}{2}$\n[/hide]", "Solution_2": "I think that the square's diagonal is strictly smaller than the radius of the circle...Just look at a correct picture...", "Solution_3": "meh, how is the square inscribed?", "Solution_4": "One vertex on each radius and the other 2 vertices on the arc....Sorry for not mentioning it from the beginning, now I see how you imagined the picture...", "Solution_5": "hmm i think I have an answer: $ \\frac {4r^{2}}{7\\sqrt{5}}$ which is almost $ \\frac{2r^{2}}{5}$", "Solution_6": "Well if it's like I think it is, then the two verticies of the square would bisect the radii of the quarter circle.\r\nMeaning that the area of the square would be $ (\\frac {r\\sqrt {2}}{2})^2 \\equal{} \\frac {2r^{2}}{4} \\equal{} \\frac {r^2}{2}$", "Solution_7": "why would the two vertices of the square would bisect the radii of the quarter circle.", "Solution_8": "In my solution I constructed the whole circle with the squars is each quarter of circle, too. Thus we get $ 5$ complet squares, $ 4$ half squares( which gives an amount of $ 7$ squares) and 8 small \"parts\" of circle...(it's easy to see on a picture)", "Solution_9": "Couldn't the square be constructed in many different ways anyways?", "Solution_10": "I can see what you're saying, although I'm not completely sure if that's a square or just something very close to one.", "Solution_11": "I think it's possible.", "Solution_12": "the diagonal is r units long divide that by square root 2 and you will have a side lenght square the side length and you will obtain your answer :alien: :starwars: :spam: is actually a type of food", "Solution_13": "I believe the area of the square is:\r\n\r\n(2R^2)/5\r\n\r\nwhere R is the radius of the semi-circle.\r\n\r\ntell me if I'm correct :D", "Solution_14": "If the vertices of the square lies on The radii or the quarter-circumference, then area of the square = half the area of the square of side equal to the radius.", "Solution_15": "[quote=\"sunnykim\"]One vertex on each radius and the other 2 vertices on the arc....Sorry for not mentioning it from the beginning, now I see how you imagined the picture...[/quote]\r\n\r\nI don't think you have the right diagram sohumdatta.", "Solution_16": "Is this a [u]Quarter Circle[/u]?\r\n[asy]draw(Arc((0,0),5,90,0));\ndraw((5,0)--(0,0));\ndraw((0,0)--(0,5));\ndraw((5/sqrt(2),5/sqrt(2))--(0,5/sqrt(2)));\ndraw((5/sqrt(2),5/sqrt(2))--(5/sqrt(2),0));\ndraw((0,0)--(5/sqrt(2),5/sqrt(2)));\nlabel(\"$r$\",(5/(2sqrt(2)),5/(2sqrt(2))),N);[/asy]\r\n\r\nIn which case, the area is:\r\n$ \\left(\\dfrac{r}{\\sqrt{2}}\\right)^2\\equal{}\\dfrac{r^2}{2}$\r\n\r\nOr a half circle?", "Solution_17": "[b]2[/b] vertices on the arc and one on each radius. :wink:" } { "Tag": [ "inequalities", "absolute value", "triangle inequality" ], "Problem": "$\\mid \\mid x \\mid - \\mid y \\mid \\mid \\ \\leq \\ \\mid x - y \\mid$\r\n\r\n\r\n(Those are absolute value signs...I wasn't sure how to make them on LaTeX....)\r\n\r\nHow would I go about proving this inequality?", "Solution_1": "Did you try to split it into cases and prove each case:\r\n\r\nLike $x$ and $y$ are both negative...", "Solution_2": "I think that is the way to go also.", "Solution_3": "Okay, so the inequality $\\vert z\\vert-\\vert y\\vert\\leq \\vert z-y\\vert $ is a direct consequence of the triangle inequality.\r\n\r\nSimilarly we have $\\vert z\\vert-\\vert y\\vert\\geq -\\vert z-y\\vert $ is a direct consequence of the triangle inequality\r\n\r\nBut $-a\\leq b\\leq a$ means that $\\vert b\\vert \\leq a$, and so we're done.", "Solution_4": "That way works too... haha *pretending like I actually saw that*" } { "Tag": [ "linear algebra", "matrix", "vector", "IMC", "college contests" ], "Problem": "Let $v_{0}$ be the zero ector and let $v_{1},...,v_{n+1}\\in\\mathbb{R}^{n}$ such that the Euclidian norm $|v_{i}-v_{j}|$ is rational for all $0\\le i,j\\le n+1$. Prove that $v_{1},...,v_{n+1}$ are linearly dependent over the rationals.", "Solution_1": "It follows that all scalar products $(v_{i},v_{j})$ are rational. Since the rational $(n+1)\\times(n+1)$ matrix with entries $(v_{i},v_{j})$ is singular, let $(\\alpha_{1},\\ldots,\\alpha_{n+1})$ be a non-zero rational vector in its nullspace. We have $\\sum_{i=1}^{n+1}\\alpha_{i}(v_{i},v_{j})=0,\\ j=\\overline{1,n+1}\\ (*)$, and this is equivalent to $\\sum\\alpha_{i}v_{i}=0$ (because $(*)$ says that in the Euclidean space spanned by the $v_{i}$, the vector $\\sum\\alpha_{i}v_{i}$ is orthogonal to a set of generators, hence it must be zero)." } { "Tag": [ "geometry" ], "Problem": "In right triangle $ ABC$, $ M$ and $ N$ are midpoints of legs $ \\overline{AB}$ and $ \\overline{BC}$, respectively. Leg $ \\overline{AB}$ is 6 units long, and leg $ \\overline{BC}$ is 8 units long. How many square units are in the area of $ \\triangle APC$?\n[asy]draw((0,0)--(8,0)--(0,6)--cycle);\ndraw((4,0)--(0,6));\ndraw((0,3)--(8,0));\nlabel(\"$A$\",(0,6),NW); label(\"$B$\",(0,0),SW); label(\"$C$\",(8,0),SE); label(\"$M$\",(0,3),W); label(\"$N$\",(4,0),S); label(\"$P$\",(8/3,2),SW);[/asy]", "Solution_1": "Hello\r\nCarpet theorem\r\n[hide=\"hint\"]\nar of quad.$ MBNP$=ar. of $ \\triangle APC$\n[/hide]\r\nThank u.", "Solution_2": "It is well-known that the three medians of a triangle (illustrated by $ \\overline{AN},\\overline{BO},\\overline{CM}$ above) divide the triangle into $ 6$ [url=http://en.wikipedia.org/wiki/Median_(geometry)#Equal-area_division]regions of equal area[/url].\r\n\r\nThus, the desired area of $ \\triangle APC$ contains $ 2$ of the small areas, or $ \\frac26\\equal{}\\frac13$ of $ \\triangle ABC$'s area, and our answer is $ \\frac13\\left(\\frac12\\cdot6\\cdot8\\right)\\equal{}\\boxed8$.", "Solution_3": "$ \\frac{AP}{PN} \\times \\frac{NC}{CB} \\times \\frac{BM}{MA} \\equal{} \\frac{AP}{PN} \\times \\frac{4}{8} \\times \\frac{3}{3} \\equal{} 1$. (by Menelaus' theorem)\r\nTherefore $ AP: PN \\equal{} 2: 1$.\r\n\r\nSo, $ \\triangle APC \\equal{} \\frac{NC}{BC} \\times \\frac{AP}{AN} \\times \\triangle ABC \\equal{} \\frac{2}{4} \\times \\frac{2}{2\\plus{}1} \\times \\left( \\frac{1}{2} \\times 8 \\times 6 \\right) \\equal{} \\boxed{8}$.", "Solution_4": "Why this when there is a much more elementary and faster, yet rigorous solution?" } { "Tag": [ "algebra", "polynomial", "induction", "trigonometry", "superior algebra", "superior algebra unsolved" ], "Problem": "Prove that trigonometric polyomial\r\n$\\sum_{k=m}^{n}a_k(sin(\\alpha))^k+b_k(sin(\\alpha))^k$ has at least 2m and at most 2n zeros when $0\\le\\alpha<2\\pi$", "Solution_1": "Do you mean $ \\sum_{k=m}^{n}a_k(sin(\\alpha))^k+b_k(cos(\\alpha))^k $ :?", "Solution_2": "oh,... i'm sorry - it was an idiot typo, i meant \r\n$\\sum_{k=m}^{n}a_ksin(k\\alpha)+b_kcos(k\\alpha)$\r\nI editted now in righ way-i hope it's ok(I'm sorry)", "Solution_3": "Upper bound can be easily proved by induction, can't it?", "Solution_4": "[quote=\"eugene\"]oh,... i'm sorry - it was an idiot typo, i meant \n$\\sum_{k=m}^{n}a_kcos(k\\alpha)+b_kcos(k\\alpha)$[/quote]\r\nThe original problem had two sines, and this problem has two cosines?", "Solution_5": "I editted it" } { "Tag": [ "geometry", "geometric transformation", "reflection", "circumcircle", "parallelogram", "BritishMathematicalOlympiad" ], "Problem": "Let $ABC$ be a triangle with $AC>AB$. The point $X$ lies on the side $BA$ extended through $A$, and the point $Y$ lies on the side $CA$ in such a way that $BX=CA$ and $CY=BA$. The line $XY$ meets the perpendicular bisector of side $BC$ at $P$. Show that\r\n$\\angle BPC + \\angle BAC = 180^{\\circ}$.", "Solution_1": "[quote=\"Saul\"]Let $ABC$ be a triangle with $AC>AB$. The point $X$ lies on the side $BA$ extended through $A$, and the point $Y$ lies on the side $CA$ in such a way that $BX=CA$ and $CY=BA$. The line $XY$ meets the perpendicular bisector of side $BC$ at $P$. Show that\n\n$\\angle BPC + \\angle BAC = 180^{\\circ}$[/quote]\r\n\r\nLet's reflect this construction in the middle of $CB$.\r\nThen $B \\to C$; $C \\to B$; $X \\to X_{1}$; $Y \\to Y_{1}$; $P \\to P_{1}$.\r\nLet $\\omega$ be the circumcircle of the triangle $ABC$; $Q$ is the point of intersection of $AX_{1}$ and $\\omega$ $\\Rightarrow$ as $\\angle BCX_{1} =\\angle CBA$ $\\Rightarrow$ $\\angle ACX_{1} = 180^\\circ - \\angle CAB$; also we have $AC=CX_{1}$ $\\Rightarrow$ $AX_{1}$ is a bissector of angle $\\angle CAB$ $\\Rightarrow$ $Q$ is a middle of arc $BC$ of $\\omega$; $T=BA_{1} \\bigcap AX_{1}$ $\\Rightarrow$ as $A_{1}B \\parallel CA$ $\\Rightarrow$ $\\angle ATB=\\angle A_{1}TX_{1} = \\angle A_{1}X_{1}T=\\angle CAX_{1}=\\angle TAB$ $\\Rightarrow$ $BT=AB$ $\\Rightarrow$ $T=Y_{1}$ and line $AT$ = line $X_{1}Y_{1}$ $\\Rightarrow$ as $Q$ lies on the perpendicular bisector of side $BC$ $\\Rightarrow$ $Q=P_{1}$ $\\Rightarrow$ $\\angle CPB=\\angle CP_{1}B=\\angle CQB=180^\\circ - \\angle CAB$ $\\Rightarrow$ $\\angle CAB +\\angle CPB = 180^\\circ$\r\nwhat we need", "Solution_2": "Let's see this:\r\nSuppose $O$ is the circumcentre of $\\triangle XYA$.\r\nYou'll be delighted to see that $O$ lies on the perpendicular bisector of $BC$,\r\nthen,it is not difficult to prove quadrilateral $OXPB$ and $OPCY$ are cyclic.\r\nThus,the rest of the problem is trivial. :D\r\n(I think these important steps are enough,so I haven't post the whole solution)", "Solution_3": "My friend [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=3230]jackhui[/url] (He's a silver medalist at IMO 2005!) has found a nice solution and I am helping him to post it here. (He seldom logs on Mathlinks.)\r\n\r\nLet $\\angle A=2\\alpha$ and so $\\angle AXY=\\angle AYX=\\alpha$. Now we construct a point $P'$ such that $P'$ is on $XY$ and $\\angle P'BC=\\alpha$. We continue to let $Q$ to be the intersection of the angle bisector of $\\angle BAC$ and the perpendicular bisector of $BC$. Then it is easy to see that $Q$ is on the circumcircle of $\\triangle ABC$ and so $\\angle QBC=\\angle QCB=\\alpha$ and $\\angle ACQ=180^\\circ -\\angle ABQ$. Now we construct a point $R$ outside $\\triangle ABC$ such that $\\triangle XBR\\cong\\triangle ACQ$. Since $A,B,Q,C$ are concyclic, $R,B,Q$ are collinear. Since $\\angle RXP'=2\\alpha =\\angle P'BQ$, we know that $R,X,P',B$ are concyclic and so $RB=P'B\\Rightarrow BQ=QC=RB=P'B$. Finally let $P$ be the intersection of $QM$ and $BP'$. $\\triangle BMP\\cong\\triangle BMQ\\Rightarrow BQ=BP$. Therefore, $P'$ and $P$ coincide and they are actually the point $P$ in the question.\r\n\r\nI apologize for any steps missing. Please ask for clarification if needed. :)", "Solution_4": "[quote=\"leepakhin\"]He's a silver medalist at IMO 2005! \n[/quote]\r\n\r\nMaybe i saw him there! :)", "Solution_5": "@GHAVR: Nice solution, but I have taken the freedom to edit your post, since writing plain text in LaTeX is pretty much a waste of space (I guess you used some program which automatically generates LaTeX code - in this case, please use this program for the formulas, and write the text normally).\r\n\r\nHere is my solution (essentially the same as GHAVR's one, but in a simpler form):\r\n\r\nLet the parallels to the lines CA and AB through the points B and C intersect at D. Then, the quadrilateral ABDC is a parallelogram, so that BD = CA, CD = BA and < BDC = < BAC. Since BD = CA and BX = CA, we get BX = BD, so that the triangle XBD is isosceles, and thus < BDX = < BXD. On the other hand, CD || AB yields < CDX = < BXD. Thus, < BDX = < CDX, so that the point X lies on the angle bisector of the angle BDC. Similarly, the point Y lies on the angle bisector of the angle BDC. Thus, the line XY is the angle bisector of the angle BDC. Hence, if this line XY meets the circumcircle of triangle BDC at a point Q (apart from D), then < BDQ = < CDQ; since in a circle, equal chordal angles correspond to equal chords, this yields BQ = CQ, and thus the point Q lies on the perpendicular bisector of the segment BC. So the point Q is the point of intersection of the line XY with the perpendicular bisector of the segment BC (thanks to AC > AB, such a point of intersection is unique). But this is exactly the way that we defined the point P. Consequently, the point Q must coincide with the point P, and since we know that the point Q lies on the circumcircle of triangle BDC, we can infer that the point P lies on the circumcircle of triangle BDC. Thus, < BPC + < BDC = 180\u00b0. Since < BDC = < BAC, this becomes < BPC + < BAC = 180\u00b0. That's all.\r\n\r\n Darij", "Solution_6": "Here is my solution(?):\r\n\r\n\r\n Line XY is symmetric to the angle bisector wrt radical line of two ex-circles.\r\n\r\n Radical line passes thru the midpoint of BC, angle bisector goes thru X - mid-arc BC,\r\n\r\n thus XY will intersect mid-perpendicular of BC in P symmetric to X, \r\n\r\n this makes BXCA inscribed.\r\n\r\n\r\n Thank you.\r\n\r\n M.T.", "Solution_7": "[b]Lemma (well-known).[/b] Let $ABC$ be a triangle. Denote: the middlepoint $M$ of the side $[BC]$; two mobile points $X\\in BA$, $Y\\in CA$ so that $BX=CY$ and the line $BC$ don't separate the points $X$, $Y$. Then the geometrical locus of the middlepoint of the segment $[XY]$ is the parallel line to the bisector of the angle $\\widehat {BAC}$ and which passes through the point $M$.\r\n\r\nUsing the notations from the proposed problem, denote: $D\\in BC$ so that $\\widehat {BAD}\\equiv \\widehat {CAD}$; the middlepoints $U$, $V$, $M$ of the segments $[AX]$, $[AY]$, $[BC]$ respectively. Then $M\\in UV$ and $AD\\parallel UV\\parallel XY$ a.s.o.", "Solution_8": "[quote=\"darij grinberg\"]@GHAVR: Nice solution, but I have taken the freedom to edit your post, since writing plain text in LaTeX is pretty much a waste of space (I guess you used some program which automatically generates LaTeX code - in this case, please use this program for the formulas, and write the text normally).\n[/quote]\r\n\r\nOK :) \r\n(But I didn't use any program :what?: I'm just a beginner in LaTeX (and i have some problems with my English :blush: ).\r\nI'm realy sorry if it made you feel uncomfortable .\r\nI'll try to follow your advice.", "Solution_9": "I just found a delightful solution to this problem. (Well, that's my opinion of course, but it's neat :))\r\n\r\nTake point $Q$ on the angle bisector of $\\angle A$ such that $AQ = PY$. It is easy to see that $AQ \\parallel PY$, so $AYPQ$ is a parallellogram. Since $CY = AB$ and $\\angle CYP = \\angle QAB = \\angle A/2$, we have $\\Delta PCY \\cong \\Delta QBA$. Hence $BQ = CP = BP$, so $\\Delta BPQ$ is isosceles and $\\angle BPQ = \\angle BQP$. Therefore,\r\n$\\angle BPC = 360^{\\circ} - \\angle CPY - \\angle YPQ - \\angle QPB = 360^{\\circ} - \\angle AQB - \\angle YAQ - \\angle PQB$\r\nwhich is equal to $\\angle AQP - \\angle YAQ = (180^{\\circ} - \\angle A/2) - \\angle A/2 = 180^{\\circ} - \\angle A.\\ \\blacksquare$", "Solution_10": "Let $(O_{B}),\\ (O_{C})$ be 2 circles congruent to the triangle circumcircle (O) with circumradius R, $(O_{B})$ tangent to BC at B and $(O_{C})$ tangent to BC at C, their centers on the same side of the line BC as the point A. Let $(O_{B})$ meet AB again at X and let $(O_{C})$ meet AC again at Y. Since $(O_{B})$ is tangent to BC at B, the angles $\\angle AOB = 2\\angle ABC = \\angle XO_{B}B$ are equal and the isosceles triangles $\\triangle AOC \\cong XO_{B}B$ with $OA = OC = O_{B}X = O_{B}B = R$ are congruent by SAS, so that BX = CA. In exactly the same way, CY = BA. Since $BC \\ge 2R$ (with equality only for $\\angle A = \\frac{\\pi}{2}$), the congruent circles $(O_{B}),\\ (O_{C})$ intersect at P, Q (or at least touch at P) on the perpendicular bisector of BC. Let M be the midpoint of BC and let N be the midpoint of the arc BC of (O) opposite to A. The segments MP, MQ are geometric solutions of a quadratic equation $x^{2}-2Rx+\\frac{a}{2}= 0$ (Descartes' method), hence\r\n\r\n$x = R \\pm \\sqrt{R^{2}-\\frac{a^{2}}{4}}= R \\pm \\sqrt{R^{2}-R^{2}\\sin^{2}A}= R(1 \\pm \\cos A)$\r\n\r\nChoose the signs, so that $MP = R(1-\\cos A),\\ MQ = R(1+\\cos A).$ It follows that P is a reflection of N in BC and $Q \\in (O).$ The perpendicular diagonals BC, NP of the quadrilateral BNCP cut each other in half at M, hence it is a rhombus, $BP = CP = BN = CN$ and $\\angle BPC = \\angle BNC.$ Since $(O_{B}) \\cong (O)$ are congruent, the angles $\\angle BXP = \\angle BAN = \\frac{\\angle A}{2}$ are equal. Similarly, the angles $\\angle CYP = \\angle CAN = \\frac{\\angle A}{2}$ are also equal. Thus the lines XP, YP are both parallel to the internal bisector AN of the angle $\\angle A.$ Since they pass through the same point P, they are identical, and XY cuts the perpendicular bisector of BC at P. Since the quadrilateral ABNC is cyclic,\r\n\r\n$\\angle CAB+\\angle BPC = \\angle CAB+\\angle BNC = \\pi$" } { "Tag": [ "USAMTS", "search", "AMC" ], "Problem": ":? I was wondering if there are any individual math competitions, i.e. where you can participate even if your school does not have a math team?\r\n\r\nI am planning to take the AMC-12 next year, but I wanted to know if there are any others that I can take part in.", "Solution_1": "USAMTS -> http://www.nsa.gov/programs/mepp/usamts.html\r\n\r\nThough it is more of a take-home thing than most other competitions like the AMC/AIME/USAMO/ARML/Mandelbrot/blah\r\n\r\nFierytycoon", "Solution_2": "Here's two more, although these are more international:\r\n\r\nTournament of Towns (IMToT): [url]http://www.amt.canberra.edu.au/imtot.html[/url] (the official site, although I find [url]http://www.math.toronto.edu/oz/turgor/index.php[/url] more helpful).\r\n\r\nMath Olympiad Correspondence Program (MOCP): [url]http://www.cms.math.ca/Competitions/MOCP/[/url]\r\n\r\nOh, and one more. Did I say two? I lied. The International Mathematics Talent Search, somewhat like the USAMTS, is at [url]http://www.cms.math.ca/Competitions/IMTS/[/url]\r\n\r\nI'm not sure if everyone here has heard of the last two, so they may interest other people, also.", "Solution_3": "Would I be mistaken in saying the IMTS problems are the same as USAMTS problems? (Yes, exactly the same problems.)", "Solution_4": "No you wouldn't be mistaken. It says so right on the IMTS site itself.", "Solution_5": "Heehee...except the IMTS doesn't have any new problems on that site. The only advantage is that it has problems from longer ago than the USAMTS site, which I believe only has back to Year 8." } { "Tag": [], "Problem": "Find 5 square numbers using each non zero digit once. \r\n\r\nI gave up on this one.", "Solution_1": "[hide]7, 25, 36, 49, 81[/hide]", "Solution_2": "[hide=\"Hint\"]\nLook for a digit that doesn't appear often in squares.\n[/hide]\n\n[hide=\"Solution\"]\n1 9 25 36 784\n[/hide]", "Solution_3": "Yes, your solution is correct, and I thank you for your time. Congrats on your first post! Welcome! \r\n\r\nxtrememath8, 7 isn't a perfect square :roll:", "Solution_4": "[quote=\"236factorial\"]\n\nxtrememath8, 7 isn't a perfect square :roll:[/quote]\r\n\r\n\r\nWow, that was a stupid mistake. :blush:", "Solution_5": "[hide=\"Another way\"]9, 25, 361, 784\nIt's not very different from Ritty's.[/hide]", "Solution_6": "[quote=\"chunkypotpan\"][hide=\"Another way\"]9, 25, 361, 784\nIt's not very different from Ritty's.[/hide][/quote]\r\n\r\nYeah, but he needs 5 squares", "Solution_7": "[quote=\"xtrememath8\"][quote=\"chunkypotpan\"][hide=\"Another way\"]9, 25, 361, 784\nIt's not very different from Ritty's.[/hide][/quote]\n\nYeah, but he needs 5 squares[/quote]\r\nOh. Oops. I guess Ritty's solution is the only possible one.", "Solution_8": "what about $1,4,9,25,36$?", "Solution_9": "You're missing the digit 7.\r\n\r\nBasically, it's five squares using nine digits, and the square with a 7 in it must be a 3-digit number, as $7$ is not a perfect square, no numbers from 70-79 are perfect squares, and a perfect square can never end in a $7$. It isn't too hard to get from there; the combination of four squares and six digits must either be three one-digits and a three-digit, or two one-digits and two two-digits, and since the former is more restrictive (only three possible one-digit squares), you should start with the latter first.", "Solution_10": ":oops: :blush: :blush: sorry \r\nI forgot." } { "Tag": [ "algebra", "polynomial", "number theory unsolved", "number theory" ], "Problem": "Show that the polynomial \r\n\r\n$ (x^2 \\minus{} 13)(x^2 \\minus{} 17)(x^2\\minus{} 221)$ has no solutions in integers whereas it has solutions modulo p for all primes $ p$", "Solution_1": "It obviously has no integer solutions.\r\nHowever, for mod p, consider Legendre symbol.\r\nIf $ \\left(\\frac{13}{p}\\right)$ or $ \\left(\\frac{17}{p}\\right)$ = 1, we are done. Otherwise, both symbols equal -1, and \r\n$ \\left(\\frac{221}{p}\\right) \\equal{} \\left(\\frac{13}{p}\\right) \\left(\\frac{17}{p}\\right) \\equal{} 1$, so we are done as well." } { "Tag": [], "Problem": "", "Solution_1": "" } { "Tag": [ "algebra", "polynomial", "geometry", "3D geometry", "quadratics" ], "Problem": "The polynomial $ f(x) \\equal{} x^{4} \\plus{} ax^{3} \\plus{} bx^{2} \\plus{} cx \\plus{} d$ has real coefficients, and $ f(2i) \\equal{} f(2 \\plus{} i) \\equal{} 0.$ What is $ a \\plus{} b \\plus{} c \\plus{} d?$\r\n\t\r\n$ \\textbf{(A)}\\ 0 \\qquad \\textbf{(B)}\\ 1 \\qquad \\textbf{(C)}\\ 4 \\qquad \\textbf{(D)}\\ 9 \\qquad \\textbf{(E)}\\ 16$", "Solution_1": "[hide]\nIf a polynomial with real coefficients has a complex roots, that root's conjugate is a root as well, so the roots are $2i$, $-2i$, $2+i$, and $2-i$\n\nSo the polynomial is\n$f(x)=(x-2i)(x+2i)(x-2-i)(x-2+i)$\nThe sum of the coefficients is the polynomial evaluated at 1\n$(1-2i)(1+2i)(1-2-i)(1-2+i)$\n$(1-2i)(1+2i)(-1-i)(-1+i)$\n$(1-4i^{2})(1-i^{2})$\n$(5)(2)$\n$10$\n\nSince we aren't counting the coefficient of $x^{4}$, the answer is $10-1=\\boxed{9}$\n[/hide]", "Solution_2": "f(x)=(x+2i)(x-2i)(x-(2+i))(x-(2-i))=(x^2+4)(x^2-4x+5)\r\n=x^4-4x^3+9x^2-16x+20. Therefore, a+b+c+d are the coefficients=9. \r\n\r\nD.", "Solution_3": "how did such a trivial problem get to #18...wtf...it is like: do you know conjugate root theorem? have you taken algebra 2?", "Solution_4": "you'd be surprised how many people missed this question.", "Solution_5": "[quote=\"mdk\"]you'd be surprised how many people missed this question.[/quote]\r\n\r\n... hey. It's not my fault I can't expand.", "Solution_6": "yes it is.", "Solution_7": "I believe it was a joke.", "Solution_8": "I know that.", "Solution_9": "lol.. I TI-89ed it, used EXPAND(f(x)) lol", "Solution_10": "[quote=\"Allan Z\"]lol.. I TI-89ed it, used EXPAND(f(x)) lol[/quote]\r\n\r\nHaha, darn. Didn't think of that. I used it on the cube truncating problem though :wink:.", "Solution_11": "That's what I did too =D", "Solution_12": "[quote=\"paladin8\"][quote=\"Allan Z\"]lol.. I TI-89ed it, used EXPAND(f(x)) lol[/quote]\n\nHaha, darn. Didn't think of that. I used it on the cube truncating problem though :wink:.[/quote]\r\n\r\nYou are so lucky that tii-89 can do radicals.", "Solution_13": "this is a fun problem", "Solution_14": "[quote=\"Allan Z\"]lol.. I TI-89ed it, used EXPAND(f(x)) lol[/quote]\r\n\r\nThat's what I did...after the test. If I had answered that problem and the problem before it using my calculator, I would have qualified and I wouldn't have needed to take the B during my break... It was a really easy question. Or, TI-89's can do too much for you. :P", "Solution_15": "It's actually not that hard to expand if you use Vieta's.\r\n\r\n[hide=\"My Solution\"]\nThe roots of the polynomial are $ 2i$, $ \\minus{}2i$, $ 2 \\plus{} i$, and $ 2 \\minus{} i$. Note that these pairs of roots must be the roots of two quadratics with a leading coefficient of $ 1$ because $ f(x)$ has a leading coefficient of $ 1$. Consider the roots $ 2i$ and $ \\minus{}2i$. The sum of these roots is $ 0$ and the product is $ 4$. Therefore, the quadratic $ x^2 \\plus{} 4$ has these roots. Consider the roots $ 2 \\plus{} i$ and $ 2 \\minus{} i$. The sum of the roots is $ 4$ and the product of the roots is $ 2^2 \\minus{} (i^2) \\equal{} 4 \\plus{} 1 \\equal{} 5$. Therefore, the quadratic $ x^4 \\minus{} 4x \\plus{} 5$ has these roots. Thus, we have\n\n$ f(x) \\equal{} (x^2 \\plus{} 4)(x^4 \\minus{} 4x \\plus{} 5)\\implies f(1) \\equal{} (5)(2) \\equal{} 10$.\n\nWe are looking for $ f(1) \\minus{} 1 \\equal{} \\boxed{9}$.\n[/hide]", "Solution_16": "I solved it like tjhance but why are all the answer choices perfect squares? Makes me wonder if there is a slicker way to solve this...or it could be just there to throw people off. Anyone know if it's obvious it'll be a perfect square?", "Solution_17": "Here's my solution.\n\n$\\longrightarrow$ Since $f(x) \\in \\mathbb{R} [X]$ , we know that all complex roots exist in conjugate pairs.\nHence the roots of $ f(x) $ are $(2i), (2+i), (2-i)$ and $(-2i)$.\nNow, let's expand $f(2i)=0$\n$\\implies f(2i) = 16 - 8ai - 4b +2ci + d = 0 \\iff (16 +d) + 2ci = 4b + 8ai$\nHence we conclude that, $ 4b = 16 +d $ and $ c = 4a$.\n Now, by vieta \n$ -a = (2i) + (-2i) + (2+i) + (2-i) \\implies a = -4$\n$ d = (2i)(-2i)(2+i)(2-i) = 20$.\n\nThus, $ a= -4 , c= -16 , d = 20 , b =9$ and so $\\sum_{cyc} a = 9$.", "Solution_18": "[quote=mathgeniuse^ln(x)][quote=\"paladin8\"][quote=\"Allan Z\"]lol.. I TI-89ed it, used EXPAND(f(x)) lol[/quote]\n\nHaha, darn. Didn't think of that. I used it on the cube truncating problem though :wink:.[/quote]\n\nYou are so lucky that tii-89 can do radicals.[/quote]\n\nA ti-89 can do radicals? I never knew that.", "Solution_19": "[quote=Electron_Madnesss]INCONSEQUENTIAL[/quote]\n\nWhy? Why did you post another same solution and necro'ing an old post? Just solve and check your answer and move on! Why!!!!!" } { "Tag": [ "MATHCOUNTS", "Gauss" ], "Problem": "And if you are then say your name (unless your not comfortable doing so), if you've done MC before, what grade and school you go to.\r\n\r\nI'll go first:\r\n\r\nI'm Nick Sparkman\r\nYes i've done MC before\r\n8th grade\r\nChallenger", "Solution_1": "gahh!\r\n\r\nSTALKER!!!\r\n\r\nlol", "Solution_2": "Um, if your going to post here, please at least answer the question.", "Solution_3": "ohh I will answer\r\n \r\nlet see\r\n\r\ni am not competing in Alabama mathcounts, cause i don't live in mathcounts, but I did wish I competed in Alabama. Massachusetts is so hard to get into Nats. \r\n\r\nLets see\r\n\r\nMy name is Someone someone. Don't feel like releasing my name in the Alabama forum. \r\nI have done MC for 2 years already\r\n8th grade\r\nClarke", "Solution_4": "I'm not telling my name. However, I'll tell you i'm 14 and im a girl and im asian.\r\nI've done MC for 2 years already and never made nats.\r\nI'm from IL, in 8th grade.\r\n\r\n*Actually, if I don't make Nationals, you'll never know my name. Ever.", "Solution_5": "WOAH, gauss is a girl....\r\nanyways\r\nim gene yu\r\nim competing in alalalalalalabama mathcounts\r\nnever even made school before XD\r\ntrying for state team\r\n12 \r\nin 8th grade" } { "Tag": [ "counting", "distinguishability", "combinatorics unsolved", "combinatorics" ], "Problem": "Fifteen freshmen are sitting in a circle around a table, but the course assistant (who\r\nremains standing) has made only six copies of today\u2019s handout. No freshman should\r\nget more than one handout, and any freshman who does not get one should be able to\r\nread a neighbor\u2019s. If the freshmen are distinguishable but the handouts are not, how\r\nmany ways are there to distribute the six handouts subject to the above conditions?", "Solution_1": "I think the result is 4320, but the solution says 125. :?:" } { "Tag": [ "sphere", "Gauss", "function", "integration", "calculus" ], "Problem": "Find the net force that the southern hemisphere of a uniformly charged, insulating sphere exerts on the northern hemisphere. Express your answer in terms of the radius R and the total charge Q.", "Solution_1": "Well, this was kind of an interesting problem...\r\nLook say that force that they exert on each other is $ F$ by magnitude. If you shrink down the radius of the southern hemisphere a little bit\r\nthe force between hemispheres won't be affected essentially. Now you complete the northern hemisphere to a full sphere, with the same surface charge density. Since the electric field in such a sphere is $ 0$ then it means that the force between the \"shrunk-down\" hemisphere and the completion of the northern hemisphere is $ F$. If we shrink the northern hemisphere to the same size as we did for the southern one then we will have that a sphere of charge $ Q$ pushes the hemisphere of charge $ \\frac{Q}{2}$ just on top it by force $ 2F$. It is easy to calculate that $ F=\\frac{Q^{2}}{32\\pi \\epsilon_{0}R^{2}}$", "Solution_2": "can u explain what u have done a bit", "Solution_3": "OK. So let's denote northern hemisphere $ N$ and southern hemisphere $ S$. I take $ S$ and make its radius $ r$, where $ R-r<1 \\ we \\ have \\ 2m=m^2, \\ therefore \\ m=2.$ \r\nIf $m=1 we \\ have \\ left \\ deg \\ge2 \\ right \\deg \\le1.$ \r\nCase 0. $m=0 \\ \\Longrightarrow P(x)=0 \\ or \\ P(x)=1.$\r\nCase 1. $m=2 \\ P(x)=ax^2+bx+c \\Longrightarrow P(x)=x^2+bx+c, %Error. \"have\" is a bad command.\n\\ a=1 \\ b \\ and \\ c \\ any.$" } { "Tag": [], "Problem": "Anyone watching hockey at the winter olympics? I think what is going on is crazy, bad teams are beating favorites, etc. How can the Swiss beat Canada and the Czecks, how can USA tie Latvia?", "Solution_1": "Ahhh i can't believe Canada lost....so sad...they'll come back, just watch! :nhl:", "Solution_2": "Canada had over 17 power-play minutes. Evidently, they were sloppy.", "Solution_3": "Are they playing now? Against what country? What's the result? :P", "Solution_4": "Canada beat. by i believe Finland.", "Solution_5": "puuhuki, I cant believe Finland last today after such a great tournament! They were supposed to dominate Sweden. Oh well, congratulations to Finland for having a great tourney, I wish they would win the final game even though I wanted Russia to win the thing.", "Solution_6": "[quote=\"hello\"]puuhuki, I cant believe Finland last today after such a great tournament![/quote]\r\nWell, that is ice-hockey. Sweden is always a hard team to win and many times Sweden have won Finland. But anything can happen in sports. Before the tournament I thought Finland will be about fifth and before the final I thought Finland will be number one. So it is good that betting is not a hobby of mine. ;)", "Solution_7": "*sigh* Finland played some great games. Canada on the other hand...oh so sad. Oh well, we'll see what happens in Vancouver 2010!" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let x,y,z>0,prove that\r\n\r\n$ \\frac{(z\\plus{}x)(x\\plus{}y)}{z\\plus{}2x\\plus{}y}\\plus{}\\frac{(x\\plus{}y)(y\\plus{}z)}{x\\plus{}2y\\plus{}z}\\plus{}\\frac{(y\\plus{}z)(z\\plus{}x)}{y\\plus{}2z\\plus{}x} \\geq \\frac{1}{2}(x\\plus{}y\\plus{}z)\\plus{}\\frac{yz}{y\\plus{}z}\\plus{}\\frac{zx}{z\\plus{}x}\\plus{}\\frac{xy}{x\\plus{}y}.$", "Solution_1": "[quote=\"xzlbq\"]Let x,y,z>0,prove that\n\n$ \\frac {(z \\plus{} x)(x \\plus{} y)}{z \\plus{} 2x \\plus{} y} \\plus{} \\frac {(x \\plus{} y)(y \\plus{} z)}{x \\plus{} 2y \\plus{} z} \\plus{} \\frac {(y \\plus{} z)(z \\plus{} x)}{y \\plus{} 2z \\plus{} x} \\geq \\frac {1}{2}(x \\plus{} y \\plus{} z) \\plus{} \\frac {yz}{y \\plus{} z} \\plus{} \\frac {zx}{z \\plus{} x} \\plus{} \\frac {xy}{x \\plus{} y}.$[/quote]\r\n\r\nS.O.S can solve it easy.\r\n\r\n$ \\sum{\\frac {(z \\plus{} x)(x \\plus{} y)}{z \\plus{} 2x \\plus{} y}} \\ge \\frac {1}{2}(x \\plus{} y \\plus{} z) \\plus{} \\sum{\\frac {xy}{x \\plus{} y}}$\r\n\r\n$ \\iff \\sum{\\left(\\frac {x \\plus{} y}{4} \\minus{} \\frac {xy}{x \\plus{} y}\\right)} \\ge \\sum{\\left(\\frac {y \\plus{} z}{2} \\minus{} \\frac {(z \\plus{} x)(x \\plus{} y)}{z \\plus{} 2x \\plus{} y}\\right)}$\r\n\r\n$ \\iff \\sum{\\frac {(x \\minus{} y)^2}{4(x \\plus{} y)}} \\ge \\sum{\\frac {y^2 \\plus{} z^2 \\minus{} 2x^2}{2(z \\plus{} 2x \\plus{} y)}}$\r\n\r\n$ \\iff \\sum{\\frac {(x \\minus{} y)^2}{4(x \\plus{} y)}} \\ge \\sum{\\left( \\frac {y^2 \\minus{} x^2}{2(z \\plus{} 2x \\plus{} y)} \\plus{} \\frac {x^2 \\minus{} y^2}{2(x \\plus{} 2y \\plus{} z)} \\right)}$\r\n\r\n$ \\iff \\sum{\\frac {(x \\minus{} y)^2}{4(x \\plus{} y)}} \\ge \\sum{\\frac {(x \\minus{} y)^2(x \\plus{} y)}{2(z \\plus{} 2x \\plus{} y)(x \\plus{} 2y \\plus{} z)}}$\r\n\r\n$ \\iff \\sum{\\left(\\frac {1}{4(x \\plus{} y)} \\minus{} \\frac {x \\plus{} y}{2(z \\plus{} 2x \\plus{} y)(x \\plus{} 2y \\plus{} z)}\\right)(x \\minus{} y)^2} \\ge 0$\r\n\r\n$ \\iff \\sum{\\frac {z^2 \\plus{} 3z(x \\plus{} y) \\plus{} xy}{4(x \\plus{} y)(z \\plus{} 2x \\plus{} y)(x \\plus{} 2y \\plus{} z)}(x \\minus{} y)^2} \\ge 0,$\r\n\r\nclear.", "Solution_2": "But \r\n$ {\\frac {x \\plus{} y \\plus{} z}{t}} \\plus{} 2\\, \\left( 3 \\minus{} 2\\,{t}^{ \\minus{} 1} \\right) \\left( {\\frac {yz }{y \\plus{} z}} \\plus{} {\\frac {zx}{z \\plus{} x}} \\plus{} {\\frac {xy}{x \\plus{} y}} \\right) \\left( t \\plus{} 2 \\right) ^{ \\minus{} 1}\\leq {\\frac {\\left( z \\plus{} x \\right) \\left( x \\plus{} y \\right) }{t x \\plus{} y \\plus{} z}} \\plus{} {\\frac {\\left( x \\plus{} y \\right) \\left( y \\plus{} z \\right) }{ty \\plus{} z \\plus{} x}} \\plus{} {\\frac {\\left( y \\plus{} z \\right) \\left( z \\plus{} x \\right) }{tz \\plus{} x \\plus{} y}}.$\r\n\r\n$ 0.8071131713\\equal{}1/12\\,\\sqrt [3]{199 \\plus{} 24\\,\\sqrt {249}} \\minus{} {\\frac {47}{12}}\\,{\\frac {1}{ \\sqrt [3]{199 \\plus{} 24\\,\\sqrt {249}}}} \\plus{} {\\frac {7}{12}}\\leq t.$", "Solution_3": "and,please:\r\n\r\n$ 2\\, \\left( x\\plus{}y\\plus{}z \\right) ^{2} \\left( {\\frac {zy}{y\\plus{}z}}\\plus{}{\\frac {zx}{z\\plus{}x\r\n}}\\plus{}{\\frac {xy}{x\\plus{}y}} \\right) \\leq \\left( {\\frac { \\left( z\\plus{}x \\right) \r\n \\left( x\\plus{}y \\right) }{z\\plus{}2\\,x\\plus{}y}}\\plus{}{\\frac { \\left( x\\plus{}y \\right) \\left( y\r\n\\plus{}z \\right) }{x\\plus{}2\\,y\\plus{}z}}\\plus{}{\\frac { \\left( y\\plus{}z \\right) \\left( z\\plus{}x\r\n \\right) }{y\\plus{}2\\,z\\plus{}x}} \\right) ^{3}\r\n.$\r\nBQ" } { "Tag": [ "calculus" ], "Problem": "Let $ f(x) \\equal{} \\frac {2x}{x^2 \\minus{} 5x \\plus{} 6}$\r\n\r\nFind $ f^{(n)}(x)$ ?", "Solution_1": "This is the wrong forum for calculus problems. (Announcements: \"PLEASE do not post calculus problems in this forum!\")", "Solution_2": "Assuming $ f^{(n)}(x) \\equal{} \\frac {d^n}{dx^n}f(x)$\r\n[hide=\"Big Hint on making it simpler\"]$ f(x) \\equal{} \\frac {2x}{x^2 \\minus{} 5x \\plus{} 6} \\equal{} \\frac {2x}{(x \\minus{} 3)(x \\minus{} 2)} \\equal{} \\frac {2x}{x \\minus{} 3} \\minus{} \\frac {2x}{x \\minus{} 2} \\equal{} \\frac {6}{x \\minus{} 3} \\minus{} \\frac {4}{x \\minus{} 2}$[/hide]" } { "Tag": [ "geometry", "LaTeX" ], "Problem": "If the height of the right triangle is 48 units larger than the base and the hypotenuse is 60 find the area", "Solution_1": "[hide=\"hint\"]\nThe two legs are $x$ and $48+x$. Then you use pythagorean theorum to solve\n[/hide]", "Solution_2": "[hide][quote=\"linzk3\"]If the height of the right triangle is 48 units larger than the base and the hypotenuse is 60 find the area[/quote]\n\na^2+(a+48)^2=60\nsolve, then multiply a and a+48 and divide by 2[/hide]", "Solution_3": "[hide]$x^{2}+(x+48)^{2}=3600$\n\n$2x^{2}+96x+2304=3600$\n\n$x^{2}+48x-648=0$\n\n$x=\\frac{-48\\pm\\sqrt{2304+2592}}{2}$\n\n$x=\\frac{-48\\pm12\\sqrt{34}}{2}$\n\n$x=-24+6\\sqrt{34}$\n\n$A=-24+6\\sqrt{34}(-24+6\\sqrt{34}+48)$\n\n$A=-576-144\\sqrt{34}+144\\sqrt{34}+36\\cdot34$\n\n$A=648$\n\nIn step 6, $\\pm$ became $+$ beacaue $x$ can't be negative.\nEveryone else is too lazy to write out all the steps. :P [/hide]", "Solution_4": "[quote=\"i_like_pie\"][hide]$x^{2}+(x+48)^{2}=3600$\n\n$2x^{2}+96x+2304=3600$\n\n$x^{2}+48x-648=0$\n\n$x=\\frac{-48\\pm\\sqrt{2304+2592}}{2}$\n\n$x=\\frac{-48\\pm12\\sqrt{34}}{2}$\n\n$x=-24+6\\sqrt{34}$\n\n$A=-24+6\\sqrt{34}(-24+6\\sqrt{34}+48)$\n\n$A=-576-144\\sqrt{34}+144\\sqrt{34}+36\\cdot34$\n\n$A=648$\n\nIn step 6, $\\pm$ became $+$ beacaue $x$ can't be negative.\nEveryone else is too lazy to write out all the steps. :P [/hide][/quote]\r\n\r\nit takes forever to write in latex :rotfl:" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "let a,b,c be the 3 sides of a triangle\r\n\r\nprove that [color=red]a/(b+c)+b/(a+c)+c/(a+b)<2[/color]", "Solution_1": "$ \\frac{a}{b\\plus{}c}\\equal{}\\frac{2a}{(b\\plus{}c)\\plus{}(b\\plus{}c)}<\\frac{2a}{a\\plus{}b\\plus{}c}$, etc.", "Solution_2": "[quote=\"m1ght\"]let a,b,c be the 3 sides of a triangle\n\nprove that [color=red]a/(b+c)+b/(a+c)+c/(a+b)<2[/color][/quote]\r\nWe have: $ \\frac{a}{b\\plus{}c} <\\frac{2a}{a\\plus{}b\\plus{}c} \\Leftrightarrow a [133 -> 55 -> 250] -> (133...)\n2004 = 0 (mod 3) = 133?\n[/hide]", "Solution_3": "[quote=\"catcurio\"]Source: 2005 AMC-10B #11\n\nThe first term of a sequence is $2005$. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the $2005^{\\text{th}}$ term of the sequence?\n\n$\\text{(A)} \\ 29 \\qquad \\text{(B)} \\ 55 \\qquad \\text{(C)} \\ 85 \\qquad \\text{(D)} \\ 133 \\qquad \\text{(E)} \\ 250$[/quote]\r\n\r\nI got 133 (D).", "Solution_4": "[hide]Terms: $\\mbox{2005, 133, 55, 250, 133...}$\n\nIf we ignore $\\mbox{2005}$ then the 2005th term is really the 2004th. This divides fully into 3, making the 2005th term equal to the 4th term:\n\n$\\Rightarrow\\mbox{250 (E)}$[/hide]", "Solution_5": "2005-> [133 -> 55 -> 250] -> (133...) \r\n2004 = 0 (mod 3) = 133?", "Solution_6": "[quote=\"coffeym\"][hide]Terms: $\\mbox{2005, 133, 55, 250, 133...}$\n\nIf we ignore $\\mbox{2005}$ then the 2005th term is really the 2004th. This divides fully into 3, making the 2005th term equal to the 4th term:\n\n$\\Rightarrow\\mbox{250 (E)}$[/hide][/quote]\r\n\r\nIt clearly says 2005 is the first term :)", "Solution_7": "He means the 2004th term of the 133, 55, 250,133, 55, 250,... sequence since we don't consider 2005 fue to teh fact that it is not a part of teh repeating sequence.", "Solution_8": "[hide]The sequence goes : \n\n2005, 133, 55, 250, 133, 55, 250, 133, ... \n\nEvery three terms, after the first, repeat. \n\nSince $2005\\equiv1\\pmod3$, it should be the same thing as the fourth term, since it is also 1 (mod 3). \n\nSo the 2005th term is 250. [/hide]" } { "Tag": [ "ratio" ], "Problem": "The geometr progress is $ \\\\S_{20}\\equal{}230$ and $ \\\\S_{45}\\equal{}1080$ then $ \\\\S_{80}\\equal{}?$", "Solution_1": "this looks messy.\r\nsuppose the first term is a and the common ratio is r\r\nthen the terms are $ a, ar, ar^2, ar^3 \\ldots$\r\nwe have $ S_{20}\\equal{}ar^{19}\\equal{}230$ and $ S_{45}\\equal{}ar^{44}\\equal{}1080$\r\ndividing the first equation into the second gives\r\n$ \\frac{ar^{44}}{ar^{19}}\\equal{}\\frac{1080}{230} \\implies r^{25}\\equal{}\\frac{1080}{230} \\implies r\\equal{}\\sqrt[25]{\\frac{108}{23}}$\r\nwe want $ S_{80}\\equal{}ar^{79}\\equal{}\\left(ar^{44}\\right) \\cdot r^{35}\\equal{}S_{45} \\cdot r^{35}\\equal{}1080 \\cdot \\left(\\sqrt[25]{\\frac{108}{23}}\\right)^{35}\\equal{}1080 \\cdot \\frac{108}{23}^{\\frac75}$\r\n\r\nthis may be the final answer depending on what form you want it in" } { "Tag": [ "function", "complex numbers", "complex analysis", "complex analysis unsolved" ], "Problem": "Let $ f: D(0,1)\\rightarrow G$ be a biholomorph function and let $ A_r\\equal{}\\{z\\in \\mathbb{C}, |z|162u^5w^3-180u^3v^2w^3+18u^2w^6=$\n$=18u^2w^3(9u^3-10uv^2+w^3)\\geq18u^2w^3(9u^3-10uv^2+4uv^2-3u^3)=108u^3w^3(u^2-v^2)\\geq0$.\nId est, $f$ is an increasing function, which says that $f$ gets a minimal value for a minimal value of $u$, \nwhich happens for equality case of two variables and $b=c=1$ gives $(a-1)^2(a^2+a+4)\\geq0$.\n2. Let $ab+ac+bc\\leq(a+b+c)\\sqrt[3]{abc}$ or $uw\\geq v^2$.\nHence, we need to prove that $\\frac{9u^2v^2-3v^4-2uw^3}{27v^6+27u^3w^3-54uv^2w^3+8w^6}\\leq\\frac{3u^2+v^2}{w(9uv^2-w^3)}$ or\n$81u^2v^6+27v^8+27uv^6w+81u^5w^3-81u^3v^4w-135u^3v^2w^3-54uv^4w^3+27u^2v^2w^4-3v^4w^4+24u^2w^6+8v^2w^6-2uw^7\\geq0$.\nSince $uw\\geq v^2$ we obtain:\n$81u^2v^6+27v^8+27uv^6w+81u^5w^3-81u^3v^4w-135u^3v^2w^3-54uv^4w^3+27u^2v^2w^4-3v^4w^4+24u^2w^6+8v^2w^6-2uw^7\\geq$\n$\\geq81u^2v^6+27v^8+27v^8+81u^4v^2w^2-81u^3v^4w-135u^3v^2w^3-54uv^4w^3+27u^2v^2w^4-3v^4w^4+3v^4w^4+21u^2w^6+8v^2w^6-2uw^7\\geq$\n$\\geq81u^2v^6+54v^8+81u^4v^2w^2-81u^3v^4w-135u^3v^2w^3-54uv^4w^3+27u^2v^2w^4+27v^2w^6$.\nThus, it remains to prove that $g(v^2)\\geq0$, where \n$g(v^2)=3u^2v^6+2v^8+3u^4v^2w^2-3u^3v^4w-5u^3v^2w^3-2uv^4w^3+u^2v^2w^4+v^2w^6$.\nBut $g'(v^2)=9u^2v^4+8v^6+3u^4w^2-6u^3v^2w-5u^3w^3-4uv^2w^3+u^2w^4+w^6>3u^2v^4+3u^4w^2-6u^3v^2w\\geq0$ by AM-GM.\nThus, $g$ gets a minimal value for a minimal value of $v^2$, which happens for equality case of two variables.\nSince $g(v^2)\\geq0$ is a homogeneous inequality, we can assume $b=c=1$, which gives\n$$\\tfrac{(a+2)^2(2a+1)^2}{27}+\\tfrac{2(2a+1)^3}{27}+\\tfrac{(a+2)^4\\sqrt[3]{a^2}}{27}-\\tfrac{(a+2)^3(2a+1)\\sqrt[3]a}{27}-\\tfrac{5(a+2)^3a}{27}-\\tfrac{2(a+2)(2a+1)a}{9}+\\tfrac{(a+2)^2\\sqrt[3]{a^4}}{9}+a^2\\geq0$$\nand after assuming $a=x^3$ we obtain \n$(x-1)^2(x^{12}-2x^{10}+4x^9-10x^7+4x^6-10x^4+12x^3+18x^2+4x+6)\\geq0$, which is obvious. Done! :)", "Solution_8": "[quote=ductrung]Prove that for any positive real numbers $a,b$ and $c$,\n\\[\\frac{1}{a^{2}+bc}+\\frac{1}{b^{2}+ca}+\\frac{1}{c^{2}+ab}\\le\\frac{1}{\\sqrt[3]{abc}}\\left(\\frac{1}{a+b}+\\frac{1}{b+c}+\\frac{1}{c+a}\\right) \\][/quote]\n\nThis problem is easy. It is equivalent to\n\\[\nf\\left(a,b,c\\right)=\\frac{1}{abc}\\left(\\frac{1}{a^3+b^3}+\\frac{1}{a^3+c^3}+\\frac{1}{b^3+c^3}\\right) -\\frac{1}{a^6+b^3 c^3}-\\frac{1}{a^3 c^3+b^6}-\\frac{1}{a^3 b^3+c^6} \\geq 0.\n\\]\n\nInequality is symmetric so we may assume that $a \\geq b \\geq c$. Then there exist non-negative reals $x$ and $y$ such that $b=c+x$ and $a=c+x+y$. After doing some algebra it is easy to verify that \n\\[\nf\\left(a,b,c\\right) = f\\left(c+x+y,c+x,c\\right) \\geq 0.\n\\]\nExpression that we get when taking everything to the common fraction is a bit longer so I rather not post it here. \n", "Solution_9": "Dear [b]math_science[/b]! It does not help here even with computer. ;) ", "Solution_10": "[quote=arqady]Dear [b]math_science[/b]! It does not help here even with computer. ;)[/quote]\n\nI take it back that it was obvious. I have not seen some negative terms but I believe we can somehow deal with them. Anyway, solutions based on computers might be used just as a demonstration of the accuracy. Such proofs are not so interesting. ", "Solution_11": "[quote=math_science]\nbut I believe we can somehow deal with them. [/quote]\nI could not. :maybe: \n\n" } { "Tag": [ "AMC", "AIME" ], "Problem": "I realize that this question may be premature, but after the AIME, how long will I have to wait for answers, and where will they be located? Will some one post them the next day on this forum, like with the AMC, or will I have to wait until they are on the AMC website a couple days later. Last year I didn't know about this site, so I had to wait in suspense for days until the answers were on the AMC websites. Also, when will we be allowed to discuss the problems and solutions. Thanks in advance for any responses!", "Solution_1": "I know that there's a math jam one to two days after the AIME that goes over all of the questions and answers to the test", "Solution_2": "And, of course, the Math Jams are almost always correct.\r\nCorrect me if I'm wrong, but this forum will remain closed for a while the AIME is in session, yes? When the forum gets unlocked, discussion of the problems are allowed." } { "Tag": [ "geometry", "geometric transformation", "rotation" ], "Problem": "Suppose that each pair of eight tennis players (rotations) either played exactly one game last week or did not play at all. Each individual player participated in all but 12 games. How many games were played among the eight players.", "Solution_1": "[hide]Let the number of games played be $k$. Each player plays $k-12$ games, so in total, there are $8k-96$ total games. However, we double count because two people play in each game, so there are only $4k-48$ games in total. Hence, we have $k=4k-48\\Rightarrow k=16$.[/hide]" } { "Tag": [ "inequalities", "IMO Shortlist", "inequalities proposed" ], "Problem": "Let $a,b,c$ be positive real numbers such that: $ab+bc+ca=1$. Prove the inequality:\r\n\\[ 3 \\cdot \\sqrt[3]{\\frac{1}{abc}+6(a+b+c)} \\le \\frac{\\sqrt[3]{3}}{abc} \\]", "Solution_1": "I remember an IMO shortlist problem like this:\r\nshow that if $ab+bc+ca=1$ then\r\n$\\sqrt[3]{\\frac{1}{a}+6b}+\\sqrt[3]{\\frac{1}{b}+6c}+\\sqrt[3]{\\frac{1}{c}+6a}\\le \\frac{1}{abc}$ :D \r\n\r\nthe solution uses power mean and the above result ;) \r\n\r\nnotice that $ab+bc+ca\\geq \\sqrt{3abc(a+b+c)}$so $a+b+c\\le \\frac{1}{3abc}$ then call $abc=x\\le \\frac{1}{3\\sqrt{3}}$ \r\nfrom here its an easy one-variable inequality...", "Solution_2": "Yeah, that's almost the same with the one in IMO2004 shortlist" } { "Tag": [ "ratio" ], "Problem": "A basketball player averaged 17 points in his first eight games and 24 points in his next\nsix games. What is his overall average for the 14 games?", "Solution_1": "The sum of his scores is $ 17\\cdot 8\\plus{}24\\cdot 6\\equal{}280$ (this follows from the definition of average), so the mean is $ \\frac{280}{14}\\equal{}\\boxed{20}$.\r\n\r\n\r\nAlternatively, the ratio of the number of elements in the first set to the second is $ \\frac{8}{6}\\equal{}\\frac{4}{3}$, so the distance from the average of the second set is $ \\frac{4}{3}$ is much as the distance from the average of the first set.\r\n\r\nThe total distance is just $ 24\\minus{}17\\equal{}7$, so $ x\\plus{}\\frac{4}{3}x\\equal{}7 \\Rightarrow x\\equal{}3$, so $ 17\\plus{}3\\equal{}20$.\r\n\r\nThe second part was rather hard to explain?" } { "Tag": [ "algebra", "polynomial", "trigonometry", "algebra unsolved" ], "Problem": "Trying to check something about a recurrence I found this (I'm not completely sure about its validity):\r\nFor any $ n \\in \\mathbb{Z}$, $ n \\ge 2$ consider the polynomial\r\n$ P_n : \\equal{} \\prod_{k\\equal{}1}^{n\\minus{}1} \\left( t \\minus{} \\cos \\frac{k\\pi}{n} \\right) \\in \\mathbb{R}[t]$,\r\nand also $ P_1 \\equiv 1$. Prove that the following identity holds for any positive integer $ n$:\r\n$ P_{n\\plus{}2} \\equal{} t \\cdot P_{n\\plus{}1} \\minus{} \\frac{P_n}{4}$.", "Solution_1": "Yes your recurrence is true! The polynomial you defined $ P_n$ is in fact equal to $ 2^{n\\minus{}1}U_{n\\minus{}1}$ where $ U$ are the Chebyshev's polynomials of second kind.\r\nThey are special cases of the Jacobi polynomials, and are thus an othogonal sequence.\r\nsee for example [url=http://en.wikipedia.org/wiki/Chebyshev_polynomials]here[/url]" } { "Tag": [ "induction", "algebra unsolved", "algebra" ], "Problem": "$ \\text{Given the sequence}$ $ \\{a_n\\}$ $ \\text{which is defined by}$ \\[ a_1\\equal{}1,\\quad a_{n\\plus{}1}\\equal{}2a_n\\plus{}2^n\\ (n\\equal{}1,2,3),\\] $ \\text{find the}$ $ n$$ \\text{th term}$ $ a_n$ $ \\text{and sum}$ $ \\sum_{k\\equal{}1}^na_k.$", "Solution_1": "hello, prove by induction that $ a_n\\equal{}n\\cdot2^{n\\minus{}1}$, for the sum i have\r\n$ \\sum_{i\\equal{}1}^na_i\\equal{}2^n(n\\plus{}1)\\minus{}2^{n\\plus{}1}\\plus{}1$.\r\nSonnhard." } { "Tag": [ "group theory", "abstract algebra", "factorial", "calculus", "integration", "algebra", "function" ], "Problem": "3)We have a group with 2002 elements, which has only one subgroup with 26 elements and only one subgroup with 77 elements, and they both are commutative. Show that the group is commutative.\r\n4) Question How do you call an eucledian ring, a factorial ring... I saw notations life UFD... what do they mean?\r\nThanks", "Solution_1": "3) Since the two groups are the unique subgroups of a given order, they are both characteristic subgroups, hence in particular normal subgroups. Since they have trivial intersection and their product is the group, the group is the direct product of the two subgroups. The direct product of two abelian groups is abelian, so we are done.\r\n\r\nEDIT: For 3), we use the following:\r\n\r\nIf $ M$ and $ N$ are normal subgroups with trivial intersection, and $ m \\in M$ and $ n \\in N$, then $ mn \\equal{} nm$. Proof: $ mnm^{\\minus{}1}n^{\\minus{}1} \\equal{} (mnm^{\\minus{}1}n^{\\minus{}1} \\in N$, similarly $ m(nm^{\\minus{}1}n^{\\minus{}1}) \\in M$. Thus $ mnm^{\\minus{}1}n^{\\minus{}1} \\equal{} e$, so $ mn \\equal{} nm$.\r\n\r\n4) A Euclidean ring is a commutative ring with a norm satisfying certain nice properties that make it similar to the ring of integers. If you want the full list of properties of the norm, you can almost certainly find them on wikipedia or mathworld.\r\n\r\nA factorial ring is a commutative integral domain where each element factors uniquely as a product of irreducibles, down to units (the integral domain part ensures that $ 0$ has \"unique factorization\", as well). Thus it is sometimes called a UFD, or unique factorization domain. Also, it should be noted that an integral domain is sometimes called an entire ring instead. It is a ring where $ ab \\equal{} 0$ implies that $ a \\equal{} 0$ or $ b \\equal{} 0$.", "Solution_2": "$ 0$ is no product of irreducibles; and integral domains are assumed to be $ \\neq 0$." } { "Tag": [ "AMC" ], "Problem": "Hello,\r\nOn my website http://www.matharena.net/ I tried to host a mock AMC about a month ago, but no one participated. I was just curious to see if anyone might be interested now.\r\nBasically it would just be a one weekend type thing. You have 75 minutes to solve the problems, but you choose a time over that weekend when you want to start. The scores would then be emailed out a few days after. I just thought it might be kind of fun if anyone is interested..", "Solution_1": "[quote=\"SnowStorm\"]Hello,\nOn my website http://www.matharena.net/, I tried to host a mock AMC about a month ago, but no one participated. I was just curious to see if anyone might be interested now.\nBasically it would just be a one weekend type thing. You have 75 minutes to solve the problems, but you choose a time over that weekend when you want to start. The scores would then be emailed out a few days after. I just thought it might be kind of fun if anyone is interested..[/quote]\r\n\r\nI would be interested if you get rid of that comma in your url :D", "Solution_2": "I'd be interested.", "Solution_3": "I'd be interested.", "Solution_4": "What they said", "Solution_5": "Sure.", "Solution_6": "I would be interested!", "Solution_7": "I would be interested. Noone participated in the last one? :huh: I would have, but I was in NY at the time... :(", "Solution_8": "definitely interested. already registered. :D so when is this starting?", "Solution_9": "ditto :D", "Solution_10": "I'm other admin of that site by the way.\r\n\r\nAlthough this site and contest he mentioned is done all by himself, I did contributed some questions (that weren't in the contest I think).\r\n\r\nHopefully, I can get to add some problems. Maybe, we'll see. :)", "Solution_11": "I'm interested! Just make sure to do it on the weekend, because otherwise 6 PM EST (which is when most of the Mock AMCs have been held) will be during school for those of us who live on the west coast.", "Solution_12": "I'd be interested if we didn't have to give out all that information.", "Solution_13": "as usual:\r\ni'm in :)", "Solution_14": "[quote=\"sirorange\"]I'd be interested if we didn't have to give out all that information.[/quote]\r\n\r\nIt's not that much information... but if you are really scared of someone stalking you because you live in the US and your title is Mister, you could BS something.", "Solution_15": "the link doesn't work...but i would like to participate, and if i had to give info, then i would just give fake stuff, because as a general rule, i never give out personal info\r\n\r\nEDIT: i got the link to work...you ought to remove that extra comma", "Solution_16": "ok,\r\nIt still needs a little more work, but I'll keep you guys updated. And if you don't want to give out that info, just make it up...", "Solution_17": "i'll do it. :D", "Solution_18": "ok, it looks like a lot of people are interested.\r\nI've made another post about it including the date. Good luck." } { "Tag": [ "inequalities", "algebra open", "algebra" ], "Problem": "Do I have enough information here to solve for the x, y, and z variables? Is there even an answer? If so, what are some possible ones? All of these conditions must be met for the correct answer, if there is one. Please help!\r\n\r\nx, y, and z, are natural numbers\r\nx is greater than or equal to 1\r\ny is greater than or equal to 1\r\nz is greater than (x +y)\r\n(5x-y-z) is greater than or equal to 0\r\n(2y-x-z) is greater than or equal to 1\r\n(z-x-y) is greater than or equal to 1", "Solution_1": "I think not....\r\n\r\nSuppose there is a solution.\r\n\r\nFrom your last three inequalities, add the first and last to give \r\n\r\n$4x - 2y \\ge 1 \\Rightarrow 4x - 2y \\ge 2$ (since $4x -2y$ is even) \r\n$\\Rightarrow 2x - y \\ge 1$ \r\n\r\nAdding the second of the inequalities gives :-\r\n\r\n$x + y - z \\ge 1 \\Rightarrow z - x - y \\le -1$ \r\n\r\nwhich contradicts the last one.\r\n\r\nTherefore, there is no solution." } { "Tag": [ "induction", "geometry", "3D geometry", "linear algebra", "matrix", "algebra", "factorization" ], "Problem": "[img]http://www.artofproblemsolving.com/Community/Images/Problems/P4.gif[/img]", "Solution_1": "An (classic) result about the Fibonacci sequence (not difficult to prove by induction) :\r\n\r\n1) F(m+n) = F(m)F(n+1) + F(m-1)F(n)\r\n\r\n\r\nEasy to deduce from this (or by induction) :\r\n\r\n2) F(n)^2 + F(n+1)^2 = F(2n+1)\r\n\r\n3) F(n+1)^2 - F(n)^2 = F(2n)\r\n\r\n\r\nFrom 1) :\r\n\r\n4) F(3n) = F(2n)F(n+1) + F(2n-1)F(n)\r\n\r\n\r\nInjecting 2) and 3) in 4) gives the given formula, after some algebraic gymnastics.", "Solution_2": "I see... but, is there a way to do it without using those other identities???\r\n\r\nPS I love Algebraic gymnastics... God I have no life...", "Solution_3": "[color=cyan]Well, I'm fairly sure you can do it using the fact that \nF(n) = (:phi:^n - (-:phi:)^(-n)) / :rt5: \nwhere :phi: = (:rt5: + 1) / 2\nbut I'm also sure that you really wouldn't want to unless you have, say, mathematica to simplify everything for you and then you can just check at the end that it works. \n\nYou can also mess around with the left side of the equation first and then use that substitution at the end to get it more nicely. For example, if you factor (F(n+1):^3: - F(n-1):^3:) as a difference of cubes, then factor an \nF(n) out of everything, then write (F(n+1):^2: + F(n-1):^2:) as \n(F(n):^2: + 2F(n+1)F(n-1)) and then do the substitution, it should work out. It's still very messy, though, and I suspect that you can keep manipulating it, without using much more than the definition of the numbers given, and it should work out. Maybe.\n\nIncidentally, this should work no matter what the values of F(1) and F(2) are, shouldn't it?[/color]", "Solution_4": "[quote=\"JBL\"]Incidentally, this should work no matter what the values of F(1) and F(2) are, shouldn't it?[/quote]\r\n\r\nNo. That's not true. Take F_1 = a and F_2 = b. Then F_3 = a + b, F_4 = a + 2b, F_5 = 2a + 3b, F_6 = 3a + 5b. And the formula does not hold for n = 2. Recurrence relations can be tricky, they can depend greatly on what the initial conditions are.", "Solution_5": "[quote=\"Idiot_without_a_village\"]I see... but, is there a way to do it without using those other identities???...[/quote]\r\n\r\nWe have a matrix identity [tex]\\displaystyle \\begin{pmatrix} F_{3n+1} & F_{3n} \\\\ F_{3n} & F_{3n-1}\\end{pmatrix}=\\displaystyle \\begin{pmatrix} F_{n+1} & F_n \\\\ F_n & F_{n-1}\\end{pmatrix}^3[/tex], since [tex]\\displaystyle \\begin{pmatrix} F_{n+1} & F_n \\\\ F_n & F_{n-1}\\end{pmatrix} = \\begin{pmatrix} 1 &1\\\\1&0\\end{pmatrix}^n[/tex].\r\n\r\nTherefore [tex]F_{3n}=(F_{n+1}^2+F_n^2)F_n + (F_{n+1}F_n+F_nF_{n-1})F_{n-1}[/tex] after multiplying the matrices on the right hand side.\r\n\r\nHence [tex]F_{3n} + F_{n-1}^3 = [/tex] [tex]F_n^3 + (F_{n+1}^2F_n + (F_{n+1}F_nF_{n-1}+(F_nF_{n-1}^2 + F_{n-1}^3)))= F_{n}^3 + F_{n+1}^3[/tex].\r\n\r\n(Well, we still need some kind of identity, a matrix identity here :-)).", "Solution_6": "An interesting question: I've solved several Fibonacci problems by converting them to counting problems on a \"zig-zag ladder,\" for lack of a better description. For example, one such problem is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13654]here[/url]. Do you think that there is a reasonable way to use this method to construct a model for questions such as this one, or the other question fuzzylogic posted recently?" } { "Tag": [], "Problem": "do you know how can we draw a line such that it is perpendicular to a fix line?(only with ruler and compasses)", "Solution_1": "[hide=\"Solution\"]\nwhat you do is,\ntake a point on the line and using the compass intersect the line twice when you get those points, putting the tip of the compass of both those points make arcs on both sides of the line so that the arcs intersect.\nJoin the intersectio. That is your perpendicular line.\n[/hide]", "Solution_2": "nice solution,realy nice i enjoyed it so much. :D" } { "Tag": [ "calculus", "integration", "function", "induction", "real analysis", "real analysis unsolved" ], "Problem": "7. Prove that if the real-valued function $f$ on the interval $[a,b]$ is bounded and is continuous except at a finite number of points, then $\\int^{b}_{a}f(x)\\,dx$ exists.\r\n\r\nHere's my start: \r\n\r\nSuppose that $f(x)$ is not continuous at $h$ number of points ${x_{1}}',{x_{2}}', \\ldots,{x_{n}}'$. Let $a ={x_{0}}' <{x_{1}}' < \\ldots <{x_{n}}' = b$. I don't really know what to do from here. It's pretty obvious that a finite number of discontinuities should not affect the existence of the integral, but how can I show this using precise mathematical notation? \r\n\r\nANy help is greatly appreciated.", "Solution_1": "Prove it for one discontinuity and use induction . :)", "Solution_2": "Can you explain in more detail. Ok let $\\eta$ be a fixed point of $[a,b]$, let $c \\in R$, and set $f(x) = 0$ if $x\\ne\\eta$ and $f(x) = c$ if $x=\\eta$. \r\n\r\nThen we can prove for any Riemann Sum $S$ corresponding to a partition of $[a,b]$ of width less than $\\delta$ we have $|S|< 2|c|\\delta$. So clearly, $f$ is integrable on $[a,b]$ with one point discontinuity. \r\n\r\nBut how can I use induction to extend this argument?", "Solution_3": "Using the upper/lower sums definition of the Riemann integral, we need to find a partition such that the upper and lower sums differ by at most $\\epsilon$ (for any positive $\\epsilon$).\r\n\r\nSuppose that $|f(x)|p-1+2 p-1+...+(p-1) p-1\n-p-(p-1)! [/b]divisible by [b]p 2 [/b]", "Solution_1": "@ehsan2004! \r\nRead http://www.mathlinks.ro/viewtopic.php?p=138915#138915\r\nIt will help you to name topics in right way next time." } { "Tag": [ "logarithms", "calculus", "calculus computations" ], "Problem": "$1.$ $d^{n}u ?$\r\n$u = e^{ax+by}$\r\n\r\n$2.$ $d^{5}u ?$\r\n$u= ln(x-y)$", "Solution_1": "Do u have any idea?", "Solution_2": "[quote=\"Nea\"]$1.$ $d^{n}u ?$\n$u = e^{ax+by}$\n\n$2.$ $d^{5}u ?$\n$u= \\ln (x-y)$[/quote]\r\n\r\nIf $u=f(x,y)$ then $du=f_{x}(x,y)\\, dx+f_{y}(x,y)\\, dy$, where here the differential operator $d$ is given by $d=\\frac{\\partial}{\\partial x}\\, dx+\\frac{\\partial}{\\partial y}\\, dy$ so that $d^{n}=\\left(\\frac{\\partial}{\\partial x}\\, dx+\\frac{\\partial}{\\partial y}\\, dy\\right)^{n}$ and hence we have\r\n\r\n$1.$ For $u = e^{ax+by}$ we have \r\n\r\n\\[\\frac{\\partial^{n}\\left(e^{ax+by}\\right)}{\\partial x^{n-k}\\partial y^{k}}=a^{n-k}b^{k}e^{ax+by}\\]\r\n\r\nand this gives\r\n\r\n\\[d^{n}u = \\left(\\frac{\\partial}{\\partial x}\\, dx+\\frac{\\partial}{\\partial y}\\, dy\\right)^{n}e^{ax+by}=\\sum_{k=0}^{n}\\binom{n}{k}\\frac{\\partial^{n-k}}{\\partial x^{n-k}}\\, dx^{n-k}\\cdot\\frac{\\partial^{k}}{\\partial y^{k}}\\, dy^{k}\\, \\left(e^{ax+by}\\right) \\]\r\n\\[=\\sum_{k=0}^{n}\\binom{n}{k}\\frac{\\partial^{n}\\left(e^{ax+by}\\right)}{\\partial x^{n-k}\\partial y^{k}}\\, dx^{n-k}dy^{k}= \\sum_{k=0}^{n}\\binom{n}{k}a^{n-k}b^{k}e^{ax+by}\\, dx^{n-k}dy^{k}\\]\r\n\\[= e^{ax+by}\\sum_{k=0}^{n}\\binom{n}{k}a^{n-k}dx^{n-k}\\cdot b^{k}dy^{k}= e^{ax+by}(a\\, dx+b\\, dy)^{n}\\]\r\n\r\n$2.$ Find $d^{5}u$ if $u= \\ln (x-y)$ for which \r\n\r\n\\[\\frac{\\partial^{n}\\ln (x-y)}{\\partial x^{n-k}\\partial y^{k}}=(-1)^{k+1}\\frac{(n-1)!}{(x-y)^{n}}\\]\r\n\r\nand this gives\r\n\r\n\\[d^{n}u = \\left(\\frac{\\partial}{\\partial x}\\, dx+\\frac{\\partial}{\\partial y}\\, dy\\right)^{n}\\ln (x-y) =\\sum_{k=0}^{n}\\binom{n}{k}\\frac{\\partial^{n-k}}{\\partial x^{n-k}}\\, dx^{n-k}\\cdot\\frac{\\partial^{k}}{\\partial y^{k}}\\, dy^{k}\\, \\left(\\ln (x-y)\\right) \\]\r\n\\[=\\sum_{k=0}^{n}\\binom{n}{k}\\frac{\\partial^{n}\\ln (x-y)}{\\partial x^{n-k}\\partial y^{k}}\\, dx^{n-k}dy^{k}= \\sum_{k=0}^{n}\\binom{n}{k}(-1)^{k+1}\\frac{(n-1)!}{(x-y)^{n}}\\, dx^{n-k}dy^{k}\\]\r\n\\[=-\\frac{(n-1)!}{(x-y)^{n}}\\sum_{k=0}^{n}\\binom{n}{k}\\, dx^{n-k}\\cdot (-1)^{k}dy^{k}=-\\frac{(n-1)!}{(x-y)^{n}}\\left( dx-dy\\right)^{n}\\]\r\n\\[=-(n-1)!\\left(\\frac{dx-dy}{x-y}\\right)^{n}\\]\r\n\r\nset $n=5$ to get \r\n\r\n\\[d^{5}u =-4!\\left(\\frac{dx-dy}{x-y}\\right)^{5}\\]" } { "Tag": [ "geometry", "exterior angle" ], "Problem": "The sides of a regular polygon with $n$ sides $(n>4)$ are extended to form a star. The number of degrees in the angle at each point of the star in terms of $n$ is?", "Solution_1": "For a regular n-gon, the general formula for each interior angle is $(n-2)180$. This can be easily proved, by sub-dividing each polygon into triangles. If we extend the sides of a regular n-gon, we form a triangle at each side, which is an isosceles triangle. This is because by using the geometry reasoning that the exterior angle equals the sum of the two interior angles, we see that the angle at each point of the star is $\\frac{(n-2)180}{2n}$", "Solution_2": "Yeah... that's not what I get. Use one point of the star and the two closest vertices to form an isosceles triangle. The two exterior angles are both $\\frac{360}{n}$ so the angle of the star is $180-\\frac{720}{n}$. Notice that for $n = 4$ this is $0$ because they never meet, and it's easily verified for $n = 6$ (in case you have doubts).\r\n\r\neddie77, try plugging in $n = 5$ into your equation; it's clearly not true." } { "Tag": [ "AMC", "AIME" ], "Problem": "C\u2019est moi! Je m\u2019appelle Joseph. J\u2019ai trente-trois ans, et je suis un plombier. J\u2019aime manger la pizza et regarder la t\u00e9l\u00e9vision. Le week-end, je faire du jogging au parc aussi. En g\u00e9n\u00e9ral, je suis amusant, content, fort, et brun. Je ne suis ni grand ni petit. J\u2019aime John McCain.\r\n\r\nC\u2019est ma femme, Chantal. Elle a trente-deux ans. Elle est gentille, intelligente, et brune. Chantal est grande et mince aussi. Elle est une joueur de golf. Elle aime aller au centre comm\u00e9rcial surtout.\r\n\r\nC\u2019est mon fils, Jean. Il est brun, timide, et petit. Jean a dix ans. Il aime jouer \u00e0 des jeux vid\u00e9o et faire du velo. Le week-end, il aime sortir avec ses amis.\r\n\r\nC\u2019est ma fille, Am\u00e9lie. Elle est petite, blonde, et mignonne. Elle est curieuse aussi. Am\u00e9lie a sept ans. Elle aime beaucoup jouer. Elle aime la MJC aussi.\r\n\r\nC\u2019est mon p\u00e8re , Charles. Il a soixante-huit ans. Il est brun, aimable, et diligent. Il est patient et sensuel aussi. Il est un architecte. Le week-end, il aime faire de l\u2019 \u00e9quitation.\r\n\r\nC\u2019est ma m\u00e8re, Elizabeth. Elle a soixante-cinq ans. Elle est blonde, s\u00e9rieuse, efficace, et pratique. Le week-end, elle aime voyager et parler au telephone.", "Solution_1": "Is that your frecnh homework?", "Solution_2": "Do you think I would write in French if it weren't my homework?", "Solution_3": "Idk, maybe. Are you part French?", "Solution_4": "Just check it.", "Solution_5": "No hablo frances. :(", "Solution_6": "[quote=\"tinytim\"]No hablo frances. :([/quote]\n\nShouldn't that be \"No hablo franc\u00e9s\"?\n\nThen again, I haven't taken Spanish for over two years.\n\n\n[quote]En g\u00e9n\u00e9ral, je suis...brun.[/quote]\r\nI think this means \"I generally have brown hair.\"", "Solution_7": "It means \"in general...\" which can also be interpreted as when I'm talking about the ordinary features.", "Solution_8": "haha, i'm in french also.... i suck at it, i speak \"franglais\" too much in class and my teacher gets pissed at me. elle me deteste parce-que je suis noir. je ne suis pas noir, je suis brun (?indien?).", "Solution_9": "Is this French 1?\r\n\r\nAlso, why do you say aime so much in ur french?", "Solution_10": "Why do we use like so much in our English?", "Solution_11": "Ahh...I get it :D\r\n\r\nSo my usename is \"LIKE15\"?" } { "Tag": [ "geometry", "3D geometry", "sphere" ], "Problem": "What is the surface area of a sphere [color=green][mod edit: CUBE!][/color] with a volume of 2197? No calculators, as this is a sprint round question.", "Solution_1": "[hide]i got 6591/r, where r is the radius of the sphere. this is a pretty ugly problem for a sprint.[/hide]", "Solution_2": "I got:\r\n3rdsqrt(6591/4pi)^2", "Solution_3": "[hide]\n$\\displaystyle (4/3)(\\pi)(r^3)=2197\\\\r=\\frac{3\\sqrt[3]{2197-\\pi}}{4}$\n\nphew...\nSA=$4\\pi\\frac{3\\sqrt[3]{2197-\\pi}}{4}=\\pi3\\sqrt[3]{2197-\\pi}\n\n[/hide]", "Solution_4": "[quote=\"IntrepidMath\"][hide]\n$\\displaystyle (4/3)(\\pi)(r^3)=2197\\\\r=\\frac{3\\sqrt[3]{2197-\\pi}}{4}$\n\nphew...\nSA=$4\\pi\\frac{3\\sqrt[3]{2197-\\pi}}{4}=\\pi3\\sqrt[3]{2197-\\pi}$\n\n[/hide][/quote]\r\n\r\nThis is wrong.\r\nFirst of all, it's $r=\\sqrt[3]{\\frac{3\\cdot2197}{4\\pi}}$\r\n\r\nThen, SA should $4\\pi{r}^{2} $", "Solution_5": "I was thinking this was ugly, and last night I suddenly realized I read the problem wrong :rotfl: \r\n\r\nI meant a cube. Easier? :D", "Solution_6": "Ahahaha, that does make it much easier. You can easily tell that the cube root of 2197 is 13, so surface area:\r\n\r\n$S = 6 \\cdot 13^2 = 1014$", "Solution_7": "I think that most of us were confused at the difficulty of this problem before there was cube in there..." } { "Tag": [ "function" ], "Problem": "Let $ f(x^2 \\plus{} 1) \\equal{} x^4 \\plus{} 5x^2 \\plus{} 3$ What is $ f(x^2 \\minus{} 1)$?", "Solution_1": "$ f(x^{2} \\plus{}1) \\equal{} x^{4} \\plus{}5x^{2}\\plus{}3$\r\n\r\n$ \\equal{}(x^{2}\\plus{}1)^{2}\\plus{}3(x^{2}\\plus{}1) \\minus{}1$.\r\n\r\nTherefore,\r\n\r\n$ f(x^{2}\\minus{}1) \\equal{} (x^{2}\\minus{}1)^{2} \\plus{}3(x^{2}\\minus{}1)\\minus{}1$\r\n\r\n$ \\equal{}x^{4}\\plus{}x^{2}\\minus{}3$.", "Solution_2": "$ f(x^{2}\\minus{}1)\\equal{}f(((x^{2}\\minus{}2)^{\\frac{1}{2}})^{2}\\plus{}1)\\equal{}(x^{2}\\minus{}2)^{2}\\plus{}5(x^{2}\\minus{}2)\\plus{}3\\equal{}x^{4}\\plus{}x^{2}\\minus{}3$", "Solution_3": "[quote=\"Truffles\"]Let $ f(x^2 + 1) = x^4 + 5x^2 + 3$ What is $ f(x^2 - 1)$?[/quote]\r\n\r\nGiven the information you have, we try to derive $ f(x)$. We can see that $ x$ is a second degree expression. We can experiment a bit by assuming that the first term of $ f(x)$ is of the form $ ax^2$. We square $ (x^2+1)$ and get $ x^4+2x^2+2$. Compensating for the $ 2x^2$, we need to add $ 3x^2$. Therefore, we have to add $ 3(x^2+1)$.\r\n\r\nSo far we have $ (x^2+1)^2+3(x^2+1)$. To compensate for the $ 3$, we simply subtract $ -1$. \r\n\r\nWe now have: $ f(x^2+1)=(x^2+1)^2+3(x^2+1)-1 \\implies f(x)=x^2+3x-1$.\r\n\r\nNow we substitute $ (x^2-1)$ into $ f(x)$ and get $ \\boxed{x^4+x^2-3}$.", "Solution_4": "Let $ f(x^2 \\plus{} 1) \\equal{} x^4 \\plus{} 5x^2 \\plus{} 3$ What is $ f(x^2 \\minus{} 1)$?.IN THIS FUNCTION f(x^2 + 1) = x^4 + 5x^2 + 3.F(x)=(x^2 + 1)^4+5(x^2 + 1)+3.then f(x)=x^4-x^2-3.then \t\r\nf(x^{2}-1)=f(((x^{2}-2)^{\\frac{1}{2}})^{2}+1)=(x^{2}-2)^{2}+5(x^{2}-2)+3=x^{4}+x^{2}-3" } { "Tag": [ "trigonometry" ], "Problem": "A crate with mass 10 kg is connected by a massless rope to a cylinder. The massless rope is wrapped around the cylinder several times, en the cilinder has a radius of 40 cm and a mass of 100 kg.\r\nThe co\u00ebfficient of friction between the crate and the inclined plane is 0,38.\r\n\r\nFind the angular acceleration of the cylinder, the acceleration of the crate, and the tension in the rope.", "Solution_1": "[hide=\"Answer\"]Let $\\theta$ be the inclination of the plane (37\u00ba), M the mass of the cylinder, m the mass of the crate, $\\mu$ the coefficient of friction, g = 9.8 m/s2 the acceleration of gravity, $\\alpha$ the angular acceleration, a the liner acceleration, and T the tension. Then:\n\n$a = \\frac{g(\\sin \\theta-\\mu \\cos \\theta)}{1+M/2m}= 0.487 m/s^{2}$\n$\\alpha = \\frac{a}{r}= 1.22 rad/s^{2}$\n$T = \\frac{Ma}{2}= 24.4 N$\n[/hide]" } { "Tag": [ "articles" ], "Problem": "http://story.news.yahoo.com/news?tmpl=story&cid=514&e=20&u=/ap/20041010/ap_on_sc/math_against_terror_4", "Solution_1": "Hmmm... I\"m not too sure about using abstract mathematics to gain ground on the war on terror is plausible, but reading the article, I now believe even more that the Patriot Act is unjust.", "Solution_2": "what is this Patriot Act that everybody is talking about?", "Solution_3": "[url]http://www.epic.org/privacy/terrorism/hr3162.html[/url]", "Solution_4": "Interesting that I didn't see [i]any[/i] mention of the NSA in the article -- seems like it deserves a mention in this article.\r\n\r\nAs for the Patriot act (if you don't want to read the several hundred pages that white_horse linked to), I can give you the short (and a-political) version:\r\n\r\nThe U.S.A. PATRIOT Act (the name is actually an acronym for something or other) was a massive security bill passed by the United States Congress and signed into law by President Bush shortly after September 11. It was passed by an overwhelming majority in both the House and Senate (95 or more of the 100 senators voted in favor of it). Since then, it's come under increasing attack by various civil liberty organizations, especially the ACLU, for allowing various actions which allegedly are unconstitutional. Actually, some of the first groups to vocally oppose it were library and librarian's associations, interestingly enough. (One portion of the bill obligated libraries to turn over customer records to Federal agents without a warrant and without being allowed to inform anyone, especially the individual whose records were seized, of what had happened.)" } { "Tag": [], "Problem": "Provide a synthesis of m-nitrotoluene starting with benzene.", "Solution_1": "Again, I am just starting so lots of help needed:\r\n\r\nMaybe:\r\n\r\n1. Nitrating Mixture to create nitrobenzene (because $ \\minus{}NO_2$ is a $ 3,5$ directing group)\r\n2. Friedel-Craft Alkylation using $ AlCl_3$ and $ CH_3Cl$ to obtain the product.", "Solution_2": "Some comments:\r\n\r\n1. -NO2 is called a meta directing group, not 3,5 directing. However, in your initial reaction that doesn't matter because the aromatic ring is unsubstituted.\r\n\r\n2. The Fridel-Crafts reactions (alkylation and acylation) only work with activated aromatic rings, that is, they don't work in aromatic rings bearing a deactivating group (like the nitro group, which is a very powerfull deactivating group). So your second reaction certainly wouldn't work." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "I like it:\r\nLet $G$ be infinite group,and we know that for any subgroup $H$ of $G$ ,$|H|\\neq 1$ ,we have $[G:H]<\\infty $prove that $ G$ is cyclic.", "Solution_1": "It follows from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43092]this one[/url].\r\n\r\nTake any element $a\\ne 1$ of $G$ and consider the cyclic group $H=\\langle a\\rangle$ generated by $a$. It has finite index, so let $x_1=1,x_2,\\ldots,x_n$ be representatives of all the left cosets of $H$. Since for each $k\\in\\overline{2,n}$ there is an $m_k$ s.t. $a^{m_k}\\in\\langle x_k\\rangle$, it means that $a'=a^{m_2\\ldots m_n}$ commutes with $x_1,\\ldots,x_n$, and it's clear now that $a'$ is a non-trivial element of $G$ belonging to the center of $G$.\r\n\r\nWe have thus shown that $Z(G)\\ne \\{1\\}$. This means that it has finite index in $G$, so, according to the problem cited above, $G'$ is finite. Given the imposed conditions on $G$, this is impossible unless $G'$ is trivial, which is equivalent to $G$ being abelian.\r\n\r\nWe can now use the structure theorem for finitely generated abelian groups ($G$ is clearly finitely generated) to conclude that $G$ is $\\mathbb Z^n$ for some $n$, and it's easy to see that $n$ must be $1$.\r\n\r\nAll that's left now is for us to solve that other problem :)." } { "Tag": [ "ratio", "geometry" ], "Problem": "One side of a square A is a diagonal of square B. What is the ratio of the area of square B to the area of square A? Express your answer as a common fraction.", "Solution_1": "[asy]draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);\ndraw((0,0)--(1,1)--(2,0)--(1,-1)--cycle);[/asy]\r\n\r\nWe can clearly see that $ \\frac{1}{2}$ of square $ B$ is the same as $ \\frac{1}{4}$ of square $ A$. Thus, the desired ratio is $ \\frac{\\frac{1}{4}}{\\frac{1}{2}}\\equal{}\\boxed{\\frac{1}{2}}$.\r\n\r\nAdditionally, let the side length of square $ B$ be $ x$. Then the side length of square $ A$ is $ \\sqrt{x^2\\plus{}x^2}\\equal{}x\\sqrt{2}$. The area of square $ B$ is $ x^2$, and the area of square $ A$ is $ (x\\sqrt{2})^2\\equal{}2x^2$. The desired ratio is thus $ \\frac{x^2}{2x^2}\\equal{}\\boxed{\\frac{1}{2}}$." } { "Tag": [ "geometry", "parallelogram", "trigonometry", "rectangle", "geometry solved" ], "Problem": "Let ABCD be a parallellogram. Let Q be the midpoint of [DA], and let F be the foot of the perpendicular on QC trough B. Show that |AF| = |AB|.\r\n\r\nYou know, this problem is just easy using cosine rules and other trigonometric stuff ... But I was wondering, is there an elegant - purely geometric - solution for this problem ? The only solution I found is the trig solution.", "Solution_1": "I think I have a ?pure? soln :) :\r\n\r\nLet the foot of the perpendicular from A to CQ be E, the midpoint of BC be P and the pt of intersection between BF and AP be M. AE parallel to BF (they are both perpendicular to CQ), AP parallel to CE (obvious), so you get AMFE is a rectangle, so AM perpendicular to BF and AE=MF. Since 2AQ=BC and AQ || BC (|| means parallel) we also have 2AE=BF. Because AE=MF we get 2MF=BF, and since AM perpendicular to BF the median AM is also an altitude in triangle ABF, so it is obviously an isosceles triangle.", "Solution_2": "Thanks a lot !", "Solution_3": "Let E be the midpoint of BC and G the intersection of AE with BF.Since AE || CQ (obviously) we have that AG is perpendicular to BF. But also , since AE || CQ , then EG || CF so in the triangle BCF , EG is a middle line . Thus BG=GF . From here it follows that ABF is isosceles since BG is a height and a median.", "Solution_4": "AB intersects QC in T. A is the mid-point of BT.\r\n AF is the median from right angle BFT in triangle BFT.\r\n\r\n\r\n Maj. Pestich\r\n\r\n\r\n Polish olympiad 2001, 2nd round.", "Solution_5": "Translate $ \\triangle DQC$, so CD is on AB, Q goes to E. Then QEBF is a rectangle, so since DQ=AQ=AE, AF=BF." } { "Tag": [ "inequalities", "IMO Shortlist", "number theory unsolved", "number theory" ], "Problem": "Let $p$ be an odd prime. Determine the positive integers $x$ and $y$ with $x\\leq y$ for which the number $\\sqrt{2p}-\\sqrt{x}-\\sqrt{y}$ is non-negative and as small as possible.", "Solution_1": "This is actually IMO Shortlist 1995 problem N8 ( http://www.kalva.demon.co.uk/short/soln/sh95n8.html ).\r\n\r\nCan anyone check my solution below? It looks very different from the proposed solution, it has less ugly computations than the proposed solution, and it uses the primality of p only at the very end (it wouldn't use it at all if we would replace \"non-negative\" by \"positive\" in the condition of the problem), so I am wondering whether it can be correct...\r\n\r\n[color=blue][b]Problem.[/b] Let p be an odd prime. Determine the positive integers x and y with $x\\leq y$ for which the number $\\sqrt{2p}-\\sqrt{x}-\\sqrt{y}$ is non-negative and as small as possible.[/color]\r\n\r\n[i]Solution.[/i] We claim that the required integers x and y are $x=\\frac{p-1}{2}$ and $y=\\frac{p+1}{2}$.\r\n\r\nIn fact, first note that\r\n\r\n$\\left(\\sqrt{\\frac{p-1}{2}}+\\sqrt{\\frac{p+1}{2}}\\right)^{2}=\\frac{p-1}{2}+\\frac{p+1}{2}+2\\cdot\\sqrt{\\frac{p-1}{2}}\\cdot\\sqrt{\\frac{p+1}{2}}$\r\n$=p+\\sqrt{\\left(p-1\\right)\\left(p+1\\right)}=p+\\sqrt{p^{2}-1}$.\r\n\r\nNow, $\\frac{p-1}{2}$ and $\\frac{p+1}{2}$ are integers (since p is odd, and thus p - 1 and p + 1 are even), and $\\sqrt{2p}-\\sqrt{\\frac{p-1}{2}}-\\sqrt{\\frac{p+1}{2}}$ is indeed non-negative (this is equivalent to\r\n\r\n$\\sqrt{2p}\\geq\\sqrt{\\frac{p-1}{2}}+\\sqrt{\\frac{p+1}{2}}\\ \\ \\ \\ \\ \\Longleftrightarrow\\ \\ \\ \\ \\ 2p\\geq\\left(\\sqrt{\\frac{p-1}{2}}+\\sqrt{\\frac{p+1}{2}}\\right)^{2}$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ 2p\\geq p+\\sqrt{p^{2}-1}\\ \\ \\ \\ \\ \\Longleftrightarrow\\ \\ \\ \\ \\ p\\geq\\sqrt{p^{2}-1}\\ \\ \\ \\ \\ \\Longleftrightarrow\\ \\ \\ \\ \\ p^{2}\\geq p^{2}-1$,\r\n\r\nwhat is true).\r\n\r\nSo it remains to prove that $\\sqrt{2p}-\\sqrt{\\frac{p-1}{2}}-\\sqrt{\\frac{p+1}{2}}$ is indeed the smallest possible nonnegative sum of the form $\\sqrt{2p}-\\sqrt{x}-\\sqrt{y}$ for positive integers x and y with $x\\leq y$ and achieved only for $x=\\frac{p-1}{2}$ and $y=\\frac{p+1}{2}$.\r\n\r\nIn order to prove this, we assume that x and y are positive integers with $x\\leq y$ and\r\n\r\n$0\\leq\\sqrt{2p}-\\sqrt{x}-\\sqrt{y}\\leq\\sqrt{2p}-\\sqrt{\\frac{p-1}{2}}-\\sqrt{\\frac{p+1}{2}}$.\r\n\r\nIf we succeed to show that $x=\\frac{p-1}{2}$ and $y=\\frac{p+1}{2}$, then we will be done.\r\n\r\nThe inequality chain $0\\leq\\sqrt{2p}-\\sqrt{x}-\\sqrt{y}\\leq\\sqrt{2p}-\\sqrt{\\frac{p-1}{2}}-\\sqrt{\\frac{p+1}{2}}$ rewrites as\r\n\r\n$\\sqrt{\\frac{p-1}{2}}+\\sqrt{\\frac{p+1}{2}}\\leq\\sqrt{x}+\\sqrt{y}\\leq\\sqrt{2p}$.\r\n\r\nWe square this inequality (it's not the time to worry about positive and negative yet ;) ):\r\n\r\n$\\left(\\sqrt{\\frac{p-1}{2}}+\\sqrt{\\frac{p+1}{2}}\\right)^{2}\\leq\\left(\\sqrt{x}+\\sqrt{y}\\right)^{2}\\leq 2p$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ p+\\sqrt{p^{2}-1}\\leq x+y+2\\cdot\\sqrt{x}\\cdot\\sqrt{y}\\leq 2p$.\r\n\r\nNow, we have $x+y\\sqrt{p^{2}-2p+1}=p-1$, this yields $x+y>\\frac12\\left(p+\\left(p-1\\right)\\right)=p-\\frac12$, and thus, since x and y are integers, this entails $x+y\\geq p$. So we have $p\\leq x+y<2p$. Thus, x + y = p + s for some integer s satisfying $0\\leq s\\leq p-1$. Denoting x - y = k, we note that $k\\leq 0$ (since $x\\leq y$) and $x=\\frac{x+y}{2}+\\frac{x-y}{2}=\\frac{p+s}{2}+\\frac{k}{2}$ and $y=\\frac{x+y}{2}-\\frac{x-y}{2}=\\frac{p+s}{2}-\\frac{k}{2}$. Hence, $p+\\sqrt{p^{2}-1}\\leq x+y+2\\cdot\\sqrt{x}\\cdot\\sqrt{y}\\leq 2p$ becomes\r\n\r\n$p+\\sqrt{p^{2}-1}\\leq p+s+2\\cdot\\sqrt{\\frac{p+s}{2}+\\frac{k}{2}}\\cdot\\sqrt{\\frac{p+s}{2}-\\frac{k}{2}}\\leq 2p$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ p+\\sqrt{p^{2}-1}\\leq p+s+\\sqrt{\\left(p+s+k\\right)\\left(p+s-k\\right)}\\leq 2p$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ \\sqrt{p^{2}-1}-s\\leq\\sqrt{\\left(p+s+k\\right)\\left(p+s-k\\right)}\\leq p-s$ (we subtracted p + s from the last inequality chain)\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ \\sqrt{p^{2}-1}-s\\leq\\sqrt{\\left(p+s\\right)^{2}-k^{2}}\\leq p-s$.\r\n\r\nSince $\\sqrt{p^{2}-1}>\\sqrt{p^{2}-2p+1}=p-1\\geq s$, the left hand side of this inequality, $\\sqrt{p^{2}-1}-s$, is positive; thus, we can square this inequality:\r\n\r\n$\\left(\\sqrt{p^{2}-1}-s\\right)^{2}\\leq\\left(p+s\\right)^{2}-k^{2}\\leq\\left(p-s\\right)^{2}$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ \\left(p^{2}-1\\right)+s^{2}-2\\sqrt{p^{2}-1}\\cdot s\\leq p^{2}+s^{2}+2ps-k^{2}\\leq p^{2}+s^{2}-2ps$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\;-1-2\\sqrt{p^{2}-1}\\cdot s\\leq 2ps-k^{2}\\leq-2ps$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ 1+2\\sqrt{p^{2}-1}\\cdot s\\geq k^{2}-2ps\\geq 2ps$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ 1+2\\sqrt{p^{2}-1}\\cdot s+2ps\\geq k^{2}\\geq 4ps$.\r\n\r\nSince $\\sqrt{p^{2}-1}<\\sqrt{p^{2}}=p$, we even have\r\n\r\n$1+2p\\cdot s+2ps>k^{2}\\geq 4ps\\ \\ \\ \\ \\ \\Longleftrightarrow\\ \\ \\ \\ \\ 1+4ps>k^{2}\\geq 4ps$.\r\n\r\nSince $k^{2}$ is an integer, we thus have $k^{2}=4ps$. Thus, $p\\mid k^{2}$, what, since p is prime, yields $p\\mid k$, so that $p^{2}\\mid k^{2}$. Thus, $p^{2}\\mid 4ps$, so that $p\\mid 4s$. Since p is odd, this yields $p\\mid s$. Since $0\\leq s\\leq p-1$, this leads to s = 0, and thus $x=\\frac{p+s}{2}+\\frac{k}{2}=\\frac{p}{2}+\\frac{k}{2}$ and $y=\\frac{p+s}{2}-\\frac{k}{2}=\\frac{p}{2}-\\frac{k}{2}$. Now,\r\n\r\n$\\left(\\sqrt{\\frac{p-1}{2}}+\\sqrt{\\frac{p+1}{2}}\\right)^{2}\\leq\\left(\\sqrt{x}+\\sqrt{y}\\right)^{2}$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ \\left(\\sqrt{\\frac{p-1}{2}}+\\sqrt{\\frac{p+1}{2}}\\right)^{2}\\leq\\left(\\sqrt{\\frac{p}{2}+\\frac{k}{2}}+\\sqrt{\\frac{p}{2}-\\frac{k}{2}}\\right)^{2}$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ \\left(\\sqrt{p-1}+\\sqrt{p+1}\\right)^{2}\\leq\\left(\\sqrt{p+k}+\\sqrt{p-k}\\right)^{2}$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ \\left(p-1\\right)+\\left(p+1\\right)+2\\cdot\\sqrt{p-1}\\cdot\\sqrt{p+1}$\r\n$\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\leq\\left(p+k\\right)+\\left(p-k\\right)+2\\cdot\\sqrt{p+k}\\cdot\\sqrt{p-k}$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ 2p+2\\sqrt{p^{2}-1}\\leq 2p+2\\sqrt{p^{2}-k^{2}}\\ \\ \\ \\ \\ \\Longleftrightarrow\\ \\ \\ \\ \\ 1\\geq k^{2}$.\r\n\r\nSince k is an integer and $k\\leq 0$, this yields either k = 0 or k = -1. For k = 0, we have $x=\\frac{p}{2}+\\frac{k}{2}=\\frac{p}{2}$, what is impossible, since p is odd and x must be an integer. Thus, k = -1, so that $x=\\frac{p}{2}+\\frac{k}{2}=\\frac{p-1}{2}$ and $y=\\frac{p}{2}-\\frac{k}{2}=\\frac{p+1}{2}$, and the problem is solved.\r\n\r\nOh dear, you are still reading this here? I apologize for being even more boring than usual, but I tried to write every single step to find a mistake. I don't see any. Do you?\r\n\r\n Darij", "Solution_2": "Denote $s=\\sqrt{y}+\\sqrt{x}, t=\\sqrt{y}-\\sqrt{x}$\r\n\r\nCase #1: $x=y$, we will get the answer easily: $\\sqrt{2p}-\\sqrt{2(p-1)}$\r\n\r\nCase #2: $x d$, it seems that $ g$ can come much closer to $ a$, though. E.g., $ [6,1,0] \\plus{} [3,3,1]\\ge 2[5,1,1]$ seems to hold. \r\nA more precise, interesting question would thus be if there is a minimum for $ \\frac {a \\minus{} g}{a \\minus{} d}$. :maybe:" } { "Tag": [ "function", "vector", "integration", "calculus", "geometry", "3D geometry", "sphere" ], "Problem": "Stokes theorem is usually stated as follows.\r\nLet $S$ be a smooth surface in $\\mathbb{R}^{3}$ parameterized by a $C^{2}$,$1-1$ function $\\phi : T\\rightarrow \\mathbb{R}^{3}$, where $T$ is a plane region bounded by a piecewise $C^{1}$ Jordan curve$\\Gamma$. Denote by $C$ the image of $\\Gamma$ under$\\phi$. Let $F$ be a $C^{1}$ vector field on an open set containing $S$ and $C$. Then we have$\\int \\int_{S}(\\nabla\\times F).ndS=\\oint_{C}F.d\\alpha$,\r\nwhere the line integral is taken in the direction inherited via$\\phi$ from the positive direction of$\\Gamma$. \r\n[b]a[/b])What is the injectivity of $\\phi$ good for?\r\n[b]b[/b])How do we conclude, rigorously, from (the extended version) of this theorem that for every closed orientable object, say a sphere, both integrals are equal to $0$?\r\nPS For the former I guess it has something to do with orientablity.\r\nAnd for the latter I have read that the boundary of closed objectes is empty, so the result follows!! But this does not content me.\r\nThank you in advanced for your reply.[A typo fixed in edit.]", "Solution_1": "If we're integrating over the image, we want that image to be traced only once. If $\\phi$ were not injective, the image might not even be a surface.\r\n\r\nFor the second question, we need a topological theorem:\r\nEvery compact smooth orientable surface can be dissected into finitely many consistently oriented pieces, each parametrized as in the statement.\r\n\r\nThen we integrate over the whole surface by adding up the pieces. The boundary integrals will directly cancel (each part of the boundary is traced twice in opposite directions), and the whole integral becomes zero.\r\n\r\nIn the general manifold treatment of Stokes' theorem, there are no tricks needed for the no-boundary special case. We need tricks here because we are restricted in the original statement to surfaces diffeomorphic to a compact subset of $\\mathbb{R}^{2}$.", "Solution_2": "[quote=\"jmerry\"]If we're integrating over the image, we want that image to be traced only once. If $\\phi$ were not injective, the image might not even be a surface.[/quote]\r\nDear, jmerry. We have a definition for a [i]smooth curve [/i]as follows; Let $W$ be a region in the $uv$ plane, we call the function $\\phi : W\\rightarrow \\mathbb{R}^{3}$ a smooth curve when these two hold;\r\n[b]a[/b])$\\phi$ is $C^{1}$ [b]b[/b]) $\\forall (u, v) \\in W, \\phi_{u}\\times \\phi_{v}$ is non zero.\r\nOn the other hand we have defined integration over a smooth curve(ie density integral) without mentioning the condition of surjectivity. So the first question is still open for me. :)", "Solution_3": "Please check if this satisfies Stokes' conditions (and/ or its results)?\r\n$\\phi (u, v)=\\rho (\\sin u\\cos v,\\sin u\\sin v, \\cos u)$, where $0\\leq u <\\pi$, $0\\leq v<2\\pi$ with $F=d\\alpha$.\r\nAnd this one is about the second question.", "Solution_4": ":help: Hey,It's also my problem.Has anyone a solution?" } { "Tag": [], "Problem": "What is the last 3 digits (hundreds, tens and units digits) of 3^1983. \r\n\r\nThanks.", "Solution_1": "I can find the units digit very easily, and the tens digit with a little more work, but hundreds is difficult... what you need to do, though, is find a pattern in the final digits. For units, we have the pattern 3,9,7,1 repeated. For tens and units together, we have 03, 09, 27, 81, 43, 29, 87, 61, 83, 49, 47, 41, 23, 69, 07, 21, 63, 89, 67, 01, at which point it repeats. So, I can tell you that $3^{1983}$ has tens and units digits of $27$.\r\n\r\nIf we look at hundreds as well, we have 003, 009, 027, 081, 243, 729, 187, 561, 683, 049, 147, 441, 323, 969, 907, 721, 163, 489, 467, 401, 203, 609, 827, 481, 443, 329, 987, 961, 883, 649, 947, 841, 523, 569, 707, 121, 363, 089, 267, 801, 403, 209, 627, 881, 643, 929, 787, 361, 083, 249, 747, 241, 723, 169, 507, 521, 563, 689, 067, 201, 603, 809, 427, 281, 843, 529, 587, 761, 283, 849, 547, 641, 923, 769, 307, 921, 763, 289, 867, 601, 803, 409, 227, 681, 043, 129, 387, 161, 483, 449, 347, 041, 123, 369, 107, 321, 963, 889, 667, 001, and from there it repeats. (I used a calculator to do this; it would probably take half an hour to find this by hand.) This pattern repeats after 100 terms, and the 83rd is $\\boxed{227}$, which is your answer.", "Solution_2": "Usually they won't ask you for the hundreds digit, just the tens and the ones (they're easy too).", "Solution_3": "I see the pattern way works, but can't you use mods which would make it faster and easier.", "Solution_4": "Okay... show me how that would work.", "Solution_5": "I suppose a combinatorics/number theory approach might work for this... but it gets awfully complicated... (read AoPS vol 2)\r\n\r\n$3^{1983}=3*9^{991}$\r\n\r\nI'm guessing that if we find the last three digits of $9^{991}$ , and multiply that by 3, then we will have the last three digits for $3^{1983}$ .\r\n\r\nSo...\r\n\r\n$9^{991}={(10-1)}^{991}={991\\choose991}{10^{991}}-{991\\choose990}{10^{990}}+{991\\choose989}{10^{989}}\\ldots-{991\\choose2}{10^2}+{991\\choose1}{10}-{991\\choose0}$\r\n\r\n\r\nTo find the last three digits, we take ${(mod 1000)}$ since we only want the last three terms, everything else will be modded out.\r\n\r\nThe last three terms are:\r\n\r\n$-{991\\choose2}*{10^{2}}+{991\\choose1}*{10}-{991\\choose0}$\r\n\r\nNow... \r\n\r\n$-{991\\choose2}*{10^{2}}=-\\frac{(991)(990)(100)}{2}\\equiv{500 (mod 1000)}$\r\n\r\n\r\n${991\\choose1}*{10}=-(991)(10)={9910}\\equiv{910(mod1000)}$\r\n\r\n\r\n$-{991\\choose0}\\equiv{-1(mod1000)}$\r\n\r\n\r\n\r\nwe add it all up: \r\n${{(500)}+{(910)}+{(-1)}=1409}$\r\n${1409}\\equiv{409(mod1000)}$\r\n\r\nMultiply 409 by 3, because we took out the 3 at the beginning to make it a power of 9, \r\nwe get ${1227}\\equiv{227(mod1000)}$\r\n\r\nTherefore the last three digits are 227.", "Solution_6": "Nice solution, I like it. :D", "Solution_7": "Here is another solution from Andrew Westerdale (Stoneman Douglas HS, Florida):\r\n\r\nso close too far: i did it using euler's theorem, but it happens to be really messy/ugly\r\nso close too far: just becasue of the numbers\r\nso close too far: m^phi(n) = 1 mod n\r\nso close too far: m^phi(n) = 1 mod n is really effective\r\nso close too far: for n = 1000\r\nso close too far: phi(n) = 1000 (1-1/2)(1-1/5) = 400\r\nso close too far: so 3^400 = 1 mod 1000\r\nso close too far: so 3^1600 = 1 mod 1000\r\nso close too far: 3^1983 = 3^1600 * 3^383 mod 1000\r\nso close too far: =3^383 mod 1000\r\nso close too far: the way you get phi(n) is by evaluating the product of n and (1-1/f) for all factors f\r\nso close too far: which is why phi(1000) = 1000(1-1/2)(1-1/5)\r\nso close too far: anyways\r\nso close too far: so we want to find 3^383mod1000\r\nso close too far: break 383 into \"binary\":\r\nso close too far: 2^8+2^6+2^5+2^4+2^3+2^2+2^1+2^0\r\nso close too far: so 3^383 = 3^(2^8)*3^(2^6)*...*3^(2^0)\r\nso close too far: the good thing about the sequence we created is that each term is the square of the one before it, with some terms, 3^(2^7), left out\r\nso close too far: but this makes the it easier to do the arithmatic\r\nso close too far: since we just keep squaring the term, from 3 to 9 to 81, etc. only keeping the last 3 digits\r\nso close too far: we get 3^383 = 3 * 9 * 81 * 561 * 721 * 841 * 281 * 521 mod 1000\r\nso close too far: where you break out the crazy arithmatic skills, combine pairs of them and keep the last 3 digits as you go\r\nso close too far: and you end up with 227 mod 1000\r\nso close too far: it usually works out nicer because the problem will be like 3^1220 so you can take out the 3^400's and are left with something more manageable" } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "i need a way to solve this integral:\r\nintegral[(x^2-2)/(x^5-x^3+x)]dx\r\nthanks.", "Solution_1": "Hint\r\n$ \\frac {x^2 \\minus{} 2}{x^5 \\minus{} x^3 \\plus{} x} \\equal{} \\frac {\\frac {1}{2}(4x^3 \\minus{} 2x)}{x^4 \\minus{} x^2 \\plus{} 1} \\minus{} \\frac {2}{x}$", "Solution_2": "and then how do i integrate the left one?", "Solution_3": "$ \\int\\frac{f'(x)}{f(x)}\\,dx \\equal{} \\ln|f(x)| \\plus{} C$." } { "Tag": [], "Problem": "Sa se determine toate morfismele de la (Q , +) la (Sn , o), unde Sn este grupul permutarilor de grad n\r\n\r\nde la definitivat 2001", "Solution_1": "$\\mathbb Q$ e grup divizibil (adica pentru orice $x\\in\\mathbb Q$ si orice numar natural nenul $n$ exista $y\\in\\mathbb Q$ astfel incat $ny=x$), si un morfism de la un grup divizibil la unul finit trebuie sa fie trivial." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "SFFT", "special factorizations" ], "Problem": "Find all ordered triples of real numbers $ (x,y,z)$ that satisfy $ \\{\\begin{array}{c} xy = z - x - y \\\\\r\nyz = x - y - z \\\\\r\nzx = y - z - x \\end{array} \\;$\r\n\r\n\r\n[u][b]Source:[/b][/u] [url=http://web.mit.edu/rwbarton/Public/mop/tc4.pdf]Shamelessly borrowed from here[/url]", "Solution_1": "EDIT: Blah, this solution was both ugly and had a subtle flaw.\r\n\r\nTo answer the below question about this being too easy for MOP, team contests have plenty of very easy problems on them. This because of the way team contests are scored, which is a little unorthodox.", "Solution_2": "Unless I'm doing it wrong, isn't this way too easy for MOP?\r\n[hide]\nWe can rearrange the first equation to get $ xy\\plus{}x\\plus{}y\\equal{}z$, and then use Simon's Favorite Factoring Trick to get \n\n$ (x\\plus{}1)(y\\plus{}1)\\equal{}z\\plus{}1$.\n\nSimilarly, using the other equations, \n\n$ (y\\plus{}1)(z\\plus{}1)\\equal{}x\\plus{}1$\n\nand\n\n$ (z\\plus{}1)(x\\plus{}1)\\equal{}y\\plus{}1$.\n\nSubstitute $ a\\equal{}x\\plus{}1$, $ b\\equal{}y\\plus{}1$, and $ c\\equal{}z\\plus{}1$ for convenience. Multiplying the three equations together, we get\n\n$ a^2b^2c^2\\equal{}abc$.\n\nFirst, we consider what happens when one of the variables, WLOG $ a$, is $ 0$. Since $ bc\\equal{}a$, either $ b$ or $ c$, WLOG $ b$, is $ 0$. Since $ ab\\equal{}c$, we then have $ c\\equal{}0$. So $ (a,b,c)\\equal{}(0,0,0)$, which corresponds to $ (x,y,z)\\equal{}(\\minus{}1,\\minus{}1,\\minus{}1)$, works.\n\nNow, assume that all variables are nonzero. We can divide through by $ abc$ to get $ abc\\equal{}1$. Since $ ab\\equal{}c$, we get $ c^2\\equal{}1$. Similarly, $ a^2\\equal{}1$ and $ b^2\\equal{}1$. Since $ abc\\equal{}1$, we can either have $ 0$ or $ 2$ of $ a,b,c$ be $ \\minus{}1$. So we find that $ (a,b,c)\\equal{}(1,1,1)$, which corresponds to $ (x,y,z)\\equal{}(0,0,0)$, is a solution, along with all permutations of $ (a,b,c)\\equal{}(1,\\minus{}1,\\minus{}1)$, which gives all permutations of $ (x,y,z)\\equal{}(0,\\minus{}2,\\minus{}2)$ as solutions.\n[/hide]", "Solution_3": "Zuton Force, how'd you find that file?", "Solution_4": "I think it's from Reid Barton's compilation:\r\nhttp://web.mit.edu/rwbarton/Public/mop/", "Solution_5": "[quote=\"MellowMelon\"]To answer the below question about this being too easy for MOP, team contests have plenty of very easy problems on them. This because of the way team contests are scored, which is a little unorthodox.[/quote]\r\n\r\nOK, but isn't this a little too easy to serve any purpose at MOP? I mean, this is a 5 minute problem at most for anyone of MOP caliber.", "Solution_6": "[quote=\"towersfreak2006\"]I think it's from Reid Barton's compilation:\nhttp://web.mit.edu/rwbarton/Public/mop/ [/quote]I know that, but how'd he find that?", "Solution_7": "It caught my eye because I made up an identical system for a math team lecture 2 years ago while teaching factoring.\r\n\r\nI found it really funny that something like that could be found around MOP, let alone, one that is coincidentally identical to an exercise used at a high school with no USAMO qualifiers.\r\n\r\nTo answer the above question, I found it clicking random links on Wikipedia. I somehow found myself at [i]\"Frank and Brennie Morgan Prize for Outstanding Research in Mathematics by an Undergraduate Student\"[/i], from where I clicked a link to go to [i]\"Reid W Barton\"[/i], which had a link to his collection of files.", "Solution_8": "[quote=\"The Zuton Force\"]Find all ordered triples of real numbers $ (x,y,z)$ that satisfy $ \\{\\begin{array}{c} xy = z - x - y \\\\\nyz = x - y - z \\\\\nzx = y - z - x \\end{array} \\;$\n\n\n[u][b]Source:[/b][/u] [url=http://web.mit.edu/rwbarton/Public/mop/tc4.pdf]Shamelessly borrowed from here[/url][/quote]\r\nHere's my solution, which is slightly different than E^(pi-1)'s solution above:\r\nPlus all of three equations we have: xy+yz+xz=-(x+y+z). Therefore:xy+x+y=-(yz+xz+z)\r\nReplace it into 1st equation we have: z=-(yz+xz+z)\r\nIf z=0 we easily solve the rest of systems\r\nNow z#0 then divide both sides by z, we get: y+x=-2\r\nDoing similarly for two more equations we rewrite the original system:\r\nxy=z+2(1)\r\nyz=x+2\r\nzx=y+2\r\nNow, just substitute y=-2-x and z=-2-x into(1) we get: (-2-x)(x)=-x\r\nThe rest is simple :)" } { "Tag": [ "inequalities", "inequalities open" ], "Problem": "Prove or disprove:\r\n\r\n$a,b>0$\r\n\r\n$\\frac{a^2}{b^3}+\\frac{b^2}{a^3}\\ge2$", "Solution_1": "[hide]\n$\\frac{a^2}{b^3}+\\frac{b^2}{a^3}\\ge2$\nCounterexample: $a=b=2$ $\\Longrightarrow$ $\\frac{2^2}{2^3}+\\frac{2^2}{2^3}\\le2$\n[/hide]", "Solution_2": "[quote=\"minsoens\"][hide]\n$\\frac{a^2}{b^3}+\\frac{b^2}{a^3}\\ge2$\nCounterexample: $a=b=2$ $\\Longrightarrow$ $\\frac{2^2}{2^3}+\\frac{2^2}{2^3}\\le2$\n[/hide][/quote]\r\n\r\nYou are right minsoens, maybe I should write $\\frac{a^2}{b^3}+\\frac{b^2}{a^3}\\ge2$ for $a,b\\le1$. Do you think it is true?\r\n\r\nOr the other one $\\frac{a^2}{b^3}+\\frac{b^2}{a^3}\\le2$ for $a,b\\ge1$...?", "Solution_3": "$\\frac{a^2}{b^3}+\\frac{b^2}{a^3}\\le2$ for $a,b\\ge 1$\r\nisn't true because if you make one variable really big and the other really low, like:\r\n$a=100$ and $b=1$..\r\n\r\n$\\frac{a^2}{b^3}+\\frac{b^2}{a^3}\\ge2$ for $a,b\\le 1$\r\nOkay, we use the AM-GM inequality:\r\n$\\frac{\\frac{a^2}{b^3}+\\frac{b^2}{a^3}}{2}\\ge \\sqrt{\\frac{a^2}{b^3}\\cdot\\frac{b^2}{a^3}}$\r\n$\\frac{a^2}{b^3}+\\frac{b^2}{a^3}\\ge \\frac{2}{\\sqrt{ab}}$\r\nSince $a,b\\le 1$, $\\frac{1}{\\sqrt{ab}}\\ge 1$ $\\Longrightarrow$ $\\frac{2}{\\sqrt{ab}}\\ge 2$\r\n$\\frac{a^2}{b^3}+\\frac{b^2}{a^3}\\ge \\frac{2}{\\sqrt{ab}} \\ge 2$", "Solution_4": "I found that if $b$ is kept constant, then the minimum value of $\\frac{a^2}{b^3} + \\frac{b^2}{a^3}$ is $\\approx 1.96 \\cdot \\frac{1}{b}$." } { "Tag": [], "Problem": "I am uneasy about approaching a professor for a recommendation for an REU? Have you ever heard of a professor refusing to recommend a student when asked?\r\n\r\nThanks", "Solution_1": "I say \"yes\" the majority of the time but I have said \"no.\" I write different letters for different people - if I didn't, my letters would be pretty much worthless. Why would I refuse? Perhaps because I don't think I know the student very well. (I may very well suggest, \"Why don't you ask Professor A whom you had for course B?\") But perhaps because I think the letter I would write wouldn't do the student much good.\r\n\r\nAll you can do is ask. But if you encounter reluctance, don't push or try to persuade - you might be pushing for a lukewarm letter.", "Solution_2": "When you ask for a recommendation letter one of the following is likely to occur:\r\n\r\n1. The professor writes an enthusiastic letter full of details that are helpful in evaluating the application. A significant effort should be made by all serious students to cultivate relationships with a few professors who will write such letters on their behalf. \r\n\r\n2. The professor agrees to write a letter but it is either poorly written, \"lukewarm,\" or lacking in specific details that tend to be helpful in evaluating the application.\r\n\r\n3. The professor forgets to write the letter. Students often need to follow up or provide gentle reminders.\r\n\r\n4. The professor says \"no.\" This is far better than #2 or #3 above. \r\n\r\nWhen you ask the professor for a recommendation use language that makes it more likely that your letter will fall into category #1. \"Professor X, I'm applying for (whatever), it's quite competitive, and I need a very strong recommendation by (date). Would you be able to write such a recommendation?\" \r\n\r\nIf the professor says \"yes\" then the next step is to make sure the right details get into the letter. Don't assume that the professor will remember details of projects or work that you've done in classes or extracurriculars. Have copies of your resume and old projects/work that you've done for the professor on hand to jog his/her memory." } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "Compute the Taylor series of $ \\frac{1}{1\\plus{}x\\plus{}x^2}$ about the point $ \\minus{}\\frac{1}{2}$ and find the interval of convergence.\r\n\r\nCan someone please help. Thank you very much. :P", "Solution_1": "$ \\dfrac{1}{1\\plus{}x\\plus{}x^2}\\;\\equal{}\\;\\dfrac{1}{\\dfrac{3}{4}\\left(1\\;\\plus{}\\;\\dfrac{4}{3}\\left(x\\plus{}\\dfrac{1}{2}\\right)^2\\right)}\\;\\equal{}\\;\\dfrac{4}{3}\\left(1\\;\\plus{}\\;\\dfrac{4}{3}\\left(x\\plus{}\\dfrac{1}{2}\\right)^2\\right)^{\\minus{}1}\\;\\equal{}\\;\\;\\sum_{n\\equal{}0}^\\infty\\;(\\minus{}1)^n\\;\\left(\\dfrac{4}{3}\\right)^{n\\plus{}1}\\;\\left(x\\plus{}\\dfrac{1}{2}\\right)^{2n}$\r\n\r\n\r\nnow, \r\n$ \\dfrac{|(n\\plus{}1)^{th}\\;\\;term|}{|n^{th}\\;\\;term|}\\;\\equal{}\\;\\dfrac{\\bigg|(\\minus{}1)^{n\\plus{}1}\\;\\left(\\dfrac{4}{3}\\right)^{n\\plus{}2}\\;\\left(x\\plus{}\\dfrac{1}{2}\\right)^{2(n\\plus{}1)}\\bigg|}{\\bigg|(\\minus{}1)^n\\;\\left(\\dfrac{4}{3}\\right)^{n\\plus{}1}\\;\\left(x\\plus{}\\dfrac{1}{2}\\right)^{2n}\\bigg|}\\;\\equal{}\\;\\bigg|\\minus{}\\dfrac{4}{3}\\left(x\\plus{}\\dfrac{1}{2}\\right)^{2}\\bigg|$\r\n\r\nthe series will converge absolutely (and hence converge) [b]iff[/b] $ \\bigg|\\minus{}\\dfrac{4}{3}\\left(x\\plus{}\\dfrac{1}{2}\\right)^{2}\\bigg|\\;<\\;1\\;\\;\\;\\quad\\Rightarrow\\;\\;\\;\\bigg|x\\plus{}\\dfrac{1}{2}\\bigg| \\;<\\; \\dfrac{\\sqrt 3}{2}$\r\n\r\nhence, interval of convergence is $ \\left(\\dfrac{\\minus{}\\sqrt 3\\;\\minus{}\\;1}{2}, \\;\\;\\dfrac{\\sqrt 3\\;\\minus{}\\;1}{2}\\right)$", "Solution_2": "The radius of convergence of a series for a \"usual\" function (meaning analytic function) should always be the distance between the center point for the series and the nearest singular point for the function.\r\n\r\nThe singular points for $ \\frac{1}{z^2\\plus{}z\\plus{}1}$ are the poles at $ \\minus{}\\frac12\\pm\\frac{i\\sqrt{3}}{2}.$ These points lie at a distance of $ \\frac{\\sqrt{3}}2$ from $ \\minus{}\\frac12,$ hence the radius of convergence should be $ \\frac{\\sqrt{3}}2.$\r\n\r\nOf course, this is exactly what misan found.\r\n\r\nHad we expanded this in a power series centered at zero, we would have had a radius of convergence of $ 1.$" } { "Tag": [ "inequalities", "search", "inequalities solved" ], "Problem": "Let $a,b,c>0$ such that $a+b+c+abc=4$.Prove that:\r\n\r\n $\\frac{a}{\\sqrt{b+c}}+\\frac{b}{\\sqrt{c+a}}+\\frac{c}{\\sqrt{a+b}}\\geq\\frac{\\sqrt{2}}{2}\\cdot (a+b+c) $ ;)", "Solution_1": "IMHO, naming each inequality cool and new and nice and magnificient is a LITTLE proof of narcisism, but after all everyone with his taste. Anyway, again IMHO the purpose of an inequality is to be strong and quite relevant. This one does not meet the last two points and about new, absolutely no (I will prove that immediately). So it remains the cool quality. :D \r\n Now, let's solve this cool inequality. The method? Obviously, Cauchy (by the way, now it's really simple to solve your inequalities, since they are all with this method). In fact the problem is un unhappy combination of two ugly inequalities and one that I am fond of (appeared in Belarussian olympiad or India). The first two are Cauchy. Actually, after applying twice Cauchy we reduce it to the very nice inequality $a+b+c\\geq ab+bc+ca$. For that one you can search the forum. Although it follows immediately from Schur.", "Solution_2": "Sorry!!!\r\nI am not good in English. So Harazi can you write clearly your solution!!!!! :blush:", "Solution_3": "Ok, by Cauchy we have $\\sum{\\frac{a}{\\sqrt{b+c}}}\\geq\\frac{(a+b+c)^2}{\\sum{a\\sqrt{b+c}}}$. Next, $\\sum{a\\sqrt{b+c}}\\leq\\sqrt{2(a+b+c)(ab+bc+ca)}$ again by Cauchy. Thus it remains to prove that $ab+bc+ca\\leq a+b+c$. Suppose that $a+b+c(a+b+c)(4-(a+b+c))$ and thus $ a+b+c<3$. But then $abc<1$ and this is impossible since $a+b+c+abc=4$.", "Solution_4": "It follows that \\frac{9abc}{a+b+c}>(a+b+c)(4-(a+b+c)) and thus a+b+c<3. But then abc<1 and this is impossible since a+b+c+abc=4.\r\n\r\nHarazi I really don't understand that I think I get some troubles from this.\r\nMore clearly, can you?\r\n(Sorry!! :blush: )", "Solution_5": "I don't understand you, man! Schur inequality is $\\frac{9abc}{a+b+c}\\geq 4(ab+bc+ca)-(a+b+c)^2$. Assume that $ab+bc+ca>a+b+c$ then $ 4(ab+bc+ca)-(a+b+c)^2>4(a+b+c)-(a+b+c)^2=(a+b+c)(4-(a+b+c))=abc(a+b+c)$ since $ a+b+c+abc=4$. Thus, it follows that $a+b+c<3$ and by AM-GM we have $abc<1$ impossible. Is it finally clear?", "Solution_6": "Oh yes ti is really great , I misunderstand your solution!!!!!!! It's right. ( Because of some English words that I misunderstand).\r\nHow about you \"ceza lupu\" do you have any solution for this problem. I think this solution is not a natural one.", "Solution_7": "Harazi can you help me to give a nice solution for this one:\r\n\r\n(a^2 + b^2 + c^2)^2>3(a^3b + b^3c + c^3a)\r\nI have seen a solution in one post of you with Sasha( his picture is a cat ) but it is not really nice. Please help me!!!! \r\nI wait you....\r\nLoading...............", "Solution_8": "Do you mean that his solution is not nice enough? :?", "Solution_9": "What is the ugly problem you are reffering to, Arne? Because I think Vasc's inequality is one of the most elegant and hard and nice inequalities I have ever seen (and I like it so much since Muierhead is useless there!!!!). Unfortunately, the problem is doomed to have a bloody computational solution and I'm sorry to say this, but I really don't think a nice solution will ever be given for this problem." } { "Tag": [ "Princeton", "college", "HCSSiM" ], "Problem": "I was trying to figure out which teams at PUMA-C had NYC representation on them, but they all have funny names and so it wasn't clear -- anyone care to enlighten me?", "Solution_1": "Team Colbert, Team Pig, and Team Goat were all from NYC. Team Pig was named as such because everybody on the team except for me went to Hampshire this past summer. Team Colbert was named as such in an attempt to get them on the show. I think Mr C. had a reason other than Mr. Novikoff's obsession with Goats for that name, but I'm not sure. \r\n\r\nIn addition, Florida had registered a full B-team when they only had 2 people for it, 6 NYC kids were on Florida B.\r\n\r\nHere are the team rosters. Florida B was formed after Mr. C sent these out, so I'm not sure of everyone that was on it.\r\n\r\n\r\nTEAM COLBERT\r\n\r\n\r\n\r\nKAM CHAN\u2014BROOKLYN TECH\r\n\r\nTAO RAN CHEN\u2014BAYSIDE\r\n\r\nABE CHIEN\u2014STUYVESANT\r\n\r\nBEN LERNER\u2014STUYVESANT\r\n\r\nKRISHANU SANKAR\u2014HORACE MANN\r\n\r\nKEN SUZUKI\u2014STUYVEANT\r\n\r\nALLAN ZHAO\u2014BROOKLYN TECH\r\n\r\nDANNY ZHU\u2014STUYVESANT\r\n\r\n\r\n\r\n \r\nTEAM GOAT\r\n\r\n\r\n\r\nLOUIS EVANS\u2014HUNTER\r\n\r\nARTUR DMOWSKI\u2014STUYVESANT\r\n\r\nELIOT KIM\u2014STUYVESANT\r\n\r\nYOUN KIM\u2014STUYVESANT\r\n\r\nKEVIN KWOK\u2014HUNTER\r\n\r\nJENNY KWUN\u2014STUYVESANT\r\n\r\nCONNIE PARK\u2014HUNTER\r\n\r\nGABE ZUCKER\u2014HUNTER\r\n\r\n\r\n \r\nTEAM PIG\r\n\r\n\r\n\r\nBENJAMIN HIRSCH\u2014STUYVESANT\r\n\r\nRACHEL KOGAN\u2014BREARLEY\r\n\r\nHARRIET LI\u2014HUNTER\r\n\r\nBENNO MIRABELLI\u2014POLY PREP\r\n\r\nDAVID MOON\u2014HORACE MANN\r\n\r\nADAM SEALFON\u2014STUYVESANT\r\n\r\nDANNY YUAN\u2014STUYVESANT\r\n\r\nJIN WAN\u2014BROOKLYN TECH", "Solution_2": "Thanks! And many congratulations are due, since you all managed to do very well (rocking the power test in particular)! Unrelatedly, I'm pleased to say that it only took me a minute or two to figure out who you were from the team roster ... otherwise, I might have gone on for ever believing you were from Nebraska." } { "Tag": [ "number theory" ], "Problem": "One incorrectly remembered Fermat's little theorem as follows :\r\n\"For any prime n, an+1 - a is divisible by n\". (*)\r\nDescribe the set of integers (not just primes) for which (*) is true.", "Solution_1": "A (not really convinced :() try. This problem proved rather challaging. I'm surprised there are only finite number of such numbers :? \r\n \r\n\r\nFirst let's notice that n=1 (and n=2) works !\r\n\r\nLet's suppose n != 1.\r\n\r\n. Let n = prod pi^ai with pi prime.\r\nIf ai > 1, applying (*) with pi gives a contradiction so n = prod pi.\r\n\r\n. \"True\" Fermat theorem : a^(pi-1) = 1 [pi] if (a,pi) = 1.\r\na^n = 1 [n] implies pi-1 | n for all i.\r\n\r\n. So we have to solve pi-1 | prod pj for all i.\r\nWolog p1 < p2 .. < pj\r\n. p1-1 < p1 so p1-1=1 or p1=2\r\n. if j > 1, p1 <= p2-1 < p2 so p2-1=p1 or p2 = 3\r\n. if j > 2 or 3, continuing gives p3 = 7, p4 = 43. \r\n. if j > 4, p5 should divide 1806 but there are no divisor 'a' of 1806, greater than 43, such that a+1 is prime so there are no p5.\r\n\r\nSo the numbers satisfying the incorrect Fermat theorem are :\r\n1, 2, 2*3=6, 2*3*7=42, 2*3*7*43=1806.", "Solution_2": "That is entirely correct !\r\n\r\nI really enjoyed this problem." } { "Tag": [ "AMC", "AIME", "blogs", "geometry", "probability", "trigonometry", "floor function" ], "Problem": "[b]2006 Mock AIME 1 Results[/b]\r\n\r\nTotal # of Test Takers:\r\n\r\n16\r\n\r\nTop Scorers:\r\n\r\n13 - gotztahbeazn, Pakman2012\r\n12 - scorpius119\r\n11 - Elemennop\r\n10 - krustyteklown\r\n\r\nHardest Questions:\r\n\r\n1 correct answer - 14, 15\r\n2 correct answers - 13\r\n\r\nEasiest Questions:\r\n\r\n14 correct answers - 3\r\n13 correct answers - 4\r\n\r\nMedian Score:\r\n\r\n8\r\n\r\nMean Score:\r\n\r\n8\r\n\r\nMode Score:\r\n\r\n8\r\n\r\nThere you go. Special congratulations to gotztahbeazn and scorpius119 to be the only people with correct answers to #15 and #14, respectively.\r\n\r\nThe solution to #15 can be found on my [url=http://wangsblog.com/jeffrey/]blog[/url].", "Solution_1": "[b]Answers and Topic Distribution[/b]\r\n\r\n1. 668 (G)\r\n2. 050 (NT/PC)\r\n3. 470 (G)\r\n4. 107 (NT)\r\n5. 391 (A/NT)\r\n6. 990 (NT)\r\n7. 112 (NT)\r\n8. 368 (SS)\r\n9. 516 (A/SS)\r\n10. 900 (G)\r\n11. 702 (G)\r\n12. 267 (G)\r\n13. 193 (PC)\r\n14. 322 (A)\r\n15. 096 (G/T)\r\n\r\nTopic Distribution:\r\nAlgebra - 3\r\nGeometry - 6\r\nNumber Theory - 5\r\nProbability & Counting - 2\r\nSequences and Series - 2\r\nTrigonometry - 1", "Solution_2": "Hmmm looks like I got an 8.\r\n\r\nOn number 1:\r\nWhat I did was the floor(2006/6)=334 to find the number within on radius on one side. But this include the orginal point, so there are actually 333. So to find the number within the certain radius on the other side, multiply by 2 gives 666.\r\n\r\nIt looks like in the answer it didn't subtract 1. Why is this not necessary?", "Solution_3": "[quote]Score: 6 \n\nDistribution: \n\n1 1 1 1 1 0 0 1 0 0 1 0 0 0 0[/quote]\r\n\r\n :( \r\nExactly as I feared\r\nGot 6 and 7 wrong due to arithmetic errors, and apparently 9 and 12, too. :( \r\nI feel dumb.", "Solution_4": "[quote=\"eryaman\"]Hmmm looks like I got an 8.\n\nOn number 1:\nWhat I did was the floor(2006/6)=334 to find the number within on radius on one side. But this include the orginal point, so there are actually 333. So to find the number within the certain radius on the other side, multiply by 2 gives 666.\n\nIt looks like in the answer it didn't subtract 1. Why is this not necessary?[/quote]\r\n\r\nIt actually doesn't include the original point; for example, try it with 7 points. The answer would be 2, or $2 \\cdot \\left\\lfloor \\frac{7}{6} \\right\\rfloor$.", "Solution_5": "I don't understand why my method is incorrect. :? I think I'm misreading the problem...\r\n\r\nLet the radius of the circle be 1, so that the circumference is $2\\pi$. The distance between each consecutive point is $\\frac{2\\pi}{2006}=\\frac{\\pi}{1003}$. The number of points $k$ that have distance $\\le 1$ from one point are $\\frac{\\pi}{1003} \\cdot k \\le 1$, so $k = 319$...", "Solution_6": "I missed #1. I loved my logic on that one; It needs to be [b]six[/b]ty degrees, so the angle is pi over [b]six[/b].\r\n\r\nAnd for #7, Is the answer not 448? f(n) is the number of divisors of n, which includes both negative and positive divisors. That would thus double the answer twice, and make it four times your given answer. I believed that you probably meant positive divisors only, but the problem didn't state it so I solved it as such.\r\n\r\nGreat test over all; I generally liked the questions, though some like the Collatz one (#8) is just plain computation. Yay for brute-forcing #13.", "Solution_7": "wow, i am really could have done better, i got 8, and for problems 2,5,6,8, i made some tiny computation error that messed me up, i have to work on my careless errors, if i had a cheap handheld calculator, i would have been fine... :wallbash_red:, but at least i did this before anything official, also i am thinking about making a mock aime...but it will be a while before i actually do anything with it, it takes a while to think of problems", "Solution_8": "Anyone have an elegant way to do #10? I had nasty computation, which I ultimately made an arithmetic error on. \r\n\r\nHow abouut a solutions thread?", "Solution_9": "I really want to see a solution for #8. :dry:", "Solution_10": "hmm, i got a 3. i got 2, 3, and 9. missed 1 and 4. :(", "Solution_11": "[quote=\"chess64\"]I don't understand why my method is incorrect. :? I think I'm misreading the problem...\n\nLet the radius of the circle be 1, so that the circumference is $2\\pi$. The distance between each consecutive point is $\\frac{2\\pi}{2006}=\\frac{\\pi}{1003}$. The number of points $k$ that have distance $\\le 1$ from one point are $\\frac{\\pi}{1003} \\cdot k \\le 1$, so $k = 319$...[/quote]\r\n\r\nIt's not distance along the arc, it's distance from one point to another.", "Solution_12": "[quote=\"dakyru\"][quote=\"chess64\"]I don't understand why my method is incorrect. :? I think I'm misreading the problem...\n\nLet the radius of the circle be 1, so that the circumference is $2\\pi$. The distance between each consecutive point is $\\frac{2\\pi}{2006}=\\frac{\\pi}{1003}$. The number of points $k$ that have distance $\\le 1$ from one point are $\\frac{\\pi}{1003} \\cdot k \\le 1$, so $k = 319$...[/quote]\n\nIt's not distance along the arc, it's distance from one point to another.[/quote]Haha chess64, I did that exact same thing at first...along with the correct method (with the -1 part wrong), then I started wondering why the answers were so different. I finally figured out what was going wrong, though, luckily. :D \r\n\r\nLooks like everyone made dumb mistakes. \r\n\r\nI answered 1-12, missed 1 from that dumb mistake, 4 and 8 from arithmetic errors.\r\n\r\nAnyone have a solution for number 11?", "Solution_13": "[quote=\"eryaman\"]Anyone have a solution for number 11?[/quote]\r\n\r\nJust use the Pythagorean Theorem (and know the centroid divides the medians in a $2: 1$ ratio) to get the height of the tetrahedron. Then use Pythagoras again to find the circumradius. Height - circumradius = inradius, which we get circumradius/inradius = $3$. So the ratio of volume of the spheres is $27$.\r\n\r\nEasier with a cube. Circumradius is half the space diagonal = $\\frac{s\\sqrt{3}}{2}$. Then the inradius is half of a side = $\\frac{s}{2}$. So circumradius/inradius = $\\sqrt{3}$. Ratio of volumes is $3\\sqrt{3}$.\r\n\r\n$27^2 - (3\\sqrt{3})^2 = 729-27 = 702$.", "Solution_14": "[quote=\"dakyru\"][quote=\"chess64\"]I don't understand why my method is incorrect. :? I think I'm misreading the problem...\n\nLet the radius of the circle be 1, so that the circumference is $2\\pi$. The distance between each consecutive point is $\\frac{2\\pi}{2006}=\\frac{\\pi}{1003}$. The number of points $k$ that have distance $\\le 1$ from one point are $\\frac{\\pi}{1003} \\cdot k \\le 1$, so $k = 319$...[/quote]\n\nIt's not distance along the arc, it's distance from one point to another.[/quote]\r\n\r\nOMG I should have realized that :wallbash:", "Solution_15": "[quote=\"eryaman\"]Anyone have a solution for number 11?[/quote]\r\n\r\nThe altitudes to the faces of the tetrahedron intersect at $\\frac{1}{4}$ their length from each face, so the radius of the inscribed circle is $\\frac{1}{3}$ if the radius of the circumscribed circle is $1$." } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "Find all$P(x)\\in R[x]$such that\r\n $P(x^2-1)=P(x)^2-1$ $\\forall x\\in R$\r\nI have a nice solution :blush: \r\n[u][b]another question[/b][/u]:\r\nLet $Q(x)\\in R[x]$;$degQ\\ge 1$.Find all $P(x)\\in R[x]$such that\r\n $P(Q(x))=Q(P(x))\\forall x\\in R$\r\nI can solve it with :$Q(x)=x^n-1$", "Solution_1": "Oh,nobody. :( Why?\r\n\r\n[color=red]Edit by Megus: because you posted it in wrong section ;) - moved[/color]", "Solution_2": "Thank you very much,[b]Megus[/b].", "Solution_3": "I think you should post your nice solutions for everyone.It's very nice :D", "Solution_4": "Discussed before for some special cases.\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?p=376258#p376258", "Solution_5": "[quote=\"Omid Hatami\"]Discussed before for some special cases.\n\nhttp://www.mathlinks.ro/viewtopic.php?p=376258#p376258[/quote]\r\nThe answer of my problem with$Q(x)=x^n-1$and $degQ\\ge 1$:\r\n$P(x)\\in (P_k(x))_{k=0}^{\\infty}: P_0(x)=x;P_{k+1}(x)=P_k^n(x)-1$ :)", "Solution_6": "can you post your solution nthd", "Solution_7": "[quote=\"Bluesea\"]can you post your solution nthd[/quote]\r\nPlease wait some days! :)", "Solution_8": "[b][u]Lemma 1[/u][/b]:If$P(x)\\in R[x],degP>1$ satisfy this equality$\\rightarrow\\exists P_1(x)$ such that:$P(x)=P_1(x^2-1)$\r\n[b][u]Lemma 2[/u][/b]: there is only one polynomial $P(x)$ with $degP$ is odd satisfied is $P(x)=x$\r\n :) ;) good luck." } { "Tag": [ "logarithms", "symmetry", "calculus", "calculus computations" ], "Problem": "If $ u\\plus{}\\ln{u}\\equal{}xy$, find $ \\frac{\\partial^2 u}{\\partial y\\partial x}$.", "Solution_1": "$ 0\\equal{}\\frac{\\partial }{\\partial x} \\left[u\\plus{}\\ln u \\minus{}xy\\right]\\equal{}u_x\\plus{}\\frac{u_x}{u}\\minus{}y$\r\n\r\n$ 0\\equal{}\\frac{\\partial }{\\partial y}\\left[ u_x \\plus{}\\frac{u_x}{u}\\minus{}y\\right]\\equal{}u_{xy}\\plus{}\\frac{u_{xy}}{u}\\minus{}\\frac{u_xu_y}{u^2}\\minus{}1$\r\n\r\n$ u_{xy}(1\\plus{}u^{\\minus{}1})\\equal{}u_xu_yu^{\\minus{}2}\\plus{}1 \\to u_{xy}\\equal{}\\frac{u_xu_yu^{\\minus{}2}\\plus{}1}{1\\plus{}u^{\\minus{}1}}$\r\n\r\n$ u_x(1\\plus{}u^{\\minus{}1})\\equal{}y \\to u_x\\equal{}\\frac{yu}{u\\plus{}1}$\r\nand by symmetry of the original, we know we will get $ u_y\\equal{}\\frac{xu}{u\\plus{}1}$\r\nso\r\n$ u_{xy}\\equal{}\\frac{\\frac{xy}{(u\\plus{}1)^2}\\plus{}1}{1\\plus{}u^{\\minus{}1}}\\equal{}\\frac{u[xy\\plus{}(u\\plus{}1)^2]}{(u\\plus{}1)^3}\\equal{}\\frac{u[u\\plus{}\\ln u\\plus{}(u\\plus{}1)^2]}{(u\\plus{}1)^3}$" } { "Tag": [ "ratio", "trigonometry", "geometry unsolved", "geometry" ], "Problem": "In a triangle $ABC$ with $AB\\neq AC$, the internal and external bisectors of angle $A$ meet the line $BC$ at $D$ and $E$ respectively. If the feet of the perpendiculars from a point $F$ on the circle with diameter $DE$ to $BC,CA,AB$ are $K,L,M$, respectively, show that $KL=KM$.", "Solution_1": "The circle with diameter DE is the A-Apollonius circle of $\\triangle ABC,$ the locus of points F with constant ratio of distances $\\frac{FB}{FC}= \\frac{AB}{AC}$ from the vertices B, C. The pedal triangle $\\triangle KLM,\\ K \\in BC, \\L \\in CA,\\ M \\in AB$ of any point F on the A-Apollonius circle is isosceles with KL = KM. For example, quadrilaterals FKCL, FKMB are both cyclic and FC, FB are diameters of their circumcircles. Therefore, $KL = FC \\sin C,\\ KM = FB \\sin B.$ Then\r\n\r\n$\\frac{KL}{KM}= \\frac{FC}{FB}\\cdot \\frac{\\sin C}{\\sin B}= \\frac{AC}{AB}\\cdot \\frac{\\sin C}{\\sin B}= 1$\r\n\r\nby the sine theorem for $\\triangle ABC.$", "Solution_2": "[b]An interesting remark.[/b] $m(\\widehat{LKM})=\\left|A-m(\\widehat{BFC})\\right|$ and the points $F,K,L,M$ never are concyclically.", "Solution_3": "In fact, it is the solution of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3459&sid=a2e3248bc6b2689da51f66d7b8f98b16#p3459]IMO 1996/2[/url].\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=644&t=356195&start=20]IMO 2010/4[/url] is also similar to the problem above. All uses Apollonius circle for angle bisectors." } { "Tag": [], "Problem": "This site is really cool, it has tons of ebooks (pdf's) for free, check it out.\r\n\r\n[url]http://www.freecomputerbooks.com/[/url]", "Solution_1": "Hmm... It doesn't have what I am taking. T.T", "Solution_2": "just be thankful that you have a lot of free books." } { "Tag": [ "function", "integration", "inequalities", "calculus", "derivative", "calculus computations" ], "Problem": "does this hold for convex functions $ f$ in $ [a,b]$\r\n$ (\\int _{a}^{b}f^{n}(x) dx)^{\\frac{1}{n}}$ is increasing in $ n$, where $ n \\geq 1$", "Solution_1": "If $ f$ is constant and the interval is of size larger than 1, it fails. Maybe if you fix the interval as $ [0, 1]$ or something?", "Solution_2": "Excuse my arrogance, but shouldn't a function satisfy $ f''(x)<0$ on an interval to be convex on in? A constant doesn't satisfy this condition. Or am I wrong?", "Solution_3": "How could such a question be arrogant? :) I don't know whether the \"usual\" definitions of concave and convex involve strict or weak inequalities on the second derivative. Even if it is strict, we could add a very very very tiny perturbation to a constant function (e.g. $ f(x) \\equal{} 2 \\minus{} \\frac {x^2}{1000}$ over $ [ \\minus{} 1, 1]$) to get a function with negative second derivative that violates the condition for small enough $ n$ -- I don't know whether the resultant sequence will be decreasing for all $ n$. (For the particular example I've given, it decreases until $ n \\equal{} 17000$ or so, but then begins to increase, at least according to Mathematica.)", "Solution_4": "I was going to give you $ f(x)\\equal{}x^2$ on $ [0,10].$ The sequence decreases for $ n<14$ and then increases.\r\n\r\nIf you add in the stipulation that $ [a,b]$ have length $ 1,$ then the sequence is nondecreasing (the proof uses H\u00f6lder's inequality) and you don't need the hypothesis of convexity.\r\n\r\nQuestion: does the sequence have to be [i]eventually[/i] increasing? And would convexity matter in that case?", "Solution_5": "Well, the $ L^p$-norm of $ f$ is a convex function of $ 1/p$ and it tends to the $ L^\\infty$ norm as $ p\\to\\plus{}\\infty$, so the only chance for it not to be eventually increasing is to be decreasing all the way. But this would mean that we can have a function bounded by $ 1$, say, such that $ \\int_a^b{f^n}\\ge 1$ for all $ n$, which is possible if and only if the measure of the set where $ f\\equal{}1$ is at least $ 1$. This, of course, is never the case for strictly convex functions.", "Solution_6": "sorry fedja...i am not competent enuf to understand what u wrote...\r\nyes Dr.Kent will it be evnetually increasing" } { "Tag": [ "function", "limit", "integration", "real analysis", "real analysis unsolved" ], "Problem": "This is a fun little problem. Prove or come up with a counterexample.\r\n\r\nIf $ f\\in L^2(\\mathbb{R})$, then $ \\lim_{n\\to\\infty}\\sqrt{n}\\int_n^{n\\plus{}1}|f|dx\\equal{}0$.", "Solution_1": "Let $ f(x) \\equal{} \\frac1k$ on the interval $ [k^2,k^2 \\plus{} 1]$ and zero otherwise. Then $ \\|f\\|_2^2 \\equal{} \\sum_{k \\equal{} 1}^{\\infty}\\frac1{k^2} < \\infty.$\r\n\r\nBut if we let $ a_n \\equal{} \\sqrt {n}\\int_n^{n \\plus{} 1}|f|,$ then $ \\limsup_{n\\to\\infty}a_n \\equal{} 1.$\r\n\r\nFurther challenge: If we require $ |f|$ to be monotone decreasing on $ [1,\\infty),$ does that change the answer?\r\n\r\n(One comment: this all seems to really be about $ \\ell^2$ rather than $ L^2$ - at least I haven't yet seen any features of the problem that I couldn't answer with a sequence.)" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "let $a.b.c$ real numbers and $x,y,z>1$ and $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=2$ .what's the maximal value of $xyz$? :?:", "Solution_1": "$x,y\\to1, z\\to\\infty$ :)", "Solution_2": "no! \r\nif $z\\to\\infty$ so $(\\frac{1}{x}+\\frac{1}{y})\\to2$ and we have $x>1,y>1$ ;)", "Solution_3": "Yes! :P \r\nto be very explicit, put $x=y=1+\\frac1{2N-1},z=N$ :D", "Solution_4": "can you explain? i think that you are mistaked :oops:", "Solution_5": "I think he isn't...\r\nYou should review your problem's statement... What/where are $a, b, c$?", "Solution_6": "real numbers positifs $> 1$", "Solution_7": "first , what's the maximal value of $x+y+z$?", "Solution_8": "There is no maximal value.", "Solution_9": "can you explain why?", "Solution_10": "I guess I should not interfer in such an intelligent and meaningful conversation, but if someone is interested in proofs or something like that:\r\n\r\nOkay, I assume the problem is to find the maximum of xyz, where x, y, z are real numbers which are > 1 and satisfy $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=2$. In fact, such a maximum does not exist, because we can always take $x=y=\\frac{1}{1-s}$ and $z=\\frac{1}{2s}$ for a small enough $s<\\frac12$ (yes, these x, y, z are > 1 and do satisfy $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=2$) and get $xyz=\\frac{1}{\\left(1-s\\right)^2\\cdot 2s}$ which can become as large as you want. The same counterexample shows that x + y + z has no maximum.\r\n\r\nThe question about the minimum of xyz makes more sense, but follows trivially from AM-GM:\r\n\r\n$\\frac{1}{x}\\cdot\\frac{1}{y}\\cdot\\frac{1}{z}\\leq\\left(\\frac{\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}}{3}\\right)^3=\\left(\\frac23\\right)^3$\r\n$\\Longrightarrow\\ \\ \\ \\ \\ xyz=\\frac{1}{\\frac{1}{x}\\cdot\\frac{1}{y}\\cdot\\frac{1}{z}}\\geq\\frac{1}{\\left(\\frac23\\right)^3}=\\frac{27}{8}$,\r\n\r\nwith equality for $x=y=z=\\frac32$.\r\n\r\nThe minimum of x + y + z is an AM-HM application:\r\n\r\n$\\frac{x+y+z}{3}\\geq\\frac{3}{\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}}=\\frac32\\ \\ \\ \\ \\ \\Longrightarrow\\ \\ \\ \\ \\ x+y+z\\geq 3\\cdot\\frac32=\\frac92$,\r\n\r\nwith equality for $x=y=z=\\frac32$.\r\n\r\n Darij", "Solution_11": "thanks :lol: \r\nnow i want the maximal value of $x+y+z$ no exist???" } { "Tag": [ "trigonometry", "function", "geometry", "3D geometry", "complex numbers" ], "Problem": "Rewrite the complex numbers in the form $ a\\plus{}bi$\r\n\r\n1. $ 4\\sin^2\\frac{\\pi}{18}$\r\n\r\n2. $ \\ 4\\sin^2\\frac{5\\pi}{18}$\r\n\r\n3. $ \\ 4\\sin^2\\frac{7\\pi}{18}$\r\n\r\nshow steps.", "Solution_1": "I don't follow. All of those numbers are real.", "Solution_2": "Yes, they are (and I guess we only have to find a since b is 0), but in the post \"Neat Cubic\", Dr Sonnhard said that $ 4\\sin^2(\\pi/18) \\equal{} \\left(\\frac{1}{2}(\\minus{}1\\minus{}i\\sqrt{3})(\\frac{1}{2}(\\minus{}1\\plus{}i\\sqrt{3}))^{\\frac{1}{3}}\\plus{}\\frac{1}{2}(\\frac{1}{2}(\\minus{}1\\minus{}i\\sqrt{3}))^{\\frac{1}{3}}(\\minus{}1\\plus{}i\\sqrt{3})\\right)^2$, but do you know how they were derived, and probably simplifying the expression into a single real number with no $ i$? How about the other two? I asked that in the post but no one has replied yet, and so I decide to start a new topic. :)", "Solution_3": "hello, to derive $ \\sin\\left(\\frac{\\pi}{18}\\right)$ you will need this formula\r\n$ \\sin(3x)\\equal{}3\\sin(x)\\minus{}4\\sin(x)^3$\r\nSonnhard.", "Solution_4": "I guess you meant to substitute $ \\pi/18$ for $ x$ and then solve for $ \\sin(\\pi/18)$, but is there a clean way to solve the resulting cubic?", "Solution_5": "That depends on what you mean by \"clean.\" Dr. Sonnhard is referring to the [url=http://mathworld.wolfram.com/CubicFormula.html]cubic formula[/url], which here is a tautology: think about what the cube root of $ \\frac{\\minus{}1 \\plus{} i \\sqrt{3}}{2}$ is in terms of complex exponentials. The identity Dr Sonnhard stated is essentially a trigonometric identity.", "Solution_6": "hello, yes, we have $ \\sin(3\\frac{\\pi}{18})\\equal{}\\frac{1}{2}\\equal{}3\\sin(\\frac{\\pi}{18})\\minus{}4\\sin(\\frac{\\pi}{18})^3$, substituting $ t\\equal{}\\sin(\\frac{\\pi}{18})$ then we have\r\nto solve $ \\frac{1}{2}\\equal{}3t\\minus{}t^3$.\r\nSonnhard.", "Solution_7": "Nice link. [url=http://mathforum.org/dr.math/faq/faq.cubic.equations.html]Here[/url] is another one.\r\nOne quick question: There should be three complex roots to the cubic, but how do we determine the desired real root?", "Solution_8": "There are two cube roots in the expression. As long as you take the \"same\" cube root in both of them, the end result will be real three times (in this case) and will give you all three real roots." } { "Tag": [ "function", "trigonometry", "calculus", "derivative", "integration", "analytic geometry", "real analysis" ], "Problem": "I got a question of a friend about defining the sin en cos as the solution of 2nd order diff. equation\r\ny''(t) = z(y), the function y is continuous with z:R-->R\r\nSo i worked it out a bit, but I don't know how to prove the identities\r\nsin'(x)=cos x, sin(x)^2+cos(x)^2=1 and cos(a+b)=cos(a)(cosb)+sin(a)sin(b) \r\n\r\nThe hint was to use Picard iteration and banah Theorem\r\n\r\nCan someone help plz", "Solution_1": "What are you asking? In particular, why do you have both $ y$ and $ z$ in what you wrote? As written, it doesn't make sense.", "Solution_2": "I defined the cos and sin as solutions of the differential eq.\r\ny''(t)=z(y(t), with f continuous and f:R->R\r\nI mean an equation like y''(t)=y(t), but then in a general form. It is easy to see that sin(t) is a solution\r\n\r\nNow I have to prove the identities 1. sin'(t)=cos(t) 2. sin(t)^2+cos(t)^2=1 and 3. cos(a+b)=cos(a)cos(b)+sin(a)+sin(b)\r\nBut the difficult thing is, because I defined the sin as a solution of the diff eq. I have to use the definition to show this simple identities. I can use the banah theorem to show that the solution is unique\r\n\r\nI don't know how to use microsoft object or something on this site, or else I can can give more explanation if needed", "Solution_3": "You have to use the solution to that differential equation as your definition of the sine function? Most texts I have seen prove Taylor's Theorem first, define sine and cosine by their Taylor series, and prove that the derivative of sine is cosine once some properties of uniform convergence of sequences of functions are established. As for proving that $ sin^{2}\\theta\\plus{}cos^{2}\\theta \\equal{} 1$, this is fairly easy once you can show that $ cos\\theta \\equal{}\\frac{e^{i\\theta}\\plus{}e^{\\minus{}i\\theta}}{2}$ and $ sin\\theta \\equal{}\\frac{e^{i\\theta}\\minus{}e^{\\minus{}i\\theta}}{2i}$.", "Solution_4": "Yeah but the problem is to define it in this way\r\n\r\nSo i have the diff equation y''(t)=z(y(t), I take the integral on both sides and get y(t)=y0+y1(t-t0)+double integral z(y(t)). so it is easy so far.\r\n\r\nI define sin as the solution of this equation with solution y(t,0,0,1)\r\nso this means the solution y(t, t0, y0,z0) with this points as boundary values. so it means that the sin starts in the origin, in (0,0) . and the cos gives y(t,0,1,0), so it has coordinates (0,1) with t zero and y 1.\r\nNow I have to prove\r\n1. sin'(t)=cos(t)\r\n2. cos(t)^2+sin(t)^2=1\r\n3. cos(s+t)=cos(s)cos(t)+sin(s)sin(t)\r\n\r\nI know that the solution is unique, so cos(s+t) and cos(s)cos(t)+sin(s)sin(t) are the same solutions for the equation y''(t)=z(y(t), . But how do I prove that they have the same solutions, so that they are equal to each other. \r\n\r\nPlz help,\r\n\r\nCause I am a bit stuck here :(", "Solution_5": "I think this helps" } { "Tag": [ "quadratics", "function" ], "Problem": "Show that if $ a|b$ and $ a>0$, $ b>0$, then $ a<\\equal{}b$", "Solution_1": "$ a|b\\Rightarrow b \\equal{} ka$\r\n\r\n$ k \\ge 1$\r\n$ ka\\ge a$\r\n$ b \\equal{} ka \\ge a$", "Solution_2": "[quote=\"SimonM\"]$ k \\ge 1$[/quote]\r\n\r\nNext, try to prove that $ k > 0 \\implies k \\ge 1$. Use the fact that any subset of the natural numbers has a least element.", "Solution_3": "If $ 0 z.", "Solution_5": "[quote=\"Santamaria\"]Well,\n\nwe can assume that sinA/cosA = tanA\n\nThen, we know that e^pi = i^i\n\n0 < 1 < x for all z values in the z/y polar cylinder, when finding the antiderivative of a polar-quadratic function with height sinBpi.\n\nIf you work out the problem using pi, and euler's theory, we can assume that the answer will be a factor of 9, iff and only iff y > z.[/quote]\r\n\r\nNow that is an example of how to kill a fly with a super laser bazooka! :P" } { "Tag": [ "logarithms", "Pascal\\u0027s Triangle" ], "Problem": "Find $(x,y,z,r)$ satisfying \r\n\r\n${\\{ \\begin{array}{lll} {\\displaystyle{ {r+y}\\choose r}} = \\log_yx \\\\ \\\\{{ \\log_yz} = 4 + \\log_xz} \\\\ \\\\ {\\displaystyle {{r+y}\\choose y}} = \\log_xz + \\log_zz \\end{array}.$", "Solution_1": "[hide]$log_yz=4+log_xz \\implies log_yz=4+\\frac{log_yz}{log_yx}$ but since: $log_yx=C_{r+y}^{r}$ we get:$log_yz=\\frac{4C_{r+y}^{r}}{C_{r+y}^{r}-1}$ and by the last system's term:\n\n$C_{r+y}^{y}=C_{r+y}^{r}=log_xz+log_zz \\implies C_{r+y}^{r}=\\frac{log_yz}{log_yx}+1=\\frac{4}{C_{r+y}^{r}-1}+1 \\implies {(C_{r+y}^{r}-1)}^2=4 \\implies C_{r+y}^{r}=3$ \nwe can check the Pascal's triangle and get:$C_{3}^{1}=C_{3}^{2}=3$ in consequence:\n\n$y+r=3 \\implies \\{y=1$ and $r=2\\}$ or $\\{y=2$ and $r=1\\}$ note that the first set is not possible because $y$ is a log base (ie. y>1), now we just need to substitute the values in the system:\n\n$C_{r+y}^{r}=log_28=log_2x \\implies x=8$ \n\n$\\frac{4C_{r+y}^{r}}{C_{r+y}^{r}-1}=log_264=log_2z \\implies z=64$\n\nAnswer:r=1, y=2, x=8, z=64[/hide]" } { "Tag": [], "Problem": "Compute: $ \\frac{5! 4!}{5! \\plus{} 4!}$", "Solution_1": "$ \\frac {5!4!}{5! \\plus{} 4!} \\equal{} \\frac {5!4!}{4!(5 \\plus{} 1)} \\equal{} \\frac {5!}{6} \\equal{} \\boxed{20}$." } { "Tag": [], "Problem": "how do you factor this completely??\r\n\r\nx^8-y^8\r\n\r\n\r\npls and thx!!", "Solution_1": "see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=108585]here[/url]", "Solution_2": "[hide] \\begin{eqnarray*}x^{8}-y^{8}&=&\\left(x^{4}-y^{4}\\right)\\left(x^{4}+y^{4}\\right) \\\\ &=&\\left(x^{2}-y^{2}\\right)\\left(x^{2}+y^{2}\\right)\\left(x^{4}+y^{4}\\right) \\\\ &=&\\left(x-y\\right)\\left(x+y\\right)\\left(x^{2}+y^{2}\\right)\\left(x^{4}+y^{4}\\right)\\end{eqnarray*} Also, if you wanted to factor this further (with coefficients in $\\mathbb{R}$ instead of $\\mathbb{Z}$), use the fact that \\[x^{4}+y^{4}=\\left(x^{2}+y^{2}\\right)^{2}-2x^{2}y^{2}=\\left(x^{2}+y^{2}-\\sqrt{2}xy\\right)\\left(x^{2}+y^{2}+\\sqrt{2}xy\\right).\\] [/hide]", "Solution_3": "[hide]$x^{8}-y^{8}=(x^{4}+y^{4})(x^{4}-y^{4})$\n\n$=(x^{4}+y^{4})(x^{2}+y^{2})(x^{2}-y^{2})$\n\n$=(x^{4}+y^{4})(x^{2}+y^{2})(x+y)(x-y)$[/hide]", "Solution_4": "[hide]$x^{8}-y^{8}=(x^{4}+y^{4})(x^{4}-y^{4})$\n\n$=(x^{4}+y^{4})(x^{2}+y^{2})(x^{2}-y^{2})$\n\n$=(x^{4}+y^{4})(x^{2}+y^{2})(x-y)(x+y)$[/hide]" } { "Tag": [ "probability", "Putnam", "advanced fields", "advanced fields unsolved" ], "Problem": "Suppose that n+1 points are chosen uniformly and independently from inside a square. Show that the probability that the points are in convex position is [C_n/(n!)]^2\r\n\r\n(C_n is the nth Catalan number)\r\n\r\nThis sounds a bit like Putnam 2006 A6, but it's probably nothing like it. Help?", "Solution_1": "I'm pretty sure that the square is easier to work with than the circle, but it's still very hard. I have no idea how to approach many points.", "Solution_2": "But the case seems a lot more general, so it may turn out to be harder overall?", "Solution_3": "So I emailed Daniel Kane about this problem and he gave me a pretty nice solution, so I thought it'd be useful to share it: \r\n\r\n[quote] \n\nIn order to produce a set of points in convex position, look at the\nvectors along the edges of the convex hull. These are some n vectors\nthat sum to 0 and are sorted by slope. You can produce set a set of\npoints by picking n vectors that sum to 0, shorting by slope and using\nthem as differences of the points. One question is whether they fit\nin the square and if so, how much \"area\" do such configurations cover.\n It is not hard to show that they fit iff the sum of the positive x\ncoordinates (say X) and the sum of the positive y coordinates (say Y)\nare both at most 1, and in this case, the area of the configuration is\n(1-X)(1-Y). But now we are looking at n random vectors summing to 0\nand looking at some function of the x-coordinates times some function\nof the y-coordinates. Hence we can compute this expectation by\nhandling each coordinate separately. In other words we look at n\nnumbers that add to 1 and see how often the sum of the positive ones\nare at most 1 and look at the average value of 1-X in this case. This\nintegral is not too hard. It involves splitting into cases based on\nhow many of the numbers are positive and gives a combinatorial sum\nyielding the desired answer.\n\n[/quote]", "Solution_4": "refer to this journal:\r\n[b]P. Valtr, Discrete Comput. Geom. 13 (1995), 637-643.[/b]\r\n\r\nand \r\n[b]Combinatorica 16 (1996), 567-573[/b]\r\nfor the case of triangle...\r\n\r\nOr please read this:\r\n[url]http://edocs.fu-berlin.de/docs/servlets/MCRFileNodeServlet/FUDOCS_derivate_000000000261/1994_01.pdf;jsessionid=6F9D8493BB7F228747429A4CB486F0FD?hosts=[/url]" } { "Tag": [ "inequalities", "algebra", "polynomial", "Cauchy Inequality", "inequalities unsolved" ], "Problem": "$ a,b,c,d$ are positive. prove that :\r\n\r\n$ \\frac{a\\plus{}b\\plus{}c\\plus{}d}{4}\\geq\\sqrt[3]{\\frac{abc\\plus{}abd\\plus{}acd\\plus{}bcd}{4}}$", "Solution_1": "if u have Problem-solving strategies then see its 3rd problem in inequalities,it's the same", "Solution_2": "maclaurin inequality! :o :wink:", "Solution_3": "You are right letmast but it can be addressed by simple Cauchy inequality\r\n\r\nHere are steps\r\n\r\n$ \\frac {abc \\plus{} bcd \\plus{} acd \\plus{} bcd}{4} \\equal{} \\frac {ab(c \\plus{} d) \\plus{} cd(a \\plus{} b)}{4} \\le \\frac {(a \\plus{} b)(c \\plus{} d)(a \\plus{} b \\plus{} c \\plus{} d)}{16} \\le \\left(\\frac {a \\plus{} b \\plus{} c \\plus{} d}{4}\\right)^3$", "Solution_4": "It is just AM-GM\r\n[hide=\"solution\"]\nIt is equivalent with $ (a\\plus{}b\\plus{}c\\plus{}d)^3 \\geq 16(abc \\plus{}abd \\plus{}acd \\plus{}bcd)$\n\n$ 16(abc \\plus{}abd \\plus{}acd \\plus{}bcd ) \\equal{}16ac(b\\plus{}d) \\plus{}16bd(a\\plus{}c) \\leq 4(b\\plus{}d)(a\\plus{}c)^2 \\plus{}4(a\\plus{}c)(b\\plus{}d)^2 \\equal{}4(a\\plus{}c)(b\\plus{}d)(a\\plus{}b\\plus{}c\\plus{}d) \\leq (a\\plus{}b\\plus{}c\\plus{}d)^3$ [/hide]", "Solution_5": "[quote=\"mahanmath\"]$ a,b,c,d$ are positive. prove that :\n\n$ \\frac {a \\plus{} b \\plus{} c \\plus{} d}{4}\\geq\\sqrt [3]{\\frac {abc \\plus{} abd \\plus{} acd \\plus{} bcd}{4}}$[/quote]\r\n\r\nDefine polynomial $ P(x) \\equal{} (x \\plus{} a)(x \\plus{} b)(x \\plus{} c)(x \\plus{} d)$ , so $ P(x)$ deegre 4\r\n\r\nDefine $ s_i$ be the coefficient of $ x^i$ , so $ s_1 \\equal{} abc \\plus{} bcd \\plus{} cda \\plus{} dab$ , and $ s_3 \\equal{} a \\plus{} b \\plus{} c \\plus{} d$, so by maclaurin inequality we have\r\n\r\n$ \\frac {abc \\plus{} bcd \\plus{} cda \\plus{} dab}{4}\\ge\\sqrt [3]{\\frac {a \\plus{} b \\plus{} c \\plus{} d}{4}}$\r\n\r\nI think there are some mistakes here, can somebody tell me please ?" } { "Tag": [ "geometry", "geometric transformation", "rotation" ], "Problem": "By catalytic hydrogenation of (+)-3-buten-2-ol $ \\left([\\alpha]^{27}_{D}=+33.2^{o}\\right)$, (+)-butan-2-ol $ \\left([\\alpha]^{27}_{D}=+13.5^{o}\\right)$ is obtained. Classify each of the following statements as true or false:\r\n\r\n(a) As both compounds have the same sign of rotation (they rotate the plane of polarized light in the same direction), then they have necessarily the same absolute configuration\r\n\r\n(b) Both compounds have the same sign of rotation because they have the same absolute configuration.\r\n\r\n(c) Both compounds have configuration R.\r\n\r\n(d) Both compounds have configuration S.", "Solution_1": "1.false\r\n2.false\r\n3,4 since nothing about one configuration is given can by just lookinga t alpha can one tell?\r\nbut one thing is sure since the bond at chiral centre is not broken both have same absolute configuration", "Solution_2": "That's all correct, because...\r\n\r\n[quote=\"Pardesi\"]since nothing about one configuration is given can by just lookinga t alpha can one tell?[/quote]\r\n\r\n...there is no relation between sign of rotation and absolute configuration." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove the following. I know for certain (a) is true, and hopefully (b) is true also. The inequality is for positive reals $a,b$. \r\n(a)\r\n$a^7 b^3 + a^5 b^8 + b + a^5 b^2 + a^2 b^3 + a^2 + 8 a^3 b^3 \\geq 2 a^5 b^5 + 2 a^3 b^4 + 2 a^4 b^3 + 4 a^6 b^4 + 4 a^4 b^6$\r\n(b)\r\n$c + c^4 b + b^4 + b^3 + b^2 c^3 + c^2 + 6 ( b^2 c + b c^2 + b^2 c^2 ) \\geq 8 (bc + b^3 c + b c^3 )$\r\n\r\nTo Davron: I've changed the question statement. Hope you can solve it.", "Solution_1": "Sorry what are the numbers : {real, postitive real, integer... ?}.\r\n\r\ndavron" } { "Tag": [], "Problem": "Hi I was wondering if anyone was familiar with the nav4all navigation system for cell phones. I have been told it is pretty good, any comments on that?", "Solution_1": "[quote=\"jjefwfef\"]Hi I was wondering if anyone was familiar with the nav4all navigation system for cell phones. I have been told it is pretty good, any comments on that?[/quote]\r\n\r\n[url=http://ar.search.yahoo.com/search?p=nav4all+navigation+system&sm=Buscar&fr=FP-tab-web-t&toggle=1&ei=ISO-8859-1&meta=all%3D1]This has been modified by the moderator because it is extremely annoying to paste long urls directly into your message as they cause the screen to stretch horizontally. Please don't do it.[/url]" } { "Tag": [ "function", "algebra", "polynomial", "Euler", "limit", "number theory", "totient function" ], "Problem": "Let $ \\Phi_{n}(x)$ be the nth cyclotomic polynomial in x, $ \\phi(n)$ be the Euler totient function, and $ \\mu(n)$ be the M\u00f6bius function. Call $ P(n)=\\prod_{prime}\\frac{p^{\\phi(n)}}{\\Phi_{n}(p)}$. Prove that:\r\n\r\n$ \\mu(n)=1$ iff $ P(n)$ diverges.\r\n$ \\mu(n)=0$ iff $ P(n)$ converges to a positive finite real.\r\n$ \\mu(n)=-1$ iff $ P(n)$ converges to 0.", "Solution_1": "Say $ \\mathbb{P}= \\{p \\in \\mathbb{N}: p \\mbox{ is prime}\\}$, $ \\mathbb{P}(x) = \\{p \\in \\mathbb{P}: p \\le x\\}$ and $ \\mathbb{N}(x) = \\{q \\in \\mathbb{N}: \\mbox{gpf}(q) \\le x\\}$, for any $ x \\in \\mathbb{R}$. Here $ \\mathbb{N}= \\{1, 2, \\ldots\\}$ and $ \\mbox{gpf}(q) = \\max\\{p \\in \\mathbb{P}: p \\mid q\\}$, for any $ q \\in \\mathbb{N}$. \r\n\r\nTrivially, $ P(1) = \\prod_{p \\in \\mathbb{P}}\\frac{p}{1-p}= \\infty$, and indeed $ \\mu(1) = 1$. So in the following let $ n \\ge 2$ and set $ r = \\sigma_{0}(n)$ and $ d_{1}, d_{2}, \\ldots, d_{r}\\in \\mathbb{N}$ to be the (pairwise distinct) divisors of $ n$, with $ 1 = d_{1}< \\ldots < d_{r}= n$. Then $ \\Phi_{n}(x) = (1-x)^{\\mu(n)}\\cdot \\prod_{k=2}^{r}(1-x^{d_{k}})^{\\mu(n/d_{k})}$ and $ n = \\sum_{k=1}^{r}\\phi(d_{k})$, yielding $ \\phi(n) = \\sum_{k=1}^{r}d_{k}\\cdot \\mu(n/d_{k})$ through the M\u00f6bius inversion formula. Now fix $ x \\in \\mathbb{R}$ and let $ P_{n}(x) = \\prod_{p \\le x}\\frac{p^{\\phi(n)}}{\\Phi_{n}(p)}$, where the product is over any $ p \\in \\mathbb{P}$ such that $ p \\le x$. Then obviously $ P_{n}(x) = (-1)^{M(x)}\\,Q_{n}(x) \\cdot \\prod_{p \\le x}\\!\\left(\\frac{p}{p-1}\\right)^{\\!\\mu(n)}$, where $ M(x) = \\pi(x) \\cdot \\sum_{k=1}^{r}\\mu(n/d_{k}) = 0$ (prove it!) and $ Q_{n}(x) = \\prod_{p \\le x}\\prod_{k=2}^{r}\\!\\left(\\frac{p^{d_{k}}}{p^{d_{k}}-1}\\right)^{\\!\\mu(n/d_{k})}= \\prod_{k=2}^{r}\\!\\left(\\prod_{p \\le x}\\sum_{i=0}^\\infty \\frac{1}{p^{id_{k}}}\\right)^{\\!\\mu(n/d_{k})}= \\prod_{k=2}^{r}\\!\\left(\\sum_{m \\in \\mathbb{N}(x)}\\frac{1}{m^{d_{k}}}\\right)^{\\!\\mu(n/d_{k})}$. Since $ d_{k}> 1$, for any $ k = 2, \\ldots, r$, it follows the limit $ \\lim_{x \\to+\\infty}Q_{n}(x)$ exists, and particularly $ Q(n) = \\prod_{k=2}^{r}\\zeta(d_{k})^{\\mu(n/d_{k})}$.\r\n\r\nHence the behaviour of $ P_{n}(x)$ as $ x \\to+\\infty$ is [i]dominated[/i] by the factor $ \\prod_{p \\le x}\\left(\\frac{p}{p-1}\\right)^{\\mu(n)}$. Actually, here are the cases:\r\n[list]1) $ n$ is squareful. Then $ \\mu(n) = 0$ and $ P(n) = \\lim_{x \\to+\\infty}P_{n}(x) = Q(n)$;\n2) $ n$ is squarefree and $ \\omega(n)$ is odd. Then $ \\mu(n) =-1$ and $ P(n) = \\lim_{x \\to+\\infty}P_{n}(x) = Q(n) \\cdot \\prod_{p \\in \\mathbb{P}}\\frac{p-1}{p}= 0$;\n3) $ n$ is squarefree and $ \\omega(n)$ is even. Then $ \\mu(n) = 1$ and $ P(n) = \\lim_{x \\to+\\infty}P_{n}(x) = Q(n) \\cdot \\prod_{p \\in \\mathbb{P}}\\frac{p}{p-1}= 0$,\n[/list]\r\nwhere $ \\omega(n)$ denotes the number of distinct primes $ p \\in \\mathbb{P}$ appearing in the Euclidean decomposition of $ n$, for any $ n \\in \\mathbb{N}$. Then the proof is complete." } { "Tag": [], "Problem": "[color=blue] [size=150]$16{th}$ Iberoamerican Olympiad[/size]\nMinas, URUGUAY. [2001]\n\nEdited by djimenez\n\nCarlos Bravo ;)\n\n[/color]", "Solution_1": "[quote][b]Problema 2.[/b] El incirculo $ (I)$ de $ \\triangle ABC$ es tangente a $ BC,CA,AB$ en $ X,Y,Z.$ Las rectas $ BI,CI$ cortan a la recta $ YZ$ en $ P,Q.$ Mostrar que $ \\triangle ABC$ es is\u00f3sceles si y solo si $ XP \\equal{} XQ.$[/quote]\nEs bien sabido que la paralela media a $ AB,$ la recta $ YZ$ y la bisectriz $ BI$ de $ \\angle ABC$ concurren en $ P.$ As\u00ed mismo la paralela media a $ AC,$ la recta $ YZ$ y la bisectriz $ CI$ de $ \\angle ACB$ concurren en $ Q.$ Una sencilla demostraci\u00f3n puede apreciarse en el t\u00f3pico [i]Another unlikely concurrency[/i] en el sitio [url]http://pagesperso-orange.fr/jl.ayme/[/url]. \n\nConsecuentemente los \u00e1ngulos $ \\angle BPC$ y $ \\angle CQB$ son rectos, es decir que los cuadril\u00e1teros $ PQBC,PIXC$ y $ QIXB$ son c\u00edclicos. As\u00ed se tienen las siguentes relaciones angulares:\n\n$ \\angle QPX \\equal{} \\angle QPI \\plus{} \\angle IPX \\equal{} \\angle ICX \\plus{} \\angle ICX \\equal{} \\angle ACB$ \n\n$ \\angle PQX \\equal{} \\angle PQI \\plus{} \\angle IQX \\equal{} \\angle IBX \\plus{} \\angle IBX \\equal{} \\angle ABC$ \n\nPor lo tanto $ \\triangle ABC \\sim \\triangle XQP$ y evidentemente $ \\triangle ABC$ es is\u00f3sceles si y solo si $ \\triangle XQP$ es is\u00f3sceles.", "Solution_2": "[quote][b]Problema 6.[/b] Probar que es imposible cubrir un cuadrado de lado 1 con 5 cuadrados iguales de lado < 1/2.[/quote]\nConsid\u00e9rese los siguientes 9 puntos en el plano cartesiano $(0,0),$ $ (\\frac{_1}{^2}, 0),$ $ (1,0),$ $ (0,\\frac{_1}{^2}),$ $ (\\frac{_1}{^2},\\frac{_1}{^2}),$ $ (1,\\frac{_1}{^2}),$ $ (0,1),$ $ (\\frac{_1}{^2},1),$ $ (1,1).$ Cualquier cuadrado de lado menor que $\\frac{_1}{^2}$ solo podr\u00e1 cubrir uno de de estos puntos. Asi, ello es cierto para 8 cuadrados." } { "Tag": [], "Problem": "What is the value of $ ((\\frac{1}{2} \\div \\frac{1}{2})\\div \\frac{1}{2}) \\div \\frac{1}{2}$?", "Solution_1": "hello, we have $ \\left(\\left(\\frac{1}{2}: \\frac{1}{2}\\right): \\frac{1}{2}\\right): \\frac{1}{2}\\equal{}\\left(1: \\frac{1}{2}\\right): \\frac{1}{2}\\equal{}2: \\frac{1}{2}\\equal{}4$.\r\nSonnhard." } { "Tag": [ "calculus", "derivative", "logarithms", "calculus computations" ], "Problem": "One that I got by guessing by wasn't able to figure out exactly:\r\n\r\nGiven that ln(x*e^y) = e^(x*ln(y)), find dy/dx. Assume x and y are both positive.", "Solution_1": "${\\log \\left(x e^{y}\\right)=e^{x \\log (y)}}$\r\n${\\frac{d\\log \\left(x e^{y}\\right)}{dx}=\\frac{de^{x \\log (y)}}{dx}}$\r\n${\\frac{e^{-y}\\left(e^{y}x \\frac{dy}{dx}+e^{y}\\right)}{x}=y^{x}\\left(\\frac{x \\frac{dy}{dx}}{y}+\\log (y)\\right)}$\r\n${\\frac{dy}{dx}+\\frac{1}{x}=y^{x}\\left(\\frac{x \\frac{dy}{dx}}{y}+\\log (y)\\right)}$\r\n${\\frac{dy}{dx}=-\\frac{y \\left(x y^{x}\\log (y)-1\\right)}{x \\left(x y^{x}-y\\right)}}$\r\n\r\nLog = Ln" } { "Tag": [ "Duke", "college" ], "Problem": "So, early decision comes out very soon (my Duke ED comes out monday). Anyone excited/anxious?", "Solution_1": "I don't know any numbers for this year but last year, I heard that Duke received a really small ED pool so they got nervous and accepted more than usual. Then, during RD, they got a record number so their RD acceptance rate was really low. Not that I care about Duke anyway (go Blue Jays!!!).", "Solution_2": "ya, i got really happy when i checked out their class of 2011 profile, they accepted 40% ED and 20% RD. Then my happiness got ruined when i checked out their 2010 profile, which was 20% ED and 20% RD.\r\n\r\nO well, lets just hope Duke freaks out this year again!" } { "Tag": [ "linear algebra", "linear algebra unsolved" ], "Problem": "maximize Z=15x+10y\r\n\r\nsubject to \r\n\r\n4x+2y=100\r\n\r\nCan you determine a solution to this model that will achieve objective? :maybe:", "Solution_1": "Without any additional constraints on the variables, $ Z$ can be made arbitrarily large. Just put $ y\\equal{}50\\minus{}2x$ to make $ Z\\equal{}500\\minus{}5x$ which tends to infinity as $ x$ tends to minus infinity. If the variables are supposed to be non-negative, then $ x\\equal{}0, y\\equal{}50$, $ Z\\equal{}500$ is obviously the maximum.\r\n\r\nIn either case this doesn't seem to belong in this forum.", "Solution_2": "[b]randomgraph[/b]\r\n\r\n\r\nThank you very much for the reply :)\r\n\r\nI thought the solution couldn't be found without any other information\r\n\r\nyes, variables are non-negative... and (0 50) is a maximum yes?and there is no another point that would maximize the equation. Z canit be more than 500\r\n\r\n sorry if I've opened the topic in the wrong place\r\nI am a new girl here and have loved getting to know this forum :roll:" } { "Tag": [ "search", "geometry theorems", "geometry" ], "Problem": "Dear mathlinker. i would like to know which is the [b]Cristea's transversal theorem[/b] ?.I've been searching on several websites but i cannot find it.I know that this theorem is similar to Van Aubel's theorem.", "Solution_1": "Where I can find material about of Cristea\u00b4s theorem?\r\nThanks.", "Solution_2": "Dear Mathlinkers,\r\nsearching among the archive,\r\nsee\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?search_id=2065798600&t=15764\r\nSincerely\r\nJean-Louis" } { "Tag": [ "inequalities", "trigonometry", "inequalities proposed" ], "Problem": "Prove the inequality $ \\begin{equation*} \\tan \\left( \\frac{\\pi \\sin x}{4\\sin \\alpha }\\right) +\\tan \\left(\\frac{\\pi \\cos x}{4\\cos \\alpha }\\right) >1\\end{equation*}$ for all values of $x$ and $\\alpha $ satisfying the conditions $\\begin{equation*}\r\n0\\leq x\\leq \\frac{\\pi }{2},\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\frac{\\pi }{6}<\\alpha <\\frac{%\r\n\\pi }{3}.\\end{equation*} $", "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] :)" } { "Tag": [], "Problem": "John has 12 marbles of different colors. In how many ways can he choose 4 marbles, if he chooses exactly one of the red, green, and blue marbles?", "Solution_1": "After taking those three, there are 9 left, and one to choose so its 9c1=9.", "Solution_2": "... no there are 252 ways, 9C3 for 3 marbles aside from the red green and blue ones, and then *3 for the color choice", "Solution_3": "There is 9, because order doesn't matter. He has a red, green, blue, and (9 other choices).", "Solution_4": "... read the question ... and i got this right ... so ...", "Solution_5": "Please make your solution clearer then, because your method doesn't make sense to me. Thanks!", "Solution_6": "[hide=\"Correct Solution\"]We will divide the $12$ distinct marbles into $2$ groups: the first group will contain $1$ red, $1$ green, and $1$ blue marble. The second group will contain the other $9$ marbles. Among the $4$ selected marbles, we need to have exactly $1$ of them to be from the first group. There are $\\dbinom{3}{1}=3$ ways. We need to select exactly $3$ marbles from the second group. There are $\\dbinom{9}{3}=84$ ways here. Since the $2$ groups do not interfere with each other/affect the order that the marbles are chosen, the answer is $3\\times84=\\boxed{252}$.[/hide]", "Solution_7": "I am so dumb. I added instead of multiplyed.", "Solution_8": ":( \nLost my 1625 problem streak along with 0.022 C&P rating\n", "Solution_9": "[quote=pinkpig]:( \nLost my 1625 problem streak along with 0.022 C&P rating[/quote]\n\nrip\nwait how is this lvl 22\n[hide = sol]There are 9 C 3 ways to choose the 3 other marbles, and 3 ways to choose one of the marbles that need to be either red, green, or blue. 3 * 84 = 252[/hide]", "Solution_10": "but how do you know that there aren't more red marbles? Nowhere does it say there is only one red marble. \"INCLUDING\"", "Solution_11": "What if ten of them were red, and the others were green and blue? This is possible, given that it just says there is at least one red, one blue, and one green" } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "$ ABC$ is right-angled at $ A$ the incircle $ (I)$ touches the hypotenuse $ BC$ at $ X$ ,the perpendicular to $ BC$ at $ X$ cut the circumcircle of $ ABC$ in $ P$, $ P$ can be in the same o different side of $ A$. Prove that the square that has side $ PX$ has the same area that $ ABC$", "Solution_1": "we have $ BX\\equal{}(BC\\plus{}AB\\minus{}AC)/2$ and $ CX\\equal{}(BC\\plus{}AC\\minus{}AB)/2$\r\nwe know also that $ CX.BX\\equal{}PX^{2}$ \r\nso it's done" } { "Tag": [ "summer program", "MathPath", "modular arithmetic", "number theory", "puzzles" ], "Problem": "For what number can you move the 1's digit to the front and create the same effect as doubling the number?\r\n\r\nFor example: 234 becomes 423, but we need one that gets doubled. \r\n\r\nAll I can think of is zero, but there must be a better answer.", "Solution_1": "[quote=\"mfs50\"]For what number can you move the 1's digit to the front and create the same effect as doubling the number?\n\nFor example: 234 becomes 423, but we need one that gets doubled. \n\nAll I can think of is zero, but there must be a better answer.[/quote]\r\n\r\n[hide=\"hint\"]This was one of the MathPath qualifying problems. What can the ones digit be? Now that you know that, what do you know about the tens digit? And so on.[/hide]", "Solution_2": "The ones digit should be twice the first digit. The tens digit should be twice the ones digit. Or could the ones digit be one more than twice the first digit, as in 15......3 changing to 315......?\r\nI'm still stuck.", "Solution_3": "[hide=\"Hint\"]It cannot be 0,1 only. Try any others.[/hide]", "Solution_4": "As the first digit or the ones digit?", "Solution_5": "The ones digit.\r\nThe answer is quite long.", "Solution_6": "1.....842\r\n\r\nOn the right track? I don't know what comes next.", "Solution_7": "Yeah, I did it from 2->4->8->6->...\r\nWrite it like this\r\n[code]......6842\n 2\n 684[/code]", "Solution_8": "...6842 * 2 = ...13684, so far so good. \r\nBut ...136842 * 2 = ...273684, doesn't work.", "Solution_9": "No, you need to do until you get another 2.", "Solution_10": "I understand that. But the digits are not the same anymore. I can 't start with .....136842 and end up with .....273684. It won't be the right digits. \r\n\r\nI appreciate you helping me, but I'm super busy and I've already spent hours on this. Do you think you could give me the answer if you have it?", "Solution_11": "[hide=\"One of the answers\"]105263157894736842[/hide]", "Solution_12": "Cool. Did you do it like this? \r\n2\r\n42\r\n842\r\n1684\r\n336842\r\n6736842\r\n134736842\r\n2694736842\r\n53894736842\r\n\r\nHow can you do this with a standard calculator or even a TI-82? Is this just dog work to be done by hand? Is there a faster way to get it?", "Solution_13": "I did it by hands.\r\n[hide=\"Another way of doing it\"]This may be harder, but it gives some information.\nDo it by writing that number as $10x+y$ where $x$ is a natural number and $y$ is $0,1,...,9$.\nSo $2(10x+y)=10^{k}y+x$ where $k$ is the number of digits of $x$. Therefore, $19x=(10^{k}-2)y$.\nBut $19$ does not divide $y$, so it divides $10^{k}-2$.\nBy using calculator OR trying to divide $99999....9998$ by $19$ by hands OR using the information from the answer we got above, $k$ can be $17$.\nSo $x=5263157894736842y$ which works for $y=2,3,...,9$ because we need $x$ with $17$ digits.[/hide]", "Solution_14": "[quote=\"OHO\"]\nBut $19$ does not divide $y$, so it divides $10^{k}-2$.\nBy using calculator OR trying to divide $99999....9998$ by $19$ by hands OR using the information from the answer we got above, $k$ can be $17$.\n[/quote]\r\nWhat we want to know the value of $k$ such that $10^{k}\\equiv 2 \\bmod 19$ \r\nwhich is same as $10^{k+1}\\equiv 20 \\equiv 1 \\bmod 19$ \r\nSince 19 is prime, Fermat's little theorem tells us $10^{18}\\equiv 1 \\bmod 19$\r\n(k=17 is one solution, there are other numbers with k=35 or 53 etc..)", "Solution_15": "Yeah, and there are no solution else other than $17, 35, 53,...$", "Solution_16": "[quote=\"OHO\"]Yeah, and there are no solution else other than $17, 35, 53,...$[/quote]\r\n\r\nYeah,and each of them give multiple solutions, like:\r\n$105263157894736842$\r\n$157894736842105263$\r\n$210526315789473684$ \r\n\r\nAlso please notice $1/19 = .526315789473684210 \\ 526315789473684210\\dots$\r\n\r\netc....", "Solution_17": "Oh! Yeah...", "Solution_18": "[quote=\"Gyan\"]Yeah,and each of them give multiple solutions, like:\n\n$105263157894736842$\n$157894736842105263$\n$210526315789473684$ \n\nAlso please notice $1/19 = .052631578947368421 \\ 052631578947368421\\dots$[/quote]Nice :)\r\n\r\n[size=150][color=blue]Summary[/color][/size]\r\n\r\n$\\text{number}=10x+y$, where $0\\leq y\\leq 9$\r\n\r\n$\\text{obtained number}=10^{k}y+x$, where $k$ is the number of digits of $x$.\r\n\r\n$19x=(10^{k}-2)y \\Rightarrow k\\equiv-1 \\pmod{18}$\r\n\r\n\r\nNow, take the case $k=17$.\r\n\r\n$10^{18}\\equiv 1 \\pmod{19}\\Rightarrow$\r\n\r\n$10^{18}-1 = 19m\\Rightarrow \\frac{1}{19}= \\frac{m}{10^{18}-1}= \\frac{m}{999999999999999999}$\r\n\r\nIt's well known that the last number is a periodic decimal number at the form $0.mmm...$, where $m$ fills 18 decimal places in each time it appears.\r\n\r\nAccording to the previous posts, $m=052631578947368421$\r\n\r\n\r\n\r\nNow we have $10^{17}-2 = \\frac{10^{18}-20}{10}= \\frac{10^{18}-1-19}{10}= \\frac{19m-19}{10}= 19\\cdot \\frac{m-1}{10}$ \r\n\r\n$19x=(10^{k}-2)y \\Rightarrow 19x = 19\\cdot \\frac{m-1}{10}y \\Rightarrow x = \\frac{m-1}{10}y \\Rightarrow$\r\n\r\n$10x = (m-1)y\\Rightarrow$\r\n\r\n$10x+y = my$\r\n\r\nSo, all the solutions for $k=17$ are $my$, where $m=\\frac{10^{18}-1}{19}$ and $y$ is some digit of $m$. But we can see that $y$ takes all the values in $\\{0,1,...,9\\}$\r\n\r\n\r\nNotice that $10x+y = \\frac{10^{18}-1}{19}y = \\left(10^{18}-1\\right)\\frac{y}{19}$\r\n\r\n[size=150][color=blue]Conclusion[/color][/size]\r\n[color=brown]So there is a quick way to find the solutions:\n$\\blacktriangleright$ Take all the numbers $\\frac{y}{19}$, where $y\\in\\{0,1,...,9\\}$ (we can include $0$ of course)\n$\\blacktriangleright$ Find the repeating part (the 18 first decimal digits)[/color]" } { "Tag": [ "integration", "function", "calculus", "calculus computations" ], "Problem": "The problem is $ \\Gamma{(1)} = \\int{t^{z-1}e^{-t}dt}$ on the interval 0 to $ \\infty$.\r\n\r\nAlso:\r\nA. Calculate $ \\Gamma{(1)}$.\r\nB. Determin $ \\Gamma{(z+1)}$ in terms of $ \\Gamma{(z)}$.\r\nC. Based on your answer, explain how $ \\Gamma{(z)}$ relates to another function you already know about.\r\n\r\n\r\nI did:\r\nA. $ \\Gamma{(1)} = \\int{t^{z-1}e^{-t}dt}$ = 1\r\n\r\nB. (z+1) = n\r\n $ \\Gamma{(z)}$ = (n-1)!\r\n n(n-1) = n!\r\n $ \\Gamma{(z+1)}$ = (z+1)($ \\Gamma{(z)}$)\r\n\r\nC. f(z+1) = f(z)(z+1)\r\n\r\nParts A and C I believe are right, but I think somewhere in B I made a mistake? Thanks for any help.", "Solution_1": "$ \\Gamma(z\\plus{}1)\\equal{}\\int_{0}^{\\infty}e^{\\minus{}t}t^{z}\\ dt\\equal{}\\minus{}e^{\\minus{}t}t^{t}\\bigg|_{0}^{\\infty}\\plus{}\\int_{0}^{\\infty}ze^{\\minus{}t}t^{z\\minus{}1}\\ dt$\r\n$ \\equal{}z\\int_{0}^{\\infty}e^{\\minus{}t}t^{z\\minus{}1}\\ dt\\equal{}z\\Gamma(z)$", "Solution_2": "You did the problem backwards. You're not supposed to already know that the Gamma function relates to factorials.; that comes out of your work in B.", "Solution_3": "In fact, your answers to B and C would make more sense switched, and you should really proofread what you wrote for A. You should devote some space to the actual integral calculations here, not just repeating stuff you know from somewhere.", "Solution_4": "[quote=\"JRav\"]$ \\Gamma(z \\plus{} 1) \\equal{} \\int_{0}^{\\infty}e^{ \\minus{} t}t^{z}\\ dt \\equal{} \\minus{} e^{ \\minus{} t}t^{t}\\bigg|_{0}^{\\infty} \\plus{} \\int_{0}^{\\infty}ze^{ \\minus{} t}t^{z \\minus{} 1}\\ dt$\n$ \\equal{} z\\int_{0}^{\\infty}e^{ \\minus{} t}t^{z \\minus{} 1}\\ dt \\equal{} z\\Gamma(z)$[/quote]\r\n\r\nSo does the $ \\minus{} e^{ \\minus{} t}t^{t}\\bigg|_{0}^{\\infty}$ drop out since it is undefined ($ 0^0$ is undefined?) and that is why you are left with $ z\\int_{0}^{\\infty}e^{ \\minus{} t}t^{z \\minus{} 1}\\ dt$?", "Solution_5": "[quote=\"jmerry\"]In fact, your answers to B and C would make more sense switched, and you should really proofread what you wrote for A. You should devote some space to the actual integral calculations here, not just repeating stuff you know from somewhere.[/quote]\r\n\r\nFor part A\r\n\r\n$ \\Gamma(1) \\equal{} \\int_{0}^{\\infty}e^{ \\minus{} t}t^{z}\\ dt \\equal{} \\int_{0}^{\\infty}e^{ \\minus{} t}t^{1}\\ dt$\r\nor $ \\minus{} (t \\plus{} 1)e^{ \\minus{} t}\\bigg|_{0}^{\\infty}$\r\nso $ \\Gamma(1)$ = 1\r\n\r\n\r\nI'm not just repeating, I just don't completely understand this problem.", "Solution_6": "$ \\Gamma(1)$ is actually $ \\int_0^{\\infty}t^0e^{\\minus{}t}\\,dt\\equal{}\\int_0^{\\infty}e^{\\minus{}t}\\,dt.$\r\n\r\nThe calculation you did, of $ \\int_0^{\\infty}te^{\\minus{}t}\\,dt,$ is $ \\Gamma(2).$\r\n\r\nOf course, $ \\Gamma(1)\\equal{}\\Gamma(2)\\equal{}1.$", "Solution_7": "[quote=\"Kent Merryfield\"]$ \\Gamma(1)$ is actually $ \\int_0^{\\infty}t^0e^{ \\minus{} t}\\,dt \\equal{} \\int_0^{\\infty}e^{ \\minus{} t}\\,dt.$\n\nThe calculation you did, of $ \\int_0^{\\infty}te^{ \\minus{} t}\\,dt,$ is $ \\Gamma(2).$\n\nOf course, $ \\Gamma(1) \\equal{} \\Gamma(2) \\equal{} 1.$[/quote]\r\n\r\n :blush: oops...yeah, that's what it should be. Guess I had a problem on part A as well. Thanks for pointing that out.", "Solution_8": "[quote=\"butterz\"]So does the $ \\minus{} e^{ \\minus{} t}t^{t}\\bigg|_{0}^{\\infty}$ drop out since it is undefined ( is undefined?) and that is why you are left with $ z\\int_{0}^{\\infty}e^{ \\minus{} t}t^{z \\minus{} 1}\\ dt$?\n[/quote]\r\nIt's a typo. That should be $ t^z$, and the value is zero for $ z>0$.", "Solution_9": "OK, that makes sense. And thanks for explaining my mistakes.", "Solution_10": "Now, can you prove that for $ z\\in\\mathbb{N}$,\r\n$ \\Gamma(z)\\equal{}(z\\minus{}1)!$?\r\n\r\nHow about $ \\Gamma\\left(\\frac{1}{2}\\right)$?\r\n[hide=\"You May Need This\"]\n$ \\int_{0}^{\\infty}e^{\\minus{}t^{2}}\\ dt\\equal{}\\frac{\\sqrt{\\pi}}{2}$[/hide]", "Solution_11": "I got:\r\n\r\n$ \\Gamma({1/2}) \\equal{} \\int_{0}^{\\infty}e^{ \\minus{} t}t^{z}\\ dt \\equal{} \\int_{0}^{\\infty}e^{ \\minus{} t}t^{1/2}\\ dt$\r\n\r\n$ u \\equal{} e^{\\minus{}t}$\r\n$ du \\equal{} \\minus{}e^{\\minus{}t} dt$\r\n$ dv \\equal{} t^{1/2} dt$\r\n$ v \\equal{} {2/3}t^{3/2}$\r\n\r\n$ ({2/3})t^{3/2}(e^{\\minus{}t})\\minus{}{1/2}e^{\\minus{}2t}\\bigg|_{0}^{\\infty}$\r\n\r\nso $ \\Gamma(1)$ = .5", "Solution_12": "$ \\Gamma(1)\\neq\\frac{1}{2}$. It's already been pointed out that $ \\Gamma(1)\\equal{}1$. And your method for $ \\Gamma\\left(\\frac{1}{2}\\right)$ does not seem correct. You didn't even use my hint.", "Solution_13": "The starting point is wrong.\r\n\r\n$ \\Gamma(z) \\equal{} \\int_0^\\infty e^{\\minus{}t} t^{z\\minus{}1}dt$.\r\nNote that -1 in the exponent of t.\r\nWe get that:\r\n\r\n$ \\Gamma(\\frac{1}{2}) \\equal{} \\int_0^\\infty e^{\\minus{}t} t^{\\minus{}\\frac{1}{2}} dt \\neq \\int_0^\\infty e^{\\minus{}t} t^{\\frac{1}{2}} dt.$", "Solution_14": "[hide=\"Here is a Hint\"]\nYou do indeed need to make a substitution. The substitution will be of the form $ u\\equal{}t^{x}$, so $ du\\equal{}xt^{x\\minus{}1}dt$. What is that $ x$? I'll let you try to figure that out.[/hide]" } { "Tag": [ "calculus", "integration", "trigonometry", "function", "limit", "derivative", "real analysis" ], "Problem": "Guys, can you help me with this improper integral?\r\n\r\n$ \\int^{\\pi}_{0} \\frac{dt}{\\sqrt{t} \\plus{} \\sin t}$ thanks very much.\r\n\r\ni tried integration by parts, substitution...but i seem to not find a way through this.", "Solution_1": "I tried Mathematica for $ \\int \\frac {dt}{\\sqrt {t} \\plus{} \\sin t}$ and indeed it seems to have no simple antiderivative. I wonder how can you solve the integral when limits of integration are set up? :maybe:\r\n\r\nBy the way, I think this post should be moved to the Computations and Tutorials folder :)", "Solution_2": "well, since the function is undefined at $ x\\equal{}0$, we can change it into: $ \\lim_{x\\minus{}>0^{\\plus{}}} \\int^{\\pi}_{x} \\frac{dt}{\\sqrt{t}\\plus{}\\sin{t}}$ but i dont know what's after this...", "Solution_3": "Well, what exactly do you want of it? :? It certainly converges by the comparison test (compare with $ 1/\\sqrt t$) and almost certainly its value cannot be expressed in terms of $ e$, $ \\pi$, $ \\sqrt 2$, and such.", "Solution_4": "hello, i have got $ \\int_{0}^\\pi \\frac{1}{\\sqrt{t}\\plus{}\\sin(t)}dt \\approx 2.40899073296947082180$\r\nSonnhard.", "Solution_5": "[quote=\"fedja\"]Well, what exactly do you want of it? :? It certainly converges by the comparison test (compare with $ 1/\\sqrt t$) and almost certainly its value cannot be expressed in terms of $ e$, $ \\pi$, $ \\sqrt 2$, and such.[/quote]\n\nI know that it converges...but to what value? Is it possible to do so? thanks!\n\n[quote=\"Dr Sonnhard Graubner\"]hello, i have got $ \\int_{0}^\\pi \\frac {1}{\\sqrt {t} \\plus{} \\sin(t)}dt \\approx 2.40899073296947082180$\nSonnhard.[/quote]\r\n\r\nDid you use an approximation? Can you please show me how you dealt with $ x\\equal{}0$ and the limits involved? thanks!!!:D", "Solution_6": "[quote]Did you use an approximation? Can you please show me how you dealt with and the limits involved?[/quote]\r\n\r\nSince the function has a singularity at zero, some of the usual techniques of approximate integration (the ones that require a value at an endpoint) are problematic to define, and the worst case error estimates for all methods, which depend on norm estimates of certain derivatives, tell you nothing at all. There are several ways to deal with this. One possibility is to relieve the singularity through a change of variables.\r\n\r\nLet $ x\\equal{}\\sqrt{t}$ so that $ t\\equal{}x^2$ in the integral:\r\n\r\n$ \\int_0^{\\pi}\\frac{dt}{\\sqrt{t}\\plus{}\\sin t}\\equal{}\\int_0^{\\sqrt{\\pi}}\\frac{2x\\,dx}{x\\plus{}\\sin(x^2)} \\equal{}2\\int_0^{\\sqrt{\\pi}}\\frac{1}{1\\plus{}\\frac{\\sin(x^2)}{x}}\\,dx.$\r\n\r\nNote now that $ f(x)\\equal{}\\frac{1}{1\\plus{}\\frac{\\sin(x^2)}{x}}$ has a removable singularity at zero so we may safely set $ f(0)\\equal{}1$ and use that value in any system that uses it (composite Simpson's rule, for instance, or the method I'd probably use myself if I were sitting at my computer with only a spreadsheet open, namely Romberg integration.) In fact, $ f(x)$ is a $ C^{\\infty}$ function, so most methods should be well-behaved.\r\n\r\n[Sonnhard Graubner isn't in the habit of explaining his methods here. It's likely he put this into some large general package, like Mathematica or Maple.]", "Solution_7": "[color=green][Unnecessary quote removed by moderator.][/color]\r\n\r\nthanks for the solution mr kent merryfield. I was also about to use approximation because wolfram online integrator couldn't find a function for this....anyway, do you think there's no antiderivative for this? thanks very much!!!:D", "Solution_8": "As a default assumption, I just assume that a function does not have an elementary antiderivative unless I have some good reason to assume it does. Here, I have no good reason to assume it does.\r\n\r\nThat's not a proof - and there's a whole subject out there that others can comment on - but my main point is that you should regard [i]not[/i] having an elementary antiderivative as being the normal state of affairs, not some sort of surprising betrayal of the fabric of the universe." } { "Tag": [], "Problem": "$ \\binom{n}{r\\minus{}1}\\equal{} \\frac{n!}{(r\\minus{}1)!(n\\minus{}r\\plus{}1)!} \\equal{} r(r\\plus{}1)$\r\n\r\n$ \\binom{n}{r}\\equal{} \\frac{n!}{r!(n\\minus{}r)!}\\equal{} (r\\plus{}1)(n\\minus{}r\\plus{}1)$\r\n\r\n$ \\binom{n}{r\\plus{}1}\\equal{} \\frac{n!}{(r\\plus{}1)!(n\\minus{}r\\minus{}1)!}\\equal{}(n\\minus{}r)(n\\minus{}r\\plus{}1)$\r\n\r\nI don't get the last steps of these equations.", "Solution_1": "Clearly something's missing. I don't think any of those work for $ n\\equal{}2$, $ r\\equal{}1$.", "Solution_2": "None of the last steps are correct in general. It is true that the second column of each equation is proportional to the third. Are you trying to solve a problem?" } { "Tag": [ "ratio", "geometric sequence" ], "Problem": "Suppose that four consecutive terms of a geometric sequence with common ratio r are the sides of a quadrilateral.What is the range of all posible values for r?", "Solution_1": "i think $\\frac{1}{x'} \\le r \\le x'$ where $x'$ is the real root of the equation $1+x+x^2=x^3$. If one of the equalities hold, the quadrilateral will degenerates to a line segment." } { "Tag": [], "Problem": "MAML is going to be Oct. 28, 2008. Tues.\r\n\r\nHow are people going to prepare for it?", "Solution_1": "[quote=\"Smartguy\"]MAML is going to be Oct. 28, 2008. Tues.\n\nHow are people going to prepare for it?[/quote]\r\n\r\nAm I going to do it? Yes.\r\nAm I going to prepare for it? Probably not.", "Solution_2": "I'm gonna get a 10 hr. sleep :|", "Solution_3": "It's about AMC 10-12 difficulty so if you prepare for other math comps you should be fine...", "Solution_4": "They should totally let Connecticut people do it.", "Solution_5": "[quote=\"worthawholebean\"]They should totally let Connecticut people do it.[/quote]\r\nI don't think it's outside the realm of possibility for the MAML board to let Connecticut students take the first level MMO. You could probably have your scoresheets machine-scored with the Massachusetts schools'. But you wouldn't be ranked with Massachusetts students and you wouldn't be eligible for the second level MMO.", "Solution_6": "Mr. Yanco,\r\n\r\nHi, do you have any other MAMLs in your possession?\r\nCan you post some online?", "Solution_7": "What is MAML? \r\nim from Rhody- could I do it?", "Solution_8": "[quote=\"Smartguy\"]Hi, do you have any other MAMLs in your possession?[/quote] I put ten online last year, which I think is enough for you to get a feel for what the contest might be like. It varies from year to year, because a different person selects the questions each year, and each contest director has a different idea of what a contest should be like.\r\nThere's also a book of MAML Olympiads that you can buy from MAML; I think they charge 10 dollars or so. It has the oldest ~10 years of contests. If you're in the state your coach should be able to order it for you.", "Solution_9": "do you have the answers to years 1972-1975 for MAML?", "Solution_10": "has anyone done the 2005-2006 MMO round one? I need the answers for it...", "Solution_11": "aaaaaaaaaa", "Solution_12": "wow, MAML is tomorrow!", "Solution_13": "I'm looking forward to it :)", "Solution_14": "haha...i need to redeem myself after my phail at the first mml meet", "Solution_15": "How high of a score do you need to go to the next round?", "Solution_16": "You guys should wait until Mr. Yanco says we can until discussion. I'm pretty sure you're allowed to take it any time today, and some boarding schools (Deerfield, Andover, probably some others) may do it in the evening.", "Solution_17": "Thank you Catalyst!! I'd rather not have to go through deleting messages all the time, but you can't talk about a test until AFTER the testing window is over!! Thanks.", "Solution_18": "Hello,\r\nSorry, I realized that i wasn't supposed to post after i had already posted it. :oops: \r\n\r\nOnce again, sorry.", "Solution_19": "Hmm...why do people need to do this all the time?", "Solution_20": "The \"official\" testing window is over. If the contest was not begun between 7:30 and 9:30 AM today, the results are not considered official. However, at least one school was planning to take the contest unofficially after school. Their results will be machine scored by MAML and ranked separately; so posting information about the contest could affect that ranking.\r\n\r\nThanks.", "Solution_21": "[quote=\"perfect628\"]Yeah, we didn't finish up until around 7:15...[/quote]\r\nMake that at least [i]two[/i] schools that participated unofficially.", "Solution_22": "Wait...Phillips Academy's results aren't going to be considered official?", "Solution_23": "[quote=\"Phelpedo\"]Wait...Phillips Academy's results aren't going to be considered official?[/quote]\r\nOnly if PA started after 9:30 AM. MAML's rule that all Olympiad administrations across the state need to begin within a fairly tight window in the morning has been around for quite some time; but this is the first year that a consequence for late administration was stated in advance." } { "Tag": [ "inequalities" ], "Problem": "Consider an ordered set of six consecutive integers in increasing order. To create a new set of six integers, the first, third and fifth elements are each multiplied by two, and the second, fourth, and sixth elements are each increased by two. The median of this new set is four more than the median of the first set. What is the sum of the median of the first set and the median of the second set?", "Solution_1": "[quote=\"mgao\"]Consider an ordered set of six consecutive integers in increasing order. To create a new set of six integers, the first, third and fifth elements are each multiplied by two, and the second, fourth, and sixth elements are each increased by two. The median of this new set is four more than the median of the first set. What is the sum of the median of the first set and the median of the second set?[/quote]\r\n\r\n\r\n[hide]Okay, so the first thing I did to answer this question was I started with\n\n$1,2,3,4,5,6$\n\nNow we do the rule that corresponds to the list.\n\n$3,4,5,8,7,12$\n$3,4,5,7,8,12$\n\nThe median of the first one is 3.5 and the median of the second is 6, so we have to make the numbers bigger. Of course this means that our answer will be in the form $2n+4$ because it is the same median but with 4 more added on to one of them.\n\nNext, let's try it starting with 2 so we can maybe find a pattern.\n\n$2,3,4,5,6,7$\n\n$4,6,6,10,8,14$\n$4,6,6,8,10,14$\n\nThe median ofthe first is 4.5 and the median of the second is 7, so the sum didn't change.\n\n$3,4,5,6,7,8$\n\n$5,8,7,12,9,16$\n$5,7,8,9,12,16$\n\nThe median of the first one is 5.5, and the median of the second is 8.5. So for each increasing odd number, the median of the second set is .5 greater than the median of the first set. So to get from a difference of 3 to a difference of 4, we need to skip to 7,8,9,10,11,12 as the set of numbers. The median of this set of numbers is 9.5, so in the form $2n+4$, we see that the answer is 23.[/hide]\r\n\r\nI think I'm right, if not, I hope this helps you get started.", "Solution_2": "i did this problem before and i remember the answer isn't 23 but i forgot how to do it.", "Solution_3": "Somebody posted this problem recently-using inequalities is the most effective strategy, I believe." } { "Tag": [], "Problem": "Which is the fifth term of the sequence: $ 1,2,4,8,...?$", "Solution_1": "Each number is twice the value of the previous number, so $ 8*2\\equal{}16$.", "Solution_2": "This reminds me of a counterexample to assuming that a sequence follows the obvious pattern: the maximum number of regions inside a circle obtained when $ n$ points around the circle are joined by chords is $ 1, 2, 4, 8, 16, 31, 57, 99, ...$, as far as I remember.\r\n\r\nAlso if we consider the number of factors of $ n!$ for $ n\\equal{}1, 2, 3,4, ...$ we get $ 1, 2, 4, 8, 16, 30, 96, ...$.", "Solution_3": "Each number is added by itself to make the next number.\r\nTherefore,\r\n\r\n[b]8+8=16[/b]", "Solution_4": "The formula is 2^n-1 \r\nTherefore, the answer is 16\r\n\r\nSolution: input 5 for n= 2^4=16", "Solution_5": "Please don't post a solution when many have been already posted :wink: \r\n\r\n[quote=\"AndrewTom\"]This reminds me of a counterexample to assuming that a sequence follows the obvious pattern: the maximum number of regions inside a circle obtained when $ n$ points around the circle are joined by chords is $ 1, 2, 4, 8, 16, 31, 57, 99, ...$, as far as I remember.\n\nAlso if we consider the number of factors of $ n!$ for $ n \\equal{} 1, 2, 3,4, ...$ we get $ 1, 2, 4, 8, 16, 30, 96, ...$.[/quote]\r\n\r\nThis is possible, we don't know unless the problem states it for sure" } { "Tag": [], "Problem": "Zuleica's mother Wilma picks her up at the train station when she comes home from school, then Wilma drives Zuleica home. They always return home at 5:00 p.m. One day Zuleica left school early and got to the train station an hour early. She then started walking home. Wilma left home at the usual time to pick Zuleica up, and they met along the route between the train station and their house. Wilma picked Zuleica up and then drove home, arriving at 4:48 p.m. For how many minutes had Zuleica been walking before Wilma picked her up?", "Solution_1": "How would you approach this problem? There are so many things to deal with, it makes it so confusing. :(", "Solution_2": "Rather than focusing on Zuleica, let's think about Wilma. Wilma leaves at the same time as usual, but gets home 12 minutes earlier. So, she has saved 6 minutes each way. Therefore, the point where she picks up Zuleica early is 6 minutes driving time from the train station. In other words, she picks up Zuleica 6 minutes earlier than she usually does. Because Zuleica reached the train station one hour earlier than usual, she reached the train station 60 minutes earlier than she usually gets picked up. She was picked up 6 minutes earlier than usual, so she must have reached the train station $60-6= 54$ minutes before she was picked up. Therefore, she was walking for $\\boxed{54} minutes$ .", "Solution_3": "BRUh\n\noK AT THIS POINT I dont even care this is like a 10 year bump because this problem is STUPID\n\nI legit bashed this with d=rt for like 3 hours\nno joke !!!!!!!!\n\nAND THEN THE ANSWER IS SOME ez logic thing\nWHAT\nI HATE THIS PROBLEM\noops caps \nI feel everyone should know how bad this problem is \n>:( :censored: :furious: :furious: :wallbash_red:", "Solution_4": "[quote=sub_math]BRUh\n\noK AT THIS POINT I dont even care this is like a 10 year bump because this problem is STUPID\n\nI legit bashed this with d=rt for like 3 hours\nno joke !!!!!!!!\n\nAND THEN THE ANSWER IS SOME ez logic thing\nWHAT\nI HATE THIS PROBLEM\noops caps \nI feel everyone should know how bad this problem is \n>:( :censored: :furious: :furious: :wallbash_red:[/quote]\n\nLOL yeah this was a pretty hard problem", "Solution_5": "We only need to consider when things happen to solve the problem. No need to worry about distances or speeds.\n\n[hide]\nLet's define our terms:\n Tw: when Wilma leaves home\n Th: when Wilma and Zuleica arrive home\n Tm: when Wilma and Zuleica meet each other\n Ts: when Zuleica arrive at the train station\nWe use these terms as is for the usual case and primed versions (T'w, T'h, T'm, T's) for this specific case.\nWe are asked to find x, i.e., for how many minutes Zuleica walked after arriving at the station until meeting with Wilma. So\n1) x = T'm - T's\nLet's state what we know:\n2) Tw = T'w (Wilma left home at the usual time)\n3) Th = 5.00 pm (They usually arrive home at 5.00 pm)\n4) T'h = 4.48 pm (They arrived home at 4.48 pm this time)\n5) Ts = Tm (They usually meet as soon as Zuleica arrives at the \n train station)\n6) Ts - T's = 1 h. (Zuleica arrived at the station 1 hour earlier this time)\nUsing 5 and 6 we get:\n7) T's = Ts - 1h = Tm - 1h\nUsing 7 in 1 we get:\n8) x = T'm - (Tm - 1h) = 1h - (Tm - T'm)\nSo now we need to find (Tm - T'm). To do so we use the information that both legs of Wilma's ride (from home to meeting place and from meeting place to home) take the same time in all cases. So for the usual case we have:\n9) Tm - Tw = Th - Tm\nand organizing we get:\n10) 2Tm = Th + Tw\nAnd for this specific case we have:\n11) T'm - T'w = T'h - T'm\nand organizing we get:\n12) 2T'm = T'h + T'w\nSubtracting equation 12 from 10 we get:\n13) 2Tm - 2T'm = Th + Tw - T'h - T'w\nUsing 2, 3, and 4 in 11 we get:\n14) 2(Tm - T'm) = 5.00 pm + Tw - 4.48 pm - Tw\nand organizing we get:\n15) Tm - T'm = 6 minutes\nUsing 15 in 8 we get\n16) x = 1h - 6 minutes = [b]54 minutes[/b]\n[/hide]", "Solution_6": "Sad I put 66", "Solution_7": "[quote=tenebrine]Sad I put 66[/quote]\n\n66 as in [hide][hide][hide][size=50]bestzack[b]66[/b]?[/size][/hide][/hide][/hide]\n\nand yeah im kinda too scared to even read this problem because it looks very scary and i SUCK at rate problems so yeah no comment\n", "Solution_8": "[url=https://artofproblemsolving.com/videos/mathcounts/mc2013/385]The MATHCOUNTS Mini with this problem.[/url]\n[url=https://youtu.be/QHce_s90Ivg?t=381]6:21[/url] is when he starts on this problem. That's okay to say right? Because this is the solution page anyways. \n\n", "Solution_9": "[hide = Solution]Rather than focusing on Zuleica, let's think about Wilma. Wilma leaves at the same time as usual, but gets home 12 minutes earlier. So, she has save$6$ minutes each way. Therefore, the point where she picks up Zuleica early is $6$ minutes earlier than she usually does. Because Zuleica reached the train station one hour earlier than usual, she reached the train station $60$ minutes earlier than she usually gets picked up. She was picked up $6$ minutes earlier than usual, so she must have reached the train station $60 - 6 = 54$ minutes before she was picked up. Therefore, se was walking for $\\boxed{54}$ minutes.[/hide]", "Solution_10": "[quote=wovenfractal]Rather than focusing on Zuleica, let's think about Wilma. Wilma leaves at the same time as usual, but gets home 12 minutes earlier. So, she has saved 6 minutes each way. Therefore, the point where she picks up Zuleica early is 6 minutes driving time from the train station. In other words, she picks up Zuleica 6 minutes earlier than she usually does. Because Zuleica reached the train station one hour earlier than usual, she reached the train station 60 minutes earlier than she usually gets picked up. She was picked up 6 minutes earlier than usual, so she must have reached the train station $60-6= 54$ minutes before she was picked up. Therefore, she was walking for $\\boxed{54} minutes$ .[/quote]\n\n[quote=wovenfractal]Rather than focusing on Zuleica, let's think about Wilma. Wilma leaves at the same time as usual, but gets home 12 minutes earlier. So, she has saved 6 minutes each way. Therefore, the point where she picks up Zuleica early is 6 minutes driving time from the train station. In other words, she picks up Zuleica 6 minutes earlier than she usually does. Because Zuleica reached the train station one hour earlier than usual, she reached the train station 60 minutes earlier than she usually gets picked up. She was picked up 6 minutes earlier than usual, so she must have reached the train station $60-6= 54$ minutes before she was picked up. Therefore, she was walking for $\\boxed{54} minutes$ .[/quote]\n\nhmmm I think thats bascially the solution in the intro to algebra solutions manual" } { "Tag": [ "percent", "geometry" ], "Problem": "What percent of circle B is shaded? Express your answer to the nearest tenth of a percent.\n[asy]import olympiad; import geometry_dev; import graph; size(150); defaultpen(linewidth(0.8)); import patterns;\npath largeArc = Arc((0,0),1,90,405); draw(unitcircle); draw(dir(90)--origin--dir(45));\nadd(\"hatchback\",hatch(NW));\nfill(largeArc--origin--cycle,pattern(\"hatchback\")); dot(Label(\"$B$\",align=S),origin);[/asy]", "Solution_1": "Checking the Asy code, the unshaded portion is $ \\frac{45^\\circ}{360^\\circ}\\equal{}\\frac18$ of the circle, so the answer is $ 1\\minus{}\\frac18\\equal{}\\frac78\\equal{}\\boxed{87.5\\%}$.", "Solution_2": "[quote=\"math154\"][b]Checking the Asy code[/b], the unshaded portion is $ \\frac {45^\\circ}{360^\\circ} \\equal{} \\frac18$ of the circle, so the answer is $ 1 \\minus{} \\frac18 \\equal{} \\frac78 \\equal{} \\boxed{87.5\\%}$.[/quote]\r\n\r\nLOL? Is there another solution?", "Solution_3": "[quote=\"GameBot\"]What percent of circle B is shaded? Express your answer to the nearest tenth of a percent.[/quote]\r\n\r\nUh, no?", "Solution_4": "wait...... but how do you know it is 45 degrees?\r\nlike on the real problem without looking at the Asy code", "Solution_5": "There is no way. Evidently, unless I, math154, BOGTRO, you, etc., are missing something huge, some key amount of information is missing. I suppose they(as in the problem writers) want you to assume it's to scale and estimate about 7/8 are shaded.", "Solution_6": "So in real games it is impossible to find out for sure? I wonder how some people actually got it.", "Solution_7": "Well, when I first encountered the problem, i guessed it was 7/8 of the circle and I put in 87.5%...\r\nAnd I got IT right!!\r\n\r\nThats right guessage rulez!!!!" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "One (well-known?) theorem in my introduction to analysis textbook states:\r\n\r\nLet $E\\subset\\mathbb{R}$. If $E$ has a finite supremum (respectively, a finite infimum), then there is a sequence $x_n\\in E$ such that $x_n \\rightarrow \\sup{E}$ (respectively, $x_n \\rightarrow \\inf{E}$) as $n\\rightarrow\\infty$.\r\n\r\nCan I have a clarification of the definition?\r\nI am assuming that $E$, a set, is not ordered, so a sequence of $E$ consists of elements from $E$ in arbitrary order. Of these sequences, at least one most converge to $\\sup{E}$. I am also assuming that terms used in the sequence can be repeated elements from the set, or else a set like $\\{1\\}$ would hamper the construction of a sequence...\r\nIs this all correct?", "Solution_1": "Yes, that is correct. We do want to say that $\\sup\\{-1,3,5,8\\}=8$ and the only way that a sequence in that set could converge to 8 would be for it to eventually equal 8.\r\n\r\nOn the other hand, suppose you have a nonempty bounded set that does not have a greatest element. In that case, the set must be infinite, and if you wanted to, you could choose a strictly increasing sequence that tends to the supremum.\r\n\r\nYour book singled out the case of a finite supremum. Suppose we adopt the following as a notational convention: if $E$ is not bounded above, then $\\sup E=\\infty$ and if $E$ is not bounded below, then $\\inf E=-\\infty.$ Then this theorem still works: if $\\sup E=\\infty$ we can find $a_n\\in E$ such that $\\lim a_n=\\infty.$", "Solution_2": "Unlike lots of other analysis topics, which tend to cunfuse me with symbols and weird words, I actually understand this topic. I had to post because I'm glad I understand. lol" } { "Tag": [], "Problem": "How many 3-digit numbers are there in which the digits are such that if you create all the different 3-digit numbers from them, and you add these numbers, the sum is divicible by 111?\r\n\r\n[hide=\"hint\"] Think about ALL the values. [/hide]", "Solution_1": "[hide]729??\nabc+cab+bac+cba+acb+bca=222(a+b+c)\na,b,c are not 0.because 0bc is not a 3-digit number.\na,b,c can be 9 different digits.\n9*9*9=729[/hide]", "Solution_2": "how did you do that?", "Solution_3": "[hide=\"Solution\"] The sum of all the possible three digit numbers made using the digits a, b, and c is 222(a+b+c), which is a multiple of 111 no matter what. There are 9*10*10 three digit numbers, so our answer is [b]900[/b].[/hide]" } { "Tag": [ "analytic geometry" ], "Problem": "In an infinite chessboard, each square is filled with a real number such that it is the arithmetic mean of the four adjacent squares. Prove that every number in the chessboard must be the same.", "Solution_1": "it's not true.fill the $ i$th infinite row with the number $ i$.\r\ni think your numbers must be from an interval...it was a problem in the mediterranian mathematical olympiad, just with $ 3D$ coordinates and the numbers were from the interval $ [0,1]$ it was not concidered an easy problem...", "Solution_2": "It has been posted before, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5327 for example. There are several other versions floating around, but this is the only one I could find with any solutions. The case with positive integers has a substantially more elementary solution than what is posted." } { "Tag": [ "quadratics", "algebra", "polynomial", "factoring polynomials" ], "Problem": "I think what you said is true but I'm not a great prover so I can't prove it to you but from what the math that I've done before, you can't factor unless there is a common factor.\r\n\r\nYou can however find the roots by using the quadratic formula.", "Solution_1": "Split it up does not always work\r\n\r\nTo have [tex]a^2-b^2[/tex]\r\n[tex]a^2-b^2+ab-ab[/tex]\r\n=[tex]a(a-b)+b(a-b)[/tex]\r\n[tex]=(a+b)(a-b)[/tex]\r\nwhere you added ab-ab to the expression.", "Solution_2": "There will always be SOME way to split it up into two things which are factorable -- however, as Beta pointed out, that might require you to add and subtract something, split a term (for example, write 3x^2 as (4x^2 - x^2) or something), or combine various different terms with each other. To see why this might be, just imagine a polynomial in factored form -- when you multiply it out, you can distribute one factor across each of the terms of the other. Then you get a sum of things, each with a common factor. However, when you combine all of them, you will probably get terms combining with each other or cancelling each other out.", "Solution_3": "Or use synthetic division. I find it easier for me to use synthetic division when factoring polynomials with the greatest term having a degree of more than 3. Granted, there has to be some guessing of the factor in the first steps of synthetic division, but after some practice you can often determine a factor in less than 3 tries. And besides, it solves the problem of [tex]x^2-1[/tex]. Just use 1 0 -1 as the dividend.\r\n\r\nThis also works for [tex]a^3-1[/tex]." } { "Tag": [ "trigonometry", "function", "limit", "probability", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "Prove that $\\lim_{n \\to \\infty}{\\cos^{n}\\Bigg(\\frac{x}{\\sqrt{n}}\\Bigg)}= e^{-x^{2}/2}$ for any $x \\geq 0$", "Solution_1": "$\\lim_{n \\to \\infty}{\\cos^{n}\\Bigg(\\frac{x}{\\sqrt{n}}\\Bigg)}= \\lim_{n\\to\\infty}([1+(\\cos{\\frac{x}{\\sqrt{n}}}-1)]^{\\frac{1}{\\cos{\\frac{x}{\\sqrt{n}}}-1}})^{n(\\cos{\\frac{x}{\\sqrt{n}}}-1)}=e^{\\frac{-x^{2}}{2}}$", "Solution_2": "Suppose $f(x)$ is positive, $C^{2},$ $f(0)=1,$ and $f(x)<1$ for $x\\ne0.$ Suppose also that $f''(0)=-a\\ne0.$\r\n\r\nWhat is $\\lim_{n\\to\\infty}\\left[f\\left(\\frac{x}{\\sqrt{n}}\\right)\\right]^{n}$ for all $x\\in\\mathbb{R}?$\r\n\r\n[This is actually an intermediate step in some proofs of the Central Limit Theorem in probability theory.]", "Solution_3": "That should be raised to the $n$th power, not squared. If it's squared, it just goes to 1. [color=green][Fixed now][/color]", "Solution_4": "We have that\r\n\\begin{eqnarray*}\\lim_{n \\to \\infty}\\left[ f \\left( \\frac{t}{\\sqrt n}\\right) \\right]^{n}&=& \\exp \\left[ \\lim_{n \\to \\infty}\\frac{\\log \\left( f \\left( \\frac{t}{\\sqrt n}\\right) \\right)}{\\frac1{n}}\\right] .\\end{eqnarray*}\r\nWe will compute $L = \\lim_{x \\to \\infty}\\frac{\\log \\left( f \\left( \\frac{t}{\\sqrt x}\\right) \\right)}{\\frac1{x}}$ instead.\r\nThis a $\\frac00$ case, so we can apply L'Hospital (the other conditions are also satisfied):\r\n\\[L = \\lim_{x \\to \\infty}\\frac{\\frac1{f \\left( \\frac{t}{\\sqrt x}\\right)}\\cdot f^\\prime \\left( \\frac{t}{\\sqrt x}\\right) \\cdot \\frac{-t}{2 x \\sqrt x}}{-\\frac1{x^{2}}}= \\lim_{x \\to \\infty}\\frac{f^\\prime \\left( \\frac{t}{\\sqrt x}\\right) \\cdot \\frac{t}{2}}{\\frac1{\\sqrt x}}. \\]\r\nIf $f$ is defined on a neighbourhood of $0$, then $f^\\prime (0) = 0$, since $0$ is a point of maximum. Hence, we can apply again L'Hospital:\r\n\\[L = \\lim_{x \\to \\infty}\\frac{f^{\\prime \\prime}\\left( \\frac{t}{\\sqrt x}\\right) \\cdot \\frac{-t^{2}}4 \\cdot \\frac1{x \\sqrt x}}{-\\frac1{2 x \\sqrt x}}= \\lim_{x \\to \\infty}\\left[ \\frac{t^{2}}2 \\cdot f^{\\prime \\prime}\\left( \\frac{t}{\\sqrt x}\\right) \\right] = \\frac{-a t^{2}}2 . \\]" } { "Tag": [ "function", "integration", "real analysis", "real analysis unsolved" ], "Problem": "$K: [0;1] X [0;1] \\to \\mathbb{R}^{*+}$, a strictly positive continuous function.\r\nFor every function $u \\in C^0([0;1], \\mathbb{R})$, one defines:\r\n$A(u)(x)=\\int_{[0;1]}K(y,x)u(y)dy$.\r\n\r\nProve that there exists $c>0$ and a function $u$ such that: $A(u)=c.u$", "Solution_1": "---------------------", "Solution_2": "$A$ would only be self-adjoint if $K(x,y)=K(y,x).$ That wasn't given as a hypothesis.", "Solution_3": "Oops! Had to erase all the stupidities I had written :).", "Solution_4": "[quote=\"alekk\"]$K: [0;1] X [0;1] \\to \\mathbb{R}^{*+}$, a strictly positive continuous function.\nFor every function $u \\in C^0([0;1], \\mathbb{R})$, one defines:\n$A(u)(x)=\\int_{[0;1]}K(y,x)u(y)dy$.\n\nProve that there exists $c>0$ and a function $u$ such that: $A(u)=c.u$[/quote]\r\nPerron-Frobenius. Set $u_0(x) = 1$ and $u_n(x) = \\frac{v_n(x)}{v_n(1)}$ with $v_n(x) = A(u_{n-1})(x)$. Then show that $u_n$ converges to a continuous limit." } { "Tag": [], "Problem": "If $ a, b, c ,d$ are, in some order, $ 1, 2, 3 , 4$ . What is the greatest possible value of:\r\n\r\n$ ab \\plus{} bc \\plus{} cd \\plus{} da$ ?\r\n\r\n\r\n\r\nAnd how to do this ? What I need to do ?", "Solution_1": "[hide=\"Easy solution\"]Note that the expression is equivalent to $ (a \\plus{} c)(b \\plus{} d)$.\n\nThen, we note that $ (a \\plus{} c) \\plus{} (b \\plus{} d) \\equal{} 10$; by AM-GM, $ \\sqrt {(a \\plus{} c)(b \\plus{} d)} \\le \\frac {(a \\plus{} c) \\plus{} (b \\plus{} d)}{2} \\equal{} 5$.\n\nHence the maximum value is no greater than $ 25$, where equality holds if $ a \\plus{} c \\equal{} b \\plus{} d \\equal{} 5$. Solutions are $ (a,c)$ and $ (b,d)$ are permutations of $ (1,4)$ and $ (2,3)$.[/hide]", "Solution_2": "Ah ok ;)\r\nIt's a nice solution. Thanks for your help :)" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "In triangle ABC, AB= 85, BC= 75 and CA = 40. A semicircle is tangent to AB and AC. It diameter lies on BC. Find the radius of the semi circle. \r\n\r\n[hide]My friend got 24, but I don't know how to do it[/hide]", "Solution_1": "This is not the right place for this kind of problems. Next time you should post at most into Intermediate.\r\nBtw, by Heron the area is $ 1500$. Also $ 1500 \\equal{}\\frac{85r}{2}\\plus{}\\frac{40r}{2}$", "Solution_2": "Sorry I got the problem from the Australian Intermediate Olympiad Exam, so I thought Olympiad was the correct section. Thanks :)" } { "Tag": [], "Problem": "What sort of approximations does one make when calculating concentrations of ions in water solutions where the original compound only partially ionizes?\r\n\r\ne.g. Calculate the concentration of $ C_{8}H_{4}O_{4}^{2\\minus{}}$ in $ a)$ a $ 0.010 M$ solution of $ H_{2}C_{8}H_{4}O_{4}$, $ b)$ a solution which is $ 0.010M$ with respect to $ H_{2}C_{8}H_{4}O_{4}$ and $ 0.020M$ with respect to $ HCl$.\r\n\r\nThe first and second ionization constant for phthalic acid are $ 1.3\\times 10^{\\minus{}3}$ and $ 3.9\\times 10^{\\minus{}6}$ respectively.", "Solution_1": "A very common approximation is to consider the activity coefficients of all species equal to unity, and so the thermodynamic activities equal the molar concentrations - this is equivalent to neglect the ionic strenght of the solution. \r\nWhen we have a weak diprotic acid, the first step in to evaluate $ \\frac{K_{1}}{K_{2}}$. If $ \\frac{K_{1}}{K_{2}}\\geq 10^{4}$, then we can treat the two equilibria separatly, and for $ pH <\\frac{pK_{1}\\plus{}pK_{2}}{2}$ only the first ionization is relevant, while for $ pH >\\frac{pK_{1}\\plus{}pK_{2}}{2}$ only the second equilibrium is important. This is not the case with phthalic acid, because $ \\frac{K_{1}}{K_{2}}\\equal{} 333$, and so it is more correct to consider both equilibria." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Just an Unsolved Problem:\r\n\r\nNon-negative reals $a,b,c$ satisfy $a^{2}+b^{2}+c^{2}=1$, Prove:\r\n\r\n$\\frac{1}{1-ab}+\\frac{1}{1-bc}+\\frac{1}{1-ca}\\leq \\frac{9}{2}$", "Solution_1": "Have been posted many times before.", "Solution_2": "Please give me a hint or a link, thanx.", "Solution_3": "[quote=\"libra_gold\"]Please give me a hint or a link, thanx.[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=66936\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=109029", "Solution_4": "Thank you very much. :)" } { "Tag": [], "Problem": "Aratati ca nu exista numere rationale $\\ x$ care verifica egalitatea\r\n$\\ (2n+1)x^{m}=x^{n}-2m-1,n>m$, cu $\\ m,n$ naturale nenule\r\nSorin Peligrad", "Solution_1": "Simpatica problema...\r\nFie $x=\\frac{a}{b}$, $a$ si $b$ numere intregi, $b\\neq\\ 0$, fractie ireductibila, deci $(|a|,|b|)=1$.\r\nAtunci:\r\n$(2n+1)a^{m}b^{n}=a^{n}b^{m}-(2m+1)b^{m}b^{n}$\r\n$b^{n}[(2n+1)a^{m}+(2m+1)b^{m}]=a^{n}b^{m}$\r\nDin moment ce rezolvarea se va baza doar pe paritate si pe numarul de factori de doi, nu conteaza semnul lui $a$ si $b$, deci putem presupune $a$ si $b$ pozitive.\r\nDaca $b$ este impar, iar $a$ este par, atunci membrul stang al identitatii este impar, iar cel drept este par, imposibil.\r\nDaca $b$ este impar, iar $a$ este impar, atunci membrul stang al identitatii este par, iar cel drept este impar, imposibil.\r\nDaca $b$ este par, iar $a$ este impar atunci paranteza din membrul stang este impara, deci nu intervine in stabilirea numarului de factori de $2$ prezenti. Rezulta ca $b^{n}$ si $b^{m}$ au acelasi numar de factori de doi, rezulta $m=n$, contradictie." } { "Tag": [ "geometry", "inequalities", "email", "inequalities open" ], "Problem": "Prove that : \r\n $ 3a^2 +2(b^2+c^2) \\geq 16.S $ , where $ S $ is the area of triangle $ ABC $ \r\n Question : Generalize this problem .", "Solution_1": "One of the strongest inequalities of this type is the following:\r\n\r\n $ 2(ab+bc+ca)-(a^2+b^2+c^2)\\geq 4S\\sqrt{3}$.\r\n This was given by Hadwiger-Finsler in 1937.\r\n In Arhimede Magazine,no.9-10/2003, I gave a new proof for this just using AM-GM. ;)", "Solution_2": "A 'new proof'? It is just Muirhead if we substitute $a=x+y$ and so on, or Schur if we square it and expand.", "Solution_3": "[Moderators, please leave this post here - I need to communicate with this user; as you'll see below, this is the only way.]\r\n\r\nspdf - please change your email in your profile. You have been subscribing to topics with a bad email in your profile, so all your topic watch emails are bouncing. I tried PMing, but you aren't reading your PM's either. I have deleted your current watched topics. Please do not watch any more topics or forums until you fix your email.", "Solution_4": "to Blahblahblah: No,Schur,or Muirhead. they are too heavy for the simplity of this inequality.If you want i'll post my solution.\r\n\r\n By the way what's wrong with my email? :? (I don't know if you are talking to me) ;)\r\n And by the way ,i'm reading my Private messages almost everyday. ;)\r\n\r\n\r\n I hate Muirhead. :mad:", "Solution_5": "This is what I meant:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=195906&highlight=#195906\r\n\r\nIt is 'muirhead', but really just a trivial application of AM-GM.", "Solution_6": "Excuse me, Blahblahblah, but my solution is not in all these posts and it is much more easier. ;)", "Solution_7": "Good for you :!:", "Solution_8": "Your \"well-known\" ineq doesn't have a connection with my problem , Cezar ! \r\n There are many ways to solve your ineq ( as blahblahblah showed ) . We don't discuss that ineq here .", "Solution_9": "Here is the solution : \r\n First , prove that : $ b^2 + c^2 \\geq \\frac{a^2}{2} + 2h_a^2 $ \r\n After that , prove that : $ 4(a^2 + h_a^2) \\geq 16S $ \r\n The first ineq can be proved by drawing the attitude from A and using Pythagore's theorem \r\n I don't know whether we can generalise this problem !", "Solution_10": "No one interests this problem :( ?" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Prove that for all a,b,c >0,a+b+c=1,we have\r\n$\\frac{a^{2}(2a+1)}{(4a+b)(4a+c)}+\\frac{b^{2}(2b+1)}{(4b+c)(4b+a)}+\\frac{c^{2}(2c+1)}{(4c+a)(4c+b)}\\leq{\\frac{1}{5}}$\r\n:)", "Solution_1": "$\\frac{a^{2}(2a+1)}{(4a+b)(4a+c)}+\\frac{b^{2}(2b+1)}{(4b+c)(4b+a)}+\\frac{c^{2}(2c+1)}{(4c+a)(4c+b)}\\leq{\\frac{1}{5}}\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}\\left(\\frac{a}{5}-\\frac{a^2(2a+1)}{(4a+b)(4a+c)}\\right)\\geq0\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}\\frac{a(16a^2+4a(1-a)+bc-10a^2-5a)}{5(4a+b)(4a+c)}\\geq0\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}\\frac{a(2a^2-a+bc)}{(4a+b)(4a+c)}\\geq0\\Leftrightarrow\\sum_{cyc}\\frac{a(2a^2-a(a+b+c)+bc)}{(4a+b)(4a+c)}\\geq0\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}\\frac{a(a-b)(a-c)}{(4a+b)(4a+c)}\\geq0.$\r\nLet $a\\geq b\\geq c.$ Then $\\sum_{cyc}\\frac{a(a-b)(a-c)}{(4a+b)(4a+c)}=$\r\n$=\\left(\\frac{a(a-b)(a-c)}{(4a+b)(4a+c)}-\\frac{b(a-b)(b-c)}{(4b+a)(4b+c)}\\right)+\\frac{c(c-a)(c-b)}{(4c+a)(4c+b)}=\\frac{c(c-a)(c-b)}{(4c+a)(4c+b)}+$\r\n$+\\frac{(a-b)^2(4a^2b+a^2c+4b^2a+b^2c+17abc-c^2a-c^2b)}{(4a+b)(4a+c)(4b+a)(4b+c)}\\geq0.$ :)" } { "Tag": [ "trigonometry" ], "Problem": "If $A = 20^\\circ$ and $B = 25^\\circ$, then the value of $(1+\\tan A)(1+\\tan B)$ is\r\n\r\nA. $\\sqrt{3}$\r\nB. 2\r\nC. $1 + \\sqrt{2}$\r\nD. $2(\\tan A + \\tan B)$\r\nE. None of these", "Solution_1": "[hide]\n\nnoticing that $20+25=45$, and $\\tan{45}=1$ prompts us to try this,\n(let $\\tan{x}=t$)\n\n$(1+\\tanh)(1+\\tan{45-x})=1+\\tan{x}+\\tan{45-x}+\\tan{20}\\tan{45-x}$ and using the tangent subtraction \n\nformulas, and $\\tan{x}=1$, we get,\n\n$1+\\tan{x}+\\tan{45-x}+\\tan{20}\\tan{45-x}=1+t+\\frac{1-t}{1+t}+\\frac{t(1-t)}{1+t}$ and simplifying gives\n\n$\\frac{2t+2}{t+1}$ and since we know that $\\tan{20}\\ne1$, we can cancel out the $t+1$\ngiving us $\\boxed{B=2}$\n\n[/hide]\r\n\r\nEDIT: woops, forgot to hide...", "Solution_2": "[hide]\n\nSince $A=20^{\\circ}$ and $B=25^{\\circ}$, we have:\n\n$(1 + \\tan A)(1 + \\tan B) = (1 + \\tan 20^{\\circ})(1 + \\tan 25^{\\circ})$\n\n$(1 + \\tan A)(1 + \\tan B) = 1 + \\tan 20^{\\circ} + \\tan 25^{\\circ} + \\tan 20^{\\circ} \\cdot \\tan 25^{\\circ}$ $( \\ast )$\n\nBut\n\n$\\tan (20^{\\circ} + 25^{\\circ}) = \\frac{\\tan 20^{\\circ} + \\tan 25^{\\circ}}{1 - \\tan 20^{\\circ} \\cdot \\tan 25^{\\circ}} \\Leftrightarrow$\n\n$\\tan 45^{\\circ} = \\frac{\\tan 20^{\\circ} + \\tan 25^{\\circ}}{1 - \\tan 20^{\\circ} \\cdot \\tan 25^{\\circ}} \\Leftrightarrow$\n\n$1 = \\frac{\\tan 20^{\\circ} + \\tan 25^{\\circ}}{1 - \\tan 20^{\\circ} \\cdot \\tan 25^{\\circ}} \\Leftrightarrow$\n\n${1 - \\tan 20^{\\circ} \\cdot \\tan 25^{\\circ}} = \\tan 20^{\\circ} + \\tan 25^{\\circ}$.\n\nThus \n\n${1 = \\tan 20^{\\circ} + \\tan 25^{\\circ} + \\tan 20^{\\circ} \\cdot \\tan 25^{\\circ}}$.\n\nApplying this result in $( \\ast )$, we get:\n\n$(1 + \\tan A)(1 + \\tan B) = 1 + \\tan 20^{\\circ} + \\tan 25^{\\circ} + \\tan 20^{\\circ} \\cdot \\tan 25^{\\circ} = 1 + 1 = 2$.\n\n[/hide]", "Solution_3": "[hide=\"Answer\"]$(1+\\tan 20)(1+\\tan 25)=1+\\tan 20+\\tan 25+\\tan 20(\\tan 25)=x$\n$\\tan (20+25)=\\frac{\\tan 20+\\tan 25}{1-\\tan 20(\\tan 25)}$\n$x-1-\\tan 20(\\tan 25)=\\tan 20+\\tan 25$\n\nIf $x-1=1$, then we have $1-\\tan 20(\\tan 25)=\\tan 20+\\tan 25\\Rightarrow \\frac{\\tan 20+\\tan 25}{1-\\tan 20(\\tan 25)}=\\tan 45=1$, which fits, so we have $x-1=1\\Rightarrow x=2\\Rightarrow \\boxed{B}$.[/hide]", "Solution_4": "[hide=\"Hopefully this is a different solution from everyone elses...\"]\n$(1+\\tan A)(1+\\tan B)=1+\\tan A + \\tan B + \\tan A \\cdot \\tan B$\n$\\tan (A + B) = \\frac{\\tan A + \\tan B}{1-\\tan A \\cdot \\tan B}$\n$\\tan (45^\\circ) = \\frac{\\tan A + \\tan B}{1-\\tan A \\cdot \\tan B}$\n$1 = \\frac{\\tan A + \\tan B}{1-\\tan A \\cdot \\tan B}$\n$\\tan A + \\tan B = 1-\\tan A \\cdot \\tan B$\n$\\tan A + \\tan B + \\tan A \\cdot \\tan B =1$\n$1 + \\tan A + \\tan B + \\tan A \\cdot \\tan B ={\\Huge \\textbf{2}}$ :D\n[/hide]\r\n\r\nedit: ah, same thing as everyone else :roll:", "Solution_5": "[quote=\"Silverfalcon\"]If $A = 20^\\circ$ and $B = 25^\\circ$, then the value of $(1+\\tan A)(1+\\tan B)$ is\n\nA. $\\sqrt{3}$\nB. 2\nC. $1 + \\sqrt{2}$\nD. $2(\\tan A + \\tan B)$\nE. None of these[/quote]\r\n\r\n[hide]\n$1 + \\tan{A} + \\tan{B} + \\tan{A}\\tan{B} = x$\n$\\tan{A + B} = \\frac{\\tan{A} + \\tan{B}}{1 - \\tan{A}\\tan{B}}$\n$\\tan{45}(1 - \\tan{A}\\tan{B}) = \\tan{A} + \\tan{B}$\n$(1 - \\tan{A}\\tan{B} + 1 + \\tan{A}\\tan{B} = x$\n$2$\nis the answer.[/hide]", "Solution_6": "Excuse me, but posting same solutions before checking if that was posted before seems too pathetic. :spam:" } { "Tag": [ "analytic geometry" ], "Problem": "Given a 400 by 400 chessboard and a line on that cheesboard, what is the maximum amount of squares that the line could go through?", "Solution_1": "Give the spaces coordinates ranging from (1,1) to (400,400).\r\n\r\nWe move into a new square whenever either the x coordinate changes by 1 or the y coordinate changes by 1.\r\n\r\nThere are a maximum of 399 possible changes in each of these two dimensions. Since we must include the starting square, the maximum number of squares the line can pass through is:\r\n\r\n$399+399+1=\\boxed{799}$\r\n\r\nThis is indeed doable. Connect the lower left and upper right endpoints, and move the resulting line up by half a unit.", "Solution_2": "OMG I didn't think of using the coordinate plane.", "Solution_3": "So for a N by N cheesboard, that lain may at most pass through exactly 2N-1 squares? But then what happens if the board is N by 2N or 3N?", "Solution_4": "For the most part, its $length+width-1$, because there are $length-1$ vertical changes and $width-1$ horizontal changes, and we add $1$ for the starting square." } { "Tag": [ "trigonometry" ], "Problem": "If $ \\triangle{ABC}$ and $ \\triangle{DEF}$ are such that\r\n\\[ \\sin A\\sin B\\sin C \\equal{} \\sin D\\sin E\\sin F\r\n\\]\r\n\r\n\\[ \\cos A\\cos B\\cos C \\equal{} \\cos D\\cos E\\cos F\r\n\\]\r\nthen prove the two triangles are similar.\r\n\r\nI came up with this problem while looking at another. The solution I found isn't very messy, but I was wondering whether there is a very clean, insightful solution...\r\n\r\nEDIT: Fixed! Thanks, Daan.", "Solution_1": "$ \\sin{A} \\sin{B} \\sin{C}$ right?" } { "Tag": [], "Problem": "Prove that $4x^3-2x^2-15x+9$ and $12x^3+6x^2-7x+1$ has three distinct real roots", "Solution_1": "[hide]if u mixed up the 9 and the 1 it would be easy, but nevertheless the idea is that if $p_1(x)$ is the first and $p_2(x)$ the second polynom, then $ap_1(x_0)+bp_2(x_0)=0$ for all real $a,b$, $x_0$ being the shared root. if all solutions of $x_0$ are not root of $p_1$ or $p_2$ then we have a contadiction, and there are no shared solutions.[/hide]", "Solution_2": "I don't understand what you said. Could you elaborate?", "Solution_3": "lets say they share one root $x_0$ then $p_1(x_0)=0$ and $p_2(x_0)=0$ therefore for every real $a,b$: $a\\cdot p_1(x_0)+b\\cdot p_2(x_0)=0$. then choose $a$ and $b$ so that something disappears. (e.g. $a=-1$ and $b=9$) then u only add up the polynoms and find its roots and see if any of its roots would match so that it is also a root of $p_1$ and $p_2$ ...\r\npeeta" } { "Tag": [ "LaTeX", "vector", "number theory", "\\/closed" ], "Problem": "See the second post in this thread.\r\n\r\nI got this while posting in the Number Theory section. Some part of the post was substituted by something else (I had some $ \\text{\\LaTeX}$ error first, but surely the output was not what I typed).\r\n\r\nIt also caused severe html problems, making the site look rather weird.", "Solution_1": "tjhance: this means that the mentioned field extension has degree 2, meaning that $ \\mathbb Q[quote: 375f7f5700 \\equal{} \"t0rajir0u\"]thbb Q$ vector space.\r\n\r\n[quote=\"t0rajir0u\"]Alternately, $ x^2 \\minus{} x \\minus{} 1$ is not a cyclotomic polynomial.[/quote]\r\n...", "Solution_2": "And here is the original message (see the obvious error in line 1^^):\r\n\r\n[code]\ntjhance: this means that the mentioned field extension has degree 2, meaning that $\\mathbb Q[\\phi}$ is 2-dimensional $\\mathbb Q$ vector space.\n\n[quote=\"t0rajir0u\"]Alternately, $x^2 - x - 1$ is not a cyclotomic polynomial.[/quote]\n...\n[/code]", "Solution_3": "While in this case the mistake is obviously user-generated, it might also be better handled by the computer. I will investigate this further.", "Solution_4": "It turned out to be also a parser error, but the user mistake was nice enough to point it out ;) Bug fixed now." } { "Tag": [ "abstract algebra", "vector", "search", "Galois Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "$\\mathbb{K}$ a field and $\\mathbb{L}$ an extension of $\\mathbb{K}$.\r\nSuppose there exists $x_1,..,x_n \\in \\mathbb{L}$ such that : $\\mathbb{L}=\\mathbb{K}[x_1,x_2,..,x_n]$. Prove that $dim_\\mathbb{K}(\\mathbb{L}) < +\\infty$", "Solution_1": "That's not exactlllllllllllllly waht you mean :D", "Solution_2": "Is $\\mathbb{L}$ supposed to be a field\u00bf Then it's very well know ;)", "Solution_3": "yes it is well known. But the only proof I found with a friend use Hilbert zeroes theorem. Is there a simpler way to do this ?", "Solution_4": ":? \r\nAs I see that, you just claim that adjoining finetely many algebraic numbers will give an extension of finite degree, which is rather obvious.", "Solution_5": "That's not it at all- $\\mathbb{L}$ doesn't have to be an algebraic extension. We want to show that if some of those $x_n$ are transcendental, the ring they generate isn't the same as the field they generate.", "Solution_6": "Now why wouldn't this work?\r\n\r\n$\\mathbb L=\\mathbb K[x_1,\\ldots,x_n]$ is a finitely generated module over $\\mathbb K$, which is, of course, Noetherian, so $\\mathbb L$ is a Noetherian $\\mathbb K$-module. An infinite-dimensional $\\mathbb K$-vector space cannot be a Noetherian module though, because if infinitely many elements $\\{e_n\\}_n\\subset\\mathbb L$ form a linearly independent set, then the chain $\\mathbb Ke_1\\subsetneq\\mathbb Ke_1+\\mathbb Ke_2\\subsetneq\\ldots$ is a stirctly increasing chain of $\\mathbb K$-submodules of $\\mathbb L$, contradicting $\\mathbb L$'s quality of being Noetherian.", "Solution_7": "Since $\\mathbb{K}$ is a field, a finitely generated module is a finite-dimensional vector space and you're assuming the conclusion.\r\n\r\nThis is a pretty subtle problem.", "Solution_8": "Yeah, in my reasoning, I foolishly mistook $\\mathbb K[\\{x_i\\}]$ for the submodule of $\\mathbb L$ generated by $x_i$. Sorry. :blush:", "Solution_9": "[quote=\"jmerry\"]That's not it at all- $\\mathbb{L}$ doesn't have to be an algebraic extension. We want to show that if some of those $x_n$ are transcendental, the ring they generate isn't the same as the field they generate.[/quote]\r\n\r\nI actually did the same mistake of interpretation. I remember a proof for this in the particular real case (taking $\\mathbb{L}=\\mathbb{R}$ and $\\mathbb{K}$ any subfield such that the existence of the $x_{i}$ is satisfied; in this case we can prove that $\\mathbb{K}=\\mathbb{R}$).\r\n\r\nThe proof uses some developments in Galois Theory (absolutely nontrivial). You can find it back by running a search on the french newsgroup fr.sci.maths (Put my name as an author, and look for \"dimension finie\"), and maybe you can adapt it to the more general case above." } { "Tag": [ "real analysis", "calculus", "integration", "topology", "real analysis unsolved" ], "Problem": "If f is measurable on [a,b] and $ \\int_{[a,c]} f\\equal{}0$ for all c0$ on a set $ E$ of positive measure, we can find a closed set $ F\\subset E$ with $ m(F)>0$ and $ \\int_{F}f>0$ , now let $ O\\equal{}(a,b)\\setminus F$ \r\n\r\nwe have $ 0\\equal{}\\int_{a}^{b}f\\equal{}\\int_{O}f\\plus{}\\int_{F}f$ hence $ \\int_{O}f\\equal{}\\minus{}\\int_{F}f \\neq 0$. Since $ O$ is open then $ O\\equal{}\\bigcup_{n}(a_{n},b_{n})$ of disjoint intervals \r\n\r\nsince $ f$ is integrable then $ \\int_{O}f\\equal{}\\sum_{n}\\int_{(a_{n},b_{n})}f \\neq 0$ then there exists $ n$ st : $ \\int_{(a_{n},b_{n})}f \\neq 0$\r\nbut we have $ \\int_{(a_{n},b_{n})}f\\equal{}\\int_{(a,b_{n})}f\\minus{}\\int_{(a,a_{n})}f \\neq 0$ then either $ \\int_{(a,b_{n})}f\\neq 0$ or $ \\int_{(a,a_{n})}f\\neq 0$. In any case we have if $ f$ is positive on a set of positive measure we can find $ x\\in [a,b]$ st : $ \\int_{(a,x)}f\\neq 0$ . \r\n\r\nSimilarly for the case $ f$ is negative on a set of positive measure. Thus we have a contradiction" } { "Tag": [ "number theory solved", "number theory" ], "Problem": "Prove that there exist infinitely many n such that the denominator of the irreducible fraction 1+1/2+...+1/n is not a prime power. I don't know how to solve this one. I've tried to take n a prime, so that the denominator should be divisible by p, but I don't know how to continue.", "Solution_1": "Take n do be prime >5 then 1+1/2+...+1/n=(n!/1+n!/2+...+n!/n)/n! . \r\nAll the terms, but n!/n, of the sum in the numerator are 0 mod p. So the numerator is not 0 mod p. If we want it to be a prime power it must be n, so we should have that (n-1)! divides n!/1+n!/2+...+n!/n. \r\nTake this for granted: there is a prime number n/22 , it must have been n.[/b] (it realy was not important)\r\n\r\nBut taking a prime n/2b_n] \\equal{}0$\r\n2/$ 1/b_n.\\sum_{i\\equal{}1}^n E[X_i .1_{|X_i|a}]$ tends towards $ 0$ as $ a$ tends to $ \\infty$ (umiform integrability). Then, for a fixed $ \\epsilon$, you can introduced $ X^{(a)}_n \\equal{} X 1_{|X| \\epsilon) \\leq P(| \\frac{\\sum_1^n X^{(a)}_n}{n } | > \\frac{\\epsilon}{2})\\plus{}P(| \\frac{\\sum_1^n Y^{(a)}_n}{n } | > \\frac{\\epsilon}{2})$\r\nit should not be hard to conclude. Do you see how ?", "Solution_6": "[quote=\"Mr.Doe\"] such that for any given $ n\\in\\mathbb{N}, \\lim_{a\\rightarrow \\infty}\\max_{1\\leq k\\leq n}\\int_{|x|\\geq a}|x|d\\mathbb{F}_{X_k}(x) \\equal{} 0$ [/quote]\r\nThis is not a condition at all; it merely means that each variable has finite expectation, which does not really allow you to conclude anything. You need some uniformity in $ n$ for the statement to be true." } { "Tag": [], "Problem": "Okay. So I guess I will start.\r\n\r\nDuring the 1st week of high school in early August, I remember when some upperclassmen were in the hallway, and as I was walking to class after lunch, a short, sturdy boy suddenly jumped right in front of me and let out a terribly feminine scream. As a result, I dropped my Spanish book, and it went clunk. Then, people around him burst out with laughter. \r\n\r\nMine's lame, but I'm sure a lot of you have much better ones.", "Solution_1": "I was rather late to class, because my locker was jammed (the school gives us half lockers and still expects us to store all of our textbooks, notebooks, backpacks, coats... but that's another tangent) and my locker was upstairs and my class was downstairs. So I was running, taking the stairs two at a time, and caught my heel on one or tripped over my own feet or something and suddenly I was tumbling down the stairs (of course I did this at the very top of them) and landed pretty much on my head at the bottom, with my books and papers scattered all around me. To make matters worse, a PE class was walking outside to the fields and saw it all happen... I won't ever forget that.", "Solution_2": "Ok,\r\n\r\nSo I was trying go to a math class, but the door was locked. So I knocked and waited for about 5 minutes. I tried knocking again and no one came to the door. Then I ran my head into the door, and at that moment the door opened, and I barely missed the teacher, ran into the podium and fell down.", "Solution_3": "Hmm...\r\n\r\nOnce in 4th grade, I was having recess (in case you forgot :P , that's when you take a break and go on the play ground), and my pants literally ripped in front of 15 people. I had to go up to the office and wait for my mom to bring me another pair of pants.\r\n\r\nField trips: In second grade, it sucked. We had to go to the BFI, to learn about recycling and stuff. It was stinky.\r\nLuckily, we also got to go to an ice cream factory, that was cool.\r\n\r\nScience camp was cool. I injured my leg on the last day. In my cabin, there was vandalism all over, saying stuff like \"Seth loves Cathy\" and \"Cathy is a lesbion\". Coincidentally, on one of the beds was the name of one of my cabin mates. We were all joking about how someone was stalking him. One of my cabin members was really hyper, and started rolling around in his sleeping bag and was shouting \"I'm a caterpillar!\". For if I got bored, I brought a science magazine. There was an ad with a picture of an old grandma, which I accidentally called grandpa. We were cracking up about it the whole week.", "Solution_4": "yo\r\n\r\nthe other day in PE we were doing this thing where you have to get everybody across a rope swing and get a bucket of water across without a drop spilling which kinda requires you to hold onto the rope with one hand\r\n\r\nand i'm somewhat athletic so i volunteered to take it across\r\n\r\nso i get it to the other side and somebody takes the bucket of water while i'm still on the rope, and it's all cool\r\n\r\nand i swing back to the other side while straddling the rope, and on the other side somebody gets the bright idea to pull me to safety by pulling the rope\r\n\r\nwhich is between my legs\r\n\r\npain", "Solution_5": "One of my friends (he's a user here, but I won't name him) was in China or something at the beginning of his freshman year in high school, so he missed the first week or two of our chemistry class (I was in 10th grade and in the same chem class). So he missed the introductory lab where we learned about all the lab equipment and did some basic things, and he had to come after school to do it. He didn't know what any lab equipment was, since he also missed the class. The teacher was doing something at the front of the room and told him to light the Bunsen burner (which she had left out on the counter in the lab for him). But he didn't know what it was... the teacher turned around a minute later and saw him trying to light the $ \\$$2000 digital balance on fire..." } { "Tag": [ "trigonometry", "inequalities", "geometry proposed", "geometry" ], "Problem": "Prove that\r\n\r\n\r\n$ \\frac {\\cos^2 A}{a^2} \\plus{} \\frac {\\cos^2 B}{b^2} \\plus{} \\frac {\\cos^2 C}{c^2}\\geq\\frac {1}{4R^2}$", "Solution_1": "We have $ \\sum\\frac{\\cos^2 A}{a^2}\\equal{}\\sum\\frac{1\\minus{}\\sin^2 A}{a^2}\\equal{}\\sum\\frac{1\\minus{}\\frac{a^2}{4R^2}}{a^2}\\equal{}\\sum\\frac{4R^2\\minus{}a^2}{4R^2a^2}$\r\n\r\nThus, the inequality reduces to : $ \\sum\\frac{4R^2\\minus{}a^2}{a^2}\\ge 1\\ \\Longleftrightarrow\\sum\\frac{4R^2}{a^2}\\ge 4\\ \\Longleftrightarrow\\ \\frac{1}{a^2}\\plus{}\\frac{1}{b^2}\\plus{}\\frac{1}{c^2}\\ge \\frac{1}{R^2}$\r\n\r\nThis results from [b]Cauchy-Schwarz[/b] inequality and $ a^2\\plus{}b^2\\plus{}c^2\\le 9R^2\\ :$ $ \\frac{1}{a^2}\\plus{}\\frac{1}{b^2}\\plus{}\\frac{1}{c^2}\\ge \\frac{9}{a^2\\plus{}b^2\\plus{}c^2}\\ge\\frac{1}{R^2}$ .", "Solution_2": "Prove that one stronger inequality $ \\boxed {\\ \\frac {\\cos^2A}{a^2} \\plus{} \\frac {\\cos^2B}{b^2} \\plus{} \\frac {\\cos^2C}{c^2}\\ \\ge\\ \\frac {1}{4R^2}\\cdot \\left(\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{4S\\sqrt 3}\\ \\right)^2\\ }\\ \\ge\\ \\frac {1}{4R^2}$ .", "Solution_3": "To virgil nicula.\r\n\r\n${ \\left(\\frac{a^{2}+b^{2}+c^{2}}{4S\\sqrt 3}^{2}\\ \\right)^{2}\\ }$\r\nwhere does the intermediate square work?\r\ni think you mistyped the square sign..", "Solution_4": "[quote=\"Virgil Nicula\"]Prove that one stronger inequality $ \\boxed {\\ \\frac {\\cos^2A}{a^2} \\plus{} \\frac {\\cos^2B}{b^2} \\plus{} \\frac {\\cos^2C}{c^2}\\ \\ge\\ \\frac {1}{4R^2}\\cdot \\left(\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{4S\\sqrt 3}\\ \\right)^2\\ }\\ \\ge\\ \\frac {1}{4R^2}$ .[/quote]\r\n[b][u]Proof[/u].[/b] $ 4R^2\\cdot\\sum \\frac {\\cos^2A}{a^2} \\equal{}$ $ \\sum\\cot^2A \\equal{} \\sum\\left(\\frac {b^2 \\plus{} c^2 \\minus{} a^2}{4S}\\right)^2\\ge$ $ \\frac {1}{16S^2}\\cdot\\frac 13\\cdot\\left[\\sum\\left(b^2 \\plus{} c^2 \\minus{} a^2\\right)\\right]^2 \\equal{} \\left(\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{4S\\sqrt 3}\\right)^2\\ge 1$ .", "Solution_5": "[quote=\"Virgil Nicula\"][quote=\"Virgil Nicula\"]Prove that one stronger inequality $ \\boxed {\\ \\frac {\\cos^2A}{a^2} \\plus{} \\frac {\\cos^2B}{b^2} \\plus{} \\frac {\\cos^2C}{c^2}\\ \\ge\\ \\frac {1}{4R^2}\\cdot \\left(\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{4S\\sqrt 3}^2\\ \\right)^2\\ }\\ \\ge\\ \\frac {1}{4R^2}$ .[/quote]\n[b][u]Proof[/u].[/b] $ 4R^2\\cdot\\sum \\frac {\\cos^2A}{a^2} \\equal{}$ $ \\sum\\cot^2A \\equal{} \\sum\\left(\\frac {b^2 \\plus{} c^2 \\minus{} a^2}{4S}\\right)^2\\ge$ $ \\frac {1}{16S^2}\\cdot\\frac 13\\cdot\\left[\\sum\\left(b^2 \\plus{} c^2 \\minus{} a^2\\right)\\right]^2 \\equal{} \\left(\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{4S\\sqrt 3}\\right)^2\\ge 1$ .[/quote]\r\n\r\n\r\nDear friend [color=red][b]Virgil [/b][/color]who is the author of the inequality? where is published? thanks", "Solution_6": "Thank you, [b]Liguras[/b] for your words ! This stronger inequality is mine. But I think that it is easily and at most interesting ...", "Solution_7": "[quote=\"Virgil Nicula\"]Thank you, [b]Liguras[/b] for your words ! This stronger inequality is mine. But I think that it is easily and at most interesting ...[/quote]\r\n\r\nIs very nice and stronger.... Congratulation my friend :wink:", "Solution_8": "[quote=\"Ligouras\"][quote=\"Virgil Nicula\"]Thank you, [b]Liguras[/b] for your words ! This stronger inequality is mine. But I think that it is easily and at most interesting ...[/quote]Is very nice and stronger.... Congratulation my friend :wink:[/quote]\r\n[b]Thank you ![/b] I consider that this inequality is my present for you in these days. [b]Happy New Year ![/b]", "Solution_9": "[quote=\"Virgil Nicula\"][quote=\"Ligouras\"][quote=\"Virgil Nicula\"]Thank you, [b]Liguras[/b] for your words ! This stronger inequality is mine. But I think that it is easily and at most interesting ...[/quote]Is very nice and stronger.... Congratulation my friend :wink:[/quote]\n[b]Thank you ![/b] I consider that this inequality is my present for you in these days. [b]Happy New Year ![/b][/quote]\r\n\r\n\r\nit is a honor for me dear friend.... :oops: \r\nwith great respect\r\n \r\n[i][b]\"[color=blue]T[/color][color=red]\u00e0[/color][color=blue]kis[/color]\"[/b][/i]", "Solution_10": "$ \\frac {\\cos^2 A}{a^2} \\plus{} \\frac {\\cos^2 B}{b^2} \\plus{} \\frac {\\cos^2 C}{c^2}\\geq\\frac {1}{4R^2} \\iff cotA+cotB+cotC\\ge 1.$" } { "Tag": [], "Problem": "[i]\u0395\u03bb\u03c0\u03af\u03b6\u03c9 \u03b1\u03c5\u03c4\u03ae \u03b7 \u03b1\u03c3\u03ba\u03b7\u03c3\u03bf\u03cd\u03bb\u03b1 \u03bd\u03b1 \u03b1\u03c1\u03ad\u03c3\u03b5\u03b9 :[/i]\r\n\r\n\u0395\u03be\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03ac \u03b5\u03bd\u03cc\u03c2 \u03bf\u03c1\u03b8\u03bf\u03b3\u03c9\u03bd\u03af\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 $ ABC$ \u03bc\u03b5 $ \\angle A = 90$ \u03b8\u03b5\u03c9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf $ BCD$ \u03bc\u03b5 $ AB=BD$.\r\n\r\n\u0391\u03bd $ AH \\perp BC$ \u03ba\u03b1\u03b9 $ O$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03c0\u03b5\u03c1\u03af\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf \u03c4\u03bf\u03c5 $ \\triangle BCD$ , \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9 $ DH \\perp BO$\r\n\r\n \u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2", "Solution_1": "Afou $ BH\\cdot BC \\equal{} BA^2 \\equal{} BD^2$, i $ BD$ einai efaptomeni tou perigegrammenoun kiklou tou $ \\triangle HDC$ opote $ \\widehat{BDH} \\equal{} \\widehat{BCD}$ (1).Eukola fainetai twra oti i $ \\widehat{BCD}\\leq\\widehat{BCA}$ ara einai panta okseia ara $ \\widehat{DBO} \\equal{} 90^\\circ \\minus{} \\widehat{BCD}$ (2). Apo tis (1) kai (2) exoume oti $ DH\\perp BO$.", "Solution_2": "M\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9:\r\n\r\n\u0395\u03af\u03bd\u03b1\u03b9 \u03b3\u03bd\u03c9\u03c3\u03c4\u03cc (\u03b1\u03c0\u03cc 2\u03bf \u03b8\u03ad\u03c9\u03c1\u03b7\u03bc\u03b1 \u03b4\u03b9\u03b1\u03bc\u03ad\u03c3\u03c9\u03bd) \u03cc\u03c4\u03b9 \u03b1\u03c1\u03ba\u03b5\u03af $ BD^2\\minus{}BH^2\\equal{}OD^2\\minus{}OH^2$.\r\n\r\n\u03a0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9:\r\n\r\n$ BD^2\\minus{}BH^2\\equal{}AD^2\\minus{} BH^2\\equal{}AH^2\\equal{}BH\\cdot HC\\equal{}R^2\\minus{}OH^2\\equal{}OD^2\\minus{}OH^2$. (R \u03b7 \u03b1\u03ba\u03c4\u03af\u03bd\u03b1 \u03c4\u03bf\u03c5 \u03c0\u03b5\u03c1\u03b9\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c5 \u03c4\u03bf\u03c5 $ BCD$)." } { "Tag": [ "AMC", "AIME", "MATHCOUNTS", "AMC 10", "geometry", "calculus", "3D geometry" ], "Problem": "how did the 2007 AIME go?", "Solution_1": "lol i got 3. :oops:", "Solution_2": "i hope AIME 2 is easy because i am taking that, because march 13 we are on spring break. :D", "Solution_3": "i wish we took the 2. i would do 7hrs a day of AIME and then get like 7 or 8 on the test. but no, i have to take it from 8-11(it's like 7-10 because of the time change) on March 13th. rawr", "Solution_4": "i was wondering if anyone knew,about the index. see, i here you add your 12 score with your 10 score or something and then add your AIME score. but you see i qualified for AIME through 10, i didn't even take 12, but i here something about a floor score for people under 10th grade or something so now i am really confused. could someone explain this to me.", "Solution_5": "OK, i will assume you are under 10th grade.\r\n\r\nyour index is:\r\n\r\nAMC 10 score + 10*AIME score.\r\n\r\nthe \"floor\" is suppose the lowest index that qualified was 210 and that person got 150/6.\r\n\r\nthen if you scored 6+, no matter what your AMC score was, you qualify for USAMO.\r\n\r\n\r\n(mathcountsfreak) any other scores?", "Solution_6": "Didn't make it!\r\nI was hoping on doing better on 10B, but then my dad woke me up late, and I had 30 minutes less time to do the contest, so I had to think about speed a little more, and even so I had to guess on around 9.", "Solution_7": "[quote=\"mathcountsfreak\"]Didn't make it!\nI was hoping on doing better on 10B, but then my dad woke me up late, and I had 30 minutes less time to do the contest, so I had to think about speed a little more, and even so I had to guess on around 9.[/quote]\r\n\r\nthat's gotta hurt...but at least you don't live in a retarded school district and you got to take both. What I'd give to take the 10b also, especially since i made a careless mistake on the 10a, which kept me from getting a 141...i was soooo close!\r\n\r\nP.S. i...[color=red][size=150]hate[/size][/color]...compliment....counting!!!! :cursing:", "Solution_8": "hahahahahahahahahahahaha!!!!!!!!! :rotfl:", "Solution_9": "lol -_- AIME was sucky :D", "Solution_10": "Mathcrazed please post the url of the indiana forum on your signature.", "Solution_11": "did i read something wrong or did someone say west lafayette doesn't allow people to take the 10B/12B?", "Solution_12": "omg, seriously! we r allowed to, technically, but our school hates anyone who goes against the flow. they hadted the whole Benjamin family cuz they went above and beyond. west lafayette is sooo lazy!", "Solution_13": "so they hate and don't do anything about it.. i really don't care..?", "Solution_14": "Can you take both 10 and 12 Undefined(do you mind if i use your name, everyone probably knows who you are by now.), do you need to ask Mr. A or something.", "Solution_15": "nice! u were lucky.", "Solution_16": "ok\r\nne ways, if i wanted to skip to high school, i could (easily), but that's only because my dad teaches me math (he's a college professor)... so... ya... could've easily gotten 9 on the AIME, makes me feel so stupid... :mad: :mad: :mad:", "Solution_17": "heh my dad's a nuclear engineer.. and kinda leaves me alone, so yeah. i look things up in wolfram mathworld and wikipedia and stuff. i get along..", "Solution_18": "[quote=\"undefined117\"]heh my dad's a nuclear engineer.. and kinda leaves me alone, so yeah. i look things up in wolfram mathworld and wikipedia and stuff. i get along..[/quote]\r\n\r\ni thought u said ur dad was lecturing u about calculus when i gave u the problem w/the sphere and cylinder. i'm not going to lie. my dad teaches me....", "Solution_19": "uhhuh.. your dad teaches you.. my dad talks/writes some stuff down on paper and leaves me to contemplate.", "Solution_20": "maybe that's why ur so good at contemplating :lol: and i'm not. i get frustrated really easily...", "Solution_21": "asian superiority (pwnage :D...) indian=asian, seriously... :D", "Solution_22": "hey, like everyone on the indiana forum is asian (except for owna i think). no wonder indiana owns (actually that isn't the main reason, but oh well :lol: )", "Solution_23": "owna is asian", "Solution_24": "No, actually indians do not equal asians (to be more specific, ill tell you next saturday..)", "Solution_25": "[quote=\"usaha\"]No, actually indians do not equal asians (to be more specific, ill tell you next saturday..)[/quote]\r\n\r\ni take it ur indian (ur name sounds indian)?", "Solution_26": "yes he is.", "Solution_27": "umm,.. india is in asia, am i correct there?", "Solution_28": "yeah but usually when people say asian they mean korean/filipino/japanese/chinese", "Solution_29": "OMG if india isn't part of asia then is it antarctican or something." } { "Tag": [ "limit", "logarithms", "geometric sequence", "calculus", "calculus computations" ], "Problem": "Prove that \\[ \\gamma\\equal{}\\lim_{n\\to\\infty}\\left[\\left(\\frac{2^m}{1}\\plus{}\\frac{3^m}{2}\\plus{}\\cdots\\plus{}\\frac{n^m}{n\\minus{}1}\\right)\\plus{}m\\minus{}(S_n^0\\plus{}S_n^1\\plus{}\\cdots\\plus{}S_n^{m\\minus{}1})\\minus{}\\ln (n\\minus{}1)\\right],\\] where $ S_n^m\\equal{}1^m\\plus{}2^m\\plus{}\\cdots\\plus{}n^m$ and $ \\gamma$ is Euler's constant.", "Solution_1": "And what about 'm'?", "Solution_2": "it's true for all $ m$\r\nit's easy you only need to check that $ \\sum_{0 \\leq i \\leq m \\minus{} 1} S_{n}^{i} \\equal{} m \\plus{} \\sum _{ 1 \\leq i \\leq n \\minus{} 1} \\frac {(i \\plus{} 1)^{m} \\minus{} 1}{i}$\r\n(use geometric progression)\r\nand in finnaly you have :\r\n$ \\lim ... \\equal{}\\lim _{n\\minus{}>\\plus{}oo} \\sum _{ 1 \\leq i \\leq n \\minus{} 1} \\frac {(i \\plus{} 1)^{m}}{i} \\plus{} m \\minus{} (m \\plus{} \\sum _{ 1 \\leq i \\leq n \\minus{} 1} \\frac {(i \\plus{} 1)^{m} \\minus{} 1}{i} ) \\minus{} \\ln (n \\minus{} 1)$\r\n =$ \\lim _{n\\minus{}>\\plus{}oo} H_ {n \\minus{} 1} \\minus{} \\ln (n \\minus{} 1) \\equal{} \\gamma$ (by definition)\r\nwhere $ H_{n} \\equal{} \\sum_{0 < i < n \\plus{} 1} \\frac {1}{i}$" } { "Tag": [ "geometry", "search", "AoPSwiki", "trigonometry", "function", "Pythagorean Theorem", "area of a triangle" ], "Problem": "What is the area of a equaladaral triangle :D", "Solution_1": "Please search this in AoPSWiki or Wolfram before asking it in a fourm.", "Solution_2": "The area of an equilateral triangle with side length $ s$ is given by $ \\frac { s^2\\sqrt {3}}{4}$.\r\n\r\nThis can be derived by dropping an altitude, and using your familiar sine and cosine functions to find the base and height of the triangle.", "Solution_3": "thanks alot for the answer", "Solution_4": "[quote=\"123456789\"]The area of an equilateral triangle with side length $ s$ is given by $ \\frac { s^2\\sqrt {3}}{4}$.\n\nThis can be derived by dropping an altitude, and using your familiar sine and cosine functions to find the base and height of the triangle.[/quote]\r\nI'm pretty sure all you need is the Pythagorean Theorem.\r\nIf you have an equilateral triangle $ ABC$, with side length $ s$, the area is as follows:\r\nDropping an altitude, you get two right triangles. They both have hypotenuse length $ s$ and base length $ \\frac{1}{2}s$. Using the Pythagorean Theorem, the square of the height of the altitude is $ s^2\\minus{}\\frac{1}{4}s^2$, so the altitude is $ \\sqrt{\\frac{3}{4}s^2}$ This can be simplified to $ \\frac{\\sqrt{3}}{2}s$. Since the area of a triangle is $ \\frac{1}{2}bh$, and the base is s, while the height is $ \\frac{\\sqrt{3}}{2}s$, the area is $ \\frac{s^2\\sqrt{3}}4$", "Solution_5": "This was on another math forum problem of the week and me and another person spent 2 hours of english class coming up with the proof for this... what a waste :P", "Solution_6": "Area of equilateral triangle? I tought the formula to find equilateral triangle is the length multiply by height /2 as it is the same for all triangles?", "Solution_7": "[quote=\"shaun ong\"]Area of equilateral triangle? I tought the formula to find equilateral triangle is the length multiply by height /2 as it is the same for all triangles?[/quote]\r\n\r\nThat is true for all triangles but what if you're only given three sides to an equliateral triangle? You won't know the height and therefore you'll have to apply different formulas for different situations.", "Solution_8": "[quote=\"7h3.D3m0n.117\"]That is true for all triangles but what if you're only given three sides to an equliateral triangle? You won't know the height and therefore you'll have to apply different formulas for different situations.[/quote]\r\n\r\nif it was equallateral and you knew the sidlengths couldnt you just use the pythagorean theorum to find the altitude because the altitude in an equallateral triangle bisects the base at a right angle. so you could take the square root of half of the base squared subtracted from the side length squared and you will have the altitude. This can be done without any knowlege of the formula $ \\frac{s^2\\sqrt{3}}{4}$. Now yes the formula does save valuable time in a countdown or a sprint so it is a good thing to know!" } { "Tag": [ "Ring Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Find the cardinality of the set of all Subrings of $\\mathbb{Q}$.\r\n\r\nIt seems like it should be simple, but I can't do it yet..", "Solution_1": "The cardinality is $|\\mathbb{R}|=2^{|\\mathbb{Q}|}$. Of course this is an upper bound, so we have to prove, that there are at least so much subrings.\r\n\r\nFor each subset $Q\\subseteq\\mathbb{P}$ the set\r\n$\\{\\frac{r}{s}| gcd(r,s)=1 \\wedge \\forall p\\in\\mathbb{P}: p|s \\Rightarrow p\\in Q\\}$\r\nis a subring. Of course these subrings are pairwise not equal.\r\n\r\nSo there are at least $2^{|\\mathbb{P}|}=|\\mathbb{R}|$ subrings of $\\mathbb{Q}$." } { "Tag": [], "Problem": "The sum of the digits of a positive three-digit palindrome is equal to the square of the product of the digits. What is the palindrome?", "Solution_1": "Let the number be $ \\overline{aba}$ so that $ 2a\\plus{}b\\equal{}a^4b^2$. It is easy to see that neither of $ a$ and $ b$ can be $ 0$, so $ a\\equal{}\\{1,2\\}$ are the only possibilities. If $ a\\equal{}2$, $ b\\equal{}1$, which doesn't work, so $ a\\equal{}1\\implies2\\plus{}b\\equal{}b^2\\implies b^2\\minus{}b\\minus{}2\\equal{}(b\\minus{}2)(b\\plus{}1)\\equal{}0\\implies b\\equal{}2\\implies\\boxed{121}$." } { "Tag": [ "Gauss", "linear algebra", "matrix" ], "Problem": "MATRIX (1,0,1)(2,2,6)(3,1,a)\r\n\r\nInverse the matrix using the gauss method.....\r\n\r\n\r\nsolve the system x+z=1\r\n 2x+2y+6z=2\r\n 3x+y+az=3", "Solution_1": "Is this what you mean?\r\n\r\n1 0 1 | 1 0 0\r\n2 2 6 | 0 1 0\r\n3 1 a | 0 0 1\r\n\r\n1 0 1 | 1 0 0\r\n0 2 4 |-2 1 0\r\n0 1 (a-3)|-3 0 1\r\n\r\n1 0 1 | 1 0 0\r\n0 2 4 |-2 1 0\r\n0 0 -2a+10| 4 1 -2\r\n\r\n...proceeding to convert the left matrix into the identity matrix, at which time your inveted matrix will be on the right.\r\n\r\n(Note that you will need to analyze two cases: $a=5$, for which the matrix won't have an inverse, and $a\\neq5$, for which it will.)" } { "Tag": [ "modular arithmetic" ], "Problem": "What is the units digit of $ 1997^{1997}$?", "Solution_1": "It is well known that these units digits of powers cycle in 4's.\r\n\r\n$ 1997^{1997} \\equiv 7^1 \\equiv 7 \\pmod 10$. 7." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c$ sides of a triangle.Prove the inequality: \r\n\r\n\r\n $a^3+b^3+c^3+9abc\\leq 2[bc(b+c)+ca(c+a)+ab(a+b)] $ ;)", "Solution_1": "hm. this seems straightforward, so I most likely made a mistake. \r\n\r\nthe inequality is untrue for any positive a, b, c. Take $a=1$, and then let $b, c$ tend to $0$. We get $1 \\le 0$. So presumably a, b, c are the sides of the triangle. If we substitute in a = x + y, etc, we get $\\sum_{\\text{sym}} x^3 + 12x^2y + 3xyz \\le \\sum_{\\text{sym}} 2x^3 + 10x^2y + 4xyz$, which is Schur.", "Solution_2": "yes,you are right.i forgot to say that $a,b,c$ are sides of a triangle.Thanks anyway. ;)" } { "Tag": [ "function" ], "Problem": "Let n be a positive integer; if the equation 2x+2y+z=n has 28 solutions in positive integers for x,y, and z, what is n?(Give 2 possibilities)\r\n\r\nAHSME 1989 No.27", "Solution_1": "Unless I'm making things too complicated, I think this should go in intermediate. [hide]Since $x$ and $y$ must be even natural numbers we can form the following generating functions for both:\n$(x^2+x^4+x^6+...)$. \nSince $z$ is any natural number we can form the following generating function:\n$(x+x^2+x^3+x^4+...)$.\n\nSo we want the exponent of the term with coefficient $28$ in the expansion of $(x^2+x^4+x^6+...)^2(x+x^2+x^3+x^4+...)$.\n\nFactoring, we have:\n$x^5(1+x^2+x^4+x^6+...)^2(1+x^2+x^3+...)$\n\nExpanding, we have:\n$x^5(1+2x^2+3x^4+4x^6+5x^8+...)(1+x+x^2+x^3+x^4+x^5+...)$\n$x^5(1+x+3x^2+3x^3+6x^4+6x^5+10x^6+10x^7+...)$\n\nNotice that the coefficients are the triangular numbers repeated twice.\nContinuing the pattern, we have $x^5(1+x+3x^2+3x^3+...28x^{12}+28x^{13}+...)$ Multiplying out (the $x^5$ term) we find the two terms $28x^{17}+28x^{18}$.\n\nSo our two answers are $17$ and $18$. I hope I didn't make any mistakes...[/hide]", "Solution_2": "[quote=\"eryaman\"]Unless I'm making things too complicated, I think this should go in intermediate. [hide]Since $x$ and $y$ must be even natural numbers we can form the following generating functions for both:\n$(x^2+x^4+x^6+...)$. \nSince $z$ is any natural number we can form the following generating function:\n$(x+x^2+x^3+x^4+...)$.\n\nSo we want the exponent of the term with coefficient $28$ in the expansion of $(x^2+x^4+x^6+...)^2(x+x^2+x^3+x^4+...)$.\n\nFactoring, we have:\n$x^5(1+x^2+x^4+x^6+...)^2(1+x^2+x^3+...)$\n\nExpanding, we have:\n$x^5(1+2x^2+3x^4+4x^6+5x^8+...)(1+x+x^2+x^3+x^4+x^5+...)$\n$x^5(1+x+3x^2+3x^3+6x^4+6x^5+10x^6+10x^7+...)$\n\nNotice that the coefficients are the triangular numbers repeated twice.\nContinuing the pattern, we have $x^5(1+x+3x^2+3x^3+...28x^{12}+28x^{13}+...)$ Multiplying out (the $x^5$ term) we find the two terms $28x^{17}+28x^{18}$.\n\nSo our two answers are $17$ and $18$. I hope I didn't make any mistakes...[/hide][/quote]\r\n\r\nok" } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "reflection", "parallelogram", "trigonometry", "power of a point" ], "Problem": "Let $ ABC$ be an acute-angled triangle, and let $ H$ be its orthocenter. Let $ D$ be the foot of the altitude from $ B$ to $ AC$, and let $ E$ be the reflection of $ A$ on $ D$. The circumcircle of triangle $ BCE$ intersects the median from $ A$ in an interior point $ F$. Prove that $ A, D, H$ and $ F$ are concyclic.", "Solution_1": "Denote by $ O$, $ O'$ the circumcenters of triangles $ ABC$ and $ BCE$, respectively. We have $ \\angle HBC\\equal{}\\angle HAD\\equal{}\\angle HEC$, it follows $ H$ lies on circles $ (BCE)$. On the other hand, we can easily (maybe well-known as well) prove that $ AOO'H$ is a parallelogram. Let $ H'$ be the reflection point of $ H$ in $ O'$, so we have quadrilateral $ AOH'O'$ is a parallelogram, too. It implies $ AH'$ passes through the middle point of $ OO'$ which is also the middle point of $ BC$. Therefore we conclude that $ AH'$ passes through $ F$, so $ HFH'\\equal{}90^{\\circ}$, i.e. points $ A$, $ D$, $ H$, $ F$ all lie on the circle with diameter $ AH$.\r\n\r\nEr, so complicated. Sorry for my unclear writing. :D", "Solution_2": "Muy f\u00e1cil ese problema, nada mas que no sabia dibujarlo xD...", "Solution_3": "[quote=\"campos\"]Let $ ABC$ be an acute-angled triangle, and let $ H$ be its orthocenter. Let $ D$ be the foot of the altitude from $ B$ to $ AC$, and let $ E$ be the reflection of $ A$ on $ D$. The circumcircle of triangle $ BCE$ intersects the median from $ A$ in an interior point $ F$. Prove that $ A, D, H$ and $ F$ are concyclic.[/quote]\r\nWe notice that $ B,H,E,C$ are concyclic as $ \\angle BHC\\equal{}\\angle BEC\\equal{}180^{\\circ} \\minus{}A$. It is well known that the radius of the cicumcircle of $ \\triangle ABC$ and $ \\triangle BHC$ are equal ($ \\equal{}R$).\r\nLet $ H',A'$ be the midpoint of $ AH,BC$ respectively and $ O'$ be the circumcenter of $ \\triangle BHC$. \r\nWe have $ AH'\\equal{}R \\cos A$ and also $ \\angle BO'A'\\equal{}A$. So, $ O'A'\\equal{}R\\cos A$.\r\nSo, $ A'O'\\equal{}AH'$ and $ AH' \\parallel O'A'$. So, $ H'O' \\parallel AA'$. But $ H'O'$ is perpendicular to the radical axis of the circles and so is $ AA'$. So, $ \\angle AHA\\equal{}90^{\\circ}$. So, $ F$ lies on the circumcircle of $ \\triangle ADH$ and we are done.", "Solution_4": "[quote]Let $ ABC$ be an acute-angled triangle, and let $ H$ be its orthocenter. Let $ D$ be the foot of the altitude from $ B$ to $ AC$, and let $ E$ be the reflection of $ A$ on $ D$. The circumcircle of triangle $ BCE$ intersects the median from $ A$ in an interior point $ F$. Prove that $ A, D, H$ and $ F$ are concyclic.[/quote]\r\n[b]Another approach:[/b]\r\nAt the very first, it is very easy to see that $ \\minus{} \\overline {DA}.\\overline {DC} \\equal{} \\overline {DH}.\\overline {DB}$. But $ \\overline {DA} \\equal{} \\minus{} \\overline {DE}$. Hence, it follows that $ \\overline {DH}.\\overline {DB} \\equal{} \\overline {DE}.\\overline {DC}\\Longrightarrow B,H,E,C$ are concyclic, which implies that $ F\\in (BHC)$ also. Let $ A_1,C_1$ be the projections of $ H$ onto $ BC, AB$, respectively. Consider the inversion through pole $ A$, power $ \\overline {AH}.\\overline {AA_1} \\equal{} \\overline {AC_1}.\\overline {AB} \\equal{} \\overline {AD}.\\overline {AC} \\equal{} k^2$. $ \\mathcal {I}(A,k^2): H\\mapsto A_1$, $ B\\mapsto C_1$, $ C\\mapsto D$. Hence $ \\mathcal {I}(A,k^2): (BHC)\\mapsto (C_1DA_1)$- which is the $ 9$- point circle wrt $ \\triangle ABC$. Thus $ \\mathcal {I}(A,k^2): F\\mapsto A'$. But $ A',B,C$ are collinear, $ \\mathcal {I}(A,k^2): BC\\mapsto (AHD)\\Longrightarrow F\\in (AHD)$. The result is lead as follow.\r\nOur proof is completed then.", "Solution_5": "Denote the points like my figure.\r\nLet $ \\{I \\} \\equal{} AM \\cap (BCE)$,by angle-chasing we have ABIC is parallelogram so $ \\overline{AI} \\equal{} 2 \\overline{AM}$\r\nWe have:$ 2\\overline{AF}. \\overline{AM} \\equal{} \\overline{AF}. \\overline{AI} \\equal{} \\overline{AE}. \\overline{AC} \\equal{} 2\\overline{AD}. \\overline{AC} \\equal{} 2\\overline{AH}. \\overline{AK}$ so $ H,F,M,K$ are concyclic so $ \\hat{HFM} \\equal{} \\hat{ADH} \\equal{} \\frac {\\pi}{2}$ then $ A,H,F,D$ are concyclic.\r\n\r\n[img]http://img37.imageshack.us/img37/9916/huhu.jpg[/img][/img]" } { "Tag": [ "inequalities", "rotation", "geometry", "parallelogram", "triangle inequality" ], "Problem": "For a triangle , Prove that\r\n$ b \\plus{} c \\geq 2m_a$", "Solution_1": "If $ X$ is the midpoint of side $ b$, and $ Y$ is the midpoint of side $ a$, then $ XY\\equal{}\\frac{c}{2}$ , $ m_a\\equal{}AY$ and $ AX\\equal{}\\frac{b}{2}$ so your inequality follows directly from a triangle inequality :)", "Solution_2": "Wow nice .......... :lol: \r\n\r\nAny other proofs? :wink:", "Solution_3": "yes :) if you rotate the triangle by $ 180^o$ around the midpoint of sida $ a$ you'll get a parallelogram with sides $ b$ and $ c$ and with diagonal $ 2 \\cdot m_a$ so again by applying triangle inequality you get the desired result", "Solution_4": "Exactly!!!! That was the one i found (after half an hour :maybe: )", "Solution_5": "Alternate method:\r\nTo prove: \r\n$ b\\plus{}c \\ge 2m_a$\r\n$ b\\plus{}c \\ge 2* 1/2 * \\sqrt{2b^2 \\plus{} 2c^2 \\minus{} a^2}$\r\n\r\n$ b^2\\plus{}c^2\\plus{}2bc \\ge 2b^2 \\plus{} 2c^2 \\minus{} a^2$\r\n\r\n$ 0 \\ge b^2\\plus{}c^2\\minus{}2bc \\minus{} a^2$\r\n\r\n$ a^2 \\ge (b\\minus{}c)^2$\r\n\r\n$ a \\ge (b\\minus{}c)$ which is true by triangle inequality\r\n\r\n\r\n\r\n[hide=\" Doubt\"]\nI dont think equality holds here? ( last inequality is strict ) .... can someone clarify?[/hide]" } { "Tag": [ "number theory", "relatively prime" ], "Problem": "Find, with proof, the unique square which is the product of four consecutive odd numbers", "Solution_1": "[hide=\"Solution\"]\nLet the numbers be $(2n-1), (2n-3), (2n-5),$ and $(2n-7)$. Also, let $4n^{2}-16n+7 = x$ and $a^{2}$ be the square we are looking for, for some integers $x$ and positive integer $a$. So we have\n\n\\begin{eqnarray*}(2n-1)(2n-3)(2n-5)(2n-7) &=& (4n^{2}-16n+7)(4n^{2}-16n+15) \\\\ &=& x(x+8) = a^{2}\\Rightarrow x =-4 \\pm \\sqrt{16+a^{2}}\\end{eqnarray*} \n\nSince $x$ must be an integer, set $16+a^{2}= k^{2}$, for some integer $k$. The only solutions in integers to this equation, in the form $(a,k)$, are $(2,-8), (4,-4),$ and $(8,-2)$. Solving gives that $a=0, \\pm 3$, so the only plausible value for $a$ is $3$. Hence, $x=1,-9$. \n\nPlugging in $x=1$ to solve for $n$ gives no solution in integers. Plugging in $x=-9$ shows that $n=2$ is the only solution for $n$. Therefore, the unique square is $(3)(1)(-1)(-3) = \\boxed{9}$. \n[/hide]", "Solution_2": "A different approach:\r\n\r\n[hide]\nLet $k_{n}= 2n+1$. Note that $k_{n+1}-k_{n}= 2 \\not | k_{n}, k_{n+1}$, so they are relatively prime. Similarly, $k_{n}$ and $k_{n+2}$ are relatively prime. $k_{n}$ and $k_{n+3}$ differ by 6, so they are not relatively prime only if both are divisible by 3.\n\nNow, consider $k_{n}k_{n+1}k_{n+2}k_{n+3}$\n\nThere are two cases. \n\n[b]Case 1:[/b] $3 \\not | k_{n}$, in which case all four $k$'s are relatively prime, so each term is either -1 or 1 (the product must be a square, so the product must be 1). However, this leaves 2 possibilities for 4 numbers, so clearly we can't have consecutive odd numbers.\n\n[b]Case 2:[/b] $3 | k_{n}$. Note that $k_{n}/ 3$ and $k_{n+3}/3$ differ by 2 and are odd, so they must be relatively prime.\n\nThen, $(9)(k_{n}/ 3) k_{n+1}k_{n+2}(k_{n+3}/3)$ is a square, implying that $(k_{n}/ 3) k_{n+1}k_{n+2}(k_{n+3}/3)$ is a square. However, since all 4 terms are relatively prime, they must all be either 1 or -1.\n\nThis leaves only one solution: $9(-1)(-1)(1)(1)=(-3)(-1)(1)(3)=9$\n\n[/hide]", "Solution_3": "nice solution Zuton. much easier than mine :P", "Solution_4": "[hide=\"another approach\"]Let the numbers be $2n-3,2n-1,2n+1$ and $2n+3$. Also let $k^{2}=(2n-3)(2n-1)(2n+1)(2n+3)=(4n^{2}-1)(4n^{2}-9)$. we have, $g=\\gcd(4n^{2}-1,4n^{2}-9)\\mid (4n^{2}-1)-(4n^{2}-9)=8$. so $g=1,2,4,$ or $8$. but since $4n^{2}-1$ and $4n^{2}-9$ are both odd, we conclude that $g=1$. so both $\\pm(4n^{2}-1)$ and $\\pm(4n^{2}-9)$ are perfect squares. Let $4n^{2}-1=\\pm x^{2}$. But this implies $(2n)^{2}\\pm x^{2}=1$. This is possible only if $n=0$ and $x=1$. so we have $k=(-3)(-1)(1)(3)=9$[/hide]" } { "Tag": [ "conics", "parabola", "geometry", "rectangle", "ratio", "calculus", "calculus computations" ], "Problem": "A negative parabola is inscribed in a rectangle, toughing two of its corners. What is the ratio of the area under the parabola to the area of the rectangle?", "Solution_1": "[hide=\"solution\"]\nIf the parabola is of height $ h$ and width $ 2w$, then the area underneath is $ \\frac {4hw}{3}$ (use $ y\\equal{}h\\minus{}\\frac{h}{w^2}x^2$ as the equation of the parabola by centering the origin on the midpoint of the bottom side of the rectangle).\nThe area of the rectangle is $ 2hw$.\nSo the ratio is $ \\frac {2}{3}$\n[/hide]" } { "Tag": [ "geometry" ], "Problem": "The total area of all the faces of a rectangular solid is $22 \\text{cm}^2$, and the total length of all its edges is $24 \\text{cm}$. Then the length in $\\text{cm}$ of any one of its internal diagonal is\r\n\r\nA. $\\sqrt{11}$\r\nB. $\\sqrt{12}$\r\nC. $\\sqrt{13}$\r\nD. $\\sqrt{14}$\r\nE. Not uniquely determined", "Solution_1": "[hide=\"Answer\"]$4(x+y+z)=24\\Rightarrow x+y+z=6$\n$2(xy+xz+yz)=22\\Rightarrow xy+xz+yz=11$\n$(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)=x^2+y^2+z^2+2(11)=36$\n$x^2+y^2+z^2=14\\Rightarrow \\sqrt{x^2+y^2+z^2}=\\sqrt{14}\\Rightarrow \\boxed{D}$[/hide]", "Solution_2": "[hide]Call the sides $a,b,c$, thus: $2bc+2ab+2ac=22$ and $4b+4a+4c=24$ so $bc+ab+ac=11$ and $a+b+c=6$. And therefore the diagonal would be $\\sqrt{a^2+b^2+c^2}$. Now you can see that $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac) \\Longrightarrow 36=a^2+b^2+c^2+2(11) \\longrightarrow \\sqrt{a^2+b^2+c^2}=\\boxed{\\sqrt{14}}$.[/hide]", "Solution_3": "[quote=\"Iversonfan2005\"][hide]Call the sides $a,b,c$, thus: $2bc+2ab+2ac=22$ and $4b+4a+4c=24$ so $bc+ab+ac=11$ and $a+b+c=6$. And therefore the diagonal would be $\\sqrt{a^2+b^2+c^2}$. Now you can see that $(a+b+c)^2=a^2+b^2+c^2-2(ab+bc+ac) \\Longrightarrow 36=a^2+b^2+c^2-2(11) \\longrightarrow \\sqrt{a^2+b^2+c^2}=\\boxed{\\sqrt{58}}$.[/hide][/quote]\r\n\r\nit should be $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=36$, not $-2(ab+bc+ac)$", "Solution_4": "[quote=\"dangvh\"][quote=\"Iversonfan2005\"][hide]Call the sides $a,b,c$, thus: $2bc+2ab+2ac=22$ and $4b+4a+4c=24$ so $bc+ab+ac=11$ and $a+b+c=6$. And therefore the diagonal would be $\\sqrt{a^2+b^2+c^2}$. Now you can see that $(a+b+c)^2=a^2+b^2+c^2-2(ab+bc+ac) \\Longrightarrow 36=a^2+b^2+c^2-2(11) \\longrightarrow \\sqrt{a^2+b^2+c^2}=\\boxed{\\sqrt{58}}$.[/hide][/quote]\n\nit should be $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=36$, not $-2(ab+bc+ac)$[/quote]\r\n\r\nOops..fixed it :blush:" } { "Tag": [], "Problem": "Hey, I am just desperate to know your opinions on what would be best to place on a three-fold poster that would easily explain thermodynamics in a user-friendly manner. And we're not talking hardcore thermo; just about the first three laws and then nicely explaining what heat is and a nice heat vs pressure explanation, also explaining how 100% efficiency cannot be achieved. So yeah, please give your advice on what you would place on the poster that'd be fun. And I'm not just a guy who's oblivious to thermodynamics, I just want to make a poster that is as enjoyable as possible to learn from for someone other than me.", "Solution_1": "Why don't you consult some high school books? You will find for sure some nice pictures and easy explanations.", "Solution_2": "Yeah, that's where I am currently at, I just figured someone might have something unique in their head." } { "Tag": [ "algebra theorems", "algebra" ], "Problem": "I'd like to learn something about optimization, because I've heard it can be useful sometimes and I know almost absolutely nothing about it. I'll be glad if any of you could give some links to texts about it [I've read about optimization on wikipedia, but there's nothing interesting] and could solve some problems with optimization.", "Solution_1": "I think you should read many books about this thing. Especially the book about CONVEX ANALYSIS of ROCKKAFELLER.\r\n\r\nGood luck!" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$\\{a,b,c\\}\\subset (0,\\infty ),\\ abc=1\\Longrightarrow (a^2+1)(b^2+1)(c^2+1)\\ge (a+1)(b+1)(c+1).$", "Solution_1": "Hi Ph-An.\r\n$(a^2+1)(b^2+1)=a^2+b^2+a^2b^2+1\\geq (ab+1)^2$=>\r\n$(a^2+1)(b^2+1)(c^2+1)\\ge (ab+1)(bc+1)(ac+1)=(a+1)(b+1)(c+1)$ :) \r\n\r\nPh-An also looking this\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61826]www.mathlinks.ro/Forum/viewtopic.php?t=61826[/url] :D :D", "Solution_2": "Just take the calculation and we will use $\\sum a^2b^2\\geq \\frac{(\\sum ab)^2 }{3},\\sum ab\\geq 3 , \\sum a^2b^2\\geq \\sum ab$.Simmilar we have $\\sum a^2\\geq \\sum a$.So by adding 2 we obtain the inequality.:)", "Solution_3": "Here is [u]my solution:[/u]\r\n\r\n$2(x^2+1)^3\\ge (x+1)^3(x^3+1)\\Longleftrightarrow (x-1)^4(x^2+x+1)\\ge0$, what is true.\r\n\r\nTherefore, $8\\prod (a^2+1)^3\\ge \\prod (a+1)^3\\cdot \\prod (a^3+1)$.\r\n\r\nBut ([u]Huygens's inequality[/u]) $(a^3+1)(b^3+1)(c^3+1)\\ge (abc+1)^3.$\r\n\r\nThus, $8\\prod (a^2+1)^3\\ge (abc+1)^3\\cdot \\prod (a+1)^3\\Longleftrightarrow 2\\prod (a^2+1)\\ge (abc+1)\\cdot \\prod (a+1).$\r\n\r\nFor $abc=1$ results the proposed inequality." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Determine whether the following series converges:\r\n\r\n1 + 1/2 + 1/3 - 1/4 - 1/5 - 1/6 + 1/7 + 1/8 + 1/9 - 1/10 - 1/11 - 1/12 + ... = sum_n=1 to oo { C(n)/n }, where C(n) = (-1)^j, with j = (n/3)-1, if n/3 is an integer or [n/3], if n/3 is not an integer.", "Solution_1": "This is a classical result.\r\nA more general result holds:\r\n\r\nIf you consider the alternanting harmonic sum of the form $\\frac{1}{n}$ where the first $k$ terms are considered with $+$ and the next $k$ terms with $-$ and so on, then such a series is known to converge, however if the numbers of terms with $+$ is not equal to the number of terms with $-$ then the series diverges.", "Solution_2": "Let's see what we can get from summation by parts:\r\n\r\nConsider $\\sum_{n=1}^{\\infty}s(n)b(n)$ where\r\n\r\n$s(n)=\\pm1$ and $b(n)$ is a decreasing function of $n$ that tends to zero as $n\\to\\infty.$\r\n\r\nLet $A(n)=\\sum_{k=1}^{n}s(k).$ We have that\r\n\r\n$\\sum_{n=1}^{N}s(n)b(n)=A(N)b(N)+\\sum_{n=1}^{N-1}A(n)(b(n)-b(n+1)).$\r\n\r\nNow take the limit as $N\\to\\infty.$ If we want to be able to prove convergence, then a useful sufficient condition would be that $A(n)$ is bounded. That is, there exists $M$ such that $A(n)\\le M$ for all $n.$ In the limit, we then have that\r\n\r\n$\\sum_{n=1}^{\\infty}s(n)b(n)= \\sum_{n=1}^{\\infty}A(n)(b(n)-b(n+1)).$\r\n\r\nBut the sum on the right is absolutely convergent:\r\n\r\n$\\sum_{n=1}^{\\infty}|A(n)(b(n)-b(n+1))|\\le M\\sum_{n=1}^{\\infty}(b(n)-b(n+1))=Mb(1)$\r\n\r\n(We used a telescoping series there.)\r\n\r\nWhat does it take for $A(n)$ to be bounded? Long-term balance between the plus signs and the minus signs, and no arbitrarily long runs of either kind of sign. This is more or less what didilica said. See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=121050]this[/url] recent topic by Cezanne123 in which there were arbitrarily long runs of each kind of sign." } { "Tag": [ "graph theory", "topology" ], "Problem": "[url]http://en.wikipedia.org/wiki/File:Cayley_graph_of_F2.svg[/url]\r\n\r\ncould any one please explain for me why the universal covering of the figure 8 looks like above. \r\nI read it from Hatcher 's book ( page 59) more than ten times but still don't get it . And by the way, how to find the universal covering of $ A\\bigvee B$, for example $ S^{2}\\bigvee \\mathbb{R}P^{2}$ ? what are the ideas behind? Thanks so much [/url]", "Solution_1": "The figure eight has a four-way fork, and each path loops back to the fork. The universal cover has to unfold all loops while preserving the local structure; this means that every path away from the fork leads to another copy of the fork. Iterate that, and you get the pictured tree.", "Solution_2": "thanks jmerry :)" } { "Tag": [ "calculus", "integration", "rotation" ], "Problem": "Find the volume of a doughnut with inner diameter $ 2$cm and boundary diameter $ 2$cm. Generalize.", "Solution_1": "[hide]$ V \\equal{} \\left(\\pi r^2\\right)\\left(2\\pi R\\right) \\equal{} 2\\pi^2Rr^2$\n\n$ R \\equal{} r \\equal{} 2\\implies V \\equal{} 2\\pi^2\\cdot2\\cdot2^2 \\equal{} 16\\pi^2$[/hide]", "Solution_2": "This is a special case of [url=http://mathworld.wolfram.com/PappussCentroidTheorem.html]Pappus' Centroid Theorem[/url].", "Solution_3": "i_like_pie did it quite nicely without integrals.\r\n[hide=\"With Integrals\"]\nRotate a circle at $ \\left( 1 , 0 \\right)$ about the line $ x\\equal{}\\minus{}1$.\n$ 4 \\pi \\int_{0}^{2} \\left( \\sqrt{1\\minus{}\\left( x \\minus{}1 \\right)^{2}} \\right) \\left( 1\\plus{}x \\right) \\ \\text{d} x$.\n[/hide]\r\n\r\nThat should be right.", "Solution_4": "Here's a more interesting question, then: prove Pappus' Centroid Theorem. :)", "Solution_5": "I did for the cylinder. :lol: However, I can not find a way to generalize it." } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "Find limit of the sum:\r\n $ \\lim_{ x \\to \\infty} \\sum_{n \\equal{} 1}^{\\infty} \\frac {x^2}{1 \\plus{} n^2x^2}$\r\n\r\nPS: Sorry, i edited it.", "Solution_1": "lim as n goes to infinity of the index? what? is that limit supposed to be there? do you mean as x goes to infinity?", "Solution_2": "$ \\lim_{x\\to\\infty}\\sum_{n\\equal{}1}^{\\infty}\\frac{x^{2}}{1\\plus{}n^{2}x^{2}}\\equal{}\\lim_{x\\to\\infty}\\sum_{n\\equal{}1}^{\\infty}\\frac{1}{\\frac{1}{x^2}\\plus{}n^{2}}\\equal{}\\lim_{x\\to0}\\sum_{n\\equal{}1}^{\\infty}\\frac{1}{x^2\\plus{}n^{2}}\\equal{}\\sum_{n\\equal{}1}^{\\infty}\\frac{1}{n^{2}}\\equal{}\\frac{\\pi^2}{6}$" } { "Tag": [], "Problem": "Find all entires value of $ x$ and $ y$ that sodisfy $ x^{3} \\plus{} y^{3} \\minus{} 3xy \\equal{} 2$\r\n\r\n\r\n(Note:entire I mean integer)", "Solution_1": "[hide]As \\[ x^3\\plus{}y^3\\plus{}z^3 \\minus{} 3xyz \\equal{} (x\\plus{}y\\plus{}z)(x^2\\plus{}y^2\\plus{}z^2 \\minus{} xy\\minus{}yx\\minus{}zx),\\] we know that \\[ x^3\\plus{}y^3\\plus{}1^3 \\minus{} 3xy \\equal{} (x\\plus{}y\\plus{}1)(x^2\\plus{}y^2\\minus{}xy\\minus{}x\\minus{}y\\plus{}1)\\] for all $ x,y$. Then if $ x,y$ also satisfy $ x^3\\plus{}y^3\\minus{}3xy\\equal{}2$, then we have \\[ 3 \\equal{} (x\\plus{}y\\plus{}1)(x^2\\plus{}y^2\\minus{}xy\\minus{}x\\minus{}y\\plus{}1).\\] Since $ x$ and $ y$ are integers, we must have $ x\\plus{}y\\plus{}1 \\in \\{3, 1, \\minus{}1, \\minus{}3\\}$ or $ x\\plus{}y \\in \\{2, 0, \\minus{}2, \\minus{}4\\}$. In each of these cases we find that $ xy \\not\\in \\mathbb{Z}$, so there are no integers $ x,y$ for which $ x^3\\plus{}y^3\\minus{}3xy\\equal{}2$.[/hide]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Suppose p is an odd prime,and a,b are two positive integers.We have \r\n[some image here before we changed servers]\r\nProve that a=b=1.\r\n\r\n[mod.: don't know what it was, but below it is claimed to be something more general :D]", "Solution_1": "This problem is very easy.", "Solution_2": "Have a look here for a more general proof :\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=catalan&t=28571\r\n\r\nPierre.", "Solution_3": "i have found now , this a very much easy solution to this trivial problem:\r\n\r\n\r\nwe have that $ b > a$ (if not then the problem is trivial)\r\n\r\nnow take $b>a>2$ and\r\n\r\nand we have that $[(P+1)^a=\\sum_{i=0}^a \\binom ai (p^i)] - P^b=1$ that will lead us to\r\n\r\n$[\\sum_{i=1}^a \\binom ai (P^i)] - P^b \\equiv 0 $ mode P^2 and cause $b>a>2$\r\n\r\nwe have$\\binom a{a-1}(P)=0$ (mode $P^2$) so $\\binom a{a-1}=0$ so $a-1=0$ and $a=1$ and so $b=1$\r\n\r\nbut the only thing that remains is that what if $b=2$ and the problem in these case is trivial", "Solution_4": "the relation is the Catalan conjecture (proven 2002):\r\nthe only solution of $a^b - c^d =1$ in positive integers where $b,d>1$ is $a=3, b=2 , c=2 , d=3$.", "Solution_5": "But...[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/16db3c01a540c318a70281e804b8de41.gif[/img] :D", "Solution_6": "really\r\n\r\nso when we have $\\binom a{a-1} P \\equiv 0$ mode ($P^2$) what is the meaning of it?", "Solution_7": "[quote=\"ZetaX\"]the relation is the Catalan conjecture (proven 2002):\nthe only solution of $a^b - c^d =1$ in positive integers where $b,d>1$ is $a=3, b=2 , c=2 , d=3$.[/quote]\r\n\r\ncan some one post a link for the solution of this problem, any one knows you first ask this problem?", "Solution_8": "I gave a link to a prove here:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=2002+conjecture&t=32508" } { "Tag": [ "inequalities", "logarithms" ], "Problem": "What values of x satisfy the following inequality:\r\n\r\n$\\log_{x^2 - 3}27\\ge 3$", "Solution_1": "Quote:What values of x satisfy the following inequality: [hide]\n\nWe have: logb(27) > 3 . . . Then: blogb(27) > b3 . . . and: 27 > b3 \n\n\n\nThen: b3 < 27 . . . (x2 - 3)3 < 27 . . . x2 - 3 < 3 . . . x2 < 6\n\n\n\nHence: |x| < sqrt{6}\n\n\n\n\n\nSince x2 - 3 is the base of a log, x2 - 3 > 0 . . . |x| > sqrt{3}\n\n\n\nSince x2 - 3 is the base of a log, x2 - 3 is not 1 . . . |x| is not 2\n\n\n\n\n\nTherefore, the inequality is true for values of x on:\n\n(-sqrt{6}, -2) U (-2, -\\sqrt{2}) U (sqrt{2}, 2) U (2, sqrt{6})[/hide]" } { "Tag": [], "Problem": "Find $m$ and $n$ if $m$ and $n$ are both natural numbers such that\r\n\\[(m+n)^{m}= n^{m}+1413. \\]\r\n(I don't even know the answer :( )", "Solution_1": "[hide=\"possible hint\"]Consider the equation mod m.[/hide]", "Solution_2": "[hide=\"Well...\"] $m | 1413 = 3 \\cdot 471 = 3^{2}\\cdot 157$. Since $9^{9}>> 1413$ we conclude $m = 1, 3$.\n\n[b]Case:[/b] $m = 1$. Then $1413 = 1$, contradiction. \n\n[b]Case:[/b] $m = 3$. Then \n\n$1413 = (n+3)^{3}-n^{3}= 9n^{2}+27n+27 \\implies$\n$n^{2}+3n-154 = 0 \\implies$\n$(n+14)(n-11) = 0 \\implies$\n$(m, n) = \\boxed{ (3, 11) }$ [/hide]", "Solution_3": "Question: does $>>$ mean [i]much greater than[/i]?", "Solution_4": "Yes it does DiscreetFourierTransform.", "Solution_5": "[hide]In any case, you just have to try m=1,2,3,4 only (since $5^{5}$ is already greater then $1413$. [/hide]" } { "Tag": [ "number theory" ], "Problem": "What is the result when the sum of fifteen and ninety is divided by teh positive difference between thirty-eight and twenty-three?", "Solution_1": "[quote=\"#H34N1\"]What is the result when the sum of fifteen and ninety is divided by teh positive difference between thirty-eight and twenty-three?[/quote]\r\n[hide] 15+90=105\n38-23=15\n105/15\n21/3\n7[/hide]", "Solution_2": "[hide=\"My guess.\"]$\\frac{(15+90)}{(38-23)}=\\frac{105}{15}=7$[/hide]", "Solution_3": "[hide]7[/hide]", "Solution_4": "[quote=\"kstan013\"][hide]7[/hide][/quote]\r\nIt would really help if you explained how you got this for the people who don't know how to do this problem.", "Solution_5": "[quote=\"Quevvy\"][quote=\"kstan013\"][hide]7[/hide][/quote]\nIt would really help if you explained how you got this for the people who don't know how to do this problem.[/quote][hide]15+90/38-23=105/15=7[/hide]happy?", "Solution_6": "[hide]15+90=105 \n38-23=15 \n105/15 \n21/3 \n7 \n[/hide]\r\nNot a hard question. I'm not sure why you have problems with it. And would this be considered number theory? lol...I don't think so." } { "Tag": [ "geometry", "geometric transformation", "rotation", "geometry proposed" ], "Problem": "Let $ABCD$ a square and $K$ a point in the side $AB$ and $N$ a point in the side $AD$, such that $AK*AN$=$2BK*DN$. Prove that $K, L, M, N$ lie in the same circle. Please someone now how to solve it with a rotation!? :spider:", "Solution_1": "You haven't defined points $L$ and $M$... :maybe:", "Solution_2": ":oops: \r\nLet L be the point of intersection of BD with CK and M be the point of intersection BD with CN. :blush:" } { "Tag": [ "function" ], "Problem": "The Ackermann function is\r\n\r\n$A(m,n)=$\r\n\r\n\r\n$n+1$ if m=0\r\n\r\n$A(m-1,1)$ if m>0 and n=0\r\n\r\n$A(m-1,A(m,n-1))$ if m>0 and n>0\r\n\r\nCompute $A(4,2)$.\r\n\r\nIs there an easiier way than brote force, or should I put this in the olympiad forum?[/hide]", "Solution_1": ":mad: Double post detected. Topic should be locked. For a discussion, go to the post replied by t0rajir0u\r\n\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=136369[/url]", "Solution_2": "Now this is very rare! I posted it yesterday, I only saw one of those topics, and still today I see only one of those topics, even though both topics are in this forum. :huh:" } { "Tag": [ "function" ], "Problem": "Well, here it goes :\r\n\r\nProve that :\r\n\r\n$\\sum_{k=0}^{n}k(k-1)\\binom{n}{k}^2=n(n-1)\\binom{2n-2}{n-2}$\r\n\r\nGood luck ;)\r\n\r\nEDIT : :blush: Sorry, I completely messed up, now that's alright :)", "Solution_1": "What is $p$? I think that matters.\r\nAnd for the $\\binom{k}{n}$ I think you meant $\\binom{n}{k}$, considering $n \\ge k$.", "Solution_2": "I've edited it now, so you can try it :)", "Solution_3": "Say we have $2n$ people, $n$ men and $n$ women. We have to choose a leader, a deputy leader and $n-2$ members for a society, and discard the rest of the people; we have the additional condition that the leader and the deputy leader must be men.\r\n\r\nWe have $n(n-1)\\binom{2n-2}{n-2}$ ways of doing this: first, among the $n$ men, we choose the leader ($n$) and the deputy leader ($n-1$), and then we choose $n-2$ members among the $2n-2$ people left ($\\binom{2n-2}{n-2}$).\r\n\r\nBut we can also choose them in this way: for each $k$, choose $k$ men ($\\binom{n}{k}$) and $n-k$ women ($\\binom{n}{n-k}=\\binom{n}{k}$), and among the selected men choose a leader ($k$) and then a deputy leader ($k-1$). Thus LHS=RHS.", "Solution_4": "Combinatorics proofs are so cool :) \r\n\r\n\r\nHere is an other one using generating functions :\r\n\r\nBy differentiating two times $(1+x)^n$ we get :\r\n\r\n$\\sum_{k=0}^{n}k(k-1)\\binom{n}{k} x^k = n(n-1)x^2 (1+x)^{n-2}$\r\n\r\n$\\sum_{k=0}^{n}\\binom{n}{k} x^k = \\sum_{k=0}^{n}\\binom{n}{n-k} x^k = (1+x)^n$ is well known ;) \r\n\r\n$\\sum_{k=0}^{n}k(k-1)\\binom{n}{k}^2$ is the \"Cauchy product\" of the two previous sums and we are looking for the coefficient of $x^n$ in both side. \r\nThe coefficient of $x^n$ in $n(n-1)x^2 (1+x)^{2n-2}$ is indeed $n(n-1)\\binom{2n-2}{n-2}$", "Solution_5": "[quote=\"t\u00b5t\u00b5\"]Combinatorics proofs are so cool :) \n[/quote]\r\nI strongly agree with it! :D\r\n\r\nSeverius, I love your proof\r\n\r\nT\u00b5t\u00b5, I solved it almost the same way as you :) Moreover, I've learnt one more thing : the [url=http://mathworld.wolfram.com/CauchyProduct.html]Cauchy product[/url]\r\n\r\nSo, thanks a lot ;)" } { "Tag": [ "quadratics", "algebra", "polynomial" ], "Problem": "Given that $y = 2x^{2}+px+16$ and that $y < 0$ when $2 < x < k$, find the range of values of $p$ and $k$.", "Solution_1": "[quote=\"vishalarul\"]Given that $y = 2x^{2}+px+16$ and that $y < 0$ when $2 < x < k$, find the range of values of $p$ and $k$.[/quote]\r\n\r\nCan't you just find $p$ so that $2$ is a root, and extend the ranges to negative infinity for $p$ and positive infinity for $k$? I might not be understanding the question, but here is my answer (tell me if im right, ill post my solution):\r\n\r\n[hide] $-\\infty4$)" } { "Tag": [ "inequalities", "calculus", "integration", "function", "linear algebra", "matrix", "complex numbers" ], "Problem": "I've just solved an oral problem from ENS and I saw that the conclusion is more than impressive:\r\n For all real numbers $a_1,...a_n$ we have $\\sum_{i,j=1,n}{\\frac{a_ia_j}{1+|i-j|}\\geq0}$. I really could not find an elementary solution, though I tried for a long time. Does anyone have a \"natural approach\"? PS: it is far more difficult than the classical poland problem where $1+|a_i-a_j|$ is replaced by $i+j$.", "Solution_1": "we have to find the right scalar product isn't it ? with an integral ?", "Solution_2": "Harazi - can you show this non-elementary solution ? I'll be thankful", "Solution_3": "I have found a solution and i hope it's right\r\n\r\nFirst ,the problem is obviously if a(i) =0 for all i =1,...,n .Then we can assume that not all a(i) is equal to 0\r\n\r\nLet F(x) = :Sigma: a(i)a(j)/1+/i-j/ . x^[a(i)a(j)(1+/i-j/)] for all x >0\r\n\r\nThen we have F'(x) > 0 for all x >0 ,so F(x) is increasing ,x>0\r\n\r\nBut now F is continuous for x>0 and Lim F(x) = 0 when x comes to 0(+) .We get F(1) >= 0 ,and we are done .", "Solution_4": "[quote=\"namanhams\"]I have found a solution and i hope it's right\n\nLet F(x) = :Sigma: a(i)a(j)/1+/i-j/ . x^[a(i)a(j)(1+/i-j/)] for all x >0\n\nThen we have F'(x) >=0 for all x >0 ,so F(x) is increasing ,x>0\n\nBut now F(0)=0 and F is continuous so F(x) is increasing for all x>=0 .We get F(1) >= 0 ,and we are done .[/quote]\r\n\r\nhow do you show F'(x) >= 0??! It is not trivial to me at all.", "Solution_5": "F'(x) = :Sigma: (a(i)a(j)):^2: .x^[a(i)a(j)(1+/i-j/) -1] >=0 for all x >0.\r\n\r\nI think it's obviously", "Solution_6": "[quote=\"namanhams\"]F'(x) = :Sigma: (a(i)a(j)):^2: .x^[a(i)a(j)(1+/i-j/) -1] >=0 for all x >0.\n\nI think it's obviously[/quote]\r\n\r\noh, i'm sorry that i misread your solution.... :blush:", "Solution_7": "I think it's wrong, though i can't point out the fallancy of it.\r\n\r\nConsider this,\r\n\\[\r\na_1 + ... + a_n \r\n\\]\r\nwhere $a_i$ are reals.\r\nUsing your idea, \r\nConstruct $ f(x) = \\sum {a_i x^{a_i } } $\r\nthen $ f'(x) = \\sum {a_i ^2 x^{a_i - 1} } $\r\nFor positive x, $f'(x) \\ge 0$, so f is increasing if x is positive.\r\nSince f(0) = 0, we have f(1) is positive.\r\nHowever, $f(1)=a_1 + ... + a_n$\r\nchoose all $a_i$ to be negative and it leads to an absurdity....", "Solution_8": "[quote=\"siuhochung\"]Consider this,\n\\[\na_1 + ... + a_n \n\\]\nwhere $a_i$ are reals.\nUsing your idea, \nConstruct $ f(x) = \\sum {a_i x^{a_i } } $\nthen $ f'(x) = \\sum {a_i ^2 x^{a_i - 1} } $\nFor positive x, $f'(x) \\ge 0$, so f is increasing if x is positive.\nSince f(0) = 0,[/quote]\r\n\r\nThe fallacy is that you can't claim f(0) = 0: In fact, if, at least, one of the reals $a_1$, $a_2$, ..., $a_n$ is negative, then $f\\left(x\\right)=\\sum_{i=1}^n a_i0^{a_i}$ is undefined.\r\n\r\nI don't know for sure whether Namanhams' proof of the original inequality suffers from the same lack, but it looks so.\r\n\r\n darij", "Solution_9": "Thank for your remark ,Sihochung and Darij .\r\nI've just corrected my proof .", "Solution_10": "[quote=\"namanhams\"]Thank for your remark ,Sihochung and Darij .\nI've just corrected my proof .[/quote]\r\n\r\nI think it still has the same mistake. Grinberg not only mentioned that $0^0$ is meaningless, he also meant that $0^a$ where a is negative is also meaningless. Your solution cannot ensure a positive power.", "Solution_11": "Well, this problem is far more difficult that one could think.\r\n To alekk: indeed, the scalar product is $(f,g)=\\int_{0}^{2\\pi}{conj(f)(x)g(x)F(x)}$ where $F(x)=\\frac{1-r^2}{2\\cdot\\pi\\cdot (1-2rcosx+r^2)}$, where $conj(f)$ is the complex conjugate of f.", "Solution_12": "I don't think so .In my solution ,F(x) is a function on positive real number ,not contain 0.\r\nAnd my solution doesn't use F(0) :)", "Solution_13": "consider the matrix $B(t)$ such that: $B(t)_{i,j}=t^{|i-j|}$. This matrix is positive definite for $00$ for $X \\neq 0$\r\nconclusion follows ..", "Solution_14": "to harazi: how on earth did you find $F$ ??", "Solution_15": "I just solved one oral problem from ENS and used the result from there. :D Your solution is the classical one, but in fact I prefer not to compute principal determinants. :D", "Solution_16": "[quote=\"harazi\"]Well, this problem is far more difficult that one could think.\n To alekk: indeed, the scalar product is $(f,g)=\\int_{0}^{2\\pi}{conj(f)(x)g(x)F(x)}$ where $F(x)=\\frac{1-r^2}{2\\cdot\\pi\\cdot (1-2rcosx+r^2)}$, where $conj(f)$ is the complex conjugate of f.[/quote]\nHi,Can somebody tell me what does the mean of $r$,how can we use this scalar product to solve the problem?\nHow can $\\frac{1}{1+|i-j|}$ appears in the integeral value?\n\nto alekk:\n\nHow to compute principal determinants?\n\n\nThank you in advance.\n\nEdit:Now I know harizi want to prove that the matrix $B(r)$ is positive defined..." } { "Tag": [ "geometry", "\\/closed" ], "Problem": "When you open the Site Faq, the shortcuts are distorted and in the message area. Can you fix this?\r\n\r\nI have Mozilla Firefox 2.0.0.5", "Solution_1": "Here too, Seamonkey 1.2 on Gentoo Linux.", "Solution_2": "Shortcuts? Message area? All I see is a page full of allowed extensions.", "Solution_3": "Same here. The problem is not there on the ML theme. Attached is a screenshot." } { "Tag": [ "geometry solved", "geometry" ], "Problem": "Let $ABC$ be a triangle and $O$ the center of its circumcircle. The tangent in $C$ at the circle meets $AB$ in $M$. The perpendicular on $OM$ in $M$ meets $BC$ and $AC$ in $P$ respectively $Q$. Prove that $MP=MQ$.", "Solution_1": "See this one: \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=16784" } { "Tag": [ "LaTeX" ], "Problem": "$x^{2000}+2000^{1999}=x^{1999}+2000^{2000}$\r\nFind all such integers $x$", "Solution_1": "[hide]$\\displaystyle x^{2000}+2000^{1999}=x^{1999}+2000^{2000}$\n\n$\\Rightarrow \\displaystyle x^{1999}(x-1)=2000^{1999}(2000-1)$\n\n$\\Rightarrow \\displaystyle (x-1)x^{1999}=2^{7996}3^{5997}1999$\n\nSince $1999$ is prime, either $x-1=1999$ so the only (and obvious if you look at the equation) solution is $x=2000$[/hide]", "Solution_2": "Can you prove that if $x>2000$, then $x^{2000}-x^{1999}>{2000}^{2000}-2000^{1999}$?", "Solution_3": "Yes. This is the same as showing that\r\n$(x-1)*(x^1999)>(2000^1999)(2000-1)$\r\nis true, which obviously holds if $x>2000$\r\n\r\nEDIT: Sorry, I'm really bad with latex, for some reason it won't accept more than one character in the exponent...so i'll just write it without latex...\r\n\r\n(x-1)(x^1999)>(2000^1999)(2000-1)", "Solution_4": "x^{1999} will give you x^1999.", "Solution_5": "Oh. Thanks.\r\n$(x-1)x^{1999}>(2000^{1999})(2000-1)$" } { "Tag": [ "function", "limit", "topology", "real analysis", "advanced fields", "advanced fields unsolved" ], "Problem": "consider the following PDE:\r\n$\\partial_{t}u(x,t)+\\Delta u(x,t)=f(u(x,t))$\r\nwhere $f$ is a smooth function that looks like $k.x(1-x)$, that is something positif on $(0;1)$, negatif on $\\mathbb{R}\\setminus[0;1]$, and $f'(0) > 0$. (Such an equation arises in biology, physic, etc ..). One would like to know if there exists a function $\\phi: \\mathbb{R}\\to \\mathbb{R}$ with $\\lim_{-\\infty}f=0$, $\\lim_{+\\infty}=1$ such that $u(x,t)=\\phi(ct-x.v)$, where $v \\in \\mathbb{R}^{n}$, $c>0$, is a solution of the initial PDE.That is: we are looking for solutions that are travelling on the direction $v$, with speed $c$. Pluging this into the PDE, one sees that $\\phi$ satisfies: \r\n$c \\phi'-\\phi''=f(\\phi)$\r\nThe question is: for what value of $c$ this equation has a solution satisfying the hypothesis ? The answer is for $c$ large enough, but how to prove this ?", "Solution_1": "This is a really nice problem for an intermediate (by European standards) course in ODE. Here are the steps you may try:\r\n0) Rescale the time to reduce everything to the case $f'(0)=1$\r\n1) Rewrite your second order ODE as a $2\\times 2$ system of first order ODE's setting $x=\\phi$, $y=\\phi'$\r\n2) Show that $(0,0)$ and $(1,0)$ are stationary points.\r\n3) Show that $(1,0)$ is a hyperbolic equilibrium and conclude, using the stable manifold theorem, that there exist two solutions approaching this point at $+\\infty$. \r\n4) Use the fact that the stable manifold is tangent to the stable space of the linearized equation to conclude that one of this solutions has $x(t)<1$, $y(t)>0$ for large positive $t$.\r\n5) Use the special kind of your system to show that, when you run time backwards, this solution preserves these properties until $x$ reaches $0$.\r\n6) Show that $y\\le \\max_{[0,1]}f/c$ during this time.\r\n7) Use the Lyapunov function $x^{2}+y^{2}-xy/c$ to show that the point $(0,0)$ has an attraction basin (for the time running backwards) of size bounded from below by some positive constant independent of $c$ when $c$ is large.\r\n8) Happily conclude that $(0,0)$ catches your solution at the moment when it crosses the $y$ axis and finish the problem. \r\n\r\nHope, that I didn't make a mistake anywhere and that this will be enough of a hint for you :).", "Solution_2": "thank you Fedja, this problem appeared in my PDE course with a nasty solution. I will try your approach :)", "Solution_3": "You are welcome,alekk :). By the way, I realized overnight that steps 5-7 can be replaced by showing that, for large $c$, the trajectory with time running backwards cannot leave the triangle with vertices $(0,0)$, $(1,0)$ and $(0.5,0.5)$ (just look at the direction field on the boundary). This approach would require doing step 4 more carefully but the benefit is that you can prove now that $0\\le\\phi\\le 1$ for all times." } { "Tag": [ "probability" ], "Problem": "find the probability that the birth day of six different persons will fall in exactly two calendar months", "Solution_1": "[hide=\"Basic Idea\"]Consider two arbitrary months, say they have $ x$ days and $ y$ days, respectively. The probability that the condition is satisfied is equal to the probability that all six people have their birthdays within the two months minus the probability that all six people have their birthdays within just one month.\n\nSo basically, $ \\left(\\frac{x\\plus{}y}{365}\\right)^6 \\minus{} \\left(\\frac{x}{365}\\right)^6 \\minus{} \\left(\\frac{y}{365}\\right)^6$ is the desired quantity. (This assumes people can have the same birthday. If not, then you do conditional probability).\n\nNow, sum over this with all valid combinations of months, depending on your definition of valid (two non-coinciding months, two consecutive months [the problem doesn't state this explicitly]).\n\n[/hide]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "for $a,b,c>0$ prove:\r\n$(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})\\frac{1}{3\\sqrt[3]{abc}}\\geq \\frac{1}{abc+ab+a}+\\frac{1}{abc+bc+b}+\\frac{1}{abc+ac+c}$\r\n\r\nI took it from another thread, but now it's after deadline :D", "Solution_1": "It's easy :)\r\n\r\nusing AM-GM we have\r\n\\[a+ab+abc\\geq 3\\sqrt[3]{a^{3}b^{2}c}\\]\r\nSo the stronger thesis is\r\n\\[\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)\\frac{1}{3\\sqrt[3]{abc}}\\geq \\frac{1}{3\\sqrt[3]{a^{3}b^{2}c}}+\\frac{1}{3\\sqrt[3]{b^{3}c^{2}a}}+\\frac{1}{3\\sqrt[3]{c^{3}a^{2}b}}\\]\r\n\r\n\\[\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\geq \\frac{1}{\\sqrt[3]{a^{2}b}}+\\frac{1}{\\sqrt[3]{b^{2}c}}+\\frac{1}{\\sqrt[3]{c^{2}a}}\\]\r\nby AM-GM:\r\n\\[\\frac{\\frac{1}{a}+\\frac{1}{a}+\\frac{1}{b}}{3}\\geq \\frac{1}{\\sqrt[3]{a^{2}b}}\\]\r\n\r\n\\[\\frac{\\frac{1}{b}+\\frac{1}{b}+\\frac{1}{c}}{3}\\geq \\frac{1}{\\sqrt[3]{b^{2}c}}\\]\r\n\r\n\\[\\frac{\\frac{1}{c}+\\frac{1}{c}+\\frac{1}{a}}{3}\\geq \\frac{1}{\\sqrt[3]{c^{2}a}}\\]\r\nAnd proof is complete", "Solution_2": "or see here http://www.mathlinks.ro/Forum/viewtopic.php?t=117113,but the solution I gave is totally same as robpal's :D" } { "Tag": [ "LaTeX", "algebra unsolved", "algebra" ], "Problem": "Let A and B be two subsets of the real numbers. It is known that A \u0012 B+\u000bZ and B \u0012 A+\u000bZ\r\nfor any positive real number \u000b. (Note that X +\u000bZ is the set consisting of all numbers x+\u000bn,\r\nwhere x 2 X and n 2 Z.)\r\n(a) Is it necessarily true that A = B?\r\n(b) If B is bounded is it necessarily true that A = B?\r\n\r\nBomb", "Solution_1": "please use latex, it is a bit hard to understand", "Solution_2": "Latex does not work for me. \r\n\r\nBomb", "Solution_3": "Why $\\LaTeX$ shouldn't work for you\u00bf Read Valentins introduction ( http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 ) and try !" } { "Tag": [ "function", "trigonometry", "real analysis", "real analysis solved" ], "Problem": "Use Mean-value Theorem for \r\n$f(x,y)=\\sin\\pi x+\\cos\\pi y$\r\nto prove that there exist $\\theta\\in (0,1)$ such that\r\n$\\frac{2}{\\pi} = \\cos\\frac{\\pi\\theta}{2}+\\sin\\frac{\\pi(1-\\theta)}{2}$", "Solution_1": "Take $x=\\frac{\\theta}{2}$ and $y=\\frac{1-\\theta}{2}$\r\n$\\theta=0\\Rightarrow f(x,y)=2$\r\n$\\theta=1\\Rightarrow f(x,y)=0$\r\n$f$ is continuous ..." } { "Tag": [ "quadratics", "algebra", "polynomial", "quadratic formula" ], "Problem": "how do you factor these two?\r\n\r\n5x^2-22x-40\r\n\r\nAND\r\n\r\n-2x^2+15x-9\r\n\r\nthanks!", "Solution_1": "[quote=\"gracexcor3\"]how do you factor these two?\n\n5x^2-22x-40\n\nAND\n\n-2x^2+15x-9\n\nthanks![/quote]\r\nThere had been many numerous posts on how to factor quadratics. You should simply refer to those. But here's a hint: use the quadratic formula, if you know what it is.", "Solution_2": "[quote=\"gracexcor3\"]how do you factor these two?\n\n5x^2-22x-40\n\nAND\n\n-2x^2+15x-9\n\nthanks![/quote]\r\nThey are quadratics.\r\nSo they can be factored into the form $ (x\\minus{}y)(x\\minus{}z)$\r\nRefer to other threads if you need help.", "Solution_3": "Although you can only do that with monic polynomials, Peijin.\r\n\r\nTaking a look at the discriminant of the first equation, which is $ 22^{2}\\minus{}4(5)(\\minus{}40)\\equal{}1284$, the expression can't be factored into a nice form since 1284 isn't a perfect square.\r\n\r\nIt is factorable, but you have to use irrationals.\r\n\r\nThe second one has a discriminant of $ 225\\minus{}72\\equal{}153$, so again it's not factorable unless you consider irrational numbers", "Solution_4": "you can aalways do the cross method\r\n\r\n5 times x to the 2nd power minus 22x-40\r\n1 1\r\n5 40\r\n1 2\r\n5 20\r\n1 4\r\n5 10\r\nand so on\r\n\r\nthese numbers have to equal the middle term(without the variable)", "Solution_5": "lol, just try factoring it\r\n\r\nand to do that stuff, you do \r\n[code]\n$5x^{2}-22x-40$\n[/code]\r\nWhich produces \r\n$ 5x^{2}\\minus{}22x\\minus{}40$", "Solution_6": "thx a lot :D", "Solution_7": "What was that (x-y)(x-z) stuff??", "Solution_8": "It's the form of factoring a quadratic in a way to find the solutions.", "Solution_9": "I see, thanks :lol:", "Solution_10": "Yes, using the quadratic equation would be the easiest way to solve this problem.\r\n\r\nHowever, if you want to actually factor it, you need to use completing the square.\r\n\r\nFirst, reduce (a) to 1 (The coefficient of x^2)\r\n\r\nThen, move (c) to the right side of the equation. (Number that isn't multiplied by x)\r\n\r\nDivide (b) by 2 and then square it. (Don't really do this in equation, it's off to the side)\r\n\r\nAdd the number you received previously to both sides.\r\n\r\nNow, you should be able to factor the left side.\r\n\r\nThen find the square root of each side\r\n\r\nThe rest from here should be self explanatory.", "Solution_11": "[quote=\"Ryuk\"]Yes, using the quadratic equation would be the easiest way to solve this problem.\n\nHowever, if you want to actually factor it, you need to use completing the square.\n\nFirst, reduce (a) to 1 (The coefficient of x^2)\n\nThen, move (c) to the right side of the equation. (Number that isn't multiplied by x)\n\nDivide (b) by 2 and then square it. (Don't really do this in equation, it's off to the side)\n\nAdd the number you received previously to both sides.\n\nNow, you should be able to factor the left side.\n\nThen find the square root of each side\n\nThe rest from here should be self explanatory.[/quote]\r\n\r\nYou do not [b]need[/b] to use the complete the square method. It may be helpful, but certainly not here.", "Solution_12": "Remember, Quadratic Formula<-->Completing the Square." } { "Tag": [ "floor function", "number theory", "relatively prime" ], "Problem": "How many numbers less than $ 529$ are relatively prime to $ 462$?", "Solution_1": "Uh...not too sure about this solution...\r\n\r\n[hide=\"?\"]\nSo, finding $ \\phi(462) \\equal{} 120$. So, there are 120 numbers less than $ 462$ that are relatively prime to $ 462$.\n\nNow, we have to look at the numbers from $ 462\\minus{}528$, inclusive.\n\nSuppose $ a<462$. If $ a$ and $ 462$ are coprime, or $ (462,a)\\equal{}1$ then $ a\\plus{}462$ should be coprime too, by the Euclidean Algorithm.\n\n$ (a\\plus{}462, 462) \\minus{}> (462,a) \\minus{}> 1$\n\nSo, we just need to find the numbers from $ 1\\minus{}66$ (462 is not going to be relatively prime to $ 462$, so I'll just disregard that case) that are relatively prime to $ 462$, which factors to $ 2,3,7,11$.\n\nTo find this, we'll count how many numbers are NOT relatively prime. This is just $ \\lfloor 66/2 \\rfloor \\plus{} \\lfloor 66/3 \\rfloor \\plus{} \\lfloor 66/7 \\rfloor \\plus{} \\lfloor 66/11 \\rfloor \\minus{} \\lfloor 66/(2\\times 3) \\rfloor \\minus{} \\lfloor 66/(2\\times 7) \\rfloor \\minus{} \\lfloor 66/(2\\times 11) \\rfloor \\minus{} \\lfloor 66/(3\\times7) \\rfloor \\minus{} \\lfloor 66/(3\\times 11) \\rfloor \\minus{} \\lfloor 66/(7\\times 11) \\rfloor \\plus{} \\lfloor 66/(2\\times 3\\times 7) \\rfloor \\plus{} \\lfloor 66/(2 \\times 3 \\times 11) \\rfloor \\plus{} \\lfloor 66/(2 \\times 7 \\times 11) \\rfloor \\plus{} \\lfloor 66/(3\\times 7 \\times 11) \\rfloor \\minus{} \\lfloor 66/(2\\times 3\\times 7\\times 11) \\rfloor$\n\nWhich equals to $ 33\\plus{}22\\plus{}9\\plus{}6\\minus{}11\\minus{}4\\minus{}3\\minus{}3\\minus{}2\\minus{}0\\plus{}1\\plus{}1\\plus{}0\\plus{}0\\minus{}0 \\equal{} 33\\plus{}16 \\equal{} 49$.\n\nSo, there are $ 49$ numbers from $ 1\\minus{}66$ that are NOT relatively prime to $ 462$.\n\nThat means that there are $ 17$ numbers that are coprime.\n\nSo, the answer is $ 120\\plus{}17\\equal{}\\boxed{137}$\n[/hide]", "Solution_2": "prime factors of 462 are 2,3,7,11\r\n\r\nso phi(462)=1/2*2/3*6/7*10/11*462=120", "Solution_3": "You have not answered the question.", "Solution_4": "[quote=\"CatalystOfNostalgia\"]You have not answered the question.[/quote]\r\n\r\nwow im stupid. Just like countdown yesterday :| \r\ni agree with zellex \r\n :lol:" } { "Tag": [], "Problem": "If b is a positive integer greater than 1, the base b expansion of a positive integer is denoted $ (d_kd_{k\\minus{}1}...d_1d_0)_b$ where the $ d_i$ are base b digits. Thus, the base 5 expansion of $ (54)_{10}$ is $ (204)_5$ since in base 10, $ 54\\equal{}2(5^2)\\plus{}0(5^1)\\plus{}4(5^0)$. How many of the integers between one and one thousand, inclusive, have a base 5 expansion that contains at least one zero?", "Solution_1": "Notice that $ 1000\\equal{}13000_5$. It is easier to count how many numbers have no zeros in them. 1 digit: $ 4$, 2 digit: $ 4^2$, 3 digit: $ 4^3$, 4 digit: $ 4^4$. For 5 digit numbers with no zeros, notice the first digit must be 1 and the second digit must be 1 or 2, for a total of $ (1)(2)(4)^3$ possibilities. The total amount without zeros is $ 4\\plus{}16\\plus{}64\\plus{}256\\plus{}128\\equal{}468$, so the amount of numbers that contain a zero is $ 1000\\minus{}468\\equal{}532$" } { "Tag": [], "Problem": "Pentru orice numar natural $\\ n$ definim expresia $\\ E(n)=n(n+1)(2n+1)(3n+1)...(10n+1)$.Determinati cel mai mare divizor comun al numerelor $\\ E(1),E(2),\\dots$.", "Solution_1": "Cine poate da o solutie la problema asta??", "Solution_2": "In esenta se gaseste cmdc al E(1) si E(2) si apoi se arata ca divide orice E(n.)", "Solution_3": "Pai asa m-am gandit si eu, Bianca.Dar poti da o solutie completa?As dori sa vad si alta rezolvare, diferita de a mea." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let $ 2$ triangle $ ABC,A'B'C'$ satisfy $ A\\in B'C',B\\in C'A',C\\in A'B'$ and $ \\widehat{ABC}\\equal{}\\widehat{A'B'C'},\\widehat{BCA}\\equal{}\\widehat{B'C'A'},\\widehat{CAB}\\equal{}\\widehat{C'A'B'}$.\r\nProve that: distance from $ 2$ orthocentre of $ 2$ this triangle to circumcentre of triangle $ ABC$ are equal.", "Solution_1": "Already posted. See here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=143090[/url]." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "(a) Find the maximum value of the expression $ x^{2}y-y^{2}x$ when $ 0\\le x \\le 1$ and $ 0 \\le y \\le1$.\r\n\r\n(b) Find the maximum value of the expression $ x^{2}y+y^{2}z+z^{2}x-x^{2}z-y^{2}x-z^{2}y$\r\n\r\nwhen $ 0 \\lex\\le 1, 0\\le y\\le 1, 0 \\le z \\le1$.", "Solution_1": "$ f(x) \\equal{} x^{2}y \\minus{} y^{2}x$ when $ 0\\le x \\le 1$ and $ 0 \\le y \\le1$.\r\n$ f(x)_{max} \\equal{} f(1) \\equal{} y \\minus{} y^{2} \\le 1/4$\r\n\r\n $ g(x) \\equal{} x^{2}y \\plus{} y^{2}z \\plus{} z^{2}x \\minus{} x^{2}z \\minus{} y^{2}x \\minus{} z^{2}y$\r\nif y\u2265z $ g(x)_{max} \\equal{} g(1) or g(0)$ ,\r\n$ g(0) \\equal{} y^{2}z \\minus{} z^{2}y \\le 1/4$,$ g(1) \\equal{} y \\plus{} y^{2}z \\plus{} z^{2} \\minus{} z \\minus{} y^{2} \\minus{} z^{2}y \\equal{} (1 \\minus{} y)z^{2} \\minus{} (1 \\minus{} y^{2})z \\plus{} y \\minus{} y^{2} \\equal{} h(z)$\r\n$ h(z)_{max} \\equal{} h(0) or h(1)$......", "Solution_2": "[quote=\"AndrewTom\"](a) Find the maximum value of the expression $ x^{2}y - y^{2}x$ when $ 0\\le x \\le 1$ and $ 0 \\le y \\le1$.\n\n(b) Find the maximum value of the expression $ x^{2}y + y^{2}z + z^{2}x - x^{2}z - y^{2}x - z^{2}y$\n\nwhen $ 0 \\lex\\le 1, 0\\le y\\le 1, 0 \\le z \\le1$.[/quote]\r\n\r\n(a) By $ x^2\\le x$, we get $ x^{2}y - y^{2}x\\le x(y-y^2)\\le y-y^2=\\frac 14-(y-\\frac 12)^2\\le \\frac 14$.\r\n\r\n(b) $ x^{2}y + y^{2}z + z^{2}x - x^{2}z - y^{2}x - z^{2}y=(x-y)(x-z)(y-z)$. As the expression is cyclic, we can assume $ x=max\\{x, y,z\\}$. For a maximum of it, we must then have $ y>z$, hence $ x=1$ and $ z=0$, which leaves us with $ y(1-y)$ as in (a). So the max is again $ \\frac 14$. :)" } { "Tag": [ "linear algebra" ], "Problem": "Let $ T: F^{\\infty}\\rightarrow F^{\\infty}$, $ T(x_{1}, x_{2},\\ldots)=(x_{2},x_{3},\\ldots)$. Find a basis for $ \\textrm{Ker }T$ and what is $ N(T^{k})$?", "Solution_1": "I hope I am not missing something here, because this seems straightforward. \r\n\r\n$ T$ is just a shift operator. So $ \\ker T=(k,0,0,\\cdots)$ where $ k\\in F$, so a basis can be given by $ k=1$. Similarly for $ T^{k}$ (which just shifts $ k$ times), $ \\ker T^{k}= (c_{1},c_{2},\\cdots,c_{k},0,0,\\cdots)$ where the $ c_{i}\\in F$." } { "Tag": [ "probability" ], "Problem": "There is a pile of 129 coins on a table, all unbiased except for one which has heads on both sides. Bob chooses a coin at random and tosses it eight times. The coin comes up heads every time. What is the probability that it will come up heads the ninth time as well?", "Solution_1": "There's a $ \\frac{1}{129}$ chance that he picked the heads-heads coin, and that coin has a $ \\frac{1}{1}$ chance of being heads on the next flip. The probability that he picked the heads-heads coin and the next flip will heads is $ \\frac{1}{129}\\times\\frac{1}{1} \\equal{} \\frac{1}{129}$\r\nThere's a $ \\frac{128}{129}$ chance that he picked a heads-tails coin, and those coins have a $ \\frac{1}{2}$ chance of being heads on the next flip. The probability that he picked a heads-tails coin and the next flip will be heads is $ \\frac{128}{129}\\times\\frac{1}{2} \\equal{} \\frac{64}{129}$\r\nSince these two sequences of events cannot both occur, (ie., he cannot pick both the heads-heads coin and a heads-tails coin), we simply add the probabilities of each to obtain the answer. Thus, the answer is $ \\frac{1}{129} \\plus{} \\frac{64}{129} \\equal{} \\frac{65}{129}$", "Solution_2": "[quote=\"aidan\"]There is a pile of 129 coins on a table, all unbiased except for one which has heads on both sides. Bob chooses a coin at random and tosses it eight times. The coin comes up heads every time. What is the probability that it will come up heads the ninth time as well?[/quote]\r\n\r\nthis is one of those tricky weighted probability questions.\r\n\r\nso well, let's look at the possibility that it came up heads eight times in a row. it seems pretty unlikely.\r\n\r\n$ \\frac{1}{129}\\cdot1 \\plus{} \\frac{128}{129}\\cdot\\frac{1}{256} \\equal{} \\frac32\\cdot\\frac{1}{129}$\r\n\r\nand now the probability of the coin picked being the biased one is $ \\frac{\\frac{1}{129}}{\\frac32\\cdot\\frac{1}{129}}\\equal{}\\frac23$, so the probability of the coin being the unbiased one is $ \\frac13$.\r\n\r\nso our answer is $ \\frac23\\cdot1 \\plus{} \\frac13\\cdot\\frac12 \\equal{} \\boxed{\\frac56}$", "Solution_3": "Isn't the probability of the coin coming up heads the same each time? So is the condition that it comes up heads 8 times in the first 8 throws important?", "Solution_4": "That's kinda what I was thinking...but I'm not completely sure.", "Solution_5": "Suppose we have the same setup (128 regular coins and 1 two-headed coin) and I choose one coin at random, flip it 10,000 times, and get heads every time. Given this information, is the probability that I chose the irregular coin still 1/129, or do you think it has changed?\r\n\r\nSuppose instead we have the same setup (128 irregular coins and 1 two-headed coin) and I choose one coin at random, flip it 8 times, and get heads 7 times and tails once. Given this information, is the probability that I chose the irregular coin still 1/129, or do you think it has changed?\r\n\r\nLazarus' solution-idea is correct (haven't checked the details), if not explained as thoroughly as could be. This is just a standard application of Bayes' Theorem on conditional probabilities, but \"standard\" applications of Bayes' Theorem tend to be very tricky if you aren't used to it.\r\n\r\nSee http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221979 for another recent example from the fora." } { "Tag": [ "limit", "logarithms", "algebra", "binomial theorem", "calculus", "calculus computations" ], "Problem": "Proove the following limit: \\[\\lim_{x\\to 0+}\\frac{1}{x^{x}}=1\\]", "Solution_1": "Where is $x$? Do you mean $\\frac{1}{x^x}\\to 1$?\r\n\r\n$\\lim_{x\\to 0^+} x^x = \\lim_{x\\to 0^+} e^{x\\ln x} = e^{\\lim_{x\\to 0^+} x\\ln x} = e^0 = 1$.", "Solution_2": "how do you prove \\[\\lim_{x\\to 0^{+}}e^{x\\ln x}= e^{\\lim_{x\\to 0^{+}}x\\ln x}\\]", "Solution_3": "[quote=\"mathematica\"]how do you proove \\[\\lim_{x\\to 0^+} e^{x\\ln x} = e^{\\lim_{x\\to 0^+} x\\ln x}\\][/quote]\r\n\r\nIn short,\r\n\r\n$f$ continuous, $\\lim_{x\\to x_0}g(x)=b$\r\n\r\nWe want to show that $\\lim_{x\\to x_0}f(g(x))=f(\\lim_{x\\to x_0}g(x))$\r\n\r\nTake $any$ sequence ${x_n}$ that $x_n\\to x_0$. We have $sequence$ ${g(x_n)}$, and $\\lim_{n\\to \\infty}g(x_n)=b$\r\n\r\nSo $\\lim_{n\\to \\infty}f(g(x_n))=f(b))$ because f is continuous at b.", "Solution_4": "could you please use the epsilon-delta notation?", "Solution_5": "If you're just learning analysis (which seems likely), it would be a good exercise for you to prove some of the statements used here. I know of a proof that uses only the binomial theorem, but I don't have time to post it now (it's in Walter Rudin's Principles of Mathematical Analysis)" } { "Tag": [], "Problem": "One thing I'm having a bit of a hard time getting into my cranium is why I keep running across the statement that halogens are just missing one electron to fill their shell. I do understand valence electrons but since all elements in this group have seven valence electrons and beyond the second shell, each \"shell\" is large enough to accomodate much more than just one more electron beyond the seven valence electrons.\r\n\r\nAnyone have a simple way of explaining the \"just missing one electron to fill their shell\" statement? :D", "Solution_1": "the one missing electron does not mean that the one electron is all the atom needs to complete its shell!\r\n\r\nbut rather, when they recieve the extra electron, they became more stable in Noble conficgration (because of the electron config.)", "Solution_2": "yea this also affects it's ionization energy and electron affinity...if you add one more electron, you'll have a stabler atom.." } { "Tag": [ "graph theory", "combinatorics proposed", "combinatorics" ], "Problem": "Let $m = 30030$ and let $M$ be the set of its positive divisors which have exactly $2$ prime factors. Determine the smallest positive integer $n$ with the following property: for any choice of $n$ numbers from $M$, there exist 3 numbers $a$, $b$, $c$ among them satisfying $abc=m$.", "Solution_1": "I think it's $11$. $30030=2\\cdot 3\\cdot 5\\cdot 7\\cdot 11\\cdot 13$. We regard the six prime divisors as the vertices of a graph, and turn the problem into the following, more general one:\r\n\r\nFind the minimal $e=e(t)$ with the property that every graph on $2t$ vertices with $e$ edges has a perfect matching (in our case, $2t=6$). \r\n\r\nThe answer should be $\\frac{(2t-1)(2t-2)}2+1$ (which gives $11$ when $2t=6$), I believe. We can prove this as follows:\r\n\r\n$\\frac{(2t-1)(2t-2)}2$ certainly isn't enough, since we can take a complete graph on $2t-1$ vertices, and an isolated vertex. This graph obviously has no perfect matching. On the other hand, $\\frac{(2t-1)(2t-2)}2+1$ is enough, since it has been shown [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=hamiltonian&t=45004]here[/url] that any graph on $u$ vertices with $\\frac{(u-1)(u-2)}2+2$ edges is hamiltonian. We apply his to $u=2t$ to conclude that our graph, which has $2t$ vertices but only $\\frac{(2t-1)(2t-2)}2+1$ edges, must have a hamiltonian path $x_1,x_2,\\ldots,x_{2t}$. Our perfect matching will then consist of the pairs of vertices $(x_i,x_{i+1}),\\ i\\in\\{1,3,\\ldots,2t-1\\}$.", "Solution_2": "I'm sorry to ask, Grobber, but how is relation between two vertices defind?", "Solution_3": "Two vertices=prime numbers are connected iff their product occurs in the set of that $n$ numbers." } { "Tag": [ "geometric series", "calculus", "calculus computations" ], "Problem": "If $ r>1$, then prove that $ \\sum_{n\\equal{}1}^{\\infty} \\frac{1\\plus{}r\\plus{}r^2\\plus{}\\cdots \\plus{}r^{n}}{(1\\plus{}r)^n}$ converges and find the sum $ S$ in terms of $ r$.", "Solution_1": "[quote=\"kunny\"]If $ r > 1$, then prove that $ \\sum_{n = 1}^{\\infty} \\frac {1 + r + r^2 + \\cdots + r^{n}}{(1 + r)^n}$ converges and find the sum $ S$ in terms of $ r$.[/quote]\r\n\r\nThe solution is very easy. We have\r\n\\[ S(r): = \\sum_{n = 1}^{\\infty}\\sum_{i = 0}^{n}\\frac {r^i}{(1 + r)^n} = \\sum_{n = 1}^{\\infty}\\frac {r^{n + 1} - 1}{(r - 1)\\cdot (1 + r)^n}.\\]\r\nHence, if $ r > 1$, the sum indeed converges. Next, we have for $ r > 1$\r\n\\begin{align*} S(r) & = \\frac {r}{r - 1}\\sum_{n = 1}^{\\infty}\\left (\\frac {r}{1 + r}\\right )^n - \\frac {1}{r - 1}\\sum_{n = 1}^{\\infty}\\frac {1}{(1 + r)^n} = \\\\\r\n& \\\\\r\n& = \\frac {r}{r - 1}\\cdot\\frac {r}{1 + r}\\cdot\\frac {1}{1 - \\frac {r}{1 + r}} - \\frac {1}{r - 1}\\cdot\\frac {1}{1 + r}\\cdot\\frac {1}{1 - \\frac {1}{1 + r}} = \\\\\r\n& \\\\\r\n& = r + 1 + \\frac {1}{r}. \\end{align*}", "Solution_2": "[quote=\"M@ri@n\"][quote=\"kunny\"]If $ r > 1$, then prove that $ \\sum_{n \\equal{} 1}^{\\infty} \\frac {1 \\plus{} r \\plus{} r^2 \\plus{} \\cdots \\plus{} r^{n}}{(1 \\plus{} r)^n}$ converges and find the sum $ S$ in terms of $ r$.[/quote]\n\nThe solution is very easy. We have\n\\[ S(r): \\equal{} \\sum_{n \\equal{} 1}^{\\infty}\\sum_{i \\equal{} 0}^{n}\\frac {r^i}{(1 \\plus{} r)^n} \\equal{} \\sum_{n \\equal{} 1}^{\\infty}\\frac {r^{n \\plus{} 1} \\minus{} 1}{(r \\minus{} 1)\\cdot (1 \\plus{} r)^n}.\\]\nHence, if $ r > 1$, the sum indeed converges. [/quote]\r\n\r\nWhy does the sum converge?", "Solution_3": "We can use Raabe's criterion.", "Solution_4": "That's new to me :roll: , but I examined it by Google.:", "Solution_5": "[quote=\"kunny\"]\nWhy does the sum converge?[/quote]\r\nLet $ S(r): \\equal{} \\sum_{n \\equal{} 1}^{\\infty}\\sum_{i \\equal{} 0}^{n}\\frac {r^{i}}{(1 \\plus{} r)^{n}} \\equal{} \\sum_{n \\equal{} 1}^{\\infty}\\frac {r^{n \\plus{} 1} \\minus{} 1}{(r \\minus{} 1)\\cdot (1 \\plus{} r)^{n}}$. \r\n\r\nLet $ A(r): \\equal{} \\sum_{n \\equal{} 1}^\\infty\\left(\\frac {r}{r \\plus{} 1}\\right)^n$, and $ B(r): \\equal{} \\sum_{n \\equal{} 1}^\\infty\\left(\\frac {1}{r \\plus{} 1}\\right)^n$. Becouse $ r > 1$ both $ A(r)$ and $ B(r)$ converge as geometric series, and becouse $ S(r) \\equal{} \\frac {r}{r \\minus{} 1}A(r) \\minus{} \\frac {1}{r \\minus{} 1}B(r)$, $ S(r)$ converges.", "Solution_6": "The reply what I expected is $ 0<\\frac{r}{1\\plus{}r}<1,\\ 0<\\frac{1}{1\\plus{}r}<1$ for $ r>1$, although you might say the fact is self-evident.\r\n\r\nAnyway M@ri@n7s answer to $ S$ was correct." } { "Tag": [ "MATHCOUNTS", "HCSSiM", "AMC", "AIME", "geometry" ], "Problem": "How did you do at the Chapter MATHCOUNTS competition? Feel free to list your test and countdown results as well as your team results. How many of you will be competing at the state level?", "Solution_1": "Here's our school's results for the chapter:\r\n\r\nSchool: CMS (its been abbreviated)\r\nChapter: Seacoast\r\nState: New Hampshire\r\n----------\r\nTeam Results:\r\nOur Team got indiv. places: 1,2,3,4\r\nOur team average: 38.25 +20 team round points\r\nWe got 1st in chapter by 20 points\r\n----------\r\nMy Results:\r\nI got a 37 and placed 3rd\r\n----------\r\nIndividual Results:\r\nOur Individuals got places: 5,7,10,-- (I think)\r\n----------\r\nCounddown Results (Unofficial in our chapter)\r\nOur qualifiers got: 1,3,3,...\r\n----------\r\nSheer ownage of our chapter", "Solution_2": "School: Smith Middle School\r\nTown: Glastonbury, Connecticut\r\nChapter: Greater Hartford\r\n\r\nSchool Results:\r\nOur Team placed 2nd :( :\r\nPlacings were 4, 12, 19, 123; Team Round=8 \r\nAnother individual placed 6th with a 37\r\n\r\nMy Results:\r\n1st Countdown (Unofficial with 16 competitors)\r\n4th Written (39 {8 Target, 23 :( :( :( :( :( Sprint})\r\n\r\nThree of our people made countdown, but two failed to advance past the first round\r\n\r\nI hope I do better at states... I was pretty upset with my performance at Chapters...", "Solution_3": "14th individual (State- NC)\r\nIndividual Score-39 (8 target, 23 sprint)\r\nSchool score- Dont even ask.....\r\n\r\nDid I make it to states- No :( \r\n\r\nHockey you are lucky........", "Solution_4": "[quote=\"GoBraves\"]14th individual (State- NC)\nIndividual Score-39 (8 target, 23 sprint)\nSchool score- Dont even ask.....\n\nDid I make it to states- No :( \n\nHockey you are lucky........[/quote]\r\n\r\nThat sounds like an extremely tough chapter.", "Solution_5": "We had like a 10 person tie, but they said i missed a problem earlier or something.", "Solution_6": "Ya... ours came down to a bunch of tiebreakers... that's how it always is in CT... one problem drops you like 2-10 spots at best, depending on where you are in the rankings....\r\nGo Braves... I feel bad for you... must have a tough tough chapter...", "Solution_7": "Dallas chapter: \r\n\r\ni got 29/7 (43)\r\n\r\nsecond place \r\n\r\nteam got 8 and second place", "Solution_8": "Clayton/Henry/Fayette county chapter\r\nCherish Christain Home Educators\r\nGeorgia\r\nHad 1 individual in top 6, he got #1 spot\r\nAnd he took that team to the top!\r\n\r\nIf you didn't figure it out, that's me.\r\n\r\nI don't know the exact score yet\r\n\r\nBilly", "Solution_9": "School: BCS\r\n\r\nChapter: Oakland\r\n\r\nState: Michigan\r\n\r\nMy Score: 44 (28/8), 2nd place\r\n\r\nSchool Score: 55.25 (35.25/10), 5th place\r\n\r\nCountdown (10 person Ladder): 3rd :( \r\n\r\nMade it to states", "Solution_10": "[quote=\"MCrawford\"][quote=\"GoBraves\"]14th individual (State- NC)\nIndividual Score-39 (8 target, 23 sprint)\nSchool score- Dont even ask.....\n\nDid I make it to states- No :( \n\nHockey you are lucky........[/quote]\n\nThat sounds like an extremely tough chapter.[/quote]\r\n\r\nIn my chapter, the 7th place written was 42. 2nd place team had no people in countdown, but an average written of 39.75.", "Solution_11": "Hm, Connecticut's state competition is going to be pretty interesting...", "Solution_12": "I want to get a 25/7 as my baseline for states... but that's realy baseline.. but I kno that I can get a 28 to thirty on a state sprint (was 2003 considered easy? I got a 30 on it...) and I'm hoping that I don't make too many stupid mistakes.. it just involves having 10-15 minutes to check....", "Solution_13": "I thought 2003 states was easy, but there were some weird scores on it. I took it as practice last year and got a 46, but people I knew who are really smart only got mid-30s on it. So look for mixed results. Top 4 in Connecticut 2003- 36,35,34,33\r\n\r\nEdit- oops, never mind. that's 2002. CT's 2003 had 43, 42, 40, 39", "Solution_14": "In 2003, I got a 43 (29/7). The mistakes I made were, to say the least, retarded. And CA had 2 perfects, so a 43 was 6th (in our Northern CA competition)", "Solution_15": "Defiance, is that a neopets picture?", "Solution_16": "[quote=\"uberl33tmage\"]I did a lot better this year. i made it in the 4th seat. Left_unattended, I'm surprised. BTW nice name.. WHAT HAPPENED TO YOU!!!? I was expecting a #1 or 2. I'm sorry you had to miss nationals.. :( .[/quote]\r\nWho is she? (left_unattended)", "Solution_17": "The SC Nats team has long been a joke, coming in 53rd last year. Watch Coach John Rushman turn it around with a great SC team ths year. Watch Phillip Cross in the Countdown Round. Remember, I called it! SC in the top 20!", "Solution_18": "Something cool: The Crayton Middle School team placed 1st regionally, and Hand Middle was 3rd. AC Flora High School feeds from Crayton; Dreher High feeds from Hand Middle. So the 3rd place Hand team went to a high school competition and came in 1st. That's 3rd in middle school, but 1st in high school. That shows how advanced MathCounts is!!!", "Solution_19": "[quote=\"Press alt f4\"]I'm from the Greater Houston Chapter which is probaboly (sp?) one of the hardest Chapters. I was over all disapointed with my preformance, the two individuals from our school did better than me. Oh well it was only a point, and I'm still going to state. Well here are the results.\nTeam -\n1st place team score (63.75) (yes it really is that high, they had two 46's)\n2nd place team score (61) :( \nMy Team #3 score (60.75) :( ONE QUESTION!!\nIndividuals -\n#1 - 2 (46)\n#3 (45)\n#4 - 6 (44)\n#7 - 10 (42)\n#11 ME! (41) :([/quote]\r\n\r\n\r\nWhos PressAltF4?\r\nAre you Drew?\r\nIm Dennis and Im also from that chapter. \r\nI got a perfect and was on the winning team.\r\n\r\nBtw, 2nd place team (Doerre) got a 60.75 and you guys (First Colony) got a 60.50.", "Solution_20": "Too bad this is Jenks' last year on top of both Tulsa and Oklahoma...", "Solution_21": "Got scores. 36,5 for team. Individuals got 40,22,14,14.", "Solution_22": "7th in chapter with 36 :( , Second in Countdown round :lol: Did not make it to state, countdown round unofficial :( 7th grade .beat :first: all 7th graders in chapter. I'll be coming next year, but Calafornia State looks pretty hard!", "Solution_23": "I got 40 points and my team got first:\r\n\r\n[color=darkblue][size=75][Moderator rcv's note: Off-topic discussion of the Northwest Chinese School moved to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31669]this thread[/url]][/size][/color]", "Solution_24": "I got 22 on the Sprint Round and all eight on the Target Round, giving me a score of 38 points. That placed me second overrall before the countdown round. I pulled ahead to first place in the countdown round :D and went on to state.", "Solution_25": "I'm pretty sure I got a 28/7 for 42 which made me 1st place, and I also won countdown. 2nd place was a 40 from the same team... third place was a 30-something from the same team.. I'm pretty sure we got really really close to a 60 for overal team score though. Our chapter is so easy!", "Solution_26": "Mine:\r\n\r\n1st in Countdown\r\n2nd Indivisual (Score - I think I got a 16 on target but I don't know what my sprint score was)\r\n\r\nTeam:\r\n2nd place we got team members 2nd, 3rd 4th 5th, but some random school comes up and beats us in the team round, yep that's how bad we did as a team\r\nAlso another indivisual not in the team got 6th and another 12th\r\n\r\nThis is the first year that we didn't win in our chapter :(", "Solution_27": "In our chapter competition, we got 4th place team (Out of 5 teams!)", "Solution_28": "um...\r\nHouston Chapter--\r\nMy scores were:\r\n25/8 (41)\r\n11th place.... \r\n\r\nTeam results:\r\nindividuals got 11th, 37th, 64, 68 (out of like 300 students)\r\nindivid scores were 41, 38, 29, 28\r\nTeam score of 7 \r\nOur team got 6th in chapter, advanced to state", "Solution_29": "Hmm...I actually dunno my score, but my coach does, he just hasn't told me (grr!)\r\n\r\nAnyway, I'm in the Greater Cleveland Chapter, I guess you could call it, which is about average difficulty I'd say with 15 teams and over 100 people. Our team placed first written and I took first individual and a teammate took third; an individual not on the team finished somewhere in the top 16 advancing him to states with the team. This is weird: our chapter does a team countdown round that is unoffical; we go down there as a team against 3 others at a time and face off; the winners of each bracket then go to the finals. We swept down there, getting 3 points instantly both rounds we were up. It was great!" } { "Tag": [ "SFFT", "videos", "Vieta", "special factorizations" ], "Problem": "Solve\r\n\r\n$ x\\plus{}y\\equal{}7$\r\n\r\n$ \\frac{x}{y} \\plus{} \\frac{y}{x} \\equal{} \\frac{25}{12}$", "Solution_1": "hello, hint plugg $ y\\equal{}7\\minus{}x$ in the second equation.\r\nThe solutions are $ (4,3)$ or $ (3,4)$.\r\nSonnhard.", "Solution_2": "[hide] The second equation can be rewritten as $ \\frac{x^2\\plus{}y^2}{xy}\\equal{}\\frac{25}{12}$. Squaring the first equation we find that $ x^2\\plus{}2xy\\plus{}y^2\\equal{}49 \\implies x^2\\plus{}y^2\\equal{}49\\minus{}2xy$. Substituting into our second equation, we have\n\n$ \\frac{49\\minus{}2xy}{xy}\\equal{}\\frac{25}{12}$\n\n$ \\implies \\frac{49}{xy}\\minus{}2\\equal{}\\frac{25}{12}$\n\n$ \\implies \\frac{49}{xy}\\equal{}\\frac{49}{12} \\implies xy\\equal{}12$.\n\nNow, we can use SFFT to show that $ (x\\plus{}1)(y\\plus{}1)\\equal{}20$, which when combined with the first equation, yields the solutions as $ (4,3)$ and $ (3,4)$.[/hide]", "Solution_3": "What is SFFT?", "Solution_4": "Simon's Favorite Factoring Trick (with American spelling). If you google this, you'll find two videos on it: they are excellent.\r\n\r\nSFFT is very useful.", "Solution_5": "From the second equation, $ 12x^2\\minus{}25xy\\plus{}12y^2\\equal{}0 \\implies 12(x\\plus{}y)^2\\equal{}49xy \\implies xy\\equal{}12$\r\n\r\nBy Vieta's, it is then just the solutions to $ a^2\\minus{}7a\\plus{}12\\equal{}0$ , i.e. $ (3,4) or (4,3)$" } { "Tag": [ "geometry", "algorithm", "number theory", "greatest common divisor", "analytic geometry", "graphing lines", "slope" ], "Problem": "I thought that it would be cool to make a topic about programming calculators, like the TI-84, or the TI-89(If it is possible, the TI-$ n$spire).\r\n\r\nI currently have three programs for the TI-84. One on Binet's formula, one for the area of a regular polygon, and one for the prime factorization of a number(can anyone teach me to scroll? :oops: ), if anyone needs it.", "Solution_1": "I used to have a \"great\" prime factorization program that a bit bruteforce...\r\n\r\nit gives the factors, ordered from smallest to largest and in a list :)\r\nThe problem is that its still bruteforce :(\r\nwhat's your factorization algorithm?", "Solution_2": "Basically I prompted N, let M=1, and if the gcd of M and N is greater than 1, then we divide N by M, store to N, and we post M, let M=M+1 if gcd(M,N)=1, and so on, until $ M > \\sqrt {N}$, at which we post N. So for 4759, you would wait until M is about 70, and then it posts 4759. Or for 10023, we have 3, 13, and 257.\r\n\r\nIt's the algorithm for testing for primes.\r\n\r\nEDIT: It uses 93 units of space. If anyone has a smaller program, I'd like to see it.", "Solution_3": "I find distance, slope, and midpoint formulas helpful.\r\n\r\nHere is my code (TI-84):\r\n\r\nClrHome\r\nDisp \"COORDINATES\",\"(X,Y),(A,B)\"\r\nPrompt X,Y,A,B\r\nClrHome\r\nB-Y->C\r\nA-X->D\r\nB+Y->E\r\nA+X->F\r\n(E+F)/2->H\r\nsqrt(C^2+D^2)->I\r\nDisp \"DISTANCE:\",I\r\nDisp \"DISTANCE^2:\",I^2\r\nPause\r\nClrHome\r\nF/2->J\r\nE/2->K\r\nDisp \"MIDPOINT:\",\"X VALUE\",J,\"Y VALUE\",K\r\nPause\r\nClrHome\r\nIf D=0\r\nThen\r\nDisp \"NO SLOPE\"\r\nElse\r\nDisp \"SLOPE:\",C/D->Frac,C/D->Dec\r\n\r\nUses 234 units of space. Very efficient.\r\n\r\nOh, and if you don't wanna work for your programs, go [url=http://www.ticalc.org/pub]here.[/url]", "Solution_4": "For a factors program (which I find to be more useful), I just use a while statement, testing whether $ n$ is divisible by any integers 1 to $ \\sqrt {n}$. Simply do if n mod i = 0 display it, then divide n by i and display that.\r\n\r\nFor prime factorization use the factor() function on TI-89 :) .", "Solution_5": "...anyone here program in TI-assembly code?\r\nI made Tetris on the 84 with it... :)", "Solution_6": "1=2, could you post the exact code for your factoring program? Thanks.\r\n\r\nI do have a \"last digit\" program that chess64 gave me.", "Solution_7": "[hide=\"code\"]Prompt N\n1->M\nLbl 0\nIf M>sqrt(N)\nThen\nGoto 1\nElse\nIf gcd(M,N)=1\nThen\nM+1->M:Goto 0\nElse\nDisp M:Pause\nN/M->N\nIf N=1\nThen\nGoto 1\nElse\nGoto 0\nLbl 1\nDisp N[/hide]\r\n\r\nWhat's a \"last digit\" program?", "Solution_8": "basically for my program of factoring (it got mem-deleted)\r\n[hide]\n[code]\nPrompt A\n{1}->L1\nWhile fPart(A/2)=0\nA/2->A\naugment(L1,{2})->L1\nEnd\n1->B\nWhile B <= sqrt(A)\nB+2->B\nWhile fPart(A/B)=0\nA/B->A\naugment(L1,{B})->L1\nEnd\nEnd\nA->L1(1)\nL1->L1\n[/code]\n\n-> means store variable and <= means less than or equal to\n[/hide]\r\n\r\nThis code cuts time in about half by skipping even numbers larger than 2 :)", "Solution_9": "My math teacher made this program (I think).\r\n\r\nInput B\r\n(0)->A\r\nLbl 1\r\n(A+1)->A\r\n(B/A)->C\r\nint(C)->D\r\nIf (C-D) $ \\neq$ 0 or (D-C) $ \\neq$ 0\r\nGoto 1\r\n(B/C)->E\r\nint(E)->F\r\nIf (E-F) $ \\neq$ 0 or (F-E) $ \\neq$ 0\r\nGoto 1\r\nDisp C\r\nDisp E\r\nPause\r\nIf AFrac\nIf B=0\nDisp \"BAD PROBLEM\"\nStop:End\nB^2-4AC->D\nDisp \"X=\",-B/(2A)+sqrt(D)/2A->Frac\nDisp -B/(2A)-sqrt{D}/2A->Frac[/hide]", "Solution_10": "Here is a not-very-useful program that can tell you the coefficient of a term in the expansion of (ax+by)^c.\r\n\r\nClrHome\r\nDisp \"(AX+BY)^Z\"\r\nPrompt A,B,Z\r\nClrHome\r\nInput \"X TERM?\",T <---See Note, Below.\r\nInput \"POWER?\",P\r\nIf P>Z\r\nThen\r\nDisp \"ERROR!\"\r\nElse\r\nA^P->D\r\nD * Z nCr P-> E\r\nZ-P->F\r\nClrHome\r\nIf T$ \\neq$0\r\nDisp E,\"X=\",P,\"Y=\",F\r\nIf T=0\r\nDisp I,\"X^\",F,\"Y=\",P\r\n\r\nNote: If T=0, then you know the Y term power. If T$ \\neq$0, then you know the X term power.", "Solution_11": "Just a general comment: Lbl and Goto's are bad. Like.... bad. Try not to use them. :wink: \r\nOh, and ernie, your If-Then-Else needs an End somewhere...", "Solution_12": "[quote=\"Brut3Forc3\"]Just a general comment: Lbl and Goto's are bad. Like.... bad. Try not to use them. :wink: \nOh, and ernie, your If-Then-Else needs an End somewhere...[/quote]\r\n\r\nI know.\r\n\r\nIt was my teacher's program.\r\n\r\nAnd I almost never put End's.\r\n\r\nIt still works. :P", "Solution_13": "I have a program for graphing the Mandelbrot set. And I did not make this, one of my friends did. Pretty useless though.\r\n\r\n[hide]\n\n[code]\n\n-2->Xmin\n1->Xmax\n1->Ymax\n-1->Ymin\nClrDraw\nAxesOff\na+bi\n(Xmax-Xmin)/94->H\n(Ymax-Ymin)/62->V\nFor(I,0,93)\nFor(J,0,61)\nXmin+IH->X\nYmin+JV->Y\nX+Yi->C\n0->Z\nFor(N,1,) *The larger the number, the more accurate the program is\nIf abs(Z)<=2\nC->Z\nEnd\nIf abs(Z)<=2\nPt-On(real(C),imag(C))\nDispGraph\nEnd\nEnd\nStorePic 7\n\n[/code]\n\nNote the last step is optional.\n[/hide]\n\nAlso, Shoelace algorithm\n\n[hide]\n\n[code]\n\nClrHome\nPrompt N\nFor(C,1,N)\nPrompt X\nPrompt Y\nClrHome\nX->L1(C)\nY->L2(C)\nEnd\nL1(1)L2(N)->S\nL1(N)L2(1)->T\nFor(D,2,N)\nS+L1(D)L2(D-1)->S\nT+L1(D-1)L2(D)->T\nEnd\nClrHome\nabs(S-T)/2->A\nDisp A\n\n[/code]\n\nN is the number of vertices of the polygon and X and Y are the coordinates\n\n[/hide]", "Solution_14": "I've also made MONOPOLY on my calc! It's like 3 KB though and doesn't handle chance cards...\r\n\r\nwill release code soon", "Solution_15": "[quote=\"Brut3Forc3\"]Just a general comment: Lbl and Goto's are bad. Like.... bad. Try not to use them. :wink: [/quote]\r\n\r\n1) How bad is bad?\r\n2) I don't know how to eliminate them in my factoring program. Can you tell me how?", "Solution_16": "[quote=\"james4l\"]basically for my program of factoring (it got mem-deleted)\n[hide]\n[code]\nPrompt A\n{1}->L1\nWhile fPart(A/2)=0\nA/2->A\naugment(L1,{2})->L1\nEnd\n1->B\nWhile B <= sqrt(A)\nB+2->B\nWhile fPart(A/B)=0\nA/B->A\naugment(L1,{B})->L1\nEnd\nEnd\nA->L1(1)\nL1->L1\n[/code]\n\n-> means store variable and <= means less than or equal to\n[/hide]\n\nThis code cuts time in about half by skipping even numbers larger than 2 :)[/quote]\r\n\r\nThere you go, 1=2.\r\n\r\nThe reason why Labels and Gotos are bad are that they require massive amounts of time searching for the label, so While and For loops are often MUCH faster.\r\n\r\nMy factoring program factors typical 7-digit numbers in about 3-4 seconds, and for each digit added on, the time increases about 1.5 fold", "Solution_17": "ah.\r\n\r\nnice program. I can actually scroll through too! :D\r\n\r\nIt cut the time in a fourth for finding that 4759 is prime, and has the memory capacity for $ 2^{2^5}\\plus{}1$.", "Solution_18": "Yeah, but if TI-84 doesn't SUPPORT Java/C then what's the point?\r\n\r\nRead the title. Not computers, calculators... :roll:", "Solution_19": "[quote=\"1=2\"]What's a \"last digit\" program?[/quote]\r\n\r\nLike, find the units digit of $ 3^{1337}$. I'll post later when I get home.", "Solution_20": "[quote=\"1=2\"][quote=\"Brut3Forc3\"]Just a general comment: Lbl and Goto's are bad. Like.... bad. Try not to use them. :wink: [/quote]\n\n1) How bad is bad?\n2) I don't know how to eliminate them in my factoring program. Can you tell me how?[/quote]\r\n[url]http://xkcd.com/292/[/url]\r\nThat's how bad.\r\n\r\nI've also made some programs -a snake game, a linear diophantine equation solver, a work generator for simplifying fractions in the form of $ \\frac {a \\plus{} bi}{c \\plus{} di}$ (I got lazy in class), and an epic animation that composed of for loops to generate lag, and outputing long strings that I had calculated myself... it had a bomb, snake, pong, and everything... too bad I broke my calculator... :ewpu: :bomb: :furious:\r\nEDIT: Oh, I also made a fake menu that let you fake deleting memory, so you could keep your programs on a test. But that was on assembly... BASIC is way too laggy to actually fake anything.", "Solution_21": "Tinspire has crazily limited programming capabilities.", "Solution_22": "Easy:\r\nInput screen:\r\nNO!\r\nNO!\r\nNO!\r\nNO!\r\nNO!\r\nNO!\r\nNO!\r\nCode:\r\nLbl 1\r\nDisp \"NO!(factorial)\"\r\nGoto 1", "Solution_23": "[hide=\"Pascal's Triangle\"]Input \"ROW?\", A\nIf A < 0\nThen\nClrHome\nDisp \"ROWS CANNOT BE\",\"NEGATIVE.\"\nPause\nprgmPSCLTRI\nEnd\nIf A >= 0\nThen\n-1 -> B\nLbl C\nB + 1 -> B\nIf B <= A\nThen\nDisp A nCr B\nPause\nGoto C\nEnd\nIf B > A\nThen\nStop:End\n[/hide]\n\n[hide=\"Binomial Expander\"]ClrHome\nDisp \"(AX+BY)^C\"\nPrompt A,B,C\n-1 -> D\nLbl E\nD + 1 -> D\nIf D <= C\nThen\nClrHome\nOutput (3, 1, C nCr D * A ^ (C - D) * B ^ D) \nOutput (4, 1, \"*X^\")\nOutput (4, 4, C - D)\nOutput (5, 1, \"*Y^\")\nOutput (5, 4, D)\nPause\nGoto E\nEnd\nIf D > C\nThen\nStop:End[/hide]\n\n[hide=\"Factor\"]Input A\n0 -> B\nLbl C\nB + 1 -> B\nIf B <= sqrt A\nThen\nIf A/B = int (A/B)\nThen\nDisp B, A/B\nPause\nGoto C\nEnd\nEnd\nIf B > sqrt A\nThen\nStop:End[/hide]\r\n\r\nWARNING: All three use Lbl's and Goto's. :P", "Solution_24": "[hide=\"Pascals Triangle\"]Input \"ROW?\", A\nIf A < 0\n Then\n ClrHome\n Disp \"ROWS CANNOT BE\",\"NEGATIVE.\"\n Pause\n prgmPSCLTRI\nEnd\nIf A >= 0\n Then\n -1 -> B\n While B <= A\n B + 1 -> B \n Disp A nCr B\n Pause\n End\nEnd\n[/hide]\r\nI think this works better. (Quote me to see the structure better.)", "Solution_25": "ok... all this hype about goto is ridiculous\r\nall practical programs dont change significantly\r\noh well here is an uber pwn program\r\n\r\nnote that => is the store function... too lazy to make a LaTeX code for it\r\n:Delvar z\r\n:ClrHome\r\n:For(A,1,16)\r\n:randInt(1,4,A)=>$ L_1$\r\n:For(B,1,A)\r\n:Output(1,B,$ L_1$(B))\r\n:End\r\n:For(B,1,100A)\r\n:End\r\n:ClrHome\r\n:For(B,1,A)\r\n:Delvar C\r\n:While c>4 or c<1\r\n:getKey=>C\r\n:C-91+(C=95)+13(C=82)=>C\r\n:End\r\n:If C$ \\neq L_1$(B)\r\n:Goto 1\r\n:End\r\n:Z+1=>Z\r\n:End\r\n:Lbl 1\r\n:ClrHome\r\n:Disp \"SCORE:\"\r\n:Output(1,7,Z)", "Solution_26": "It is NOT ridiculous, most bruteforce-ish programs that encounter too many Gotos and Lbls and Menus slow down considerably. Consider my RPG game, in around 500 moves, and 120 battles (battles used Labels back then) the program slowed down about 60%. In about 1200 moves the game is unplayably slow [which was why I had savegames].", "Solution_27": "but that game isnt practical; most practical programs should be like under 300 bytes\r\nanyway who has tried my program?\r\nat least it has no goto's except for 1 and that is used only once :D\r\nthe programs pretty beast", "Solution_28": "[quote=\"stevenmeow\"]but that game isnt practical; most practical programs should be like under 300 bytes\nanyway who has tried my program?\nat least it has no goto's except for 1 and that is used only once :D\nthe programs pretty beast[/quote]\r\n\r\nNot always. Many practical programs like spreadsheets, probability simulators, and \"calculator in calculator\" programs can take upward of 10kb.\r\n\r\nAnd consider this (admittedly impractical) program:\r\n\r\n0->A\r\nLbl A\r\nA+1->A\r\nDisp A\r\nGoto A\r\n\r\nThis should be obvious that it is a counter that displays 1,2,3,4,5,6,7,...\r\n\r\nBut at higher numbers [I think it is at 6000-10000, not so sure] it displays ERR:MEMORY.\r\n\r\nOf Course we can substitute this particular program with:\r\n\r\nFor(A,1,999999)\r\nDisp A\r\nEnd\r\n\r\nYes, One-Time Gotos/Lbls are generally O.K. But if the Gotos and Lbls are being used like loops that is intolerable to the calculator. Personally I don't use Gotos or Lbls, I use subroutines.", "Solution_29": "whats a subroutine? and you dont have to close your parentheses and quotes\r\nEDIT:\r\nFor(Q,1,9999999\r\nGarbageCollect\r\nEnd\r\nill put some thought into making an actual program later but now im too lazy", "Solution_30": "[quote=\"ternary0210\"]Easy:\nInput screen:\nNO!\nNO!\nNO!\nNO!\nNO!\nNO!\nNO!\nCode:\nLbl 1\nDisp \"NO!(factorial)\"\nGoto 1[/quote]\r\n\r\nA better way (the program must be called \"NO\"):\r\n\r\n[code]Disp \"NO!\"\nprgmNO[/code]", "Solution_31": "did i mistype something? because as far as im concerned, all stevenmeow's program does is see if you're patient enough to wait before you type in a number\r\nOh Wait, i missed a clrhome", "Solution_32": "I made this really beast program that cuts time in chemistry homework basically in HALF. Since its an epic pain to keep looking up atomic masses and then typing them in, it would work so much better if I could just have the masses stored in variables on my calculator. This would work great for single-letter elements, but then it fails for two letter ones. So, using omnicalc to type the lower case letters, I made a program, that, when turned on, will parse my calculations and replace any single uppercase letter of an uppercase letter followed by a lowercase letter by the corresponding atomic of that element.\r\n\r\nThis works much better than all the programs people at my school tried to make a program that just accept a string and then prints out the corresponding mass of one element at a time :P", "Solution_33": "oh, woohoo. my quartic solver failed so i just made a minigame\r\n[hide]im sure this can be shortened and improved massively but w/e\ndelvar c\nclrhome\nfor(a,1,4\nrandint(10^(A-1),10^A-1,8)-L1\nfor(B,1,8\noutput(b,1,b\noutput(b,2,\")\noutput(b,3,L1(b))\nEnd\nfor(B,1,250+50A\ndelvar d\ngetkey->d\nif D<=94 and D>=72:Then\nD-91+13(D<85)+13(D<75)->d\nfor(E,1,8\nIf L1(d)>L1(E)\ngoto 1\nend\noutput(C,1,\"*16 spaces*\n10^(A)->L1(D)\nEnd\nEnd\nfor(E,1,8\nIf L1(Q) =/= 10^(A)\ngoto 1\nend\nC+1->C\nend\nlbl 1\ndisp \"score:\noutput(1,7,c\n[/hide]", "Solution_34": "Ok, if you have TI-84+ (SE or not), then you have a polynomial root finder and simontaneous equation solver.", "Solution_35": "ono my previous program had a bunch of errors. ill try again\r\nthis one is a bit lengthy(350 bytes or so)\r\n->=sto\r\n[hide]clrhome\ndisp \"score:\ndelvar a\nrandint(0,2,5)->L1\nrandint(0,0,5)->L2\ngetkey\nFor(b,1,100\noutput(1,7,\"*4spaces*\noutput(1,7,a\nfor(C,1,5\nif L1(c)=0\noutput(8,3C-1,\"*space*\nif L1(C)=1 or L1(C)=2\noutput(8,3C-1,L1(C)-1\nend\ngetkey->c\nif int(c/10)=1:then\nC-10->C\nif L1(c)=2\nA+1->A\nif L1(C)=1\nA-1->A\nif L1(C) =/= 0\n0->L2(C)\n0->L1(C)\noutput(8,3C-1,\"*space*\nend\nfor(d,1,5\nL2(D)+1->L2(D)\nIf L1(D)=/=0 and L2(d)=4:then\noutput(8,3D-1,\"*space*\n0->L1(d)\n0->L2(D)\nend\nif L1(d)=0 and L2(D)=2>:then\nrandint(1,2)->L1(D)\n0->L2(D)\nend:end:end\ndisp \"game finished\npause:clrhome[/hide]", "Solution_36": "I made a program for simulating a risk battle (basically no max on number of soldiers),\r\nand one for calculating chances of each outcomes (max 40 soldiers on defender and offender).\r\n\r\nI also made a program that displays a word and moves it in a direction in the screen...and bounce off edges...", "Solution_37": "My factoring program(uses a Goto) uses less memory than james4l's one, which doesn't use a Goto. The difference is only 18 bytes, though, and I can't scroll up on mine. :(\r\n\r\n@lifeisacircle: Your shoelace algorithm is very practical! Only 140 bytes of space!\r\n\r\n@ernie: What does your factor program exactly do? When I input 75, it displayed 1, 75, then it ended. When I put in 144, it displayed 1,144, 2,72, 3,48, and 4,36. Then it ended.", "Solution_38": "Oh, that's I think because I typoed. :blush: \r\n\r\n[hide=\"Factor Program\"]Input A\n0->B\nLbl C\nB+1->B\nIf B <= sqrt A\nThen\nIf A/B = int(A/B)\nThen\nDisp B,A/B\nPause\nEnd\nGoto C\nEnd\nIf B > sqrt A\nThen\nStop:End[/hide]\n\nThat should work.\n\n[hide=\"Modular Arithmetic Program\"]ClrHome\nDisp \"A^B MOD N\"\nPrompt A,B,N\nIf A<0 or B<0 or N<0 or int(A) $ \\neq$ A or int(B) $ \\neq$ B or int(N) $ \\neq$ N\nThen\nClrHome\nDisp \"A,B, AND N MUST\",\"BE WHOLE NUMBERS\"\nPause\nprgmMODARITH\nElse\nLbl C\n(B/N-fPart(B/N))+(fpart(B/N)+N)->B\nIf B>N\nThen\nGoto C\nElse\nfPart(A^B/N) * N->D\nIf D>0.9\nThen\n1->D\nEnd\nIf D<0.1\nThen\n0->D\nEnd\nOutput(5,1,\"A^B MOD N=\")\nOutput(5,11,D)[/hide]", "Solution_39": "Is there anyway to display a radical symbol in an answer? (Such as the solution to the quadratic formula but preferablly using radicals instead of decimals). I believe this is possible now, thanks to James' factoring program, but I'm too lazy to think of how to make it.", "Solution_40": "[quote=\"1=2\"]I thought that it would be cool to make a topic about programming calculators, like the TI-84, or the TI-89(If it is possible, the TI-$ n$spire).\n\nI currently have three programs for the TI-84. One on Binet's formula, one for the area of a regular polygon, and one for the prime factorization of a number(can anyone teach me to scroll? :oops: ), if anyone needs it.[/quote]\r\nwhat's the code for the prime factorization progam? :)", "Solution_41": "heh sry for reviving the topic :oops: \r\n\r\nbut I just got a TI-$ n$spire and I want to know what programs I can put into it. I can program it with the TI-84+ keypad, but not with the TI-$ n$spire keypad. Does anyone know any good programs?\r\n\r\nThanks in advance :lol:" } { "Tag": [ "MATHCOUNTS" ], "Problem": "Do the competition sets have the countdown round or solutions? Here's the link:\r\n\r\n[url]http://secure.sportsawardsonline.com/applications/default/store/index.asp?product_category_id=1000009[/url]", "Solution_1": "yuppers! i know because i bought 1999 meself!", "Solution_2": "My coach has those books (at least the covers look the same) and they DO have countdown rounds." } { "Tag": [ "inequalities", "trigonometry" ], "Problem": "[quote=\"Virgil Nicula\"][color=darkred][b]Let $ ABC$ be a triangle. Prove that[/b] $ \\{\\begin{array}{cc}1\\blacktriangleright &\\boxed{\\ \\sum\\sin\\frac{A}{2}\\ \\le\\ \\sqrt{\\frac{4R+r}{2R}}\\ \\le\\ \\frac{3}{2}\\ }\\\\ \\\\ 2\\blacktriangleright &\\boxed{\\ \\sum\\cos\\frac{A}{2}\\ \\le\\ \\sqrt{\\frac{3(4R+r)}{2R}}\\ \\le\\ \\frac{3\\sqrt 3}{2}\\ }\\end{array}$[/color][/quote]\r\n\r\n[color=darkblue][b]Consequence.[/b] $ \\boxed{\\ (\\sum\\sin\\frac{A}{2})^{2}\\ +\\ (\\sum\\cos\\frac{A}{2})^{2}\\le\\ 8\\ +\\ \\frac{2r}{R}\\ \\le\\ 9\\ }\\ \\ (*)$\n\nApply the inequality $ (*)$ to the orthic triangle of an acute triangle $ ABC$ and obtain : \n\n$ \\{\\begin{array}{cc}1\\blacktriangleright &\\boxed{\\ (\\sum\\sin A)^{2}+(\\sum\\cos A)^{2}\\le 8(1+\\cos A\\cos B\\cos C)\\ }\\\\ \\\\ 2\\blacktriangleright &\\boxed{\\ \\frac{p^{2}+r^{2}+2Rr-7R^{2}}{8R^{2}}\\le\\cos A\\cos B\\cos C\\le\\frac{1}{8}\\ }\\\\ \\\\ 3\\blacktriangleright &\\boxed{\\ ab+bc+ca\\le 2R(4R+r)\\ }\\end{array}$[/color]", "Solution_1": "[size=150]Very nice, Virgil Nicula![/size]\r\n\r\n$ 1.\\blacktriangleright$ $ \\tan\\frac{A}{2}\\cdot\\sin B\\plus{}\\tan\\frac{B}{2}\\cdot\\sin A\\ge 4\\cdot\\sin\\frac{A}{2}\\sin\\frac{B}{2}$\r\n$ \\implies\\sum\\tan\\frac{A}{2}\\cdot\\left(\\sin B\\plus{}\\sin C\\right)\\ge 4\\cdot\\sum\\sin\\frac{B}{2}\\sin\\frac{C}{2}$\r\nOn the other hand, $ \\tan\\frac{A}{2}\\cdot\\left(\\sin B\\plus{}\\sin C\\right)\\equal{}\\frac{\\sin\\frac{A}{2}}{\\cos\\frac{A}{2}}\\cdot 2\\cdot\\sin\\frac{B\\plus{}C}{2}\\cos\\frac{B\\minus{}C}{2}\\equal{}\\cos B\\plus{}\\cos C$\r\nIt implies that $ \\sum\\cos A\\ge 2\\cdot\\sum\\sin\\frac{B}{2}\\sin\\frac{C}{2}$\r\n$ \\implies\\sum\\cos^{2}\\frac{A}{2}\\ge\\left(\\sum\\sin\\frac{A}{2}\\right)^{2}$\r\n$ \\implies\\frac{3}{2}\\plus{}\\frac{1}{2}\\cdot\\sum\\cos A\\ge\\left(\\sum\\sin\\frac{A}{2}\\right)^{2}$\r\n$ \\implies 2\\plus{}\\frac{r}{2\\cdot R}\\ge\\left(\\sum\\sin\\frac{A}{2}\\right)^{2}$\r\n$ \\implies\\sqrt{\\frac{4R\\plus{}r}{2R}}\\ge\\sum\\sin\\frac{A}{2}$\r\n\r\n$ 2.\\blacktriangleright$ $ \\sum\\cos\\frac{A}{2}\\le\\sqrt{3\\cdot\\left(\\sum\\cos^{2}\\frac{A}{2}\\right)}\\equal{}\\sqrt{\\frac{3\\left(4R\\plus{}r\\right)}{2R}}$", "Solution_2": "Note that the last line reduces into:\r\n\r\n$ 4Rr\\plus{}r^{2}\\plus{}s^{2}\\le 2rR\\plus{}8R^{2}\\iff s^{2}\\le 8R^{2}\\minus{}2rR\\minus{}r^{2}$\r\n\r\nNote this bound is slightly weaker than the bound given by blundon's inequality (the weaker version without all the squareroots, etc):\r\n\r\n$ 4R^{2}\\plus{}4rR\\plus{}3r^{2}\\le 8R^{2}\\minus{}2rR\\minus{}r^{2}\\iff$ \r\n$ 4R^{2}\\minus{}6rR\\minus{}4r^{2}\\equal{}2(R\\minus{}2r)(2R\\plus{}r)\\ge 0$" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "In a school,each teacher must choose $ 10$ students to join a math contest,a group student is call \"acceptable\" with teacher $ A$ if in this group has at least one student who is choosen by teacher $ A$.With any 6 teacher,exits a group is \"acceptable\" with them(all 6).Prove that we can choose $ 10$ students join the contest such that this group is \"acceptable\" with all teacher.", "Solution_1": "[quote=\"mylove\"]With any 6 teacher,exits a group is \"acceptable\" with them(all 6).[/quote]\r\n\r\nIsn't this always the case? (Just pick one distinct student from each teacher's group of ten... and then pick four more students however you wish.)", "Solution_2": "With any 6 teacher,always exits a group is \"acceptable\" with them." } { "Tag": [ "function", "geometry", "geometric transformation", "real analysis", "real analysis unsolved" ], "Problem": "Here is a problem I just cannot seem to crack:\r\n\r\n Suppose f is continuous on [0,1] and that f(0) = f(1). Prove that for any positive integer n, there exists some x in the interval [0,1-(1/n)] such that \r\n f(x) = f(x+(1/n)).\r\n\r\n [Comment:If one assumes that f > =0 (or f<= 0) the problem is straightforward enough. For example let f >= 0. Without loss of generality assume f(0) =f(1) =0 and let g(w) = f(w) - f(w+ (1/n)) for w in [0,1-1/n]. Then g(0) = -f(1/n) < 0 and g(1-(1/n)) = f(1-(1/n))-f(1) = f(1-(1/n))>0 . So g changes sign between 0 and 1-(1/n) and so by the Intermediate Value Theorem, there is an x: g(x) = 0 . That is, there is an x such that f(x)=f(x+(1/n)). The problem seems much harder to do for the case where f changes sign at least once in [0,1]. ]", "Solution_1": "Actually, if $ f\\geq 0$ in $ [0,1]$, you can't assume $ f(0)\\equal{}0$ (the function might be something like $ (2x\\minus{}1)^2$ which is nonnegative but drops below $ f(0)$).\r\n\r\nIf you let $ g(x)\\equal{}f(x)\\minus{}f\\left( x\\plus{}\\frac{1}{n}\\right)$, look at\r\n\\[ g(0),g\\left( \\frac{1}{n}\\right),g\\left( \\frac{2}{n}\\right),\\cdots ,g\\left( 1\\minus{}\\frac{1}{n}\\right)\\]\r\nThey can't all be strictly positive or strictly negative.", "Solution_2": "Thankyou for your reply. \r\n\r\nIf f is continuous on [0,1], then f - k is continuous on [0,1] where k is a constant. So if f(0) = f(1) = k then F(0) = (f-k)(0) = 0 etc.\r\nSo a translation disposes of the problem provide the resulting function >=0 or <=0. Is this reasoning faulty ??\r\n\r\nI can see that, making no assumptions about f, that g(0)+g(1/n)+g(2/n)+ . . . +g(1-1/n) = 0 and so not all terms in this sum are of the same sign. But how does one proceed from here ? How is the proof completed - ie., how does one now set out the argument from this point ?", "Solution_3": "Now you have a continuous function that either has the value 0 at all those points (in which case ...) or that has a positive value at some point and a negative value at some other point (in which case ...). Fill in the blanks :wink:", "Solution_4": "[i]Now you have a continuous function that either has the value 0 at all those points (in which case ...) or that has a positive value at some point and a negative value at some other point (in which case ...). Fill in the blanks [/i]\r\n\r\nThankyou !![/i]" } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "Let two circles $ \\Gamma_1$ and $ \\Gamma_2$ intersect at $ A$ and $ B$. Points $ P$ and $ Q$ lie on $ \\Gamma_1$ and $ \\Gamma_2$ respectively such that line $ PQ$ is the common tangent of $ \\Gamma_1$ and $ \\Gamma_2$. Line $ AB$ and $ PQ$ intersect at $ E$. Point $ F$ is the foot of perpendicular from $ A$ to line $ PQ$. Point $ G$ is such that $ EG$ is the diameter of the circumcircle of $ EBF$. Prove that $ P,Q,B,G$ lie on a circle.", "Solution_1": "Let $ AT_1 \\cap \\Gamma_1\\equal{} A_1$ and similarly define $ A_2$.$ T_1,T_2$ are the centers.\r\nLet $ G' \\equal{} (PQB) \\cap A_1A_2$.\r\nwe must prove that $ G'A \\perp PQ$.\r\nwe have $ QA \\perp PG'$ because we have $ \\angle PQA \\plus{} \\angle G'PQ \\equal{} \\angle ABQ \\plus{} \\angle QBA_2 \\equal{} \\angle ABA_2 \\equal{} 90^\\circ$.\r\nsimilarly we have the perpendicularity $ PA \\perp G'Q$.\r\nso $ A$ is the orthocenter of $ PQG'$ and we have $ G'A \\perp PQ$.", "Solution_2": "nice proof shoki. :D I made this problem by inversion from an IMO problem." } { "Tag": [], "Problem": "If $ g(x) \\equal{} x^2$ and $ f(x) \\equal{} 2x \\minus{} 1$, what is the value of $ f(g(2))$?", "Solution_1": "you have g(x)=x^2, f(x)=2x-1 and f(g(2)). Plug in 2 for the function.\r\n\r\ng(2)=2^2=4\r\n\r\nNow you know g(2) is 4. so, f(g(2))=f(4). Now plug in 4 for the other function.\r\n\r\nf(4)=2(4)-1=[b]7[/b]" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Let $ x\\in\\mathbb{R}$. Find a closed formula for the infinite sum $ \\sum_{i\\equal{}0}^{\\infty}x^{2^{\\minus{}i}}\\minus{}1$", "Solution_1": "$ \\& \\sum\\limits_{i = 0}^\\infty {{x^{{2^{ - i}}}} - 1} = \\left( {x - 1} \\right) + \\left( {{x^{\\frac {1} {2}}} - 1} \\right) + \\left( {{x^{\\frac {1} {4}}} - 1} \\right) + \\left( {{x^{\\frac {1} {8}}} - 1} \\right) \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\\\\r\n\\& = \\left( {x - 1} \\right) + \\frac {{\\left( {x - 1} \\right)}} {{\\left( {{x^{\\frac {1} {2}}} + 1} \\right)}} + \\frac {{\\left( {x - 1} \\right)}} {{\\left( {{x^{\\frac {1} {2}}} + 1} \\right)\\left( {{x^{\\frac {1} {4}}} + 1} \\right)}} + \\frac {{\\left( {x - 1} \\right)}} {{\\left( {{x^{\\frac {1} {2}}} + 1} \\right)\\left( {{x^{\\frac {1} {4}}} + 1} \\right)\\left( {{x^{\\frac {1} {8}}} + 1} \\right)}} \\cdots \\\\\r\n\\& = \\left( {x - 1} \\right)\\left[1+ {\\frac {1} {{\\left( {{x^{\\frac {1} {2}}} + 1} \\right)}} + \\frac {1} {{\\left( {{x^{\\frac {1} {2}}} + 1} \\right)\\left( {{x^{\\frac {1} {4}}} + 1} \\right)}} + \\frac {1} {{\\left( {{x^{\\frac {1} {2}}} + 1} \\right)\\left( {{x^{\\frac {1} {4}}} + 1} \\right)\\left( {{x^{\\frac {1} {8}}} + 1} \\right)}} \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots } \\right]$\r\n\r\nbut i have no idea on how to sum up this series" } { "Tag": [ "topology", "calculus", "geometry", "set theory", "Galois Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "I am not going to ask a question in this category for it is too advanced for me. However, I am curious about what ACTUALLY is the toughest math course in school.\r\n\r\nSome say topology, others set theory and still others say differential calculus plus differential geometry.\r\n\r\nHonestly, what is the toughest math course out there?", "Solution_1": "[quote=\"Interval\"]I am not going to ask a question in this category for it is too advanced for me. However, I am curious about what ACTUALLY is the toughest math course in school.\n\nSome say topology, others set theory and still others say differential calculus plus differential geometry.\n\nHonestly, what is the toughest math course out there?[/quote]\r\n\r\nSince a mathematician is only interested in the things he does not understand, anything in math is hard. But Algebra & Topology is fun yeah", "Solution_2": "I'd say everyone is naturally good at different things. Some people naturally get algebra, but think analysis is really hard. Others vice versa. I just took a class in Galois theory which is famed to be quite challenging, but I'd take a test in that class anyday over a test on probability/combinatorics which some think is below college level. So I'd say it all depends on what areas you take a liking to." } { "Tag": [ "geometry", "incenter", "circumcircle", "vector" ], "Problem": "My daughter just asked me why the orthocenter of a triangle is usually called $H$. I confessed that I didn't know why. Does anyone know?", "Solution_1": "H - the point of intersection of [b]H[/b]eights?", "Solution_2": "I always assumed it was the next letter after G, the centroid.\r\nI guess I though a,b,c are vertices, d,e,f are medial triangle points (?), g centroid, h orthocenter, i incenter, etc. now O would be circumcenter because it is a commonly used circle (?) and the circumcenter is also like the origin of the triangle in terms of vectors usually...\r\n\r\njust some crazy thoughts of mine", "Solution_3": "Don't know if this is anything helpfull, but in german the orthocenter is called \"[b]H[/b]\u00f6henschnittpunkt\", meaning that this point is the intersection of the altitudes (german: \"H\u00f6hen\").", "Solution_4": "Indeed... H stands for Horthocentre :D. At least that's what I like to think.", "Solution_5": "The German thing may be right, but what about this...\r\n\r\nOnce you've labeled the vertices a, b, c, and the endpoints of the cevians d, e, f, the leftovers are g (centroid), h (orthocenter), i (incenter). The circumcenter, which is generally more useful and intuitive (and probably well-known long before the other three), gets the default name for a circle's center.\r\n\r\nJust a guess, anyway.", "Solution_6": "The explanation with [b]h[/b]eight seems to be definitely most convincing for me.", "Solution_7": "the point of intersection of Heights=H\u00f6henschnittpunkt?\r\n\r\nMakes it more convincing, though the other argument (abcd...gh...) is good too" } { "Tag": [ "limit", "logarithms", "floor function", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Let $ a_n $ be the maximum integer such that : $ 5^{a_n} < 1^1.2^2.3^3...n^n $ \r\n Find $ \\lim_{n\\to \\infty}\\frac{a_n}{n^2} $ ?", "Solution_1": "$\\lim_{n\\to\\infty}\\frac{a_n}{n^2}=\\infty.$\r\n\r\nMore precisely, $\\lim_{n\\to\\infty}\\frac{a_n}{n^2\\ln n}=\\frac1{2\\ln5}.$", "Solution_2": "Are you sure , Kent ? My teacher said the answer is $ \\lim_{n\\to \\infty}\\frac{a_n}{n^2}=\\frac{1}{8} $ \r\n Can you write down your solution ?", "Solution_3": "$a_n = \\lfloor \\ln_5 2^2 \\cdot 3^3 \\cdot \\ldots \\cdot n^n \\rfloor = \\lfloor \\sum^n_{j = 2} j \\cdot \\ln_5 j \\rfloor \\geqslant \\lfloor \\sum^n_{j = 2} j\\cdot \\ln_5 2 \\rfloor \\geqslant 2 \\sum_{j = 2}^n j = n^2 + n - 2$, so $\\lim_{n\\rightarrow \\infty} \\frac{a_n}{n^2} \\geqslant \\lim_{n \\rightarrow \\infty} \\frac{n^2 + n - 2}{n^2} = 1$. Thus youre teacher is wrong.", "Solution_4": "In fact $a_n = \\frac{1}{\\ln 5} \\sum_{j = 1}^n j \\cdot \\ln_{} j + O ( 1 )$, and $\\sum_{j = 1}^n j \\ln j = \\int_1^n x \\ln x %Error. \"tmop\" is a bad command.\n{dx} + O ( n \\ln n ) =\r\n\\frac{n^2 \\ln n}{2} - \\frac{n^2}{4} + O ( n \\ln n )$. Thus $a_n = \\frac{n^2 \\ln n}{2 \\ln_{} 5} - \\frac{n^2}{4 \\ln 5} + O ( n \\ln n )$ and $\\lim_{n\r\n\\rightarrow \\infty} \\frac{a_n}{n^2} = + \\infty$. Also from this asymptotic result we get, just as pointed Kent, $\\lim_{n \\rightarrow \\infty} \\frac{a_n}{n^2 \\ln n} = \\frac{1}{2 \\ln 5}$." } { "Tag": [ "vector", "algebra proposed", "algebra" ], "Problem": "Let $S=(x_1, x_2, ..... , x_n)$ be a set of reals which are at least 1. We count the number $f(S,I)$ of the sums $\\sum {a_i.x_i}$, where $a_i$ is either 0 or 1, that lies on an interval $I$ with length 1. Find the maximum of $f(S,I)$ for any interval $I$ and any set $S$.", "Solution_1": "I need a precisation:\r\nIs the interval closed? Also can the numbers be $1$?\r\nIf both questions are affirmative, my answer is smth like $C_n^{[\\frac {n-1}2]}+C_n^{[\\frac n 2]}$.\r\nOtherwise it's $C_n^{[\\frac n2]}$", "Solution_2": "As far as I remember, the right statement corresponds to your second answer. It easily follows from Sperner's theorem on antichain.\r\n\r\nPierre;", "Solution_3": "The interval here is closed such as $(a, a+1)$.\r\n Pbornztein, would you mind giving me a clear solution to this problem which I 'm very interested in. Thank you.", "Solution_4": "I don't quite understand: I always thought that a closed interval is $[a;a+1]$ (so $a,a+1 \\in I$)", "Solution_5": "I checked my archives and the right statement deals with an open interval of length 1.\r\nThe desired maximum is therefore $C_n^{[ \\frac n 2 ]}$ which follows by Sperner's theorem.\r\nIndeed, for each $A \\subset S$, let $s(A)$ be the sum of the elements of $A$ (with $s( \\emptyset ) = 0$).\r\n\r\nLet $I$ be an open interval with length 1.\r\n\r\nLemma.\r\nIf $A,B$ are two distinct subsets from $S$ such that $A \\subset B$ then at least one of the numbers $s(A)$ and $s(B)$ is not in $I$.\r\n\r\nProof of the lemma.\r\nLet $x \\in B \\setminus A$. Then $x \\geq 1$ and $s(B) \\geq s(A) + x \\geq s(A) + 1$. Thus, $s(A)$ and $s(B)$ cannot both belong to a same open interval with length 1.\r\n\r\nNow, since each number of the form $\\sum_{i=1}^n a_ix_i$ is of the form $s(A)$ for some subset $A$ from $S$, we deduce that the maximum number of such numbers which belong simultanously in an open interval of length 1 is not greater than the maximum length of an antichain formed by subsets from $S$. According to Sperner's theorem, this proves that the desired maximum is not greater than $C_n^{[ \\frac n 2 ]}$.\r\n\r\nNow, it is straightforward to verify that if $x_i = 1 + \\frac 1 {2^{p+i}}$ where $[ \\frac n 2 ] \\leq p $ and $I = ] [ \\frac n 2 ] , 1 + [ \\frac n 2 ] [$ then, for each subset $A$ from $S$ with $|A| = [ \\frac n 2 ] $ then $s(A) \\in I$.\r\n\r\nPierre.", "Solution_6": "This result has been generalised by Kleitman to higher dimensions :\r\nLet $u_1, \\cdots, u_n$ be vectors in $\\mathbb {R}^d$ with $||u_i|| \\geq 1$ for all $i$.\r\nAmong the $2^n$ linear combinations of the form $\\sum_{i=1}^n a_iu_i$, where $a_i \\in \\{-1;1 \\}$, the maximum number which lie in a given open ball with radius 1 is $C_n^{[ \\frac n 2 ]}.$\r\n\r\nReferences :\r\n[1] D. Kleitman, 'On a lemma of Littlewood and Offord on the distribution of linear combinations of vectors', Advances Math., 5 (1970), p.155-157.\r\n[2] M. Aigner, G.M. Ziegler, 'Proofs from The Book', Springer.\r\n\r\nPierre." } { "Tag": [ "geometry", "topology", "advanced fields", "advanced fields unsolved" ], "Problem": "Let L/K be a field extension, and V a K-variety in L^n.\r\nA point g on V is called a \"generic point of V over K\" if the Zarisky closure of {g} is V.\r\n\r\nProve the following:\r\n\r\n-If L is algebraically closed and the trascendence degree of L over K (that is, the size of a maximal subset of L which is algebraically independent over K) is at least n, then every V has generic points.\r\n\r\n-If V has generic points, then it is irreducible (that is, it's not the union of two other proper subvarieties).\r\n\r\n-The following are equivalent:\r\ni) g is a generic point of V over K.\r\nii) Ideal_K({g}) = Ideal_K(V).\r\niii) K[V] is isomorphical to K[g_1,... g_n] \\leq L, where g = (g_1,... g_n).\r\niv) for all a \\in V, there is a homeomorphism h:K[g_1,... g_n] -> K[a_1,... a_n] such that h(g_i) = a_i for all i.\r\n\r\n-If g and g' are generic points of V over K, there is an isomorphism h:K[g_1,... g_n] -> K[g_1',... g_n'] such that h(g_i) = g_i' for all i.\r\n\r\n-Let L' be an extension of L, V a K-variety in L^n, CV its Zarisky closure in (L')^n. If g \\in V is a generic point of CV, it is of V also.\r\n\r\n-Let g be a generic point of V over K, and d:=Trascendence degree_K({g_1,...,g_n}). Then for all a \\in V such that Trascendence degree_K({a_1,...,a_n}) = d, a is a generic point of V.\r\n\r\nYep, I know it is long. Can't help it, sorry. And please forgive any grammar or spelling errors (if any); after all, English is not my mother tongue.\r\n\r\nThank you guys. Take care and good luck.", "Solution_1": "[quote=\"Fibonacci\"]Let $L/K$ be a field extension, and $V$ a $K$-variety in $L^n$.\nA point $g$ on $V$ is called a \"generic point of $V$ over $K$\" if the Zarisky closure of $\\{g\\}$ is $V$.\n\nProve the following:\n\n\\begin{enumerate}\n\\item If $L$ is algebraically closed and the trascendence degree of $L$ over $K$ (that is, the size of a maximal subset of $L$ which is algebraically independent over $K$) is at least $n$, then every $V$ has generic points.\n\n\\item If $V$ has generic points, then it is irreducible (that is, it's not the union of two other proper subvarieties).\n\n\\item The following are equivalent:\n\\begin{enumerate}\\item $g$ is a generic point of $V$ over $K$.\n\\item $\\textrm{Ideal}_K(\\{g\\}) = \\textrm{Ideal}_K(V)$.\n\\item $K[V]$ is isomorphic to $K[g_1,... g_n] \\subset L$, where $g = (g_1,... g_n)$.\n\\item for all $a \\in V$, there is a homeomorphism $h:K[g_1,... g_n] \\to K[a_1,... a_n]$, such that $h(g_i) = a_i$ for all $i$.\n\\end{enumerate}\n\n\\item If $g$ and $g'$ are generic points of $V$ over $K$, there is an isomorphism $h:K[g_1,... g_n] \\to K[g_1',... g_n']$, such that $h(g_i) = g_i'$ for all $i$.\n\n\\item Let $L'$ be an extension of $L$, $V$ a $K$-variety in $L^n$, $\\overline V$ its Zarisky closure in $(L')^n$. If $g \\in V$ is a generic point of $\\overline V$, it is also one of $V$.\n\n\\item Let $g$ be a generic point of $V$ over $K$, and $d=\\textrm{Trascendence degree}_K(\\{g_1,...,g_n\\})$. Then for all $a \\in V$ such that $\\textrm{Trascendence degree}_K(\\{a_1,...,a_n\\}) = d$, $a$ is a generic point of $V$.\n[/quote]\r\n\r\n\r\nFor questions of this sort, the best references are elementary algebraic geometry books. Start with Harris, Joe. Algebraic Geometry, a first course, Springer. (http://www.ajorza.org/papers/harris/)\r\n\r\nI will suggest the answer to your second question: If $V$ is reducible, there exist nonempty disjoint open and closed $A,B$ such that $V=A\\cup B$. If $g\\in A$, then $\\overline{\\{g\\}}\\subset\\overline A=A$, so $A=V$ which is a contradiction." } { "Tag": [ "number theory", "prime numbers", "AMC" ], "Problem": "This was question #18 on the 2005 AMC 12A. I know the answer is 100 but when I try the problem I keep on getting 101 :wallbash_red:. Anyone have a good solution?\r\n\r\n Call a number \"prime-looking\" if it is composite but not divisible by $2, 3,$ or $5$. The three smallest prime-looking numbers are $49, 77,$ and $91$. There are $168$ prime numbers less than $1000$. How many prime-looking numbers are there less than $1000$?", "Solution_1": "You can look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367462#p367462]here[/url].\r\n\r\nDid you forget that 1 is neither prime nor composite?", "Solution_2": "Use a complement argument.\r\n# of prime looking numbers=1000-(prime numbers)-(multiples of 2, 3, or 5 except 2, 3 and 5)-1, because 1 is not prime nor composite, but isn't prime looking.\r\n\r\nHence, our answer is $1000-168-(-3+500+333+200-166-100-66+16)-1$ which doesn't seems to come out correctly, but that's the jist of how I would do it...", "Solution_3": "[quote=\"K81o7\"]Use a complement argument.\n# of prime looking numbers=1000-(prime numbers)-(multiples of 2, 3, or 5 except 2, 3 and 5)-1, because 1 is not prime nor composite, but isn't prime looking.\n\nHence, our answer is $1000-168-(-3+500+333+200-166-100-66+16)-1$ which doesn't seems to come out correctly, but that's the jist of how I would do it...[/quote]\r\n\r\nI think you messed up on how many numbers <1000 are divisible by 2, 3, and 5 when you counted. It should be 33, not 16.", "Solution_4": "yeah I forgot that 1 isn't composite :oops: .", "Solution_5": "[quote=\"joml88\"][quote=\"K81o7\"]Use a complement argument.\n# of prime looking numbers=1000-(prime numbers)-(multiples of 2, 3, or 5 except 2, 3 and 5)-1, because 1 is not prime nor composite, but isn't prime looking.\n\nHence, our answer is $1000-168-(-3+500+333+200-166-100-66+16)-1$ which doesn't seems to come out correctly, but that's the jist of how I would do it...[/quote]\n\nI think you messed up on how many numbers <1000 are divisible by 2, 3, and 5 when you counted. It should be 33, not 16.[/quote]\r\n :what?: Oh yeah...haha...that's what I would mess up on the AMC...thinking [1000/30]=16 :cursing: :fool: :wallbash_red:" } { "Tag": [ "analytic geometry", "function" ], "Problem": "A robot is programmed to do the following moves in a coordinate system: start at the origin facing in the positive $ y$ direction and move $ 1$ unit. For each subsequent move, it is to turn $ 90^\\circ$ clockwise and go forward $ 1$ unit farther than it went in the previous move. What is the sum of the coordinates of the robot's position after it just moved $ 14$ units in one direction?", "Solution_1": "This is just the addition of the following:\r\n\r\n1-2-3+4+5-6-7+8+9-10-11+12+13-14\r\nGroup them in pairs:\r\n-1+1-1+1-1+1-1\r\nGroup them in pairs again:\r\n-1. Solved.", "Solution_2": "The review says 15. So...?", "Solution_3": "The answer is 15.\n\nAfter the first 2 moves, turning points for the robot lie on either the function $y=-|x|$ or $y=|x-1|$ on the coordinate plane.\n \nOn its second turn, the robot is on $(2,1)$. On its sixth turn, it is on $(4,3)$. These two points form one diagonal of a 2x2 square, and after mapping out the other turning points, diagonals of a 2x2 square can be seen between any two consecutive turning points in the same quadrant. The difference between any two consecutive turning points in the same quadrant is also 4 (as in the first quadrant will have the turning point #'s 2, 6, 10, 14, 18, 22, 26, 30,... quadrant 2 will have #'s 5, 9, 13, 17, 21,...differences of 4)\n\nAfter the walk of 14 units, the robot ends up in the 1st quadrant of the coordinate plane.\n\nThus, on its tenth turn, the robot must lie on the point $(6,5)$, and on its 14th turn, it will lie on $(8,7)$\n$8+7=15$", "Solution_4": "james4l, you're way does work, and it's much quicker than mine... (Fortunately)\n\nThe sequence would be 1+2-3-4+5+6-7-8+9+10-11-12+13+14.\n\nGrouping the 1st and 3rd, 2nd and 4th, 5th and 7th, 6th and 8th, 9th and 11th, and 10th and 12th, you get a bunch of -2's, 6 of them to be exact. The last two numbers add up to 27.\n\n$27+6(-2)=15$ as desired." } { "Tag": [ "calculus", "integration", "inequalities", "function", "real analysis", "real analysis unsolved" ], "Problem": "[i find it interesting, not sure if it's already posted]\n\nLet $ f : [0, 2] \\to R$ be a function 3 times derivable, with $x_0 = 1$ point of relative extremum and $f'''(x) \\geq 0, \\forall x \\in [0, 2]$.\nProve $\\int_1^2 f(x) dx \\geq \\int_0^1 f(x) dx$.", "Solution_1": "$f(x) = f(1) + f'(1)(x - 1) + \\frac{1}{2}f''(\\theta (x)){\\left( {x - 1} \\right)^2}$\nSo, we prove that $\\begin{array}{l}\n\\int\\limits_1^2 {f''(\\theta (x)){{\\left( {x - 1} \\right)}^2}} dx \\ge \\int\\limits_0^1 {f''(\\theta (x)){{\\left( {x - 1} \\right)}^2}} dx\\\\\n \\Leftrightarrow f''({c_1})\\int\\limits_1^2 {{{\\left( {x - 1} \\right)}^2}} dx \\ge f''({c_2})\\int\\limits_0^1 {{{\\left( {x - 1} \\right)}^2}} dx\n\\end{array}$\n\nit 's the result" } { "Tag": [ "function", "articles", "LaTeX" ], "Problem": "When we type small letters in some Commands like \r\n$\\text{mathcal}$\r\n the letters will change to some NEW symbols.\r\nFor example if you type $\\{$ abcdefghijklmnopqrstuvwxyz $\\}$\r\nin mathcal we will have \\[\\mathcal{abcdefghijklmnopqrstuvwxyx }\\] Does any one know another command like this , that can Create symbols?", "Solution_1": "It not create new symbols, this is just because mathcal command is not defined for lowercase characters.\r\nAnd if you want symbols \"already created\" :D see [url=http://www.ctan.org/tex-archive/info/symbols/comprehensive/symbols-a4.pdf]this[/url].", "Solution_2": "$\\mathcal{VERY,,,USE,,,FULL,,,ARTICLE }$\r\n\r\nI tried to type lots of them but i get the usual POTENTIOL error .\r\n\r\nFor istance i typed the command [code] \\tent [/code] page 59 in your article but it didnt work.\r\n\r\nAnyway , Thanks.", "Solution_3": "\\tent requires the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=ifsym]ifsym[/url] package, which provides extra symbol fonts, to be loaded which you can do on your own computer but not here.", "Solution_4": "Under every table of symbol's command, you will find the package required for this table. So you must put tis in the preamble:[code]\\usepackage{packagename}[/code]" } { "Tag": [ "algebra", "polynomial", "superior algebra", "superior algebra unsolved" ], "Problem": "what is the min. polynomial of 3^(1/4)+3^(1/2) over Q(sqrt(3))?", "Solution_1": "It is fairly easy to make a good first guess. If $ \\alpha \\equal{} \\sqrt[4]{3}\\plus{}\\sqrt{3}$, then $ (\\alpha\\minus{}\\sqrt{3})^2\\equal{}\\sqrt{3}$. So we guess that $ (x\\minus{}\\sqrt{3})^2\\minus{}\\sqrt{3}$ is our minimal polynomial. The only thing you have to do is show that this polynomial doesn't have any roots in $ \\mathbb{Q}(\\sqrt{3})$.\r\n\r\nEquivalently, show that $ \\sqrt[4]{3}$ is not in $ \\mathbb{Q}(\\sqrt{3})$. Do you know how to do that?" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Is set { p/q | where p, q are prime } dense in $ [0, \\infty )$ ?", "Solution_1": "Yes, you can easily reduce it to this generalisation of Bertrand's theorem: for any $ c>1$ there exist an $ N$ such that there is a prime in $ [n,cn]$ for any $ n>N$.", "Solution_2": "I am afraid that i does not follow this way. But if youe have simple proof just show it :)", "Solution_3": "It suffices to show: for any $ s > 0$ and any $ \\epsilon >0$, there exists primes $ p,q$ such that $ s \\leq \\frac pq \\leq s\\plus{}\\epsilon$.\r\n\r\nTake $ c\\equal{} 1 \\plus{} \\frac{\\epsilon}{s}>1$ and then $ N$ as in the cited lemma above (that lemma is a consequence of the prime number theorem). Take a prime $ q$ such that $ qs>N$ and a prime $ p \\in [sq,csq]$ which exists by the lemma. Then $ \\frac pq \\geq \\frac{sq}{q} \\equal{} s$ and $ \\frac pq \\leq \\frac{csq}{q} \\equal{} cs \\equal{} s \\plus{} \\epsilon$, as claimed.", "Solution_4": "Thank you very much! I have wrongly understood that theorem - just did not noticed \"for any c>1\" :blush:.\r\nCould you please give some references to that version of Bertrands theorem?", "Solution_5": "To be honest - I do not see how that version of Bertrand's theorem follows from prime number theorem. Could you (or someone) please provide more info about it ?", "Solution_6": "See my post in http://www.mathlinks.ro/viewtopic.php?t=180277 ." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra solved" ], "Problem": "If p is a prime. How can it be shown that a group with order 2p must have a normal subgroup of order p?", "Solution_1": "It has an element $x$ of order $p$. This element generates a subgroup of index $2$ and therefore is normal." } { "Tag": [], "Problem": "During a certain election campaign, $p$ different kinds of promises are made by the different political parties ($p>0$). While several political parties may make the same promise, any two parties have at least one promise in common; no two parties have exactly the same set of promises. Prove that there are no more than $2^{p-1}$ parties.", "Solution_1": "How to solve that........?", "Solution_2": "Let $A$ be some subset of promises.\nConsidering $(A,A^C)$, you cannot pick both of them because that means that those parties doesn't share a promise.\nSince there are $2^{p-1}$ of those, there are at most $2^{p-1}$ parties.\nThe inequality is tight, by choosing one promise, have all parties share it, and then taking every subset of the other $p-1$ promises." } { "Tag": [ "probability" ], "Problem": "8 regular, identical, six-sided dice are arranged in a 2x2x2 cube.Two touching dice are said to \"match\" if the side they share has the same number on both dice. If the 8 dice are arranged randomly, what is the probability that all 12 pairs of touching dice \"match.\" Note: it is known that on regular dice the opposite sides always add up to seven.", "Solution_1": "Here's a hint. The answer can be expressed in the form [(1/2^a]*[(1/3^b]. Solve for a and b." } { "Tag": [ "puzzles" ], "Problem": "There is an island with a certain number of inhabitants, 10 of whom have a red face. For some reasons, a red face problem is a topic that is never-to-talk on this island. No mirrors or anything on the island, and nobody is allowed to say to the person who has a red face that the person has a red face. And law of island says that if a person learns that he has a red face, he have to commit a suicide at 11:00pm that day. Since the topic is never-to-talk, there are no ways for red faced people to realize that he has a red face, so everyone lives happily until an alien visits the island. Right before leaving the island, an alien tells everyone: \u201cThere are some people on this island who have a red face! You should all start thinking about it now.\u201d Ten days later (that is, on the the tenth 11:00pm after this declaration), all ten red-faced inhabitants commit suicide. \r\n\r\nQuestion: How did they realize that they had a red face? \r\nWhat did an alien say to the islanders that they didn\u2019t know already? \r\nWhat was an alien's contribution? \r\n\r\n\r\nAnyone have an answer?", "Solution_1": "There have been DOZENS of questions already, almost exactly like this one:\r\n\r\nOne example: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=117110\r\n\r\nEDIT: Another. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=89127", "Solution_2": "I kind of get the main part of the question, but I still don't get\r\n\r\nWhat did an alien say to the islanders that they didn\u2019t know already? \r\n\r\nWhat was an alien's contribution? \r\n\r\nparts... Can you help me with this? I mean they all knew that there are at least 9 people before alien came..", "Solution_3": "Your question\r\n\r\n[quote]What was the alien's contribution?[/quote]\n\nis very strange. Isn't the alien's contribution what he told the red-faced people, and therefore the same as what he told them that they didn't know?\n\nAnyways, in case you don't have the main question completely figured out, here's the answer:\n\n[quote] At the beginning, they all know that there is at least 1 zero.\n\nIf 1 midnight passes without anyone leaving, everyone will know that there are at least 2 zeroes because if there were 1 zero, that person would know his number.\nIf 2 midnight pass without anyone leaving, everyone will know that there are at least 3 zeroes because if there were 2 zero, those people would know their numbers.\n\nAND SO ON UNTIL\n\nIf 99 midnight pass without anyone leaving, everyone will know that there are at least 100 zeroes because if there were 99 zero, those person would know thier numbers.\n\nNo one leaves before the 99th night because the information is not enough.\nAt the 100th midnight, the people with 0 will know their number because they know the information that there are at least 100 zeroes, and they see only 99 zeroes. \n[/quote]\r\n\r\nExcept that was for the problem with 100 people, so you just shift the numbers a bit. Plus, they commit suicide instead of \"leaving\", as that answer says. Plus, replace \"know they have a 0\" with \"know they have a red face\"\r\n\r\nAnyways, \r\nAs for what he told them... this is pretty complicated to think about, I'll try to come back to answer later." } { "Tag": [ "trigonometry", "calculus", "integration", "number theory", "relatively prime", "complex analysis", "complex analysis unsolved" ], "Problem": "Let $ k,n \\in \\mathbb{N}$, with $ n\\geq 1$. Prove that $ z\\equal{}\\cos{\\frac{2k\\pi}{n}}\\plus{}i\\sin\\frac{2k\\pi}{n}$ is a primitive $ n$th root of unity if and only if $ k$ is relatively prime to $ n$.", "Solution_1": "Let $ (k,n)\\equal{}d$, and write $ k\\equal{}da, n\\equal{}db$. then $ 1\\equal{}(\\text{cis} \\frac{2k\\pi}{n})^r \\equal{} \\text{cis} \\frac{2ar\\pi}{b}\\Leftrightarrow \\frac{2ar\\pi}{b}$ is an integral multiple of $ 2 \\pi \\Leftrightarrow \\frac{ar}{b} \\in \\mathbb{Z} \\Leftrightarrow b|r$. so the smallest possible value of $ r$ is $ b$.\r\n\r\nIf $ d\\equal{}1$, then $ b\\equal{}n$, so the smallest possible value of $ r$ is $ n$, i.e. $ \\text{cis} \\frac{2k\\pi}{n}$ is a primitive $ n$-th root.\r\n\r\nIf $ d>1$, $ b0$. Show that \\[ \\frac{a^{3}}{(8a^{2}\\plus{}b^{2})c}\\plus{}\\frac{b^{3}}{(8b^{2}\\plus{}c^{2})a}\\plus{}\\frac{c^{3}}{(8c^{2}\\plus{}a^{2})b}\\ge\\frac{1}{3}\\]\r\nBut your methods really kill it! :( :D Still if someone has other solutions please be kind to post it. :)" } { "Tag": [ "calculus", "number theory proposed", "number theory" ], "Problem": "Let the divisors of $ n$ be $ 1\\equal{}d_12$ :(", "Solution_2": "$n$ has $k$ divisors. If $k \\leq 12$ then $d_{6}^{2} + d_{7}^{2} > 2d_{6}d_{7} > n + 1$ (note that $d_{\\lfloor{\\frac{k}{2}}\\rfloor + 1}^{2} \\geq n$), contradition.\nHence $k \\geq 13$ (1)\nIf $k \\geq 16$ then $d_{6}^{2} + d_{7}^{2} < 2d_{7}^2 < n + 1$ (you can prove it). (2)\nFrom (1) & (2) we have $13 \\leq k \\leq 15$\nOn other hand, we can write $n = p_{1}^{a_{1}}.p_{2}^{a_{2}}.....p_{m}^{a_{m}}$\nIf $k = 13$ so $m = 1$ and $a_{1} = 12$, so $n = p^{12}$ and $d_{6} = p^{5} and d_{7} = p^{6}$, then $p^{10} + p^{12} = p^{12} + 1$, so $p = 1$, contradition.\nIf $k = 14$ and $k = 15$ you can do it :). :first:" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Let $M(n)=\\{-1,-2,\\ldots,-n\\}$. For every non-empty subset $T$ of $M(n)$, let $P(T)$ be the product of all elements of $T$. Find the sum of all $P(T)$.", "Solution_1": "[quote=\"Yimin Ge\"]Let $M(n)=\\{-1,-2,\\ldots,-n\\}$. For every non-empty subset $T$ of $M(n)$, let $P(T)$ be the product of all elements of $T$. Find the sum of all $P(T)$.[/quote]\r\n\r\nIf $h(n)$ is the requested sum, it's easy to see that $h(n)$ is the sum with all subsets without $-n$ (hence $h(n-1)$), plus the sum of all subsets containing $-n$ + something else (hence $-nh(n-1)$), plus the sum of subset $\\{-n\\}$). $\\Rightarrow $ $h(n)=(1-n)h(n-1)-n$\r\n\r\nAnd , since $h(1)=-1$, $h(n)=-1$ $\\forall n\\in\\mathbb{N}$\r\n\r\n-- \r\nPatrick", "Solution_2": "for any subset T of M(n) which does not contain -1, there is a unique T' which contain all the elements of T and -1. Thus the sum of every pair of T and T' (there are 2^(n-1)-1 pairs), giving the sum of P(T) equal to -1." } { "Tag": [], "Problem": "\u0391\u03bd \u03b7 $ f$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03c5\u03bf \u03c6\u03bf\u03c1\u03ad\u03c2 \u03c0\u03b1\u03c1\u03b1\u03b3\u03c9\u03b3\u03af\u03c3\u03b9\u03bc\u03b7 \u03c3\u03c4\u03bf $ R$ \u03ba\u03b1\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 $ c$ \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03cc\u03c2 : $ \\frac{f(x)\\minus{}f(y)}{x\\minus{}y}\\ne f'(c)$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $ x \\ne y$ \u03c4\u03cc\u03c4\u03b5 \u03bd\u03b1 \u03b4\u03b5\u03b9\u03c7\u03c4\u03b5\u03af \u03cc\u03c4\u03b9 $ f''(c)\\equal{}0$", "Solution_1": "[quote=\"r_boris\"]\u0391\u03bd \u03b7 $ f$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03c5\u03bf \u03c6\u03bf\u03c1\u03ad\u03c2 \u03c0\u03b1\u03c1\u03b1\u03b3\u03c9\u03b3\u03af\u03c3\u03b9\u03bc\u03b7 \u03c3\u03c4\u03bf $ R$ \u03ba\u03b1\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 $ c$ \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03cc\u03c2 : $ \\frac {f(x) \\minus{} f(y)}{x \\minus{} y}\\ne f'(c)$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $ x \\ne y$ \u03c4\u03cc\u03c4\u03b5 \u03bd\u03b1 \u03b4\u03b5\u03b9\u03c7\u03c4\u03b5\u03af \u03cc\u03c4\u03b9 $ f''(c) \\equal{} 0$[/quote]\r\n\r\n[b]\u039b\u03a5\u03a3\u0397[/b]\r\n\u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 $ f''(c)\\neq0$. \u0391\u03c2 \u03c5\u03c0\u03bf\u03b8\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c7\u03c9\u03c1\u03af\u03c2 \u03b2\u03bb\u03ac\u03b2\u03b7 \u03c4\u03b7\u03c2 \u03b3\u03b5\u03bd\u03b9\u03ba\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2 \u03cc\u03c4\u03b9 $ f''(c)>0$. \u0398\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 $ \\epsilon >0$ \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03ce\u03c3\u03c4\u03b5 $ f''(x)>0, \\forall x\\in [c\\minus{}\\epsilon,c\\plus{}\\epsilon]$.\r\n\r\n[i]\u0391\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7:[/i] \u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03b1\u03c5\u03c4\u03cc \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9, \u03bf\u03c0\u03cc\u03c4\u03b5 \u03b3\u03b9\u03b1 \u03bf\u03c3\u03bf\u03b4\u03ae\u03c0\u03c4\u03b5 \u03bc\u03b9\u03ba\u03c1\u03cc $ \\epsilon>0$ \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 $ \\xi\\in [c\\minus{}\\epsilon, c\\plus{}\\epsilon]$ \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03ce\u03c3\u03c4\u03b5 $ f''(\\xi)\\leq 0$. \u03a4\u03cc\u03c4\u03b5 \u03b3\u03b9\u03b1 $ \\epsilon\\to 0$, \u03bb\u03b1\u03bc\u03b2\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 $ f''(c)\\leq0$, \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c4\u03bf\u03c0\u03bf \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03b1\u03c1\u03c7\u03b9\u03ba\u03ae \u03bc\u03b1\u03c2 \u03c5\u03c0\u03cc\u03b8\u03b5\u03c3\u03b7. \u0386\u03c1\u03b1 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9 \u03b7 $ f$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03c5\u03c1\u03c4\u03ae \u03c3\u03b5 \u03bc\u03b9\u03b1 \u03c0\u03b5\u03c1\u03b9\u03bf\u03c7\u03ae \u03c4\u03bf\u03c5 $ c$. $ \\square$\r\n\r\n\u03a4\u03cc\u03c4\u03b5 \u03c9\u03c2 \u03b3\u03bd\u03c9\u03c3\u03c4\u03cc\u03bd,\u03b7 \u03b3\u03c1\u03b1\u03c6\u03b9\u03ba\u03ae \u03c0\u03b1\u03c1\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 \u03c4\u03b7\u03c2 $ f$ \u03b8\u03b1 \u03b2\u03c1\u03af\u03c3\u03ba\u03b5\u03c4\u03b1\u03b9 \u03c0\u03ac\u03bd\u03c9 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03b5\u03c6\u03b1\u03c0\u03c4\u03bf\u03bc\u03ad\u03bd\u03b7 \u03c4\u03b7\u03c2 \u03c3\u03c4\u03bf $ c$ \u03c3\u03c4\u03bf \u03b4\u03b9\u03ac\u03c3\u03c4\u03b7\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03b1\u03c5\u03c4\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03c5\u03c1\u03c4\u03ae, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae $ f(x)\\geq g(x)\\equal{}f'(c)(x\\minus{}c)\\plus{}f(c)$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $ x\\in [c\\minus{}\\epsilon, c\\plus{}\\epsilon]$. \u0398\u03b5\u03c9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c4\u03ce\u03c1\u03b1 \u03bc\u03b9\u03b1 \u03bc\u03b5\u03c4\u03b1\u03c4\u03cc\u03c0\u03b9\u03c3\u03b7 \u03c4\u03b7\u03c2 $ g(x)$ \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b1 \u03c0\u03ac\u03bd\u03c9 \u03ba\u03b1\u03b9 \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03b1 \u03c4\u03b7\u03bd \u03b5\u03c5\u03b8\u03b5\u03af\u03b1 $ h(x)\\equal{}g(x)\\plus{}\\delta$ \u03bc\u03b5 $ 0<\\delta 0$. \u0398\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 $ \\epsilon > 0$ \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03ce\u03c3\u03c4\u03b5 $ f''(x) > 0, \\forall x\\in [c \\minus{} \\epsilon,c \\plus{} \\epsilon]$.\n[quote]\n[i]\u0391\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7:[/i] \u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03b1\u03c5\u03c4\u03cc \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9, \u03bf\u03c0\u03cc\u03c4\u03b5 \u03b3\u03b9\u03b1 \u03bf\u03c3\u03bf\u03b4\u03ae\u03c0\u03c4\u03b5 \u03bc\u03b9\u03ba\u03c1\u03cc $ \\epsilon > 0$ \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 $ \\xi\\in [c \\minus{} \\epsilon, c \\plus{} \\epsilon]$ \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03ce\u03c3\u03c4\u03b5 $ f''(\\xi)\\leq 0$. \u03a4\u03cc\u03c4\u03b5 \u03b3\u03b9\u03b1 $ \\epsilon\\to 0$, \u03bb\u03b1\u03bc\u03b2\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 $ f''(c)\\leq0$,[/quote] \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c4\u03bf\u03c0\u03bf \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03b1\u03c1\u03c7\u03b9\u03ba\u03ae \u03bc\u03b1\u03c2 \u03c5\u03c0\u03cc\u03b8\u03b5\u03c3\u03b7. \u0386\u03c1\u03b1 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9 \u03b7 $ f$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03c5\u03c1\u03c4\u03ae \u03c3\u03b5 \u03bc\u03b9\u03b1 \u03c0\u03b5\u03c1\u03b9\u03bf\u03c7\u03ae \u03c4\u03bf\u03c5 $ c$. $ \\square$\n\n\u03a4\u03cc\u03c4\u03b5 \u03c9\u03c2 \u03b3\u03bd\u03c9\u03c3\u03c4\u03cc\u03bd,\u03b7 \u03b3\u03c1\u03b1\u03c6\u03b9\u03ba\u03ae \u03c0\u03b1\u03c1\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 \u03c4\u03b7\u03c2 $ f$ \u03b8\u03b1 \u03b2\u03c1\u03af\u03c3\u03ba\u03b5\u03c4\u03b1\u03b9 \u03c0\u03ac\u03bd\u03c9 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03b5\u03c6\u03b1\u03c0\u03c4\u03bf\u03bc\u03ad\u03bd\u03b7 \u03c4\u03b7\u03c2 \u03c3\u03c4\u03bf $ c$ \u03c3\u03c4\u03bf \u03b4\u03b9\u03ac\u03c3\u03c4\u03b7\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03b1\u03c5\u03c4\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03c5\u03c1\u03c4\u03ae, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae $ f(x)\\geq g(x) \\equal{} f'(c)(x \\minus{} c) \\plus{} f(c)$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $ x\\in [c \\minus{} \\epsilon, c \\plus{} \\epsilon]$. \u0398\u03b5\u03c9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c4\u03ce\u03c1\u03b1 \u03bc\u03b9\u03b1 \u03bc\u03b5\u03c4\u03b1\u03c4\u03cc\u03c0\u03b9\u03c3\u03b7 \u03c4\u03b7\u03c2 $ g(x)$ \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b1 \u03c0\u03ac\u03bd\u03c9 \u03ba\u03b1\u03b9 \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03b1 \u03c4\u03b7\u03bd \u03b5\u03c5\u03b8\u03b5\u03af\u03b1 $ h(x) \\equal{} g(x) \\plus{} \\delta$ \u03bc\u03b5 $ 0 < \\delta < min\\{f(c \\minus{} \\epsilon) \\minus{} g(c \\minus{} \\epsilon),f(c \\plus{} \\epsilon) \\minus{} g(c \\plus{} \\epsilon)\\}$ \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03b7 \u03b3\u03c1\u03b1\u03c6\u03b9\u03ba\u03ae \u03c0\u03b1\u03c1\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 \u03c4\u03b7\u03c2 $ f$ \u03c3\u03b5 \u03b4\u03cd\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ (a,f(a))$ \u03ba\u03b1\u03b9 $ (b,f(b))$ \u03c0\u03bf\u03c5 \u03b2\u03c1\u03af\u03c3\u03ba\u03bf\u03bd\u03c4\u03b1\u03b9 \u03b5\u03ba\u03c4\u03ad\u03c1\u03c9\u03b8\u03b5\u03bd \u03c4\u03bf\u03c5 $ c$ \u03ba\u03b1\u03b9 \u03c9\u03c2 \u03b5\u03ba \u03c4\u03bf\u03cd\u03c4\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ac \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03c4\u03bf\u03c5\u03c2. \u039f \u03c3\u03c5\u03bd\u03c4\u03b5\u03bb\u03b5\u03c3\u03c4\u03ae\u03c2 \u03b4\u03b9\u03b5\u03cd\u03b8\u03c5\u03bd\u03c3\u03b7\u03c2 \u03c4\u03b7\u03c2 $ h$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03bf\u03c2 \u03bc\u03b5 $ f'(c)$, \u03b5\u03c0\u03bf\u03bc\u03ad\u03bd\u03c9\u03c2 $ \\lambda_h \\equal{} \\frac {f(b) \\minus{} f(a)}{b \\minus{} a} \\equal{} f'(c)$. \u0397 \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03b1 \u03c3\u03c7\u03ad\u03c3\u03b7 \u03cc\u03bc\u03c9\u03c2 \u03bf\u03b4\u03b7\u03b3\u03b5\u03af \u03c3\u03b5 \u03ac\u03c4\u03bf\u03c0\u03bf \u03bb\u03cc\u03b3\u03c9 \u03c4\u03b7\u03c2 \u03b1\u03c1\u03c7\u03b9\u03ba\u03ae\u03c2 \u03c5\u03c0\u03cc\u03b8\u03b5\u03c3\u03b7\u03c2 \u03c4\u03bf\u03c5 \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03ad\u03c4\u03c3\u03b9 \u03b7 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03bf\u03bb\u03bf\u03ba\u03bb\u03b7\u03c1\u03ce\u03bd\u03b5\u03c4\u03b1\u03b9. $ \\blacksquare$[/quote]\r\n\r\n\u039d\u03af\u03ba\u03bf \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03b1\u03bd \u03bd\u03b1 \u03bb\u03b5\u03c2 \u03cc\u03c4\u03b9 $ f''$ \u03c3\u03c5\u03bd\u03b5\u03c7\u03ae\u03c2 ? \u03ae \u03bc\u03bf\u03c5 \u03b4\u03b9\u03b1\u03c6\u03b5\u03cd\u03b3\u03b5\u03b9 \u03ba\u03ac\u03c4\u03b9?\r\n\u0395\u03c0\u03b5\u03b9\u03b4\u03ae \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03b4\u03b9\u03ba\u03b9\u03ac \u03bc\u03bf\u03c5 \u03bb\u03cd\u03c3\u03b7 \u03ad\u03c7\u03c9 \u03b1\u03bd\u03ac\u03bb\u03bf\u03b3\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c0\u03b5\u03c1\u03b9\u03bc\u03ad\u03bd\u03c9 \u03bd\u03b1 \u03b4\u03c9 \u03b1\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03ac\u03bb\u03bb\u03b5\u03c2 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03ac \u03b8\u03b1 \u03c0\u03bf\u03c3\u03c4\u03ac\u03c1\u03c9 \u03c4\u03b7\u03bd \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5", "Solution_3": "[quote=\"Nick Rapanos\"]\n\u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 $ f''(c)\\neq0$. \u0391\u03c2 \u03c5\u03c0\u03bf\u03b8\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c7\u03c9\u03c1\u03af\u03c2 \u03b2\u03bb\u03ac\u03b2\u03b7 \u03c4\u03b7\u03c2 \u03b3\u03b5\u03bd\u03b9\u03ba\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2 \u03cc\u03c4\u03b9 $ f''(c) > 0$. \u0398\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 $ \\epsilon > 0$ \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03ce\u03c3\u03c4\u03b5 $ f''(x) > 0, \\forall x\\in [c \\minus{} \\epsilon,c \\plus{} \\epsilon]$.\n[/quote]\r\n\r\n\u0394\u03b5 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03b5\u03af\u03c2 \u03c0\u03bf\u03c5\u03b8\u03b5\u03bd\u03ac \u03c3\u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c4\u03bf\u03c5 \u03bb\u03ae\u03bc\u03bc\u03b1\u03c4\u03bf\u03c2 \u03c4\u03b7 \u03b4\u03b5\u03b4\u03bf\u03bc\u03ad\u03bd\u03b7 \u03c3\u03c7\u03ad\u03c3\u03b7 $ f(x) \\minus{} f(y) \\neq f'(c)(x \\minus{} y)$. \u038c\u03bc\u03c9\u03c2 \u03b7 \u03c5\u03c0\u03cc\u03b8\u03b5\u03c3\u03ae \u03c3\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03c5\u03c7\u03b1\u03af\u03b5\u03c2 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2. (\u03b3\u03b9\u03b1 \u03c4\u03bf\u03bd \u03af\u03b4\u03b9\u03bf \u03bb\u03cc\u03b3\u03bf \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03c4\u03bf: $ f'(x_0) > 0 \\Rightarrow f$ \u03b1\u03cd\u03be\u03bf\u03c5\u03c3\u03b1 \u03ba\u03bf\u03bd\u03c4\u03ac \u03c3\u03c4\u03bf $ x_0$ \u03b3\u03b9\u03b1 $ f$ \u03c0\u03b1\u03c1\u03b1\u03b3\u03c9\u03b3\u03af\u03c3\u03b9\u03bc\u03b7.) \u0394\u03b9\u03cc\u03c1\u03b8\u03c9\u03c3\u03ad \u03bc\u03b5 \u03b1\u03bd \u03ad\u03c7\u03c9 \u03bb\u03ac\u03b8\u03bf\u03c2.\r\n\r\nIn any case, here goes:\r\n\r\n\u0398\u03b5\u03c9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c4\u03b7\u03bd $ g(t) : \\equal{} f(t) \\minus{} f(c) \\minus{} f'(c)(t \\minus{} c)$, $ t \\in \\mathbb{R}$. \u0391\u03c0\u03cc \u03c4\u03b7 \u03b4\u03b5\u03b4\u03bf\u03bc\u03ad\u03bd\u03b7 \u03b1\u03bd\u03af\u03c3\u03c9\u03c3\u03b7, \u03b7 $ g$ \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 $ 1 \\minus{} 1$. \u03a9\u03c2 $ 1 \\minus{} 1$ (\u03ba\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c7\u03ae\u03c2), \u03b7 $ g$ \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03bf\u03bd\u03cc\u03c4\u03bf\u03bd\u03b7. \u0395\u03c0\u03bf\u03bc\u03ad\u03bd\u03c9\u03c2, \u03b7 $ g'$ \u03b4\u03b5\u03bd \u03b1\u03bb\u03bb\u03ac\u03b6\u03b5\u03b9 \u03c0\u03bf\u03c4\u03ad \u03c0\u03c1\u03cc\u03c3\u03b7\u03bc\u03bf. \u0386\u03c1\u03b1, \u03b7 $ g'$ \u03ad\u03c7\u03b5\u03b9 \u03b1\u03ba\u03c1\u03cc\u03c4\u03b1\u03c4\u03bf \u03c3\u03c4\u03bf $ c$ (\u03b1\u03c6\u03bf\u03cd $ g'(c) \\equal{} 0$), \u03ba\u03b1\u03b9 \u03ac\u03c1\u03b1 $ g''(c) \\equal{} 0$.", "Solution_4": "[quote=\"cmad\"][quote=\"Nick Rapanos\"]\n\u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 $ f''(c)\\neq0$. \u0391\u03c2 \u03c5\u03c0\u03bf\u03b8\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c7\u03c9\u03c1\u03af\u03c2 \u03b2\u03bb\u03ac\u03b2\u03b7 \u03c4\u03b7\u03c2 \u03b3\u03b5\u03bd\u03b9\u03ba\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2 \u03cc\u03c4\u03b9 $ f''(c) > 0$. \u0398\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 $ \\epsilon > 0$ \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03ce\u03c3\u03c4\u03b5 $ f''(x) > 0, \\forall x\\in [c \\minus{} \\epsilon,c \\plus{} \\epsilon]$.\n[/quote]\n\n\u0394\u03b5 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03b5\u03af\u03c2 \u03c0\u03bf\u03c5\u03b8\u03b5\u03bd\u03ac \u03c3\u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c4\u03bf\u03c5 \u03bb\u03ae\u03bc\u03bc\u03b1\u03c4\u03bf\u03c2 \u03c4\u03b7 \u03b4\u03b5\u03b4\u03bf\u03bc\u03ad\u03bd\u03b7 \u03c3\u03c7\u03ad\u03c3\u03b7 $ f(x) \\minus{} f(y) \\neq f'(c)(x \\minus{} y)$. \u038c\u03bc\u03c9\u03c2 \u03b7 \u03c5\u03c0\u03cc\u03b8\u03b5\u03c3\u03ae \u03c3\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03c5\u03c7\u03b1\u03af\u03b5\u03c2 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2. (\u03b3\u03b9\u03b1 \u03c4\u03bf\u03bd \u03af\u03b4\u03b9\u03bf \u03bb\u03cc\u03b3\u03bf \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03c4\u03bf: $ f'(x_0) > 0 \\Rightarrow f$ \u03b1\u03cd\u03be\u03bf\u03c5\u03c3\u03b1 \u03ba\u03bf\u03bd\u03c4\u03ac \u03c3\u03c4\u03bf $ x_0$ \u03b3\u03b9\u03b1 $ f$ \u03c0\u03b1\u03c1\u03b1\u03b3\u03c9\u03b3\u03af\u03c3\u03b9\u03bc\u03b7.) \u0394\u03b9\u03cc\u03c1\u03b8\u03c9\u03c3\u03ad \u03bc\u03b5 \u03b1\u03bd \u03ad\u03c7\u03c9 \u03bb\u03ac\u03b8\u03bf\u03c2.\n\nIn any case, here goes:\n\n\u0398\u03b5\u03c9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c4\u03b7\u03bd $ g(t) : \\equal{} f(t) \\minus{} f(c) \\minus{} f'(c)(t \\minus{} c)$, $ t \\in \\mathbb{R}$. \u0391\u03c0\u03cc \u03c4\u03b7 \u03b4\u03b5\u03b4\u03bf\u03bc\u03ad\u03bd\u03b7 \u03b1\u03bd\u03af\u03c3\u03c9\u03c3\u03b7, \u03b7 $ g$ \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 $ 1 \\minus{} 1$. \u03a9\u03c2 $ 1 \\minus{} 1$ (\u03ba\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c7\u03ae\u03c2), \u03b7 $ g$ \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03bf\u03bd\u03cc\u03c4\u03bf\u03bd\u03b7. \u0395\u03c0\u03bf\u03bc\u03ad\u03bd\u03c9\u03c2, \u03b7 $ g'$ \u03b4\u03b5\u03bd \u03b1\u03bb\u03bb\u03ac\u03b6\u03b5\u03b9 \u03c0\u03bf\u03c4\u03ad \u03c0\u03c1\u03cc\u03c3\u03b7\u03bc\u03bf. \u0386\u03c1\u03b1, \u03b7 $ g'$ \u03ad\u03c7\u03b5\u03b9 \u03b1\u03ba\u03c1\u03cc\u03c4\u03b1\u03c4\u03bf \u03c3\u03c4\u03bf $ c$ (\u03b1\u03c6\u03bf\u03cd $ g'(c) \\equal{} 0$), \u03ba\u03b1\u03b9 \u03ac\u03c1\u03b1 $ g''(c) \\equal{} 0$.[/quote]\r\n\u0392\u03b1\u03c3\u03b9\u03ba\u03ac \u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03bb\u03ad\u03bc\u03b5\r\n\u0397 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5 \u03c3\u03c4\u03bf attachment", "Solution_5": ":oops: exete dikio....vlakeia mou...", "Solution_6": "By the way, \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b1\u03c5\u03c4\u03ae \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03bc\u03b5\u03c4\u03b1\u03c6\u03b5\u03c1\u03b8\u03b5\u03af \u03c3\u03c4\u03b9\u03c2 \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03ad\u03c2. \u0394\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9 \u03c4\u03af\u03c0\u03bf\u03c4\u03b1 \u03bc\u03b7-\u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03cc \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c4\u03b7\u03c2.", "Solution_7": "[quote=\"cmad\"]By the way, \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b1\u03c5\u03c4\u03ae \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03bc\u03b5\u03c4\u03b1\u03c6\u03b5\u03c1\u03b8\u03b5\u03af \u03c3\u03c4\u03b9\u03c2 \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03ad\u03c2. \u0394\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9 \u03c4\u03af\u03c0\u03bf\u03c4\u03b1 \u03bc\u03b7-\u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03cc \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c4\u03b7\u03c2.[/quote]\r\n\u0388\u03c7\u03b5\u03b9\u03c2 \u03b4\u03af\u03ba\u03b9\u03bf \u03b1\u03c0\u03bb\u03ce\u03c2 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03b5\u03cd\u03ba\u03bf\u03bb\u03b7 :wink: \u03c4\u03bf \u03ad\u03ba\u03b1\u03bd\u03b1 \u03b3\u03b9\u03b1 \u03b8\u03ad\u03bc\u03b1 \u03c4\u03bf\u03c5 2007" } { "Tag": [ "function", "calculus", "integration", "logarithms", "inequalities", "number theory proposed", "number theory" ], "Problem": "Let ${\\mathbb N }$ be the set of positive integers.\r\n \r\n Consider a function $ F: {\\mathbb N}\\to {\\mathbb N} $ which satisfies properties :\r\n\\[ 1)\\; \\; F(1)=0 , F(2)=1\\] \r\n\\[ 2) \\; \\; F(xy)=F(x) +F(y) ,\\; \\forall (x,y) \\in {\\mathbb N}\\times {\\mathbb N} \\]\r\nand $ 3) \\; \\; F(x)= F(x-1) $ when $ x $ is a prime number from ${\\mathbb N} $ .\r\n\r\n[b]Questions:[/b]\r\n i) The function $ F:{\\mathbb N}\\to {\\mathbb N} $ is well-defined ?\r\n\r\n ii) Let [.] denotes integral part , and $a(x):=\\frac{\\ln{x}}{\\ln{3}}\\; ,\\;\r\nA(x):=\\frac{\\ln{x}}{\\ln{2}} . $\r\n\r\n[i] It's true that for $ x=3,4,... $ the inequalities\n\\[\n [a(x)] \\le F(x)\\le [A(x)] \n\\]\nare verified ? [/i]", "Solution_1": "Hi,\r\n\r\nIn your problem statement, you say the function $F : N \\rightarrow N$ where $N$ is the set of positive integers. Then you say that $F(1) = 0$, but $0$ is not a positive integer.\r\n\r\nMoreover, you say $F(p) = F(p-1)$ if $p$ is prime. It then follows that $F(2) = F(1) = 0$. But you say that $F(2) = 1$.\r\n\r\nCould you please explain this?\r\n\r\nThanks,\r\nSambit", "Solution_2": "I think you should simply understand N as it is: the set of naturals {0,1,2,...}\r\n\r\nAs for the other one, that's quite strange indeed. I think the problem will require x>2 or x>3 somehow... or the question on part 1 is just no :D", "Solution_3": "Thank you !\r\nIndeed there is a mistake . Please consider tha N={1,2,...}= the set of positive integers, N_0={0,1,...}= the set of non-negative integers, and F:N---> N_0 .\r\nMoreover, instead of 3) please consider \r\n3') F(p)=F(p-1) when p is prime > 2 .\r\nRegards,proposer.", "Solution_4": "i) let $A_n$ be the subset of $N$ of those number which are only divisible by the first n primes. if $F$ is well defined for the first n primes, then for $A_n$ by 2). but by 3), it is then well-defined for the n+1-th prime since $p_{n+1}-1$ can only be divisible by smaller primes(let $p_n$ be the n-th prime). we inductively see that $F$ is well-defined.\r\nii) (the inequalities even hold for 1 and 2) the first observation is, that if the inequalities hold for all primes, then they hold everywhere by 2). the second inequality holds for 2, thus for $A_1$, thus for 3 by propery 3), thus for $A_2$, thus for 5 and so on.\r\nthe second inequality holds for 2 and 3($F(3)=1$). we will show that equality holds iff $n$ is a power of 3. this is true in $A_2$. now, if want to go from $A_n$ to $A_{n+1}$, we have to show that $F(p_{n+1})>\\frac{ln(p_{n+1})}{ln(3)}$. but $F(p_{n+1})=F(p_{n+1}-1)=F(2)+F(\\frac{p_{n+1}-1}{2})\\geq 1+\\frac{ln(p_{n+1}-1)-ln(2)}{ln(3)}>\\frac{ln(p_{n+1})}{ln(3)}$ since $ln(x+1)-ln(x) N_0 .\nMoreover, instead of 3) please consider \n3') F(p)=F(p-1) when p is prime > 2 .\nRegards,proposer.[/quote]\r\n\r\nNo... N_0 is by definition {1,2,...} and N is by definition {0,1,2,...} that are just the international standards all around the world ;)\r\nSo it's a F: N_0 --> N\r\n\r\ngreetz, Peter", "Solution_6": "[quote=\"Peter VDD\"][quote=\"flip2004\"]Thank you !\nIndeed there is a mistake . Please consider tha N={1,2,...}= the set of positive integers, N_0={0,1,...}= the set of non-negative integers, and F:N---> N_0 .\nMoreover, instead of 3) please consider \n3') F(p)=F(p-1) when p is prime > 2 .\nRegards,proposer.[/quote]\n\nNo... N_0 is by definition {1,2,...} and N is by definition {0,1,2,...} that are just the international standards all around the world ;)\nSo it's a F: N_0 --> N\n\ngreetz, Peter[/quote]\r\n\r\nI have always seen N_0 as {0,1,2,3,...} and N as {1,2,3,4,...}\r\n\r\nso I guess that just means people are going to have to specify in their problem statements, thats all." } { "Tag": [ "inequalities", "Cauchy Inequality", "inequalities proposed" ], "Problem": "Prove that, for any positive real numbers $a,b$ and $c$,\r\n\\[ \\sum_{cycl}\\frac{a^2+bc}{a^2+(b+c)^2}\\le\\frac{18}{5}\\cdot\\frac{a^2+b^2+c^2}{(a+b+c)^2} \\]", "Solution_1": "My ugly proof:\r\n$\\sum_{cyc}\\frac{a^2+bc}{a^2+(b+c)^2}\\le\\frac{18}{5}\\cdot\\frac{a^2+b^2+c^2}{(a+b+c)^2}\\Leftrightarrow\\sum\\left(\\frac{a^2+bc}{a^2+(b+c)^2}-\\frac{2}{5}\\right)\\leq\\frac{6}{5}\\left(\\frac{3(a^2+b^2+c^2)}{(a+b+c)^2}-1\\right)\\Leftrightarrow$\r\n$\\sum_{cyc}\\frac{3a^2-3bc-2(b-c)^2}{a^2+(b+c)^2}\\leq6\\cdot\\sum_{cyc}\\frac{(a-b)^2}{(a+b+c)^2}\\Leftrightarrow$\r\n$\\Leftrightarrow3\\cdot\\sum_{cyc}\\frac{2a^2-2bc}{ a^2+(b+c)^2}\\leq\\sum_{cyc}(a-b)^2\\left(\\frac{12}{(a+b+c)^2}+\\frac{4}{c^2+(a+b)^2}\\right)\\Leftrightarrow$\r\n$\\Leftrightarrow3\\cdot\\sum_{cyc}\\left(\\frac{(a-b)(a+c)}{a^2+(b+c)^2}-\\frac{(c-a)(a+b)}{a^2+(b+c)^2}\\right)\\leq\\sum_{cyc}(a-b)^2\\left(\\frac{12}{(a+b+c)^2}+\\frac{4}{c^2+(a+b)^2}\\right)\\Leftrightarrow$\r\n$\\Leftrightarrow3\\cdot\\sum_{cyc}(a-b)\\left(\\frac{a+c}{a^2+(b+c)^2}-\\frac{b+c}{b^2+(a+c)^2}\\right)\\leq\\sum_{cyc}(a-b)^2\\left(\\frac{12}{(a+b+c)^2}+\\frac{4}{c^2+(a+b)^2}\\right)\\Leftrightarrow$\r\n$\\Leftrightarrow3\\cdot\\sum_{cyc}\\frac{(a-b)^2(a^2+b^2+3c^2+2ac+2bc)}{(a^2+(b+c)^2)(b^2+(a+c)^2)}\\leq\\sum_{cyc}(a-b)^2\\left(\\frac{12}{(a+b+c)^2}+\\frac{4}{c^2+(a+b)^2}\\right)\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}(a-b)^2S_c\\geq0,$ where\r\n$S_c=\\frac{12}{(a+b+c)^2}+\\frac{4}{c^2+(a+b)^2}-\\frac{3((a+c)^2+(b+c)^2+c^2)}{(a^2+(b+c)^2)(b^2+(a+c)^2)}=$ \r\n$=\\frac{12}{(a+b+c)^2}+\\frac{4}{c^2+(a+b)^2}-\\frac{3(b^2+(a+c)^2+a^2+(b+c)^2+c^2-a^2-b^2)}{(a^2+(b+c)^2)(b^2+(a+c)^2)}=$\r\n$=\\frac{12}{(a+b+c)^2}-\\frac{3}{a^2+(b+c)^2}-\\frac{3}{b^2+(a+c)^2}+\\frac{4}{c^2+(a+b)^2}+\\frac{3(a^2+b^2-c^2)}{(a^2+(b+c)^2)(b^2+(a+c)^2)}.$\r\n$\\frac{6}{(a+b+c)^2}-\\frac{3}{a^2+(b+c)^2}\\geq0\\Leftrightarrow2(a^2+(b+c)^2)\\geq(a+b+c)^2,$\r\n$\\frac{6}{(a+b+c)^2}-\\frac{3}{b^2+(a+c)^2}\\geq0\\Leftrightarrow2(b^2+(a+c)^2)\\geq(b+a+c)^2$ \r\n and\r\n$\\frac{4}{c^2+(a+b)^2}+\\frac{3(a^2+b^2-c^2)}{(a^2+(b+c)^2)(b^2+(a+c)^2)}\\geq0,$ which obviously true. \r\nId est, our inequality is proved. :)", "Solution_2": "(My friend & I solved it)\r\nWLOG,assume that $a+b+c=3$.We have to prove that $\\frac{2}{5}(a^2+b^2+c^2)\\geq\\sum\\frac{4a^2+(3-a)^2}{4(a^2+(3-a)^2)}$\r\nThis is an easy problem because:\r\n$\\frac{2}{5}a^2-\\frac{4a^2+(3-a)^2}{4(a^2+(3-a)^2)}\\geq\\frac{11}{25}a-\\frac{11}{25}\\Leftrightarrow(a-1)^2(80a^2-168a+171)\\geq 0\\ldots$ :lol:", "Solution_3": "Nice proof! Hiep.\r\n\r\nI found this inequality the same time as this one\r\n[url]http://www.artofproblemsolving.com/Forum/topic-79611.html[/url]", "Solution_4": "[quote=\"ductrung\"]Prove that, for any positive real numbers $a,b$ and $c$,\n\\[\\sum_{cycl}\\frac{a^{2}+bc}{a^{2}+(b+c)^{2}}\\le\\frac{18}{5}\\cdot\\frac{a^{2}+b^{2}+c^{2}}{(a+b+c)^{2}}\\]\n[/quote]\r\nMy solution:\r\nThe inequality is equivalent to\r\n\\[\\sum\\frac{(b+c)^{2}-bc}{a^{2}+(b+c)^{2}}+\\frac{18}{5}\\cdot\\frac{a^{2}+b^{2}+c^{2}}{(a+b+c)^{2}}\\ge 3 \\]\r\nSince $(b+c)^{2}\\ge 4bc$ hence it suffices to prove\r\n\\[\\sum\\frac{(b+c)^{2}}{4(a^{2}+(b+c)^{2})}+\\frac{6}{5}\\cdot\\frac{a^{2}+b^{2}+c^{2}}{(a+b+c)^{2}}\\ge 1 \\]\r\nBy Cauchy Inequality we have\r\n\\[\\sum\\frac{(b+c)^{2}}{4(a^{2}+(b+c)^{2})}\\ge \\frac{(a+b+c)^{2}}{\\sum(a^{2}+(b+c)^{2})}=\\frac{(a+b+c)^{2}}{2(a^{2}+b^{2}+c^{2})+(a+b+c)^{2}}\\]\r\nWLOG, we may assume $a+b+c=1$. Setting $x=a^{2}+b^{2}+c^{2}$ then $3x \\ge 1$, it remains to prove that \r\n\\[\\frac{1}{2x+1}+\\frac{6x}{5}\\ge 1 \\]\r\n\\[\\Leftrightarrow x(3x-1) \\ge 0 \\]\r\nWhich is true. \r\nWe are done. :)", "Solution_5": "A similar inequality that I just thought of:\r\n\r\nProve that $\\sum \\frac{a^{2}+bc}{2a^{2}+(b+c)^{2}}\\le \\frac{7}{4}-\\frac{1}{4}\\cdot \\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}$", "Solution_6": "[quote=\"probability1.01\"]A similar inequality that I just thought of:\n\nProve that $\\sum \\frac{a^{2}+bc}{2a^{2}+(b+c)^{2}}\\le \\frac{7}{4}-\\frac{1}{4}\\cdot \\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}$[/quote]\r\nWe can prove it by the same method. :)" } { "Tag": [ "integration", "trigonometry", "calculus", "calculus computations" ], "Problem": "Can anyone solve the following?\r\n\r\n$ sin 3x dx\\plus{}2y (cos 3x)^{3}dy \\equal{} 0$? \r\n\r\nI'm having trouble integratig $ \\frac{sin 3x}{(cos 3x)^{3}}$.", "Solution_1": "[quote=\"bythecliff\"]Can anyone solve the following?\n\n$ sin 3x dx\\plus{}2y (cos 3x)^{3}dy \\equal{} 0$? \n\nI'm having trouble integratig $ \\frac{sin 3x}{(cos 3x)^{3}}$.[/quote]\r\n\r\n[hide]$ \\int\\frac{\\sin3x}{\\cos^{3}3x}dx \\equal{}\\minus{}\\frac{1}{3}\\int\\frac{d(\\cos3x)}{\\cos^{3}3x}dx$[/hide]", "Solution_2": "pluricomplex:\r\n\r\n$ \\int\\frac{\\sin{3x}}{\\cos^{3}{3x}}\\,dx \\equal{}\\minus{}\\frac{1}{3}\\int\\frac{d(\\cos{3x})}{\\cos^{3}{3x}}$\r\n\r\nSolution\r\n\r\n$ \\sin{3x}\\,dx\\plus{}2y\\cos^{3}{3x}\\,dy \\equal{} 0$\r\n\r\nFor now, assume $ \\cos^{3}{3x}\\neq 0$\r\n\r\n$ \\Rightarrow\\frac{sin{3x}}{\\cos^{3}{3x}}\\,dx\\plus{}2y\\,dy \\equal{} 0$\r\n\r\n$ \\Rightarrow\\int\\frac{sin{3x}}{\\cos^{3}{3x}}\\,dx\\plus{}\\int 2y\\,dy \\equal{} c$\r\n\r\nLet $ u \\equal{}\\cos{3x}$\r\n\r\n$ \\Rightarrow du \\equal{}\\minus{}3\\sin{3x}\\,dx$\r\n\r\n$ \\Rightarrow\\minus{}\\frac{1}{3}du \\equal{}\\sin{3x}\\,dx$\r\n\r\n$ \\Rightarrow\\minus{}\\frac{1}{3}\\int\\frac{1}{u^{3}}\\,du\\plus{}\\int 2y\\,dy \\equal{} c$\r\n\r\n$ \\Rightarrow\\minus{}\\frac{1}{3}\\left(\\minus{}\\frac{1}{2u^{2}}\\right)\\plus{}y^{2}\\equal{} c$\r\n\r\n$ \\Rightarrow\\frac{1}{6 u^{2}}\\plus{}y^{2}\\equal{} c$\r\n\r\n$ \\Rightarrow y^{2}\\equal{} c\\minus{}\\frac{1}{6\\cos^{2}{3x}}$\r\n\r\nThis is not the complete solution, as we have ignored the cases for $ \\cos^{3}{3x}\\equal{} 0$.\r\n\r\nSolving this: \r\n\r\n$ \\cos{3x}\\equal{} 0$\r\n\r\n$ \\Rightarrow 3x \\equal{} (2n\\plus{}1)\\frac{\\pi}{2}$\r\n\r\n$ \\Rightarrow x \\equal{} (2n\\plus{}1)\\frac{\\pi}{6}$\r\n\r\nThese values of $ x$ are all solutions to the differential equation too (for any $ y$). This is because if you put these values into the differential equation $ dx \\equal{} 0$ is obtained." } { "Tag": [ "trigonometry", "geometry", "AMC", "AIME", "ratio", "function", "Pythagorean Theorem" ], "Problem": "If $ \\tan{x} \\plus{} \\tan{y} \\equal{}25^{\\circ}$, and $ \\cot{x} \\plus{} \\cot{y} \\equal{}30^{\\circ}$, what is $ \\tan{x\\plus{}y}$?\r\n\r\nAnother one:\r\n\r\nCompute $ \\displaystyle \\sin{\\theta}$\r\n\r\nhttp://www.math.ku.edu/mathawareness/98competion/56.pdf\r\n\r\n(Its problem 2 In the 10-12th grade division)", "Solution_1": "$ 1/\\tan x \\plus{} 1/\\tan y \\equal{} (\\tan x \\plus{} \\tan y)/(\\tan x \\tan y) \\equal{} 25/(\\tan x \\tan y) \\equal{} 30$\r\n\r\n$ \\tan x \\tan y \\equal{} 5/6$\r\n\r\n$ \\tan(x \\plus{} y) \\equal{} (\\tan x \\plus{} \\tan y)/(1 \\minus{} \\tan x \\tan y) \\equal{} 25/(1 \\minus{} 5/6) \\equal{} 25*6 \\equal{} 150$\r\n\r\n$ \\sin \\theta \\equal{} \\sin(\\pi/2 \\minus{} 2\\arcsin(1/\\sqrt {5})) \\equal{} \\cos(2\\arcsin(1/\\sqrt{5}) \\equal{} 1 \\minus{} 2/5 \\equal{} 3/5$", "Solution_2": "Um the last line you wrote seemed like a formula because I don't see how you got it from manipulation, can you please derive $ \\tan{\\left(x \\plus{} y\\right)} \\equal{} \\frac {\\tan{x} \\plus{} \\tan{y}}{1 \\minus{} \\tan{x}\\tan{y}}$?", "Solution_3": "Well, just take $ \\frac{\\sin{(\\alpha \\plus{} \\beta)}}{cos{(\\alpha \\plus{} \\beta)}}$, expand it, and divide the numerator and denominator by $ \\cos{\\alpha}\\cos{\\beta}$.", "Solution_4": "sorry, but I don't quite understand what you're saying. How do you \"expand\" these terms? (Sorry I haven't done a lot of trig)", "Solution_5": "[url]http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Angle_sum_and_difference_identities[/url]. Volume 2 has a good proof using the area of a triangle.", "Solution_6": "If I read through vol 2, will I have enough knowledge of trig and complex numbers up till the AIME level?", "Solution_7": "[quote=\"mathemagician1729\"]If $ \\tan{x} \\plus{} \\tan{y} \\equal{} 25^{\\circ}$...[/quote]\r\n\r\nI don't understand how the tangent of something can equal an angle. Tangent is a ratio of sides. Are you saying that it equals tangent of 25 degrees?", "Solution_8": "It doesn't equal an angle. It equals a degree measurement.\r\n$ 25^\\circ$ is a real number.\r\n\r\nIf it helps, converting to radians gives $ 0.436332$.", "Solution_9": "For your second problem, you could also solve it like this \r\n\r\n[hide]\n\nCall the square $ ABCD$ with midpoints $ M$ and $ N$ on $ AD$ and $ CD$, respectively. By Pythagorean Theorem, we have $ BN\\equal{}BM\\equal{}\\frac{\\sqrt{5}}{2}$ and $ MN\\equal{}\\frac{\\sqrt2}{2}$. Hence, by Law of Cosines on triangle $ BMN$, \\[ \\frac{5}{4}\\plus{}\\frac{5}{4}\\minus{}2(\\frac{5}{4})\\cos\\theta\\equal{}\\frac{1}{2} \\] Solving gives $ \\cos\\theta \\equal{}\\frac{4}{5}$. Since $ \\sin^2\\theta \\plus{} \\cos^2\\theta \\equal{}1$, we have $ \\sin\\theta \\equal{} \\frac{3}{5}$\n\n[/hide]", "Solution_10": "[quote=\"grn_trtle\"]It doesn't equal an angle. It equals a degree measurement.\n$ 25^\\circ$ is a real number.\n\nIf it helps, converting to radians gives $ 0.436332$.[/quote]\r\n\r\nI still don't understand what the question is asking for. Tangent is a function that gives you a fraction as an answer, such as 5/12 or 2. You can take the tangent of a measurement of an angle, but tangent will give you a number, and I don't understand what $ 25^\\circ$ stands for. Perhaps an example will help me.\r\n\r\nWhat does the $ \\Tan \\frac {\\pi}{4}$ equal in degrees?", "Solution_11": "$ \\frac{\\pi}{4}\\equiv 45^\\circ$.\r\n\r\nNow, realize that $ \\tan: \\mathbb{R}\\to\\mathbb{R}$ is a function that takes a real number input and gives a real number output. $ 25^\\circ$ is a numerical output (an angle measurement). It's just in some unit.\r\n\r\nNow, it might be confusing if you had to draw a line that's $ 25^\\circ$ long, but in this case it's all algebraic.\r\n\r\nNote: $ \\tan$ maps a real number to a real number, [b]not necessarily[/b] a rational number. (Simple terms: it doesn't always give you a fraction).", "Solution_12": "Um, that's not even what he's asking.\r\nAnyway, those are supposed to be 25 and 30, not 25 and 30 degrees.", "Solution_13": "I was correcting his statement \"Tangent is a function that gives you a fraction as an answer\"\r\n\r\nAnyway, the problem statement itself was wrong, so whatever :D", "Solution_14": "Oh, I see now. But I think his use of the word fraction was meant to mean opposite/adjacent, a ratio of side lengths, as opposed to a measure of angle (in degrees).", "Solution_15": "[quote=\"Brut3Forc3\"]Oh, I see now. But I think his use of the word fraction was meant to mean opposite/adjacent, a ratio of side lengths, as opposed to a measure of angle (in degrees).[/quote]\r\nThat is what I meant, but I also didn't want to use the word ratio, because that does not denote a value.\r\n\r\nMy question was supposed to read $ \\tan{\\frac{\\pi}{4}}$.\r\n\r\nWhile degrees is written as a value, the value represents a degree measurement. Tangent on the other hand is a ratio, which has a real answer. $ \\tan {\\frac{\\pi}{4}}\\equal{}1$ for example. I don't see a way to say what 1 equals in degrees. Without a definition of how to change a real number into a degree measurement, the equation made no sense. I am still trying to find out what the definition is, if such a definition exists. Seeing as the problem was incorrectly written, it does not seem to be a problem, but if there is a definition, I would like to know it." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Maybe it's well-known, but I didn't manage to find it on the forum. Let $\\varphi(\\cdot)$ be the totient function. Prove that for all positive integers $n$ we have $\\sum_{k=1}^{n}\\frac{\\varphi(2k)}{k}\\le n \\le \\sum_{k=1}^{n}\\frac{\\varphi(2k-1)}{k}$.", "Solution_1": "As $\\phi (2k)$ is not relativly prime with $2,4,...,2k$ so $\\phi(2k)\\leq k$. Obvious $\\sum_{k=1}^{n}{\\frac{\\phi(2k)}{k}}\\leq n$. For right side it is $\\phi(2k-1)\\geq k$ because $2k-1$ is relativly prime with $2,4,...,2k-2$ and $(2k-1,2k-3)=1$ (this doesn't work for $k=1$, but we have $\\phi(1)\\geq 1$). So $\\sum_{i=1}^{n}{\\frac{\\phi(2k-1)}{k}}\\geq n$.", "Solution_2": "It's not true that $2k-1$ is coprime to $2,4,6,...,2k-2$, e.g. take $k=5$.", "Solution_3": "Oh, I see :blush:" } { "Tag": [ "limit", "calculus", "derivative", "Putnam", "function", "calculus computations" ], "Problem": "Let $ p,\\ q,\\ r$ be given constants. Find the values of $ a,\\ b$ such that $ \\lim_{x\\rightarrow p} \\frac {\\sqrt {ax \\plus{} b} \\minus{} q}{x \\minus{} p} \\equal{} r$.\r\n\r\nNote that we are not allowed to use l'Hopital's theorem here.", "Solution_1": "Is it \r\n\r\n$ a \\equal{} 2qr$ and $ b \\equal{} p^2r^2 \\plus{} q^2$ provided that $ p \\neq 0$ and $ pr \\plus{} q > 0$ ?", "Solution_2": "Please show your work. :)", "Solution_3": "[quote=\"JBL\"]$ (x \\minus{} p)(x \\plus{} p) \\neq x^2 \\plus{} p^2$ ;) (Reality check: the denominator approached 0, then you multiplied it my something that approaches a constant; what must the result approach?)[/quote] \r\n\r\nYeah, I can't explain what I was thinking =p", "Solution_4": "That's incorrect. I can't understand what you did.\r\nThis problem seems to be difficult.", "Solution_5": "$ (x \\minus{} p)(x \\plus{} p) \\neq x^2 \\plus{} p^2$ ;) (Reality check: the denominator approached 0, then you multiplied it my something that approaches a constant; what must the result approach?)\r\n\r\nIf you want to multiply by something to make this look nicer, you should be trying to deal with that square root up top -- how about rationalizing the numerator?", "Solution_6": "We cannot use L'Hopital's rule, but we can use the definition of the derivative.\r\n\r\nRecall that\r\n\r\n$ f'(p) \\equal{} \\lim_{x \\to p} \\frac{f(x) \\minus{} f(p)}{x\\minus{}p}$.\r\n\r\nThus, choose $ f(x) \\equal{} \\sqrt{ax\\plus{}b}$, and we require that\r\n\r\n$ f(p) \\equal{} \\sqrt{ap\\plus{}b} \\equal{} q$\r\n\r\n$ f'(p) \\equal{} \\frac{a}{2f(p)} \\equal{} r$.\r\n\r\nThis immediately gives us $ \\frac{a}{2q} \\equal{} r$, or $ a \\equal{} 2qr$. It then follows that\r\n\r\n$ b \\equal{} q^2 \\minus{} ap \\equal{} q^2 \\minus{} 2pqr$.", "Solution_7": "Hum, your solution is almost using l'Hopital's rule, isn't it? :wink: \r\n\r\nAnyway, why does it need to be $ \\lim_{x\\rightarrow p} (\\sqrt{ax\\plus{}b}\\minus{}q)\\equal{}0?$, which is the main key of the problem.", "Solution_8": "[quote=\"kunny\"]Hum, your solution is almost using l'Hopital's rule, isn't it? :wink: \n\nAnyway, why does it need to be $ \\lim_{x\\rightarrow p} (\\sqrt {ax \\plus{} b} \\minus{} q) \\equal{} 0?$, which is the main key of the problem.[/quote]\r\n\r\nI don't think your remark is fair. I did not require L'Hopital because your expression happened to be simple enough to work the way I solved it. In other words, it was sufficiently weak and I solved it using a method any elementary calculus student who knows what a derivative is should be able to understand.\r\n\r\nIf you disagree, then I ask you, at what point in my solution did I claim that\r\n\r\n$ \\lim_{x \\to p} \\frac {f(x)}{g(x)} \\equal{} \\lim_{x \\to p} \\frac {f'(x)}{g'(x)}$?\r\n\r\nIf you want to make a problem that adequately addresses your question of why the numerator must also tend to zero if the denominator tends to zero, then construct one that is sufficiently abstract and then we can talk about epsilon-delta proofs, if that is how \"elementary\" you want to take things to.", "Solution_9": "First of all, you should answer my question.", "Solution_10": "[quote=\"kunny\"]First of all, you should answer my question.[/quote]\r\n\r\nI'm not a student. I don't [b]have[/b] to answer anything unless I choose to.", "Solution_11": "Boring, that doesn't make sense.\r\nThose who readily use l'Hopital's rule doesn't understand the principle.\r\nMy problem asks us more basic fact. :wink:", "Solution_12": "And I am going to say this only one more time. If you are trying to make a point, construct a problem that can't be solved any other way than the way you want it to be solved. Otherwise, don't complain when someone solves it using a method you did not anticipate.\r\n\r\nUsually I enjoy solving your problems, but this is ridiculous. From now on, I'm not going to play your games. I'm done.", "Solution_13": "O.K.\u3000You seem to be an expert of calculus, I dare to say, your work make 8/10 points, for not being logic mathematicaly.\r\nI dare to make a opinion \uff21nti-\uff34heze in a sense, whatever I am to blame.\r\nBecause I have been receiving sevaral private mails, \"Why do you dislike using l'Hopital?''\r\n\r\nAnyway, I do wait others opinion.", "Solution_14": "Kunny, your arrogance has lead you nowhere. \r\n\r\nGuess I was among the first few who gave up answering your questions all together. Your problems are generally elementary but you make them sound like rocket science. Anyway, happy posting and try to offend fewer people. 'Pride goes before a fall'... not sure if you understand :rotfl:", "Solution_15": "[quote=\"IndoChina\"]Kunny, your arrogance has lead you nowhere. [/quote]\r\n\r\nThank you for your advice, Indochina. Let keep me the argument a little longer.", "Solution_16": "Solution.\r\n\r\nLet $ f(x)=\\frac{\\sqrt{ax+b}-q}{x-p}.$\r\n\r\nThen $ \\lim_{x \\to p}{\\sqrt{ax+b}=\\lim_{x \\to p}(f(x)(x-p)+q)}$\r\n\r\nSo $ \\sqrt{ap+b}=q \\Longrightarrow b=q^2-ap.$\r\n\r\nSo $ \\lim_{x \\to p} \\frac{(\\sqrt{ax+q^2-ap}-q)(\\sqrt{ax+q^2-ap}+q)}{(x-p)(\\sqrt{ax+q^2-ap}+q)}=\\lim_{x\\to p}\\frac{a(x-p)}{(x-p)(\\sqrt{ax+q^2-ap}+q)}=\\frac{a}{2q}$\r\n\r\nSo $ a=2rq$ and $ b=q^2-2rpq$", "Solution_17": "[quote=\"Dimitris X\"]Solution.\n\nLet $ f(x) = \\frac {\\sqrt {ax + b} - q}{x - p}.$\n\nThen $ \\lim_{x \\to p}{\\sqrt {ax + b} = \\lim_{x \\to p}(f(x)(x - p) + q)}$\n\nSo $ \\sqrt {ap + b} = q \\Longrightarrow b = q^2 - ap.$\n\nSo $ \\lim_{x \\to p} \\frac {(\\sqrt {ax + q^2 - ap} - q)(\\sqrt {ax + q^2 - ap} + q)}{(x - p)(\\sqrt {ax + q^2 - ap} + q)} = \\lim_{x\\to p}\\frac {a(x - p)}{(x - p)(\\sqrt {ax + q^2 - ap} + q)} = \\frac {a}{2q}$\n\nSo $ a = 2rq$ and $ b = q^2 - 2rpq$[/quote]\n\nYes, that's exactly what I wanted to say. The problem seems easy for experts, but few people can answer the basic question from beginner's question: Why does it need to be $ \\lim_{x\\rightarrow p} (\\sqrt {ax + b} - q) = 0?$\n\n[quote=\"atomicwedgie\"][quote=\"kunny\"]Hum, your solution is almost using l'Hopital's rule, isn't it? :wink: \n\nAnyway, why does it need to be $ \\lim_{x\\rightarrow p} (\\sqrt {ax + b} - q) = 0?$, which is the main key of the problem.[/quote][/quote]\r\n\r\nThe thing I mentioned before is the fact: \r\nFor some constants $ \\alpha ,\\ \\beta$, $ \\boxed{\\boxed {\\lim_{x\\rightarrow a} f(x) = \\alpha ,\\ \\lim_{x\\rightarrow a} g(x) = \\beta \\Longrightarrow \\lim_{x\\rightarrow a} f(x)g(x) = \\alpha \\beta }}$. We Japanese high school students just learn the fact, but we aren't taught the proof.\r\n\r\nAfter all, for my problem, $ \\lim_{x\\rightarrow p} \\frac {\\sqrt {ax + b} - q}{x - p}\\cdot (x - p) = r\\cdot 0 = 0$, yielding \r\n\r\n$ \\lim_{x\\rightarrow p} (\\sqrt {ax + b} - q) = 0$, which is the main key point.\r\n\r\n\r\nNow the discussion is end up.\r\n\r\nFinally, I appologize for my arrogance especially for atomicwedgie.\r\n\r\nkunny", "Solution_18": "A post I made in a different thread, one that kunny responded to so there should be no excuse for him to claim that I don't know how to solve this question in the way that he expected:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=1733931#1733931\r\n\r\nNow combine that with my earlier post of the solution and the result is mathematically equivalent to the \"ACCEPTED\" solution. It is the same because in both cases one rationalizes the numerator by multiplying numerator and denominator by $ \\sqrt {ax \\plus{} b} \\plus{} q$.\r\n\r\nI come here to comment on math problems and provide solutions that I think are clean, understandable, and illustrative. I don't come here to be challenged by arrogant people. I already have my degree and that was challenge enough. I took the Putnam twice and scored in the top 200 without any preparation or training. But I have not felt the need to say anything about my background because I feel that my posts speak for themselves. Math is supposed to be enjoyable and the process of learning math should be better guided than this charade.\r\n\r\nBut thanks to you kunny I am no longer going to post answers to your questions. You can take your presumptuous attitude and shove it.", "Solution_19": "This seems to have gotten a little out of hand. I think there's been overreaction and misinterpretation of intent on both side.\r\n\r\nTo atomicwedgie: Calm down. But find the threads and topics that you want to post on and post there.\r\n\r\nTo kunny: He does have this much of a point: far too often you put fences around your problems. \"Solve by _ method,\" or \"Solve without using _,\" or \"Solve by methods that Japanese high school students should know.\" My reactions to that sometimes include, first, that I personally don't care what Japanese high school students should know, and second, I bet there are quite a few Japanese high school students who don't themselves care what they should know or not know.\r\n\r\nI almost posted on this earlier, and this is approximately what I would have said:\r\n\r\nI have a long record on this site as being one of the more anti-L'H\u00f4pital posters around, and I'm the same in my own classes. I try very hard to wean my students away from using L'H\u00f4pital as some kind of instant reflex. I try to get them to understand size - what grows or shrinks faster than what else. And I try to get them to use power series - to make power series something close to a native language when talking about limits. I dislike L'H\u00f4pital mostly because it's a \"black box\" that doesn't enhance understanding, and because of that a user may be blind to its limitations.\r\n\r\nBut I don't say never. I use L'H\u00f4pital myself when it's useful - and sometimes it is the most efficient way to do things. Do you want the residue of $ f(z)\\equal{}\\frac{1}{z^4\\plus{}4}$ at $ z\\equal{}1\\plus{}i?$ L'H\u00f4pital is as good a way as any to do that one.\r\n\r\nThat brings up this particular problem. It's one of the relative few that I might well use L'H\u00f4pital on. Thinking about size, I see that it's a function with a simple zero (yes, we have to check that it's zero) divided by a function with a simple zero. L'H\u00f4pital provides a simple-enough organization for that. Or maybe you do it algebraically, or there are other methods as well. It's just not a problem that I'd use in convincing my students not to use L'H\u00f4pital.", "Solution_20": "As long as this thread isn't about math any more, I'd like to add that it seems that the biggest communication problem in this thread is not that kunny is arrogant (I've never detected arrogance in kunny's posts) but that kunny is [i]cryptic[/i] -- it's frequently unclear what kunny is actually saying. I understand that English isn't kunny's first language, and that this has something to do with the lack of clarity, but it would certainly help to use additional words and/or sentences to make it clear what things mean. For example, in kunny's first response to atomicwedgie, it's pretty clear that there was a quip followed by a question about the math, but it's not entirely clear what the mathematical question really says (and I don't think atomicwedgie noticed or understood the question); in follow-up comments, kunny never clarified.\r\n\r\nCan anyone explain to me what the following sentence means?\r\n[quote]I dare to make a opinion \uff21nti-\uff34heze in a sense, whatever I am to blame. [/quote]", "Solution_21": "it doesn't have to start with $ \\lim_{x\\minus{}p} \\sqrt{ax\\plus{}b}\\minus{}q\\equal{}0$\r\n\r\nyou can simply starts with\r\n\r\n$ r\\equal{}\\frac{(x\\minus{}p) r}{x\\minus{}p}$\r\n\r\nthen $ \\sqrt{ax\\plus{}b}\\minus{}q\\equal{}(x\\minus{}p)r$\r\n\r\nas $ x\\to p, q^2\\equal{}ap\\plus{}b$\r\n\r\ni think atomicwedgie solution is excellent", "Solution_22": "Sorry, could you make a more detail explanation?" } { "Tag": [ "calculus", "analytic geometry", "geometry solved", "geometry" ], "Problem": "I've posted this in the calculus forum and haven't got replies, maybe here it will find it's answer.\r\n\r\nYou have n points in the plane (you know the coordinates of the points) and you have to find point A so that the sum of distances fron the n pointsto A is minimum. I know that for n = 3 point A is called Fermat's point. This is a computer programming task, and a friend of mine told me it is a classical problem in calculus and you can solve it using derivation, I couldn't solve it myself ... Any help would be appreciated. Thanks", "Solution_1": "This is the famous Fermat-Weber problem; for general values of n, the point A is not constructible with compass and ruler. There are lots of literature about this problem, and there are some approximative methods known. What do you mean by \"solve it using derivation\"?\r\n\r\n Darij", "Solution_2": "You can easily use gradient method for finding required point.\r\nAnd simply derivation gives nothing, since you obtain nonlinear system of equaitions, which unsolvable in exact way." } { "Tag": [ "quadratics", "number theory unsolved", "number theory" ], "Problem": "Let a be a non-perfect square number.Prove that ther are exist a prime number p such that p|x^2-a does not hold for all integer number x", "Solution_1": "I think you meant: there exists no $x$ such that $p|x^2-a$, the other one is trivial ;)\r\n\r\nOr equivalently: such a number is not a quadratic residue $\\mod p$.\r\n\r\nProve (but not elementary):\r\nLet $a=2^s b = 2^s p_1^{v_1} p_2^{v_2} ... p_n^{v_n}$ be the prime factorisation and $b$ it's greatest odd factor. Since $a$ is not a square, either (WLOG) $v_1$ is odd or $s$ is odd.\r\n\r\n1. case: all $v_i$ are even, so $s$ is odd.\r\nThen simply taking any prime number $q$ not dividing $a$ and $\\equiv 3\\mod 8$ gives $\\left( \\frac{a}{q} \\right) = \\left( \\frac{2^s}{q} \\right) \\left( \\frac{b}{q} \\right) = (-1)^s = -1$.\r\n\r\n2. case: $v_1$ is odd.\r\nNow let $c$ be a quadratic non-residue $\\mod p_1$ and take a prime $q$ such that $q \\equiv 1 \\mod 8$, $q \\equiv c \\mod p_1$ and $q \\equiv 1 \\mod p_i$ for all other $p_i$.\r\nThen we get:\r\n$\\left( \\frac{a}{q} \\right) = \\left( \\frac{2}{q} \\right)^s \\left( \\frac{p_1}{q} \\right)^{v_1} \\left( \\frac{p_2}{q} \\right)^{v_2} ... \\left( \\frac{p_n}{q} \\right)^{v_n} = \\left( \\frac{q}{p_1} \\right)^{v_1} \\left( \\frac{q}{p_2} \\right)^{v_2} ... \\left( \\frac{q}{p_n} \\right)^{v_n}=(-1)^{v_1}=-1$\r\nso $a$ is not a quadratic residue $\\mod q$,.", "Solution_2": "Ok, I just see that it's really easy to make the proof above elementary:\r\n\r\nJust forget that $q$ has to be prime in the 2. case (but still let it belong to that residue classes). Then we get $\\left( \\frac{a}{q}\\right) =-1$, thus $a$ is not a quadratic residue $\\mod$ at least one of $q$'s prime factors.\r\n\r\nIn case one wants to prove \"if $a$ is quadratic residue $\\mod$ all primes $p>N$, then $a$ is a square\", just take $q$ such that it is also coprime to all primes $\\leq N$.", "Solution_3": "[quote=\"ZetaX\"]Ok, I just see that it's really easy to make the proof above elementary:\n\nJust forget that $q$ has to be prime in the 2. case (but still let it belong to that residue classes). Then we get $\\left( \\frac{a}{q}\\right) =-1$, thus $a$ is not a quadratic residue $\\mod$ at least one of $q$'s prime factors.\n.[/quote]\r\n\r\ni think Legendre symbol is defined when q is prime", "Solution_4": "It's the Jacobi symbol here." } { "Tag": [ "summer program", "Mathcamp" ], "Problem": "This is a reminder to all USA/Canada Mathcamp staff that I will attend mathcamp 2010.\r\nI am chosen to be in there because I dreamed it.", "Solution_1": "okay.\r\n \r\nThe message is too small. Please make the message longer before submitting." } { "Tag": [ "algebra", "polynomial", "logarithms", "quadratics", "algebra unsolved" ], "Problem": "What complex numbers (if there exist) are the roots of some polynomial with positive coefficients?", "Solution_1": "[quote=\"james digol\"]What complex numbers (if there exist) are the roots of some polynomial with positive coefficients?[/quote]\r\n\r\n[i](nota : in the following, $ \\mathbb R^\\plus{}$ is the set of positive reals and $ \\mathbb R_0^\\plus{}\\equal{}\\mathbb R^\\plus{}\\cup\\{0\\}$[/i]\r\n\r\nIt's easy to check that $ z\\equal{}\\rho e^{i\\theta}$ is a root of $ P(x)\\equal{}0$ where $ P(x)\\equal{}x^{2^n}\\minus{}2\\rho^{2^{n\\minus{}1}}\\cos(2^{n\\minus{}1}\\theta)x^{2^{n\\minus{}1}}\\plus{}\\rho^{2^n}$ \r\n\r\nThen, for $ \\theta\\in(\\minus{}\\pi,0)\\cup(0,\\pi)$, consider $ n\\equal{}1\\plus{} \\left[\\log_2(\\frac{\\pi}{|\\theta|})\\right]$ such that $ \\log_2(\\frac{\\pi}{|\\theta|})\\ge n\\minus{}1 > \\log_2(\\frac{\\pi}{|\\theta|})\\minus{}1$ and so $ \\pi\\ge 2^{n\\minus{}1}|\\theta| >\\frac{\\pi}2$. \r\n\r\nWe clearly have $ \\cos(2^{n\\minus{}1}\\theta)<0$ and so $ P(x)\\in\\mathbb R_0^\\plus{}[X]$\r\n\r\nAnd for $ \\theta\\equal{}0$ or $ \\rho\\equal{}0$ , it obviously does not exist $ P\\in\\mathbb R^\\plus{}[X]$ such that $ P(z)\\equal{}0$\r\n\r\nThen, multiplying $ x^{2^n}\\minus{}2\\rho^{2^{n\\minus{}1}}\\cos(2^{n\\minus{}1}\\theta)x^{2^{n\\minus{}1}}\\plus{}\\rho^{2^n}$ (polynomial $ \\in\\mathbb R_0^\\plus{}[X]$) by $ (1\\plus{}x\\plus{}x^2\\plus{}...\\plus{}x^{2^{n\\minus{}1}})$, we clearly get the required polynomial $ \\in\\mathbb R^\\plus{}[X]$\r\n\r\nHence rhe result : $ P(z)\\equal{}0$ for some polynomial $ P\\in\\mathbb R^\\plus{}[X]$ $ \\iff$ $ z\\in\\mathbb C\\backslash\\mathbb R_0^\\plus{}$", "Solution_2": "[b]Solution.[/b]\r\n\r\nWe show that every complex number that does not lie on the nonnegative real line is a root of such polynomial. (If we allow coefficients of terms of degree less than the degree of the polynomial to be $ 0$, then only positive real number are excluded.)\r\n\r\nFirst, observe that if $ w$ is a nonegative real number and $ P(z)$ is a polynomial with positive coefficients, then clearly $ P(w)>0$. Thus no $ w\\geq0$ is the root of a polynomial with positive coefficients.\r\n\r\nEvery $ w=a+bi \\in \\mathbb{C}$ is a root of the real polynomial \\[ q(z)=(z-w)(z-\\overline{w})=z^2-2az+a^2+b^2.\\] If $ a<0$, then $ q(z)$ is a polynomial with positive coefficients. This shows that every complex number in the open left half-plane is the root of some such polynomial.\r\n\r\nNow assume that $ a\\geq0$ and $ b\\neq0$. Thus $ w$ lies in the right half-plane, but not on the real axis, so $ 0<\\left | \\arg(w) \\right |\\leq \\pi/2$. If $ n$ is the smallest positive integer for which $ \\pi/2<\\left | \\arg(w^n) \\right |$, then $ w^n$ lies in the open left half-plane. By our work above, there exists a quadratic polynomial $ q$ with positive coefficients such that $ q(w^n)=0$there exists, say $ q(z)=z^2+c_1z+c_0$ . We now let \\[ P(z)=(z^{2n}+c_1z^n+c_0)(z^{n-1}+z^{n-2}+\\cdots+z+1).\\] It follows that $ P(z)$ has positive coefficients and $ P(w)=0$." } { "Tag": [ "IMO" ], "Problem": "Hi,\r\n\r\nI'd like to know which IMO shortlist problems you consider most beautiful? \r\n\r\nFor me its IMO 2001 N6 and 2004 G8", "Solution_1": "2000 G6 \r\n2001 G7\r\n2007 G5", "Solution_2": "2007 N5, 2006 N5, 2004 C6, 2004 G7" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "hi,\r\n[u]problem:[/u]\r\n$x > 0,x<0 \\Rightarrow f(x)=\\frac{\\sqrt{1+x}-\\sqrt{1-x}}{x}$\r\n\r\n$f(0)=1$\r\n\r\n$g(x) = \\frac{f(x)-f(0)}{x}$\r\ncompute $\\lim_{x \\to\\ 0}g(x)$ \r\n\r\nthanks in advance.", "Solution_1": "$\\lim_{x \\to 0}g(x) = \\lim_{x \\to 0}\\frac{f(x)-1}{x}$\r\n$= \\lim_{x \\to 0}\\frac{\\sqrt{1+x}-\\sqrt{1-x}-x}{x^{2}}$\r\n$= \\lim_{x \\to 0}\\frac{0.5 \\cdot (1-x)^{-0.5}+0.5 \\cdot (1+x)^{-0.5}}{2x}\\ \\ \\ \\ \\ \\ \\ \\textnormal{(l'Hospital)}$\r\n$= \\lim_{x \\to 0}\\frac{0.25 \\cdot (1-x)^{-1.5}+0.25 \\cdot (1+x)^{-1.5}}{2}\\ \\ \\ \\ \\textnormal{(l'Hospital)}$\r\n$=0$" } { "Tag": [ "binomial coefficients", "algebra unsolved", "algebra" ], "Problem": "Prove that there are no natural $n$ such that among numbers :\r\n${{n} \\choose 0},{{n} \\choose 1},{{n} \\choose 2}, ... ,{{n} \\choose n}$\r\nthere are exactly half odd and half even numbers. \r\n\r\nSorry for my poor English", "Solution_1": "I think that\r\n[hide]You should see the cases $n$ even and $n$ odd and use the fact that \n\n${{n} \\choose 0}={{n} \\choose n}$ etcera\n[/hide]", "Solution_2": "Hint: obviously, $n$ must be odd, say $n=2^mq-1$, with odd $q$. Observe that the binomial coefficients in each of the following groups:\r\n\\[ \\{ \\binom{n}{0},\\binom{n}{1},\\ldots ,\\binom{n}{2^m-1}\\}, \\]\r\n\\[ \\{ \\binom{n}{2^m},\\binom{n}{2^m+1},\\ldots ,\\binom{n}{2\\cdot2^{m}-1}\\}, \\]\r\n\\[ \\vdots \\]\r\n\\[ \\{ \\binom{n}{(q-1)2^m},\\binom{n}{(q-1)2^m+1},\\ldots ,\\binom{n}{q2^m-1}\\}, \\]\r\nhave the same parity, hence the conclusion.", "Solution_3": "A try with elementary arguments :\r\n\r\n$\\binom{n}{k} = \\binom{n}{n-k}$, $\\sum \\binom{n}{k} = 2^n$ : so there is always an even number of odd binomial coefficients.\r\n\r\nIf n is even, there is n+1 (odd) binomial coefficients, so there is an odd number of even binomial coefficients.\r\n\r\nIf n is odd, $\\binom{n}{k} = n/k \\binom{n-1}{k-1}$ : this proves that in this case $\\binom{n}{k}$ is odd for all even k, i.e for at least (n-1)/2 value. since $\\binom{n}{1}$ and $\\binom{n}{n}$ are even there are more odd binomial coefficients then even ones.", "Solution_4": "[quote=\"t\u00b5t\u00b5\"]\n\nIf n is odd, $\\binom{n}{k} = n/k \\binom{n-1}{k-1}$ : this proves that in this case $\\binom{n}{k}$ is odd for all even k[/quote]\r\nWe have $\\binom{11}{6}=462$, for instance." } { "Tag": [ "inequalities", "calculus", "derivative", "inequalities proposed" ], "Problem": "If $ a,b,c,d\\ge 0$,and $ a \\plus{} b \\plus{} c \\plus{} d \\equal{} 4$,Prove that\r\n\\[ 3(\\sqrt {a} \\plus{} \\sqrt {b} \\plus{} \\sqrt {c} \\plus{} \\sqrt {d}) \\ge 2(ab \\plus{} bc \\plus{} ca \\plus{} ad \\plus{} bd \\plus{} cd)\\]", "Solution_1": "[quote=\"zhaobin\"]If $ a,b,c,d\\ge 0$,and $ a \\plus{} b \\plus{} c \\plus{} d \\equal{} 4$,Prove that\n\\[ 3(\\sqrt {a} \\plus{} \\sqrt {b} \\plus{} \\sqrt {c} \\plus{} \\sqrt {d}) \\ge 2(ab \\plus{} bc \\plus{} ca \\plus{} ad \\plus{} bd \\plus{} cd)\\]\n[/quote]\r\nTry $ a \\equal{} b \\equal{} c \\equal{} \\frac {4}{3}$ and $ d \\equal{} 0.$\r\n\r\nHowever, the following inequality holds and nice:\r\n$ \\frac{ab\\plus{}ac\\plus{}ad\\plus{}bc\\plus{}bd\\plus{}cd}{\\sqrt{a}\\plus{}\\sqrt{b}\\plus{}\\sqrt{c}\\plus{}\\sqrt{d}} \\le \\frac{8}{3\\sqrt{3}}.$", "Solution_2": "One can of course use half convex half concave theorem to see that we only need to check a few cases, but is there a nice (and perhaps classical) solution?", "Solution_3": "It is easy to transform this inequality into $ \\sum_{cyc} f(a) \\ge 16$, where $ f(x)\\equal{}x^2\\plus{}\\frac{8}{3\\sqrt{3}}\\sqrt{x}$. This has an inflection point at $ x\\equal{}\\frac{1}{3}$, is concave on the left and convex on the right. Therefore, by Karamata on the first interval, then the second, it suffices to show the inequality when $ a,b,c,d \\in \\{0,m,n\\}$, where $ m < \\frac{1}{3}$, $ n\\ge \\frac{1}{3}$, and at most one of $ a,b,c,d$ is $ m$.\r\n\r\nIf now 3 or 4 of $ a,b,c,d$ are 0, the result is immediate by referring back to the original inequality. If none are $ m$, there are only 4 cases to consider, (0,0,0,4), (0,0,2,2), (0,4/3,4/3,4/3), and (1,1,1,1), which are essay to do through direct computation. Therefore, the only cases left to consider are $ (0,0,m,n)$, $ (0,m,n,n)$, and $ (m,n,n,n)$, up to permutation. But the first has $ n \\ge \\frac{10}{3}$, which has $ f(n) \\ge f(\\frac{10}{3}) \\ge 16$, so the first case is done. For the second and the third, employ the substitutions $ m\\equal{}4\\minus{}2n$ and $ m\\equal{}4\\minus{}3n$ respectively, and we then have 2 one-variable inequalities which can be solved using the first derivative test, and some approximations. If anyone wants to actually write these out, be my guest :P\r\n\r\nCheers,\r\n\r\nRofler\r\n\r\nP.S. what is your solution can_hang2007?", "Solution_4": "Thank you,I am wrong,sorry.\r\n[quote=\"can_hang2007\"][quote=\"zhaobin\"]If $ a,b,c,d\\ge 0$,and $ a \\plus{} b \\plus{} c \\plus{} d \\equal{} 4$,Prove that\n\\[ 3(\\sqrt {a} \\plus{} \\sqrt {b} \\plus{} \\sqrt {c} \\plus{} \\sqrt {d}) \\ge 2(ab \\plus{} bc \\plus{} ca \\plus{} ad \\plus{} bd \\plus{} cd)\\]\n[/quote]\nTry $ a \\equal{} b \\equal{} c \\equal{} \\frac {4}{3}$ and $ d \\equal{} 0.$\n\nHowever, the following inequality holds and nice:\n$ \\frac {ab \\plus{} ac \\plus{} ad \\plus{} bc \\plus{} bd \\plus{} cd}{\\sqrt {a} \\plus{} \\sqrt {b} \\plus{} \\sqrt {c} \\plus{} \\sqrt {d}} \\le \\frac {8}{3\\sqrt {3}}.$[/quote]", "Solution_5": "[quote=\"Soarer\"]One can of course use half convex half concave theorem to see that we only need to check a few cases, but is there a nice (and perhaps classical) solution?[/quote]\r\nThere is a nice solution. We can modify a little to get an equivalence inequality:\r\n\r\nIf $ a,$ $ b,$ $ c,$ $ d$ are nonnegative real numbers such that $ a\\plus{}b\\plus{}c\\plus{}d\\equal{}3,$ then\r\n$ \\sqrt{a}\\plus{}\\sqrt{b}\\plus{}\\sqrt{c}\\plus{}\\sqrt{d} \\ge ab\\plus{}ac\\plus{}ad\\plus{}bd\\plus{}bd\\plus{}cd.$\r\n\r\nNow, take a look here to view my proof: http://can-hang2007.blogspot.com/2010/01/inequality-27-v-q-b-can.html" } { "Tag": [ "trigonometry", "MATHCOUNTS", "function" ], "Problem": "Is ther anyone here that can help me with trigonometry(the basics). I was supposed to have taken it in class last year but the teacher told me study for mathcounts instead. \r\n\r\nAll I need to know is how the Sin, Cos, And Tan work in a tringle, and what it is used for.", "Solution_1": "$ SOH,CAH,TOA$\r\n\r\nlook this up. \r\nI doubt someone can teach you in one single post.\r\n\r\nread about it around the internet if no one can help you personally.", "Solution_2": "I'm also learning trigonometry on my own.\r\n\r\nDoes any know of any great books I should read?\r\nAny books that are designed for trig problem solving?\r\n\r\n@tonypr: I found that learning just the trignometric functions was enough to figure out the rest. \r\n I was able to recreate much of the book using just these definitions on my own.", "Solution_3": "[quote=\"tonypr\"]Is ther anyone here that can help me with trigonometry(the basics). I was supposed to have taken it in class last year but the teacher told me study for mathcounts instead. \n\nAll I need to know is how the Sin, Cos, And Tan work in a tringle, and what it is used for.[/quote]\r\n\r\nThe $ \\sin$ of an angle is the opposite side length divided by the hypoteneuse side length of that angle placed in a triangle with a right angle.\r\nThe $ \\cos$ of an angle is the adjacent side length over the hypoteneuse side length of that angle placed in a triangle with a right angle.\r\nThe $ \\tan$ of an angle is the opposite side divided by the adjacent side of that angle placed in a triangle with a right angle.\r\n\r\nA good way to remember other than $ SOHCAHTOA$ ($ SOHCAHTOA$ doesn't work that well for me) is to associate $ \\sin$ with opposite, $ \\cos$ with adjacent, and $ \\tan$ with nonhypoteneuse (or hypoteneuse, whichever is easier to remember). Also remember that other than $ \\tan$, they're divided by the length of the hypoteneuse." } { "Tag": [ "modular arithmetic" ], "Problem": "Let $ f(n)$ denote the sum of the digits of n. Let $ N \\equal{} 4444^{4444}$. Find $ f(f(f(N)))$.", "Solution_1": "[hide=\"Solution\"]\n\nFirst we see that $ 4444\\equiv 7\\pmod{9}$, and $ 4444^{3k\\plus{}1}\\equiv 7\\pmod{9}$. This implies that the sum of the digits of $ 4444^{4444}$ will be congruent to $ 7\\pmod{9}$.\n\nNext we analize $ 4444^{4444}$, $ 4444^{4444}< (10^4)^{4444} \\equal{} 10^{17776}$ so $ f(N)\\le 17776 \\cdot 9 \\equal{} 159984 < 199999$, from that we get $ f(f(N))\\le 46<49$ and $ \\therefore f(f(f(N)))\\le 13$. Since $ f(f(f(n)))\\equiv 7\\pmod{9}$ there is only one answer $ \\boxed{7}$.\n\nAny questions? :ninja: \n[/hide]", "Solution_2": "I got a similar solution, and I was wondering if there were any other solutions." } { "Tag": [ "algebra", "polynomial", "number theory unsolved", "number theory" ], "Problem": "The sequence $ (x_n)_{n \\geq 1}$ is defined by: \r\n$ x_1\\equal{}3$ \r\n$ x_2\\equal{}7$ and \r\n$ x_{n\\plus{}2}\\equal{}3x_{n\\plus{}1}\\minus{}x_n$.\r\nProve that if $ x_n\\plus{}(\\minus{}1)^n$ is a prime number, then $ n\\equal{}3^m$ $ (m \\in N)$\r\n\r\nI'm sorry if it had been posted already.I can not find this problem in the forum.Thanks a lot.", "Solution_1": "Trivially $ x_1\\minus{}1\\equal{}2 \\in \\mathbb{P}$ and in $ \\mathbb{Z}/2\\mathbb{Z}$ we obtain that if $ x_n \\in \\mathbb{P}$ for some $ n>1$ then $ 3 \\mid n$. Working straighforward with charateristic polynomial $ \\epsilon^2\\minus{}3\\epsilon\\plus{}1$ (that incidentally have square of gold section as roots) we obtain $ y_n: \\equal{}x_n\\plus{}(\\minus{}1)^n\\equal{}z^n\\plus{}z^{\\minus{}n}\\plus{}(\\minus{}1)^n$, where $ z: \\equal{}(3\\plus{}\\sqrt{5})2^{\\minus{}1}\\equal{}\\left( (1\\plus{}\\sqrt{5})2^{\\minus{}1}\\right)^2$. Define $ q(n): \\equal{}3^{\\upsilon_3(n)}$ for all $ n \\in \\mathbb{N}_0$. We claim that if $ nq(n)^{\\minus{}1}>1$ then $ y_{q(n)}$ divides (strictly) $ y_n$, that is $ z^{q(n)}\\plus{}z^{\\minus{}q(n)}\\minus{}1 \\mid z^{n}\\plus{}z^{\\minus{}n}\\plus{}(\\minus{}1)^n$. Now it is [i]equivalent[/i] to show that if a real $ t$ is fixed, and if $ p(x): \\equal{}\\phi_3(x)x^{\\minus{}1}$ (where $ \\phi_i(x)$ is the i-th cyclotomic polynomial) verify $ p(t^k) \\in \\mathbb{Z}$ for all $ k \\in \\mathbb{N}_0$ then $ p(t)$ divides $ p(t^q)$ for all $ q \\in \\mathbb{N} \\setminus 3 \\mathbb{N}$. We'll show it by PMI: if $ q\\equal{}2$ then $ p(t^2)\\equal{}p(t)(p(t)\\minus{}2)$ and if $ q\\equal{}4$ then $ p(t^4)\\equal{}$ $ p(t)(p(t)\\minus{}2)(p(t^2)\\minus{}2)$. Suppose that the thesy is true up to $ q\\equal{}3j\\minus{}1$, we'll verify it for $ q\\equal{}3j\\plus{}1$ and $ q\\equal{}3j\\plus{}2$, infact $ p(t^{3j\\plus{}1})\\equal{}$ $ p(t)(p(t)\\minus{}2)(p(t^{3j\\minus{}1}\\minus{}1)\\minus{}p(t^{3j\\minus{}1})\\plus{}[p(t^{3j\\minus{}3})\\minus{}3]$ and $ p(t^{3j\\plus{}2})\\equal{}$ $ p(t)(p(t^{3j\\plus{}1})\\minus{}1)\\minus{}p(t^{3j\\plus{}1})\\plus{}[p(t^{3j}\\minus{}3]$. It mean that it is enough to show that $ p(t)$ divides $ p(t^{3v})\\minus{}3$ for all $ v \\in \\mathbb{N}_0$, and we'll show it again by PMI: in fact if $ v\\equal{}1$ then $ p(t^3)\\minus{}3\\equal{}p(t)(p^2(t)\\minus{}3p(t))$, if $ v\\equal{}2$ then $ p(t^6)\\minus{}3\\equal{}p(t)(p^2(t)\\minus{}3p(t))(p(t^3)\\minus{}2)$ and if the thesy is shown up to $ v\\minus{}1$ then $ p(t^{3v})\\minus{}3\\equal{}p(t)(p^2(t)\\minus{}3p(t))(p(t^{3v\\minus{}3})\\minus{}1)\\plus{}2(p(t^{3v\\minus{}3})\\minus{}3)\\minus{}(p(t^{3v\\minus{}6})\\minus{}3)$.", "Solution_2": "thank you very much :roll:" } { "Tag": [ "AMC", "AIME", "AMC 10" ], "Problem": "With the new scoring system of 1.5 points for every blank answer, how many questions should I answer on the AMC 10?", "Solution_1": "Assuming you need a $ 120$ to qualify for AIME, and your goal is qualifying for AIME, you must answer at least $ 19$ to qualify. I would answer at least $ 22$ questions, because stupid mistakes crop up inevitably.", "Solution_2": "I probably should have mentioned that I want to qualify for the AIME. Thanks.\r\n\r\nIf I answered 22, the maximum number of answers I could get wrong and still qualify would be 2. The same is true if I answered 21 questions. My next question is - should I answer 21 or 22? Would the extra time in solving one less problem improve my chances?", "Solution_3": "Don't try to game the test. Answer as many as you can solve - this will usually give you the best chance of making any target score.", "Solution_4": "[quote=\"nuclearbob\"]With the new scoring system of 1.5 points for every blank answer, how many questions should I answer on the AMC 10?[/quote]\r\n\r\nIf it's possible, answer every single one. :lol: \r\n\r\nHowever, in the case where you are unable to, guess on the problems that you have narrowed down to like 2 answers, and just skip the rest.", "Solution_5": "Wait, what it's only 1.5 for every blank answer now? Is this the same for the amc 12 too?", "Solution_6": "AMC 10 = 6 points per correct answer, 1.5 per blank answer, 0 points incorrect answer, qualifying score 120 and/or top 1%\r\nAMC 12 = same as AMC 10 except qualifying score is 100 and/or top 5%", "Solution_7": "[quote=\"Truffles\"]Wait, what it's only 1.5 for every blank answer now? Is this the same for the amc 12 too?[/quote]\r\n\r\nYes, that is correct. \r\n\r\nI would answer all of them, if possible." } { "Tag": [ "ARML", "email", "mathleague.org", "MATHCOUNTS" ], "Problem": "KS does not have a team, so you have to join MO (there were 6 Kansans on MO ARML last year). The site seems to be down. Please contact the head coach \"RArmstrong@stlcc.edu\" to join the mailing list. An application is sent out in the spring, and there's a fee, like 200-300 dollars. MO sends 2 teams and a few alternates, so a little more than 30 people, which in the last two years, is everyone that applied. MO does have practices about once a month in Springfield and St. Louis, though they are not required.", "Solution_1": "i am trying to revive the KS team for this year (i started the team in 98 but it kind of died when i moved to CA). i'll be recruiting from AMC and GPML results, but if you're interested send an email to mathleague@mathleague.org and i'll put you on a mailing list for more information when it becomes available..", "Solution_2": "Whats GPML? Also, what are the typical scores to get in?", "Solution_3": "Great Plains Math League: http://mathleague.org/\r\n\r\nis a competition with about 6 qualifying meets per year, then a state meet, and then a championship meet. The meets are hosted throughout MO, IA, KS, and CA.\r\n\r\nIt's kinda like ARML and/or Mathcounts, with a sprint, target, team, and relay round. (I'm not sure how they do power round)." } { "Tag": [ "function", "limit", "modular arithmetic", "algebra", "polynomial", "real analysis", "real analysis unsolved" ], "Problem": "Let $ f: [0,1]\\to\\mathbb{R}$ a continuous function .Prove that \r\n\\[ \\lim_{n\\to\\infty}\\frac{1}{2^n}\\sum _{k\\equal{}1}^n (\\minus{}1)^kC_n^k\\cdot f\\left( \\frac{k}{n}\\right)\\equal{}0\\]", "Solution_1": "[hide=\"hint\"]\nsince $ f$ is continious there exists a maxima $ M$ and a minima $ m$\nwe have then $ m\\sum_{k \\equal{} 1,k \\equiv 0 \\pmod 2}^{n}\\binom{n}{k} \\minus{} M \\sum_{k \\equal{} 1,k \\equiv 1 \\pmod 2}^{n} \\binom{n}{k} \\leq \\sum_{k \\equal{} 1}^{n} ( \\minus{} 1)^{k} \\binom{n}{k} f(\\frac {k}{n}) \\leq M\\sum_{k \\equal{} 1,k \\equiv 0 \\pmod 2}^{n}\\binom{n}{k} \\minus{} m \\sum_{k \\equal{} 1,k \\equiv 1 \\pmod 2}^{n} \\binom{n}{k}$\nalso we have if\n$ \\lim_{n\\to\\infty}\\frac {1}{2^n}\\sum _{k \\equal{} 1}^n ( \\minus{} 1)^kC_n^k\\cdot f\\left( \\frac {k}{n}\\right) \\equal{} L$\nthen \n$ \\frac {m\\sum_{k \\equal{} 1,k \\equiv 0 \\pmod 2}^{n}\\binom{n}{k} \\minus{} M \\sum_{k \\equal{} 1,k \\equiv 1 \\pmod 2}^{n} \\binom{n}{k}}{2^{n}} \\leq L \\leq \\frac {M\\sum_{k \\equal{} 1,k \\equiv 0 \\pmod 2}^{n}\\binom{n}{k} \\minus{} m \\sum_{k \\equal{} 1,k \\equiv 1 \\pmod 2}^{n} \\binom{n}{k}}{2^{n}}$\nthsi along with the know binomial sum and that as $ n \\to \\infty$ m=M can be proved...\ni am leaving will return hope i am right sorry :( [/hide]", "Solution_2": "Actually, pardesi's hint is not very useful :(. What one should notice instead is that the statement is true for every polynomial (the expression gets exactly $ 0$ as soon as $ n$ exceeds the degree of the polynomial) and that polynomials are dense in $ C[0,1]$. This is not the only possible approach, of course.", "Solution_3": "sorry...i wiil try if possible to continue :rotfl: :)" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Find maximum value of x>0 such that for every a,b,c with conditions 0=1/3+x(1-a)(1-b)(1-c)", "Solution_1": "please solve this problem i couldn't", "Solution_2": "[quote=\"Ulanbek_Kyzylorda KTL\"]Find maximum value of x>0 such that for every a,b,c with conditions 0=1/3+x(1-a)(1-b)(1-c)[/quote]\r\nIf $ a\\equal{}b\\equal{}c\\equal{}\\frac{1}{3}$ we getting $ x\\leq\\frac{9}{4}.$\r\nId est, we need to prove that $ \\frac{1}{a\\plus{}b\\plus{}c}\\geq\\frac{1}{3}\\plus{}\\frac{9}{4}(1\\minus{}a)(1\\minus{}b)(1\\minus{}c).$\r\nLet $ a\\plus{}b\\plus{}c\\equal{}t.$\r\nHence, $ (1\\minus{}a)(1\\minus{}b)(1\\minus{}c)\\leq\\left(\\frac{3\\minus{}a\\minus{}b\\minus{}c}{3}\\right)^3\\equal{}\\left(\\frac{3\\minus{}t}{3}\\right)^3.$\r\nThus, it remains to prove that $ \\frac{1}{t}\\geq\\frac{1}{3}\\plus{}\\frac{9}{4}\\left(\\frac{3\\minus{}t}{3}\\right)^3.$\r\nBut $ \\frac{1}{t}\\geq\\frac{1}{3}\\plus{}\\frac{9}{4}\\left(\\frac{3\\minus{}t}{3}\\right)^3\\Leftrightarrow4(3\\minus{}t)\\geq t(3\\minus{}t)^3$ and $ t(3\\minus{}t)^2\\equal{}4t\\cdot\\frac{3\\minus{}t}{2}\\cdot\\frac{3\\minus{}t}{2}\\leq4\\left(\\frac{t\\plus{}\\frac{3\\minus{}t}{2}\\plus{}\\frac{3\\minus{}t}{2}}{3}\\right)^3\\equal{}4.$ Done!." } { "Tag": [ "calculus", "integration", "number theory", "prime numbers" ], "Problem": "[color=darkblue][size=75][/size][/color]\r\n\r\nWe can use $n! + 1$ to prove that there are infinitely many primes (a nicer proof than the usual one, in my opinion).\r\n\r\nAs the first post noted, every prime factor of $n! + 1$ must be greater than $n$. In particular, there is a prime greater than $n$. That means there are arbitrarily large primes. That means there are infinitely many primes. QED.", "Solution_1": "Hmm... ~QED.\r\n\r\nFirst we have to prove that there are infinitely many n such that n!+1 is composite. Besides this, the proof is rather unsatsifying; it seems to have a lot of holes. I'll finish it up, while answering the first question simultaneously:\r\n\r\nSuppose n is one less than a prime (p). Then n!=pm-1 (for some integral m), a famous theorem. Add one to get pm. Therefore, not all n!+1 are prime.\r\n\r\nNow take n!+1=mp (where n=p-1 and p is the largest prime known). Since n!+1=mp is not divisible by anything less than p, there exist other prime factors greater than p. Given such a prime factor, repeat to find something larger. QED", "Solution_2": "[quote=\"jb05\"]First we have to prove that there are infinitely many n such that n!+1 is composite. Besides this, the proof is rather unsatsifying; it seems to have a lot of holes.[/quote]\r\nWhy do we have to prove that $n! + 1$ is composite for infinitely many $n$? If $n! + 1$ is prime, then it is already a prime greater than $n$.\r\n\r\nCan you point out one hole in the proof?", "Solution_3": "My math book gave a very simple explanation, starting with the fact that there are infinitely many integers.", "Solution_4": "bah. didn't euclid prove this one way back?\r\n\r\nassume $p_{1},p_{2},p_{3},...,p_{n}$ is all the prime numbers (assume a finite number)\r\n\r\n$p_{1}p_{2}p_{3}...p_{n} + 1$ is not divisible by any $p_{x}$, therefore it is prime. This contradicts the fact that there is a finite number of primes.", "Solution_5": "[quote=\"Hamster1800\"]bah. didn't euclid prove this one way back?\n\nassume $p_{1},p_{2},p_{3},...,p_{n}$ is all the prime numbers (assume a finite number)\n\n$p_{1}p_{2}p_{3}...p_{n} + 1$ is not divisible by any $p_{x}$, therefore it is prime. This contradicts the fact that there is a finite number of primes.[/quote]\r\n\r\nyep, i believe so :D", "Solution_6": "hasn't this been posted before? :huh:", "Solution_7": "[quote=\"Hamster1800\"]$p_{1}p_{2}p_{3}...p_{n} + 1$ is not divisible by any $p_{x}$, therefore it is prime.[/quote]\r\nThis step is false. For example, 2 * 3 * 5 * 7 * 11 * 13 + 1 is not prime. (It factors into 59 * 509.)", "Solution_8": "but 59 and 509 are not prime because only 2, 3, 5, 7, 11, and 13 are primes so therefore 59 and 509 are not prime factors of that number making it prime (not divisible by any primes)", "Solution_9": "I think there is a little confusion here, or would be to some reading this. What is being said is that you can prove by contradiction. Assume there is a finite number of primes, say $p_i, 1 \\le i \\le k$. Then if we consider $\\prod_{i=1}^k p_i + 1$, not one of the existing primes is a factor. This is a contradiction, since all numbers have at least one prime factor.", "Solution_10": "Look inside this book:\r\n\r\nhttp://www.amazon.com/exec/obidos/tg/detail/-/3540404600/qid=1117417130/sr=8-1/ref=pd_csp_1/103-4841712-2342228?v=glance&s=books&n=507846\r\n\r\n5 of those proofs are amazing.", "Solution_11": "http://www.amazon.com/gp/reader/3540404600/ref=sib_dp_pt/104-5819939-1591118#reader-page\r\n\r\nCool :)" } { "Tag": [], "Problem": "Is anybody there?", "Solution_1": "whom you are asking? :)", "Solution_2": "wht's up? :)", "Solution_3": "$ a,b,c,d$ \r\n$ (8a\\plus{}4b\\plus{}2c\\plus{}d\\plus{}16)(27a\\plus{}9b\\plus{}3c\\plus{}d\\plus{}81)<0$\r\n$ max\\{|a|,|b|,|c|,|d|\\}>1$" } { "Tag": [ "email", "AwesomeMath", "summer program" ], "Problem": "Hey, guys do you know precamp material is in the mail?\r\nI am really excieted!!", "Solution_1": "hmm perhaps they will send them after may 19, since that's when they send out emails for the C registration?", "Solution_2": "anybody from C registration get the materials yet?", "Solution_3": "What materials?", "Solution_4": "Hmm...Last year I took test C and got my acceptance like the end of May. Looking back at my email, I think that precamp materials were mostly sent out the first week of July. I'm not sure about that. For the time being, you have most of what you need. Take a look at the curriculum pdf on the site. The only other things that the sent out were like rules, maybe some forms to fill out, packing advice, and how to meet everybody at the airport.", "Solution_5": "Can we drive there if we live close-ish-far-ish (but mostly close-ish)?", "Solution_6": "I couldn't see the directors saying that must fly there... As long as your parents drive you, I don't see a problem. Last year everybody sorta went to this one building to say hello and all. Your parents could easily drop you off there. I think there was some opening ceremony after everybody arrived that parents could attend. \r\n\r\nOn another note, the Awesomemath website says \"AMSP Pre-camp materials are in the mail.\" so thats the answer I guess." } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "What's the most efficient rigorous way to calculate limits like\r\n$\\lim_{x\\to \\infty}[\\sqrt[2007]{x^{2007}+x^{2006}}-\\sqrt[2007]{x^{2007}-x^{2006}}]$ \r\nwithout using L'Hopital's Rule?", "Solution_1": "Well, I'm always a big fan of Newton's binomial series:\r\n\r\n$(1+u)^{\\alpha}=1+\\alpha u+\\frac{\\alpha(\\alpha-1)}2u^{2}+O(u^{3})$ as $u\\to0.$\r\n\r\nThe first rule of thumb in using this: first factor out the thing that is big. So, as $x\\to\\infty,$\r\n\r\n$(x^{2007}+x^{2006})^{1/2007}-(x^{2007}-x^{2006})^{1/2007}$\r\n\r\n$=x\\left(1+\\frac1x\\right)^{\\frac1{2007}}-x\\left(1-\\frac1x\\right)^{\\frac1{2007}}$\r\n\r\n$=x\\left[\\left(1+\\frac{1}{2007x}+O(x^{-2})\\right)-\\left(1-\\frac{1}{2007x}+O(x^{-2})\\right)\\right]$\r\n\r\n$=\\frac2{2007}+O(x^{-1}),$ which tends to $\\frac{2}{2007}$ as $x\\to\\infty.$", "Solution_2": "So basically, degree 2005 and lower is insignificant under the 2007th root and you can approximate $x^{2007}+x^{2006}\\approx (x+1/2007)^{2007}$? :maybe:", "Solution_3": "oh, allways $\\sqrt[n]{ax^{n}+bx^{n-1}+...}\\sim \\sqrt[n]{a}\\mid x+\\frac{b}{na}\\mid$." } { "Tag": [ "inequalities", "trigonometry", "logarithms" ], "Problem": "Prove that:\r\n$ sin^{2n}(\\theta)cos^{2n}(\\theta)\\le \\frac {sin^{4n}(\\theta)}{log_{2}(2 \\plus{} 2^{tan^{2n}(\\theta)})} \\plus{} \\frac {[log_{2}(2 \\plus{} 2^{tan^{2n}(\\theta)}]\\cos^{4n}(\\theta)}{4}$\r\nis true for all $ n\\in\\mathbb{N^{*}}$ and $ (\\theta)\\neq \\frac {\\pi}{2} \\plus{} k\\pi, k \\in \\mathbb{Z}$\r\nWhen equality?", "Solution_1": "[hide=\"Solution?\"]Simplify the logarithm first. \\begin{align*}\\log_2(2+2^{\\tan^{2n}\\theta})&=\\log_2[2(1+2^{\\tan^{2n}\\theta-1})]\\\\&=1+\\log_2(1+2^{\\tan^{2n}\\theta-1})\\qquad(1)\\end{align*}\nTo make the inequality look \"nicer\" and a lot easier to work with (and type), let\n\\[ a = \\sin^n\\theta\\newline b = \\cos^n\\theta\\newline x = \\log_2(2 + 2^{\\tan^{2n}\\theta}).\\]\nThen the previously complicated-looking inequality becomes the simple task of proving\n\\[ a^2b^2\\le\\frac {a^4}{x} + \\frac {b^4x}{4}.\\]\nThis can easily be shown using AM-GM since all the terms are nonnegative. This is easy to see: by the Trivial Inequality, $ a^2,b^2\\ge0$ and from $ (1),$ the quantity is $ 1+k$ with $ k$ being the logarithm of a quantity greater than $ \\frac{1}{2}$, meaning that the logarithm must be greater than $ -1$ (even though the exact value is unknown), so that is also nonnegative.\n\\[ \\frac {\\frac {a^4}{x} + \\frac {b^4x}{4}}{2}\\ge\\sqrt {\\left(\\frac {a^4}{x}\\right)\\left(\\frac {b^4x}{4}\\right)}\\newline\\frac {a^4}{x} + \\frac {b^4x}{4}\\ge2\\sqrt {\\frac {a^4b^4}{4}}\\newline\\frac {a^4}{x} + \\frac {b^4x}{4}\\ge2\\left(\\frac {a^2b^2}{2}\\right)\\]\n$ \\text{Q.E.D}$\n[/hide]" } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "let (C1),(C2),...,(Cn) be n circles .\r\n\r\nfind the locus of point M,in wich : the power of M to (C1)+the power of M to (C2)+...+the power of M to (Cn)=stable", "Solution_1": "circle?line?.....what?", "Solution_2": "I think that is that is quite hard .\r\nwhere did you find it ashegh??\r\n :o", "Solution_3": "No Kheirkhah come on it's so easy try to solve it.I will send my solution if U couldn't solve it.", "Solution_4": "the problem isn't hard. plz use this fact.\r\n\r\ndo u know what will be the locus of points like M in wich:\r\n\r\nthe power of point M to (C)+the power of point M to (C')=stable\r\n\r\ni talked about it recently in fisaghoures(pytagoras)theorem.\r\n\r\nthe locus will be a circle wich its center is the mid point of OO'. (wich OO' is the centers of the circles (C) and (C').\r\n\r\nwe can also solve it by using gravity :!:", "Solution_5": "I'll hide my solution, because there are people who want to post their own\r\n\r\n\r\n[hide]I'll start inversely\n\nLet $O_1,O_2,...,O_n$ the points of the circles and $G$ their center of gravity $\\Rightarrow \\sum_{i=1,..,n}\\overline{GO_i}=0$\n\n\nThe power of $M$ to the circle $C_i$ is $MO_i^2-r_i^2$\n\nLet $S$ the sum of all powers\n\n$S=\\sum{(MO_i^2-r_i^2)}=constant \\Leftrightarrow$\n\n$\\sum{MO_i^2}=constant$\n\n$\\overline{MO_i}=\\overline{MG}+\\overline{GO_i}$\n\n$|\\overline{MO_i}|^2=\\overline{MO_i}^2=(\\overline{MG}+\\overline{GO_i})^2$\n\n$|\\overline{MO_i}|^2=|\\overline{MG}|^2+|\\overline{GO_i}|^2+2\\overline{MG} \\cdot \\overline{GO_i}$\n\n$\\sum{|\\overline{MO_i}|^2}= \\sum{|\\overline{MG}|^2}+\\sum{|\\overline{GO_i}|^2}+2\\sum{\\overline{MG} \\cdot \\overline{GO_i}}$\n\n$\\sum{|\\overline{MO_i}|^2}= n \\cdot |\\overline{MG}|^2+\\sum{|\\overline{GO_i}|^2}+2\\overline{MG} \\cdot\\sum{ \\overline{GO_i}}$\n\n$\\sum{|\\overline{MO_i}|^2}= n \\cdot |\\overline{MG}|^2+\\sum{|\\overline{GO_i}|^2}+2\\overline{MG} \\cdot\\sum{ \\overline{0}}$\n\n$\\sum{|\\overline{MO_i}|^2}= n \\cdot |\\overline{MG}|^2+\\sum{|\\overline{GO_i}|^2}$\n\n\nIn the last equation, $\\sum{|\\overline{GO_i}|^2}$ is constant and $n \\cdot |\\overline{MG}|^2$ depends only on $R=|\\overline{MG}|$\n\nFinally we find that\n\n$S=n \\cdot R^2 +c$, where $c$ is a constant ($c=\\sum{|\\overline{GO_i}|^2}-\\sum{r_i^2}$)\n\nEach of the circles $(G,R)$ is a locus of points, which gives its own sum $S=n \\cdot R^2 +c$\n\n\n\n\n\nGiven the value of $S$, we can find the value of $R$, iff $S\\geq c$\n\n\nIf $S=c=\\sum{|\\overline{GO_i}|^2}-\\sum{r_i^2}$ then the locus is the point $G$ itself[/hide]\r\n\r\nActually, this problem was really nice, but not so easy, until ashegh gave the hint.\r\n\r\nI'd like to see other solutions, particularly with pure geometry!", "Solution_6": "excelent... :) \r\n\r\nu wrote the thing that i said..." } { "Tag": [ "trigonometry", "inequalities unsolved", "inequalities" ], "Problem": "Find the max of $sinA_{1}sinA_{2}\\cdots sinA_{n}$ such that $tanA_{1}tanA_{2}\\cdots tanA_{n}= 1$.", "Solution_1": "$\\prod_{i=1}^{n}(\\sin A_{i})^{2}= \\prod_{i=1}^{n}\\frac{(\\tan A_{i})^{2}}{1+(\\tan A_{i})^{2}}$\r\n$= \\prod_{i=1}^{n}\\frac{1}{1+(\\tan A_{i})^{2}}$\r\n$\\leq \\prod_{i=1}^{n}\\frac{1}{2 \\tan A_{i}}= 2^{-n}$\r\nHence the maximum is $2^{-0.5n}$" } { "Tag": [], "Problem": "Hello,\r\n\r\nI think this is as off topic as is possible in a forum about Mathematical Problem Solving, but hey, that's what The Round Table is for, right?\r\n\r\nI'm working on a robot design with two sets of wheels - one needs to be lowered using a motor while the robot is running. I plan to do this by building a frame to hold the second set of wheels and gluing the end of a bolt to the center of said frame. This bolt will run through a nut glued to the body of the vehicle. Now when the bolt turns the frame will be lowered or raised, thus accomplishing the goal of switching wheels. To make things easy, I'm looking for a bolt that had a head with teeth so that I can turn it with a geared motor.\r\n\r\nSo my question is, where could I buy a bolt with a toothed head? Preferred length is about an inch, but I could always cut off excess.\r\n\r\nI'm also open to any better ideas for changing wheels. Note that I'm trying to limit the size of the body to about 9 by 3 by 3 inches, with a micro-controller board coming vertically off of the top.\r\n\r\nThanks!", "Solution_1": "Probably the part of the forum that deals with engineering would be better ;) I've moved it there." } { "Tag": [ "linear algebra", "matrix", "complex numbers", "superior algebra", "superior algebra solved" ], "Problem": "Let a,b,c,d,e,f,g,h,i reals or complex numbers \r\n$A=\\left (\\begin{array}{ccc}a&b&c\\\\d&e&f\\\\g&h&i\\end{array}\\right )$\r\n\r\n$B=\\left (\\begin{array}{ccc}a&-c&b\\\\-g&i&-h\\\\d&-f&e\\end{array}\\right )$\r\n\r\n\r\n\r\n\r\nProve that A and B are similar.", "Solution_1": "What is the definition of two similar matrices, A and B?", "Solution_2": "There is an invertible matrix $X$ s.t. $A=XBX^{-1}$.", "Solution_3": "[tex]X=I_{2,3}I_{3}(-1)=\\left(\\begin{array}{ccc}\r\n1&0&0\\\\0&0&-1\\\\0&1&0\\end{array}\\right)\r\n[/tex]\r\n\r\nAnd [tex]XAX^{-1}=B[/tex] , so A and B are similar." } { "Tag": [ "trigonometry", "Olimpiada de matematicas" ], "Problem": "[color=blue][size=150] Rusia 1997[/size][/color]\r\n\r\nSea $\\Gamma$ un circulo con centro $O$ e inscrito al triangulo $ABC$,\r\nal cual toca en sus lados $AC, AB, BC$ en los puntos $K,M,N$ respectivamente. La mediana $BB_1$ del triangulo $ABC$\r\ncorta al segmento $MN$ en $D$. Mostrar que los puntos $O,D,K$ son colineales.\r\n\r\n(Sugerencia)\r\n[hide=\"Hint\"] Consideremos $L = KO \\cap MN$, y probemos que $L = D.$ Luego juega con algo de semejanza. [/hide]\r\n\r\n\r\nCarlos Bravo :)\r\nLima -PERU", "Solution_1": "he encontrado una soluci\u00f3n bastante fea para el problema. utilizando ley de senos, teorema de la bisectriz generalizada , identidades trigonom\u00e9tricas prob\u00e9 que la mediana y la linea OK dividen al segmento MN en raz\u00f3n AB : BC Y por lo tanto MN,OK Y BB_1 concurren . pero me gustar\u00eda ver una soluci\u00f3n mas bonita (la de semejanza) \r\npuedes colgarla carlos?? :) gracias :P :lol:", "Solution_2": "Bueno el Hint realmente contiene el key ..\r\naqui envio la solucion referida en la sugerencia ...\r\n\r\n [color=blue][size=150]Solucion:[/size][/color]\r\n \r\n Sea $L = KO \\cap MN$, y sea $A_1, C_1$ los puntos de interseccion de $AB$ y $BC$ respectivamente con la paralela trazada a $AC$ que pase por $L$, portanto tenemos $A_1L =LC_1$, lo cual implica que $L$ es el punto medio de $BB_1$ y entonces $L= D.$\r\n\r\nNotar que: $\\angle A_1MO$ y $\\angle A_1LO$ son rectos, entonces el cuadrilatero $MA_1OL$ es ciclico, entonces $\\angle MLA_1 = \\angle MOA_1$ como tambien $\\angle C_1LN = \\angle C_1ON$ \r\n\r\n( de manera analoga, notemos que $\\angle MLA_1$ y $\\angle C_1LN$ son rectos, entonces $\\angle MOA_1 = \\angle C_1ON$). De aqui tenemos que los triangulos $OMA_1$ y $ONC_1$ son congruentes, entonces $OA_1 = OC_1$ con lo cual concluimos que $A_1L = LC_1.$ \r\n\r\nCarlos Bravo :)\r\nLima -PERU", "Solution_3": "Supongamos que $ KO$ corta a $ MN$ en $ D.$ Tenemos en el triangulo is\u00f3sceles $ \\triangle BMN$\n\n$ \\frac { \\sin \\angle DBC}{\\sin \\angle DBA} \\equal{} \\frac {DN}{DM} \\equal{} \\frac {KN}{KM} \\cdot \\frac {\\sin \\angle DKN}{\\sin \\angle DKM} \\equal{} \\frac { \\cos (\\frac {C}{2}) \\sin (\\frac {C}{2})}{ \\cos (\\frac {A}{2}) \\sin (\\frac {A}{2})} \\equal{} \\frac { \\sin C}{ \\sin A} \\equal{} \\frac {c}{a}$ \n\n$ \\Longrightarrow D$ yace en la mediana $ BB_1.$", "Solution_4": "Denotemos $ T_{\\infty}$ el punto del infinito de la recta $ AC,$ el cual es polo de $ KD$ respecto a $ (O).$ $ B$ es el polo de $ MN$ respecto a $ (O)$ y el polo de $ BB_1$ es el punto $ P$ en el rayo $ MN$ tal que $ (M,N,D,P) \\equal{} \\minus{} 1,$ pero como $ (A,C,B_1,T_{\\infty}) \\equal{} \\minus{} 1$ $ \\Longrightarrow$ $ PB \\parallel AC.$ Como los polos $ B,P,T_{\\infty}$ estan alineados, entonces las polares $ MN,BB_1,DK$ concurren $ \\Longrightarrow$ los puntos $ O,D,K$ estan alineados." } { "Tag": [ "Euler" ], "Problem": "Let $n\\neq 2^{k}$ be a positive integer such that $\\phi(n)\\mid n$. Show that $\\frac{n}{\\phi(n)}=3$. (here $\\phi$ is the Euler function.)", "Solution_1": "We can rewrite it ass follows $\\frac{n}{\\phi (n)}=\\frac{p_{1}p_{2}...p_{k}}{(p_{1}-1)(p_{2}-1)...(p_{k}-1)}$,$p_{1}0)$ such that \r\n$ a\\plus{}b\\plus{}c\\equal{}1(a,b,c>0)$\r\nand:\r\n$ \\frac{a^2\\plus{}kb}{b\\plus{}c}\\plus{}\\frac{b^2\\plus{}kc}{c\\plus{}a}\\plus{}\\frac{c^2\\plus{}ka}{a\\plus{}c} \\ge \\frac{3k\\plus{}1}{2}$\r\nb,try with $ k\\equal{}7$", "Solution_1": "Is the denominator in the third fraction supposed to be $ a\\plus{}b$?" } { "Tag": [], "Problem": "Ahaha... Epic fail at square roots. I forget, when you have something like say $ \\sqrt5 \\plus{} 4$ Would I add the inside first THEN find the square root, or the square root of each number and then find then add? And how about multiplication? (E.g $ \\sqrt(4)(4)$) Oh by the way, the parentheses mean multiplication if you didn't know, AND ALSO; I couldn't seem to stretch the square root sign over the whole entire expression; could someone tell me how to do that?\r\nAlso, I forget what you do when you divide an expression by a number for example... $ \\frac{4\\minus{}4x}{4}$. Would it be $ 1\\minus{}x$ or what? Do you divide everything by it or...\r\nThanks; and if you think these are homework problems; they aren't I'm just studying for finals... Epic fail... I have a bad memory! :D", "Solution_1": "hello, here is what i would do\r\n$ \\sqrt{5}\\plus{}4$ at first you must calculate the square root and then you have to add 4, but it is sensless, because you can only calculate an approximate value of this term.\r\n$ \\sqrt{4}\\cdot 4\\equal{}2\\cdot 4\\equal{}8$\r\n$ \\frac{4\\minus{}4x}{4}\\equal{}\\frac{4}{4}\\minus{}\\frac{4x}{4}\\equal{}1\\minus{}x$.\r\nSonnhard.", "Solution_2": "I didn't know how to stretch the square root sign over the whole thing... it wouldn't be the same thing would it?", "Solution_3": "If you have the square root stretched over the whole thing, it's like $ \\sqrt(5\\plus{}4)$, so you solve the thing under the square root and THEN find the square root.", "Solution_4": "Say something is a variable and you don't know what \"$ x$\" is equal to? Say something like; $ \\sqrt(54 \\plus{} 3x)$?\r\nI'm sorry; my mind seems to have imploded this information...", "Solution_5": "Here is how to stretch the square root sign:\r\n\r\n[code]\n$\\sqrt{(4)(4)}$\n[/code]\r\n\r\nYou have to put curly brackets around the place where you want it to go over.\r\n\r\n\r\n[hide=\"EXPONENT LAWS\"]\n\n$ a^n \\times b^n=(a\\times b)^n$ and vice versa.\n\n$ n^a\\times n^b=n^{(a+b)}$ and vice versa.\n\n$ \\frac{a^n}{b^n}=(\\frac{a}{b})^n$ and vice versa.\n\n$ \\frac{n^a}{n^b}=n^{a-b}$ and vice versa.\n\n$ (n^a)^b=n^{a\\times b}$ and vice versa.\n\nThere is no laws concerning adding or subtraction.\n[/hide]\n\n[hide=\"YOUR QUESTIONS\"]\n\nYou can rewrite $ \\sqrt{n}$ as $ n^{\\frac{1}{2}}$. \n\nExample: You asked what is $ \\sqrt{4\\times4}$. You can do this two ways:\n\nWay 1: $ \\sqrt{4\\times4}=\\sqrt{16}=\\boxed{4}$\n\nWay 2: $ \\sqrt{4\\times4}=\\sqrt{4]\\times\\sqrt{4}=2\\times2=\\boxed{4}}$\n\nFor your last question ($ \\sqrt{54+3x}$), you can't simplify it further. If it is part of an equation, you have to get it onto its own side and then square both sides.\n[/hide]\r\n\r\nHope that helps!", "Solution_6": "Ah... That makes much more sense. Thanks! ^________________^\r\nSo what was the exponent laws about? \r\nAlso, considering the fact that I have spontaneously become idiotic and stupid; would I do the same thing for the division thing?", "Solution_7": "Exponent laws was explained because essentially, square roots ARE exponents because $ \\sqrt{n}$ can be written as $ n^{\\frac{1}{2}}$.\r\n\r\nAnd yes, it's the same for division because division is merely multiplication of the reciprocal. So $ \\sqrt{\\frac{n}{m}} \\equal{} \\sqrt{n * \\frac{1}{m}}$", "Solution_8": "The exponent laws is for anything else you have to ask. This is because square roots ARE EXPONENTS. Use the exponent laws for any more questions, such as the division thing. If you don't understand anything, just ask.\r\n\r\nEDIT: I guess two posts were posted at the same time.", "Solution_9": "[quote=\"joyofpi\"] I forget, when you have something like say $ \\sqrt5 + 4$ Would I add the inside first THEN find the square root, or the square root of each number and then find then add? [/quote]\r\n\r\nThe $ \\sqrt$ symbol counts as a grouping symbol, so follow GEMDAS (grouping symbols first) and add $ 4+5$ first, then take the Square root." } { "Tag": [ "Putnam", "induction", "inequalities", "percent", "college contests", "Putnam easy" ], "Problem": "Basketball star Shanille O'Keal's team statistician keeps track of the number, $S(N),$ of successful free throws she has made in her first $N$ attempts of the season. Early in the season, $S(N)$ was less than 80% of $N,$ but by the end of the season, $S(N)$ was more than 80% of $N.$ Was there necessarily a moment in between when $S(N)$ was exactly 80% of $N$?", "Solution_1": "Yes. At some point for say, N-1 throws S(N-1)<4/5*(N-1) and the next throw (being necessarily succesful) gives us S(N)>=4/5*N. Since the Nth throw was succesful S(N-1)=S(N)-1.\r\n\r\n4/5*(N-1)>S(N)-1>=4/5*N-1\r\n\r\n=> 4N-4>5S(N)-5>=4N-5\r\n\r\nsubtract 4N and add 5 to everything\r\n=> 1>5S(N)-4N>=0\r\n\r\nSince 5S(N)-4N is an integer the only possibility left is that \r\n\r\n 5S(N)-4N = 0 so therefore S(N)=4/5*N=.8N\r\n\r\n\r\nIf someone wants to TeX this answer go ahead.", "Solution_2": "[u][b]The author of this posting is : blahblahblah[/b][/u]\r\n____________________________________________________________________\r\n\r\n[quote=\"Kent Merryfield\"]Basketball star Shanille O'Keal's team statistician keeps track of the number, $S(N),$ of successful free throws she has made in her first $N$ attempts of the season. Early in the season, $S(N)$ was less than 80% of $N,$ but by the end of the season, $S(N)$ was more than 80% of $N.$ Was there necessarily a moment in between when $S(N)$ was exactly 80% of $N$?[/quote]\r\n\r\nYes.\r\n\r\nLet $P\\left(n\\right) = \\frac{S\\left(n\\right)}{n}$.\r\n\r\nSay that over a certain interval, we have $P\\left(n\\right) <0.8$. If there is a multiple of $5$ in this interval, we must have $P\\left(5i\\right) \\leq 0.8 - \\frac{1}{5i}$. Thus we must have $P\\left(5i+j\\right) <0.8$ for $1\\leq i\\leq 4$.\r\nBy induction, we can establish a similar result for $P\\left(5i+5\\right)$ and conclude that we can only have $P\\left(n\\right) \\geq 0.8$ if $n=5j$, and the best we can do is equality, so the answer is yes.", "Solution_3": "[u][b]The author of this posting is : Jrthedawg[/b][/u]\r\n____________________________________________________________________\r\n\r\nAnother solution:\r\n\r\nSorry, I don't know how to use that fancy font. [color=red][Moderator edit: This \"fancy font\" is probably LaTeX... :D ][/color]\r\n\r\nWe prove by contradiciton: Assume there exists some N that satisfies\r\n\r\nS(N)/N<4/5<(S(N)+1)/(N+1).\r\n\r\nThis statement basically says that there exists some N for which O'Keal's percentage is below 80%, and with one more make, her percentage jumps above 80%.\r\n\r\nThis leads to the inequality 4N-1<5S(N)<4N.\r\n\r\nWe know 5S(N) is an integer, and no integer exists between 4N-1 and 4N, thus, a contradiction, and our original assumption is false. If no such N satisfies our original inequality, then there is no way O'Keal could have progressed past 80 percent without being at exactly 80 percent.", "Solution_4": "[u][b]The author of this posting is : Ravi B[/b][/u]\r\n____________________________________________________________________\r\n\r\nAnother way to say it. Consider the quantity 5S(N) - 4N. It is an integer. Early on it is negative and later it is positive. After each throw, the quantity changes by either +1 or -4. Thus it is clear the quantity must hit 0 at some point in between. To be more precise, consider the first time (in between) that the quantity is nonnegative. The moment before it must have been negative, that is at most -1. Because jumps are at most +1, at the current time it must be at most -1 + 1 = 0. The only nonnegative number at most 0 is 0 itself, so at the current time the quantity is 0.", "Solution_5": "someone said something about this working for all percentages of the form 1-1/n, $n\\geq 2$\r\n\r\ni proved it for .9, i think (i can't remember but i'm pretty sure i scrawled it in my notebook).\r\n\r\nit would be cool to see a general proof, maybe i will do it tommorrow as it is 2 am here.", "Solution_6": "For luther_driggers:\r\n\r\nSuppose the target percentage is $p=1-\\frac1k,k\\ge2.$ (In the problem as stated, $k=5.$) $S(N)pN$.\r\n\r\nConsider the quantity $Q(N)=kS(N)-(k-1)N.$ The quantity $Q(N)$ is an integer. With each made free throw, $Q(N)$ increases by 1; with each missed free throw, $Q(N)$ decreases by $k-1.$ Since $Q(N)$ integer valued and it starts out negative and becomes positive, and since it increases by 1 each time it increases, then it must equal 0 at some point.\r\n\r\nNote that this is exactly Ravi B's proof.\r\n\r\nIt doesn't work if the target percentage has a different form. Suppose we ask if the percentage must ever be 60%. It doesn't. Suppose she misses the first one, then makes the next two. The percentages go 0$, 50%, 67% and never equal 60%.", "Solution_7": "sorry kerry i couldn't see your code, u must have an xtra dollar sign or whatever :) \r\n\r\ni was just gonna prove it the way i did the first one.\r\n\r\n[hide=\"I like hidden proofs\"]\n\n$p=\\frac{k-1}{k},k\\geq 2$.\n\n$S(N-1)< \\frac{k-1}{k} (N-1)$\n\n$S(N)-1=S(N-1)$\n\n$S(N)\\geq \\frac{k-1}{k}N$\n\ncombining and multiplying by k\n\n$(k-1)(N-1)>kS(N)-k \\geq (k-1)N -k$\n\nsubtract $(k-1)N$ and add $k$\n\n$1>kS(N)-(k-1)N \\geq 0$\n\nwhich is an integer so\n\n$kS(N)-(k-1)N=0$\n\nand \n\n$S(N)= \\frac{k-1}{k}N$\n\n[/hide]\r\n\r\nit would be cool to c the fixed version of yours thou.", "Solution_8": "Sorry, Luther. Fixed now.", "Solution_9": "Wow, that was easy.\r\n\r\nThe only Putnam question I'll probably ever get.", "Solution_10": "[quote=CrazyBeast2.0]Hey I am no expert in college math but isn't the graph of S(N) continuous. If it is, then definitely when it goes from less than 80% to more than 80% at one point it must be 80%.[/quote]\n\nNot really, since $S$ is defined over $\\mathbb{N}$, not over $\\mathbb{R}$. In fact, if we change $80\\%$ to $60\\%$ then the answer is negative (misses the fiirst one, scores the second and third, and never shoots again).", "Solution_11": "Assume the contrary. It follows that there's a $n$ such that $S(n)<80\\%$ and $S(n+1)<80\\%.$ So $$S(n)/n<80\\%80\\%$. Suppose that this occurs at $N=n$. Then, $$\\frac{S(n)}{n}<\\frac{4}{5} \\implies 5S(n)<4n, \\quad \\frac{S(n+1)}{n+1}>\\frac{4}{5} \\implies 5S(n+1)>4n+4.$$Since $\\tfrac{S(n+1)}{n+1}>\\tfrac{S(n)}{n}$, it follows that $S(n+1)>S(n)$, which is only true if the $(n+1)$th free throw was successful. Thus, $S(n+1)=S(n)+1$.\n\nThen, $5(S(n)+1)>4n+4 \\implies 5S(n)+1>4n$. We also have $5S(n)<4n$, so $$5S(n)<4n<5S(n)+1.$$ However, $5S(n)$ and $5S(n+1)$ are consecutive integers, and no integers can exist strictly between two consecutive integers, a contradiction.\n\nTo-do: Add in the vice versa case.", "Solution_14": "As mentioned by the folks above, the easiest way to go about it is contradiction. Suppose right before going from below an $80$ win streak to a above an $80$ win streak, we have $a$ wins and $b$ losses. So clearly $a < 4b$. Now after this game, we will have $a+1$ wins and $b$ losses with us purportedly obeying $a+1 > 4b \\implies a > 4b-1$. So $4b-1 < a < 4b$. But $a$ is an integer, contradiction. ", "Solution_15": "This is true, so let's give a proof my contradiction. We can test cases to see that we can't have $S(N)>0.8N$ for $N\\leq 5$, meaning at some $N$, we have $S(5N)=4N-\\epsilon_1$ and $S(5(N+1))=4(N+1)+\\epsilon_2$, with $\\epsilon_1, \\epsilon_2 > 0$, giving $4+\\epsilon_1+\\epsilon_2\\geq 6$. However, we arrive at\n\\begin{align*}\n &4 +\\epsilon_1 + \\epsilon_2\\\\\n &= 4(N+1)+\\epsilon_2 - (4N-\\epsilon_1)\\\\\n &= S(5(N+1))-S(5N)\\\\\n &\\leq 5(N+1)-5N = 5,\n\\end{align*}\ncontradiction.", "Solution_16": "The answer is yes.\n\t\n\tNotice that at $S(N)$ can only increase when Yang Liu scores, and so there has to be a point $N'$ before $S(N)\\ge 80\\%$ from which on Yang Liu keeps on scoring consecutively until $S(N)>80\\%$. Say that $S(N')=\\frac{a}{b}$, and let $\\frac{4}{5}=\\frac{4(b-a)}{5(b-a)}$, where we note that $5(b-a)>b$ as $\\frac{4}{5}>\\frac ab\\iff 4b>5a$. We note that $S(N'+m)=\\frac{a+m}{b+m}$ now, and we see that the difference between denominator and numerator stays constant. Now clearly, $S(N'+m)$ runs through all fractions with difference between denominator and numerator $b-a$ such that the denominator is bigger than $b$. Eventually, $b+m'=5(b-a)$. But now $b+m'-a+m'=b-a$ and $5(b-a)-4(b-a)=b-a$, so $S(N'+m)=\\frac{4(b-a)}{5(b-a)}=\\frac45$, as desired.", "Solution_17": "[hide]\nThe answer is $\\boxed{\\text{Yes}}$; otherwise, there are $a,b\\in\\mathbb{Z}$ so that $\\tfrac{a}{b}<\\tfrac{4}{5}$ and $\\tfrac{a+1}{b+1}>\\tfrac{4}{5}$, which by cross-multiplying we find is absurd. \\qed", "Solution_18": "[quote=OTIS Version]\nBasketball star Yang Liu\u2019s team statistician keeps track of the number, $S(N)$, of successful free throws he has made in his first $N$ attempts of the season. Early in the season, $S(N)$ was less than $80\\%$ of $S(N)$, but by the end of the season, $S(N)$ was more than $80\\%$ of $N$. Was there necessarily a moment in between when $S(N)$ was exactly $80\\%$ of $N$?\n[/quote]\n\nThe answer is $\\boxed{\\text{yes}}$. For the sake of contradiction, assume that there was never necessarily a moment when $S(N)$ was $80\\%$ of $N$. \n\nSuppose that in the moment right as the percent of successful free throws jumps from below $80\\%$ to above $80\\%$, Yang Liu had made $a$ shots and missed $b$ shots. It is clear that $\\frac{a}{b}<4 \\implies a<4b$. After making a shot, we must have $\\frac{a+1}{b}>4 \\implies a>4b-1$. Thus,\n\n\\[4b-12 +y 2 =r 2 :( \r\n\r\n[b]So can anyone please show me if there is a link between the graph of a function and the graph of its antiderivative concerning the area underneath of a curve or something else?[/b]", "Solution_1": "maybe I don't understand, but the relationship between int (f) and f is the same as the relationship between f and f'.", "Solution_2": "I understand that clearly but what i'm talking about is: when finding the area of under a curve in [a, b], you find the antiderivative of the function and substitute a and b in F(b)-F(a), though i accept the theorem, looking at the the graphs( f and F) it makes no sense to me why i substitute a and b in the antiderivative :( . \r\n\r\n[b]But can anyone do a graphical analysis for me to show why the antiderivative is used to find the area. Maybe i'll find my way out of what seems to be a labyrinth. \n[/b]\r\nPS:I'm not good at asking question so if it doesn't make sense tell me and i'll try to rephrase what i'm trying to ask", "Solution_3": "For a semicircle with radius $r$, the equation is $y=\\sqrt{r^2-x^2}$.\r\n$\\int \\sqrt{r^2-x^2} dx = \\frac{\\sin^{-1}{\\left(\\frac{x}{r}\\right)}r^2}{2} + \\frac{x\\sqrt{r^2-x^2}}{2}+C$.\r\nGraph both of these and you will see the relationship.\r\n\r\nFor a simpler example look at the graph od $y=x$, which is the derivitive of $y=\\frac{1}{2}x^2$, so the integral of a parabola is a line.\r\n\r\nThe reason we use the integral to find the area under a curve is because each strip we take has a with $dx$, a height $f(x)$, and an area $dA$, and the integral undoes the derivative, $dA = f(x)dx \\Rightarrow A=\\int_a^{b} f(x) dx$. This is equivilent to taking infinit infinitely thin strips from $a$ to $b$ and adding their areas. This concept of infinitely thin is similar to the definition of a derivative $f'(x) = \\lim_{h\\to 0}{\\frac{f(x+h)-f(x)}{h}}$.", "Solution_4": "Thanx Jimmy, i was working with it using y=x and i came up to that same conclusion. I was stupid enough :lol: trying to use a semicircle to analyze it. I completely overlooked the simple function f(x)=x thinking that it would be easier for me using a semicircle since i already know how to find its area using simple geometry.", "Solution_5": "also PLEASE DON'T WRITE YOUR TITLES IN CAPITAL LETTERS ;) thank you", "Solution_6": "[quote=\"Peter VDD\"]also PLEASE DON'T WRITE YOUR TITLES IN CAPITAL LETTERS ;) thank you[/quote]\r\n\r\nno problem" } { "Tag": [ "probability" ], "Problem": "At a certain intersection, $ 30\\%$ of the cars turn left, $ 40\\%$ turn right, and $ 30\\%$ go straight. What is the probability that the next car turns left and the following one turns right? Express your answer as a common fraction.", "Solution_1": "$ \\frac3{10}\\cdot\\frac25\\equal{}\\frac{3}{25}$", "Solution_2": "uh to clarify nike's answer...\r\n\r\n$ 30\\%\\equal{}\\frac{3}{10}$ so the probability the first car turns left is $ \\frac{3}{10}$.\r\n$ 40\\%\\equal{}\\frac{2}{5}$ so the probability the second car turns right is $ \\frac{2}{5}$.\r\n\r\nSince these are independent events, meaning one choice doesn't affect the other, the probabilities are multiplied.\r\n\r\n$ \\frac{3}{10}\\times\\frac{2}{5}\\equal{}\\boxed{\\frac{3}{25}}$ which is the same answer nike got :)" } { "Tag": [ "algebra", "polynomial", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Prove that we can't find two polynomials $f,g \\in \\mathbb C[X]$ such that $\\frac{f(n)}{g(n)}= 1+\\frac1{1!}+\\frac1{2!}+\\ldots+\\frac1{n!}$ for all integers $n \\geq 1$.", "Solution_1": "Assume that such polynomials can be found. Then we have that $degree (f)=degree (g)=k$. \r\n\r\nNow $\\frac{f(n+1)}{g(n+1)}-\\frac{f(n)}{g(n)}=\\frac{1}{(n+1)!}$.\r\nHence there is $c_{n}\\in (n, n+1)$ such that \r\n\r\n$\\frac{f'g-fg'}{g^{2}}(c_{n})=\\frac{1}{(n+1)!}$. \r\n\r\nNow let $degree (f'g-fg')=a\\leq k-1$ and let $b=k^{2}-a$. Then observe that we have \r\n\r\n${(c^{b}_{n})\\frac{f'g-fg'}{g^{2}}(c_{n})\\frac{(n+1)!}{c^{b}_{n}}}=1$.\r\nNow observe that \r\n\r\n$\\lim (c_{n})^{b}\\frac{f'g-fg'}{g^{2}}(c_{n})$ is finite and nonzero. The limit is zero provided that $f'g=fg'$ but this is excluded. \r\n\r\nHence we get that $\\lim \\frac{(n+1)!}{(c_{n})^{b}}$ is finite which is definitely false since $c_{n}\\approx n$.", "Solution_2": "$lim\\frac{f(n)}{g(n)}=e$ hence $f(x)=ex^{k}+\\dots$ and $g(x)=x^{k}+\\dots$. Moreover $e-\\frac{f(n)}{g(n)}\\leq\\frac{1}{n!}$; it's easy to show that $e-\\frac{f(n)}{g(n)}\\sim\\frac{a}{n^{r}}$, a contradiction." } { "Tag": [ "inequalities" ], "Problem": "Give x, y, z >0 and xyz=1. Prove\r\nSorry ... wait ...", "Solution_1": "[hide=\"the sign is flipped\"]\nThe correct form of it is:\n$ x^{2}y+y^{2}z+z^{2}x\\leq x^{2}y^{4}+y^{2}z^{4}+z^{2}x^{4}$ $ \\leftrightarrow$ $ xyz(x^{2}y+y^{2}z+z^{2}x)\\leq x^{2}y^{4}+y^{2}z^{4}+z^{2}x^{4}$ \nIf I denote $ xy^{2}=a$, $ yz^{2}=b$ and $ zx^{2}=c$ the inequality is equivalent to: $ ab+bc+ca\\leq a^{2}+b^{2}+c^{2}$\n[/hide]" } { "Tag": [ "videos", "counting", "distinguishability", "number theory" ], "Problem": "How do you learn to problem solve?\r\n\r\nWhen I practice math contest problems, there is a lot of running into dead ends, hair pulling, head banging, punching my chair in frustration, walking around in circles telling myself \"I don't know how to do this,\" getting angry when the problem required a trick that would have been very difficult for me to derive myself, ripping my paper up then reprinting it to try the problems again, and laying my head down on the desk in submission during hard times. \r\n\r\nJust wondering... that's how its supposed to be, right? That is, is that the basic method in learning how to problem solve in general? It isn't linear, like... section 6, now we know how to factor integers, section 7, now we know how to do long division... right?", "Solution_1": "[quote]Just wondering... that's how its supposed to be, right?[/quote]\r\n\r\nit was and it still is the same with me ... but i don't have any ilusion that i am able to solve every problem (if i havn't heard for anything simular before)... that's basicly why i like solving problems ... because i always learn something new...(and that's why i dislike homework because it doesn't teach you anything new)", "Solution_2": "aaaaaaaaaaaaaaaaa", "Solution_3": ":rotfl: Welcome to my universe.\r\nIf anyone ever tells you that they know how to solve every problem, they're either highly delusional, egotistical/kallikakian/narcissistic, or they've only been exposed to problems that are far too easy for them. Something of the sort.\r\n\r\nThis 'phase' of learning how to problem solve will never end. You will always be frustrated. You will always be triumphant. You will always be amused and amazed. You will always be bewildered and lost. You will always be found and rediscovered.\r\nThat's part of the great beauty of it. :) It never ends.\r\n\r\nJust learn your stuff. Learn the concepts, derive the formulas, grasp the concepts, and smile while you do it. Be tortured and enjoy it. \r\n\r\n\r\n...However, if you need a self-esteem boost, you could always do easy problems. :P", "Solution_4": "Haha... nice question. \r\n\r\nI won't say that I'm an expert in problem solving (because I am absolutely not) but I see where you come from. When I try problems I spend long fruitless hours on them, racking my brain, and I HATE looking at the solutions before I'm finished. It just goes against everything I am and do to look at the answers, and I sympathize with you. But I keep coming back again and again, just because it's so fun to finally say, \"Yes! I did it!\" and have that feeling of accomplishment. \r\n\r\nIt's good that you're still smashing your head against problems like that because it shows your dedication and how much you're trying. I sometimes wish I had that kind of dedication and spirit... so many times I look back and see how much time I have wasted that could have been used to do math... aieeeee....\r\n\r\nSometimes I wish I could be nearly as good as all of you posters seem. Because I'm really really intimidated with the level of math that you are at, and I wish I could be at least near that good. One of my friends is Kevin Li (budi713), and looking at his level of accomplishments makes me look up to him in awe. I just wish I could figure out what makes him so good. \r\n\r\n--\r\n\r\nAnd then, what comes after problem solving? It's great to see all these \"fun\" problems with clever answers, but unfortunately real life is a different story. Equations don't always solve with integers in real life, and the world of applied mathematics is unfortunately a lot tougher. \r\n\r\nSo what do we do? We don't weep, we don't cry, we keep plugging away and try to learn the talents used later in life. We try, fail, try and try again, and get to do things with the tools we use. It's all great. \r\n\r\nI'm just glad to see so many people that feel the way I do.", "Solution_5": "Many people (including me, and i'm not from USA) have learnt how to solve problems by solving competition problems...AoPS is another way ... i found it in december last year and i thought 'oh another site that has high level books for IMO' but when i've found the experts form Intro books i found out that i basicly know almost everything form intro books...so i realy liked that this was 'all levels' site. and then i found this forum and I love it ... probably because there are so many people here that know so much so you can talk about interesting problems. and i think the most inportant thing is to try ... cids from my school will probably never know this things because whenever they see a more difficult problem they look at it for 5 min find it boring and just to difficult and move on...", "Solution_6": "I like to take a short break when I don't understand something. This could mean playing some video games, taking a nap, or going on AoPS. It usually relaxes the mind from it's state of failure. \r\n\r\nAlso, reviewing past material might help. I was looking at the Number Theory portion of AoPS vol 2 and I a bit unsure of the material. I looked back in AoPS vol 1 and reviewed; it helped me immensely in understanding that chapter (as well as further on in the Diophantine Equations). \r\n\r\nMoved to Other Problem Solving Topics.", "Solution_7": "There are, in fact, books that discuss problem-solving technique. A short list:\r\n\r\n[url=http://www.amazon.com/How-Solve-Aspect-Mathematical-Method/dp/0691023565]Polya[/url] (Problem-solving in general: probably what you're looking for)\r\n[url=http://www.amazon.com/Problem-Solving-Strategies-Problem-Books-Mathematics/dp/0387982191]Engel[/url] (Olympiad training)\r\n[url=http://www.amazon.com/Art-Craft-Problem-Solving/dp/0471789011/ref=pd_sim_b_title_5]Zeitz[/url] (Contest training)\r\n\r\nOne of the most useful things is to do lots of problems, but that's just so you can become familiar with common strategies. What you're probably looking for is ways to figure those strategies out if you didn't know them already, which is much more difficult to teach. I don't actually know if the creative aspect of problem-solving can be taught.\r\n\r\nPart of your problem, specifically, might just be that you're letting yourself get frustrated too easily. Don't just use one method and be disappointed when it fails; try as many plausible avenues as you can, and develop the ones that look promising. Like the previous poster said, take short breaks, or even long breaks. Your brain works on problems on a subconscious level; returning to a problem after a break might give you the bright idea you need.", "Solution_8": "[quote=\"t0rajir0u\"]There are, in fact, books that discuss problem-solving technique. A short list:\n\n[url=http://www.amazon.com/How-Solve-Aspect-Mathematical-Method/dp/0691023565]Polya[/url] (Problem-solving in general: probably what you're looking for)\n[url=http://www.amazon.com/Problem-Solving-Strategies-Problem-Books-Mathematics/dp/0387982191]Engel[/url] (Olympiad training)\n[url=http://www.amazon.com/Art-Craft-Problem-Solving/dp/0471789011/ref=pd_sim_b_title_5]Zeitz[/url] (Contest training)\n\nOne of the most useful things is to do lots of problems, but that's just so you can become familiar with common strategies. What you're probably looking for is ways to figure those strategies out if you didn't know them already, which is much more difficult to teach. I don't actually know if the creative aspect of problem-solving can be taught.\n\nPart of your problem, specifically, might just be that you're letting yourself get frustrated too easily. Don't just use one method and be disappointed when it fails; try as many plausible avenues as you can, and develop the ones that look promising. Like the previous poster said, take short breaks, or even long breaks. Your brain works on problems on a subconscious level; returning to a problem after a break might give you the bright idea you need.[/quote]\r\n\r\nTo follow up on that advice, Zeitz is the best book for Problem Solvers on the AoPS level (and it follows a similar regime to what t0rajir0u was talking about).", "Solution_9": "With all due respect, y'all are kinda math maniacs who would spend all day on math problems if your schedule allows it.\r\n\r\nI probably shouldn't be speaking since I more or less live in the shadow of people like y'all, yet I must give credit to user aufha for bringing up this very interesting topic.\r\n\r\nThanks for the awesome book recommendations from the others as well, which were both practical and extremely helpful.\r\n\r\nHowever, I would strongly advise against extremist comments such as the one posted by user zhh:\r\n\r\n[quote]When I try problems I spend long fruitless hours on them, racking my brain, and I HATE looking at the solutions before I'm finished. It just goes against everything I am and do to look at the answers, and I sympathize with you. But I keep coming back again and again, just because it's so fun to finally say, \"Yes! I did it!\" and have that feeling of accomplishment. \n\nIt's good that you're still smashing your head against problems like that because it shows your dedication and how much you're trying. I sometimes wish I had that kind of dedication and spirit... so many times I look back and see how much time I have wasted that could have been used to do math... aieeeee.... \n[/quote]\r\n\r\nA crazy 14-year-old who applauds you for smashing your head in frustration, such as user zhh, poses a threat not only to himself but others as well, whether you know him on a personal level or across your computer screen......lol\r\n\r\nI personally get a lot of my solutions to math problems when I'm half-heartedly fingering through a Chopin or Richard Clayderman piece on my piano (or singing indisinguishable tunes to myself in the showers :p).\r\n\r\nI would say instead of hurting yourself the way user zhh does, if you find that a math solution comes to you whenever Randy Moss catches a Tom Brady TD, watch all the Patriots games you can find!", "Solution_10": "[quote=\"aufha\"]How do you learn to problem solve?\n\nWhen I practice math contest problems, there is a lot of running into dead ends, hair pulling, head banging, punching my chair in frustration, walking around in circles telling myself \"I don't know how to do this,\" getting angry when the problem required a trick that would have been very difficult for me to derive myself, ripping my paper up then reprinting it to try the problems again, and laying my head down on the desk in submission during hard times. \n\nJust wondering... that's how its supposed to be, right? That is, is that the basic method in learning how to problem solve in general? It isn't linear, like... section 6, now we know how to factor integers, section 7, now we know how to do long division... right?[/quote]\r\n\r\ni have the same problem", "Solution_11": "I [i]would[/i] spend close to all day on math every now and then if i had the time. Math is fun, however lately i've been kinda out of it- not sure why. Dedication is important to developing problem solving skills though there is a fine line between dedication and obsession.", "Solution_12": "well i am sorry for replying to this topic after such a long time.\r\nbut now only i came to know about this. This is an interesting question.Though i am not good in maths myself i would like to give some advice.Well the previous posters speak about problem solving and especially some of them speak about learning concepts from books.\r\ni am not telling that is wrong but i would like to do books which dont spoon feed and state some conjectures which are left as an exercise.This is because more than the theorem or concept the idea or the base on which the theorem was constructed is very important.This might help you when to use them and apply them correctly.When a constructive problem is given ,the student should not just solve the problem ,but should also look for spaces where the dynamics and boundary of the problem can be extended.By doing this the student himself might get some new concept.Ultimately given a problem it is for the student to explore the theory part of it by analyzing and working on it.It is not about doing x books and y problems ,even if one problem is done the student should analyze its beauty and should get benefited from it and you will never forget the theorem. This is what a great maths thinker told me.I would like to implement it more vigorously and i feel a student should do books which provide the above quailities" } { "Tag": [ "conics", "ellipse", "analytic geometry", "graphing lines", "slope", "function", "geometry unsolved" ], "Problem": "Let the equation of an ellipse be $\\frac{x^{2}}{4}+\\frac{y^{2}}{12}=1$ .\r\n$A(1,3),B,C$ are three points located on the ellipse such that $ABC$ is isosceles triangle with $AB=AC$ . Find the slope of $BC$ \r\n\r\n\r\np/s : I have posted this question in pre olympiad months ago but remain unsolved,so I repost again here .", "Solution_1": "[quote=\"shyong\"]Let the equation of an ellipse be $\\frac{x^{2}}{4}+\\frac{y^{2}}{12}=1$ .\n$A(1,3),B,C$ are three points located on the ellipse such that $ABC$ is isosceles triangle with $AB=AC$ . Find the slope of $BC$ \n\n\np/s : I have posted this question in pre olympiad months ago but remain unsolved,so I repost again here .[/quote]\r\n\r\nThe conditions specified do not make $\\triangle ABC$ unique. This is easy to see because for a circle of radius $0 < r < 6.557570355\\ldots$ centered at $(1,3)$, there are exactly 2 points of intersection with the given ellipse. These points of intersection $B$ and $C$, along with $A$, form the vertices of an isosceles triangle with $AB = AC$ by construction. However, it should be obvious that the slope of $BC$ is not independent of $r$, and hence there is no unique answer.", "Solution_2": "Following from the previous post, we can see that different values of r will result in a set of parallel chords BC.\r\nTheorem.The midpoints of a system of parallel chords of gradient m of an ellipse ${\\frac{x^{2}}_{a^{2}}}$+${\\frac{y^{2}}_{b^{2}}}$ =1 lie on the line y=$\\frac{-b^{2}x}_{a^{2}m}$(such a line is called a diameter).\r\nIn this particular problem the line joining A to the origin is a diameter for the system of chords BC.", "Solution_3": "Then prove that the locus of segment BC as a function of the congruent legs of triangle ABC are parallel lines. You can't, because they're NOT, as I pointed out. When r = AB = AC is small, BC approximates the tangent at A. This is obviously not the case as r increases!\r\n\r\nIf you're not convinced, solve the system\r\n\r\n$\\frac{x^{2}}{4}+\\frac{y^{2}}{12}= 1$\r\n$(x-1)^{2}+(y-3)^{2}= r^{2}$.\r\n\r\nThen compute the slope of the intersection points as a function of r, and you will immediately find that it is a variable function of r, and not at all a simple one at that." } { "Tag": [ "function", "algebra", "polynomial", "ceiling function" ], "Problem": "There exists a function so that f(0)=12, and f(x)=x for x=1-11 (integers).\r\nFind the simplest function possible, even though there are polynomials, I'd like a function using the floor/ceiling) function (perhaps dividing by 12 and taking the ceiling?).", "Solution_1": "Interesting question... but this is probably not what you're looking for:\r\n[hide=\"an answer\"] $ \\left\\lceil\\frac{x}{12}\\right\\rceil (x\\minus{}12)\\plus{}12$[/hide]", "Solution_2": "Also...\r\n[hide]\n$ 12+(x-12)\\lceil\\frac{x}{1+x}\\lceil$\n[/hide]\r\n\r\nThat would be better, as it doesn't crash after 12.\r\n\r\n@archimedes: you might want to put the denominator as 1+x.", "Solution_3": "Although putting $ x\\plus{}k$ in the denominator woud make the function not \"crash\" after $ 12$, putting $ x\\plus{}k$ in the denominator would also make $ x\\equal{}\\minus{}k$ undefined.... \r\nHow should we define \"simplest function possible\"?", "Solution_4": "You can also use fractional part function: $ f(x) \\equal{} 12\\left\\{{x\\minus{}1\\over 12}\\right\\}\\plus{}1$ (which is periodic)\r\n\r\nOr you can use absolute values: $ f(x)\\equal{}6|x\\minus{}1|\\minus{}5x\\plus{}6$ (which extends as lines over both sides)", "Solution_5": "[hide=\"The simplest function possible\"] $ \\begin{equation*}f(x) =\\begin{cases}12 &\\text{if }x=0,\\\\ x &\\text{if }x\\in\\{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\\}.\\\\ \\end{cases}\\end{equation*}$ [/hide]\r\n;)\r\n\r\nPerhaps you should be more specific in what you mean by \"function\" even before you specify what you mean by \"simplest.\" The above is a function, after all.", "Solution_6": "[quote=\"t0rajir0u\"][hide=\"The simplest function possible\"] $ \\begin{equation*}f(x) =\\begin{cases}12 &\\text{if }x = 0,\\\\ x &\\text{if }x\\in\\{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\\}.\\\\ \\end{cases}\\end{equation*}$ [/hide]\n;)\n\nPerhaps you should be more specific in what you mean by \"function\" even before you specify what you mean by \"simplest.\" The above is a function, after all.[/quote]\r\n\r\n\"Simplest\" in the sense \"which requires the least amount of time to think of it\" ;)\r\n\r\nI guess he meant \"simplest formula\", since he mentions polynomials, floor, ceiling etc.", "Solution_7": "[quote=\"t0rajir0u\"][hide=\"The simplest function possible\"] $ \\begin{equation*}f(x) =\\begin{cases}12 &\\text{if }x = 0,\\\\ x &\\text{if }x\\in\\{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\\}.\\\\ \\end{cases}\\end{equation*}$ [/hide]\n;)\n\nPerhaps you should be more specific in what you mean by \"function\" even before you specify what you mean by \"simplest.\" The above is a function, after all.[/quote]\r\nI'm trying to avoid a piecewise function.\r\nArchimedes' works for what I need, thanks.", "Solution_8": "What is it that you need, anyway? If you'd been more specific from the start, we wouldn't need to guess what function you would consider \"simplest\" :|" } { "Tag": [ "function", "inequalities", "calculus", "derivative", "number theory", "inequalities proposed" ], "Problem": "Suppose $ a,b,c$ are real numbers, with $ a\\plus{}b\\plus{}c\\equal{}3$. Prove that $ \\frac{1}{5a^2\\minus{}4a\\plus{}11}\\plus{}\\frac{1}{5b^2\\minus{}4b\\plus{}11}\\plus{}\\frac{1}{5c^2\\minus{}4c\\plus{}11} \\le \\frac{1}{4}$.", "Solution_1": "[quote=\"Raja Oktovin\"]Suppose $ a,b,c$ are real numbers, with $ a \\plus{} b \\plus{} c \\equal{} 3$. Prove that $ \\frac {1}{5a^2 \\minus{} 4a \\plus{} 11} \\plus{} \\frac {1}{5b^2 \\minus{} 4b \\plus{} 11} \\plus{} \\frac {1}{5c^2 \\minus{} 4c \\plus{} 11} \\le \\frac {1}{4}$.[/quote]\r\nLet $ f(t) \\equal{} \\frac {1}{5t^2 \\minus{} 4t \\plus{} 11}$ we have $ f\"(t) \\le\\ 0 \\rightarrow f(t)$ is inconvex fuction\r\n\r\nFollow Jensen : $ f(a) \\plus{} f(b) \\plus{} f(c) \\le\\ 3 f(\\frac {a \\plus{} b \\plus{} c}{3}) \\equal{} \\frac {1}{4}$\r\n\r\nWe have done!", "Solution_2": "[quote=\"onlylove_math\"][\nLet $ f(t) \\equal{} \\frac {1}{5t^2 \\minus{} 4t \\plus{} 11}$ we have $ f\"(t) \\le\\ 0 \\rightarrow f(t)$ is inconvex fuction\n![/quote]\r\nplease, I don't think so", "Solution_3": "[quote=\"lenhamquy\"][quote=\"onlylove_math\"][\nLet $ f(t) \\equal{} \\frac {1}{5t^2 \\minus{} 4t \\plus{} 11}$ we have $ f\"(t) \\le\\ 0 \\rightarrow f(t)$ is inconvex fuction\n![/quote]\nplease, I don't think so[/quote]\r\nBut i think so :lol: \r\n$ f\"(t)\\equal{}\\frac{\\minus{}10t^2\\minus{}120t\\minus{}78}{(5t^2 \\minus{} 4t \\plus{} 11)^3} \\le\\ 0$ :wink:", "Solution_4": "[quote=\"onlylove_math\"]\n$ f\"(t) \\equal{} \\frac { \\minus{} 10t^2 \\minus{} 120t \\minus{} 78}{(5t^2 \\minus{} 4t \\plus{} 11)^3} \\le\\ 0$ :wink:[/quote]\r\nI got $ 150x^2\\minus{}120x\\minus{}78$ instead of $ \\minus{} 10t^2 \\minus{} 120t \\minus{} 78$", "Solution_5": "That's getting pretty funny indeed! Because I found...\r\n$ f''(t) \\equal{} \\frac { 50t^2 \\minus{} 120t \\minus{} 78}{(5t^2 \\minus{} 4t \\plus{} 11)^3}$ :huh:\r\n\r\n....and by checking $ discriminant> 0$ we cannot conclude for all $ t$, $ f''(t) \\leq 0$.", "Solution_6": "lol:\r\nI am pretty sure Jensen is useless here.But I'm afraid that using derivatives is necessary,refering to [b]arqady[/b]'s solution,which he gave almost one year ago...\r\nSearching is a virtue :wink: http://www.mathlinks.ro/viewtopic.php?p=968464#968464", "Solution_7": "[quote=\"Raja Oktovin\"]Suppose $ a,b,c$ are real numbers, with $ a \\plus{} b \\plus{} c \\equal{} 3$. Prove that $ \\frac {1}{5a^2 \\minus{} 4a \\plus{} 11} \\plus{} \\frac {1}{5b^2 \\minus{} 4b \\plus{} 11} \\plus{} \\frac {1}{5c^2 \\minus{} 4c \\plus{} 11} \\le \\frac {1}{4}$.[/quote]\r\nThe $ pqr$ technique helps here. :) \r\nDenote $ q \\equal{} ab \\plus{} bc \\plus{} ca$ and $ r \\equal{} abc$ then\r\n\\[ \\sum {\\left( {5a^2 \\minus{} 4a \\plus{} 11} \\right)\\left( {5b^2 \\minus{} 4b \\plus{} 11} \\right)} \\equal{} \\minus{} 90r \\plus{} 25q^2 \\minus{} 264q \\plus{} 1089\r\n\\]\r\n\r\n\\[ \\prod {\\left( {5a^2 \\minus{} 4a \\plus{} 11} \\right)} \\equal{} 125r^2 \\minus{} \\left( {100q \\plus{} 814} \\right)r \\plus{} 275q^2 \\minus{} 1694q \\plus{} 5324\r\n\\]\r\nHence we have the inequality equalivents to\r\n\\[ f\\left( r \\right) \\equal{} 125r^2 \\minus{} \\left( {100q \\plus{} 454} \\right)r \\plus{} 175q^2 \\minus{} 638q \\plus{} 968 \\ge 0\r\n\\]\r\nActually we have $ f'\\left( r \\right) \\equal{} 250r \\minus{} 100q \\minus{} 454 \\le 0$ and $ r \\le \\frac {{q^2 }}{9}$\r\n\\[ \\Rightarrow f\\left( r \\right) \\ge f\\left( {\\frac {{q^2 }}{9}} \\right) \\equal{} \\frac {{\\left( {q \\minus{} 3} \\right)\\left( {125q^3 \\minus{} 525q^2 \\plus{} 8514q \\minus{} 26136} \\right)}}{{81}} \\ge 0\r\n\\]\r\nWe are done. Equality holds if and only if $ a \\equal{} b \\equal{} c$.\r\n\r\nFinally, for the funtion $ f\\left( t \\right) \\equal{} \\frac{1}{{5t^2 \\minus{} 4t \\plus{} 11}}$ we have $ f''\\left( t \\right) \\equal{} \\frac{{150t^2 \\minus{} 120t \\minus{} 78}}{{\\left( {5t^2 \\minus{} 4t \\plus{} 11} \\right)^3 }}$. :D", "Solution_8": "Hello Honey_S! Akashnil got the same too.\r\nCan you show the elementary steps of obtaining $ f''(t)$? I am still obtaining $ 50t^2$ instead of $ 150t^2$. :maybe:", "Solution_9": "It is just a simple calculation about derivative :) The result's honey is correct $ f''(x)\\equal{}honey$ :D\r\n\r\nBefore transforming the expression into pqr we can prove this one very easily by the following way:\r\n\\[ \\Delta\\equal{}(100q\\plus{}454)^2\\minus{}4\\times 125(175q^2\\minus{}638q\\plus{}968)\\equal{}\\minus{}4(19375q^2\\minus{}102450q\\plus{}69471)<0,\\forall q\\in \\mathbb{R}\\]", "Solution_10": "[quote=\"Sunkern_sunflora\"]Can you show the elementary steps of obtaining $ f''(t)$? [/quote]\r\n[hide]$ \\frac{d(5t^2\\minus{}4t\\plus{}11)^{\\minus{}1}}{dt}\\equal{}\\frac{d((5t^2\\minus{}4t\\plus{}11)^{\\minus{}1}}{d(5t^2\\minus{}4t\\plus{}11)}\\cdot \\frac{d(5t^2\\minus{}4t\\plus{}11)}{dt}$\n\n$ \\equal{}(\\minus{}1)(5t^2\\minus{}4t\\plus{}11)^{\\minus{}2}\\cdot (10t\\minus{}4)\\equal{}(4\\minus{}10t)(5t^2\\minus{}4t\\plus{}11)^{\\minus{}2}$\n\n\n\n$ \\frac{d((4\\minus{}10t)(5t^2\\minus{}4t\\plus{}11)^{\\minus{}2})}{dt}$\n\n$ \\equal{}\\frac{d(4\\minus{}10t)}{dt}\\cdot (5t^2\\minus{}4t\\plus{}11)^{\\minus{}2}\\plus{}\\frac{d(5t^2\\minus{}4t\\plus{}11)^{\\minus{}2}}{dt}\\cdot (4\\minus{}10t)$\n\n$ \\equal{}(\\minus{}10)(5t^2\\minus{}4t\\plus{}11)^{\\minus{}2}\\plus{}\\frac{d(5t^2\\minus{}4t\\plus{}11)^{\\minus{}2}}{d(5t^2\\minus{}4t\\plus{}11)}\\cdot \\frac{d(5t^2\\minus{}4t\\plus{}11)}{dt}\\cdot (4\\minus{}10t)$\n\n$ \\equal{}(\\minus{}10)(5t^2\\minus{}4t\\plus{}11)^{\\minus{}2}\\plus{}(\\minus{}2)(5t^2\\minus{}4t\\plus{}11)^{\\minus{}3}\\cdot (10t\\minus{}4)\\cdot (4\\minus{}10t)$\n\n$ \\equal{}(5t^2\\minus{}4t\\plus{}11)^{\\minus{}3}((\\minus{}10)(5t^2\\minus{}4t\\plus{}11)\\plus{}(\\minus{}2)(10t\\minus{}4)(4\\minus{}10t))$\n\n$ \\equal{}(5t^2\\minus{}4t\\plus{}11)^{\\minus{}3}(150t^2\\minus{}120t\\minus{}78)$[/hide]", "Solution_11": "yes, Honey_S is right \r\nto onlylove_math but $ f'' < 0\\Longleftrightarrow 0\\le t\\le 1.224621125$", "Solution_12": "[quote=\"Raja Oktovin\"]Suppose $ a,b,c$ are real numbers, with $ a \\plus{} b \\plus{} c \\equal{} 3$. Prove that $ \\frac {1}{5a^2 \\minus{} 4a \\plus{} 11} \\plus{} \\frac {1}{5b^2 \\minus{} 4b \\plus{} 11} \\plus{} \\frac {1}{5c^2 \\minus{} 4c \\plus{} 11} \\le \\frac {1}{4}$.[/quote]\r\nThere exists a nice solution with Cauchy Schwarz. Can someone try this? ;) :)", "Solution_13": "I'm eager to see the solution applying Cauchy-Schwarz Inequality which, I guess, can_hang2007 loves very much and does well in :wink:", "Solution_14": "Here's my solution. It's simple.\r\nWLOG, we can assume $ a\\equal{}max(a,b,c)$\r\nLemma: if $ x\\leq \\frac{9}{5}$ we have: $ \\frac{1}{5x^2\\minus{}4x\\plus{}11} \\leq \\frac{1}{24}(3\\minus{}x)$ (because it's equivalent to $ (9\\minus{}5x)(x\\minus{}1)^2 \\geq 0$)\r\nCase 1:\r\n $ a \\leq \\frac{9}{5}$, applying the lemma, we get $ \\sum{\\frac{1}{5a^2\\minus{}4a\\plus{}11}} \\leq \\sum{\\frac{1}{24}(3\\minus{}a) \\equal{}\\frac{1}{4}}$\r\nCase 2:\r\n$ a > \\frac{9}{5}$, it's obvious that $ \\frac{1}{5a^2\\minus{}4a\\plus{}11} < \\frac{1}{20}$\r\nNote that: $ 5t^2\\minus{}4t\\plus{}11\\equal{}5(t\\minus{}\\frac{2}{5})^2\\plus{}\\frac{51}{5} \\geq \\frac{51}{5}$, therefore\r\n$ \\frac{1}{5b^2\\minus{}4b\\plus{}11} \\plus{}\\frac{1}{5c^2\\minus{}4c\\plus{}11} \\leq \\frac{10}{51}$\r\nSo, $ \\sum{\\frac{1}{5a^2\\minus{}4a\\plus{}11}} < \\frac{1}{20}\\plus{}\\frac{10}{51} < \\frac{1}{4}$", "Solution_15": "But where is the equality case considered when $ a>\\frac{9}{5}$? You are obtaining a strict ineq.", "Solution_16": "[quote=\"Sunkern_sunflora\"]But where is the equality case considered when $ a > \\frac {9}{5}$? You are obtaining a strict ineq.[/quote]\r\nThere isn't a equality case with $ a > \\frac {9}{5}$ :wink:", "Solution_17": "But we should obtain the same condition as required by the problem everywhere! :o \r\n\r\n[hide=\"Sudhu Akashnil er jonno\"]Tomar number theory er problem topic e r kono post koroni keno? \n:) [/hide]", "Solution_18": "[quote=\"Raja Oktovin\"]Suppose $ a,b,c$ are real numbers, with $ a \\plus{} b \\plus{} c \\equal{} 3$. Prove that $ \\frac {1}{5a^2 \\minus{} 4a \\plus{} 11} \\plus{} \\frac {1}{5b^2 \\minus{} 4b \\plus{} 11} \\plus{} \\frac {1}{5c^2 \\minus{} 4c \\plus{} 11} \\le \\frac {1}{4}$.[/quote]\r\n\r\nyour ineq is equivalent to : \r\n$ \\sum{\\frac {x^2}{x^2 \\plus{} 51}} \\geq \\frac {9}{20}$ with $ x \\plus{} y \\plus{} z \\equal{} 9$ \r\nby cauchy-shwarz : $ LHS ( \\sum{ (x^2 \\plus{} 51)(x \\plus{} 1)^2} ) \\geq ( \\sum{ x(x \\plus{} 1)} )^2$\r\nso it remains to prove that $ 20 ( \\sum{ x(x \\plus{} 1)})^2 \\geq 9 ( \\sum{ (x^2 \\plus{} 51)(x \\plus{} 1)^2} )$\r\nlet $ f(x,y,z) \\equal{} LHS \\minus{} RHS$ , note that $ f(x,y,z) \\minus{} f(x,t,t) \\geq 0$ where $ t \\equal{} \\frac {y \\plus{} z}{2}$ and $ x \\equal{} max(x,y,z) \\geq 3$\r\nso it sufficies to prove the inequality when two variables are equal\r\nbut $ f(x,x,9 \\minus{} 2x) \\equal{} (x \\minus{} 3)^2(31x^2 \\minus{} 144x \\plus{} 261) \\geq 0$ , our inequality is proved", "Solution_19": "[quote=\"Raja Oktovin\"]Suppose $ a,b,c$ are real numbers, with $ a \\plus{} b \\plus{} c \\equal{} 3$. Prove that $ \\frac {1}{5a^2 \\minus{} 4a \\plus{} 11} \\plus{} \\frac {1}{5b^2 \\minus{} 4b \\plus{} 11} \\plus{} \\frac {1}{5c^2 \\minus{} 4c \\plus{} 11} \\le \\frac {1}{4}$.[/quote]\r\nHere is my second solution. :) \r\nFirst we will prove the follow inequality. :oops:\r\n\\[ 179\\sum {a^4 } \\plus{} 855\\sum {a^2 b^2 } \\minus{} 406\\sum {ab\\left( {a^2 \\plus{} b^2 } \\right)} \\minus{} 222abc\\sum a \\ge 0\r\n\\]\r\nfor $ a,b$ and $ c$ are real numbers. It is obviously true with $ m \\equal{} 179,n \\equal{} 855,p \\equal{} g \\equal{} \\minus{} 406$ \r\nThen, assume $ a \\plus{} b \\plus{} c \\equal{} 3$ and $ q \\equal{} ab \\plus{} bc \\plus{} ca,r \\equal{} abc$ then we obtain the follow result from the above inequality\r\n\\[ \\Rightarrow 1074 \\minus{} 748q \\plus{} 150q^2 \\minus{} 180r \\ge 0\r\n\\]\r\nNow come back to the problem, using the Cauchy Schwarz inequality we have\r\n\\[ \\sum {\\frac {1}{{5a^2 \\minus{} 4a \\plus{} 11}}} \\le \\frac {1}{4}\r\n\\]\r\n\r\n\\[ \\Leftrightarrow \\sum {\\frac {{5a^2 \\minus{} 4a \\plus{} 1}}{{5a^2 \\minus{} 4a \\plus{} 11}}} \\ge \\frac {1}{2}\r\n\\]\r\n\\[ \\left( {\\sum {\\frac{{5a^2 \\minus{} 4a \\plus{} 1}}{{5a^2 \\minus{} 4a \\plus{} 11}}} } \\right)\\left[ {\\sum {\\left( {5a^2 \\minus{} 4a \\plus{} 1} \\right)\\left( {5a^2 \\minus{} 4a \\plus{} 11} \\right)} } \\right] \\ge \r\n\\]\r\n\\[ \\ge \\left( {\\sum {5a^2 \\minus{} 4a \\plus{} 1} } \\right)^2 \r\n\\]\r\nHence we need to prove\r\n\\[ 2\\left( {\\sum {5a^2 \\minus{} 4a \\plus{} 1} } \\right)^2 \\ge \\sum {\\left( {5a^2 \\minus{} 4a \\plus{} 1} \\right)\\left( {5a^2 \\minus{} 4a \\plus{} 11} \\right)}\r\n\\]\r\n\r\n\\[ \\Leftrightarrow 1074 \\minus{} 748q \\plus{} 150q^2 \\minus{} 180r \\ge 0\r\n\\]\r\nwhich is known before, :).", "Solution_20": "[quote=\"Erken\"]lol:\nI am pretty sure Jensen is useless here.But I'm afraid that using derivatives is necessary,refering to [b]arqady[/b]'s solution,which he gave almost one year ago...\nSearching is a virtue :wink: http://www.mathlinks.ro/viewtopic.php?p=968464#968464[/quote]\r\nI don't think so because if $ f(x)$ is not concave at any point,the ineq is not true,I really think Jensen is very strong! :wink:", "Solution_21": "[quote=\"Math pro\"][quote=\"Erken\"]lol:\nI am pretty sure Jensen is useless here.But I'm afraid that using derivatives is necessary,refering to [b]arqady[/b]'s solution,which he gave almost one year ago...\nSearching is a virtue :wink: http://www.mathlinks.ro/viewtopic.php?p=968464#968464[/quote]\nI don't think so because if $ f(x)$ is not concave at any point,the ineq is not true,I really think Jensen is very strong! :wink:[/quote]\r\nHuh? :huh: \r\nDang...\r\nOK,you can think whatever you want,but please speak more clearly,I am not telling anyone that Jensen is not strong,I just said that Jensen inequality here is not helpful and that's all...\r\nProbably you mean that this function is concave or convex,but this function on the given segment simultaneously has \"concavity\" and \"convexity\" points...", "Solution_22": "[quote=\"shalex\"]I'm eager to see the solution applying Cauchy-Schwarz Inequality which, I guess, can_hang2007 loves very much and does well in :wink:[/quote]\r\nOk, here is the Cauchy Shwarz solution of mine. :)\r\nIt is easy to see that there exist $ a,b$ such that $ (a\\minus{}1)(b\\minus{}1) \\ge 0$, then $ a^2\\plus{}b^2 \\le 1\\plus{}(a\\plus{}b\\minus{}1)^2 \\equal{}c^2\\minus{}4c\\plus{}5$. Now, rewrite our inequality as\r\n\\[ \\left( 5 \\minus{}\\frac{51}{5a^2\\minus{}4a\\plus{}11}\\right) \\plus{}\\left( 5\\minus{}\\frac{51}{5b^2\\minus{}4b\\plus{}11}\\right) \\ge \\frac{51}{5c^2\\minus{}4c\\plus{}11} \\minus{}\\frac{11}{4}\\]\r\nor\r\n\\[ \\frac{(5a\\minus{}2)^2}{5a^2\\minus{}4a\\plus{}11}\\plus{}\\frac{(5b\\minus{}2)^2}{5b^2\\minus{}4b\\plus{}11} \\ge \\frac{ 83\\plus{}44c\\minus{}55c^2}{4(5c^2\\minus{}4c\\plus{}11)}\\]\r\nBy Cauchy Schwarz Inequality, we have\r\n\\[ \\frac{(5a\\minus{}2)^2}{5a^2\\minus{}4a\\plus{}11}\\plus{}\\frac{(5b\\minus{}2)^2}{5b^2\\minus{}4b\\plus{}11} \\ge \\frac{(5(a\\plus{}b)\\minus{}4)^2}{ 5(a^2\\plus{}b^2)\\minus{}4(a\\plus{}b) \\plus{}22} \\ge \\frac{ (5c\\minus{}11)^2}{5c^2\\minus{}16c\\plus{} 35}\\]\r\nIt suffices to prove that\r\n\\[ \\frac{ (5c\\minus{}11)^2}{5c^2\\minus{}16c\\plus{} 35} \\ge \\frac{ 83\\plus{}44c\\minus{}55c^2}{4(5c^2\\minus{}4c\\plus{}11)}\\]\r\nwhich is just equivalent to\r\n\\[ (c\\minus{}1)^2( 775c^2 \\minus{}2150c\\plus{}2419) \\ge 0\\]\r\nwhich is obviously true. :)", "Solution_23": "[quote=\"Math pro\"][quote=\"Erken\"]lol:\nI am pretty sure Jensen is useless here.But I'm afraid that using derivatives is necessary,refering to [b]arqady[/b]'s solution,which he gave almost one year ago...\nSearching is a virtue :wink: http://www.mathlinks.ro/viewtopic.php?p=968464#968464[/quote]\nI don't think so because if $ f(x)$ is not concave at any point,the ineq is not true,I really think Jensen is very strong! :wink:[/quote]\r\nYou didnt understand all of the idea of Jensen ineq, Okey ,you can try the ineq in Poland 1991 to find you are wrong (the function is not convex with all $ x$ but the ineq is still true, :lol: )and of course,If it had been able to be killed by only Jensen(simply),it would have appeared in China TST prblem\r\nYour proof is very nice,Can_hang :o :o", "Solution_24": "To [b]MathPro[/b]:\r\nIt is really annoying me to dispute with you about who of us is right...\r\nIn order to clarify the point I would like to reply,but for the last time.\r\nThere was some kind of discussing whether Jensen solves this problem or not.Some member said that it does and he/she gave some calculation of second derivative,but as it was observed later, there was a mistake.\r\nI said that this problem can not be solved by the use of Jensen inequality and in order to show the solution(I must admit that this solution is really good) I put a link.That's all.\r\nThen,I don't know why,you were starting to make some senseless replies,like \"I really think that Jensen is very strong\".OK,I agree with you,but what is the point of telling a such things?\r\nYou said:\"I don't think so\" \r\nDid I say anything about concavity in my first post?\r\nDid I say that Jensen is \"b*llsh*t\"?\r\nMan...Do you understand how senseless you replies look like?And if really you don't want to be a spamer then stop doing a such\r\nthings.", "Solution_25": "it's really nice can_hang2007, and I think you are always \"playing\" Cauchy-Swartz Inequality very well!", "Solution_26": "[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=175288[/url]", "Solution_27": "[quote=\"kunny\"][url]http://www.mathlinks.ro/Forum/viewtopic.php?t=175288[/url][/quote]\r\nWhat about to have a look at the first page of this thread?\r\nEdited:\r\nNo,you misunderstood me,I guess,I meant that I have already meantioned it before you,in the first page of this thread.", "Solution_28": "Ah, sorry, you have already posted it. :blush: \r\n\r\n[quote=\"Erken\"]lol:\nI am pretty sure Jensen is useless here.But I'm afraid that using derivatives is necessary,refering to [b]arqady[/b]'s solution,which he gave almost one year ago...\nSearching is a virtue :wink: http://www.mathlinks.ro/viewtopic.php?p=968464#968464[/quote]", "Solution_29": "I think this is nicer.\r\n\r\nLet $ a,b,c$ be real numbers such that $ a \\plus{} b \\plus{} c \\equal{} 3$. Prove that\r\n\r\n$ \\frac 1{9 \\minus{} 10a \\plus{} 9a^2} \\plus{} \\frac 1{9 \\minus{} 10b \\plus{} 9b^2} \\plus{} \\frac 1{9 \\minus{} 10c \\plus{} 9c^2} \\le \\frac 3{8}$.", "Solution_30": "[quote=\"Vasc\"]Let $ a,b,c$ be real numbers such that $ a \\plus{} b \\plus{} c \\equal{} 3$. Prove that\n\n$ \\frac 1{9 \\minus{} 10a \\plus{} 9a^2} \\plus{} \\frac 1{9 \\minus{} 10b \\plus{} 9b^2} \\plus{} \\frac 1{9 \\minus{} 10c \\plus{} 9c^2} \\le \\frac 3{8}$.[/quote]It's really nicer in respect that\r\n\r\n$ \\sum{\\frac {4}{11 \\minus{} 4a \\plus{} 5a^2}}\\leq\\sqrt [3]{\\frac {8}{3}\\sum{\\frac {1}{9 \\minus{} 10a \\plus{} 9a^2}}}\\leq 1$\r\n\r\nhold for $ a \\plus{} b \\plus{} c \\equal{} 3$.", "Solution_31": "[quote=\"Vasc\"]I think this is nicer.\n\nLet $ a,b,c$ be real numbers such that $ a \\plus{} b \\plus{} c \\equal{} 3$. Prove that\n\n$ \\frac 1{9 \\minus{} 10a \\plus{} 9a^2} \\plus{} \\frac 1{9 \\minus{} 10b \\plus{} 9b^2} \\plus{} \\frac 1{9 \\minus{} 10c \\plus{} 9c^2} \\le \\frac 3{8}$.[/quote]\r\n\r\nmy method that I used to the original inequality works here :wink:", "Solution_32": "[quote=\"Vasc\"]I think this is nicer.\n\nLet $ a,b,c$ be real numbers such that $ a \\plus{} b \\plus{} c \\equal{} 3$. Prove that\n\n$ \\frac 1{9 \\minus{} 10a \\plus{} 9a^2} \\plus{} \\frac 1{9 \\minus{} 10b \\plus{} 9b^2} \\plus{} \\frac 1{9 \\minus{} 10c \\plus{} 9c^2} \\le \\frac 3{8}$.[/quote]\r\nSee here: http://can-hang2007.blogspot.com/2010/02/inequality-120-m-rozenberg.html", "Solution_33": "Suppose $ a,b,c$ are real numbers, with $ a+b+c=3$. [url=https://math.stackexchange.com/questions/509580/how-prove-this-frac12a2-6a9-frac12b2-6b9-frac12c2-6c9-le-f?rq=1]Prove that[/url] $$\\dfrac{1}{2a^2-6a+9}+\\dfrac{1}{2b^2-6b+9}+\\dfrac{1}{2c^2-6c+9}\\le\\dfrac{3}{5}$$" } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "reflection", "symmetry", "trapezoid", "angle bisector" ], "Problem": "The points $A, B, C, D$ lie in this order on a circle $o$. The point $S$ lies inside $o$ and has properties $\\angle SAD=\\angle SCB$ and $\\angle SDA= \\angle SBC$. Line which in which angle bisector of $\\angle ASB$ in included cut the circle in points $P$ and $Q$. Prove $PS =QS$.", "Solution_1": "Let $O$ be the circumcenter of the quadrilateral $ABCD$. The triangles $\\triangle SDA \\sim \\triangle SBC$ are directly similar, having 2 internal angles equal. Reflect one of these triangles in the line $SO$, say the triangle $\\triangle SBC$ into the oppositely congruent triangle $\\triangle SB'C'$. The reflections $B', C'$ of the points $B, C \\in (O)$ in the line $SO$ passing through the circle center also lie on the circle $(O)$. Thus we obtain a cyclic quadrilateral $AC'B'D$ with the triangles $\\triangle SDA \\sim \\triangle SC'B'$ oppositely similar. This implies that the common vertex $S$ of these 2 triangles is the diagonal intersection of the quadrilateral $AC'B'D$, i.e., the points $D, S, C'$ are collinear and the points $A, S, B'$ are also collinear. Since the symmetry axis $SO$ bisects one angle $\\angle BSB'$ formed by the lines $AB' \\equiv SB'$ and $SB$, the normal to this symmetry axis at the point $S$ bisects the complemetary angle $\\angle ASB$ formed by these 2 lines. This normal is thus the defined angle bisector and it meets the circumcircle $(O)$ at points $P, Q$. The line $SO$ passing through the circle center $O$ and perpendicular to the chord $PQ$ cuts this chord at its midpoint $S$, hence, $PS = QS$.", "Solution_2": "Here's my a little bit long proof. \r\n\r\nLet $P$ is on arc $AB$, and $Q$ is on arc $CD$. and $X$ is the intersection of $AD$ and $BC$. \r\n\r\nWLOG $X$ is on right side of $CD$\r\n\r\nIt's self-confident that $ASCX$, $BXDS$ is cyclic respectively. \r\n\r\nso we can know that the three common cords of circles $AC, BD, XS$ is concurrent. \r\n\r\nLet's denote $R$ by the intersection of three cords. \r\n\r\nHere, $\\angle CSX= \\angle CAX=\\angle CBD=\\angle DSX$. So we get $\\angle CSX=\\angle DSX$. \r\n\r\nTherefore, $P, Q$ are on line $SX$. \r\n\r\nAnd here, we can find that from $\\angle CAX=\\angle XSD$, $ASRD$ is cyclic. similarly, $BSCR$ is also cyclic. \r\n\r\nNow, we almost get it!\r\n\r\n$XQ\\times XP=XD\\times XA= XS\\times XR$ $\\cdots (Y)$\r\n\r\n$XR\\times SR = AR\\times RC= PR\\times QR$ $\\cdots (T)$\r\n\r\nNow, Let $a=XQ, b=QR, c=SR, d=SP$. \r\n\r\nThen, $(Y)$: $a(a+b+c+d)=(a+b)(a+b+c) \\Leftrightarrow ad=b(a+b+c)$. \r\n\r\n$(T)$: $c(a+b)=(c+d)b \\Leftrightarrow ac=bd$. \r\n\r\nfrom $(Y)$, we get $acd=bc(a+b+c)$ $\\Rightarrow bd^2 = b(bd+bc+c^2 )$ (by $(T)$).\r\n\r\n$\\Leftrightarrow d^2- c^2 = b(c+d)$ $\\Leftrightarrow d=b+c$\r\n\r\nTherefore, $PS=QS$. \r\n\r\n$Q.E.D.$", "Solution_3": "These solutions ([u]Yetti[/u]'s and [u]Chang Woo-Jin[/u]'s) are fine. I can add only my congratulations and a little remark: if $U\\in AC'\\cap PQ,\\ V\\in DB'\\cap PQ$ then $SU=SV$ (the problem of \"butterfly\").", "Solution_4": "And here is my solution:\r\n\r\n[color=blue][b]Problem.[/b] Let ABCD be a cyclic quadrilateral with the circumcircle k, and let S be a point inside this quadrilateral such that the triangles SBC and SDA are directly similar.\nLet the internal angle bisector of the angle ASB meet the circle k at the points P and Q. Prove that PS = QS.[/color]\r\n\r\n[i]Solution.[/i] We will work with directed angles modulo 180\u00b0.\r\n\r\nLet W be the point of intersection of the lines BC and DA, let R be the point of intersection of the lines AC and BD, and let O be the center of the circle k.\r\n\r\nSince the triangles SBC and SDA are directly similar, we have < SBC = < SDA, or, in other words, < SBW = < SDW. Hence, the point S lies on the circumcircle of the triangle BDW. Hence, < BWS = < BDS. Similarly, the point S lies on the circumcircle of the triangle ACW, and < CAS = < CWS. Thus,\r\n\r\n < RAS = < CAS = < CWS = < BWS = < BDS = < RDS.\r\n\r\nThus, the point S lies on the circumcircle of triangle RDA. Similarly, the point S lies on the circumcircle of triangle RBC.\r\n\r\n[Note that thus, the point S is the Miquel point of the complete quadrilateral formed by the lines BC, DA, AC and BD.]\r\n\r\nSince the point S lies on the circumcircle of the triangle ACW, we have < ASC = < AWC, and since the point S lies on the circumcircle of the triangle RBC, we have < BSC = < BRC. Thus,\r\n\r\n < ASB = < ASC - < BSC = < AWC - < BRC = < (DA; BC) - < (BD; AC)\r\n = (< (DA; AC) + < (AC; BC)) - < (BD; AC) = (< (DA; AC) - < (BD; AC)) + < (AC; BC)\r\n = < (DA; BD) + < (AC; BC) = < ADB + < ACB.\r\n\r\nSince the points A, B, C, D lie on one circle (namely, the circle k), we have < ADB = < ACB, and since O is the center of this circle, the central angle theorem yields $\\measuredangle AOB=2\\cdot\\measuredangle ACB$. Hence,\r\n\r\n$\\measuredangle ASB=\\measuredangle ADB+\\measuredangle ACB=\\measuredangle ACB+\\measuredangle ACB=2\\cdot\\measuredangle ACB=\\measuredangle AOB$.\r\n\r\nThus, the point S lies on the circumcircle of triangle OAB. Similarly, the point S lies on the circumcircle of triangle OCD.\r\n\r\nSince the point S lies on the circumcircle of triangle OAB, we have < ASO = < ABO and < OSB = < OAB. But since AO = BO (in fact, the point O is the center of the circle k which passes through the points A and B), the triangle AOB is isosceles, and thus < ABO = < OAB. Hence, < ASO = < OSB, so that the line OS bisects the angle ASB. From the arrangement of the points, it follows that the line OS is the external angle bisector of the angle ASB. Hence, the internal angle bisector of the angle ASB is perpendicular to the line OS (in fact, the internal angle bisector of an angle is always perpendicular to the external angle bisector). Thus, the triangles PSO and QSO both are right-angled at S. Also, these two right-angled triangles share a common catet (OS) and have the same hypotenuse (OP = OQ, since both points P and Q lie on the circle k with the center O). Thus, these two right-angled triangles are congruent, so that PS = QS, and the problem is solved.\r\n\r\n darij", "Solution_5": "Heeeere's my solution (?):\r\n\r\n\r\n Point S is on a circle built on diagonal AC and at the same time on a circle\r\n\r\n built on diagonal BD. Thus it is on their intersection. Angle bisector of ASB \r\n\r\n is the radical axis of these two circles. Their centers line is pependicular to radical\r\n\r\n axis. Thus line OS where O is the center of circle(ABCD) is perpendicular to PS\r\n\r\n and as diameter cuts chord PS in two equal segments.\r\n\r\n\r\n \r\n Looks very much like a Butterfly Theorem.\r\n\r\n\r\n Thank you.\r\n\r\n M.T.", "Solution_6": "This is Exactly same as China 1992!!!!", "Solution_7": "$Problem$: let $ABCD$ be a cyclic quadrilateral which is scribed in circle $(C)$.$S$ is a point in circle,inwhich:\r\n\r\n$\\angle SAD=\\angle SCB$ and $\\angle SDA=\\angle SBC$.\r\n\r\nthe bisector of the angle $\\angle ASB$,intersects circle $(C)$ in points $P$ and $Q$.\r\n\r\nprove that:$SP=SQ$.\r\n\r\nok...\r\n\r\nlets solve the problem by the most power full tool in geometry,$INVERSION$.\r\n\r\ntriangles $SAD$ and $SBC$ are simmilar to each other ,because:\r\n\r\n$\\angle SAD=\\angle SCB$ and $\\angle SDA=\\angle SBC$ and $\\angle ASD=\\angle BSC$.\r\n\r\nand the conclusion is :$SA.SB=SC.SD$.\r\n\r\nconsider an $INVERSION$ with the center $S$ and the radius $SA.SB$ or $SC.SD$.\r\n\r\nlets invert the shape:\r\n\r\n$A$ goes to $A'$ $(SA'=SB)$.\r\n\r\n$B$ goes to $B'$ $(SB'=SA)$.\r\n\r\n$C$ goes to $C'$ $(SC'=SD$.\r\n\r\n$D$ goes to $D'$ $(SD'=SC)$.\r\n\r\n$thorem$:the $INVERSION$ of a circle which doesnt go through the center of the $INVERSION$,is a circle.\r\n\r\nthen circle $(C)$,after invert changes to circle $(C')$,because it doesnt go through the center.\r\n\r\nand $A'B'C'D'$ is a cyclic quadrilateral which is scribed in circle $(C')$.\r\n\r\nNow we should use the fact:$SP$ is the bisector of $\\angle ASB$ or $SQ$ is the bisector of $\\angle DSC$.\r\n\r\npay attention to quadrilateral $A'B'C'D'$.$A',B',C',D'$, with a reflect to $PQ$,changes to $B,A,D,C,$, respectively.\r\n\r\nthen its clear that :the circles $(C)$ and $(C')$,also by reflecting to $PQ$,will change to each other.\r\n\r\nand because of the smmetry of the shape,we can claime that:\r\n\r\ncircle $(C')$,goes through points $P$ and $Q$.\r\n\r\n(note:dears ,this fact will be simply proved.and i think it doesnt need to disscus. ;).u can disscus it with ure self )\r\n\r\nthen,in the $INVERSION$,which we used the points $P$ and $Q$,are fixed.i mean $P=P'$and $Q=Q'$. then we \r\n\r\nhave:$SA.SB$=$SP.SP'$=$SP^2$\r\n\r\n\r\n\r\nand also:\r\n\r\n$SD.SC$=$SQ.SQ'$=$SQ^2$.\r\n\r\nbut:\r\n\r\n$SA.SB$=$SD.SC$. (because the simmilaity of the triangles $SAD$ and $SBC$)\r\n\r\nthen:$SP^2=SQ^2$ and $SP=SQ$.\r\n\r\nthe probelm is really nice and u saw that,its proved very nice and simple by $INVERSION$.\r\n\r\ni think the resourse of the problem is $INVERSION$.and the teacher created it by seeng and using inversive facts.\r\n\r\ni say again.really a nice problem.\r\n\r\ni can enjoy it during my life and up to the time my life stops. ;)", "Solution_8": "the blue circle is the inverted of circumcircle of$ABCD$,\r\n\r\nand the green circle is the inversion circle!\r\n\r\ni should say, this diagram is proposed by my dear friend$Amir.S$ :wink:", "Solution_9": "[quote=\"yetti\"]Thus we obtain a cyclic quadrilateral $AC'B'D$ with the triangles $\\triangle SDA \\sim \\triangle SC'B'$ oppositely similar. This implies that the common vertex $S$ of these 2 triangles is the diagonal intersection of the quadrilateral $AC'B'D$, i.e., the points $D, S, C'$ are collinear and the points $A, S, B'$ are also collinear.[/quote]Why? :huh:", "Solution_10": "First, the posted problem is trivial if S is the center of the given circle (O), so we assume that S is not the circle center. Extend AS, BS, CS, DS to meet the circle (O) again at B', A', D', C' (keepeing the notation of my figure). The cyclic quadrilaterals AC'B'D, A'CBD' have the common circumcircle (O), the same diagonal intersection S, the same angle between the diagonals (because $\\triangle SBC \\sim \\triangle SDA$), and the same opposite sides AC' = A'C, B'D = BD' (because $\\angle AB'C' = \\angle A'BC,$ $\\angle B'C'D = \\angle BCD'$), therefore they are congruent. In particular, their diagonals are equal, AB' = A'B, CD' = C'D, so that we obtain 2 isosceles trapezoids ABB'A' and CDD'C' with the same diagonal intersection S. The opposite sides of all these cyclic quadrilaterals (AC'B'D, A'CBD' and trapezids ABB'A', CDD'C') meet on the polar s of the diagonal intersection S with respect to the circle (O) and $s \\perp OS.$ Because of symmetry, the opposite sides of the isosceles trapezoids ABB'A', CDD'C' meet at the foot P of the normal OS to this polar s. Thus these 2 isosceles trapezoids are symmetrical with respect to OS, so is the circumcircle (O), which means that A', B', C', D' ( originally obtained as the other intersections of BS, AS, DS, CS with (O)) are reflections of A, B, C, D in the line OS. Obviously, D, S, C' are collinear, A, S, B' are collinar, etc.", "Solution_11": "[quote=\"armpist\"][color=darkblue][b]Proof.[/b] Point $S$ is on a circle built on diagonal $AC$ and at the same time on a circle built on diagonal $BD$. Thus it is on their intersection. Angle bisector of $\\widehat{ASB}$ is the radical axis of these two circles. Their centers line is pependicular to radical axis. Thus line $OS$ where $O$ is the center of circle $(ABCD)$ is perpendicular to $PS$ and as diameter cuts chord $PS$ in two equal segments.\n\n[b]Remark.[/b] Looks very much like a Butterfly Theorem.[/color][/quote]\r\nYes Armpist, I like much the Butterfly Theorem ! Very nice your proof, Armpist ! In the my opinion, is the best.\r\nPlease, forgive me if I could afford to write otherwise (more coloured) the your solution in \"quote\" !", "Solution_12": "[quote=\"Virgil Nicula\"]... Very nice your proof, Armpist ! In the my opinion, is the best...[/quote]\r\n\r\nAnd of course, the purpose of opinions is to vary.", "Solution_13": "[quote=\"Virgil Nicula\"][quote=\"armpist\"][color=darkblue][b]Proof.[/b] Point $S$ is on a circle built on diagonal $AC$ and at the same time on a circle built on diagonal $BD$. Thus it is on their intersection. Angle bisector of $\\widehat{ASB}$ is the radical axis of these two circles. Their centers line is pependicular to radical axis. Thus line $OS$ where $O$ is the center of circle $(ABCD)$ is perpendicular to $PS$ and as diameter cuts chord $PS$ in two equal segments.\n\n[b]Remark.[/b] Looks very much like a Butterfly Theorem.[/color][/quote]\nYes Armpist, I like much the Butterfly Theorem ! Very nice your proof, Armpist ! In the my opinion, is the best.\nPlease, forgive me if I could afford to write otherwise (more coloured) the your solution in \"quote\" ![/quote]Could any one clarify me why angle bisector of $\\widehat{ASB}$ is the radical axis of these two circles, and why the line joining the centers is parallel to $OS$?\r\n\r\nThanks.", "Solution_14": "let $AD$ and $BC$ intersect at $T$ ;$AC,BD$ at $R$,then $ASCT;BSDT$ concyclic;hence in the 3 circles,equal-power axes $AC,BD,ST$are concurrent\nso $PSRQT$ collinear.what follows is just a well-known conclusion in polar theory.", "Solution_15": "[quote=armpist]Thus line OS where O is the center of circle(ABCD) is perpendicular to PS\n[/quote]\n\nI still don't get it. Why $OS$ is parallel to line connecting centers of circles $ASC, BSD$?" } { "Tag": [ "function", "inequalities", "induction", "algebra unsolved", "algebra" ], "Problem": "$ f: N\\to N$ satisfies i) and ii)\r\n\r\ni) $ f(93)\\equal{}116$\r\nii) $ f(f(n)) \\plus{} 7f(n) \\equal{} 8n \\plus{} 207$\r\n\r\nfind $ f(n)$", "Solution_1": "[b]I will only apply property ii) to show $ \\forall m: f(m)\\equal{}m\\plus{}23$.[/b]\r\n\r\nSince $ f(f(m))>0$ we get $ f(m)\\leq\\frac{8m\\plus{}206}{7}$ and after resubstitution $ 7f(m)\\equal{}8m\\plus{}207\\minus{}f(f(m))\\geq 8m\\plus{}207\\minus{}\\frac{8f(m)\\plus{}206}{7}$, which is equivalent to $ f(m)\\geq\\frac{56m\\plus{}1243}{57}$. \r\n\r\nNow define the sequences, $ \\{a_n\\},\\{b_n\\},\\{c_n\\},\\{d_n\\}$ via $ a_0\\equal{}56,b_0\\equal{}8,c_0\\equal{}1243,d_0\\equal{}206$ and recursively \r\n\r\n$ a_{n\\plus{}1}\\equal{}8b_n,b_{n\\plus{}1}\\equal{}8a_n\\plus{}7,c_{n\\plus{}1}\\equal{}207b_n\\minus{}d_n,d_{n\\plus{}1}\\equal{}207(a_n\\plus{}1)\\minus{}c_n$\r\n\r\nAssume we observe at some point the inequality (*) $ \\forall m: \\frac{ma_n\\plus{}c_n}{a_n\\plus{}1}\\leq f(m)\\leq\\frac{m(b_n\\plus{}1)\\plus{}d_n}{b_n}$. Substitution into ii) yields $ \\forall m: \\frac{ma_{n\\plus{}1}\\plus{}c_{n\\plus{}1}}{a_{n\\plus{}1}\\plus{}1}\\leq f(m)\\leq\\frac{m(b_{n\\plus{}1}\\plus{}1)\\plus{}d_{n\\plus{}1}}{b_{n\\plus{}1}}$. \r\n\r\n[b]Claim[/b]: (1) $ \\forall n>0: 22.3\\leq\\frac{c_n}{a_n\\plus{}1}\\leq 23\\leq\\frac{d_n}{b_n}\\leq 23.2$\r\n(2) $ \\forall m: f(m)\\equal{}m\\plus{}23$\r\n\r\n[i]Proof[/i]: As for (1) we verify the assertion for $ n\\equal{}1$. For $ n>1$ using $ a_n,b_n>0$ and an induction argument yields $ \\frac{c_{n\\plus{}1}}{a_{n\\plus{}1}\\plus{}1}\\equal{}\\frac{207b_n\\minus{}d_n}{8b_n\\plus{}1}$ $ \\leq \\frac{207b_n\\minus{}23b_n}{8b_n}\\equal{}23$ and $ \\frac{d_{n\\plus{}1}}{b_{n\\plus{}1}}\\equal{}\\frac{207(a_n\\plus{}1)\\minus{}c_n}{8a_n\\plus{}7}$ $ \\geq\\frac{207(a_n\\plus{}1)\\minus{}23(a_n\\plus{}1)}{8a_n\\plus{}8}\\equal{}23$. The inequalities for the outer bounds follow similarly taking into account the growth of $ a_n,b_n$.\r\n\r\nFor (2) note that $ f$ fulfills (*) for $ n\\equal{}0$ and therefore for all $ n\\in\\mathbb{N}$. Due to (1) this implies for all $ m\\in\\mathbb{N}$ the inequality:\r\n\r\n$ m\\plus{}22.3\\leq\\limsup_{n\\rightarrow\\infty}\\frac{ma_n\\plus{}c_n}{a_n\\plus{}1}\\leq f(m)\\leq\\liminf_{n\\rightarrow\\infty}\\frac{m(b_n\\plus{}1)\\plus{}d_n}{b_n}\\leq m\\plus{}23.2$ and with $ f(m)\\in\\mathbb{N}$ the assertion follows." } { "Tag": [ "real analysis", "real analysis solved" ], "Problem": "Let u_n:u_0=a>0;u_(n+1)=u_n/(1+(u_n) 2 ).\r\nFind lim u_n*\\sqrt n.\r\nThis is a base problem for lim .", "Solution_1": "easily to have lim u_n=0\r\nlim 1/(u_(n+1) 2 - 1/(u_n) 2 \r\n=lim (2+ u_n 2 ) = 2\r\nBy cesaro :\r\n lim 1/((u_n) 2 *n) =2\r\nso lim u_n* \\sqrt n = 1/ \\sqrt 2" } { "Tag": [ "probability" ], "Problem": "If $ x$ is chosen at random from $ \\{\\minus{}4.\\minus{}3,\\minus{}2,\\minus{}1,0,1,2,3,4\\}$, what is the probability that it is a solution to $ x(x \\minus{} 3)(x \\plus{} 2) < 0$? Express your answer as a common fraction.", "Solution_1": "Plug them all in and see if they work.", "Solution_2": "Or we could solve because we learn more that way\r\n\r\nFor $ x(x\\minus{}3)(x\\plus{}2)$ to be less than $ 0$, all of the factors must be negative or exactly one of them must be negative. If all are negative, then the condition $ x\\plus{}2<0$ suffices, so $ x<\\minus{}2$. There are $ 2$ values that work for this case.\r\n\r\nIf exactly one of them is negative, then it must be $ x\\minus{}3$, so $ x>0$ and $ x\\minus{}3<0\\Rightarrow x<3$. There are another $ 2$ values that work.\r\n\r\nThere are $ 9$ elements in the set, so $ \\boxed{\\frac{4}{9}}$" } { "Tag": [ "inequalities", "geometry", "circumcircle", "inequalities solved" ], "Problem": "Show that in every triangle we have\r\n\r\n1) (a^4b^4+b^4c^4+c^4a^2)/a^2b^2c^2 >= 9R^2 >= a^2 + b^2 + c^2\r\n\r\n2) (r_a/a)^2 + (r_b/b)^2 + (r_c/c)^2 >=9/4\r\n\r\nwhere a, b, c are sides, R is circumradius, r_a, r_b, r_c are escribed radii.\r\n\r\nNamdung", "Solution_1": "9R^2 >= a^2 + b^2 + c^2\r\n\r\nwe know that sum(sin^2(A))=sum(1-cos^2(A))=3-sum(cos^2(A))=3-(1-2prod cos(A))=2(1+prod cos(A))<=2(1+1/8)=9/4\r\nand we know that a/sin(A)=2R and replacing it in sum(sin^2(A)) we get the desired ineq!\r\n\r\ncheers!", "Solution_2": "For the second one put a=y+z, etc and reduce it to sum (yz*(x+y+z)/(x(y+z)^2)>=9/4. Put m=yz, n=zx and p=xy. It becomes (mn+np+mp)(1/(m+n)^2+1/(n+p)^2+1/(m+n)^2)>=9/4, which holds by Iran (in fact another Crux) inequality.", "Solution_3": "Can you please write in detail how the inequality reduces to\r\n\r\nsum yz*(x+y+z)/(x(y+z)^2)>=9/4.\r\n\r\nThank you very much.", "Solution_4": "well it's like so:\r\n\r\nwe know that r_a=S/(p-a)=sqrt((x+y+z)xyz)/x using the notation harazi just made: a=y+z; b=z+x and c=x+y.", "Solution_5": "I am very very very very stupid. I checked the first inequality and not the second because I thought Harazi referred to:\r\n\r\n(a^4b^4+b^4c^4+c^4a^2)/a^2b^2c^2 >= 9R^2\r\n\r\nThank you very much Lagrangia for your clarification.", "Solution_6": "[quote=\"Namdung\"]Show that in every triangle we have\n\n[b]1[/b]) $ \\frac {b^2c^2}{a^2} \\plus{} \\frac {c^2a^2}{b^2} \\plus{} \\frac {a^2b^2}{c^2} \\geq9R^2;$\n\n[b]2[/b]) $ \\frac {r_{a}^{2}}{a^2} \\plus{} \\frac {r_{b}^{2}}{b^2} \\plus{} \\frac {r_{c}^{2}}{c^2}\\geq\\frac {9}{4},$\n\nwhere $ a, b, c$ are sides, $ R$ is circumradius, $ r_a, r_b, r_c$ are escribed radii.[/quote]A counterexample $ a \\equal{} 3,b \\equal{} c \\equal{} 2$ of the first one :\r\n\r\n$ \\frac {b^2c^2}{a^2} \\plus{} \\frac {c^2a^2}{b^2} \\plus{} \\frac {a^2b^2}{c^2} \\equal{} \\frac {178}{9} < \\frac {144}{7} \\equal{} 9R^2.$\r\n\r\nThe following inequality \r\n\r\n$ \\left(\\frac {r_{a}^{2}}{a^2} \\plus{} \\frac {r_{b}^{2}}{b^2} \\plus{} \\frac {r_{c}^{2}}{c^2}\\right)\\left(\\frac {b^2c^2}{a^2} \\plus{} \\frac {c^2a^2}{b^2} \\plus{} \\frac {a^2b^2}{c^2}\\right)\\geq\\frac {81}{4}R^2$\r\n\r\nholds for all triangle.\r\n\r\nThe second one see also here : http://www.mathlinks.ro/viewtopic.php?t=130999", "Solution_7": "let:p-a=x,...\r\nwe need to prove that:\r\n$ \\frac{x\\plus{}y\\plus{}z}{xyz}(\\sum\\frac{x^2y^2}{(x\\plus{}y)^2}) \\geq \\frac{9}{4}$\r\nbut:\r\n$ LHS \\geq \\frac{x\\plus{}y\\plus{}z}{xyz}*\\frac{(\\sum xy)^2}{\\sum (x\\plus{}y)^2}$\r\nremember that:\r\n$ \\sum(x\\plus{}y)^2 \\leq \\frac{4}{3}(x\\plus{}y\\plus{}z)^2$\r\nso $ LHS \\geq \\frac{x\\plus{}y\\plus{}z}{xyz}*\\frac{3}{4}*\\frac{(\\sum xy)^2}{(\\sum x)^2}\\equal{}\\frac{3}{4}*\\frac{(\\sum xy)^2}{xyz(x\\plus{}y\\plus{}z)} \\geq \\frac{9}{4}$" } { "Tag": [ "LaTeX" ], "Problem": "Hi! I need help with the following problem:\r\n\r\nA ship is steaming parallel to a straight coastline, distance D offshore, at speed V. A coastguard cutter, whose speed is u 0$, and let $ f(x) = o(g(x))$ as $ x \\rightarrow 0$. As $ x \\rightarrow 0$ prove or give a counterexample to:\r\n\r\n(i) $ \\int_{0}^{x}f = o \\left( \\int_{0}^{x}g \\right)$\r\n\r\n(ii) $ f'(x) = o(g'(x))$", "Solution_1": "You can prove (i) using L'H\u00f4pital's rule.\r\n\r\nHere's an alternative argument: Given $ \\epsilon>0,$ there exists $ \\delta>0$ such that $ |f(x)|<\\epsilon g(x)$ for $ 0a$, so the ellipse has a vertical major axis. We know that the ellipse's center is at the origin due to the fact that $ h$ and $ k$ are both $ 0$. Now we have to find $ c$ to find out the coordinates of the foci.\n\\begin{align*}\nc&=\\sqrt{b^2-a^2}\\\\\nc&=\\sqrt{4-1}\\\\\nc&=\\sqrt{3}\\\\\n\\end{align*} \nSo the foci are $ (h,k\\pm c) \\Longrightarrow (0,0\\pm \\sqrt{3}) \\Longrightarrow \\boxed{(0,\\sqrt{3}),(0,-\\sqrt{3})}$[/hide]", "Solution_2": "[hide=\"Solution using definition of focus\"]\nThe foci are points $ A,B$ such that for any point $ X$ on the ellipse, $ AX\\plus{}XB \\equal{} c$ for some $ c$.\n\nEasily see that if $ X$ lies on the y-axis, then we must have $ A(0,f)$ and $ B(0,\\minus{}f)$, so that $ AX\\plus{}XB\\equal{}(2\\plus{}f)\\plus{}(2\\minus{}f)\\equal{}4$\n\nNow, let $ X$ be the point $ (1,0)$ on the ellipse. We have that $ 2 \\sqrt{1\\plus{}f^2} \\equal{} 4 \\implies f \\equal{} \\pm \\sqrt{3}$\n\nThus, the foci are $ (0, \\pm \\sqrt{3})$[/hide]" } { "Tag": [ "algorithm", "number theory", "Diophantine equation" ], "Problem": "This problem is actually on the final challenge of the class \"Introduction to Number Theory\" given on here!\r\n\r\n\"Given any solution to the Diophantine equation 41x+53y=12, find the residue of x in modulo 53.\"\r\n\r\nEnjoy!\r\n\r\n-interesting_move", "Solution_1": "[hide]52[/hide] i hope im right", "Solution_2": "Correct, but how did you do it?", "Solution_3": "Modulo 53 :\r\n41x+53y=12 => 41x = 12 => -12x = 12 => x = -1 = 52\r\n\r\nFor :\r\n. 41 = -12\r\n. since (12,53)=1 we can divide by 12", "Solution_4": "wanna have fun try proving that little fact\r\n\r\nyou wouldnt believe how much is behind the \"simple statement\"\r\n\r\na|bc, (a,b)=1 => a|c.\r\n\r\nits crazy.", "Solution_5": "I don't see why that's so hard to prove. Let b = p1_e1*p2^e2*p3^e3... and c = q1_f1*q2^f2*q3^f3*..., where ps and qs are primes. Since (a,b)=1, none of the ps are factors of a. Thus, if a != 1, then a must be divisible by some of the qs, and then we're done -- that is, if the Fundamental Theorem of Arithmetic can be proved without this fact, which may or may not be possible.", "Solution_6": "thats exactly the problem.\r\n\r\nyou need it to prove the fundamental theorem of arithmetic.\r\n\r\nBasically you need to go\r\n\r\nWOP -> Div. Algorithm -> Linear Diophantine Eq -> thm i just stated -> fundamental thrm of arithmetic.\r\n\r\n\r\nLinear Dio Eq:\r\nGiven a,b integers\r\n ax + by = 1 is solvable in integers x,y iff (a,b) = 1." } { "Tag": [ "IMO 2004", "Social" ], "Problem": "Can any Mathlinks forum member who is going to take part as a contestant in IMO 2004 in Athens leave a message here? :) It would be nice to see who will be going.", "Solution_1": "I'll be going :)", "Solution_2": "wow, finally someone admitted that he/she is gonna be present at IMO :D ;)", "Solution_3": "I think I'll go too.\r\nthe same with DusT", "Solution_4": "I suppose this will end up being a huge \"looking for MathLinks people\" session :) Maybe the Romanian members should coordinate it ;)\r\n\r\nIt'll probably be pretty obvious to spot me, being a girl and all... :blush:", "Solution_5": "The post above was by me...hm, how did I get logged out :?", "Solution_6": "[quote=\"Anonymous\"]I suppose this will end up being a huge \"looking for MathLinks people\" session :) Maybe the Romanian members should coordinate it ;)\n\nIt'll probably be pretty obvious to spot me, being a girl and all... :blush:[/quote] :) yeah well if I manage to go too (afterall) I will probably be wearing a nice MathLinks t-shirt :P and who knows maybe I'll bring some spares :D", "Solution_7": "[quote=\"Valentin Vornicu\"][quote=\"Anonymous\"]I suppose this will end up being a huge \"looking for MathLinks people\" session :) Maybe the Romanian members should coordinate it ;)\n\nIt'll probably be pretty obvious to spot me, being a girl and all... :blush:[/quote] :) yeah well if I manage to go too (afterall) I will probably be wearing a nice MathLinks t-shirt :P and who knows maybe I'll bring some spares :D[/quote]\r\nQuestion : \r\nHOW many girls have username on mathlinks and how many of them will go to IMO this year ? :lol: Don't be :blush: !!!", "Solution_8": "valiowk is a girl\r\n\r\nArne, iura, dust aren't\r\n\r\nso that makes a 75%-25% balance for now... not bad :)", "Solution_9": "me too i'm going", "Solution_10": "I feel so old and tired... :) I cannot go to IMO and so I will never participate at this competition. Too bad. But I wish to all of you goooooood luck.", "Solution_11": "I won't go too :( but I agree to wear the mathlinks t-shirt at home :D \r\nHey Valentin, do not forget to send some t-shirts with you book when it will be available.\r\n\r\nPierre.", "Solution_12": "[quote=\"pbornsztein\"]I won't go too :( but I agree to wear the mathlinks t-shirt at home :D \nHey Valentin, do not forget to send some t-shirts with you book when it will be available.\n\nPierre.[/quote] well the t-shirt idea depends whether I can find a sponsor or not :P but hopefully I will.", "Solution_13": "[quote=\"Valentin Vornicu\"][quote=\"pbornsztein\"]I won't go too :( but I agree to wear the mathlinks t-shirt at home :D \nHey Valentin, do not forget to send some t-shirts with you book when it will be available.\n\nPierre.[/quote] well the t-shirt idea depends whether I can find a sponsor or not :P but hopefully I will.[/quote]\r\nI WANT A ML T-SHIRT TOO !!!! And what book is Pierre refering to ???... :?", "Solution_14": "ok, let's not deviate too much from this IMO topic, ok? :) people going to IMO should post here :P \r\n\r\nalex please do not use caps locks anymore, ok? :) he is probably talking about the translation (and add-ups) of my book in English ... which I didn't have time to complete yet :( ... but it will be done in the near future :P :)", "Solution_15": "i'm leaving tomorrow morning(just a 2 hours flight :D ). hope to see you as well.\r\n\r\nPeter", "Solution_16": "Amazingly, we (moldavian team) are going also tommorow, but we have a 3 :? hours flight, and Moldova is closer to Greece than Germany. I guess you will be flying on something very fast :D:)", "Solution_17": "hmm...maybe it were 3 hours, but anyway not too long(we're taking a normal plane, nothing special :D ).\r\n\r\nPeter", "Solution_18": "Well, our plain departs in Frankfurt at 12:55, and we arrive in Athens at 16:40. Taking into account the time difference of 1 hour, the flight dures 2.45 h.\r\n\r\n Darij", "Solution_19": "I'm at the airport in Athens right now. Our flight arrived one hour early (!!), so I have a lot of time to burn while waiting for the UK and Israel teams to arrive.\r\n\r\nUnfortunately there is also a 10-minute Internet limit and the connection is really slow :(\r\n\r\nTo those who haven't gotten on the plane yet, have a safe trip!", "Solution_20": "I am leaving in 7 hours only. We're one of the last teams to arrive. :?", "Solution_21": "Vietnamese team arrived on 8 July.\r\nI hope i'll meet the European Champions :D :D", "Solution_22": "[quote=\"pbornsztein\"]Do not forget to take some pictures of all of you...\n\nPierre.[/quote]\r\n\r\nYeah, I completely agree with Pierre. Something like the IMO2001 page with lots of contestants' pictures and some vita/preferences/interests stuff would be very nice. :)\r\n\r\nBTW will Valentin distribute any AOPS/ML T-Shirts at the IMO ? :D", "Solution_23": "Hi Valentin,\r\n\r\nI would like to have a mathlink T-shirt, large size.\r\n\r\nThx oubi", "Solution_24": "So am I :) \r\n\r\nPierre.", "Solution_25": "So any pictures of all of you?\r\n\r\nSome are available here (french team) :\r\n http://membres.lycos.fr/imo04/Photos%20Antony/\r\n\r\nPierre.", "Solution_26": "it looks as if there's an Asian in French team. Which country is he from?", "Solution_27": "I think you are talking about Antony Lee. As far as I know, his family comes from Taiwan.\r\n\r\nPierre.", "Solution_28": "we (romanian team) will also post some pictures later :)", "Solution_29": "yeah, I'm from Taiwan, and actually, I'm now in Taiwan (for vacation)" } { "Tag": [ "AMC", "AIME" ], "Problem": "On one of the proof writing tutorial pages, there was a problem to find S(S(S(4444^4444))) where S(x) is the sum of the digits of x. I don't understand the proof for why S(S(S(4444^4444))) is congruent 7 mod 9. Can someone please explain this to me?\r\n\r\nAlso, is there any place I can find a copy of all the 2005 AIME questions?\r\n\r\nThanks!", "Solution_1": "Can't help you with your problems, but I can tell you where to find this year's AIME. If you click \"math jams\" at the top of the page, you'll find a list of upcoming math jams. Math jams are where the admins give a list of problems from a test, ie. AIME, and walk people through detailied solutions. Click \"trancripts\" to find transcripts of old math jams. Both AIMEs of this year are there.", "Solution_2": "compute $4444^4444$ modulo $9$ ..", "Solution_3": "What do you mean?\r\n\r\ndeej21, thanks!", "Solution_4": "[quote=\"alekk\"]compute $4444^4444$ modulo $9$ ..[/quote]\r\n\r\nalekk, Ithink you forgotr to put {} around your 4444; it should look like this:4444^{4444}", "Solution_5": "The proof relies on two results: \r\n\r\n1)If $x=amod9$, then $S(x)=amod9$. \r\n\r\n2)$4444^{4444} mod9=(4446-2)^{4444}mod9=(-2)^{4444}mod9=(-8)^{1479}(-128)mod9=-128mod9=7$", "Solution_6": "The problem is actually IMO 1975 problem 4... See http://www.mathlinks.ro/Forum/viewtopic.php?t=31409 .\r\n\r\n darij" } { "Tag": [ "Euler", "geometry", "circumcircle", "geometry proposed" ], "Problem": "(V.Protasov, 9--10) The Euler line of a non-isosceles triangle is parallel to the bisector of one of its angles. Determine this\r\nangle (There was an error in published condition of this problem).", "Solution_1": "The angle is 120", "Solution_2": "Let $ O,H,N$ be the circumcenter, orthocenter and 9-point center of $ \\triangle ABC.$ Assume that $ OH$ is parallel to the internal bisector of $ \\angle BAC.$ Which implies that $ HO$ bisects $ \\angle BHC,$ since $ HB,HC$ and the external bisector of $ \\angle BAC$ bound an isosceles triangle with apex $H.$ Hence if $ U,V$ denote the midpoints of $ HC,HB,$ then the quadrilateral $ HUNV$ is cyclic, where $ N$ is the midpoint of the arc $ UV$ of its circumcircle. By simple angle chase we obtain then\n\n$ \\angle UNV \\plus{} \\angle BHC \\equal{} 180^{\\circ} \\Longrightarrow 2\\angle BHC \\plus{} \\angle BHC \\equal{} 180^{\\circ}$ \n\n$ \\Longrightarrow \\angle BHC \\equal{} 60^{\\circ} \\Longrightarrow \\angle BAC \\equal{} 120^{\\circ}.$", "Solution_3": "easy problem.just see that $ HO$ bisects $ BHC$ and $ BO\\equal{}CO$ so $ BHCO$ is cyclic.so $ 360\\minus{}2A\\plus{}180\\minus{}A\\equal{}180$ so $ A\\equal{}120$" } { "Tag": [ "Pascal\\u0027s Triangle", "combinatorics unsolved", "combinatorics" ], "Problem": "Show that\r\n\\[\\sum_{i=0}^{n}\\binom{n}{i}^{2}=\\binom{2n}{n}\\]", "Solution_1": "Consider the coefficient of $x^{n}$ in the expansion\r\n\r\n$(1+x)^{2n}= \\left( (1+x)^{n}\\right)^{2}$\r\n\r\nAnd the result follows immediately. \r\n\r\nAlternately, a path-counting solution down Pascal's triangle to ${2n \\choose n}$ is equivalent.\r\n\r\nFinally, using ${n \\choose i}={n \\choose n-i}$, consider the union $U$ of two disjoint sets $A, B$ of $n$ elements each. Then count the number of ways to create $n$-subsets of $U$ by selecting $k$ elements from $A$ and $n-k$ elements from $B$.", "Solution_2": "Hmmm... just posted about 2 days ago most recently...\r\n\r\n(you might as well have proven Vandermonde's)" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let x and y be distinct positive reals. Prove:\r\n\r\n1/2*(x+y) < (x-y)/(2*arctan((x-y)/(x+y))) < (1/2*(x^2+y^2))^0.5 < (x^x*y^y)^(1/(x+y))", "Solution_1": "hello orl, i have found this inequality, do you meant\r\n${ \\frac{1}{2}(x+y)<\\frac{x-y}{2\\arctan\\left(\\frac{x-y}{x+y}\\right)}}<\\left(\\frac{1}{2}(x^2+y^2)\\right)^{\\frac{1}{2}}<\\left(x^xy^y\\right)^{\\frac{1}{x+y}}$?\r\nIs this true?\r\nSonnhard." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "For a given integer $ n\\geq 2,$ determine the necessary and sufficient conditions that real numbers $ a_{1},a_{2},\\cdots, a_{n},$ not all zero satisfy such that there exist integers $ 00$. For necessity, note that we can write\r\n\\[ \\sum a_ix_i \\equal{} x_1(a_1\\plus{}\\dots\\plus{}a_n)\\plus{}(x_2\\minus{}x_1)(a_2\\plus{}\\dots\\plus{}a_n)\\plus{}(x_3\\minus{}x_2)(a_3\\plus{}\\dots\\plus{}a_n)\\plus{}\\dots\\plus{}(x_n\\minus{}x_{n\\minus{}1})a_n.\\] \r\nSince the differences of the $ x_i$ are by assumption positive, at least one of the sums of the $ a_i$ must be positive. \r\n\r\nFor sufficiency, let us suppose that the statement holds for some $ k$. Then let $ x_1\\equal{}1, x_2\\equal{}2, \\dots x_{k\\minus{}1}\\equal{}k\\minus{}1, x_k\\equal{}m\\plus{}k, x_{k\\plus{}1}\\equal{}m\\plus{}k\\plus{}1, \\dots x_n\\equal{}m\\plus{}k\\plus{}n$, where $ m$ is to be determined later. \r\n\r\nThen $ \\sum a_ix_i$ has the form $ m(x_{k}\\plus{}x_{k\\plus{}1}\\plus{}\\dots\\plus{}x_n)\\plus{}c$, where $ c$ is some constant not depending on $ m$. For large enough $ m$, this is positive." } { "Tag": [ "linear algebra", "matrix", "linear algebra solved" ], "Problem": "Prove that two similar matrix have the same rank.", "Solution_1": "the two matrices represent the same endomorphism (in different basis), sothey have same rank.", "Solution_2": "Suppose $A$ is similar to $B$ and $A$ has rank $k.$ $A$ had rank $k$ iff there exists an invertible matrix $E$ such that $EA=M,$ where $M$ is in row echelon form and has $k$ pivots, hence exactly $k$ nonzero rows.\r\n\r\nThere exists an invertible $P$ such that $A=PBP^{-1}.$\r\n\r\nThen $EA=EPBP^{-1}=M$\r\n\r\n$(EP)B=MP$\r\n\r\nSo $B$ can be row reduced to $MP$, but if we examine the definition of matrix multiplication, we see that $MP$ has no more than $k$ nonzero rows. That means that the rank of $MP$ and hence the rank of $B$ is no more than $k.$\r\n\r\nThus $\\text{rank}(B)\\le\\text{rank}(A).$\r\n\r\nSince similarity is a symmetric relation, we have by the same argument that\r\n\r\n$\\text{rank}(A)\\le\\text{rank}(B).$", "Solution_3": "Actually, it's pretty well-known that multiplication with an invertible matrix does not change the rank: whenever $T$ is invertible, $r(AT)\\le r(A)$ on the one hand, and, on the other hand, $r(A)=r(AT\\cdot T^{-1})\\le r(AT)$.", "Solution_4": "i think that this question was giving in the class :D", "Solution_5": "[quote=\"grobber\"]Actually, it's pretty well-known that multiplication with an invertible matrix does not change the rank: whenever $T$ is invertible, $r(AT) = r(A)$.[/quote]\r\n\r\nIs there an elementary proof of this?", "Solution_6": "See Kent Merryfield's post in this thread, and modify appropriately." } { "Tag": [ "geometry", "circumcircle", "national olympiad" ], "Problem": "Download [url=http://www.mathlinks.ro/Forum/download.php?id=2268]PDF Version[/url] of the problems.\r\n\r\n[b]Day 1[/b]\r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=242547]Problem 1.[/url]\r\nLet $x$, $y$ be two positive integers such that $3x^2+x=4y^2+y$.\r\nProve that $x-y$ is a perfect square.\r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=242550]Problem 2.[/url]\r\nTwo right angled triangles are given, such that the incircle of the first one is equal to the circumcircle of the second one.\r\nLet $S$, respectively $S'$, be the areas of the first triangle and respectively of the second triangle.\r\n\r\nProve that $\\displaystyle \\frac{S}{S'}\\geq 3+2\\sqrt{2}$.\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=242552]Problem 3.[/url]\r\nIn an international meeting of $n \\geq 3$ participants, 14 languages are spoken. We know that:\r\n\r\n- Any 3 participants speak a common language.\r\n- No language is spoken more that by the half of the participants.\r\n\r\nWhat is the least value of $n$?\r\n\r\n[b]Day 2[/b]\r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=243242]Problem 4.[/url]\r\nDefine $[u]$ as the integer part of any real $u$. Let $X$ be a non empty subset of $\\mathbb{N}^*$.\r\nSuppose that for all $x \\in X$, $4x \\in X$ and $[\\sqrt{x}] \\in X$.\r\n\r\nProve that $X=\\mathbb{N}^*$.\r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=243246]Problem 5.[/url]\r\nLet $ABC$ be a triangle such that $BC=AC+\\frac{1}{2}AB$. Let $P$ be a point of $AB$ such that $AP=3PB$.\r\n\r\nShow that $\\widehat{PAC} = 2 \\widehat{CPA}$.\r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=243253]Problem 6.[/url]\r\nLet $P$ be a polynom of degree $n \\geq 5$ with integer coefficients:\r\n\\[ P(x)=a_{n}x^n+a_{n-1}x^{n-1}+\\cdots+a_0 \\quad \\textrm{ with } \\ a_i \\in \\mathbb{Z},\\ a_n \\neq 0. \\]\r\nSuppose that $P$ has $n$ different integer roots: $0,\\alpha_2,\\ldots,\\alpha_n$. Find all integers $k \\in \\mathbb{Z}$ such that $P(P(k))=0$.", "Solution_1": "Is it a day with 21 points Igor? :P \r\n\r\nPierre.", "Solution_2": "No ... I expect (12-)13-14.\r\n\r\nTomorrow will be a 21 point day :P", "Solution_3": "good luck tomorrow Igor!", "Solution_4": "Erf .... a 14 (maybe 15) day :oops:", "Solution_5": "Wow, so a total of 28-30 points. I don't know if it's good or bad in the French TST :) These ones seem a little bit more difficult than last year's tests, isn't that true Pierre? :P :)", "Solution_6": "Bruno will get 42 :roll:", "Solution_7": "So I guess 30 is that bad? :(", "Solution_8": "I will be something like 2nd (with a lot of luck) or 4th-5th (with no luck) I think ...", "Solution_9": "That's good :) I have edited your first post to include the problems and the pdf version of them! Nice choice of problems there!", "Solution_10": "Thanks !\r\n\r\nBy the way, the working time is 4.5 hours and not 4 hours ;)", "Solution_11": "Oh OK :) I will rebuild the .pdf file :P :)", "Solution_12": "Isn't Problem 4, from Japan 1990 :) ?" } { "Tag": [], "Problem": "evaluate $\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{...}}}}}}$", "Solution_1": "[hide=\"Hint\"]What happens when you square it and subtract 2?[/hide]", "Solution_2": "[hide][quote=\"daermon\"]evaluate $\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{...}}}}}}$[/quote]\n\n2\njust keep subtracting and minusing 2[/hide]", "Solution_3": "[quote=\"moogra\"][hide][quote=\"daermon\"]evaluate $\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{2+...}}}}}}$[/quote]\n\n2\njust keep subtracting and minusing 2[/hide][/quote]\n :huh: Your solution doesn't make too much sense. Can you clarify?\n[hide=\"a hint\"]Solve this like an infinite summation[/hide]\n[hide=\"a solution\"]$\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{...}}}}=x$\nSquaring gives you $2+x=x^{2}$\n$x^{2}-x-2=0$\n$(x-2)(x+1)$\n-1 is clearly extraneous :|\ntherefore the answer is 2.[/hide]\n[hide=\"a hint for the VERY unlikely case that this is a target problem seeing as a version of this was on a sprint before\"]If by some unlikely case this is a target problem then you can find the answer by punching in the first five or so sqrts and see what it tends to.[/hide]", "Solution_4": "[quote=\"daermon\"]evaluate $\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{...}}}}}}$[/quote]\r\n\r\n[hide=\"solution\"]\nLet $x$ = $\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{...}}}}}}$\n$x^{2}=2+\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{...}}}}}}$\n$x^{2}=2+x$\n$x^{2}-x-2=0$\n$(x+1)(x-2)=0$\nprincipal sq rt cannot be negative\n$x=2$\n[/hide]\nEdit-(I know ur gonna ask for generalization...)\n[hide=\"Generalization\"]\nfor $a \\geq-\\frac{1}4$\n$x^{2}-x-a=0$\n$x^{2}-x+\\frac{1}4=a+\\frac{1}4$\n$x-\\frac{1}2=\\sqrt{a+\\frac{1}4}$\n$x-\\frac{1}2=\\frac{\\sqrt{4a+1}}2$\nif $a>0$ then\n$x=\\frac{1+\\sqrt{4a+1}}2$\nElseif ($a\\leq 0$) $AND$ ($a\\geq-\\frac{1}4$) then\n$x=\\frac{1+\\sqrt{4a+1}}2$ or $x=\\frac{1-\\sqrt{4a+1}}2$\nElse ($a\\leq-\\frac{1}4$) then\nNo Solution\n[/hide]", "Solution_5": "bpms and protestanT are correct! :D", "Solution_6": "[hide=\"solution\"]We can call that $\\omega$. Therefore,\n\n$\\omega=\\sqrt{2+\\sqrt{2+\\sqrt{2+\\cdot\\cdot\\cdot}}}$\n\nWe notice that\n\n$\\omega=\\sqrt{2+\\omega}$\n\nSo we can solve:\n\n$\\omega^{2}=2+\\omega$\n$\\omega^{2}-\\omega-2$\n$(\\omega-2)(\\omega+1)=0$\n$\\omega=[2,-1]$\n\nSince we want the positive value, $\\boxed{2}$ is the answer.[/hide]\r\n\r\nDo you want a generalzation? :gleam:", "Solution_7": "[hide=\"Generalization\"]\n\\begin{eqnarray*}\\sqrt{n+\\sqrt{n+\\sqrt{n+\\sqrt{\\ldots}}}}&=&x \\\\ \\sqrt{n+x}&=& x \\\\ n+x&=&x^{2}\\\\ x^{2}-x-n&=&0 \\\\ \\frac{-(-1)\\pm\\sqrt{(-1)^{2}-4(1)(n)}}{2(1)}&=&\\boxed{\\frac{1\\pm \\sqrt{1+4n}}{2}}\\\\ \\end{eqnarray*}\nOf course, we are taking the positive value.[/hide]", "Solution_8": "that is correct as well :D", "Solution_9": "[quote=\"ragnarok23\"][hide=\"Generalization\"]\n\\begin{eqnarray*}\\sqrt{n+\\sqrt{n+\\sqrt{n+\\sqrt{\\ldots}}}}&=&x \\\\ \\sqrt{n+x}&=& x \\\\ n+x&=&x^{2}\\\\ x^{2}-x-n&=&0 \\\\ \\frac{-(-1)\\pm\\sqrt{(-1)^{2}-4(1)(n)}}{2(1)}&=&\\boxed{\\frac{1\\pm \\sqrt{1+4n}}{2}}\\\\ \\end{eqnarray*}\nOf course, we are taking the positive value.[/hide][/quote]For this generilazation is protestanT's distributed form better or is ragnarok's?", "Solution_10": "oops i forgot to rationalize denominator" } { "Tag": [ "analytic geometry", "graphing lines", "slope", "calculus", "derivative", "calculus computations" ], "Problem": "I was assigned a problem by my Calc teacher that requires me to model a ski jump first with a single curve and then with multiple.\r\n\r\nI have been working out the details for further along in the process, but I got stuck at the very first part. I have deduced the following:\r\n\r\nthe curve at the takeoff point, (0,0), must have a slope (derivative) of tan(10.5)=.1853\r\nthe curve must intersect the point (144, 100) (slope does not matter there)\r\nat some point (x, y) the curve must have slope (derivative)= tan(39)=.809784\r\n\r\nI can take it from there, but I am struggling with this part. Any ideas?", "Solution_1": "[quote=\"aveld\"]I was assigned a problem by my Calc teacher that requires me to model a ski jump first with a single curve ...\n[/quote]\r\n\r\nWhat single curve are you planning on using?\r\n\r\nHow about a Lorentz, 1/(1 + x*x) ?" } { "Tag": [], "Problem": "For a natural number n, n^2 gives a remainder of 4 when divided by 5, and n^3 gives a remainder of 2 when divided by 5. What is the remainder when n is divided by 5?", "Solution_1": "[quote=\"easyas3.14159...\"]For a natural number n, n^2 gives a remainder of 4 when divided by 5, and n^3 gives a remainder of 2 when divided by 5. What is the remainder when n is divided by 5?[/quote]\r\n\r\n[hide]i got \n[size=200]2[/size][/hide]", "Solution_2": "[hide]One way of doing this is to substitute real numbers into $n$. One number that fits the conditions is 3. So, the answer is $3$ because $3 \\div 5=0 \\ R3$.[/hide]", "Solution_3": "Well actually, the answer is [hide]3[/hide]", "Solution_4": "[quote=\"robinhe\"][hide]One way of doing this is to substitute real numbers into $n$. One number that fits the conditions is 3. So, the answer is $3$ because $3 \\div 5=0 \\ R3$.[/hide][/quote]\r\n\r\ncan you solve without guess and check? >.> i tried without guess and check but i got the answer wrong", "Solution_5": "[quote=\"easyas3.14159...\"]For a natural number n, n^2 gives a remainder of 4 when divided by 5, and n^3 gives a remainder of 2 when divided by 5. What is the remainder when n is divided by 5?[/quote]\r\n\r\n[hide=\"Answer\"]\n\n $n^2 = 4$ (mod 5)\n\n $n^3 = 2$ (mod 5)\n\n Substitue the value of $n^2$ into the second equation:\n\n $n^3=(n^2) \\cdot n=4\\cdot n=4n=2$ (mod 5)\n\n $4n=2=12$ (mod 5)\n\n $n=3$ (mod 5)\n\n Hence, $n$ yields a remainder 3 on division by 5. [/hide]" } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "algebra unsolved", "algebra" ], "Problem": "$ a_{1} \\equal{} 1,a_{2} \\equal{} 2,a_{n}a_{n \\plus{} 2} \\equal{} 1 \\plus{} a_{n \\plus{} 1}^{3} (n\\geq1)$\r\n(1) proof that all $ a_(n) \\in Z^{ \\plus{} }$\r\n(2) find $ a_(n) (n\\geq1)$", "Solution_1": "[quote=\"admire9898\"]$ a_{1} \\equal{} 1,a_{2} \\equal{} 2,a_{n}a_{n \\plus{} 2} \\equal{} 1 \\plus{} a_{n \\plus{} 1}^{3} (n\\geq1)$\n(1) proof that all $ a_(n) \\in Z^{ \\plus{} }$\n(2) find $ a_(n) (n\\geq1)$[/quote]\r\n\r\nI think it is not a real olympiad or olympiad preparation problem.\r\n\r\nSee http://www.research.att.com/~njas/sequences/A003818\r\n\r\nIt seems nobody yet gave there a closed formula for this sequence ... \r\n\r\nMaybe admire9898 may give us the origin of this problem, and how does he know there is a solution.", "Solution_2": "[quote=\"pco\"]\n\nI think it is not a real olympiad or olympiad preparation problem.\n\nMaybe admire9898 may give us the origin of this problem, and how does he know there is a solution.[/quote]\r\n\r\nthis problem is from a chinese contest like AIME in USA.BUT NOT equivalent to USAMO\r\nthe (1) is only a lamma to solve that problem\r\nbut (2) is my guess to this problem :P :D I'm interested to this sequence if it really have a the formula of general term", "Solution_3": "[quote=\"admire9898\"]this problem is from a chinese contest ... [/quote]\r\n\r\nThanks for this answer. So we now are sure there is a general closed formula for this sequence. \r\n\r\nDont hesitate to post this formula when the solutions of the contest will be published.", "Solution_4": "Did the solutions of the contest publish ?" } { "Tag": [ "function", "induction", "search", "complex analysis", "real analysis", "real analysis unsolved" ], "Problem": "Easy one but interesting:\r\nlet $f$ be an increasing (no continuity etc ..) function on $\\mathbb{R}$.\r\nProve that there is a power series $g(x)=a_0+a_1x+..+a_kx^k+..$, with infinite radius of convergence that satisfies:\r\n$g(x)>f(x)$ for all $x \\in \\mathbb{R}$", "Solution_1": "What's your solution ? Mine is kind of tedious to write: on each interval $[n,n+1]$ (where $n \\in \\mathbb{Z}$), $f$ is bounded. Note $M_{n} \\in \\mathbb{Z}$ one of its upper bounds on this interval. Now we construct a power series $S(x) = \\sum a_{n}\\frac{x^{2n}}{n!}$ with coefficients $(a_{n})$ as follows:\r\n$\\forall n \\in [0,M_{1}],a_{n}=M_{0}$.\r\n $\\forall n \\in [M_{1},M_{2}+T_{2}],a_{n}=M_{1}$, where $T_{2}$ is large enough in order to guarantee that $min(S^{*}(x),x \\in [1,2]) \\geq M_{1}$, where $S^{*}$ has the coefficients previously defined on $[0,M_{1}]$ and coefficients stationary to $M_{1}$ for $n \\geq M_{1}$.\r\nAnd so on by induction on $[M_{2}+T{2},M_{3}+T_{3}]$, etc ... Using also the parity of $S$ and its relation to $e^{x^{2}}$, the radius of convergence is clear and the majoration too.", "Solution_2": "it is easy to check that the problem can be reduced to the following:\r\nLet $b_0,b_1,..,b_k,..$ be arbitrary real numbers\r\nShow that there exists an even entire function $f$ (i.e. $f(x)=f(-x)$) that satisfies $f(n) \\geq b_n$ for $n \\geq 0$.\r\n\r\nWe will search $f$ under the shape $f(x)=a_0+a_1x^{p_1}+..+a_nx^{p_n}+..$ where $a_k > 0$, $p_{n+1}>p_n$, $p_i$ is an even integer.\r\n*To assure the convegence, we will take $(a_n, p_n)$ such that $|a_nx^{p_n}| < \\frac{1}{n^2}$ for all $x \\in [0;\\sqrt{n}]$.\r\n*to assure the hypothesis we will also take $(a_n, p_n)$ such that $a_nn^{p_n} > b_n$\r\n\r\nIt suffices that : $a_nn^{\\frac{p_n}{2}}<\\frac{1}{n^2}$ and $a_nn^{p_n} > b_n$.\r\ni.e. $a_n<\\frac{1}{n^{\\frac{p_n}{2}+2}}$ and $a_n>\\frac{b_n}{n^p_n}$.\r\n\r\nBut for $p_n$ large: $\\frac{b_n}{n^{p_n}}<\\frac{1}{n^{\\frac{p_n}{2}+2}}$. So we can find $a_n$.\r\nConclusion follows." } { "Tag": [ "function", "LaTeX", "national olympiad" ], "Problem": "\\[\\text{Day I}\\]\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=182624#p182624]Problem 1 Discussion[/url]\r\nLet $x,y$ be real numbers satisfying the condition:\r\n\\[x-3\\sqrt{x+1}=3\\sqrt{y+2}-y \\]\r\nFind the greatest value and the smallest value of:\r\n\\[P=x+y \\]\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=182627#p182627]Problem 2 Discussion[/url]\r\nLet $(O)$a fixed circle with the radius $R$.Let $A$ and $B$ be fixed points in $(O)$ such that $A,B,O$ are not collinear.\r\nConsider a variable point $C$ lies on $(O)(C\\neq A,B)$.Construct two circles $(O_{1}),(O_{2})$ passed through $A,B$ and tangent $BC,AC$ at $C$,respectively.The circle $(O_{1})$ intersect the circle $(O_{2})$ in $D(D\\neq C)$.Prove that:\r\na)\r\n\\[CD\\leq R \\]\r\nb)The line $CD$ passed through point independent of $C$(i.e there exist fixed point on the line $CD$ when $C$ lies on $(O)$.\r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=182628#p182628]Problem 3 Discussion[/url]\r\nLet $A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}A_{7}A_{8}$ be convex 8-gon (no three diagonals concruent).\r\nThe intersection of arbitrary two diagonals will be called \"button\".Consider the convex quadrilaterals formed by four vertices of $A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}A_{7}A_{8}$ and such convex quadrilaterals will be called \"sub quadrilaterals\".Find the smallest $n$ satisfying:\r\nWe can color n \"button\" such that for all $i,k \\in\\{1,2,3,4,5,6,7,8\\},i\\neq k,s(i,k)$ are the same where $s(i,k)$ denote the number of the \"sub quadrilaterals\" has $A_{i},A_{k}$ be the vertices and the intersection of two its diagonals is \"button\".\r\n\\[\\text{Day II}\\]\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=183462#183462]Problem 4 Discussion[/url]\r\nFind all function $f: \\mathbb R\\to \\mathbb R$ satisfying the condition:\r\n\\[f(f(x-y))=f(x)\\cdot f(y)-f(x)+f(y)-xy \\]\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=183463#183463]Problem 5 Discussion[/url]\r\nFind all triples of natural $(x,y,n)$ satisfying the condition:\r\n\\[\\frac{x!+y!}{n!}=3^{n}\\]\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=183465#183465]Problem 6 Discussion[/url]\r\nLet $\\{x_{n}\\}$ be a real sequence defined by:\r\n\\[x_{1}=a,x_{n+1}=3x_{n}^{3}-7x_{n}^{2}+5x_{n}\\]\r\nFor all $n=1,2,3...$ and a is a real number.\r\nFind all $a$ such that $\\{x_{n}\\}$ has finite limit when $n\\to+\\infty$ and find the finite limit in that cases.", "Solution_1": "kunny: why? :D", "Solution_2": "[quote=\"chavez\"]kunny: why? :D[/quote]\r\n\r\nIt's secret. :D", "Solution_3": "kunny: Nice.", "Solution_4": "Please, everyone which posted usefull information in this thread, repost these in the first topics (the ones in which the problems are discussed). They will be deleted if you do not do that... Thanks!", "Solution_5": "Wow! You are the fastest one today,Pigfly! :D \r\nThe exam has just taken this morning and in the evening, I see it'problem here\r\nHow do you do in teh exam? :D", "Solution_6": "I have just posted 3 problems come from Day II :) .See #1\r\n@Minh Khoa:How did you do in the exam? :)", "Solution_7": "Wow!You did 5 problems\r\nXin chuc mung\r\nI did'nt solve problem 3 and problem 6 in case 0\\frac{\\sqrt{3}}{2}$, then $P$ is contained inside $ORIQ$, which is obviously a contradiction." } { "Tag": [ "algebra", "partial fractions" ], "Problem": "My friend gave me this problem and I was able to solve it, but I thought I should post it here so that others can learn from it.\r\n\r\nEvaluate $\\sum_{k=1}^{\\infty}\\frac{k}{(k+1)(k+2)^2}$\r\n\r\nThe only snag on this problem is that you will have to know a reasonably well known sum.", "Solution_1": "[hide]\n\n\n\nUsing partial fractions, the equation becomes \n\n\n\nThe first pair of terms telescope leaving simply . The second, using the fact that , it becomes . The sum of these two is the answer, which is .[/hide]", "Solution_2": "Excellent!" } { "Tag": [ "geometry", "circumcircle", "incenter", "angle bisector", "geometry proposed" ], "Problem": "[b][size=100][color=DarkBlue]A triangle $ \\bigtriangleup ABC$ is given and let $ A'$ be, the reflexion of $ A,$ with respect to the midpoint $ M$ of its side-segment $ BC.$ The line through $ A'$ and parallel to the internal angle bisector of $ \\angle A,$ intersects the circumcircle $ (K)$ of the triangle $ \\bigtriangleup BIC,$ where $ I$ is the incenter of $ \\bigtriangleup ABC,$ at point so be it $ P.$ Prove that $ BD \\equal{} DE \\equal{} EC,$ where $ D\\equiv AB\\cap CP$ and $ E\\equiv AC\\cap BP.$[/color][/size][/b]\r\n\r\nKostas Vittas.", "Solution_1": "Assume there exists the points $ D$ and $ E$ on $ BA$ and $ AC$ such that $ BD \\equal{} DE \\equal{} EC$ and $ P' \\equiv BE \\cap CD.$ We'll prove that $ P$ and $ P'$ are necessarily identical. It's easy to see that $ P'$ lies on $ \\odot(IBC),$ due to\n\n$ \\angle DP'B \\equal{} \\angle EDC \\plus{} \\angle DEB \\equal{} \\frac {_1}{^2}(\\angle AED \\plus{} \\angle ADE) \\ \\Longrightarrow$\n\n$\\angle DP'B \\equal{} \\frac {_1}{^2}(\\pi \\minus{} \\alpha) \\equal{} \\pi \\minus{}\\angle BIC$ $ \\Longrightarrow \\ BIP'C$ is cyclic.\n\nOn the other hand, it is well-known that the locus of the intersection of the diagonals of the convex quadrilateral $ BD'E'C$ satisfying $ BD' \\equal{} E'C$ is the A'-angle bisector of the antimedial triangle $ \\triangle A'B'C',$ which is the parallel line to $ AI$ from $ A'.$ This was discussed in the topic [url=http://www.artofproblemsolving.com/viewtopic.php?t=300736]plane geometry 2[/url], hence $ P' \\equiv P$ and the proof is completed.", "Solution_2": "[b]Remark:[/b]\r\n\r\n[i][b][color=blue]If $ \\widehat {BAC} \\equal{} 60^\\circ$, then $ P \\equiv O$, where $ O$ is the circumcenter of $ \\triangle ABC$[/color][/b][/i].\r\n\r\nBest regards,\r\nsunken rock" } { "Tag": [ "topology", "advanced fields", "advanced fields unsolved" ], "Problem": "Prove that two metric spaces are homeomorphic iff their [url=http://planetmath.org/encyclopedia/StoneVCechCompactification.html]Stone-Cech compactifications[/url] are homeomorphic.", "Solution_1": "The same is true for completely regular spaces, wich satisfy the first axiom of countability at all points." } { "Tag": [ "Putnam", "quadratics", "geometry", "3D geometry", "function", "Stanford", "college" ], "Problem": "Show that the curve $x^{3}+3xy+y^{3}=1$ contains only one set of three distinct points, $A,B,$ and $C,$ which are the vertices of an equilateral triangle.", "Solution_1": "This curve is Folium of Decarte. This problem is interesting to me. :)", "Solution_2": "But this particular curve is degenerate. Factor it!", "Solution_3": "$x^{3}+3xy+y^{3}= 1$\r\n$(x+y)^{3}-3x^{2}y-3xy^{2}+3xy-1=0$\r\n$(x+y)^{3}-1^{3}-3xy(x+y-1) = 0$\r\n$(x+y-1)((x+y)^{2}+(x+y)+1)-3xy(x+y-1)=0$\r\n$(x+y-1)(x^{2}-xy+y^{2}+x+y+1) = 0$\r\n$(x+y-1)((x-y)^{2}+(x+1)(y+1)) = 0$", "Solution_4": "[quote=\"Kent Merryfield\"]But this particular curve is degenerate. Factor it![/quote]\r\n\r\nI know, $x^{3}+y^{3}+3xyz=1\\Longleftrightarrow x^{3}+y^{3}+(-1)^{3}-3xy(-1)=0$\r\n$\\Longleftrightarrow (x+y-1)\\{x^{2}+y^{2}+(-1)^{2}-xy-x(-1)-y(-1)\\}=0$$\\Longleftrightarrow (x+y-1)(x^{2}+y^{2}-xy+x+y+1)=0.$", "Solution_5": "And what is the set $x^{2}+y^{2}-xy+x+y+1=0?$", "Solution_6": "[quote=\"Kent Merryfield\"]And what is the set $x^{2}+y^{2}-xy+x+y+1=0?$[/quote]\r\nIt's the point (-1,-1). Consider the equation as a quadratic in x. The discriminant is $-3(y+1)^{2}$, so x is real-valued when y=-1. Then x=-1.", "Solution_7": "[quote=\"Bictor717\"][quote=\"Kent Merryfield\"]And what is the set $x^{2}+y^{2}-xy+x+y+1=0?$[/quote]\nIt's the point (-1,-1). Consider the equation as a quadratic in x. The discriminant is $-3(y+1)^{2}$, so x is real-valued when y=-1. Then x=-1.[/quote]\r\n\r\nAlternatively, note $(x-y)^{2}+(x+1)^{2}+(y+1)^{2}= 2(x^{2}+y^{2}-xy+x+y+1$. This one I found very annoying -- I don't think it was transparent enough to be B1.", "Solution_8": "[quote=\"Kent Merryfield\"]And what is the set $x^{2}+y^{2}-xy+x+y+1=0?$[/quote]\r\n\r\nCompleting square yields what you want.\r\n\r\n$x^{2}+y^{2}-xy+x+y+1=0$\r\n\r\n$\\Longleftrightarrow x^{2}-(y-1)x+y^{2}+y+1=0.$\r\n\r\n$\\Longleftrightarrow \\left(x-\\frac{y-1}{2}\\right)^{2}+\\frac{3}{4}(y+1)^{2}=0.$\r\n\r\n$\\Longleftrightarrow (x,y)=(-1,-1).$", "Solution_9": "I wound up making headway on it and figuring out what was going on by making the substitutions $u=x+y,v=x-y.$\r\n\r\nStart by moving the $3xy$ over to the right, then adding in the middle terms needed to complete the cube:\r\n\r\n$x^{3}+3x^{2}y+3xy^{2}+y^{3}=3x^{2}y+3xy^{2}-3xy+1$\r\n\r\n$(x+y)^{3}-1=3xy(x+y-1)$\r\n\r\nNow make the substitution I was talking about, and note that $u^{2}-v^{2}=4xy:$\r\n\r\n$u^{3}-1=\\frac34(u^{2}-v^{2})(u-1).$\r\n\r\nThere's still more to go through from here, but I found that much to be useful.", "Solution_10": "I noticed very quickly that the equation looked a lot like $(x+y)^{3}$ with $x+y=1$. In fact, I should have factored $x+y-1$ immediately when I saw this! But instead (I wasn't able to focus much yesterday anyway) I messed around with it and managed to factor it. Then I used the discriminant on the second factor as did Bictor.\r\n\r\nI really like the problem.", "Solution_11": "This should make things apparent:\r\n\r\n$a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$\r\n$= \\frac12(a+b+c)[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]$\r\n\r\nThe second factor on the RHS above directly yields the point $(-1,-1)$ when we substitute $a = x,\\ b = y$ and $c =-1$.\r\n\r\nThe other factor of course gives $x+y = 1$ from which it is easy to see that there is only one set of three points, which form an equilateral triangle, whose area is $\\sqrt6$.\r\n\r\nSadly, this is the only problem I got correct. :(", "Solution_12": "I thought it actually looked like a folium of descartes (you know, with the loop), and had no idea as to how to proceed on such a problem.\r\n\r\nI moved on to B2 in about 2 minutes.", "Solution_13": "[quote=\"FieryHydra\"] which form an equilateral triangle, whose area is $\\sqrt6$.\n\n this is the only problem I got correct. :([/quote]\r\n\r\nUM.....\r\nif i recall correctly,\r\nNo, the side length of the triangle is $\\sqrt{6}$ the area is $\\frac{3\\sqrt{3}}{2}$\r\n\r\ni just cant believe i missed the easy factoring solution, but i did it long way with (-1.-1) and y=1-x and got it.\r\nhopefully a 10 on this one.", "Solution_14": "[quote=\"amirhtlusa\"][quote=\"FieryHydra\"] which form an equilateral triangle, whose area is $\\sqrt6$.\n\n this is the only problem I got correct. :([/quote]\n\nUM.....\nif i recall correctly,\nNo, the side length of the triangle is $\\sqrt{6}$ the area is $\\frac{3\\sqrt{3}}{2}$[/quote]\r\n\r\nYou are correct amirhtlusa. My bad. I do remember getting $\\sqrt6$ as the side length and $\\frac{3\\sqrt3}{2}$ as the area. Thanks!", "Solution_15": "[quote=\"APSStudent\"]I had a question about the cubic formula in general, is there a closed form for the cubic formula--in the same way there is a closed form of the quadratic formula?[/quote]\r\n1) There was no need to quote that very long post before asking your question (you could have just quoted the first couple of lines or something). \r\n\r\n2) Yes. But you'll rarely see the whole general solution written anywhere explicitly because it's way too unwieldy. There's also a general solution for quartic equations, which is similarly enormous. There's nothing in general for higher order equations, although certain special cases may be solvable. You can see the method for solving cubics in that post you quoted. It works for any equation. \r\n\r\nGenerally, in contests, if you need to solve an equation of degree 3 or higher, you'll be able to use the rational root theorem to find one of the roots by hand, or find some other clever way to guess at one of the factors (as in this problem).", "Solution_16": "[url=http://planetmath.org/encyclopedia/CardanicFormulae.html]See[/url]", "Solution_17": "[quote=\"Xevarion\"]\n\n1) There was no need to quote that very long post before asking your question (you could have just quoted the first couple of lines or something). \n[/quote]\n\nAgreed, I quoted a bit too much, I've edited it--so it should look more compact.\n\n[quote=\"Xevarion\"]\n\n2) Yes. But you'll rarely see the whole general solution written anywhere explicitly because it's way too unwieldy. There's also a general solution for quartic equations, which is similarly enormous. There's nothing in general for higher order equations, although certain special cases may be solvable. You can see the method for solving cubics in that post you quoted. It works for any equation. \n\nGenerally, in contests, if you need to solve an equation of degree 3 or higher, you'll be able to use the rational root theorem to find one of the roots by hand, or find some other clever way to guess at one of the factors (as in this problem).[/quote]\r\n\r\nThanks for the reply, while I am familiar with the methods of solving cubics, such as the Cardano-Tartaglia method and Tschirnhaus transformation.\r\n\r\nI have not even seen a closed form solution ever written down in a textbook or online. By closed form--I mean a solution to,\r\n\r\n$ {{ax^3} \\plus{} {bx^2} \\plus{} {cx} \\plus{} {d}} \\equal{} {0}$\r\n\r\nin terms of, \r\n\r\n $ a, b, c, d$ \r\n\r\nThat is why (since you have mentioned it exists)--what it is?\r\n\r\nThanks, \r\n\r\n-APSStudent", "Solution_18": "I don't know. I've never seen it, I don't want to work it out myself, and I don't really want to search around online either (you can try this if you want). You can get it by taking the Cardano solution (in kunny's link) and doing the reverse of that transformation. It will be a huge expression. The reason no one ever writes it is that it's easier to do the transformation and then use the Cardano solution.", "Solution_19": "The key is to notice the convenient factorization. The equation can be written as\n$$x^3 + 3xy + y^3 -1 = (x+y)^3 - 3xy(x+y-1) -1 = (x+y-1)((x+y)^2+(x+y)+1-3xy)=(x+y-1)(x^2+y^2-xy+1+x+y) = 0$$\nIf $x+y=1$, we get a line. If $x^2 + y^2 -xy+1+x+y=0$, then $$2x^2+2y^2-2xy+2+2x+2y = 0 \\Longleftrightarrow (x+1)^2+(y+1)^2 + (x-y)^2 = 0$$\nClearly, this only happens when $x=y=-1$. Therefore, we have a line and a point. Clearly, only one equilateral triangle can be drawn.", "Solution_20": "Good Laugh.\n\nWe use the key identity that $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$. Let $a = x$, $b = y$, $c = -1$, which gives us our problem statement. Hence, it factors to\\[(x + y - 1)(x^2 + y^2 + 1 - xy + x + y) = 0\\]so either $x + y = 1$ or $(x+1)^2 + (y+1)^2 + (x - y)^2 = 0$. Hence the locus of all points satisfying the problem is the line $x + y = 1$ and the point $(-1, -1)$. The other two vertices of this triangle are clearly unique so the triangle is unique.\n\nIn fact, the distance from $(-1, -1)$ to $x + y = 1$ can be calculated to be $\\tfrac{3}{\\sqrt{2}}$, which also happens to be the height of the unique equilateral triangle. Hence, the equilateral triangle has area $\\tfrac{h^2}{\\sqrt{3}} = \\tfrac{3\\sqrt{3}}{2}$.", "Solution_21": "[i]Solved with andyxpandy99.[/i]\n\n[hide=Solution]\nNote that \\begin{align*}\nx^3+3xy+y^3&=1\\\\\n\\leftrightarrow x^3+3xy(x+y)+y^3&=1+3xy(x+y)-3xy\\\\\n\\leftrightarrow (x+y)^3-1&=3xy(x+y-1)\\\\\n\\leftrightarrow (x+y-1)(x^2+xy+y^2+x+y+1)&=0\\\\\n\\leftrightarrow (x+y-1)((x-y)^2+(x+1)^2+(y+1)^2)&=0\\\\\n\\end{align*}\n\nThus, either $x+y=1$, or $(x,y)=(-1,-1)$. So, the curve is one line and one other point not on the line. The only equilateral triangle with all vertices on the curve must have one vertex be $(-1,-1)$, and the two other vertices be on the line. The point $(-1,-1)$ has distance $\\frac{3\\sqrt{2}}{2}$ from the line $x+y=1$. The side length of the triangle is $\\frac{2}{\\sqrt{3}}$ times that, which is $\\sqrt{6}$. Thus the area of the equilateral triangle is $\\frac{(\\sqrt{6})^2\\cdot\\sqrt{3}}{4}=\\boxed{\\frac{3\\sqrt{3}}{2}}$.\n[/hide]", "Solution_22": "oh titu...\n\nWe have\n\\begin{align*}\nx^3+3xy+y^3&=1\\\\\nx^3+y^3+(-1)^3-3xy(-1)&=0\\\\\n(x+y-1)((x-y)^2+(x+1)^2+(y+1)^2)&=0,\n\\end{align*}\nso the graph of the curve is the union of $x+y=1$ and $(x,y)=(-1,-1)$. From here the result is immediate. $\\blacksquare$", "Solution_23": "Consider the behavior of the curves $C_b$: $f(x,y)=x^3+y^3+bxy-1=0$, when $b\\in (-3-\\epsilon,3+\\epsilon)$ in a neighborhood of the point $A=(-1,-1)$.\nThe system $\\dfrac{\\partial f}{\\partial x}=0,\\dfrac{\\partial f}{\\partial y}=0,f(x,y)=0$ give the singular points of $C_b$.\nFor every $b\\not= 3$, there are no singular points; for $b=3$, we obtain the sole point $A$ which is an isolated point of $C_3$.\nWhen $b=3-h,h>0$ small , $C_b$ has no points close to $A$. \nWhen $b=3+h$, the curve $C_b$ presents a kind of small ellipse close to $A$.\n\nIn fact, when $b<3$ the complex set $C_b$ contains two conjugate curves that don't intersect; yet, they intersect -when $b=3$- in the isolated singular real point $A$. \nA simpler example is as follows:\nConsider the curves $C_a$: $x^2+y^2-a=0$, where $a\\leq0$. If $a<0$, then $C_a$ contains the conjugate complex curves\n $\\{(x,y);x\\in\\mathbb{R},y=i\\sqrt{x^2-a}\\}\\},\\{(x,y);x\\in\\mathbb{R},y=-i\\sqrt{x^2-a}\\}$ that don't intersect.\nWhen $a=0$, these $2$ curves intersect in the isolated point $(0,0)$.\nOf course, when $a>0$ small, we obtain a little circle close to $(0,0)$.", "Solution_24": "The A1 of Putnam 2006 is more cool.", "Solution_25": "The curve can be factored to $.5((x+1)^2+(y+1)^2+(x-y)^2).$ It follows that the curve consists of a point $(-1,-1)$ and the line $x+y=1.$ We need one of the vertices to be $(-1,-1)$ for the triangle with three points on this curve to be formed. Moreover, the distance from $(-1,-1)$ to $(.5,.5)$ is $3\\sqrt{2}/2$, in fact this is the altitude from $(-1,-1).$ So the area of the triangle is $\\boxed{3\\sqrt{3}/2}.$ ", "Solution_26": "Zeta, read my post. \nI don't want to solve the problem; moreover the same solution has already been written at least 4 times above (5 times with yours).\nI explain how an isolated point of an algebraic curve can be dynamically created.", "Solution_27": "[quote=loup blanc]Zeta, read my post. \nI don't want to solve the problem; moreover the same solution has already been written at least 4 times above (5 times with yours).\nI explain how an isolated point of an algebraic curve can be dynamically created.[/quote]\n\nI see. I guess my solution happens to be \"the most obvious one.\"", "Solution_28": "We see $x^3+3xy+y^3-1$ factors as $(x+y-1)((x+1)^2+(y+1)^2-(x+1)(y+1))=0$ and we may notice that the second factor is zero only at $x=y=-1$ since it is the average of $(x+1)^2+(y+1)^2$ and $(x-y)^2.$ Since the curve is a point and a line we see that one of $A,B,C$ must be $(-1,-1)$ and the other two are on $x+y=1.$ Their midpoint must be the foot from $(-1,-1)$ to $x+y=1,$ and now it is easy to finish. The triangle has height $\\frac{3\\sqrt2}2$ so it has area $\\frac{3\\sqrt3}2.$", "Solution_29": "Notice that $$x^3+y^3+3xy-1=(x+y)^3-3x^2y-3xy^2+3xy-1=(x+y)^3-3xy((x+y)-1)-1$$ so the given equation is equivalent to $(x+y)^3-1=3xy((x+y)-1)$. Notice that this implies that $x+y=1$ is a solution. Since this is a line, an equilateral triangle can be made from any point such that $x+y\\ne 1$ and $2$ points on $x+y=1$. Therefore, it suffices to prove that there is only one solution to $(x+y)^3-1=3xy((x+y)-1)$ such that $x+y\\ne1$. Since we've already considered $x+y=1$, we can divide by $x+y-1$ to get $(x+y)^2+(x+y)+1=3xy$ or $x^2-xy+y^2+x+y+1=0$. We can rewrite this as $(x+1)^2+(y+1)^2-(x+1)(y+1)=0$. Clearly, $(-1, -1)$ is a solution to this. Plugging in $x=-1+a$ and $y=-1-a$ into the original equation gives $a=0$, and we have already considered $x=y=-1$, so we multiply both sides by $x+y+2=(x+1)+(y+1)$ to get that $(x+1)^3+(y+1)^3=0$ so $x+y+2=0$, which we've already shown only has solution $(-1, -1)$. Now, we calculate the area of this unique equilateral triangle to be $\\frac{3\\sqrt2}{2}\\cdot\\frac{3\\sqrt\\frac23}{2}=\\frac{3\\sqrt3}{2}$." } { "Tag": [ "inequalities" ], "Problem": "Let be $ n\\in \\mathbb{N}\\ ,\\ n\\ge 2$ and $ a_1,a_2,...,a_{2n}\\in \\mathbb{R}_\\plus{}$ such that $ s\\equal{}a_1\\plus{}a_2\\plus{}...\\plus{}a_{2n}$ . Show that :\r\n\r\n$ \\frac{a_1}{s\\plus{}a_{n\\plus{}1}\\minus{}a_1}\\plus{}...\\plus{}\\frac{a_n}{s\\plus{}a_{2n}\\minus{}a_n}\\plus{}\\frac{a_{n\\plus{}1}}{s\\plus{}a_1\\minus{}a_{n\\plus{}1}}\\plus{}...\\plus{}\\frac{a_{2n}}{s\\plus{}a_n\\minus{}a_{2n}}\\ge 1$\r\n\r\nWhen does the equality occurs ?", "Solution_1": "nice, By Cauchy-Scwarz,\r\n$ \\left(\\frac{a_1}{s\\plus{}a_{n\\plus{}1}\\minus{}a_1}\\plus{}...\\plus{}\\frac{a_{2n}}{s\\plus{}a_n\\minus{}a_{2n}}\\right)\\left(a_1s\\plus{}a_1a_{n\\plus{}1}\\minus{}a_1^2\\plus{}...\\plus{}a_{2n}s\\plus{}a_{2n}a_{n}\\minus{}a_{2n}^2\\right) \\geq s^2$\r\nand we have that,\r\n$ a_1s\\plus{}a_1a_{n\\plus{}1}\\minus{}a_1^2\\plus{}...\\plus{}a_{2n}s\\plus{}a_{2n}a_{n}\\minus{}a_{2n}^2\\equal{}2(a_1a_{n\\plus{}1}\\plus{}...\\plus{}a_{2n}a_n)\\plus{}s^2\\minus{}\\sum_{i\\equal{}1}^{2n}a_i^2$\r\nand it's easy to probe that, $ 2(a_1a_{n\\plus{}1}\\plus{}...\\plus{}a_{2n}a_n) \\leq \\sum_{i\\equal{}1}^{2n}a_i^2$\r\ntherefore $ a_1s\\plus{}a_1a_{n\\plus{}1}\\minus{}a_1^2\\plus{}...\\plus{}a_{2n}s\\plus{}a_{2n}a_{n}\\minus{}a_{2n}^2 \\leq s^2$\r\nthen $ \\frac{a_1}{s\\plus{}a_{n\\plus{}1}\\minus{}a_1}\\plus{}...\\plus{}\\frac{a_n}{s\\plus{}a_{2n}\\minus{}a_n}\\plus{}\\frac{a_{n\\plus{}1}}{s\\plus{}a_1\\minus{}a_{n\\plus{}1}}\\plus{}...\\plus{}\\frac{a_{2n}}{s\\plus{}a_n\\minus{}a_{2n}} \\geq 1$\r\nequality occurs when, $ a_1\\equal{}a_2\\equal{}...\\equal{}a_{2n}$", "Solution_2": "[quote=\"peine\"]nice, By Cauchy-Scwarz,\n$ \\left(\\frac {a_1}{s \\plus{} a_{n \\plus{} 1} \\minus{} a_1} \\plus{} ... \\plus{} \\frac {a_{2n}}{s \\plus{} a_n \\minus{} a_{2n}}\\right)\\left(a_1s \\plus{} a_1a_{n \\plus{} 1} \\minus{} a_1^2 \\plus{} ... \\plus{} a_{2n}s \\plus{} a_{2n}a_{n} \\minus{} a_{2n}^2\\right) \\geq s^2$\nand we have that,\n$ a_1s \\plus{} a_1a_{n \\plus{} 1} \\minus{} a_1^2 \\plus{} ... \\plus{} a_{2n}s \\plus{} a_{2n}a_{n} \\minus{} a_{2n}^2 \\equal{} 2(a_1a_{n \\plus{} 1} \\plus{} ... \\plus{} a_{2n}a_n) \\plus{} s^2 \\minus{} \\sum_{i \\equal{} 1}^{2n}a_i^2$\nand it's easy to probe that, $ 2(a_1a_{n \\plus{} 1} \\plus{} ... \\plus{} a_{2n}a_n) \\leq \\sum_{i \\equal{} 1}^{2n}a_i^2$\ntherefore $ a_1s \\plus{} a_1a_{n \\plus{} 1} \\minus{} a_1^2 \\plus{} ... \\plus{} a_{2n}s \\plus{} a_{2n}a_{n} \\minus{} a_{2n}^2 \\leq s^2$\nthen $ \\frac {a_1}{s \\plus{} a_{n \\plus{} 1} \\minus{} a_1} \\plus{} ... \\plus{} \\frac {a_n}{s \\plus{} a_{2n} \\minus{} a_n} \\plus{} \\frac {a_{n \\plus{} 1}}{s \\plus{} a_1 \\minus{} a_{n \\plus{} 1}} \\plus{} ... \\plus{} \\frac {a_{2n}}{s \\plus{} a_n \\minus{} a_{2n}} \\geq 1$\nequality occurs when, $ a_1 \\equal{} a_2 \\equal{} ... \\equal{} a_{2n}$[/quote]\r\n\r\n\r\nCan you explain it clearly,\r\n\r\n$ \\boxed{2(a_1a_{n \\plus{} 1} \\plus{} ... \\plus{} a_{2n}a_n) \\leq \\sum_{i \\equal{} 1}^{2n}a_i^2}$\r\n\r\n\r\nThanks.", "Solution_3": "[quote=\"PKMathew\"]\n\nCan you explain it clearly,\n\n$ \\boxed{2(a_1a_{n \\plus{} 1} \\plus{} ... \\plus{} a_{2n}a_n) \\leq \\sum_{i \\equal{} 1}^{2n}a_i^2}$\n\n\nThanks.[/quote]\r\n\r\n$ \\Leftrightarrow (a_1\\minus{}a_{n\\plus{}1})^2\\plus{}...\\plus{}(a_{2n}\\minus{}a_n)^2\\ge 0$", "Solution_4": "This seems like a special case of http://www.mathlinks.ro/viewtopic.php?t=168801. Which I don't remember how to prove... I think I was in my \"discovering cauchy\" time.... I'll try to remember and post it.\r\n\r\n:)" } { "Tag": [ "LaTeX", "algebra", "polynomial", "Rational Root Theorem" ], "Problem": ":) :D \r\nShow that the number x given by \r\n$x=1+\\sqrt{2+\\sqrt{1+\\sqrt{2+...}}}$\r\nis irrational.", "Solution_1": "[Deleted by Farenhajt]", "Solution_2": "Hrmm...I wouldn't have expected that to turn up in Pre-Olympiad.\r\n\r\nOn a similar note:\r\n\r\nProve $\\pi$ is transcendential.", "Solution_3": "@Farenhajt\r\nwouldn't it be $((x-1)^2-2)^2=x$ instead", "Solution_4": "Yes, you're right. My bad.\r\n\r\nDeleted.", "Solution_5": "$x=1+\\sqrt{2+\\sqrt{x}}$\r\n\r\nsquare, and rearrange, square, and use rrt to show x is not rational", "Solution_6": "sorry.. but what's rrt? ( i'm n00b T.T)", "Solution_7": "The rational root theorem.", "Solution_8": "ah.. thanks man... i gotta look it up on the net to find out what it is.. :blush:\r\n\r\nContinuing from above, you'll get $x^4-4x^3+2x^2+3x+1=0$.. Suppose a rational root $x_0=p/q$ exists, where $p,q$ are integers. From rational root theorem, $p|1$ and $q|1$, which implies $p=+-1$, $q=+-1$, and thus $x_0=+-1$.\r\nSubbing $x_0=1$ to $((x-1)^2-2)^2=x$, you'll get a contradiction.\r\nSubbing $x_0=-1$ to $((x-1)^2-2)^2=x$, you'll get another contradiction.\r\nThus, x can't be a rational number ( x is irrational ).. \r\n\r\nis my working correct?", "Solution_9": "That looks right :) Also, you can use \\pm in $\\LaTeX$ to get $\\pm$." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all f:R->R such that\r\nf(f(x))=x and f(x+y)=f(x)+f(y)\r\nfor all x,y\u2208R", "Solution_1": "Last is Cauchy's equation with solution $ f(x)\\equal{}ax$. Put this into first we get $ a^2\\equal{}1$ and two solutions $ f(x)\\equal{}x$ and $ f(x)\\equal{}\\minus{}x$.", "Solution_2": "No, Cauchy equation usually requires additional assumption on continuity, monotonicity, measurability or smth. else. Given just $ f(x\\plus{}y)\\equal{}f(x)\\plus{}f(y)$, we can have a lot of \"weird\" solutions. This problem seems to be closer to real analisys(i think i have a solution), but i would like to see something algebraic.", "Solution_3": "In fact, there are a lot of functions satisfying $ f(x \\plus{} y) \\equal{} f(x) \\plus{} f(y), \\ f(f(x)) \\equal{} x$. Let $ \\{a_{\\alpha}, \\alpha\\in \\mathbb{A} \\}$ be a basis of $ \\mathbb{R}$ over $ \\mathbb{Q}$, let $ b\\in \\{ \\plus{} 1, \\minus{} 1\\}^{\\mathbb{A}}$ and define $ f(\\sum_{\\alpha}q_{\\alpha}a_{\\alpha}) \\equal{} \\sum_{\\alpha} q_{\\alpha}a_{\\alpha}\\cdot b_{\\alpha}$. It is clear that $ f$ satisfies both conditions. But, we say something about the functions $ \\mathbb{R}^{ \\plus{} }\\to \\mathbb{R}^{ \\plus{} }$, there will be only one solution $ f(x) \\equal{} x$.\r\nThe following problem is somewhat similar: $ f: \\mathbb{R}^{ \\plus{} }\\to\\mathbb{R}^{ \\plus{} }, \\ f(x \\plus{} y) \\equal{} f(x) \\plus{} f(y), \\ f(z)f(\\frac 1z) \\equal{} 1$." } { "Tag": [ "function", "topology" ], "Problem": "Let (X,d)be a compact metric space and f_n be a sequence of countinuous functions from X to R.We assume that for all \u03b5>o ,for all x in X,there is a positive \u03b4_x such that for all n in N and for all y in X, |f_n (x)- f_n (y) |< \u03b5,if |x-y|< \u03b4_x.\r\na.If for all x in X,there is a positive M_x such that |f_n (x) |\u2264M_x for all n in N,show that there is a positive M such that |f_n (x) |\u2264M, for all x in X and n in N\r\nb. Show that there is a positive \u03b4,such that for all x and y in X and all n in N, |f_n (x)- f_n (y) |<\u03b5,if |x-y|< \u03b4", "Solution_1": "That property is called [url=http://en.wikipedia.org/wiki/Equicontinuity]equicontinuity[/url], and the point of this problem is to show that an equicontinuous family on a compact space is uniformly equicontinuous.*\r\n\r\nThe basic idea: pick an $ \\epsilon$, take all of those $ \\delta_x$ balls, and find a finite subcover of the space. There is some small $ \\delta$ such that every point has a $ \\delta$-ball around it contained entirely in one of that finite collection of $ \\delta_x$ balls. That's (b).\r\nNow, what about (a)? Cover the space with finitely many $ \\delta$-balls, and use the bound at the center together with uniform equicontinuity to extend to a bound on the whole ball. The maximum of those finitely many bounds is still finite.\r\n\r\n* A side note on terminology: you presumably know what \"uniform\" or \"uniformly\" means. \"Equi\" is uniform in a different variable, when we already have a direction to reasonably call uniform. This appears most often in dealing with sequences of functions $ f_n(x)$, where variation over $ x$ can be \"uniform\" and variation over $ n$ can be \"equi\".", "Solution_2": "[color=green]Moderator edit: redundant quote deleted[/color]\r\n\r\nThanks. I'm just not quite sure how to do b though, as I am having difficulty choosing delta.", "Solution_3": "The maximum size ball you can fit around $ x$ in one of those open sets is a continuous function of $ x$. It has a minimum.", "Solution_4": "http://en.wikipedia.org/wiki/Lebesgue's_number_lemma\r\n\r\nIt's in Munkres, and should be in any intro topology book, because it is very useful when dealing with compact metric spaces." } { "Tag": [ "function", "limit", "inequalities", "algebra", "functional equation", "algebra solved" ], "Problem": "Find all continuous functions $f: R \\rightarrow R$ satisfying: \r\n$f(f(x))=f(x)+2x$; $\\forall x \\in R$.", "Solution_1": "There are only two solutions, namely $f(x)=-x$ and $f(x)=2x$.\r\n\r\nSuppose $f$ is a solution.\r\nClearly $f$ is injective. Thus, $f(0) = 0$ and since $f$ is continuous, $f$ is monotonic. Therefore it has limits in $+ \\infty$ and $-infty$. Since, the range of $fof - f$ is $R$, we deduce that the range of $f$ is $R$ too.\r\nNow, $f$f is a bijective function from $R$ onto $R$, and we denote $f^{-1}$ its inverse.\r\n\r\nLet $f^0 = f$, and $f^{n+1} = fof^n$ for all integers $n$.\r\nLet $g(x) = 2x-f(x)$ and $h(x) = f(x)+x$.\r\nLet $x$ be fixed.\r\nThen, using the given functional equation, we easily deduce that $f^{n+2}(x) = f^{n+1}(x) + 2f^{n}(x)$.\r\nSolving the recurrence relation, we deduce that $f^n(x) = \\frac {g(x)} 3 (-1)^n + \\frac {h(x)} 3 2^n$.\r\n\r\n- First case : if $f$ is increasing.\r\nThen, from $f(0)=0$ we deduce that $f^n(x) >0$ for all $x > 0$ and all integers $n$.\r\nLet $x_0 > 0$ be given.\r\nIf $g(x_0) \\neq 0$ then lim $2^n \\frac {h(x)} x = 0$ when $n$ tends to $- \\infty$ thus according to the sign of $g(x_0)$ we may found $n$ sufficientely large such that $f^n(x_0) < 0$. A contradiction.\r\nThus $g(x_0) = 0$ and $f(x) = 2x$ for all $x \\geq 0$.\r\n\r\nWe deduce that $f(x)= 2x$ for all $x < 0$ in the same manner.\r\nThus $f(x) = 2x$ for all $x$.\r\n\r\n- Second case : $f$ is decreasing.\r\nIn a similar manner, we deduce that $f(x)=-x$.\r\n\r\nPierre.", "Solution_2": "Alternative solution. We can show as pbornsztein did that [tex]f[/tex] is injective, monotonic, and the image [tex]f(\\mathbb{R})=\\mathbb{R}[/tex]. Let [tex]S=\\sup_{x\\in \\mathbb{R}}(\\frac{f(x)}{x})[/tex] and [tex]I=\\inf_{x\\in \\mathbb{R}}(\\frac{f(x)}{x})[/tex]. Take two sequences [tex]\\{x_i\\}^{\\infty}_{i=0}[/tex] and [tex]\\{y_i\\}^{\\infty}_{i=0}[/tex] such that [tex]\\lim_{i\\to \\infty}\\frac{f(x_i)}{x_i}=S[/tex] and, similarly, [tex]\\lim_{i\\to \\infty}\\frac{f(y_i)}{y_i}=I[/tex]. If [tex]S=0[/tex], then [tex]\\lim_{i \\to \\infty}\\frac{f(f(x_i))}{f(x_i)}=\\lim_{i \\to \\infty}\\left(1+\\frac{2x_i}{f(x_i)}\\right)[/tex] implies that [tex]I=-\\infty[/tex], but [tex]\\lim_{i \\to \\infty}\\frac{f(f(y_i))}{f(y_i)}=\\lim_{i \\to \\infty}\\left(1+\\frac{2y_i}{f(y_i)}\\right)=1[/tex], so [tex]S\\geq1[/tex]. Contradiction. Similarly, the following equalities and inequalities take place: [tex]-\\infty < S,I < +\\infty[/tex] and .[tex]SI \\neq 0[/tex]. Using [tex]\\lim_{i \\to \\infty}\\frac{f(f(x_i))}{f(x_i)}=\\lim_{i \\to \\infty}\\left(1+\\frac{2x_i}{f(x_i)}\\right)[/tex], we arrive at [tex]I=1+\\frac{2}{S}[/tex], since [tex]S=\\sup_{x\\in \\mathbb{R}}(\\frac{f(x)}{x})=\\sup_{x\\in \\mathbb{R}}(\\frac{f(f(x))}{f(x)})[/tex] cause of [tex]f(\\mathbb{R})=\\mathbb{R}[/tex]. Similarly [tex]S=1+\\frac{2}{S}[/tex], so [tex]S+2=SI=2+I[/tex], so [tex]S=I[/tex] and [tex]f(x)=kx[/tex] for all [tex]x\\in \\mathbb{R}[/tex]. It is easy to find [tex]k[/tex] which is equal to [tex]2[/tex] or [tex]-1[/tex]." } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "circumcircle", "analytic geometry" ], "Problem": "In $ xyz$ space, given 3 points $ A(1,\\ 0,\\ 0),\\ B(0, \\minus{} 2,\\ 0), C(0,\\ 0,\\ 4)$. Let $ P$ be the cenre of circumcircle of $ \\triangle{ABC}$.\r\n\r\nDraw a line passing through the pont $ P$ and perpendicular to the plane $ ABC$. Take the point $ Q$ on the line.\r\n\r\n(1) Find the $ x$ coordinate of $ P$.\r\n\r\n(2) Let $ R$ be one of the point of the circumcircle of $ \\triangle{ABC}$ such that $ \\angle{PRQ} \\equal{} 60^\\circ$. Find the $ x$ coordinate of $ Q$.\r\n\r\n(3) Under the condition of question (2), find the volume of tetrahedron $ QABC$.", "Solution_1": "Do you mean the circumcircle of the triangle?", "Solution_2": "No, I'm pretty sure he means [url=http://en.wikipedia.org/wiki/Incenter]excircle[/url].\r\n\r\nOkay, I lied.", "Solution_3": "[quote=\"brightzhu\"]Do you mean the circumcircle of the triangle?[/quote]\r\n\r\nYes. :blush:", "Solution_4": "I'm very sorry that I made a typo. I have just edited it.\r\n\r\nkunny" } { "Tag": [ "combinatorics", "algorithm", "game strategy", "search algorithms", "IMO Shortlist" ], "Problem": "The lock of a safe consists of 3 wheels, each of which may be set in 8 different ways positions. Due to a defect in the safe mechanism the door will open if any two of the three wheels are in the correct position. What is the smallest number of combinations which must be tried if one is to guarantee being able to open the safe (assuming the \"right combination\" is not known)?", "Solution_1": "[quote=\"orl\"]The lock of a safe consists of 3 wheels, each of which may be set in 8 different ways positions. Due to a defect in the safe mechanism the door will open if any two of the three wheels are in the correct position. What is the smallest number of combinations which must be tried if one is to guarantee being able to open the safe (assuming the \"right combination\" is not known)?[/quote]\n\nSomeone who knows the solution of this?", "Solution_2": "The problem belongs to the chapter of Combinatorics called \"Packings and Coverings\". \nA typical question would be: given a set $X$ of $n$ elements, what is the minimal number of triplets in a family $\\mathcal{F}$ of subsets of $X$ of cardinality $3$, such that any pair of elements of $X$ is contained in at least one triplet of $\\mathcal{F}$ (triplets covering doubletons)? This question has been fully answered, sometimes towards the end of the XX century - but I lack a reference.\nIn the problem at hand, the situation is complicated by needing also that the pair be in the right position (in both doubletons and triplets, position matters).", "Solution_3": "The answer is $32$\n\n[b]Lower Bound[/b]\nAssume there exists a set $S$ of $31$ triplets that cover all combinations. Let $S_i$ be the triplets in $S$ with first component equal to $i$, for $i=1,2,...,8$. There must exist some $i$ such that $|S_i| \\le \\textstyle\\lfloor\\frac{31}{8} \\rfloor = 3$. wlog $i=1$ \n\nConsider the second and third components in the triplets of $S_1$. Suppose $k_1\\le 3$ numbers appear in some triplet as a second component and $k_2\\le 3$ as a third component. Then there are $(8-k_1)(8-k_2)\\ge 25$ triplets $(1,y,z)$ that don't share two components with any triplet in $S_1$. Let the set of triplets be $K$. So among the other sets $S_i$ there exists a triplet of the form $(i,y,z)$ that meets $(1,y,z)$ in the last two components for each $(1,y,z)\\in K$. \n\nIf each $S_i$ for $i\\ge 2$ contained a triplet $(i,j,k)$ that doesn't meet a triplet in $K$ in two components then $|S| \\ge |S_1| + |K| + 7 \\ge 0 + 25 + 7 = 32$ contradiction. So there exists a set, say $S_2$, such that every triplet in $S_2$ meets an element of $K$ in two components (i.e. the last two). But now there are at least $k_1$ elelemts are that aren't a second component of a triplet in $S_2$ and at least $k_2$ elements that aren't a third component. So there are at least $k_1k_2$ triplets $(2,y,z)$ that don't meet any element of $S_2$ in two components. Call these triplets $K'$. No triplet can meet a triplet in both $K$ and $K'$ because triplets in $K$ and $K'$ have distinct components. Meaning that $|S| \\ge |K| + |K'| = (8-k_1)(8-k_2) + k_1k_2 = 32 + 2(4-k_1)(4-k_2)$ where $k_1,k_2 \\le 3$. This gives $|S| \\ge 34$ and we have our contradiction.\n\n\n[b]Upper Bound[/b] \nFor all $0\\le i,j \\le 3$ take the triplets $(i,j,i+j \\mod 4)$ and $(4+i,4+j,4 +(i+j \\mod 4))$ (which is $2\\cdot 4^2=32$ triplets in total). Now for any triplet $(a,b,c)$ one of the pairs $(a,b),(b,c)$ or $(c,a)$ will be in the same interval $\\{1,2,3,4\\}$ or $\\{5,6,7,8\\}$. So in one of $(i,j,i+j \\mod 4)$ or $(4+i,4+j,4 + (i+j\\mod 4))$ we will be able to select $i,j$ to get the desired result." } { "Tag": [ "algebra", "polynomial", "factorial", "abstract algebra", "function", "domain", "calculus" ], "Problem": "Let G be the additive group of all polynomails in x with integer coefficients. Show that G is isomorphic to the group Q+ of all positive rationals (under multiplication). (Hint: Use the Fundamental Theorem of Arithmetic to construct an isomorphism).\r\n\r\nAttempted solution:\r\n\r\nI wanted to do something like $ \\phi \\left(\\frac{p_1^{t_1}\\cdots p_n^{t_n}}{q_1^{s_1}\\cdots q_m^{s_m}} \\right) \\equal{} \\sum_{k\\equal{}0}^n x^{p_k} t_k \\minus{} \\sum_{k\\equal{}0}^m x^{q_k} s_k$ but of course then you only get polynomials with prime powers.", "Solution_1": "Then why don't you take $ p_k$ to $ k \\minus{} 1$?", "Solution_2": "since $ \\mathbb{Z}$ is factorial, $ \\mathbb{Q}_{>0}$ is free abelian with basis consisting of all prime numbers. and $ \\mathbb{Z}[X]$ is also free abelian with an infinite countable basis.", "Solution_3": "-oo-, I don't know exactly what is by \"Z is factorial\", but I see how that idea works.", "Solution_4": "A unique factorization domain is any integral domain in which every nonzero noninvertible element has essentially unique decomposition as the product of prime elements or irreducible elements." } { "Tag": [ "modular arithmetic" ], "Problem": "Find the last two digits of a sum of eightth powers of $100$ consecutive integers.", "Solution_1": "[hide]$n^{8}+(n+1)^{8}+\\cdots+(n+99)^{8}\\equiv 0^{8}+1^{8}+2^{8}+\\cdots+99^{8}\\equiv 2(1^{8}+2^{8}+3^{8}\\cdots+49^{8})+50^{8}$ \n(but since, $k^{2}\\equiv (50-k)^{2}\\pmod 100$ )\n$\\implies \\equiv 4(1^{8}+2^{8}+\\cdots+24^{8})+2(25^{8})\\equiv 80 \\pmod 100$, from manual calculations. how do you do this properly?[/hide]" } { "Tag": [ "ARML", "email" ], "Problem": "I can't come this year due to other commitments, but how do I join in the future? Do I just email Mr. Yang? What pre-requisite tests do I need to take?\r\n\r\nAny help in this matter would be appreciated.", "Solution_1": "I don't know how your region does it but at least where I live, there is a high school math olympiad that extends for a few weeks. The top scorers are then invited to training sessions and the ARML team is built from those people. I would imagine your region has a similar process.", "Solution_2": "From the FAQ at [url]http://www.cscc.edu/docs/math/faculty/ayang2.htm[/url]\r\n\r\nQ: How can I join Ohio ARML?\r\nA: Mostly, if you score high on OCTM, AMC, or Ohio Math League, you will be invited to join Ohio ARML. Among these competitions, only OCTM provides Ohio ARML with participants\u2019 personal contact information. Therefore, if your school does not participate in OCTM, it is suggested that you contact Prof. Alan Yang at Columbus State Community College. \r\n\r\nSome students get in through their school's or district's recommendations. Teachers, Gifted Coordinators, or Gifted Services teachers, especially from districts (or schools) that do not actively participate in math competitions, may also recommend students. Those wishing to nominate students should submit the appropriate information to Prof. Alan Yang at: Math Department, Columbus State Community College, 550 E. Spring Street, Columbus, Ohio 43215.\r\n\r\nOhio ARML members who still are eligible will get the first round of invitations in February. Sometime in mid March, the next round of invitations will be sent out. If needed, another round of invitations will be sent out in early April.", "Solution_3": "Cool. So I just send him AMC scores next year? I'm not sure if we do OCTM.", "Solution_4": "If you are interested in joining ARML '09 shoot me an e-mail at\r\nayang@cscc.edu\r\n\r\nAlan", "Solution_5": "If you are interested in joining ARML '09, shoot me an e-mail at\r\nayang@cscc.edu\r\n\r\nAlan", "Solution_6": "darn, I continue to fail at math while under time pressure", "Solution_7": "Darn, so do I That's why I phailed nats...must do better this year." } { "Tag": [ "geometry", "angle bisector", "geometry unsolved" ], "Problem": ":D :D :D", "Solution_1": "this problem is equivalent to this.we have $ a$ and $ h_a$ and $ l\\minus{}a$ and i think its open.", "Solution_2": "\"this problem is equivalent to this.\" ? :D", "Solution_3": "Let $ V,D,M$ denote the unknown feet of the altitude, angle bisector and median on $ BC,$ respectively. The angle between the altitude $ h_a$ and the angle bisector $ AV$ is presicely half of the given difference $ |\\beta - \\gamma|.$ Thus, choose $ D$ on a fixed line $ a$ (the sideline BC), construct the A-vertex on the perpendicular to $ a$ at $ D$ with the given distance $ h_a$ and draw the angle bisector of $ \\angle BAC,$ which is a A-ray $ v$ such that $ \\angle(v,h_a) = \\frac {_1}{^2}|\\beta - \\gamma|$ $ \\Longrightarrow$ $ V \\equiv v \\cap a.$ Perpendicular to $ AV$ through $ A$ is the external bisector of $ \\angle BAC$ meeting $ a$ at $ V'.$ The circle with diameter $\\overline{ VV'}$ is the A-Apollonius circle, which is orthogonal to the circle $ (M)$ with diameter $\\overline{ BC},$ since $ (B,C,V,V') = - 1.$ Hence, if $ N$ is the midpoint of $ VV'$ and $ P$ is one of the intersections of $ (N)$ and $(M)$ $ \\Longrightarrow$ $ NP \\perp MP.$ The right $ \\triangle MPN$ is constructible since its catheti $ NP$ and $ MP$ are known as half of the distances $ VV'$ and $ BC$ $\\Longrightarrow$ length of the hypotenuse gives the position of $ M$ on the sideline $a.$ Circle centered at $ M$ with radius $ \\frac {_1}{^2}BC$ cuts $ a$ at $ B,C$ and $\\triangle ABC$ is completed.", "Solution_4": "\"construct the A-vertex on the perpendicular to $ a$ at $ D$ with the given distance $ h_{a}$ and draw the angle bisector of $ \\angle BAC$ which is a A-ray $ v$ such that $ \\angle(v,h_{a}) \\equal{}\\frac{_{1}}{^{2}}(\\beta\\minus{}\\gamma)$ $ \\Longrightarrow$ $ V\\equiv v\\cap a$ \"\r\n\r\nThere's two feet of altitude onto BC: $ D$ and $ V$ ?\r\n\r\n :D", "Solution_5": "[quote=\"Luis Gonz\u00e1lez\"] choose $ D$ on a fixed line $ a$ (the sideline BC), construct the A-vertex on the perpendicular to $ a$ at $ D$ with the given distance $ h_a$ and draw the angle bisector of $ \\angle BAC,$ which is a A-ray $ v$ such that $ \\angle(v,h_a) \\equal{} \\frac {_1}{^2}|\\beta \\minus{} \\gamma|$ $ \\Longrightarrow$ $ V \\equiv v \\cap a.$ [/quote]\nRead carefully, we have to fix a sideline $ a$ and then choose a point $ D$ as the foot of the A-altitude. But if you mean that vertices $ B$ and $ C$ are previously fixed, then there's no problem either, because we can construct $ \\triangle A'B'C' \\cong \\triangle ABC.$ Then $AB=A'B'$ and $AC=A'C'.$", "Solution_6": "thanks\r\n :D" } { "Tag": [ "function", "blogs", "\\/closed" ], "Problem": "For the threads containing problems in all my class message boards, the solutions are shown and I recall that all of them were hidden before. It appears the hidden function was tinkered with and needs to be fixed.", "Solution_1": "Yes, this also happens to me. It's not just in classes, either.\r\n\r\nFor example:\r\n\r\n[hide]\nIt's not hidden, is this?\n[/hide]\r\n\r\nAnother note: It doesn't show up in preview function either.", "Solution_2": "Yeah I know I will fix it :)", "Solution_3": "Now it's hidden, but when you click it, nothing happens.", "Solution_4": "[quote=\"AIME15\"]Now it's hidden, but when you click it, nothing happens.[/quote]Guys I'm working on it, I will let you know when it's fixed :)", "Solution_5": "OK main functionality is fixed. Please report any other bugs you find (I've made a major update to the javascript on the site so there are bound to be some things broken.)", "Solution_6": "It appears that, when you click on hide tags in blogs, it redirects to the top of the page instead of revealing the hidden content.", "Solution_7": "Also, when I'm posting this message, under \"Topic Review\", when I click my hide tag, it doesn't open and automatically scrolls up to the top, like cf249 said.", "Solution_8": "Hide tags in blogs are fixed.", "Solution_9": "It doesn't work @ http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=237933", "Solution_10": "Works for me -- refresh the page.", "Solution_11": "It doesn't work for me in the forum..." } { "Tag": [], "Problem": "$ a_{1},a_{2},...$ is a sequence of real numbers where $ a_{n}$ is the arithmetic mean of the previous $ n-1$ terms for $ n>3$ and $ a_{2004}=7$. $ b_{1},b_{2},...$is a sequence of real numbers in which b_n is the geometric mean of the previous n-1 terms for $ n>3$ and $ b_{2005}=6$. If $ a_{i}=b_{i}$ for $ i=1,2,3$ and $ a_{1}=3$, then compute the value of $ a_{2}^{2}+a_{3}^{2}$", "Solution_1": "$ a_{4}=a_{5}=a_{6}=\\cdots$, as $ k\\cdot a_{k+1}= a_{1}+a_{2}+\\cdots+a_{k}= a_{1}+a_{2}+\\cdots+a_{k-1}+\\frac{a_{1}+a_{2}+\\cdots+a_{k-1}}{k-1}$\r\n\r\n...which equals $ k\\cdot a_{k}$.\r\n\r\nTherefore, $ a_{4}=7$, so $ a_{2}+a_{3}=18$.\r\n\r\nYou can similarly prove that that $ b_{4}=b_{5}=b_{6}=\\cdots$. It follows that $ b_{2}\\cdot b_{3}=a_{2}\\cdot a_{3}=72$\r\n\r\n$ a_{2}^{2}+a_{3}^{2}=18^{2}-2\\cdot72=180$." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "proof that for $ a,b,c>0$ we have this inequality:\r\n$ \\forall t ,2(a^2\\plus{}b^2\\plus{}c^2\\minus{}ab\\minus{}bc\\minus{}ac)\\plus{}\\sum_{cyc} a^{2t}(a\\minus{}b)^2 \\ge \\sum a^t(a\\minus{}b)(a\\minus{}c)$ with equality if and only if $ a\\equal{}b\\equal{}c$", "Solution_1": "waitin for a solution :(", "Solution_2": "no one found a solution ?! am still waiting :(", "Solution_3": "I just analyze it as SOS:\r\nineq$ \\Leftrightarrow\\sum(a^{2t}\\plus{}1\\minus{}\\frac{a^t\\plus{}b^t\\minus{}c^t}{2})(a\\minus{}b)^2\\geq 0$\r\nwho can evaluate it", "Solution_4": "did you continue this SOS analysis ? coz actually i have a very simple solution, i want just to see a simpler one", "Solution_5": "[quote=\"othman\"]proof that for $ a,b,c > 0$ we have this inequality:\n$ \\forall t ,2(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} bc \\minus{} ac) \\plus{} \\sum_{cyc} a^{2t}(a \\minus{} b)^2 \\ge \\sum a^t(a \\minus{} b)(a \\minus{} c)$ with equality if and only if $ a \\equal{} b \\equal{} c$[/quote]\r\nWe have\r\n\\[ LHS \\equal{}\\sum (a\\minus{}b)^2\\plus{}\\sum a^{2t} (a\\minus{}b)^2 \\equal{}\\sum [(a\\minus{}c)^2\\plus{}a^{2t}(a\\minus{}b)^2] \\ge 2\\sum a^t \\left| (a\\minus{}b)(a\\minus{}c)\\right| \\ge \\sum a^t (a\\minus{}b)(a\\minus{}c)\\]\r\n:)" } { "Tag": [ "probability" ], "Problem": "A given sequence $r_1, r_2, \\dots, r_n$ of distinct real numbers can be put in ascending order by means of one or more ``bubble passes''. A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the third term with the second term and exchanging them if and only if the third term is smaller, and so on in order, through comparing the last term, $r_n$, with its current predecessor and exchanging them if and only if the last term is smaller. \r\n\r\nThe example below shows how the sequence 1, 9, 8, 7 is transformed into the sequence 1, 8, 7, 9 by one bubble pass. The numbers compared at each step are underlined.\r\n\r\n\\[ \\begin{array}{c} \\underline{1 \\quad 9} \\quad 8 \\quad 7 \\\\ 1 \\quad \\underline{9 \\quad 8} \\quad 7 \\\\ 1 \\quad 8 \\quad \\underline{9 \\quad 7} \\\\ 1 \\quad 8 \\quad 7 \\quad 9 \\end{array} \\]\r\n\r\nSuppose that $n = 40$, and that the terms of the initial sequence $r_1, r_2, \\dots, r_{40}$ are distinct from one another and are in random order. Let $p/q$, in lowest terms, be the probability that the number that begins as $r_{20}$ will end up, after one bubble pass, in the $30^{\\text{th}}$ place. Find $p + q$.", "Solution_1": "[hide]Notice that in order for $r_{20}$ to reach $r_{30}$, it must be true that $r_{20}$ is the greatest of all the terms $r_1,r_2,\\ldots,r_{30}$. This is because if $r_{20}$ is less than one of those terms, it won't be passed to the next position. This can happen with $\\frac{1}{30}$ probability.\n\nWe also see that if $r_{30}>r_{31}$, it will be passed to the $31^{\\text{st}}$ position. Therefore, we must have $r_1,r_2,\\ldots,r_{30}a_{21}$ in comparing those two, and then another $ 1/2$ probability that $ a_{20}>a_{22}$ and so on... so shouldn't the probability be $ (1/2)^{11}$? (I'm pretty sure there's an equal $ 1/2$ probability for either element being greater in each comparison.) What's wrong with this reasoning?\r\n\r\nNote that the above is equivalent to saying that we if pick a random element $ a_i$ out of $ \\{1,2,3,4,5\\}$, find the probability that that element is the biggest, which is clearly $ 1/5$. However, if we consider this in another way of first picking $ a_i$ and another element and compare them, and then eliminate the element that's smaller. So there's $ 1/2$ probability of $ a_i$ being greater than the first element from the rest four, and then another $ 1/2$ probability of $ a_i$ being greater than the element from the rest three, and so on; so continuing we should get $ 1/2^4$, but those two answers aren't the same... :huh:", "Solution_4": "EDIT: Just realized that my solution was wrong :). 4everwise is right.\r\n\r\n@cdymdcool Indexes determine starting place of the numbers.", "Solution_5": "[quote=me for the past month]How do I manage to solve the hardest problems and yet fail on the easiest ones?[/quote]\n\n[hide=Solution]We only care about the first $31$ terms of the sequence. The condition rewrites itself as $r_{31}$ and $r_{20}$ being the largest and second-largest elements of the sequence, respectively. There's $\\frac{1}{31}$ probability $r_{31}$ is the largest, and given that, there's $\\frac{1}{30}$ probability $r_{20}$ is the second largest, so the answer is $\\frac{1}{31\\cdot30}=\\frac{1}{930}\\implies\\boxed{931}$.", "Solution_6": "Only the first $31$ number matter; It is necessary and sufficient to have\n\\[ r_{31} > r_{20} > r_{30}, \\dots, r_{21}, r_{19},\\dots, r_1.\\]\nFor any $31$ element subset of $\\{1,2,\\dots, 40\\}$, there are $31!$ ways to assign the values to the $r_i$, and $29!$ of these will have the above property ($r_{31}, r_{20}$ have to be largest, second largest resp. and the remaining 29 can be permuted freely). The answer is\n\\[ \\frac{29!}{31!} = \\frac{1}{31\\cdot 30} = \\frac{1}{930}\\to\\boxed{931}.\\]" } { "Tag": [ "calculus", "integration", "function", "Ring Theory", "superior algebra", "superior algebra theorems" ], "Problem": "Show that ring $\\mathbb{Z}[\\sqrt{2}]$ is euclidean.Show furthermore that its units are given by $(1+\\sqrt{2})^n;-(1+\\sqrt{2})^n(n\\in\\mathbb{Z})$ and determine its prime elements.", "Solution_1": "A norm you can use is $N: a+\\sqrt 2 b \\mapsto a^2-2b^2$, and then you have only to show that for every $z \\in \\mathbb{Q}[\\sqrt 2]$, there are $a,b \\in \\mathbb{Z}$ with $|N(z-a-b\\sqrt 2)| < 1$ (this commes from dividing $N(c-qd)=N(r)0$. Then take the greatest $n$ such that $(1- \\sqrt 2)^n (x+\\sqrt 2 y) \\geq 1$ and we want to prove that equality occurs. Well, assume the contraposition, then we have a unit $u=x+\\sqrt 2 y$ with $11 \\implies u<1$, contradiction.\r\nIn general, you can reduce every oterh case to this one by taking inverses and multiplying with $-1$.\r\n\r\nWell, every prime either remains prime or splits into a product of two primes since $N(p)=p^2$.\r\nSo when there are no $a,b$ with $a^2-2b^2=p$, the prime $p$ remains prime. But that is at least unsolvable when $c^2 \\equiv 2 \\mod p$ is unsolvable.\r\nThis are also all primes that remain prime:\r\nFor all others, take such an $c$ (as an integer) and $p$ divides $(c+ \\sqrt 2)(c- \\sqrt 2)$, but none of the factors, thus can't be prime anymore.", "Solution_2": "[quote=\"ZetaX\"]A norm you can use is $N: a+\\sqrt 2 b \\mapsto a^2-2b^2$\n[/quote]\r\n\r\nBut i think.Norm to be function nonnegative? :roll:", "Solution_3": "Take $|N|: a+b\\sqrt 2 \\mapsto |a^2-2b^2|$ instead ;)", "Solution_4": "You are right!This is a problem other:Find all free square positive integer number $d>1$ such that $\\mathbb{Z}[d]$ is euclidean.", "Solution_5": "Do you mean $\\mathbb{Z}[\\sqrt{d}]$\u00bf Then the problem should be very hard...", "Solution_6": "[quote=\"ZetaX\"]Do you mean $\\mathbb{Z}[ \\sqrt{d}]$\u00bf Then the problem should be very hard...[/quote]\r\n\r\nYes,i think so.", "Solution_7": "I think that is an open problem :) However one can prove that if $d<0$ the only euclidean rings of integers $\\mathbb{Z}[\\sqrt d]$ are for $d=-1,-2,-3,-7,-11$. \r\n\r\nActually you can look up Sloane's [url=http://www.research.att.com/~njas/sequences/A048981]A048981[/url].", "Solution_8": "I think my teacher said if $\\mathbb Z[\\sqrt d]$ was PID for each $d$ then Fermat's last theorem would be proved easily", "Solution_9": "[quote=\"Omid Hatami\"]I think my teacher said if $\\mathbb Z[\\sqrt d]$ was PID for each $d$ then Fermat's last theorem would be proved easily[/quote]\r\nAre you sure about that\u00bf I think you need every number field to be PID for that (than UFD follows and the standard arguments work)..." } { "Tag": [], "Problem": "For what real values of $ x$ is\r\n\r\n$ \\sqrt {x \\plus{} \\sqrt {2x \\minus{} 1}} \\plus{} \\sqrt {x \\minus{} \\sqrt {2x \\minus{} 1}} \\equal{} A,$\r\n\r\ngiven (a) $ A \\equal{} \\sqrt {2}$, (b) $ A \\equal{} 1$, (c) $ A \\equal{} 2$, where only non-negative real numbers are admitted for square roots?\r\n\r\nwow....\r\n\r\nold IMO's are SO trivial...\r\n\r\nit's pretty sad i guess\r\n\r\nthis is WAY too trivial for IMO", "Solution_1": "its not too trivial, its kinda according to the people in those times. anyways the imo`s have become tougher and tougher since the 90`s .\r\n\r\n\r\nthis is not the case only with imo`s . we also can see that the old admisiion or enterance tests of many famous insittes which have world wide reputation are pretty easy. :blush:" } { "Tag": [], "Problem": "\u6709\u4ec0\u4e48\u7ebf\u6027\u4ee3\u6570\u7684\u4e60\u9898\u96c6\u6bd4\u8f83\u597d\u7684\u554a,\u4ecb\u7ecd\u4e00\u4e0b\u5427.", "Solution_1": "You certainly mean quatre-vingt-treize ...", "Solution_2": "quatre-vingt-treize???\r\nIt is 93 in French... :|", "Solution_3": "\u8fd9\u662f\u96e8\u679c\u5199\u7684\u4e00\u90e8\u5c0f\u8bf4\u53eb\u505a\u300a\u4e5d\u4e09\u5e74\u300b\u2026\u2026" } { "Tag": [ "The Mole" ], "Problem": "[size=150]Welcome to the very first, and hopefully not the last, [b]The Mole: AoPS Deception[/b]![/size]\r\n\r\n[u]PLAYER LIST[/u]:\r\n2. xpmath\r\n5. miyomiyo\r\n6. BOGTRO [size=59]replacing kimmystar94 [/size]\r\n7. hm29168 [size=59]replacing woser456[/size]\r\n8. CatalystOfNostalgia\r\n\r\n\r\nEliminated:\r\nTemperal (Fraud)\r\nHamster1800 (Test 1)\r\nerabel (Test 2)\r\nmnmath (Test 3)\r\nzapi2007 [size=59]replacing Cyrazeno[/size] (Test 4)\r\nKlebian (Test 5)\r\n\r\nSorry to those that applied and didn't make the cut; I tried to make the game as diverse, interesting, and sure to last as possible. You're now making up the replacement list, though! \r\n\r\nFirst, let\u2019s get some logistics out of the way. Exactly 9 of you are playing the game, trying (not really) to win hundreds of thousands of dollars. However, one of you is THE MOLE, who is trying to thwart your efforts and stop you from winning money. Each round, you will participate in several tasks, in which you shall work together to earn money for the pot. The mole, detesting all of you, will sabotage these tasks, while evading all of your watchful eyes. At the end of each round, there will be a test, questions about the mole. Whoever receives the lowest score will be executed, and lose all chances of winning the game. The final two players, along with the mole, will take part in a final test, and whichever player scores the highest will win the game! Well, that is, if the mole doesn\u2019t defeat all of you\u2026\r\n\r\nYou see, the mole can win too. If (s)he prevents you from reaching your target pot(known only to myself and the mole), and prevents both of the final two players from getting a perfect score on the last test, then the mole will steal the cash and be the victor! But that certainly won\u2019t happen\u2026\r\n\r\nTHE RULES:\r\n1. There are ten of you, and some more temporary players will not be here long at all. Get over it.\r\n2. There will be outside communication, of any form, allowed at all times in the game UNLESS STATED OTHERWISE.\r\n3. All deadlines are final, and will not for any reason be changed.\r\n4. To be clear, lying and deception are a part of the game, and are encouraged.\r\n5. Don\u2019t disappear: replacements aren\u2019t fun.\r\n6. All players, including the mole, can be replaced. \r\n7. Tasks will test your abilities, and are not meant to be hurtful, nor easy. Don\u2019t take anything personally.\r\n8. I reserve the right to(but don\u2019t plan to) change the tasks or rules at any times.\r\n9. Tests will proceed as follows: I will post the questions, with a three day time limit. The lowest scorer will be executed. In the case of a tie, whoever turned it in later will be executed. \r\n10. There will be several possible exemptions offered during the game. If a player wins one, they will not be required to fill out the round\u2019s test, and will automatically advance to the next round.\r\n11. Expect surprises along the way: I, like The Mole, am unpredictable.\r\n12. This is a game. Have fun, and don\u2019t ruin it for yourself or others.\r\n\r\nWell, here we go. Good luck, make a lot of money, and don\u2019t let the mole take advantage of you.\r\n\r\n----------------\r\n\r\nQUID PRO QUO", "Solution_1": "[color=blue][size=150][b][u]TASK 1: The Invader[/u][/b][/size][/color]\r\n\r\nHey, didn\u2019t I say that there were only 10 players? I think that I did anyway\u2026Well, yes I did. That means that one of you eleven is a fraud! Your first task will be to discover this culprit, and reveal him/her for the scoundrel they truly are. For clues, well, see this and the first post of the topic\u2026To submit, one of you must post, in bold and word for word, [b]\"The fraud is _______\"[/b]. Communication is allowed during this task.\r\n\r\nReward: $ \\$20,000$\r\n\r\nDeadline: 2/06/08 8:00 EST(About 24 hours from this edit-this is a quick one!)\r\n\r\n----------------\r\n\r\nQUID PRO QUO", "Solution_2": "To clarify - we only get one submission, yes?", "Solution_3": "That is correct.", "Solution_4": "[quote=\"perfect628\"]2. There will be outside communication, of any form, allowed at all times in the game UNLESS STATED OTHERWISE.[/quote]Is this what you meant?\r\n\r\nEDIT: Do we posts clues we find in the thread?", "Solution_5": "I suppose this wasn't clear...You can say whatever you want in thread, and out of the thread, at any time, unless otherwise stated. It will be clear at times when you can not talk to each other.", "Solution_6": "Well then,\r\n[quote=\"perfect628\"]Sorry to those that applied and didn't make the cut; I tried to make the game as diverse, interesting, and sure to last as possible. You're now making up the replacement list, though! [/quote]The fraud would be somebody who \"didn't make the cut\" and would thus be making up the \"replacement list\". This may be referring to the mafia replacement list....\r\n\r\nJust a thought.", "Solution_7": "Not possible - unless the answer is Klebian. There's currently no mafia replacement list so the only person who that could possibly apply to is Klebian. I'd rather look for some other clues.", "Solution_8": "True. If you compare the \"The Mole\" thread with signups with the player list, a bunch of people didn't post in that thread but are on the list.", "Solution_9": "I'M THE FRAUD!\r\n\r\n(since jokes like this have gotten me erroneously lynched in mafia games *cough*Peter game 31*cough*, this is a joke for the record)\r\n\r\nI'm suspicious of the first few posters in this thread; they seem to have noticed the topic awfully quickly...", "Solution_10": "well...\r\nokay...\r\ndo we know if the fraud knows that they are a fraud?\r\n\r\noh. and read perfect's signature/last statement of first post..\r\n\r\nQUID PRO QUO..\r\n\r\nany guesses?\r\n\r\nEDIT: I read wiki's version, and it means \"something for something\"...\r\nit might be a clue..", "Solution_11": "[quote=\"Temperal\"]I'm suspicious of the first few posters in this thread; they seem to have noticed the topic awfully quickly...[/quote]\r\n(a) I have no life\r\n(b) I'm currently watching Super Tuesday with cnnpoliticis.com and AoPS open\r\n(c) I like posting", "Solution_12": "[quote=\"miyomiyo\"][quote=\"Temperal\"]I'm suspicious of the first few posters in this thread; they seem to have noticed the topic awfully quickly...[/quote]\n(a) I have no life\n(b) I'm currently watching Super Tuesday with cnnpoliticis.com and AoPS open\n(c) I like posting[/quote]\r\n\r\neh... same here...\r\nwell.. not the first part of b..", "Solution_13": "21 hours until deadline.\r\n\r\n----------------\r\n\r\nQUID PRO QUO", "Solution_14": "Pro Nose=Carl.\r\nBUt I don't know.", "Solution_15": "Well, it was inevitable that I'd be knocked out. It was very fun nonetheless, great job perfect!\r\n\r\nI might join the second one but you might want to give us a little break between games.", "Solution_16": "Yay. Moleish behavior that I noticed:\r\n\r\n1. In the challenge where the mole had to pick the truth out of three statements, the only one the mole got correct was miyomiyo's. It was about miyomiyo's age and grade. Well of course woser456 was going to get it right, they know each other!\r\n\r\n2. In the challenge where you all (final 5) picked me to survive, I claimed (yes, I was lying, and I do not have a brother), that we would win 10,000 if I survived. hm29168 said that he should die, because we would win more if I survived. But that was a lie, as he knew we would win 40,000 if I survived.\r\n\r\n3. Three Questions task: hm29168 put me down as all three answers. That makes absolutely no sense, as it would mean I was both meaner and nicer than miyomiyo. In addition, when guessing my answers, he put himself as all three, that also makes no sense and would never be right.\r\n\r\n4. Final task: he was uberstalling us.\r\n\r\n\r\n\r\nAlso perfect: hm29168 wins 337,000, not 377,000.\r\n\r\nI think I was trying to act like the mole too much. Eh.", "Solution_17": "[quote=\"miyomiyo\"]I might join the second one but you might want to give us a little break between games.[/quote]\r\nHmm...Well the few weeks between now and USAMO should suffice...Any longer and it would probably fail. As it is, I'm going to have to go really fast.", "Solution_18": "[quote=\"BOGTRO\"]Psst... If I AM the mole in the next game I'll probably drop clues about that[/quote]\r\n\r\nGOOD PLAN. Now maybe if you're clever you'll use REVERSE PSYCHOLOGY. oh man i cannot WAIT to see what happens.", "Solution_19": "Just as a tip of advice for the next game. SamL told me that most of you didn't split answers (miyomiyo doesn't count), and that was your downfall. I split answers on every single test other than the last two. If you don't understand what \"splitting answers\" means, even if you think person X is the mole, you shouldn't necessarily answer the questions as if person X is actually the mole. Because you can never be sure (I knew I had nothing to lose on the last test because I knew miyomiyo wasn't the mole, and probably wasn't going to beat me).\r\n\r\nIt just so happened that lots of miyomiyo and hm29168 fell in lots of the same categories on execution quiz 7 (the one I got 10/10), so I would have still gotten like 8 had I been wrong.", "Solution_20": "[quote=\"BOGTRO\"]\nThanks to [b]kimmystar94[/b], for backing down, and allowing me\n\nto step in, effectively crushing kimmy :P jk\n[/quote]\r\n\r\nthanks so much... :mad: hehe..\r\ndon't worry. i'm gonna apply for the next mole... teehee!", "Solution_21": "I don't think the Mole should win if only one person figured out who he is. I mean, hypothetically (obviously this didn't happen), miyomiyo could have gotten like 10 on the test because my play was so amazing that he was thoroughly convinced that I was the mole.\r\n\r\nThe Mole should win if both players score <27 on the test and the pot is not reached. I mean, isn't that what the original rules (first post) said? Except instead of <27, non-perfect scores.\r\n\r\nSo according to the original rules, don't I win the game?", "Solution_22": "[quote=\"CatalystOfNostalgia\"]Yay. Moleish behavior that I noticed:\n\n3. Three Questions task: hm29168 put me down as all three answers. That makes absolutely no sense, as it would mean I was both meaner and nicer than miyomiyo. In addition, when guessing my answers, he put himself as all three, that also makes no sense and would never be right.\n\n4. Final task: he was uberstalling us.[/quote]\n\nComments: For point 3, I don't know ANYBODY here, and I didn't even pay attention to the questions, so I just put down the first player name that popped into my head for that.\n\nFor point 4, I was extremely busy on the day when all of the Final Task information got posted. I wasn't even intentionally trying to stall you guys.\n\nBut, whatever, I'm still happy I won. :P \n\nCan I still play in mole game 2?[/quote]", "Solution_23": "Just as my evidence:\r\n\r\n[quote=\"SamL\"]You see, the mole can win too. If (s)he prevents you from reaching your target pot(known only to myself and the mole), and prevents both of the final two players from getting a perfect score on the last test, then the mole will steal the cash and be the victor! But that certainly won\u2019t happen\u2026 [/quote]", "Solution_24": "Defense:\r\n\r\n[quote=\"perfect628\"] 11. Expect surprises along the way: I, like The Mole, am unpredictable\n\n8. I reserve the right to(but don\u2019t plan to) change the tasks or rules at any times[/quote]", "Solution_25": "Well, as BOGTRO pointed out, I was allowed to change them, and I did. The point of that rule is to make The Mole sabotage(pot total), while remaining unknown. I wasn't expecting a player to not try on the quizzes...This was an unpredictable flaw--And one that shouldn't have happened. Look at it this way: Catalyst, you won, but the players lost.", "Solution_26": "[quote=\"SamL\"]...This was an unpredictable flaw--And one that shouldn't have happened.[/quote]\r\n\r\n :P :D\r\n\r\nAnyway, does this occur in the actual Mole? Like the one on TV?", "Solution_27": "No, but it's different online...In real life, they're playing for money, and ratings. Online, everyone needs a win condition. So it has to be implemented, as otherwise The Mole is just there, and doesn't have to try very hard. He can just blatantly sabotage everything, so everyone knows who it is, or sabotage nothing, so the ot skyrockets, but no one knows who it is.", "Solution_28": "[quote=\"perfect628\"]I wasn't expecting a player to not try on the quizzes...This was an unpredictable flaw--And one that shouldn't have happened.[/quote] :lol: \r\n\r\nThe really unpredictable thing that messed up the game was that my random quizzes got me this far.", "Solution_29": "Yes...I suppose...If you play in The Mole 2, let's hope it doesn't happen again." } { "Tag": [ "search", "email", "MATHCOUNTS", "\\/closed" ], "Problem": "Is there a good way to encourage many other people to post on this forum? IT is a good way to discuss with other people. :)", "Solution_1": "Get your yellowbook and call all 1 million numbers in there.\r\n\r\n :roll: \r\n\r\nSigh* I was just kidding.\r\n\r\nThe best way is by word of mouth. Tell your friends, and even better, tell your teachers.", "Solution_2": "Tell them that it increases brainpower :D", "Solution_3": "Don't forget to mention that you get to meet cool people like me, the greatest loser on the face of the earth!", "Solution_4": "Submit this website to search engines.", "Solution_5": "There is already quite a large number of members - over 23000!\r\n\r\nHowever, telling people you know, especially at math club meetings, etc, will greatly help spreading popularity.", "Solution_6": "[quote=\"b-flat\"]There is already quite a large number of members - over 23000!\n\nHowever, telling people you know, especially at math club meetings, etc, will greatly help spreading popularity.[/quote]\r\nOnly about 1200 of them have 50+ posts. I myself had <20 posts until about 6 months ago.", "Solution_7": "And only 8685 of them have any posts, 6390 at least two.\r\n\r\nBut do we really want more spamming people here... :roll:", "Solution_8": "So what is the point of 20000+ people, when only like 1000 of them are active. \r\n\r\nSecondly, some people just sign up, then just don't come back again from sheer disappointment of not being able to solve any problems. I know a couple of kids like that from my school.", "Solution_9": "[quote=\"anirudh\"]So what is the point of 20000+ people, when only like 1000 of them are active. \n\nSecondly, some people just sign up, then just don't come back again from sheer disappointment of not being able to solve any problems. I know a couple of kids like that from my school.[/quote]\r\n\r\nHmm. Well I got the feeling that all of the forums are going and creeping a bit higher than the difficulty level. For example, I find it strange when a pre-olympiad forum has olympiad+ level questions.", "Solution_10": "[quote=\"anirudh\"]So what is the point of 20000+ people, when only like 1000 of them are active. \n\nSecondly, some people just sign up, then just don't come back again from sheer disappointment of not being able to solve any problems. I know a couple of kids like that from my school.[/quote]\r\nyea, but some people decide that they need to improve and start spamming furiously for some unknown reasons(*cough anirudh cough moogra cough*)", "Solution_11": "[quote=\"anirudh\"]So what is the point of 20000+ people, when only like 1000 of them are active. [/quote]What do you understand by active? Some people register accounts and subscribe to topics so that they receive email notifications of when the topic has been posted into. Others check the forum regularly and use the unread messages feature.", "Solution_12": "[quote=\"anirudh\"]Secondly, some people just sign up, then just don't come back again from sheer disappointment of not being able to solve any problems. I know a couple of kids like that from my school.[/quote]\r\nI was one of those people until about 6 months ago when I needed practice for state and I saw that I was at the level of the mathcounts forum, so I started posting like crazy.", "Solution_13": "Most forums have a very low percentage of people who post actively; I think AoPS actually has a higher percentage.\r\n\r\nWith every so many users that sign up, you will get a new user who posts, so you want lots of people to sign up. Like VV said, not everyone wants to post, but still visit, etc." } { "Tag": [], "Problem": "[color=darkblue]Sa se detemine numerele naturale $n, p$ unde $p$ este prim si\n$n^{3}+np=p^{2}+1.$[/color]", "Solution_1": "Nu zice nimeni nimic?", "Solution_2": "solutia mea se incadreaza la categoria brute-force\r\nimi e destul de lene sa o postez, dar e foarte straight-forward\r\nexista vreo solutie interesanta?\r\n[hide=\"raspuns\"]\n37,10\n[/hide]", "Solution_3": "[color=darkblue]Hm... nu stiu daca solutia mea e gen brute-force. Iat-o:\n[hide]Avem $(n-1)(n^{2}+n+1)=p(p-n)$ de unde $p>n$\nCum $p$ este prim rezulta ca din $p|(n-1)(n^{2}+n+1)$ avem\n$p|n-1$ sau\n$p|n^{2}+n+1$\nPrima varianta este exclusa, dat fiind ca ar rezulta ca $p\\leq n-13$ \nVerificandu-se valorile lui $k=$1,2,3, vom obtine pentru $k=3$ solutiile $(n,p)=(10,37)$\n[/hide]\nOk, asa te-ai gandit, sau altfel?[/color]", "Solution_4": "tot geva in genu\r\n[hide=\"click\"]\ndin cauza ca $p|n^{2}+n+1$, rezulta ca $n-1 | p-n$, deci $p-n=a(n-1)$, si de aici il inlocuiesc pe $p$ si apoi pun conditia ca $p|n^{2}+n+1$ si tot umblu la divizibilitate pe acolo pana cand ajung la $a=3$ sau ceva, ca da la sfarsit un polinom de gradu 3 in $a$.\n[/hide]" } { "Tag": [ "function", "induction", "calculus", "floor function", "algebra proposed", "algebra" ], "Problem": "Find all functions $ h: \\mathbb{N}\\to\\mathbb{N}$ satisfying the functional relation $ h(h(n))+h(n+1)=n+2$ for all $ n\\in\\mathbb{N}$.", "Solution_1": "By induction $ h(n)=[\\phi n+c]$, were $ \\phi=\\frac{\\sqrt 5-1}{2},c=\\frac{3(3-\\sqrt 5)}{2}=3-3\\phi$.", "Solution_2": "[quote=\"Rust\"]By induction $ h(n)=[\\phi n+c]$, were $ \\phi=\\frac{\\sqrt 5-1}{2},c=\\frac{3(3-\\sqrt 5)}{2}=3-3\\phi$.[/quote]\r\n\r\nCan you post your solution .Thanks :)", "Solution_3": "[quote=\"Rust\"]By induction $ h(n)=[\\phi n+c]$, were $ \\phi=\\frac{\\sqrt 5-1}{2},c=\\frac{3(3-\\sqrt 5)}{2}=3-3\\phi$.[/quote]\r\n\r\nI'm sorry, Rust, but your result is wrong again (calculus error, maybe).\r\nWith your formula, : \r\n$ h(2)=[2\\phi+c]=[\\sqrt{5}-1+\\frac{9-3\\sqrt 5)}{2}]=[\\frac{7-\\sqrt 5)}{2}]=2$\r\n$ h(3)=[3\\phi+c]=3$\r\n\r\n$ h(h(2))+h(3)=5\\neq 2+2$", "Solution_4": "[quote=\"pco\"][quote=\"Rust\"]By induction $ h(n)=[\\phi n+c]$, were $ \\phi=\\frac{\\sqrt 5-1}{2},c=\\frac{3(3-\\sqrt 5)}{2}=3-3\\phi$.[/quote]\n\nI'm sorry, Rust, but your result is wrong again (calculus error, maybe).\nWith your formula, : \n$ h(2)=[2\\phi+c]=[\\sqrt{5}-1+\\frac{9-3\\sqrt 5)}{2}]=[\\frac{7-\\sqrt 5)}{2}]=2$\n$ h(3)=[3\\phi+c]=3$\n\n$ h(h(2))+h(3)=5\\neq 2+2$[/quote]\r\n\r\n$ h(n)=[n\\alpha]-n+1$", "Solution_5": "[quote=\"quangpbc\"]Find all functions $ h: \\mathbb{N}\\to\\mathbb{N}$ satisfying the functional relation :\n\n$ h(h(n))+h(n+1)=n+2$ for all $ n\\in\\mathbb{N}$[/quote]\r\n\r\nThe unique solution is $ h(n)=1+\\lfloor(\\phi-1)n\\rfloor$ where $ \\phi=\\frac{1+\\sqrt{5}}{2}$\r\n\r\n$ 1).$ $ h(n)=1+\\lfloor(\\phi-1)n\\rfloor$ is a solution.\r\nLet $ a=\\phi-1$. We have $ a^{2}=1-a$\r\n$ h(h(n))=1+\\lfloor ah(n)\\rfloor=1+\\lfloor a+a\\lfloor an\\rfloor\\rfloor$\r\n$ h(h(n))=1+\\lfloor a+a(an-\\{an\\})\\rfloor$\r\n$ h(h(n))=1+\\lfloor a+a^{2}n-a\\{an\\})\\rfloor$\r\n$ h(h(n))=1+\\lfloor a+n-an-a\\{an\\})\\rfloor$\r\n$ h(h(n))=n+1+\\lfloor a-an-a\\{an\\})\\rfloor$\r\n\r\nThen, we must notice that $ a-an-a\\{an\\}<0$ and that $ -a+an+a\\{an\\}$ can't be an integer : $ -a+an+a\\{an\\}=$ $ -a+an+a(an-\\lfloor an\\rfloor)$ $ =-a+n-a\\lfloor an\\rfloor$ and this number can't be integer, else $ a$ would be a rational number.\r\n\r\nSince $ a-an-a\\{an\\}<0$ and is not an integer, then $ \\lfloor a-an-a\\{an\\})\\rfloor=-1-\\lfloor-a+an+a\\{an\\})\\rfloor$ and so :\r\n$ h(h(n))=n-\\lfloor-a+an+a\\{an\\})\\rfloor$\r\n$ h(h(n))=n+2-\\lfloor 2-a+an+a\\{an\\})\\rfloor$\r\n$ h(h(n))=n+2-(1+\\lfloor an\\rfloor+\\lfloor 1-a+(a+1)\\{an\\})\\rfloor)$\r\n\r\nOn the other hand, we have :\r\n$ h(n+1)=1+\\lfloor an+a\\rfloor$\r\n$ h(n+1)=1+\\lfloor an\\rfloor+\\lfloor \\{an\\}+a\\rfloor$\r\n\r\nSo it remains to show that $ \\lfloor 1-a+(a+1)\\{an\\})\\rfloor=\\lfloor \\{an\\}+a\\rfloor$ :\r\nIf $ \\{an\\}<1-a$, RHS is $ 0$ and $ 0<1-a+(a+1)\\{an\\}<1-a+1-a^{2}=1$. So LHS is $ 0$ too.\r\nIf $ \\{an\\}\\geq 1-a$, RHS is $ 1$ (since $ 1\\leq\\{an\\}+a<2$) and $ 2>1-a+(a+1)\\{an\\}\\geq 1-a+1-a^{2}=1$. So LHS is $ 1$ too.\r\n\r\nAnd so $ h(h(n)=n+2-h(n+1)$\r\nQ.E.D.\r\n\r\n$ 2).$ $ h(n)=1+\\lfloor(\\phi-1)n\\rfloor$ is the [u][b]unique [/b][/u]solution\r\nLet $ h(1)=u$\r\nThen $ h(u)+h(2)=3$ which allow two cases : $ h(2)=1$ or $ h(2)=2$\r\n\r\nCase 1 : $ h(2)=1$ and $ h(u)=2$\r\nThen $ h(h(2))+h(3)=4$ gives $ h(3)=4-u$ and $ u$ can only be $ 1,2$ or $ 3$ but :\r\n$ u=1$ implies $ h(1)=u=1$ but we already have $ h(u)=h(1)=2$\r\n$ u=2$ implies $ h(u)=h(2)=2$ but we already have $ h(2)=1$\r\n$ u=3$ implies $ h(u)=h(3)=2$ but we have $ h(3)=4-u=1$\r\nSo no solution exist in case 1.\r\n\r\nCase 2 : $ h(2)=2$ and $ h(u)=1$\r\nThen $ h(h(2))+h(3)=4$ gives $ h(3)=2$\r\nThen $ h(h(3))+h(4)=5$ gives $ h(4)=3$\r\nIts then easy to show that we must have $ u=1$ : consider $ m$ as the first positive integer for which $ h(m)=1$\r\nThen, if $ m>1$, $ h(h(m-1))=m+1-h(m)=m$\r\nBut, from $ h(h(n))+h(n+1)=n+2$ we have $ h(n+1)1$\r\n$ h(m-1)>1$ ($ m$ is the lowest) and so $ h(h(m-1))\\leq h(m-1)$. But $ h(2)=2$ and so $ m-1\\neq 2$ and $ m-1>1$ and so $ h(m-1)\\leq m-1$ and so $ h(h(m-1))\\leq m-1$, in contradiction with $ h(h(m-1))=m$\r\n\r\nSo $ m=u=1$\r\nSo $ h(1)=1$\r\nSo $ h(n)\\leq n$ $ \\forall n$\r\nSo $ h(n+1)=n+2-h(h(n))$ and $ h(n)$ is totally defined as soon as $ h(1)$ is defined\r\nSo there is a unique solution to the required equation.\r\n\r\nAnd so this unique solution is $ h(n)=1+\\lfloor(\\phi-1)n\\rfloor$" } { "Tag": [ "arithmetic sequence", "AMC" ], "Problem": "How many non-similar triangle have angles whose degree measures are distinct positive integers in arithmetic progression?\r\n\r\n$ \\textbf{(A) } 0 \\qquad \\textbf{(B) } 1 \\qquad \\textbf{(C) } 59 \\qquad \\textbf{(D) } 89 \\qquad \\textbf{(E) } 178$", "Solution_1": "[hide]Call the three angles $a-d, a, a+d$ with $d>0$. Then $(a-d)+a+(a+d)=3a=180^\\circ$ so $a=60^\\circ$. Thus $d$ is any integer from $1$ to $59$ so $C$[/hide]", "Solution_2": "[hide=\"Answer\"]Since we have three angles, the middle angle must be $60$ so that the three sum to $180$. The other angles can range from $(59,61)$ to $(1,119)$, so the answer is $\\boxed{C}$.[/hide]", "Solution_3": "I basically did the same thing, but I did it in a longer way. I made the three angles angles in an arithmetic progression and used the sum rule. The sum rule states that the second and third angle have to equal 120, so the second angle is 60. If we let x be the first angle, then 0 1, k only occurs if k-1 occurs before the last occurrence of k. How many full sequences are there for each n?", "Solution_1": "The answer is $ n!$. This is surprising since just looking at the \"full\" sequences, they don't look much like permutations. For example when $ n\\equal{}3$ we have the following sequences:\r\n\r\n111\r\n112\r\n121\r\n122\r\n123\r\n212\r\n\r\nA very nice and clear exposition of a bijection between the set of \"full\" sequences and permutations can be found [url=http://www.rose-hulman.edu/mathjournal/archives/2005/vol6-n2/paper6/v6n2-6pd.pdf]here[/url]." } { "Tag": [ "function", "algebra", "polynomial", "geometry", "perimeter" ], "Problem": "Let $ f$ be a polynomial function such that, for all real $ x$, $ f(x^2 \\plus{} 1) \\equal{} x^4 \\plus{} 5x^2 \\plus{} 3$.\r\n\r\nFor all real $ x$, $ f(x^2 \\minus{} 1)$ is:\r\n\r\n(1983 AHSME #18)\r\n\r\n\r\nHow many non-congruent [b]obtuse/acute/right[/b] triangles such that the perimeter in cm and the area in square centimetres are numerically equal?\r\n\r\n(Solve for each case, i.e. obtuse, acute and right)\r\n\r\nI know the answer is quite clearly [b]infinitely many,[/b] but could someone supply a proof?\r\n\r\nThis is similar to question 24 on the 1983 AHSME[/b]", "Solution_1": "[quote=\"ThetaPi\"]Let $ f$ be a polynomial function such that, for all real $ x$, $ f(x^2 \\plus{} 1) \\equal{} x^4 \\plus{} 5x^2 \\plus{} 3$.\n\nFor all real $ x$, $ f(x^2 \\minus{} 1)$ is:\n\n(1983 AHSME #18)\n[/quote]\r\n\r\n$ x^4 \\plus{} 5x^2 \\plus{} 3 \\equal{} (x^2 \\plus{} 1)^2 \\plus{} 3(x^2 \\plus{} 1) \\minus{} 1$ so $ f(y) \\equal{} y^2 \\plus{} 3y \\minus{} 1$\r\n\r\nThen $ f(x^2 \\minus{} 1) \\equal{} (x^2 \\minus{} 1)^2 \\plus{} 3(x^2 \\minus{} 1) \\minus{} 1 \\equal{} x^4 \\plus{} x^2 \\minus{} 3$\r\n\r\nand for the second question...\r\nRight Triangles:\r\nThere is a general form for pythagorean triples so we are done.\r\n\r\nummm for the other cases, we only have to find one triangle and we can just scale it. 13-14-15 triangle is acute and has area 84.\r\n\r\ntake a look at this link:\r\nhttp://mathworld.wolfram.com/IntegerTriangle.html", "Solution_2": "Just plug in $ \\sqrt{x^2\\minus{}2}$ for x. Then note that if you let $ g(x)\\equal{}f(x^2\\plus{}1)$ then $ g(x)\\equal{}x^4\\plus{}5x^2\\plus{}3$ for infinitely many x, so it is identically equal to it. This means that the relation holds for all x, real or complex, so whether or not $ \\sqrt{x^2\\minus{}2}$ is real does not effect the final answer. Then just multiply it out.", "Solution_3": "[quote=\"jli\"][quote=\"ThetaPi\"]Let $ f$ be a polynomial function such that, for all real $ x$, $ f(x^2 \\plus{} 1) \\equal{} x^4 \\plus{} 5x^2 \\plus{} 3$.\n\nFor all real $ x$, $ f(x^2 \\minus{} 1)$ is:\n\n(1983 AHSME #18)\n[/quote]\n\n$ x^4 \\plus{} 5x^2 \\plus{} 3 \\equal{} (x^2 \\plus{} 1)^2 \\plus{} 3(x^2 \\plus{} 1) \\minus{} 1$ so $ f(y) \\equal{} y^2 \\plus{} 3y \\minus{} 1$\n\nThen $ f(x^2 \\minus{} 1) \\equal{} (x^2 \\minus{} 1)^2 \\plus{} 3(x^2 \\minus{} 1) \\minus{} 1 \\equal{} x^4 \\plus{} x^2 \\minus{} 3$\n\nand for the second question...\nRight Triangles:\nThere is a general form for pythagorean triples so we are done.\n\nummm for the other cases, we only have to find one triangle and we can just scale it. 13-14-15 triangle is acute and has area 84.\n\ntake a look at this link:\nhttp://mathworld.wolfram.com/IntegerTriangle.html[/quote]\r\n\r\nUmm... You can't scale it. Then it would NOT be numerically equal anymore.", "Solution_4": "[quote=\"herefishyfishy1\"][quote=\"jli\"][quote=\"ThetaPi\"]Let $ f$ be a polynomial function such that, for all real $ x$, $ f(x^2 \\plus{} 1) \\equal{} x^4 \\plus{} 5x^2 \\plus{} 3$.\n\nFor all real $ x$, $ f(x^2 \\minus{} 1)$ is:\n\n(1983 AHSME #18)\n[/quote]\n\n$ x^4 \\plus{} 5x^2 \\plus{} 3 \\equal{} (x^2 \\plus{} 1)^2 \\plus{} 3(x^2 \\plus{} 1) \\minus{} 1$ so $ f(y) \\equal{} y^2 \\plus{} 3y \\minus{} 1$\n\nThen $ f(x^2 \\minus{} 1) \\equal{} (x^2 \\minus{} 1)^2 \\plus{} 3(x^2 \\minus{} 1) \\minus{} 1 \\equal{} x^4 \\plus{} x^2 \\minus{} 3$\n\nand for the second question...\nRight Triangles:\nThere is a general form for pythagorean triples so we are done.\n\nummm for the other cases, we only have to find one triangle and we can just scale it. 13-14-15 triangle is acute and has area 84.\n\ntake a look at this link:\nhttp://mathworld.wolfram.com/IntegerTriangle.html[/quote]\n\nUmm... You can't scale it. Then it would NOT be numerically equal anymore.[/quote]\r\n\r\nright i read it wrong...again" } { "Tag": [ "modular arithmetic" ], "Problem": "if n is natural and 3n+1 and 4n+1 are perfect squares then prove that 56 divides n", "Solution_1": "[hide=\"2 facts that could be useful\"]squares can only be $ 0,1\\pmod{8}$ and $ 0,1,2,4\\pmod{7}$ :wink: [/hide]", "Solution_2": "[quote=\"Albanian Eagle\"][hide=\"2 facts that could be useful\"]squares can only be $ 0,1\\pmod{8}$ and $ 0,1,2,4\\pmod{7}$ :wink: [/hide][/quote]\r\n\r\n$ 0,1,4\\pmod{8}$ :)", "Solution_3": "yes :lol: , but we won't need the $ 0$ nor the $ 4$" } { "Tag": [ "USAMTS", "LaTeX" ], "Problem": "When I compile the PDF on my USAMTS solutions, the first page overruns and I can't see most of my solution. I don't know what happened. Could someone give me some pointers on how to fix this problem?", "Solution_1": "Does it run off the side of the page or the bottom of the page?\r\n\r\nBecause if it is the side, add some \"\\\\\"s to bring the LaTeX down to the next line.", "Solution_2": "Thanks. :)" } { "Tag": [ "algebra", "polynomial", "number theory", "greatest common divisor", "complex numbers", "Galois Theory", "complex analysis" ], "Problem": "Prove that it is impossible for the root of unity to have a real part in the form $ \\sqrt{k\\plus{}1}\\minus{}\\sqrt{k}$ for some positive integers $ k$.", "Solution_1": "Note that either $ k \\plus{} 1$ or $ k$ is not a perfect square. Let us say $ k$ is not a perfect square. Then $ \\sqrt {k \\plus{} 1} \\minus{} \\sqrt {k}$ is conjugate to $ \\sqrt {k \\plus{} 1} \\plus{} \\sqrt {k}$ over $ \\mathbb Q$ (since $ k$ and $ k \\plus{} 1$ are coprime). Therefore, there exists an automorphism $ \\phi: \\bar {\\mathbb Q} \\to \\bar {\\mathbb Q}$ that maps $ \\sqrt {k \\plus{} 1} \\minus{} \\sqrt {k}$ to $ \\sqrt {k \\plus{} 1} \\plus{} \\sqrt {k}$. \r\n\r\nNow assume that $ \\xi$ is a root of unity and $ \\mathrm{Re}(\\xi) \\equal{} \\sqrt {k \\plus{} 1} \\minus{} \\sqrt {k}$. Then\r\n\\[ \\frac {\\xi \\plus{} \\xi^{ \\minus{} 1}}{2} \\equal{} \\mathrm{Re}(\\xi) \\equal{} \\sqrt {k \\plus{} 1} \\minus{} \\sqrt {k},\r\n\\]\r\nand\r\n\\[ \\frac {\\phi(\\xi) \\plus{} \\phi(\\xi)^{ \\minus{} 1}}{2} \\equal{} \\phi(\\sqrt {k \\plus{} 1} \\minus{} \\sqrt {k}) \\equal{} \\sqrt {k \\plus{} 1} \\plus{} \\sqrt {k}.\r\n\\]\r\nOn the other hand, since $ \\xi$ is a root of unity, $ \\phi(x)$ is also a root of unity and $ |\\phi(x)| \\equal{} 1$. Therefore,\r\n\\[ \\left|\\frac {\\phi(\\xi) \\plus{} \\phi(\\xi)^{ \\minus{} 1}}{2}\\right| \\leq 1.\r\n\\]\r\nWe get a contradiction.", "Solution_2": "[quote=\"Yury\"]Note that either $ k \\plus{} 1$ or $ k$ is not a perfect square. Let us say $ k$ is not a perfect square. Then $ \\sqrt {k \\plus{} 1} \\minus{} \\sqrt {k}$ is conjugate to $ \\sqrt {k \\plus{} 1} \\plus{} \\sqrt {k}$ over $ \\mathbb Q$ (since $ k$ and $ k \\plus{} 1$ are coprime). Therefore, there exists an automorphism $ \\phi: \\bar {\\mathbb Q} \\to \\bar {\\mathbb Q}$ that maps $ \\sqrt {k \\plus{} 1} \\minus{} \\sqrt {k}$ to $ \\sqrt {k \\plus{} 1} \\plus{} \\sqrt {k}$. \n[/quote]\r\n\r\nhi yury, thanks for replying.. i have questions, what do you mean conjugate [b]over[/b] $ \\mathbb{Q}$. and also why there is an automorphism.. what theorem did you use?\r\n\r\nthanks.", "Solution_3": "Two numbers are conjugate over $ \\mathbb Q$ if they are roots of the same irreducible polynomial (over $ \\mathbb Q$) with rational coefficients (see http://en.wikipedia.org/wiki/Conjugate_element ).\r\n\r\nThe theorem I am using is a basic theorem from the Galois theory ( http://en.wikipedia.org/wiki/Galois_theory ):\r\n\r\n[b]Theorem.[/b] For every two conjugate numbers $ \\alpha$ and $ \\beta$ there exists an automorphism of $ \\mathbb C$ that maps $ \\alpha$ to $ \\beta$.[size=75] (The statement applies to other fields as well.)[/size]\r\n\r\nWe can prove the statement \"explicitly\" without using the Galois theory (but essentially this will be the same proof). \r\n\r\nFor simplicity, suppose that $ k \\plus{} 1$ is a perfect square and $ k$ is not. Denote $ \\alpha \\equal{} \\sqrt {k \\plus{} 1} \\minus{} \\sqrt {k}$ and $ \\beta \\equal{} \\sqrt {k \\plus{} 1} \\plus{} \\sqrt {k}$. Observe that $ \\alpha, \\beta \\notin \\mathbb Q$ but $ \\alpha \\plus{} \\beta$ and $ \\alpha\\beta$ are in $ \\mathbb Q$. Let $ \\xi$ be a root of unity as in my post above; say, $ \\xi^d \\equal{} 1$. We have $ \\xi^2 \\plus{} 1 \\minus{} 2\\xi \\alpha \\equal{} 0$ (why?). Thus $ (\\xi^2 \\plus{} 1 \\minus{} 2\\xi \\alpha)(\\xi^2 \\plus{} 1 \\minus{} 2\\xi \\beta) \\equal{} 0$. That is, $ \\xi$ is a root of the polynomial\r\n\\[ f(x) \\equal{} (x^2 \\plus{} 1 \\minus{} 2\\alpha x) (x^2 \\plus{} 1 \\minus{} 2\\beta x).\r\n\\]\r\nObserve that $ f(x)$ has rational coefficients (use the fact that $ \\alpha \\plus{} \\beta$ and $ \\alpha\\beta$ are in $ \\mathbb Q$). We get that $ \\xi$ is a root of $ x^d \\minus{} 1$ and a root of $ f(x)$. Therefore, $ \\xi$ is a root of $ g(x) \\equiv \\mathrm{gcd}(f(x), x^d \\minus{} 1)$ (see http://en.wikipedia.org/wiki/Greatest_common_divisor_of_two_polynomials ). Note that $ g(x)$ is a polynomial with rational coefficients since both $ x^d \\minus{} 1$ and $ f(x)$ have rational coefficients; also every root $ r$ of $ g(x)$ is a root of unity (why?) and a root of $ f(x)$. Therefore, $ |r| \\equal{} 1$ and $ \\mathrm{Re} (r) \\in \\{\\alpha, \\beta\\}$. Since $ \\beta > 1$, $ \\mathrm{Re}(r) \\equal{} \\alpha$. Thus $ r\\in \\{\\xi, \\bar\\xi \\}$; and $ g(x) \\equal{} x \\minus{} \\xi$ or $ g(x) \\equal{} (x \\minus{} \\xi)(x \\minus{} \\bar \\xi)$. But neither of these two polynomials have rational coefficients! We get a contradiction.\r\n\r\nIn general, if $ k \\plus{} 1$ is not a perfect square we need to consider the polynomial $ f(x) \\equal{} \\prod(x^2 \\plus{} 1 \\minus{} 2 (\\pm\\sqrt {k \\plus{} 1} \\pm\\sqrt {k}))$. The rest of the argument is very similar.", "Solution_4": "Thank you very much yury. I understand your nice proof :)" } { "Tag": [ "geometry", "rectangle", "ARML", "analytic geometry", "vector" ], "Problem": "I think there might be a way to do this analytically but I can't see a nice way.\r\nProblem: In rectangle ABCD, AB=6 and BC=8. Equilateral triangles ADE and DCF are drawn on the exterior of ABCD. Find the area of triangle BEF.\r\n\r\nSource: 1996 Chicago ARML tryouts #24", "Solution_1": "You could easily find BE, EF and BF. Then, use Herons formula.\r\nOr I guess you could partition BEF into three sections so that the center point where they're partitionedis 3 from BC and 4 from AB. Then you could find the areas.", "Solution_2": "[quote=\"K81o7\"]Or I guess you could partition BEF into three sections so that the center point where they're partitioned is 3 from BC and 4 from AB. Then you could find the areas.[/quote]\r\n\r\nAnother way to approach this is to introduce coordinates; put $B$ at the origin, $A$ and $C$ on the axes, and find $E=(x_1,y_1)$ and $F=(x_2,y_2)$. The area is then $\\frac12|x_1y_2-x_2y_1|.$ This formula can be derived in general by dissections like the one quoted above.\r\n\r\nThe area will have a $\\sqrt{3}$ in it; don't expect a rational answer with the irrational coordinates.", "Solution_3": "Thanks-that works very well. And there was a rad3 in the area. Is that the same as one of those proof without words formulas running in the boxes on the left part of the screen?", "Solution_4": "Is the answer by any chance 36 + 25 :rt3: ? I just found the area of ABCFE and then subtracted the areas of triangles ABE and BCF.", "Solution_5": "That's what I got, by setting up the picture in coordinates and finding the area the way jmerry described it. My coordinates had A at (0, 0) and the verticies of the triangle at (0, 6), (4, -4sqrt(3)), (8+3sqrt(3), 3)", "Solution_6": "[quote=\"WarpedKlown1335\"]Is the answer by any chance 36 + 25 :rt3: ? I just found the area of ABCFE and then subtracted the areas of triangles ABE and BCF.[/quote]\r\n\r\nThat is what i got :D \r\n[hide]I drew the rectangle like:\nA-------D\nB-------C\n\nThen I assigned (0,0) to point B, the point E in relation to B becomes (4, 6 + 4 :rt3:) and F becomes (8 + 3 :rt3:) Then using the shoelace method the area of the triangle is 36 + 25 :rt3:[/hide]\r\n\r\nedit - Wumbate beat me to it :blush:", "Solution_7": "[ABCFE] = [ABCD] + [ADE] + [CDF] + [DEF] = 48 + 9 :rt3: + 16 :rt3: + 12 = 60 + 25 :rt3: .\r\n\r\n[ABE] + [BCF] = 24.\r\n\r\n[BEF] = [ABCFE] - [ABE] - [BCF] = 60 + 25 :rt3: - 24 = 36 + 25 :rt3: .\r\n\r\nUsing analytic geometry for this problem is kind of overkill don't you think? ;)", "Solution_8": "Yeah, you are all right that is the official answer. I don't think analytical geometry is really that much overkill...most of the time i'm more likely to figure out a problem like this with coordinates and vectors, etc. than with \"pure\" geometrical approaches. But either way, you get the same answer. :)" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "find max of \\[ \\frac{15}{2}a^2 + 4ab \\]\r\nif \\[ a^2 + b^2=2 \\]\r\n\r\nwhat i did was \\[ \\frac{15}{2}a^2 + 4ab\\leq k \\]\r\n\\[ k=(a^2+b^2)\\frac{k}{2} \\]\r\n\\[ \\frac{k-15}{2}a^2 - 4ab + \\frac{k}{2}b^2 \\geq 0 \\]\r\nI assumed that it should be in this form \\[ (1) (\\sqrt{\\frac{k-15}{2}}a - \\sqrt{\\frac{k}{2}}b)^2\\geq 0 \\]\r\nand thus \\[ 2\\sqrt{\\frac{k-15}{2}}\\sqrt{\\frac{k}{2}}=4 \\]\r\nso the answer is\\[ k= 16 \\]\r\n\r\nI would like to know if i can assume (1) or is there another way to prove it...\r\nthanks", "Solution_1": "yes you can. In fact when you write your solution, you only need to write\r\n$\\frac{15}{2}a^2+4ab \\leq \\frac{15}{2}a^2+\\frac{a^2+16b^2}{2} = 8(a^2+b^2) =16$" } { "Tag": [ "vector", "algebra", "system of equations", "linear algebra" ], "Problem": "Prove that every linear system of equations falls into one of the three categories:\r\nno solution, one solution, or infinitely many solutions.", "Solution_1": "Given a linear system $ \\mathbf{Ax} \\equal{} \\mathbf{b}$, if $ \\mathbf{b}$ is not in the column space of $ \\mathbf{A}$ then no solutions are possible. Otherwise, the set of solutions is a translate of the nullspace of $ \\mathbf{A}$ (which is a much stronger statement than the one desired) and in particular is a vector space. Dimension zero is one solution; dimensions one and higher are infinite (in a vector space over an infinite field)." } { "Tag": [ "number theory open", "number theory" ], "Problem": "Here is my conjecture.\r\n\r\nLet $ b\\in\\mathbb{Z}$ such that $ b > 1$. Define, for each $ n\\in\\mathbb{N}$, $ S_{b}(n)$ and $ P_{b}(n)$ to be the sum and the product of digits of $ n$, when expressed in the base-$ b$ representation.\r\nFor each positive integers $ u$ and $ v$, there exists $ n$ for which\r\n\\[ uS_{b}(n)\\plus{}vP_{b}(n) \\equal{} n\\,.\\]", "Solution_1": "It's false.\r\nDoesn't exist n such that:\r\n$ S_{10}\\plus{}2P_{10}\\equal{}n$\r\nIn this case you can prove $ n$ must have 4 or 5 digit.", "Solution_2": "[quote=\"Vinxenz\"]It's false.\nDoesn't exist n such that:\n$ S_{10}\\plus{}2P_{10}\\equal{} n$\nIn this case you can prove $ n$ must have 4 or 5 digit.[/quote]\r\n\r\n$ n\\equal{}669$\r\n$ S_{10}(669)\\equal{}21$\r\n$ P_{10}(669)\\equal{}324$\r\n\r\n$ 21\\plus{}2*324\\equal{}669$" } { "Tag": [ "limit", "algebra unsolved", "algebra" ], "Problem": "Let $ a_n $ be a sequence satisfying : $ a_n = 1 + \\frac{n+1}{2n}.a_{n-1} $ for n >=2 . Find $ \\lim_{n\\to \\infty} a_n $ ?", "Solution_1": "[quote=\"romano\"]Let $ a_n $ be a sequence satisfying : $ a_n = 1 + \\frac{n+1}{2n}.a_{n-1} $ for n >=2 . Find $ \\lim_{n\\to \\infty} a_n $ ?[/quote]\r\n\r\nThe answer seems to be 2, no matter what a_1 is. But I don't know why.", "Solution_2": "$\\lim_{n \\to \\infty} \\frac{n+1}{2n} = \\frac 12$, so if $\\lim_{n \\to \\infty} = a$ then $a = 1+\\frac a2$, from which $a = 2$.", "Solution_3": "[quote=\"pleurestique\"]$\\lim_{n \\to \\infty} \\frac{n+1}{2n} = \\frac 12$, so if $\\lim_{n \\to \\infty} = a$ then $a = 1+\\frac a2$, from which $a = 2$.[/quote]\r\n\r\nThis isn't a complete proof, is it? All it shows is that if the limit exist, the limit is 2.", "Solution_4": "yeah, that's right, *if* it exists...", "Solution_5": "and it seems to be true that it always exists. i'll post with something rigorous tomorrow, unless someone else posts in the meantime. gotta sleep.." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ x_1,x_2,\\cdots,x_n$ such that $ x_1x_2\\cdots x_n \\equal{} 1$,prove that\r\n\\[ \\frac {1}{1 \\plus{} (n \\minus{} 1)x_1} \\plus{} \\frac {1}{1 \\plus{} (n \\minus{} 1)x_2} \\plus{} \\cdots \\plus{} \\frac {1}{1 \\plus{} (n \\minus{} 1)x_n} \\ge 1\\]\r\n\r\n\\[ \\ge \\frac {1}{1 \\plus{} x_2 \\plus{} \\cdots \\plus{} x_n} \\plus{} \\frac {1}{1 \\plus{} x_1 \\plus{} x_3 \\plus{} \\cdots \\plus{} x_n} \\plus{} \\cdots \\plus{} \\frac {1}{1 \\plus{} x_1 \\plus{} \\cdots \\plus{} x_{n \\minus{} 1}}\\]\r\n\r\n\\[ \\ge \\frac {1}{n \\minus{} 1 \\plus{} x_1} \\plus{} \\frac {1}{n \\minus{} 1 \\plus{} x_2} \\plus{} \\cdots \\plus{} \\frac {1}{n \\minus{} 1 \\plus{} x_n}\\]", "Solution_1": "for \\[ 1 \\ge\\frac{1}{1\\plus{}x_{2}\\plus{}\\cdots\\plus{}x_{n}}\\plus{}\\frac{1}{1\\plus{}x_{1}\\plus{}x_{3}\\plus{}\\cdots\\plus{}x_{n}}\\plus{}\\cdots\\plus{}\\frac{1}{1\\plus{}x_{1}\\plus{}\\cdots\\plus{}x_{n\\minus{}1}}\\]\r\n\r\nwe have by AM-GM $ x_{2}\\plus{}\\cdots\\plus{}x_{n} \\ge n\\minus{}1$\r\n\r\nby the similary way we are going to find $ \\frac{1}{1\\plus{}x_{2}\\plus{}\\cdots\\plus{}x_{n}}\\plus{}\\frac{1}{1\\plus{}x_{1}\\plus{}x_{3}\\plus{}\\cdots\\plus{}x_{n}}\\plus{}\\cdots\\plus{}\\frac{1}{1\\plus{}x_{1}\\plus{}\\cdots\\plus{}x_{n\\minus{}1}} \\leq n.\\frac{1}{n} \\equal{} 1$", "Solution_2": "why?$ x_{2}\\plus{}\\cdots\\plus{}x_{n}\\ge n\\minus{}1$", "Solution_3": "[quote=\"zhaobin\"]why?$ x_{2} \\plus{} \\cdots \\plus{} x_{n}\\ge n \\minus{} 1$[/quote]\r\nHe is wrong because by AM-GM and the condition, we would have $ x_{2} \\plus{} \\ldots \\plus{} x_{n}\\geqslant (n \\minus{} 1)\\sqrt[n\\minus{}1]{\\frac{1}{x_1}}$...", "Solution_4": "[quote=\"Martin N.\"][quote=\"zhaobin\"]why?$ x_{2} \\plus{} \\cdots \\plus{} x_{n}\\ge n \\minus{} 1$[/quote]\nHe is wrong because by AM-GM and the condition, we would have $ x_{2} \\plus{} \\ldots \\plus{} x_{n}\\geqslant (n \\minus{} 1)\\sqrt [n \\minus{} 1]{\\frac {1}{x_1}}$...[/quote]\r\nBut why $ (n \\minus{} 1)\\sqrt [n \\minus{} 1]{\\frac {1}{x_1}} \\ge n\\minus{}1$..." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all functions $ f: \\mathbb{N}\\rightarrow\\mathbb{N}$ such that for all $ m,n\\in\\mathbb{N}$,\r\n\r\n$ (2^m\\plus{}1)f(n)f(2^mn)\\equal{}2^mf(n)^2\\plus{}f(2^mn)^2\\plus{}(2^m\\minus{}1)^2n$", "Solution_1": "The equation equivalent:\r\n$ (2^mf(n) \\minus{} f(2^mn))(f(2^mn) \\minus{} f(n)) \\equal{} n(2^m \\minus{} 1)^2$\r\nBecause $ 2^mf(n) \\minus{} f(2^{m}n)\\equiv f(n) \\minus{} f(2^{m}n) (\\mod 2^m \\minus{} 1)$ so easy to check that :\r\n$ 2^m \\minus{} 1|f(2^{m}n) \\minus{} f(n)$\r\nExist $ (g_n,h_n)\\in N$ such that:\r\n$ 2^mf(n) \\minus{} f(2^{m}n) \\equal{} g_n(2^m \\minus{} 1)$ and $ f(2^{m}n) \\minus{} f(n) \\equal{} h_n(2^m\\minus{}1)$ where $ g(n).h(n) \\equal{} n$\r\n$ \\Rightarrow f(n) \\equal{} g_n \\plus{} h_n,f(2^{m}n) \\equal{} 2^{m}h_n \\plus{} g_n,f(n).g(n) \\equal{} n$\r\nWe can find $ f(m)$ if we know all $ f(n)$ where n is an odd number.\r\nEasy to check that this function satisfy condition . For example $ g_n \\equal{} 1,h_n \\equal{} n$ then $ f(n) \\equal{} n \\plus{} 1$" } { "Tag": [ "algebra", "polynomial", "calculus", "integration", "algebra proposed" ], "Problem": "Let $f(X) = a_{0}+a_{1}X+\\hdots+a_{n-1}X^{n-1}+X^{n}\\in\\mathbb{Q}[X]$ be a monic polynomial with integral coefficients. Show that if $\\overline{a_{0}}+\\overline{a_{1}}X+\\hdots+\\overline{a_{n-1}}X^{n-1}+X^{n}\\in\\mathbb{Z}_{p}[X]$ is irreducible (for some prime $p$), then $f(X)$ is irreducible over $\\mathbb{Q}$.\r\n\r\nEDIT: Added something.", "Solution_1": "Isn't it obvious? If $f(x)$ would be reducible over $\\mathbb{Z}$, let's say $f(x) = p(x)q(x)$ then of course also $f(x) \\equiv p(x)q(x) \\mod{p}$. So $f(x)$ is irreducible over $\\mathbb{Z}$ and the irreducibility over $\\mathbb{Q}$ follows from Gauss's Lemma.", "Solution_2": "$f(x)=(px+1)(x+1)$?", "Solution_3": "Is your example monic ?? :wink:" } { "Tag": [], "Problem": "A telephone call costs $ \\$1.19$ for the first three minutes and $ \\$0.28$ for each additional minute. If a call started at 10:55:00 AM and ended at 11:06:00 AM, how much did the call cost?", "Solution_1": "the elapsed time is 9 min. so the price is $ \\$1.19\\plus{}\\$0.28*(9\\minus{}3)\\equal{}\\$2.87$", "Solution_2": "I'm pretty sure the elapsed time is $ 11$ minutes so $ 1.19\\plus{}0.28(11\\minus{}3)\\equal{}\\boxed{3.43}$?" } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "Let $ P(x)$ be a polynomial of degree n with real coefficients and let $ a \\geq 3$. Prove that $ \\max_{0 \\leq j \\leq n\\plus{}1}|a^j \\minus{} P(j)| \\geq 1$", "Solution_1": "We can write $ P(x) \\equal{} \\sum_{i \\equal{} 0}^n (x_i\\plus{}(a\\minus{}1)^i)C_x^i.$ Thent $ b_j\\equal{}P(j)\\minus{} a^j \\equal{}\\sum_{i\\equal{}0}^jC_j^ix_i, j 3$", "Solution_3": "I am sorry, I edit my post." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "f is continuous on [a,b] \r\n\u2200x\u2208[a,b] x is local minimum.\r\nshow that f is a constant function\r\nthanks~", "Solution_1": "Hardly an unexpected result, but, it turns out, not as trivial as one may think. :D (I'm not saying it's hard... it's just not *that* trivial.)\r\n\r\nAnyway, since $ f$ is continuous, the intermediate value theorem applies and thus there exist $ x_m\\in [a,b]$ and $ x_M\\in [a,b]$ such that $ m \\equal{} f(x_m)$ is a global minimum of $ f$ and $ M \\equal{} f(x_M)$ is a global maximum of $ f$. (So, $ m\\leq M$. We want to show $ m \\equal{} M$ - then we are obviously done.)\r\n\r\nWithout loss of generality, we assume $ x_m\\leq x_M$. If $ x_m \\equal{} x_M$ we have $ m \\equal{} f(x_m) \\equal{} f(x_M) \\equal{} M$ and we're done. So, $ x_m < x_M$. Now we'll take a look at $ S$ defined with $ S \\equal{} \\{x\\in [a,b]: x > x_m \\wedge f(x) \\equal{} M\\}$. Since $ x_M\\in S$, $ S$ is not empty and since $ S\\subseteq [a,b]$, there exists $ t \\equal{} \\inf S$.\r\n\r\nObviously, $ t\\geq x_m$. If $ t \\equal{} x_m$, then we have a sequence $ (r_k)_{k\\in\\mathbb{N}}\\subseteq S$ such that $ r_k\\rightarrow x_m$. Since $ f$ is continuous, $ f(r_k)\\rightarrow f(x_m)$ and therefore, $ M \\equal{} m$. \r\n\r\nSo, the only case left is $ t > x_m$. We'll prove that isn't possible. Since $ f$ is continuous, $ t\\in S$ (again, the same argument with the sequence as in the previous paragraph). So, $ f(t) \\equal{} M$. We know that $ t$ is a local minimum, so there exists $ \\delta > 0$ such that $ f(x)\\geq f(t) \\equal{} M$ for all $ x\\in\\langle t \\minus{} \\delta,t]$. However, since $ M$ is a global maximum, $ f(x)\\leq M$ for all $ x$. So, $ x\\in\\langle t \\minus{} \\delta,t]\\rightarrow f(x) \\equal{} M$. Now, since $ t > x_m$, $ t > \\max\\{x_m,t \\minus{} \\delta\\}$. So, there exists $ p$ such that $ t > p > \\max\\{x_m,t \\minus{} \\delta\\}$. (For instance, you can take $ p \\equal{} \\frac {t \\plus{} \\max\\{x_m,t \\minus{} \\delta\\}}{2}$.) In that case, then, $ p > x_m$ and $ f(p) \\equal{} M$. So, $ p\\in S$. However, since $ p < t \\equal{} \\inf S$, that is a contradiction.", "Solution_2": "Hmmm to complicated. But the approach is right. Take $ M \\equal{} \\max_{[a,b]} f$ and $ x' \\equal{} \\inf\\{x\\in[a,b]: f(x) \\equal{} M\\}$, $ x'' \\equal{} \\sup\\{x\\in[a,b]: f(x) \\equal{} M\\}$. Obviously, $ f(x') \\equal{} f(x'') \\equal{} M$ and we have $ f(x)iai, piai+1)\r\n\r\nWe choose n s.t. n+piai is divisible with piai+ci but not with piai+ci+1. It seems to me right now that this works, but there might, of course, be a mistake. Can someone please check it?" } { "Tag": [], "Problem": "For each $n\\ge 1$, we call a sequence of $n$ (s and $n$ )s \"legal\" if the parentheses match up, somehow. For example, if $n=4$, the sequence (()()()) is legal, but ()())(() is not. Let $l_{n}$ denote the number of legal arrangements for the $2n$ parentheses. Find a recurrence relation for $l_{n}$.", "Solution_1": "It's simply the $n$th catalan number." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "$f$:$R^n\\rightarrow R^n$ and $n\\geq 2$\r\n$f$ is $C^{1}$ and $C>0$ lipschitzienne \r\n\r\nIf exist a solution $x$ of $x'=f(x)$, $T$-periodic \r\n\r\nProve $T\\geq \\frac{2\\pi}{C}$", "Solution_1": "Nice one. Hint: Wirtinger's inequality." } { "Tag": [ "calculus", "integration", "analytic geometry", "trigonometry", "geometry", "calculus computations" ], "Problem": "Find the volume of the region between surfaces $ x^2\\plus{}z^2\\equal{}b^2$, $ x^2\\minus{}y^2\\minus{}z^2\\equal{}0$, where $ x>0$.", "Solution_1": "[quote=\"tmrfea\"]Find the volume of the region between surfaces $ x^2 \\plus{} z^2 \\equal{} b^2$, $ x^2 \\minus{} y^2 \\minus{} z^2 \\equal{} 0$, where $ x > 0$.[/quote]\r\n\r\nCylinder/Cone intersection Volume: is not nice... The volume integral in modified cylindrical coordinates (swap x and z in regular cyl coords) is\r\n\r\n\\[ V\\equal{}\\int_{0}^{2\\pi}\\int_{0}^{{\\scriptstyle \\frac{b}{\\sqrt{1\\plus{}\\cos^2\\theta}}}}\\int_{r}^{{\\scriptstyle \\sqrt{b^2\\minus{}r^2\\cos^2\\theta} }}r\\, dx\\, dr\\, d\\theta \\equal{} {\\scriptstyle \\frac{1}{3}}b^3\\int_{0}^{2\\pi} \\frac{1\\minus{}\\left( 1\\plus{}\\cos^2\\theta \\right) ^{{\\scriptstyle \\minus{}\\frac{1}{2}}}}{\\cos^2\\theta}d\\theta\\]\r\n\r\nMaple give this in terms of Elliptic integrals as \r\n\r\n\\[ V\\equal{} 4b^3 \\left( \\mbox{EllipticE}( \\i ) \\minus{} \\mbox{EllipticK}( \\i )\\right)\\]\r\n\r\nIt is no wonder that this is the case when you study the geometry at hand...\r\n\r\n[img]http://i306.photobucket.com/albums/nn249/arinkovsky/intersectionCylandcone.jpg[/img] [img]http://i306.photobucket.com/albums/nn249/arinkovsky/intersectionCylandcone_back.jpg[/img]\r\n\r\n[img]http://i306.photobucket.com/albums/nn249/arinkovsky/intersectionCylandcone_elipse.jpg[/img]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Give $ a,b,c$ are non-negative real number.Prove that \r\n$ \\frac3{4}\\le \\frac{a^2}{(a\\plus{}b)(a\\plus{}c)}\\plus{}\\frac{b^2}{(b\\plus{}a)(b\\plus{}c)}\\plus{}\\frac{c^2}{(c\\plus{}a)(c\\plus{}b)}\\le 1$", "Solution_1": "its easy...just multyply out..", "Solution_2": "[quote=\"dduclam\"]Give $ a,b,c$ are non-negative real number.Prove that \n$ \\frac3{4}\\le \\frac {a^2}{(a \\plus{} b)(a \\plus{} c)} \\plus{} \\frac {b^2}{(b \\plus{} a)(b \\plus{} c)} \\plus{} \\frac {c^2}{(c \\plus{} a)(c \\plus{} b)}\\le 1$[/quote]\r\nThe inequality equaivalent to $ \\sum_{sym}a^2b \\ge 6abc$\r\nand $ 2abc \\ge 0$.Which is obviously true.", "Solution_3": "Yes,very easy. A similar problem: \r\nFind the min and max value of $ P\\equal{}\\sum\\frac{a^3}{(a\\plus{}b)(a\\plus{}c)(a\\plus{}d)}$ for $ a,b,c,d\\ge0$", "Solution_4": "it was appeared in Mathlink Forum, I remember that pvthuan purposed above problem and Japanese (Naoki Sato) solved it by AM-GM :) \r\nThe inequality is equivalent to:\r\n\\[ \\sum \\left(\\frac{a}{a\\plus{}2}\\right)^3\\ge \\frac{4}{27}\\]\r\nUsing the following result:\r\n \\[ \\left(\\frac{x}{x\\plus{}2}\\right)^3\\ge \\frac{2x\\minus{}1}{27}\\]\r\nIt is really nice but not hard too :)", "Solution_5": "You are right,Hoang. Find max problem already in pvthuan'book. Now,can we find the min value of P? \r\nMean find the max value of k such that:\r\n $ \\sum_{cyc}a^3(b\\plus{}c)\\ge k(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)(a\\plus{}b)$", "Solution_6": "[quote=\"dduclam\"]\nMean find the max value of k such that:\n $ \\sum_{cyc}a^3(b \\plus{} c)\\ge k(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)(a \\plus{} b)$[/quote]\r\nAre you sure? Maybe find the maximum value of $ k$ for which en inequality\r\n\\[ \\sum_{cyc}a^3(b \\plus{} c)\\ge k(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)(a \\plus{} b\\plus{}c)\\]\r\nholds for all non-negative $ a,$ $ b$ and $ c.$ :wink:", "Solution_7": "Sory,it's better is: find the maximum value of k such that:\r\n $ \\sum_{cyc}^{a,b,c,d}a^3(b \\plus{} c)(b\\plus{}d)(c\\plus{}d)\\ge k (a\\plus{}b)(a\\plus{}c)(a\\plus{}d)(b\\plus{}c)(b\\plus{}d)(c\\plus{}d)$\r\nfor all non-negative real numbers $ a,b,c$." } { "Tag": [ "absolute value" ], "Problem": "Given 187 integers, both the sum and product of these integers are 188. \r\n\r\nWhat's the sum of the absolute value of these integers?", "Solution_1": "236.\r\n\r\n[hide=\"rationale\"]\nfor the product to be 288, there must be a few factors of 288, and the rest are either 1 or -1.\n\nusing the prime factors 2, 2, and 47 doesn't work because an even amount of remaining 1's and -1's (284 of them) can't make the extra 188-2-2-27=odd number that we need\n\ncombining factors into 4 and 47 gives the odd amount needed.\n\nthe numbers, then, are:\n\n4, 47, 24 -1's, and 161 1's\n[/hide]", "Solution_2": "I think you accidentally said 288 at the beginning of your solution when you meant 188 ;) . Other than that it looks good to me." } { "Tag": [ "Putnam", "AMC", "USA(J)MO", "USAMO" ], "Problem": "Does anyone know the specific sources (or inspirations) for this year's Putnam and USAMO?", "Solution_1": "The solutions on the AMC page state the problem authors for the USAMO.", "Solution_2": "Oops, I didn't realize they had posted solutions. Sorry." } { "Tag": [ "function", "logarithms", "number theory", "totient function" ], "Problem": "find the first 3 and the last 3 digits of 2^5127. I was able to find the last three digits using Euler's totient function and mods, but I have no idea how to find the first three digits. :?:", "Solution_1": "I think you can do the first three digits this way.\r\n[hide]Take the log of 2 and multiply it by 5127. Then subtract the whole number part of your answer and let the result be $x$. Then $10^x$. The first three digits should be your answer... The reason this should work is because when you take the log of two and multiply it by 5127, you get the value of $x$ such that $10^x = 2^{5127}$ But the whole number part is just a bunch of zeros, so subtracting that part and then doing the $10^x$ of that should result in the $\\frac {2^{5127}}{10^{1543}}$ because 1543 is the whole number part of $5127 * \\log 2$.[/hide]", "Solution_2": "Can you use a calculator?", "Solution_3": "[quote=\"probability1.01\"]Can you use a calculator?[/quote]\r\n\r\nWell, I learned to use one back almost twenty years ago. I suppose you can if you know how to use it, if you don't, you can just learn :D :D :D .\r\n\r\nJust kidding. Really, I don't have an idea about how to compute the first three digits without a calculator that can handle logarithms!" } { "Tag": [], "Problem": "hi im a new member and i want know some moroccain friend\r\nbecause i'm it :roll:", "Solution_1": "Ain't Moroccan but your country is very good :D" } { "Tag": [ "geometry", "AoPS Books" ], "Problem": "I read posts in this forum and it makes me go \"WOW! :lol: \" and there are lots of good problems (hard!!!) especially proof ones.\r\n\r\nSo, here is my question:\r\n\r\nHow do you solve or at least practice your skill on proof? I know there is an article about it in this site but I want to hear from this forum since there are lots of good problems in here.\r\n\r\nI'm hoping to take some proof writing competitions but I'm not sure where to start.\r\n\r\nAny suggestion or your idea is GREAT! :)", "Solution_1": "Do you mean proof by actual writing-out part, or solving problems that require you to prove something? \r\n\r\nWhen I first started on math at Canadian National camp, the organizers made participants write out many messy proofs, and gave feedbacks. Usually logical ones and geometry problems are hard to write out.\r\n\r\nIf you mean the latter, I guess you have to solve lots of problems.. :lol:", "Solution_2": "Mathematical circles differs from the AoPs books in that it has tons of clever proof exercises instead of \"finding this\" problems. Maybe that's what your looking for?", "Solution_3": "Thanks.\r\n\r\nI have that book.", "Solution_4": "Um...what book?" } { "Tag": [], "Problem": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=274251]Puerto Rico PRMO Team Selection Test 2009[/url]\r\n\r\nThe last three digits of $ N$ are $ x25$. For how many values of $ x$ can $ N$ be the square of an integer?", "Solution_1": "[hide][quote=\"sunehra\"]We can rewrite it as $ 100x \\plus{} 25 \\equal{} k^2$ for some integer $ k$. Factoring, we have $ 5^2(4x \\plus{} 1) \\equal{} k^2$. So $ 4x \\plus{} 1$ must also be a perfect square.\n\n$ 4x \\plus{} 1 \\equal{} 9 \\implies x \\equal{} 2 \\\\\n4x \\plus{} 1 \\equal{} 25\\implies x \\equal{} 6$\n\nTherefore, $ N$ is a perfect square for $ \\boxed{x \\in \\{2,6\\}}$.[/quote][/hide]\r\n\r\n:)", "Solution_2": "Note that $ N$ doesn't necessarily have to equal $ \\overline{x25}$; only the last two digits of $ N$ are $ x25$. Thus, $ 0$ would also be a solution, and the actual answer is $ x\\in \\{0,2,6\\}$. :wink:", "Solution_3": "wow, what an ez question, 25squared= 625. 15squared=225, 45sqared=2025. :rotfl:", "Solution_4": "Don't revive old threads, especially if you're not going to contribute to the discussion. Anyway, you provided no proof at all of your answer.", "Solution_5": "AIME15, $ \\frac{1}{6}^\\text{th}$ of a month is certainly not a revival, though the post was indeed totally unnecessary. Also, I don't see how the $ 2025$ answer can count.", "Solution_6": "Are the last three digits in the form $ \\overline{x25}$ where $ x$ is a digit? Note that $ N$ doesn't need to be only three digits. :wink:" } { "Tag": [ "abstract algebra", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "How do i show that the non-abelian groups of order 8 are either $D_{4}$ or $Q_{8}$? Similarly, for order 12, either $D_{6}$, $A_{4}$ or $\\mathbb{Z}/3\\mathbb{Z} \\rtimes \\mathbb{Z}/4\\mathbb{Z}$.", "Solution_1": "Use Sylow theorems to determine the number and the structure of groups of order 8 and 12. Obviuosly the direct product of cyclic groups is an abelian group, so it excludes from non-abelianity C(8), C(4)xC(2), C(2)xC(2)xC(2). Show that D(4) and Q(8) are non-abelian giving a pair of elements which do not commute (this is simple, one is a dihedral group, the other is the quaternions' group). For the other case, in a silmilar way, C(12) and C(4)xC(3) are abelian. To determine the non-abelianity of the others use the same procedure used before, a little bit more complex is to find the incriminated elements in the semidirect product.\r\n\r\nFor the first part you know that the order of an element in G is a divisor of the order of group, for 8 we have 2, 4, 8 (if it's 8 you have done) and for 12 we have 2, 3, 4, 12. Now use the third Sylow theorem to determine the number of Sylow p-groups, and so on... it's a little bit tedious, i know :|" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Let $ a\\equal{}\\sqrt[1992]{1992}$. Which number is greater\r\n\\[ \\underbrace{a^{a^{a^{\\ldots^{a}}}}}_{1992}\\quad\\text{or}\\quad 1992?\r\n\\]", "Solution_1": "$ \\underbrace{a^{a^{a^{\\ldots^{a}}}}}_{1992}>1992\\equal{}(\\sqrt[1992]{1992})^{1992}\\equal{}a^{1992} \\Leftrightarrow \\underbrace{a^{a^{a^{\\ldots^{a}}}}}_{1991}>1992 \\Leftrightarrow \\dots \\dots \\Leftrightarrow a>1992$, which is impossible, so $ 1992$ is larger.", "Solution_2": "Fix integer $m>1$, let $A_0=10$ we have:\r\n$ \\sum (x+y)^{2}z \\leq \\frac{3}{2} \\prod(x+y)$.\r\n\r\nmade by Mnich W.\r\n\r\ncheers! :D :D", "Solution_1": "x (y+z)^2 + y (z+x)^2 + z (x+y)^2 <= 3/2 (y+z)(z+x)(x+y)\r\n<==> (yz^2 + zx^2 + xy^2 + zy^2 + xz^2 + yx^2) + 6 xyz <= 3/2 (yz^2 + zx^2 + xy^2 + zy^2 + xz^2 + yx^2 + 2 xyz)\r\n<==> (yz^2 + zx^2 + xy^2 + zy^2 + xz^2 + yx^2) + 6 xyz <= 3/2 (yz^2 + zx^2 + xy^2 + zy^2 + xz^2 + yx^2) + 3 xyz\r\n<==> 3 xyz <= 1/2 (yz^2 + zx^2 + xy^2 + zy^2 + xz^2 + yx^2)\r\n<==> 6 xyz <= (yz^2 + zx^2 + xy^2 + zy^2 + xz^2 + yx^2)\r\n<==> 0 <= (yz^2 - xyz) + (zx^2 - xyz) + (xy^2 - xyz) + (zy^2 - xyz) + (xz^2 - xyz) + (yx^2 - xyz)\r\n<==> 0 <= yz(z-x) + zx(x-y) + xy(y-z) + zy(y-x) + xz(z-y) + yx(x-z)\r\n<==> 0 <= y(z-x)(z-x) + z(x-y)(x-y) + x(y-z)(y-z)\r\n<==> 0 <= y(z-x)^2 + z(x-y)^2 + x(y-z)^2,\r\n\r\nwhat is trivial. Equality if and only if x = y = z.\r\n\r\nBut there should be a simpler proof.\r\n\r\n Darij", "Solution_2": "Let 2a = x+y, 2b = y+z and 2c = z+x.\r\nThen, the inequality reduces to \r\n \\sum a 3 - 2 \\sum a 2 b + \\sum abc \\geq 0, (where the sums are symetrics) which is Schur.\r\n\r\nPierre.", "Solution_3": "Or we observe that $(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)-xyz$ and after multiplications our inequality becomes $(x+y+z)(xy+yz+zx)+3xyz \\le \\frac{3}{2}(x+y)(y+z)(z+x)$.If we substitute the identity in the inequality we have to prove that $(x+y)(y+z)(z+x) \\ge 8xyz$ which is obvious by AM-GM." } { "Tag": [ "analytic geometry", "graphing lines", "slope", "calculus", "calculus computations" ], "Problem": "A curve in the plane is defined parametrically by the equations $ x\\equal{}t^3\\plus{}t$ and $ y\\equal{}t^4\\plus{}2t^2$. Find the equation of the line tangent to the curve at $ t\\equal{}1$.\r\n\r\nI figured out that at $ t\\equal{}1$, it is $ (2,3)$, but then I don't know how to find the derivative...", "Solution_1": "[hide=\"Hint\"]\n$ \\frac {dy}{dt}\\equal{}4t^3\\plus{}4t$ and $ \\frac {dx}{dt}\\equal{}3t^2\\plus{}1$. For $ t\\equal{}1$, the values, respectively, are $ 8$ and $ 4$. Note also that $ \\frac {dy}{dx}\\equal{}\\frac {dy}{dt}.\\frac {dt}{dx}\\equal{}8.\\frac {1}{4}\\equal{}2$. Then the gradient of the tangent line is $ 2$. The equation will be $ y\\equal{}2x\\minus{}1$. I am sorry if I am wrong... :( \n[/hide]", "Solution_2": "[quote=\"Qruni\"][hide=\"Hint\"]\n$ \\frac {dy}{dt} \\equal{} 4t^3 \\plus{} 4t$ and $ \\frac {dx}{dt} \\equal{} 3t^2 \\plus{} 1$. For $ t \\equal{} 1$, the values, respectively, are $ 8$ and $ 4$. Note also that $ \\frac {dy}{dx} \\equal{} \\frac {dy}{dt}.\\frac {dt}{dx} \\equal{} 8.\\frac {1}{4} \\equal{} 2$. Then the gradient of the tangent line is $ 2$. The equation will be $ y \\equal{} 2x \\minus{} 1$. I am sorry if I am wrong... :( \n[/hide][/quote]\nUsing that, I think I solved it.\n\n[hide=\"Am I right?\"]took my point (2,3) and my slope 2 to get the equation $ y\\minus{}3\\equal{}2(x\\minus{}2)$\n$ y\\minus{}3\\equal{}2x\\minus{}4$\n$ y\\equal{}2x\\minus{}1$[/hide]", "Solution_3": "Correct." } { "Tag": [ "AMC", "AMC 12", "AIME", "USA(J)MO", "USAMO", "MIT", "college" ], "Problem": "What is the policy on having food/water for the AMC12, AIME, and USAMO? I am pretty sure they all allow water, but I am not sure about the food.", "Solution_1": "i've seen people eating during the AIME...i dunno if they're supposed to, but they say it helps", "Solution_2": "I'm pretty sure it's OK since so many do, but\r\na.) It's not OK if there's any math-related stuff on it, and I'm not sure if Nutritional Facts pass that requirement since it has percentage*\r\nb.) It will probably get checked to make sure it's not a cheat-sheet...er, food :lol: \r\nc.) What's the point of having food anyways? I mean candies help but a full-course dinner...? :lol: (an insider joke to all who visited/went to MIT)", "Solution_3": "1) You may eat and drink on AMC/AIME/USAMO. If you go to a school that cares, it might even be provided. Food and water were provided at my high school on the AIME and USAMO.\r\n\r\n2) The content was only what happened in 2001. The AMC did not make any guarantee that a similar breakdown will occur at any time in the future.", "Solution_4": "Ok, thanks for the feedback. I know there was a thread a long while back about drinking water during competitions, but do any of you eat food as well?", "Solution_5": "ive heard that eating andes or mint stuff helps you think", "Solution_6": "During AIME, only candies and water were provided. I had a candy or two and my own sparkling water. On the USAMO, we had full lunches since there were only two of us rather than 21 or something for the AIME.", "Solution_7": "But do you think the candy actually helped?", "Solution_8": "what kind of candy are we talking about here?\r\n\r\nWe take the test in the nicest place in the school, so they dont allow food / drink there.", "Solution_9": "Mmm, both statistically and biochemically, yes. For me, I've never tried it.", "Solution_10": "Given my score on the AIME, I don't think the candy helped.", "Solution_11": "im not exactly sure why it would... for me it'd probablly be more of a distraction. I could see it being helpful if it's something you always do when you do math, but otherwise...", "Solution_12": "I suggest eating the food before the test, preferrably around 5 minutes beforehand. If you eat after that, then it can be a distraction and/or it won't get digested in time, and if after you might have the urge to go to the bathroom :) \r\nThe reason why candy is better than other food is because it contains lots of sugar in forms of glucose or fructose, which is very easy to digest. They are also small in volume so it won't cause any urge to go to the bathroom.\r\nAs I said before, the temporal blood-sugar boost will make you excited and sugar-high, which will in turn quicken your thinking, kind of like amphetamine except its effects are miniscule compared to the real thing. It is a double-edged sword though, for the excitement can lead to a lot of calculation error/careless miss.", "Solution_13": "maybe i'm mixing somethings up, but I think that chewing gum helps.", "Solution_14": "yea\r\ni think its scientifically shown that gum helps us think", "Solution_15": "As much as I hate to edit things, we don't want to spread misinformation. So I appended a little note to the end of Tare's \"Question-Analysis\" post. If anyone feels the note should be omitted, I will do so.", "Solution_16": "Er, if it concerns you that much I'll just delete it.", "Solution_17": "It's fine.", "Solution_18": "i think sugar in great amounts helps a lot. dont eat a donut like i did though. i spent half an hour during the AIME trying to clean up the mess i caused by dropping it onto my desk (and my test, paper, etc...).", "Solution_19": "I'm quite sure it helps. I always eat a lot of candies, chocolate, ... during an exam. And some coke may help also. At the IMO you really need to have much sugar since the exam lasts for 4.5 hours, which is long ; sugar helps you to keep your concentration. It's also no bad idea to go to the toilet during the exam, even if you don't really have to. At the IMO, Kaloyan Slavov, seated very close to me in the exam room, went more than 5 times to the toilet each exam. I think it helped since he got 41/42 :)", "Solution_20": "...\r\n.....\r\n...er, right...\r\n\r\n[quote=\"tetrahedr0n\"]dont eat a donut like i did though. i spent half an hour during the AIME trying to clean up the mess i caused by dropping it onto my desk (and my test, paper, etc...).[/quote]\r\n\"That donut's haunted! It turned on me, on[b] me[/b]!!!\" :lol:", "Solution_21": "Hey, I know Kaloyan -- he's the roommate of a good friend of mine. Did you meet Dobrimir (not sure I spelled this correctly), also from the Bulgarian team? He's a pretty cool guy -- we're in the same math class.", "Solution_22": "I know Dobromir Rahnev was in the Bulgarian team but I don't think I talked to him (maybe a few words).", "Solution_23": "Cool. Small world. :)" } { "Tag": [ "limit", "inequalities proposed", "inequalities" ], "Problem": "Let $a;b$ are real numbers satisfying $a \\geq \\frac{-1}{2}$ and $\\frac{a}{b}>1$.Find min:$A=\\frac{2a^{2}+b}{b(a-b)}$", "Solution_1": "Set $a=-\\frac{1}{4}-h$, $b=-\\frac{1}{4}$.\r\n$\\frac{a}{b}= 1+4h> 1 \\Rightarrow h>0$\r\n$a=-h-\\frac{1}{4}\\geq-\\frac{1}{2}\\Rightarrow h \\leq \\frac{1}{4}$\r\n\r\n$A=\\frac{2(h+\\frac{1}{4})^{2}-\\frac{1}{4}}{-\\frac{1}{4}(-h)}= \\frac{8(h+\\frac{1}{4})^{2}-1}{h}= 8h+4-\\frac{1}{2h}$\r\n\r\n$\\lim_{h \\to 0^{+}}8h+4-\\frac{1}{2h}=-\\infty$ \r\n\r\nHence, there is no minimum value of A." } { "Tag": [ "MATHCOUNTS", "geometry", "3D geometry", "probability", "AMC", "AIME", "number theory" ], "Problem": "Was it easy or what?", "Solution_1": "It was really easy. I couldn't doubt anyone getting a good score on it.", "Solution_2": "It was waaaay too easy for even a chapter level test. In my opinion, the school target was even harder.", "Solution_3": "guess wat i voted. :rotfl: :rotfl: :rotfl:", "Solution_4": "Who didn't think that it was the easiest target round that they had ever seen? If you did say that, then which one did you take that was easier?", "Solution_5": "Uh, maybe an old school target round?", "Solution_6": "i got perfect on it, and im horrible", "Solution_7": "Yeah, I agree. Most of the questions were simple calculations. The hardest problem was probably the ammonia one, but that one was still pretty easy.\r\n3 of my 4 teammates got a perfect score, and the 4th one messed up on the ammonia problem and the handshake problem.", "Solution_8": "lol i thought u were talking about nats for a sec..... but yes i got a lower score on school target than chapter. the ammonia one (the only one i missed)i had narrowed it down to the right answer and a wrong one. wrote in the right, erased it, wrote in the wrong, and wrote the right at the bottom with a question mark.", "Solution_9": "lol I got the exact same score on both tests (26/16) with like so many careless mistakes on both (well, 4...)\r\n\r\nAnd then I phailed states with--a really bad score. :blush: (oh well, 5th written 1st cd)", "Solution_10": "Argh. I thought this was about the Nationals Target and voted Extremely Difficult, only to see that it was about the Chapter Target, for which I should've voted \"Easy.\" :mad:", "Solution_11": "It was really easy, and if it weren't for me and misreading problems, then, well...", "Solution_12": "[quote=\"Yongyi781\"]Argh. I thought this was about the Nationals Target and voted Extremely Difficult, only to see that it was about the Chapter Target, for which I should've voted \"Easy.\" :mad:[/quote]\r\n\r\nthe nats target wasn't hard :) I just made tons of stupid mistakes. The only hard one there was 5 (if you didn't know how to do it) and 8 was semi-hard.", "Solution_13": "Are you kidding me? 5 was super easy (if you mean the triangles one), but you know I stink at counting/probability problems. And there were 4 of them. #8 was really hard at the moment, but after the test I beat myself realizing how easy it was. The dominoes one was pure brute force, the cube one was evil, and #1 was evil as well. Not to mention the ultimate evilness of #2, when I thought \"sales\" meant profit. I thought #3 and #4 were the easiest, though. Thus for the reason I failed miserably at target (only 3 out of 8).", "Solution_14": "I stink at counting/probability problems too! thats why i missed the last problem and some others on the state sprint :mad: :mad: :mad:", "Solution_15": "hmph....i didnt think the chapter target was that easy...", "Solution_16": "well now i am stupid, i was far away from even making nats. I didn't even reach cd in chapter. But then i got Aops books and got 18th at state :) but still didn't make nats :wallbash_red: :wallbash_red:", "Solution_17": "[quote=\"levans\"]Hint: (I'm giving a lot of hints today) - If you finished number 62 among all middle school students in the U.S., you aren't stupid but rather the exact opposite of stupid.[/quote]\r\n\r\nHint: The same could be said of the person who got 228th :D \r\n\r\nBut really, 62nd!!! That's better than me, at least, and all the AoPSers think I'm good at math for some reason (:D)\r\n\r\nAnyways, Nationals this year really focused on counting and probability (19 out of 38 questions were Counting and Probability) and if you're like me, you're bad at those questions and you bomb it. But that was just one test, and if you didn't do well because... well... yeah.", "Solution_18": "I suck at c and p problems too :mad: \r\n\r\nI would phail with you :wink:", "Solution_19": "I'm pretty good at counting and probability, I just make a bunch of careless mistakes.", "Solution_20": "Wow...you're opposite of me...Then which subject is hardest for you?", "Solution_21": "The topic that is hardest for me is probably geometry. But usually I do pretty good at that too. I just miss the hard AIME ones or the really hard ones at Nats.", "Solution_22": "My weakest topic is geometry as well. I'm okay at counting/probability, but I'm not so great at Number Theory. My best topic is algebra. :)", "Solution_23": "[quote=\"(^_^)\"]hmph....i didnt think the chapter target was that easy...[/quote]\r\n[sarcasm]Really?! I didn't know that![/sarcasm]\r\n\r\nThe target basically decided if you were in countdown or not. The target hated me.", "Solution_24": "My worst topic is Number Theory, but I don't see why I got the ammonia problems wrong on target if I do welll in algebra.\r\n\r\nBut I PWN IN GEOMETRY :D", "Solution_25": "I think I'll start another poll for this topic.", "Solution_26": "[quote=\"Poincare\"]well now i am stupid, i was far away from even making nats. I didn't even reach cd in chapter. But then i got Aops books and got 18th at state :) but still didn't make nats :wallbash_red: :wallbash_red:[/quote]\r\n\r\nHow did you do MathCounts for two years if you were in 5th grade this past school year?", "Solution_27": "I was in the sixth grade this past year.", "Solution_28": "[quote=\"isabella2296\"][quote=\"Poincare\"]well now i am stupid, i was far away from even making nats. I didn't even reach cd in chapter. But then i got Aops books and got 18th at state :) but still didn't make nats :wallbash_red: :wallbash_red:[/quote]\n\nHow did you do MathCounts for two years if you were in 5th grade this past school year?[/quote]\r\nI think he was just talking about if he really puts his shoes in the feelings of an actual State competitor. :)", "Solution_29": "Oh I thought she was asking me that question. I didn't see the quote from Poincare." } { "Tag": [ "abstract algebra", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Suppose we have a finite group $ G$ with exactly one element $ f$ of order 2, then $ \\prod_{g\\in G}g\\equal{}f$.\r\n\r\nThis seems like a really easy problem but I just can't figure it out.", "Solution_1": "$ \\prod_{g \\in G} g \\equal{} \\left(\\prod_{g \\in G\\setminus \\{e,f\\}} g \\right)( f)$.\r\n\r\nIn the first factor we can pair every element with its inverse. Since $ e$ and $ f$ are the only elements of $ G$ that are equal to their inverses, we can assure that this pairing lists every element of $ G \\setminus \\{e,f\\}$ only once. Hence $ \\left(\\prod_{g \\in G\\setminus \\{e,f\\}} g \\right) \\equal{} e$ and we're done.", "Solution_2": "[quote=\"coquitao\"]$ \\prod_{g \\in G} g \\equal{} \\left(\\prod_{g \\in G\\setminus \\{e,f\\}} g \\right)( f)$.\n\nIn the first factor we can pair every element with its inverse. Since $ e$ and $ f$ are the only elements of $ G$ that are equal to their inverses, we can assure that this pairing lists every element of $ G \\setminus \\{e,f\\}$ only once. Hence $ \\left(\\prod_{g \\in G\\setminus \\{e,f\\}} g \\right) \\equal{} e$ and we're done.[/quote]\r\n\r\nThis was exactly my argument but I wasn't convinced. We are not assuming the group is abelian and the product is in any arbitrary order you can think of so why does pairing each element with its inverse work?", "Solution_3": "Never mind. I thought about it some more and you're right. $ gh\\equal{}hx$ where $ x\\equal{}(h^{\\minus{}1}gh)$, so we can keep moving around $ h$ in the product $ \\prod_{g\\in G}g$ until it is next to its inverse at which point we can cancel it and move on to the next element. Since every $ h\\in G$ other than $ f$ has an inverse not equal to itself we can cancel all of them except $ f$.", "Solution_4": "[quote=\"davidk01\"]Never mind. I thought about it some more and you're right. $ gh \\equal{} hx$ where $ x \\equal{} (h^{ \\minus{} 1}gh)$, so we can keep moving around $ h$ in the product $ \\prod_{g\\in G}g$ until it is next to its inverse at which point we can cancel it and move on to the next element. Since every $ h\\in G$ other than $ f$ has an inverse not equal to itself we can cancel all of them except $ f$.[/quote]\r\n\r\nWell, try that on abAB where capitals are inverses. You get aAb^AB = b^A B, but there is no reason to think that b^A and B match up as inverses in general (b^A's inverse could have been a!). Of course {a,b,A,B} is not the entire group (minus the identity and the order 2 guy), but I think if you check the group SL(2,3) of order 24, you'll find the product depends on the order. The nonabelian group of order 12 with a single element of order 2 is also an example, .", "Solution_5": "Wonder why it was your teacher decided to drop the condition $ G \\in \\mathbf{Ab}$...", "Solution_6": "[quote=\"JackSchmidt\"]\n\nWell, try that on abAB where capitals are inverses. You get aAb^AB = b^A B, but there is no reason to think that b^A and B match up as inverses in general (b^A's inverse could have been a!). Of course {a,b,A,B} is not the entire group (minus the identity and the order 2 guy), but I think if you check the group SL(2,3) of order 24, you'll find the product depends on the order. The nonabelian group of order 12 with a single element of order 2 is also an example, .[/quote]\r\n\r\nI'm not saying that they will match up but $ gh \\equal{} hx$ for some $ x\\in G$ and the product doesn't change when we perform this shuffle. Since all the elements of $ G$ appear in the product $ \\prod_{g\\in G}g$ at some point $ h$ must meet its inverse and be annihilated at which point we are left with fewer elements and can perform the same procedure. In the end only $ f$ will be left since it is its own inverse and every other element has an inverse that does not coincide with itself. I don't see anything wrong with this argument.", "Solution_7": "You change a lot of factors on the way, which may cause some to be there twice whereas others are missing.\r\nBy the way, it's simpler to just show that the product _must_ change if you swap two neighbouring noncommuting elements.", "Solution_8": "[quote=\"ZetaX\"]You change a lot of factors on the way, which may cause some to be there twice whereas others are missing.\nBy the way, it's simpler to just show that the product _must_ change if you swap two neighbouring noncommuting elements.[/quote]\r\n\r\nOk, I see that commuting elements changes what happens among the other factors but it does not change the product. If $ s \\equal{} ght$ and $ gh \\equal{} hx$ then swapping $ gh$ with $ hx$ will not change the product since $ hxt \\equal{} (hx)t \\equal{} (gh)t \\equal{} s$. I still think this line of reasoning will work, unless someone has an actual counterexample. All the groups I could think of worked.", "Solution_9": "I gave you a counterexample: any nonabelian group. And just the product not changing doesn't give any reason why this should actually work.", "Solution_10": "[quote=\"ZetaX\"]I gave you a counterexample: any nonabelian group. And just the product not changing doesn't give any reason why this should actually work.[/quote]\r\n\r\nSaying any non-abelian group is not a counter-example. The conditions are: $ G$ is a finite group, $ G$ has exactly one element of order 2. So you need to produce a non-abelian $ G$ with exactly one element of order 2 and you need to show that $ \\prod_{g\\in G}g\\neq f$. You did no such thing, hence you didn't produce a counter-example.", "Solution_11": "[quote=\"davidk01\"]unless someone has an actual counterexample. All the groups I could think of worked.[/quote]\r\n\r\nI did give two explicit counterexamples, SL(2,3) of order 24 and 4|x3 of order 12.\r\n\r\n(edited to add the following)\r\n\r\nHere is the too explicit (parental caution advised) version: Let G=. Then G is the set { 1, b, a, bb, ab, abb, aa, aab, aabb, aaa, aaab, aaabb }. If I multiply this set in that particular order, I get aabb as the final reduced answer. This is an element of order 6. However, G has a unique element of order 2, aa = 1*b*bb*a*ab*abb*aa*aab*aabb*aaa*aaab*aaabb. In one order you get the unique elment of order 2, but in another you get an element of order 6.\r\n\r\nIn some sense ZetaX is exactly right, if I change the order ( b,a was in the first list, but a,b in the second list), then I introduce a commutator. If the product of all the elements of the group does not depend on the order, then every commutator must be the identity, and so the group must be abelian. In other words, every nonabelian group with a unique element of order 2 is a counterexample.\r\n\r\nFor instance, the quaternion group of order 8 is { 1, -1, i, -i, j, -j, k, -k } and 1*-1*(i*-i)*(j*-j)*(k*-k) = 1*-1*1*1*1 = -1 is the unique element of order 2. However, 1*-1*(i*-i)*(j*k)*(-j*-k) = 1*-1*1*i*i = -1*-1 = 1 is not of order 2.\r\n\r\nI just happened to try SL(2,5) first and then simplified the example down SL(2,3) and then to 4:3.", "Solution_12": "[quote=\"JackSchmidt\"]I did give two explicit counterexamples, SL(2,3) of order 24 and 4|x3 of order 12.\n\n(edited to add the following)\n\nHere is the too explicit (parental caution advised) version: Let G=. Then G is the set { 1, b, a, bb, ab, abb, aa, aab, aabb, aaa, aaab, aaabb }. If I multiply this set in that particular order, I get aabb as the final reduced answer. This is an element of order 6. However, G has a unique element of order 2, aa = 1*b*bb*a*ab*abb*aa*aab*aabb*aaa*aaab*aaabb. In one order you get the unique elment of order 2, but in another you get an element of order 6.\n\nIn some sense ZetaX is exactly right, if I change the order ( b,a was in the first list, but a,b in the second list), then I introduce a commutator. If the product of all the elements of the group does not depend on the order, then every commutator must be the identity, and so the group must be abelian. In other words, every nonabelian group with a unique element of order 2 is a counterexample.\n\nFor instance, the quaternion group of order 8 is { 1, -1, i, -i, j, -j, k, -k } and 1*-1*(i*-i)*(j*-j)*(k*-k) = 1*-1*1*1*1 = -1 is the unique element of order 2. However, 1*-1*(i*-i)*(j*k)*(-j*-k) = 1*-1*1*i*i = -1*-1 = 1 is not of order 2.\n\nI just happened to try SL(2,5) first and then simplified the example down SL(2,3) and then to 4:3.[/quote]\r\nOk, I'm convinced. Thanks.", "Solution_13": "[quote=\"davidk01\"][quote=\"ZetaX\"]I gave you a counterexample: any nonabelian group. And just the product not changing doesn't give any reason why this should actually work.[/quote]\n\nSaying any non-abelian group is not a counter-example. The conditions are: $ G$ is a finite group, $ G$ has exactly one element of order 2. So you need to produce a non-abelian $ G$ with exactly one element of order 2 and you need to show that $ \\prod_{g\\in G}g\\neq f$. You did no such thing, hence you didn't produce a counter-example.[/quote]\r\nI just didn't want to chew it for you. Finding such a group is really not hard and showing that this product is not constant (especially not always $ f$) was already explained by me." } { "Tag": [ "geometry", "calculus" ], "Problem": "one more for ya.....can't use calculus and have to use geometry. \r\n\r\nA rectangular sheet of paper is to contain 72 square inches of printed matter with 2-inch margins at the top and bottom and 1-in margis on each side. What dimensions for the sheet will use the least paper?", "Solution_1": "[hide=\"Hint\"]\nAM-GM\n[/hide]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let x,y,z >0 such that x+y+z=1.Prove that\r\n$ \\sum\\frac{x}{x\\plus{}yz}\\leq\\frac{9}{4}$", "Solution_1": "[quote=\"DuongChinh_K41SPHN\"]Let x,y,z >0 such that x+y+z=1.Prove that\n$ \\sum\\frac {x}{x \\plus{} yz}\\leq\\frac {9}{4}$[/quote]\r\n$ x\\plus{}yz\\equal{}(x\\plus{}y)(x\\plus{}z).$ :wink:", "Solution_2": "[quote=\"arqady\"][quote=\"DuongChinh_K41SPHN\"]Let x,y,z >0 such that x+y+z=1.Prove that\n$ \\sum\\frac {x}{x \\plus{} yz}\\leq\\frac {9}{4}$[/quote]\n$ x \\plus{} yz \\equal{} (x \\plus{} y)(x \\plus{} z).$ :wink:[/quote]\r\n and Hint: \r\n$ (x \\plus{} y)(y \\plus{} z)(z \\plus{} x) \\geq \\frac {8(xy \\plus{} yz \\plus{} zx)}{9}$\r\n$ xyz \\leq \\frac {xy \\plus{} yz \\plus{} zx}{9}$\r\ni'm sorry i cut my post.\r\nbut i don't understand why everyone don't post solution, and only ...vie to denote small mistake.", "Solution_3": "[quote=\"Hong Quy\"] \n$ xyz \\geq \\frac {xy \\plus{} yz \\plus{} zx}{9}$[/quote]\r\nI think you want to say $ xyz \\le \\frac {xy \\plus{} yz \\plus{} zx}{9}$ :)", "Solution_4": "you are right.\r\n\r\n$ (x \\plus{} y \\plus{} z)(xy \\plus{} yz \\plus{} zx)\\geq 9xyz$ by $ AM \\minus{} GM$. Isn't it?", "Solution_5": "[quote=\"kunny\"]you are right.\n\n$ (x \\plus{} y \\plus{} z)(xy \\plus{} yz \\plus{} zx)\\geq 9xyz$ by $ AM \\minus{} GM$. Isn't it?[/quote]\r\nYes :)" } { "Tag": [ "SuMAC", "Stanford", "college", "geometry" ], "Problem": "I want to attend a math camp over summer and I thought SUMaC would be a good choice because I would like to go to Stanford in the future...but it seems like the curriculum is too specific. My intention is to get a solid understanding of all of the precalculus topics, such as algebra, combinatorics, geometry, etc. Is SUMaC a good idea, or do you recommend something else?\r\n\r\nThanks.", "Solution_1": "hm.. so I don't think you should pick a camp simply because you'd like to attend the university later on, but thats just me. obviously, stanford is great place and sumac is a great program probably because its in such a great place, but the college shouldnt determine it.\r\n\r\nyou say you want to get a good feel for precalculus topics like combinatorics and such... but SUMaC program I is pretty much Abstract Algebra (groups, rings, fields, and such) and program II is Algebraic Topology... both topics are pretty difficult to completely grasp. i wouldn't say there's much of a pre-requisite to sumac, but there are some precalculus topics you're expected to know. What i'm trying to get at is that summer camps like SUMaC are to learn stuff youll never encounter in high school, and most people dont encounter in college. If you're trying to get ahead for next year by doing precalculus over the summer consider EPGY (which is also stanford) or self-studying. If you want math that's above and beyond, go to a camp.", "Solution_2": "[quote=\"Chion\"]What i'm trying to get at is that summer camps like SUMaC are to learn stuff youll never encounter in high school, and most people dont encounter in college. If you're trying to get ahead for next year by doing precalculus over the summer consider EPGY (which is also stanford) or self-studying. If you want math that's above and beyond, go to a camp.[/quote]\r\n\r\nI agree. The national-level summer mathematics programs are all about enrichment rather than acceleration. That is, you won't learn high-school math; you'll learn parts of the college and graduate mathematics curriculum along with mathematics rarely taught in any college/graduate courses. \r\n\r\nAlso, all of the national-level programs require doing well on an exam of some sort to get in. I don't know that any of the programs require precalculus before entrance, but they all require that level of mathematical maturity; that is, you'd have to be at least ready for precalculus-level ideas." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Assume [tex]x>1[/tex], [tex]y>1[/tex],[tex]x^xy^y=z^z[/tex], p is prime, p|x, show that p|y.\r\ngoodluck", "Solution_1": "How about $x=z=2,y=1$ ?", "Solution_2": "[quote=\"nguyenquockhanh\"]Assume [tex]x^xy^y=z^z[/tex], p is prime, p|x, show that p|y.\ngoodluck[/quote]\r\n\r\nGive an example of such x,y and z. Do they exists? If they do not exist then the problem is wrong.", "Solution_3": "[quote=\"mozilla\"][quote=\"nguyenquockhanh\"]Assume [tex]x^xy^y=z^z[/tex], p is prime, p|x, show that p|y.\ngoodluck[/quote]\n\nGive an example of such x,y and z. Do they exists? If they do not exist then the problem is wrong.[/quote]\r\nyes,\r\n[b]PRO[/b]:[b]there exist infinite triples satisfying that condition[/b]\r\nHINT:for pro \"exist\" we use form $2^a\\cdot{(2^n-1)^b}$", "Solution_4": "if you worry for \"exist\", can see:\r\n$(2^{12}\\cdot{3^6},2^8\\cdot{3^8},2^{11}\\cdot{3^7})$\r\nokie?", "Solution_5": "x < z, p|z => p^z|z^z => p^z|(x^x).(y^y) => p|y^y => p|y\r\n\r\nSorry, but i don\u00b4t speak english very well and i don\u00b4t know TeX :(", "Solution_6": "[quote=\"Lopes\"]x < z, p|z => p^z|z^z => p^z|(x^x).(y^y) => p|y^y => p|y\n\nSorry, but i don\u00b4t speak english very well and i don\u00b4t know TeX :([/quote]\r\n\r\n[b]why $p|y^y$[/b][/b]", "Solution_7": "i think this pro very hard\r\nfor \"[b]exist infinite[/b]\": can see solutions:\r\n$x=2^{2^{n+1}\\cdot({2^{n+1}-n-1})+2n}\\cdot{(2^n-1)^{2\\cdot({2^n-1})}}$\r\n$y=2^{2^{n+1}\\cdot({2^{n+1}-n-1)}}\\cdot{(2^n-1)^{2\\cdot({2^n-1})+2}}$\r\n$z=2^{2^{n+1}\\cdot({2^{n+1}-n-1})+n+1}\\cdot{(2^n-1)^{2\\cdot({2^n-1})+1}}$", "Solution_8": "sorry..\r\nmy conclusion is not correct :(", "Solution_9": "okie\r\nin my solution, i have some mistake and i've corrected it\r\nand now, check its" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "I was unsure of four problems that was on my test. The first three is finding the derivatives. The last one is finding the limit. I want to make sure that I had the right answer.\r\n\r\n1.\r\n\r\n(3x+3) ^5\r\n_______\r\nx-2\r\n\r\n2.\r\n\r\n xy+y=2\r\n\r\n3 \r\n\r\n1\r\n_____\r\n(5x-3)^5\r\n\r\n4. sin x\r\n_______\r\n 3x\r\n\r\nlimit is 0.", "Solution_1": "Here is the answers that I went along with.\r\n\r\n1. (3x+3)^9 (9)\r\n _________ ___\r\n (x-2) (x-2)^2\r\n\r\n2. 1+x \r\n _____\r\n y\r\n\r\n3. 5\r\n _____\r\n\r\n (5x-3)^6\r\n\r\n4. 1\r\n ___\r\n 3" } { "Tag": [ "inequalities", "AMC", "AIME", "pigeonhole principle", "modular arithmetic", "number theory", "relatively prime" ], "Problem": "I have observed the lessons on inequalities and think that it would be nice to extend the concept to another less than mainstream topic: Number Theory. Problems and theorems useful for >= AIME level math would be appreciated.", "Solution_1": "One of my favorites is Wilson's Theorem which states that if we have a prime $p$ then $(p-1)!+1$ is divisible by $p$ while if $p$ is composite then $(p-1)!+1$ is not divisible by $p$.", "Solution_2": "When you state the theorem or whatever, could you also give a problem using it, if possible? Thanks. That would help to understand it.\r\n\r\nFor example, what are the applications of Wilson's Thoerem?\r\n\r\nCan someone please explain the Chinese Remainder Theorem? I looked it up but I couldn't understand it really.", "Solution_3": "[quote=\"chess64\"]When you state the theorem or whatever, could you also give a problem using it, if possible? Thanks. That would help to understand it.\n\nFor example, what are the applications of Wilson's Thoerem?\n\nCan someone please explain the Chinese Remainder Theorem? I looked it up but I couldn't understand it really.[/quote]\r\nApperently some usage of that relies on tables, so I don't think you will find many problems using it.", "Solution_4": "Prove Wilson's Theorem!\r\n\r\nLet $p$ be a prime number which yields the remainder $1$ upoin division by $4$. Prove that there exists an integer $x$ such that $x^2+1$ is divisible by $p$.", "Solution_5": "This is not quite going how I expected. I guess I'll start it off. Perhaps you can provide a proof DPopov?\r\nProblem:\r\nfind all n such that\r\n13|(n^5+5^n)", "Solution_6": "Proof of Wilson's theorem: (tell me if this works, I'm a little sleepy)\r\nLemma: if (x,p) =1, then there exists some integer y such that \r\nxy = 1 mod p, and y is one of the smallest p-1 remainders mod p\r\nProof:\r\nContradiction. Suppose there does not exist such a y. Then consider the first p-1 multiples of x mod p\r\nx, 2x, ...., (p-1) x. In some order these are congruent to 2, 3, ... but not 1. Thus by pigeonhole principle, two of these must be congruent mod p. so ax = bx mod p. Factoring, we find x(a-b)=0 mod p, and thus \r\np|(a-b). But since (a-b)x is one of the multiples of x under consideration, p does not divide a-b. ><\r\nSince we have proven the first part of the lemma, we can find an integer <= p-1 congruent to y, so that it is one of the first p-1 remainders mod p.\r\nNow we apply our lemma to the product\r\n1 * 2 * ... (p-1) mod p\r\nEach of 2, 3, must be paired with another integer in the product. Also, p-1 = -1 mod p\r\nCombining these two facts, we find that the middle p-3 integers 'pair' with each other (they can't pair with 1 or -1) to produce a product of 1, leaving a total product of -1.", "Solution_7": "Heh, I am also sleepy but I am 99.9% sure that you have given a correct proof.", "Solution_8": "[quote=\"DPopov\"]Let $p$ be a prime number which yields the remainder $1$ upoin division by $4$. Prove that there exists an integer $x$ such that $x^2+1$ is divisible by $p$.[/quote]\r\n\r\nLemma: If $GCD(a,p) = 1$, $a^{\\frac {p-1}2} \\equiv 1 \\Leftrightarrow a$ is a square mod $p$ (quadradic residue)\r\n\r\nIf $a$ is a square mod $p$ then there exists some $u$ with $GCD(u,p)=1$ such that $u^2 \\equiv a \\pmod p$. By Fermat's Little, $u^{p-1} \\equiv 1 \\pmod p \\Rightarrow a^{\\frac {p-1}2} \\equiv 1 \\pmod p$.\r\n\r\nIf $a^{\\frac {p-1}2} \\equiv 1$ then there exists some primitive root $g$ such that $g^k \\equiv a \\pmod p$ (A primitive root is a number relatively prime to the mod, $p$ in this case, so that when the number is raised to all the powers 1 to $\\phi (p-1)$, all of the possible remainders relatively prime to the mod are obtained. There is a proof that there always exists a primitive root mod $p$ where $p$ is a prime, but I will not explain it right now.)\r\n\r\nThus $g^{\\frac {k(p-1)}2} \\equiv 1 \\pmod p$. Because $g$ is a primitive root, $(p-1)|\\frac {k(p-1)}2$ (I will not say why right now). Therefore $\\frac {k(p-1)}2 = (p-1)q$ for some integer $q$ and we find that $k = 2q \\Rightarrow a$ is a square since $g^{2k} \\equiv a \\pmod p$.\r\n\r\nNow if $-1$ is a square mod $p$, $(-1)^{\\frac {p-1}2} \\equiv 1 \\pmod p$ and the only way this can happen is if $\\frac {p-1}2$ is even, $\\frac {p-1}2 = 2r \\Rightarrow p= 4r+1 \\Rightarrow p \\equiv 1 \\pmod 4$. \r\n\r\nEdit: Oops I went the wrong way. The acutal proof starts at the last \"paragraph\". Reverse the steps.", "Solution_9": "[quote=\"chess64\"]When you state the theorem or whatever, could you also give a problem using it, if possible? Thanks. That would help to understand it.\n\nFor example, what are the applications of Wilson's Thoerem?\n\nCan someone please explain the Chinese Remainder Theorem? I looked it up but I couldn't understand it really.[/quote]\r\n\r\nThe Chinese Remainder Theorem is a method of finding $x$ when\r\n\r\n$x \\equiv r_1 \\pmod {m_1}$\r\n\r\n$x \\equiv r_2 \\pmod {m_2}$\r\n\r\n...\r\n\r\n$x \\equiv r_n \\pmod {m_n}$ \r\n\r\nwhere $m_1, m_2, ..., m_n$ are pairwise relatively prime.", "Solution_10": "The problem I gave has a proof using an application of Wilson's Theorem. However, I don't have time now but if no one has posted a similar proof I will post one later tonight.", "Solution_11": "[quote=\"DPopov\"]The problem I gave has a proof using an application of Wilson's Theorem. However, I don't have time now but if no one has posted a similar proof I will post one later tonight.[/quote]\r\n\r\n$\\left(\\frac {p-1}2\\right)! \\equiv (-1)^{\\frac {p-1}2} \\cdot \\frac {(p-1)!}{\\left(\\frac {p-1}2\\right)!} \\pmod p$\r\n\r\n$\\left(\\frac {p-1}2\\right)!^2 \\equiv (p-1)! \\cdot (-1)^{\\frac {p-1}2} \\equiv (-1)^{\\frac {p+1}2} \\pmod p$\r\n\r\nThus if $p \\equiv 1 \\pmod 4$, $\\frac {p+1}2$ is odd and $\\left(\\frac {p-1}2 \\right)!^2 \\equiv -1 \\pmod p$ and$\\left(\\frac {p-1}2 \\right)!^2 + 1 \\equiv 0 \\pmod p$", "Solution_12": "Some important numerical exercises.\r\nSolve:\r\n1235x+45=9090(mod24)\r\n1235x+45=9090(mod11)\r\n\r\n(separately)\r\n\r\nsolve the system of congruences:\r\nx=1(mod3)\r\nx=4(mod5)\r\nx=2(mod7)\r\nx=9(mod11)\r\nx=3(mod13)\r\n\r\nalso, the problem \r\nfind all n such that\r\n13|(5^n+n^5)\r\nremains.", "Solution_13": "Here's another proof.\r\n\r\nLet $x=\\left(\\frac {p-1}{2}\\right)!$, and $p=4k+1$\r\n\r\nThen in $\\mathbb{Z}_p$,\r\n\r\n$x^2=((2k)!)^2=(2k)!\\prod^{2k}_{i=1}(-1)(p-i)=(-1)^{2k}(4k)!=(p-1)!=-1$" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Prove that:\r\n$\\frac{a}{\\frac{1}{b}+\\frac{1}{c}+1}+\\frac{b}{\\frac{1}{c}+\\frac{1}{a}+1}+\\frac{c}{\\frac{1}{b}+\\frac{1}{a}+1}\\leq 1$\r\nwhere $a;b;c$ are non-negative", "Solution_1": "$a=b=c=2$ :huh:", "Solution_2": "[quote]Prove that:\n$\\frac{a}{\\frac{1}{b}+\\frac{1}{c}+1}+\\frac{b}{\\frac{1}{c}+\\frac{1}{a}+1}+\\frac{c}{\\frac{1}{b}+\\frac{1}{a}+1}\\leq 1$\nwhere a;b;c are non-negative[/quote]\r\num, let $a, b, c \\to \\infty$ then $LHS \\to \\infty$" } { "Tag": [ "AMC", "AIME" ], "Problem": "how do you delete posts", "Solution_1": "1. 58\r\n2. 8\r\n3. 3\r\n4. 30\r\n5. 919\r\n6. 18/343\r\n7. 2032\r\n8. 17\r\n9. 126\r\n10. 120/17\r\n\r\n:)", "Solution_2": "thanks so much :lol: aime 15" } { "Tag": [ "function", "calculus", "integration", "trigonometry", "algebra unsolved", "algebra" ], "Problem": "Find the necessary and sufficient conditions on values$a$ and $b$ if the real function($R$ to $R$)$f(x)$ is periodic with an integral period(non-zero) and\r\n$f(x+1)$ $=$ $af(x)$ $+$ $bf(x-1)$ for all real $x$", "Solution_1": "For any a,b, exist periodic f, satisfyed your condition, but f(x) can be is not continiosly and had irrational period. If f(x) continiosly or had rational period, then b=-1 $a=2$ or $a=0$ or $a=\\pm 1$.", "Solution_2": "Not quite.\r\n$f(x)=\\sin{2\\pi\\over n}x$ is continuous with period $n$, satisfying $f(x+1)=2cos{2\\pi\\over n}f(x)-f(x-1)$.\r\n\r\nAnd $f\\equiv 0$ satisfies the conditions and is periodic for any $a,b\\in\\mathbb{R}$.\r\nAnd $f\\equiv 1$ satisfies the conditions and is periodic for any $a,b\\in\\mathbb{R}$ with $a+b=1$.\r\nAnd $f(x)=\\{{1\\atop-1}{\\text{for }[x] \\text{ even}\\atop \\text{for }[x] \\text{ odd}}$ satisfies the conditions and is periodic for any $a,b\\in\\mathbb{R}$ with $b-a=1$.\r\nAnd $f(x)=\\sin\\pi x$ also satisfies the conditions and is periodic for any $a,b\\in\\mathbb{R}$ with $b-a=1$.", "Solution_3": "Yes. \r\n1. If f(x) had period 1, then work a+b=1.\r\n2. If f(x) had period 2, and f(x+1)=-f(x), then work b-a=1.\r\n3. If b=-1 work $|a|\\le 1$, for example f(x)=sin(cx), a=2cos c, b=-1." } { "Tag": [ "email", "USAMTS" ], "Problem": "1 Unable to move uploaded file. Please try again.\r\n\r\n\r\nIs what my page tells me.\r\n\r\nThe regular upload link isn't showing up, and the alternate page loads, then turns completely blank. Is this supposed to happen? I uploaded my round 1 solutions without any problems.", "Solution_1": "Apparently the website crashed last night shortly before the deadline.\r\n\r\nAnyone who tried to upload last night and failed can email their solutions to usamts@usamts.org before 6 PM today and we will still accept them. [b]You are on the honor system not to continue to work on your solutions but to email the same file that you tried to upload last night.[/b]", "Solution_2": "I uploaded my solutions last night, but when I look at the PDF file I uploaded (on the My USAMTS page), it seems that the file only includes the top-left corner of each page.\r\nIs there something wrong?\r\n\r\nEdit: Oh, and I haven't received a checkmark either.", "Solution_3": "[quote=\"andersonw\"]I uploaded my solutions last night, but when I look at the PDF file I uploaded (on the My USAMTS page), it seems that the file only includes the top-left corner of each page.\nIs there something wrong?\n\nEdit: Oh, and I haven't received a checkmark either.[/quote]Hard to tell, but if you uploaded and didn't get a checkmark, it probably didn't go through properly. Email us the file so that we can be sure." } { "Tag": [ "inequalities", "floor function", "number theory proposed", "number theory" ], "Problem": "Proove That:\r\n\r\n\\[ \\frac{3084}{10^{4}}\\leq \\sum_{k=1}^{\\infty }\\frac{\\underbrace{123456789123456 \\ldots }_{k}}{10^{k}k^{3/2}}\\leq \\frac{3086}{10^{4}}\\]\r\n\r\n\r\nNote:\r\n$ k=5, \\underbrace{123456789123456 \\ldots }_{k}=12345$\r\n$ k=12, \\underbrace{123456789123456 \\ldots }_{k}=123456789123$", "Solution_1": "[quote=\"lordykelvin\"]Proove That:\n\\[ \\frac{3084}{10^{4}}\\leq \\sum_{k=1}^{\\infty }\\frac{\\underbrace{123456789123456 \\ldots }_{k}}{10^{k}k^{3/2}}\\leq \\frac{3086}{10^{4}}\\]\nNote:\n$ k=5, \\underbrace{123456789123456 \\ldots }_{k}=12345$\n$ k=12, \\underbrace{123456789123456 \\ldots }_{k}=123456789123$[/quote]\r\n\r\nSo, we have to compute (with approximation) $ S=\\sum_{k=1}^{\\infty }\\frac{\\lfloor 10^{k}A\\rfloor}{10^{k}k^{3/2}}$ where $ A=\\frac{123456789}{999999999}$\r\n\r\nIt is rather easy to establish that,$ \\forall k\\geq p$, $ 2\\frac{\\lfloor 10^{p}A\\rfloor}{10^{p}}(\\frac{1}{\\sqrt{k}}-\\frac{1}{\\sqrt{k+1}})$ $ < \\frac{\\lfloor 10^{k}A\\rfloor}{10^{k}k^{3/2}}$ $ <2A(\\frac{1}{\\sqrt{k-1}}-\\frac{1}{\\sqrt{k}})$\r\n\r\nAnd so $ 2\\frac{\\lfloor 10^{p}A\\rfloor}{10^{p}}\\frac{1}{\\sqrt{p}}$ $ <\\sum_{k=p}^{\\infty}\\frac{\\lfloor 10^{k}A\\rfloor}{10^{k}k^{3/2}}$ $ <2A \\frac{1}{\\sqrt{p-1}}$\r\n\r\nAnd so $ \\sum_{k=1}^{p-1}\\frac{\\lfloor 10^{k}A\\rfloor}{10^{k}k^{3/2}}+2\\frac{\\lfloor 10^{p}A\\rfloor}{10^{p}}\\frac{1}{\\sqrt{p}}$ $ 2 +y 2 = radius 2 . So in this case, it would be x 2 +y 2 = 25 2 .", "Solution_12": "To FermatPrime: \r\n\r\nNumber 19) You have 12 quarters and you give them amongst A, B, C, D, and E. That is 5 different people. Set A = 2 quarters to start. So, there are still 5 people but 10 quarters left. There is a clever formula that I memorized for item distribution. Suppose you have n items to be distributed amongst m people. It is [tex]\\binom{n + m - 1}{n}[/tex]. So, you have 10 items, 5 people, so [tex]\\binom{14}{10} = 1001[/tex] using the calculator. (D). :lol:\r\n\r\nTo add on - and answer the question. Visit [url]http://mathworld.wolfram.com/PythagoreanTriple.html[/url]", "Solution_13": "[quote=\"Fermatprime\"]How'd you do 19 and 21?[/quote]\r\n\r\n21)Let RS=x, 17+x=21+16-x, yields x=10, BS=6. Now you can use Stewart\r\ns Theorem or Law of Cosines. Since I forgot Stewart's Theorem, I have to do it with Law of Cosines. Let $RSH= \\theta$, $HSB=180-\\theta$\r\n$10^2+HS^2-20HS\\cos\\theta=17^2$\r\n$6^2+HS^2+12HS\\cos\\theta=21^2$\r\n\r\n5 times the second equation+3 times the first yields\r\n$480+8HS^2=3072$\r\nHS^2=324, HS=18", "Solution_14": "Could someone explain(or provide a link) for why white_horse_king's formula works?\r\n\r\n\r\n\r\n12. Suppose that we choose 12 integers at random, without replacement, from the set {1, 3, 5, ..., 2005}. The probability that 1979 is chosen can be written as , where (m, n)=1. Find mn. \r\n(a) 1015 (b) 13039 (c) 12024 (d)10030 (e) Not Here \r\n\r\nSince 2005=2(1003)-1 there are 1003 numbers in that set. The total number of ways to choose 12 from it is C(1003,12). The total number of ways to choose 12 with 1979 is C(1002,11) because we already know that 12 is one of them so we have to choose the other 11 from the remaining 1002. Thus the probability is \\[\\frac{{1002\\choose 11}}{{1003\\choose 12}}\\] Using a calulator this comes out to $\\frac{12}{1003}$ and since it is looking for mn and not m+n :-x the answer is 12*1003=12036 (E)", "Solution_15": "[quote=\"joml88\"]Could someone explain(or provide a link) for why white_horse_king's formula works?[/quote]\r\nLook at this problem, 10 quarters to distribute to 5 people. \r\nLook at the sequence \r\nQQQQQQQQQQ and you have 4 dividers to divide the sequence into 5 part, the number of Q's in each part correponds to the number of quarters each person has.\r\nFor example Q_Q_QQQ_QQQ_QQ. So our problem corresponds to how many ways can we divide the sequence into 5 parts(it's possible that someone can have 0 quarters). Thus the total number of ways correponds to ${14 \\choose 4}$.", "Solution_16": "I got (A) too because I thought Rep was looking for m+n. :D\r\n\r\n\r\nCould someone inform me on how to do number 16? I tried that one, but I resorted to lots of trig - and somewhere along the way I got disorganized.", "Solution_17": "[quote=\"beta\"][quote=\"joml88\"]Could someone explain(or provide a link) for why white_horse_king's formula works?[/quote]\nLook at this problem, 10 quarters to distribute to 5 people. \nLook at the sequence \nQQQQQQQQQQ and you have 4 dividers to divide the sequence into 5 part, the number of Q's in each part correponds to the number of quarters each person has.\nFor example Q_Q_QQQ_QQQ_QQ. So our problem corresponds to how many ways can we divide the sequence into 5 parts(it's possible that someone can have 0 quarters). Thus the total number of ways correponds to ${14 \\choose 4}$.[/quote]\r\n\r\nYeah, that's right. I should've put 1 as an answer choice (because someone has to get at least 2 quarters. :P )\r\n\r\nHeh, everyone got the same solutions I got when I wrote the questions. Yeah, those two were mine. I was sort of hoping for something completely unexpected.\r\n\r\n(BTW, Stewart's Theorem is c2 n + b 2 m = d 2 a + mna)", "Solution_18": "Suppose we have a triangle with vertices at points A, B, and C. Let a, b, and c be the lengths of sides BC, AC, and AB. Then the angle bisectors of the triangle meet at the point (aA + bB + cC) / (a + b + c). \r\n\r\nWas that your question?", "Solution_19": "24. \r\n\r\nI think this way works.\r\nThere's a theorem that states that if a is a root of f'(x) and a is a root of f(x)\r\nthen a is a multiple root of the polynomial f(x).\r\nNow if we find roots of the derivative of the original polynomial, we find that 1.5 is a root. Then we can check to see if it works for the original polynomial.\r\n\r\nI'm not completely sure it works though", "Solution_20": "[quote=\"Rep123max\"]Anybody have a solution to #22 without using $\\sqrt{R(R-2r)}$? I guess you might be able to do it analytically.?[/quote]\r\n\r\nwhat formula is this? i guess it's pretty well known, but i've never encountered it >_<\r\n\r\ni did this the tedious coordinates way- expressing the perpendicular bisectors of 2 sides as linear equations, seeing where the two equations intersected, then pythagorean's formula to find the distance between that and the incenter.", "Solution_21": "Can someone please show how to do questions 23 and 24? Thanks!", "Solution_22": "Also, can someone tell how to do number 16?\r\n\r\nIt said:\r\n\r\nTriangle ABC has =100 on the AMC12? I'm taking it tomorrow... :P", "Solution_24": "Can anyone show me how to do #24?? The only way I can think of is graphing it or plugging in the answers. :? Oh, and how do you do #17?\r\n\r\nthanks!" } { "Tag": [ "recurrence relation", "combinatorics", "system of equations", "IMO", "IMO 1967" ], "Problem": "In a sports meeting a total of $m$ medals were awarded over $n$ days. On the first day one medal and $\\frac{1}{7}$ of the remaining medals were awarded. On the second day two medals and $\\frac{1}{7}$ of the remaining medals were awarded, and so on. On the last day, the remaining $n$ medals were awarded. How many medals did the meeting last, and what was the total number of medals ?", "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": "[b]Solution.[/b] Let $ m_{k}$ be the number of medals awarded on the $ k^{\\text{th}}$ day. Notice that after the $ \\ell^{\\text{th}}$ day, $ 6(m_{\\ell}-\\ell)$ medals are left for day $ \\ell+1$. By the given information in the problem,\r\n\\[ \\begin{eqnarray*}m_{x+1}&=&x+1+\\frac{6(m_{x}-x)-x-1}{7}\\\\ &=&\\frac{6}{7}(m_{x}+1)\\\\ &=&\\left(\\frac{6}{7}\\right)^{x}(m_{1}-6)+6\\\\ &=&\\frac{6^{x}}{7^{x+1}}(m-36)+6,\\]\r\nusing $ m_{1}=1+\\frac{m-1}{7}$.\r\n\r\nSince $ m_{n}=n$, $ 7^{n}(n-6)=6^{n-1}(m-36)\\implies 6^{n-1}|n-6\\implies n=6\\wedge m=36$.\r\n\r\nAs a result, $ (m,n)=(36,6)$. $ \\Box$", "Solution_3": "[hide=\"small correction\"]\nAssume that $(m, n) = (36, 6)$. Then there will be $\\frac{36-1}{7}=5$ medals after the first day, and then $\\frac{5-2}{7} = \\frac{3}{7}$ medals after the second day, impossible as there must be an integer number of medals. \n\nThus, we conclude the only answer to be $(m, n) = (1, 1)$[/hide]\n\nEDIT: not sure what I was thinking, the above is utter nonsense. Apologies.", "Solution_4": "[quote=\"johnmichaelwu\"]\nAssume that $(m, n) = (36, 6)$. Then there will be $\\frac{36-1}{7}=5$ medals after the first day, and then $\\frac{5-2}{7} = \\frac{3}{7}$ medals after the second day, impossible as there must be an integer number of medals. \n\nThus, we conclude the only answer to be $(m, n) = (1, 1)$[/quote]\nYou read it wrong. With $(m,n)=(36,6)$, there will be $1 + \\frac{36-1}{7}=6$ medals awarded the first day, thus $36-6 = 30$ left. Then there will be $2 + \\frac{30-2}{7}=6$ medals awarded the second day, thus $30-6 = 24$ left. And so on, everything working fine.", "Solution_5": "Let [i]K(i)[/i] denote the number of medals awarded on [i]ith[/i] day.\n\nSo, medals awarded on [i]tth[/i] day, K(t) = t + 1/7[m - {t + \\sum\\limits_{i=1}^t-1 K(i)}]\n\nK(t+1) = (t+1) + 1/7 [m - {t+1 + \\sum\\limits_{i=1}^t K(i)}]\n\nK(t+1) - K(t) = 1 - 1/7 - 1/7K(t)\n\nK(t+1) = 6/7 (K(t) +1)\n\nTherefore, K(n-1) = 6/7 (K(n-2) +1)\n\n\\sum\\limits_{i=1}^n-1 K(i) = m - n\n\nK(1) * (\\sum\\limits_{i=1}^n-2 (6/7)^i-1) + n * (\\sum\\limits_{i=1}^n-2 (6/7)^i) - (\\sum\\limits_{i=1}^n-2 (i+1)*(6/7)^i) = m-n\n\nm*(6/7)^n-1 - 7n = 36*(6/7)^n-1 - 7(6)\n\nTherefore, [i]m=36; n=6[/i]", "Solution_6": "We can let $f(a)$ be the number of medals awarded on day $a$.\n\nWe know that after day $x$ there will be $6(f(x)-x)$ medals left.\n\nWe know that if there are $6(f(x)-x)$ after day $x$ then\n\\begin{align*}\nf(x+1)&=x+1+\\frac{6(f(x)-x)-x-1}7 \\\\\n&=\\frac{6f(x)-6x-x-1+7x+7}7 \\\\\n&=\\frac{6f(x)+6}7 \\\\\n&=\\frac67f(x)+\\frac67\n\\end{align*}\nso we can prove with induction that $f(x+1)=(\\frac{6}{7})^x(f(1)-6)+6$.\n\nWe know that $f(1)=1+\\frac{m-1}{7}$ so\n\\begin{align*}\nf(x+1)&=(\\frac{6}{7})^x(1+\\frac{m-1}{7}-6)+6\\\\\n&=(\\frac{6}{7})^x(\\frac{m-36}7)+6\\\\\n&=\\frac{6^x}{7^{x+1}}(m-36)+6.\n\\end{align*}\n\nWe know that $f(n)=n$ and $f(n)=\\frac{6^{n-1}}{7^n}(m-36)+6$ so\n\\begin{align*}\nn&=\\frac{6^{n-1}}{7^n}(m-36)+6\\\\\n7^nn&=6^{n-1}(m-36)+6\\cdot7^n\\\\\n7^n(n-6)&=6^{n-1}(m-36)\\\\\n\\end{align*}\nand $n=6$ and $m=36$.\n" } { "Tag": [], "Problem": "I heard of this game in a magazine. You define a word in a funny, although somewhat true way. For example, for \"water\" you could say \"a colorless odorless chemical that can cause death if inhaled\" or you could just describe someone. WHen someone else gives the \"daffynition\" of a word, they have to provide a new word as well. Let's start with(and this may get interesting):\r\n[b]politicians[/b]", "Solution_1": "a synonym of actors\r\n\r\n[b]statisticians[/b]", "Solution_2": "someone who fumbles with the government and SafeAuto\r\n\r\n[b]athletes[/b]", "Solution_3": "Please, don't let this game die so early.........", "Solution_4": "Fine: people who get paid to exercise\r\n\r\n[b]CNN[/b]", "Solution_5": "great story tellers....\r\n\r\n[b]Valentin Vornicu[/b]", "Solution_6": "I think this is going down the wrong path...........", "Solution_7": "I don't get it and it is pretty confuisng...", "Solution_8": "New Daffynition: [b]Zimbabwe[/b]", "Solution_9": "A word that may be used in spelling contests\r\n\r\n\r\n\r\n[b]maybach[/b]", "Solution_10": "a car rich people buy and don't drive(they get chaueffered)\r\n\r\n[b]Ted Stevens[/b]" } { "Tag": [], "Problem": "If X dollars will buy Y pounds of chopped meat, how many dollars are need to buy Z pounds of the same meat?\r\n\r\na XZ/Y b XYZ c) Z/XY d) (X+Y)/Z e) XY/Z''.", "Solution_1": "Price is $ \\frac{X}{Y}$ dollars per pound. There are $ z$ pounds of meat, so the price for $ Z$ pounds of meat is $ \\frac{XZ}{Y}$." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": ":D I have a money distribution calculation problem. I know someone is up to the challenge. I have $44,564.00 inheritance to split between 4 people. Person G borrowed (in the past) $11,800.00 Person Z borrowed (in the past) $ 4,300.00 Person T and person B have never borrowed any. I have tried to add dedections to the inheritance amount, then deduct the amount loaned. I have tried to split up each way possible. I cannot come up with a fair distribution that pleases my brothers. Can anybody help in the most fair way to split up the remaining money 4 ways, taking into consideration past loans? Thank you", "Solution_1": "[quote=\"rewritten to avoid TeX display problems from dollar sign...\"] \nI have USD 44,564.00 inheritance to split between 4 people. Person G borrowed (in the past) USD 11,800.00 Person Z borrowed (in the past) USD 4,300.00 Person T and person B have never borrowed any. I have tried to add dedections to the inheritance amount, then deduct the amount loaned. I have tried to split up each way possible. I cannot come up with a fair distribution that pleases my brothers. Can anybody help in the most fair way to split up the remaining money 4 ways, taking into consideration past loans? Thank you[/quote]\r\n\r\nYour calculation sounds right.\r\nThere were originally 44564 + 11800 + 4300 dollars = 70664. \r\nEveryone is (I assume) entitled to a 1/4 share of that = 17666. \r\nThe ones who took advances on their shares through borrowing, receive the remainder (i.e. their 1/4 share, minus amount borrowed), as you calculated. \r\n\r\nWhat are the brothers' objections? A similar computation could be done with interest on the borrowings, if that is an issue. The parties that didn't borrow could of course decide to cancel some or all of the loans due to them, but (legally speaking) that is not a decision that the borrowers can make unilaterally. The loan doesn't die with the lessor, generally it is an asset that passes to the heirs. There may be questions of taxation that are affected by the difference between past loans and present dollars, which may make it more favorable to the inheritors to carry out some other arrangement than the above." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "If $x_{0}= 5$ and $x_{n+1}= x_{n}+\\frac{1}{x_{n}}$\r\n, show that\r\n$45 < x_{1000}< 45.1$.", "Solution_1": "This problem you have to post in box Algebra, i think :wink:", "Solution_2": "we need to proove inequality. $inequality \\in algebra$ anyway", "Solution_3": "[quote=\"Beat\"]If $x_{0}= 5$ and $x_{n+1}= x_{n}+\\frac{1}{x_{n}}$\n, show that\n$45 < x_{1000}< 45.1$.[/quote]\r\n\r\nI will prove $x_{1000}>45$.\r\nFrom hypothesis we have $u_{n+1}^{2}=u_{n}^{2}+\\frac{1}{u_{n}^{2}}+2>u_{n}^{2}+2$. By telescopic sum we have $u_{n}^{2}>2n+25$. In particular $u_{1000}^{2}>2025$ hence $u_{1000}>45$." } { "Tag": [ "inequalities", "Cauchy Inequality", "inequalities proposed" ], "Problem": "Let $ a$, $ b$, $ c$ be positive real numbers satisfying $ a\\plus{}b\\plus{}c\\equal{}1$. Prove that:\r\n$ \\left(bc\\plus{}ca\\plus{}ab\\right)\\left(\\left(\\frac{2a\\plus{}b}{b}\\right)^{2}\\plus{}\\left(\\frac{2b\\plus{}c}{c}\\right)^{2}\\plus{}\\left(\\frac{2c\\plus{}a}{a}\\right)^{2}\\right)\\geq 9$.", "Solution_1": "I have a solution by brute force but it is quite long.Anyone has a nice solution for this.Thank you", "Solution_2": "$2a+b=a+a+b$, then by AM-GM $(\\frac{2a+b}{b})^2\\geq \\frac{9\\sqrt[3]{a^4b^2}}{b^2}$, $(\\frac{2b+c}{c})^2\\geq\\frac{9\\sqrt[3]{b^4c^2}}{c^2}$ and $(\\frac{2c+a}{a})^2\\geq\\frac{9\\sqrt[3]{c^4a^2}}{a^2}$.\r\n\r\n$(\\frac{2a+b}{b})^2+(\\frac{2b+c}{c})^2+(\\frac{2c+a}{a})^2\\geq \\frac{9\\sqrt[3]{a^4b^2}}{b^2}+\\frac{9\\sqrt[3]{b^4c^2}}{c^2}+\\frac{9\\sqrt[3]{c^4a^2}}{a^2}$\r\nUsing AM-GM again yields $\\frac{9\\sqrt[3]{a^4b^2}}{b^2}+\\frac{9\\sqrt[3]{b^4c^2}}{c^2}+\\frac{9\\sqrt[3]{c^4a^2}}{a^2}\\geq 27\\sqrt[3]{\\frac{a^2b^2c^2}{a^2b^2c^2}}\\geq 27 (*)$\r\n\r\nIt is known that $\\frac{1}{2}((a-b)^2+(b-c)^2+(c-a)^2)\\geq 0$, then $a^2+b^2+c^2\\geq ab+bc+ca$ or $(a+b+c)^2\\geq 3(ab+bc+ca)$, from the last inequality follows that $ab+bc+ca\\leq \\frac{1}{3} (*1)$\r\n\r\n$(*).(*1)$ yields $\\sum_{cyc}^{a,b,c}(\\frac{2a+b}{b})^2\\geq 81\\sum_{cyc}^{a,b,c} ab$", "Solution_3": "Oh, sorry I proved something else. I will try to solve later your inequality again.", "Solution_4": "[quote=\"Beat\"]$2a+b=a+a+b$, then by AM-GM $(\\frac{2a+b}{b})^2\\geq \\frac{9\\sqrt[3]{a^4b^2}}{b^2}$, $(\\frac{2b+c}{c})^2\\geq\\frac{9\\sqrt[3]{b^4c^2}}{c^2}$ and $(\\frac{2c+a}{a})^2\\geq\\frac{9\\sqrt[3]{c^4a^2}}{a^2}$.\n\n$(\\frac{2a+b}{b})^2+(\\frac{2b+c}{c})^2+(\\frac{2c+a}{a})^2\\geq \\frac{9\\sqrt[3]{a^4b^2}}{b^2}+\\frac{9\\sqrt[3]{b^4c^2}}{c^2}+\\frac{9\\sqrt[3]{c^4a^2}}{a^2}$\nUsing AM-GM again yields $\\frac{9\\sqrt[3]{a^4b^2}}{b^2}+\\frac{9\\sqrt[3]{b^4c^2}}{c^2}+\\frac{9\\sqrt[3]{c^4a^2}}{a^2}\\geq 27\\sqrt[3]{\\frac{a^2b^2c^2}{a^2b^2c^2}}\\geq 27 (*)$\n\nIt is known that $\\frac{1}{2}((a-b)^2+(b-c)^2+(c-a)^2)\\geq 0$, then $a^2+b^2+c^2\\geq ab+bc+ca$ or $(a+b+c)^2\\geq 3(ab+bc+ca)$, from the last inequality follows that $ab+bc+ca\\leq \\frac{1}{3} (*1)$\n \n \n$(*).(*1)$ yields $\\sum_{cyc}^{a,b,c}(\\frac{2a+b}{b})^2\\geq 81\\sum_{cyc}^{a,b,c} ab$[/quote]\r\n\r\n Sorry , I think your solution is wrong because $ab+ac+bc \\leq \\frac{1}{3}$", "Solution_5": "I think ab+bc+ca<=1/3 is true because a+b+c=1.Can you have the full solution,Beat and Romano", "Solution_6": "[quote=\"bui_haiha2000\"]I think ab+bc+ca<=1/3 is true because a+b+c=1.Can you have the full solution,Beat and Romano[/quote]\r\n Yes, It is true that $ab+bc+ca \\leq \\frac13$ but from the sols of Beat, he can't prove your inequality because the reverse sign at the end :D .\r\n Probably, you should give us more time for the nice inequality! :lol:", "Solution_7": "Yes,of couse because i need the second solution for this very nice ine.My solution is quite long.I hope every body will help me :D", "Solution_8": "No one help me?????????? :o", "Solution_9": "[quote=\"bui_haiha2000\"]No one help me?????????? :o[/quote]\r\n\r\n I'm always here to help you...especially if u are a girl :D ...\r\nOK i solve this problem in three steps:\r\n1) $( \\sum ab ) ( \\sum ( \\frac{2a+b}{b} )^2 ) \\geq 9$ is equivalent with $( \\sum ab )( 3+ 4 \\sum (( \\frac{a}{b})^2+ \\frac{a}{b} ) \\geq 9$\r\nDue to Cauchy inequality we have\r\n$\\sum (( \\frac{a}{b})^2+ \\frac{a}{b}) \\geq \\sum ( \\frac{a}{b} + \\frac{b}{a} )$ ( self prove )\r\nBut we also have:\r\n$\\frac{a}{b} + \\frac{b}{a} = \\frac{ a^2+b^2 }{ab}$\r\nSo the RHS is equal to : $\\sum ( \\frac{a^2+b^2}{ab})$\r\n2)From Bunhiakowski inequality: $RHS( \\sum ab ) \\geq ( \\sum( \\sqrt{a^2+b^2}))^2$ expanding the LHS\r\nContinue by using Minkowski inequality to prove that:\r\n$\\sum ( \\sqrt{a^2+b^2} \\sqrt{a^2+c^2} ) \\geq \\sum (a^2) + \\sum(ab)$\r\n3)It's very simple now,u pair $\\sum (ab)$ and $\\sum(a^2)$ to have $( \\sum(a))^2$ and remember that $\\sum a =1$ and $\\sum(a^2) \\geq \\frac{1}{3}$....\r\n\r\nFor such long time that i've not touched inequality, hope my solution not too bad...." } { "Tag": [ "videos", "linear algebra", "matrix", "geometry", "geometric transformation", "reflection", "Gauss" ], "Problem": "POst why you think the person above you chose their avatar (it doesn't have to make sense!)", "Solution_1": "Because he is mewto?", "Solution_2": "HAAHAHAHA ITS A DECAPITATED PACMAN SKULL! IT HAS NO EYEHOLES!", "Solution_3": "because mewto sucks, just like mewto55555.", "Solution_4": "[quote=\"mewto55555\"]HAAHAHAHA ITS A DECAPITATED PACMAN SKULL! IT HAS NO EYEHOLES![/quote]\r\n\r\n1st of all: Pacman doesn't have body at the original game\r\n2nd of all: Pacman have small eyes\r\n\r\n1=2 chose a fake avatar because his username is fake", "Solution_5": "pacman got eaten by the evil pacman eating things in the game pacman. then they took a picture of him and made him use it as his avatar :D", "Solution_6": "guineapiggies has a guinea piggy because (no duh) SHES GUINEAPIGGIES\r\n\r\no and dont try mine trust me there was no reason for my avatar just that i wanted one and i got a fractal\r\n\r\nedit: just changed my avatar\r\n\r\n<-- guess who this is (hes from a video game not main character but ally)", "Solution_7": "Err you think he resembles your personality.", "Solution_8": "Umm... you made that avatar because you were bored and you still use MS Paint, the worst image editor ever.", "Solution_9": "herefishyfishy1 chose their pic 'cause...um...they...couldnt come up with an idea for the pic.\r\n\r\nHINT for my avatar: i like people to realize that im a girl! :mad:", "Solution_10": "Because she needed a flower to make herself feel better because she hadn't yet signed up for [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=220270]BB2[/url] in the games forum, duh.", "Solution_11": "whats BB2???", "Solution_12": "My new game.", "Solution_13": "o...THAT. what r the rules??? MAYBE i'll join.", "Solution_14": "because flowers make him/her sneeze", "Solution_15": "And you just have a simple avatar. too simple. you are a simple person.", "Solution_16": "You have an undefined avatar. Nobody knows what it is except you. You can only understand yourself.", "Solution_17": "it is a colorful blob of blobbing blob blobbiness", "Solution_18": "it is too small for me to see.", "Solution_19": "Lstar please stop reviving old threads, the last post was in 2009\nPhirekaLk6781 lock this now", "Solution_20": "Because you like baseball", "Solution_21": "Lstar I have counted 16 revivals now.\nSTOP IT!\nthis has gotten really annoying for PhirekaLk6781 now so stop it.\nIF YOU LIKE A GAME MAKE A NEW TOPIC SAYING \"game name(revival)\"\nDON't REVIVE OLD THREADS!\nALSo if a mod locks one of your old posts, then make sure that you don't try it again, they have reasons to lock threads.\nSo do Phirekalk6781 a favor and stop making SPAM!\nJust stop reviving old threads.\nA majority of AOPS WOULD BENEFIT IF YOU DID!\nsorry for yelling :wink:", "Solution_22": "Yes. Lstar, you cannot post in a thread if the last post was more than a month ago. Please stop reviving, or you will be banned. Mods, please lock.\n\nQ.E.D.", "Solution_23": "what about firebubble", "Solution_24": "Firebubble revived it by about 4 hours\nYOu revived it by 10 months", "Solution_25": "I REVIVED IT FIRST!!!\n\nthis is making me angry", "Solution_26": "Nice to know. Just make sure you return to the place from which you came and detonate yourself where you won't be hurting others, please. \n\nAlso, reporting another user for \"being mean to me,\" especially when they're just stating the rules is the worst abuse report I've seen so far, even among the majority of Fun Factory abuse reports being improper.", "Solution_27": "Where did this pop up from?\nFrom what i've seen, this should be locked.\nAlso, um, Lstar, are you trying to take credit for reviving something when you ought not to?", "Solution_28": "[quote=\"PhireKaLk6781\"]Nice to know. Just make sure you return to the place from which you came and detonate yourself where you won't be hurting others, please. \n\nAlso, reporting another user for \"being mean to me,\" especially when they're just stating the rules is the worst abuse report I've seen so far, even among the majority of Fun Factory abuse reports being improper.[/quote]\nWhat the heck, my post got reported for me stating the rules?", "Solution_29": "Don't worry, there won't be any discrediting to you for that. \n\nYeah, time to lock this thread before it starts dancing on its belly." } { "Tag": [ "geometry", "3D geometry", "sphere", "calculus", "integration", "function", "conics" ], "Problem": "I know that the equation of the cone is Z = R, the bounds of my integral relative to (theta) are 0 and 2 pi, but I do not have the clue as to what is the function to integrate or the bounds relative to R.\r\n\r\nHowever, I do know that an element of a polar surface is R DR D(theta). And the sphere is radius 1.", "Solution_1": "The familiar sno-cone problem: Let the sphere have radius $ R$ and the cone be the usual $ z^2 \\equal{} x^2 \\plus{} y^2$ so that we have the sphere being $ r^2 \\plus{} z^2 \\equal{} R^2$ and the cone is $ z \\equal{} r$. The upper bound on $ r$ is the projection of the curve of intersection of the sphere and the cone in the $ xy$-plane, namely $ r^2 \\plus{} r^2 \\equal{} R^2\\Rightarrow r \\equal{} \\frac {1}{\\sqrt {2}}R$. The lower bound of $ r$ is $ 0$. Also, $ z$ ranges from the cone to the sphere, which is $ r\\leq z\\leq\\sqrt {R^2 \\minus{} r^2}$\r\n\\[ V \\equal{} \\int_{\\theta \\equal{} 0}^{2\\pi}\\int_{r \\equal{} 0}^{\\frac {1}{\\sqrt {2}}R}\\left(\\sqrt {R^2 \\minus{} r^2} \\minus{} r\\right)r\\, dr\\, d\\theta\r\n\\]\r\n[hide=\"Click for calculation of integral\"]\\[ V \\equal{} \\int_{\\theta \\equal{} 0}^{2\\pi}\\int_{r \\equal{} 0}^{\\frac {1}{\\sqrt {2}}R}\\left(\\sqrt {R^2 \\minus{} r^2} \\minus{} r\\right)r\\, dr\\, d\\theta \\equal{} 2\\pi\\cdot \\int_{u \\equal{} \\frac {1}{2}R^2}^{R^2}\\sqrt {u}\\, \\frac {du}{2} \\minus{} 2\\pi\\cdot\\int_{r \\equal{} 0}^{\\frac {1}{\\sqrt {2}}R}r^2\\, dr\n\\]\n\\[ \\equal{} \\frac {2\\pi}{3}\\left[ u^{\\frac {3}{2}}\\right|_{u \\equal{} \\frac {1}{2}R^2}^{R^2} \\minus{} \\frac {2\\pi}{3}\\left[ r^3\\right|_{r \\equal{} 0}^{\\frac {1}{\\sqrt {2}}R} \\equal{} \\frac {2\\pi}{3}R^3\\left( 1 \\minus{} \\frac {1}{2\\sqrt {2}}\\right) \\minus{} \\frac {2\\pi}{3}\\cdot\\frac {1}{2\\sqrt {2}}R^3\n\\]\n\n\\[ \\equal{} \\boxed{\\frac {\\pi}{3}R^3\\left( 2 \\minus{} \\sqrt {2}\\right) }\n\\]\nI did, BTW, verify this calculation by adding the volume formulas for a cone and a spherical cap appropriate to our situation.[/hide]", "Solution_2": "(i) Does Z = R means the generatrix makes 45\u00b0 with the axis?\r\n\r\n(ii) The sphere has unitary radius? \r\n\r\n(iii) The sphere is tangent (i.e. inscribed ) to the conic surface?\r\n\r\n(iv) What is sno-cone?", "Solution_3": "(i) Yes.\r\n\r\n(ii) Yes.\r\n\r\n(iii) No, the sphere is not tangent to the cone. Its center is the vertex of the cone, so it's actually orthogonal rather than tangent.\r\n\r\n(iv) A sno-cone is an old-fashioned American snack food, associated with fairs, carnivals, and outdoor gatherings in general. A ball of shaved ice is placed into the top of a paper cone, and flavored (and colored) syrup is poured onto the ball of shaved ice. The angular opening of the cone is probably only about half of the 45 degrees of this problem, and the ball in the sno-cone is pretty close to tangent to the cone, so it isn't all that accurate an image for this problem.", "Solution_4": "[quote=\"Kent Merryfield\"](iv) A sno-cone is an old-fashioned American snack food, associated with fairs, carnivals, and outdoor gatherings in general. A ball of shaved ice is placed into the top of a paper cone, and flavored (and colored) syrup is poured onto the ball of shaved ice. The angular opening of the cone is probably only about half of the 45 degrees of this problem, and the ball in the sno-cone is pretty close to tangent to the cone, so it isn't all that accurate an image for this problem.[/quote] This put me in a really, really good mood.", "Solution_5": "Oh - did you also go for the cotton candy? (Since I've had a beard for my entire adult life, that would make cotton candy a really bad idea for me - but I didn't particularly like the stuff when I was a kid, anyway.)", "Solution_6": "Thank you... now I can eat a sno-cone knowing what its volume will be." } { "Tag": [ "IMO", "IMO 2007" ], "Problem": "Kazakhstan IMO team:\r\n\r\nDaliyev Asset(10th grade)\r\nBayev Alen(11th)\r\nStassiy Igor(11th)\r\nIsmailov Adilzhan(10th)\r\nDiyarov Arman(11th)\r\nand me-Mussagaliyev Asset(11th)\r\nMojno(ili daje nujno) zdes pisat po-russki", "Solution_1": "Otkuda pishesh ACET? Vi ushe vo Vietname? Udachi na olympiade. :)Ne zabudte zavtra napisat kak proshel 1 tur.", "Solution_2": "Segodnia napisali 1-iy tur:\r\nPredvaritelnie rezultati(sootvetstvenno #1,#2,#3):\r\nAset Junior:+/-,-,-;\r\nAlen:+,+,-:\r\nIgor:+/-,+,-;\r\nAdiljan:+/-,-,-;\r\nArman:+/-,-,-;\r\nYa:+,-,-.", "Solution_3": "Pochemu ne vse reshili 2 zadachu :| ?Ona ved ne slojnaja.Udachi vo 2 ture :)", "Solution_4": "[quote=\"ACET\"]Segodnia napisali 1-iy tur:\nPredvaritelnie rezultati(sootvetstvenno #1,#2,#3):\nAset Junior:+/-,-,-;\nAlen:+,+,-:\nIgor:+/-,+,-;\nAdiljan:+/-,-,-;\nArman:+/-,-,-;\nYa:+,-,-.[/quote]\r\n\r\nCould you please continue to post [url=http://www.mathlinks.ro/viewtopic.php?p=893786#893786]your rumours in this thread.[/url]" } { "Tag": [ "Diophantine equation", "number theory solved", "number theory" ], "Problem": "equation $x^2+y^2+1=2z^2$. how many integer solutions?", "Solution_1": "Take y=0 and an infinite number of solutions of the Pell equation x^2+1=2z^2. You will find that your equation has an infinite number of solutions.", "Solution_2": "[u][b]The author of this posting is : fgomezr[/b][/u]\r\n____________________________________________________________________\r\n\r\n[quote=\"harazi\"]Take y=0 and an infinite number of solutions of the Pell equation x^2+1=2z^2. You will find that your equation has an infinite number of solutions.[/quote]\r\n\r\nWhat with solutions in Z+. I havent try it but maybe then, it is harder.", "Solution_3": "[u][b]The author of this posting is : Anh Cuong[/b][/u]\r\n____________________________________________________________________\r\n\r\nUm,I think if y=1 and we still have a Pell equation x^2+2=2z^2 with the smallest solution is (4,3).So it has an infinite number of solutions too. :)" } { "Tag": [ "function", "\\/closed" ], "Problem": "I think it would be cool if we had a graphing function in AoPS so we could solve graphing problems and graph things .....dahdahdahdah.", "Solution_1": "[quote=\"shinwoo\"]I think it would be cool if we had a graphing function in AoPS so we could solve graphing problems and graph things .....dahdahdahdah.[/quote]\r\n\r\nIt would be and hopefully we will one day. But it's not near the top of our priority list right now.", "Solution_2": "what is on the top of the priority list? :) \r\njust wondering", "Solution_3": "[quote=\"math92\"]what is on the top of the priority list? :) \njust wondering[/quote]That is a classified information ;) :lol:" } { "Tag": [], "Problem": "Two cyclists start at the same time and travel in opposite directions. One travels at 3 miles per hours faster than the other. In 9 hours, they are 171 miles apart. What is the speed of the slowest cyclist?", "Solution_1": "[hide]Call the rate of the slower cyclist $x$. We can set up the equation $9(x+x+3)=171$. Solving we find the slower cyclist travels at $\\boxed{\\text{8 mph}}$.[/hide]", "Solution_2": "[hide]the two cyclists together are going $\\frac{171}{9}$mph. \n\nSo $x+y=19$\n$x+3=y$\n$2x+3=19$\n$x=8$\n\n$\\boxed{\\text{8 mph}}$[/hide]", "Solution_3": "[hide=\"Here is the answer\"]Let x = speed of slow cyclist, y=speed of fast cyclist. \ny=x+3\n\nEvery hour, their distance increases by 171/9 = 19 miles. Therefore, \nx+y=19\n\nSubstituting y=x+3,\n\n2x+3 = 19\n\n2x = 16\n[b]x = 8 miles per hour[/b][/hide]", "Solution_4": "Good job. You are all correct. :winner_first: :)", "Solution_5": "[quote=\"nonie\"]Two cyclists start at the same time and travel in opposite directions. One travels at 3 miles per hours faster than the other. In 9 hours, they are 171 miles apart. What is the speed of the slowest cyclist?[/quote]\r\n\r\n[hide]\ny=x+3\n9x+9y=173\nsolve from heere[/hide]", "Solution_6": "[hide]$9(x+x+3)=171\\longrightarrow9(8+8+3)=171\\longrightarrow x=8$[/hide]" } { "Tag": [ "function", "integration", "probability", "calculus", "calculus computations" ], "Problem": "How would you solve this?\r\nLet $X$ be a random variable that is uniformly distributed on $(a,b)$ with $ac)$\r\n[hide=\"what i have\"]\n$P(X>c)=\\frac{b-c}{b-a}$ and $P(X\\cap X>c)=\\frac{1}{b-c}$\nWhat did I do wrong since it is clearly wrong.\n[/hide]", "Solution_1": "The density function is not a constant- it's a constant times the characteristic function of a set. Include that, and you'll be right.", "Solution_2": "I dont follow what you are saying. :(", "Solution_3": "According to what you've written $P(X > c)$ tends to infinity as $c$ tends to negative infinity.", "Solution_4": "I think the answer is right. The density function for a Uniformly distributed random variable is constant and is given by $\\frac{1}{b-a}$. A univariate conditional distribution is given by: \r\n$f(X|A) = \\frac{f(x)}{P(A)}$. In your case $P(A) = P(X>c) = \\int_{c}^{b}\\frac{1}{b-a}dx = \\frac{b-c}{b-a}$. The probability becomes $\\frac{\\frac{1}{b-a}}{\\frac{b-c}{b-a}}= \\frac{1}{b-c}$. \r\nThis is the conditional density of $X$, which is also Uniformly distributed.", "Solution_5": "ok but how do you get that $P(X|A)=\\frac{P(x)}{P(A)}$?", "Solution_6": "It's a definition. See this [url=http://www.math.uah.edu/stat/dist/Conditional.xhtml]link.[/url]" } { "Tag": [ "calculus" ], "Problem": "I've been accepted to the University of Rhode Island (with a full tuition scholarship). I've also been accepted to these local Massachusetts schools:\r\n\r\nUMass Lowell\r\nUMass Dartmouth\r\nStonehill College\r\nWorcester Polytechnic Institute (WPI)\r\n\r\nI will be hearing from Wheaton College in Massachusetts.\r\n\r\nI've received good scholarships to all the college I've been accepted to so far. My number one choice is WPI because I really like it. My brother majored in chemical engineering there; he's now going for a Ph.D in the same field at the University of California at Santa Barbera. WPI has offered me their Chemistry and Biochemistry Scholarship; it provides me with cool benefits like research experience over the summer.\r\n\r\nDoes anyone have any opinion on any of these colleges? Do you have a story to share about them? Are any of you going or have any of you gone to any of them for any time? Did you consider them at some point? I'd love to hear any information or advice that any of may have.\r\n\r\nI really like chemistry, and I really like math (particularly calculus so far)... so I'm interested in majoring in either chemistry or chemical engineering. A double major is out of the question because I'm perfectionistic, unfortunately (that's also why I didn't apply to colleges like MIT). Regardless of my college choices or regarding my college choices, do any of you have any information or advice about these majors? Are majoring in or have you majored in any of them?\r\n\r\nThanks!!!", "Solution_1": "w00t!!! I got accepted to Wheaton College with a good scholarship! I've been accepted to all 6 colleges I applied to!\r\n\r\nDoes ANYONE have ANY comments on ANY of these schools? Please share any insights you may have!\r\n\r\nThanks!", "Solution_2": "Does anybody have anything to say about any of these colleges and/or majors?", "Solution_3": "I thought other participants would reply by now. I think WPI is the one best known outside its region (where I am) for technical disciplines.", "Solution_4": "Well, I decided to go to WPI! w00t!\r\n\r\nPlease let me know if you have been to WPI, are going to WPI, or will be going to WPI! Maybe we can discuss all things WPI...", "Solution_5": "Nice :) That would have been my choice...even though I don't know much about any of those schools :oops: \r\n\r\nHave fun in college. It'll be awesome, I'm pretty sure." } { "Tag": [], "Problem": "I have a couple of topics for my language arts class that I have to write on. I wanted to hear your opinions.\r\n\r\nThese are summaries of the prompts:\r\n\r\n1) Is it necessary to have guidelines on student participation in after school sports (or other extracurricular activities), concerning the students GPA? To what extent should those guidelines go?\r\n\r\n2) Does a office referral deter future troublemaking? What type of punishment should an administrator employ? What is the most effective means a school has to encourage good behavior amongst students?\r\n\r\n3) There are numerous shows on television and radio designed to embarass a person for the purpose of entertaining an audience that is in on the joke. Is this a bonafide form of entertainment?\r\n\r\nThanks in advance for your input!", "Solution_1": "ugh. these are tricky. the office referral one really depends on the ways of the people, like the environment.\r\n\r\npersonally, i think that there should be a GPA requirement, and i dont think that a 2.0 is very hard. and you need skills like writing to write applications, and speeches. you should be able to understand the fine print on a contract, understand what 'compounded interest' means, things like that. \r\n\r\nthink about if a student can't pass all their classes, hows that going to affect them later on? \r\n\r\ni think that a kid getting bad enough grades not to make the team, and then caring about it should work harder.\r\n\r\nhope this helps.", "Solution_2": "thanks! \r\n\r\nif anyone has experience relating to one of these things, it would be appreciated.", "Solution_3": "1. There does need to be GPA requrements to protect the students. Absent these types of protections, students will be pushed into spending the time required for sports at the expense of their studies. (See lawsuit where 5 students that played basketball for USC and graduated saying that they couldn't even read and write their name.) The students tried to take advantage of the system, but they can't make these types of decisions.\r\n\r\n2. Hard to tell what will encourage proper behavior, but usually it is only imposed after classroom methods have failed or take too much class time.\r\n\r\n3. I don't think so. I think this is cruel, similar but more moral than the killing in the arena in Rome.", "Solution_4": "thanks, wizard2378, but i cant find a link for the USC basketball players. sounds a little far-fetched to me, not being able to read their own name.\r\n\r\ni don't watch that much TV, so can anyone give me some examples of the entertainment that is mentioned in #3?\r\n\r\nThanks so much, star99 and wizard2378! You are a big help!", "Solution_5": "1) No. These guidelines shouldn't go anywhere.\r\n\r\n2) What is an office referral? Punishment should be given depending on the severity of the offense committed. The most effective (but not most practical) means a school has to encourage good behavior is to turn private and only accept goody two shoes.\r\n\r\n3) BORAT!!!!!! What does bonafide mean?", "Solution_6": "[quote=\"webspiral\"]1) Is it necessary to have guidelines on student participation in after school sports (or other extracurricular activities), concerning the students GPA? To what extent should those guidelines go?[/quote]\r\n\r\nSchool is there to give you an education - not to provide sports teams, debate clubs, etc. These can be viewed as 'perks' - that is why they are called [i]extra[/i]curricular activities. I see it that you should not be allowed the reward - sports, clubs, and the like unless you are proficient at what you are really there to do - learn. \r\n\r\nHow many people would misuse the system if there were no guidelines? I think my school has a system where you cannot be failing any classes. This makes sure that you actually do the work in the classes, even if you are struggling, just so long as you try you should be able to maintian a passing grade. \r\n\r\nFundamentally, I would argue that schools are in place to make sure that you get an education, and limiting extras unless you are working hard makes sense - ever had the no TV/computer/candy or whatever reward until you're homework is done pulled on you when you were younger? I view it as essentially the same concept. Maybe schools should go further to make sure that you're catching up - provide mandatory tutoring for a class you're failing, weekly parent/councelour/student conferences, and so on. But these cost money - something that many schools are lacking." } { "Tag": [ "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $G$ be a finite group of order $p+1$ with $p$ prime. Show that $p$ divides the order of $Aut(G)$ if and only if $p$ is Mersenne prime, that is, of the form $2^{n}-1$, and $G$ is isomorphic to $(Z/2)^{n}$.\r\n\r\nProbably something for grobber, harazi or zetax ;) \r\n\r\nI once tried this, but I couldn't find it. Since it was part of a competition, which is now over, I couldn't ask for hints! So... surprise me :)", "Solution_1": "The problem is wrong- it misses a case.\r\n\r\nThe automorphism group of $G$ naturally injects into the permutation group on the non-identity elements of $G$. The only elements of order $p$ in that group are $p$-cycles which move every element; we therefore have that the automorphisms of $G$ act transitively on the non-identity elements if $p$ divides the order of the automorphism group.\r\nAutomorphisms preserve order, so every non-identity element of $G$ has order $q$. This $q$ must be prime, and $|G|$ is a power of $q$. There are two cases; either $p=2$ and $G$ is cyclic with three elements, or $p$ is odd and $q=2$. In the second case, all elements of $G$ other than the identity have order 2. By [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=68000]standard arguments[/url], $G$ is isomorphic to $(\\mathbb{Z}/2\\mathbb{Z})^{n}$." } { "Tag": [ "function" ], "Problem": "is there any possible way to solve for x albegriacally for the following problem?: log base 2x 216 = x ?? thanks!! =]", "Solution_1": "Note that if $x=y$ then $b^{x}=b^{y}$. You're given an equation, so try to find b.\r\n\r\nThis probably isn't the best solution, but it's been a while since I've worked with logs.", "Solution_2": "[hide=\"Well...\"] We have $216 = (2x)^{x}$. You should know that $216 = 6^{3}$, so $x = 3$.\n\nI do not think there is a way to generally solve $a = x^{x}$. [/hide]", "Solution_3": "$a = x^{x}$ has no general elementary solution, but it can be given a closed solution using lambert's omega function. [url]http://en.wikipedia.org/wiki/Lambert%27s_W_function[/url] :maybe: \r\nbut one's probably better off using numerical methods to solve this general form.", "Solution_4": "thanks everyone!! =]...i'm new to this forum...don't really know how to use the fancy math symbols, but i'll start looking into it....and i think this problem should've gone into the beginners section...=P." } { "Tag": [], "Problem": "Find the number of ordered triples $(x,y,z)$ ($x$, $y$, $z$ are real) such that $x=yz$, $y=xz$ and $z=xy$.", "Solution_1": "multiply them to get\r\n\r\n$xyz=(xyz)^2\\implies xyz(xyz-1)=0$\r\n\r\nthen $x=0 \\implies y=z=0$\r\n\r\n$xyz=1\\implies yz=\\frac{1}{x}$, substitute\r\n\r\n$x=\\frac{1}{x} =\\implies x=\\pm 1$, then we get\r\n\r\n$(-1,-1,1)$ (with permutations), $(1,1,1)$ and $(0,0,0)$", "Solution_2": "Altheman, it asks for the number of solutions, which is \r\n[hide]5[/hide]", "Solution_3": "[quote=\"b-flat\"]Altheman, it asks for the number of solutions, which is \n[hide]5[/hide][/quote]\r\n\r\nWho cares." } { "Tag": [ "calculus", "AoPS Books" ], "Problem": "I'm just wondering how much you guys spend on math each day for many of you to get as good as you are. Not sure if this is the right forum for this to be in, so move it if necessary.\r\nFor the most part, I spend between 1/2-1 hour each day doing old problems and going through some textbooks (excluding homework for math at school)", "Solution_1": "I've been doing 8-10 hours for the past two weeks :lol: Don't know about my chances of getting into IMO, though.", "Solution_2": "8-10 hours when it is break...\r\nabout 3-4hours in regular school days", "Solution_3": "I have calculus for two periods but I usually don't spend the entire time working. Usually about an hour and a half for calc on math. Then about 30 more minutes interspersed throughout the school day on math. When I get home after track/XC I would say I average about 3 hrs more on math. So about 4.5 hrs a day. [Much] More on weekends and during break though.", "Solution_4": "Depends on the day, on average 2 hours, because sometimes I'll come home and just sleep, other times I'll spend 6-8 hours on math stuff for math team. So it's pretty sparatic, but I usually get my couple of hours in.", "Solution_5": "Usually about $3$.", "Solution_6": "Depends for what, if its for school usually only the time I spend at class. If it's something else like a competition or something for fun, a couple of hours i guess.", "Solution_7": "wow you guys do alot of math... it's very inspirational to me, on average i only get in 1 hour... :(", "Solution_8": "I try to do as much as I can. Calculus class is hard to pretend to pay attention in, unfortunately. Calculus homework then takes a while, since we have to write out a lot of stuff. We had to do a full set of AP free response per day for a while, which is like 90 minutes. Of course I only spent around 45 :P. Remaining time minus maybe 1 hour a day of other homework is probably half math and half whatever else.", "Solution_9": "about $5$ for me, since i'm pretty busy in school now...", "Solution_10": "Too much :D At least if browsing the Fun & Games forum counts as math... after all, it is on a math forum ;)", "Solution_11": "about 1 to 2 hours... i have heavy school work and i got physic olympiad....", "Solution_12": "About an hour a day...", "Solution_13": "About 2 hours at home and there's always time at school to do math.", "Solution_14": "The trick is to have a few maths books lying on your table, when your bored in your room you just pick one up, flick through it, see something interesting and read that chapter! Tis what i do, and im going to the IMO this year with ireland, so im doing something right!", "Solution_15": "Well, Practice makes perfect, I used to be okay in Math but then I started doing about 3-4 hours of math every day for like the last year, going through the AOPS books and got a whole lot better. :)", "Solution_16": "When I have school, I do around 2 hours.\r\n\r\nWhen I have break, I do a lot more. (I've been on AoPS all day toady except for eating :D )", "Solution_17": "When in school I've been doing a lot of math cause of AP Calc... though i'm really lame when it comes to math contests (can't even get into AIME). Probably an hour a day :D" } { "Tag": [ "function", "logarithms" ], "Problem": "Find the value of $ x$ for which the function $ f(x)\\equal{}2^x\\minus{}4^x$ reaches of a maximum.", "Solution_1": "[quote=\"PhireKaLk6781\"]Find the value of $ x$ for which the function $ f(x) \\equal{} 2^x \\minus{} 4^x$ reaches of a maximum.[/quote]\r\n\r\n[hide]Note that $ (2^x)^2 \\equal{} 2^{2x} \\equal{} (2^2)^x \\equal{} 4^x$\n\n\nLet $ u \\equal{} 2^x$.\nThen $ u^2 \\equal{} 4^x$.\n\n$ f(u) \\equal{} u \\minus{} u^2$\n\n$ f(u) \\equal{} \\minus{} u^2 \\plus{} u$\n\nCoefficients:\n---------------\n$ a \\equal{} \\minus{} 1$\n$ b \\equal{} 1$\n\n$ u \\equal{} \\frac { \\minus{} b}{2a}$\n\n$ u \\equal{} \\frac12$\n\n$ u \\equal{} 2^x$\n\n$ \\frac12 \\equal{} 2^x$\n\n$ 2^{ \\minus{} 1} \\equal{} 2^x$\n\n$ \\minus{} 1 \\equal{} x$\n$ x \\equal{} \\minus{} 1$\n\n$ f(x) \\equal{} 2^x \\minus{} 4^x$ reaches a maximum at $ x \\equal{} \\minus{} 1$.\n\n\n\n\n[/hide]", "Solution_2": "[hide=\" Alternatively\"]\n\n[Wrong forum, but we can check by]\n\ndifferentiating, f(x) = $ 2^{x} \\ln 2 \\minus{} 4^{x} \\ln 4 \\equal{} 0$\n\n$ \\implies \\frac {4^{x}}{2^{x}} \\equal{} \\frac {\\ln 2}{\\ln 4}$\n\n$ \\implies 2^{x} \\equal{} \\frac {1}{2}$\n\n$ \\implies x \\equal{} \\minus{} 1$ and so $ f(x) \\equal{} \\frac {1}{4}$\n\n[/hide]" } { "Tag": [ "modular arithmetic", "combinatorics unsolved", "combinatorics" ], "Problem": "Let n and k are naturals numbers such that $ 1\\le k\\le n$.Write a n-digits natural number by digits 0 or 1. How many that numbers which such that exist k digits 1 consecutive in its?\r\nhope a nice solution!", "Solution_1": "[hide=\"click\"]lets denote k consecutive 1's by letter x, so the question reduces to find all numbers of length n-k+1 , having one letter x and other places filled by 0 or 1.\n\nHence required number = $ \\dbinom{n\\minus{}k\\plus{}1}{1} * 2^{n\\minus{}k}$[/hide]", "Solution_2": "No, that massively over-counts: any word with several different runs of $ k$ 1s is counted many times in that expression.\r\n\r\nIt's easy to write a recursion for the complement of what you're counting (i.e., the number of sequences of 0s and 1s of length $ n$ with no $ k$ consecutive 1s): break every such word up according to its leftmost occurrence of 0 to see that $ a_n \\equal{} a_{n \\minus{} 1} \\plus{} a_{n \\minus{} 2} \\plus{} \\ldots \\plus{} a_{n \\minus{} k}$ with $ a_j \\equal{} 2^j$ for $ 0 \\leq j \\leq k \\minus{} 1$. This assumes that an $ n$-digit natural number may have leading 0s, but it's not hard to adjust for the other situation.\r\n\r\n\r\nTo the original poster: please give your threads more descriptive names.", "Solution_3": "[quote=\"JBL\"]To the original poster: please give your threads more descriptive names.[/quote]\r\nsorry, I will attention more.\r\ncan you post a solution for it?thanks", "Solution_4": "if there are no runs of $ k$ ones in the number then it has form\r\n\\[ 1^{s_1} 0 1^{s_2} 0 1^{s_3} 0 \\dots 0 1^{s_t}\r\n\\]\r\nwhere $ t$ is the number of runs of ones, $ 0\\leq s_i < k$ is the length of i-th runs, and\r\n\\[ t \\minus{} 1 \\plus{} s_1 \\plus{} s_2 \\plus{} \\dots \\plus{} s_t \\equal{} n.\r\n\\]\r\nFor a fixed $ t$, the number of such numbers is the coefficient of $ x^{n \\minus{} t \\plus{} 1}$ in\r\n\\[ (1 \\plus{} x \\plus{} \\dots \\plus{} x^{k \\minus{} 1})(1 \\plus{} x \\plus{} \\dots \\plus{} x^{k \\minus{} 1})\\dots(1 \\plus{} x \\plus{} \\dots \\plus{} x^{k \\minus{} 1}) \\equal{} \\left(\\frac {x^k \\minus{} 1}{x \\minus{} 1}\\right)^t.\r\n\\]\r\nor equivalently the coefficient of $ x^{n \\plus{} 1}$ in $ \\left(\\frac {(x^k \\minus{} 1)x}{x \\minus{} 1}\\right)^t.$\r\n\r\nThe total number of such numbers is the coefficient of $ x^{n \\plus{} 1}$ in\r\n\\[ \\sum_{t \\equal{} 0}^{\\infty} \\left(\\frac {(x^k \\minus{} 1)x}{x \\minus{} 1}\\right)^t \\equal{} \\frac {1}{1 \\minus{} \\frac {(x^k \\minus{} 1)x}{x \\minus{} 1}} \\equal{} \\frac {1 \\minus{} x}{1 \\minus{} x \\plus{} x^{k \\plus{} 1}}\r\n\\]\r\nwhich equals to $ c_{n \\plus{} 1} \\minus{} c_n$ where\r\n\\[ c_{m,k} \\equal{} \\sum_{t\\equiv m\\pmod{k}} \\binom{t}{\\frac {m \\minus{} t}{k}} ( \\minus{} 1)^{\\frac {m \\minus{} t}{k}}\\qquad\\text{is the coefficient of}\\,x^m\\,\\text{in}\\qquad\\frac {1}{1 \\minus{} x \\plus{} x^{k \\plus{} 1}}.\r\n\\]\r\nTherefore, the answer to the original question is\r\n\\[ 2^n \\minus{} c_{n \\plus{} 1,k} \\plus{} c_{n,k}.\r\n\\]" } { "Tag": [ "induction", "combinatorics proposed", "combinatorics" ], "Problem": "Let $ a_1,a_2,\\dots,a_n$ be a sequence of $ n$ integers so that their sum equals to $ 1$. Among the $ n$ cyclic sequences,\r\n\\[ \\{a_1,\\dots,a_n\\},\\{a_2,\\dots,a_1\\},\\dots,\\{a_n,\\dots,a_{n\\minus{}1}\\}\\] there is exactly one sequence whose every partial sum is positive. \r\n(A $ k$- partial sum of a sequence $ \\{b_1,b_2,\\dots,b_n\\}$ is $ \\sum_{i\\equal{}1}^k b_i$)", "Solution_1": "I think you mean at least one. Cause, for example, {1/n, 1/n... 1/n} has all n cyclic sequences being postive. \r\n\r\nNow suppose there exist a sequence of size n such that none of its cyclic sequences has sum over n. Then we prove we can find a sequence of size n-1 with the same property. There must be one element that is less than 0 in the sequene, call it a. Now add a to the element in front of it (i.e. b). There are now n-1 elements in the sequence with the above property (easily verified by checking cases) Now apply induction on n, we get that we can find a sequence of length 1 with the same property, but a = 1. Contradiction. So there can be no sequences with this property. \r\n\r\nInstead of = 1, we can make it = any postive integer.", "Solution_2": "[quote=\"srikanth\"]Let $ a_1,a_2,\\dots,a_n$ be a sequence of $ n$ [b]integers[/b]...[/quote]\r\nSo there is a little more work to be done.\r\n\r\nConsider the following visualization. Extend the sequence $ a_k$ to be a periodic sequence with period $ n$, let $ s_n \\equal{} \\sum_{k \\equal{} 1}^{n} a_k$, and consider the plot of the points $ (n, s_n)$ (with broken lines between points for effect). This plot is \"almost\" periodic in the sense that $ (k, s_k) \\equal{} (k \\plus{} n, s_k \\plus{} 1)$. We call a period an interval of the form $ (pn, (p \\plus{} 1)n]$. This plot clearly attains a minimum value some number of times on $ (0, n]$; we will refer to the points such that this occurs as minima. \r\n\r\nNow the claim that a given permutation contains only positive partial sums is precisely the claim that if we draw a horizontal line from a given point on this graph of length $ n$, it will not intersect the graph, and we claim that this is true only of the [b]rightmost minimum[/b] on $ (0, n]$. \r\n\r\nIf we draw the line from any point to the left of the rightmost minimum, it will intersect the graph around the rightmost minimum, and if we draw the line from any point to the right of the rightmost minimum, it will intersect the graph around the rightmost minimum of the next period (since that minimum is exactly $ 1$ unit higher than the previous, and any points to the right of the rightmost minimum are at least $ 1$ unit higher than the rightmost minimum). \r\n\r\nOn the other hand, if we draw the line from the rightmost minimum, in one period the rest of the points will be strictly above the line (by construction; this minimum is rightmost) and in the next period the rest of the points are higher by $ 1$ than they were in the previous period, so the minima (the lowest points) will also be strictly above the line.", "Solution_3": "[hide]\nSince the sum of the sequence is positive, there must exist at least two adjacent terms $ a_k, a_{k + 1}$ such that $ a_k > 0$ and $ a_k + a_{k + 1} > 0$ (where we define $ a_{n + 1} = a_1$). We then replace the the two terms $ a_k, a_{k + 1}$ with the positive term $ a_k + a_{k + 1}$. \n\nWe repeat this process with the $ n - 1$ numbers and so forth until there are only two numbers, one of which is positive, and the other which is non-positive (since the sum is always $ 1$). Obviously the non-positive term was also a term $ a_j$ in the initial sequence, since it was never the result of a replacement. Then we claim that the cyclic sequence starting with term $ a_{j + 1}$ is the unique desired sequence with positive partial sums. \n\nSince in the process we eventually replaced the term $ a_{j + 1}$ with the term $ a_{j + 1} + a_{j + 2}$ (here $ a_{j + 2}$ does not necessarily denote the original term there), it follows that $ a_{j + 1} > 0$ and $ a_{j + 1} + a_{j + 2} > 0$, and so forth. It now follows that the partials sums are positive by induction; indeed, if at some point $ a_{j + 2}$ had been replaced by terms $ a_{j + 2} + a_{j + 3}$, then by the definition of our construction, $ a_{j + 2} > 0$, and so forth. Uniqueness follows also as in our construction, there eventually exists only one $ a_k$ satisfying $ a_k>0, a_k + a_{k+1} > 0$, and the term $ a_j$ was never collapsed.\n[/hide]", "Solution_4": "Woops sorry didn't see the integers.\r\n\r\nIf it is integers then it's exactly one. Suppose we have a contradiction and there's 2 or more sequences. Let us say that the first element of the second sequence start in the ith position of the first sequence. Then the first element of sequence 1 up to i-1 is positive. But i to the last element must also be positive (due to the property of second sequence). Both sums are integer, and add up to 1. Thus one of them must be not positive. Contradiction. So there's at most one." } { "Tag": [ "ARML", "geometry", "perimeter", "inradius", "circumcircle" ], "Problem": "Define angle A to be \"nice\" if both SinA and CosA are rational.\r\nDefine a convex polygon \"nice\" if all of its interior angles are \"nice\"\r\n\r\nA convex quadrilateral has the following properties:\r\n1. Its sides are integers whose product is a square\r\n2. It can be inscribed in a circle and can be circumscribed about another circle.\r\n\r\nProve this quadrilateral is nice.", "Solution_1": "wierd, i did this set of power questions but i dont remember this one. i suppose from the difficulty it was part C?", "Solution_2": "The sum of two nice angles is again nice (easy to prove).\r\nHowever I don't know whether that may help for the solution.", "Solution_3": "[hide]i missed the word quadrilateral. now everything is easier \n\nit has inscribed circle, so divide its primeter into a,a,b,b,c,c,d,d \n\nso its sides are a+b, a+d, d+c, b+c \n\n\n\nits area is sqrt((a+b)(c+d)(b+c)(a+d)) (cyclic quad) \n\nhence its area is an integer \n\n\n\nthe area can be expressed as: \n\n0.5((a+d)(a+b)sin(k)+(b+c)(d+c)sin(180-k)) \n\nand similarly another way \n\nsin(180-k) is obviously sin (k) \n\nso sin of the angles is rational \n\n\n\nfor the cos, for simplicity let the sides be w,x,y,z\n\nthen w^2+x^2-2wxcos(k)=y^2+z^2-2cos(180-k)\n\ncos(180-k)=-cosk\n\nsolving for cos(k)\n\nhence cos(k) is alo rational[/hide]", "Solution_4": "[color=cyan]There may have been an earlier part of this question that made that easier. One earlier part, for example, was that a triangle with rational sides and one nice angle has all nice angles. You can also show that the area, inradius and circumradius of such a triangle are rational. I don't know if any of these things is very helpful, but on the actual question they may have given you tools to work with.[/color]", "Solution_5": "I also tried to prove it for any polygon ...\r\nThat's probably why I didn't manage to prove it :? \r\nThe word quadrilateral makes it easier !" } { "Tag": [ "factorial", "algebra", "system of equations", "number theory proposed", "number theory" ], "Problem": "Let $ a,b,c,x,y,z$ satisfy followings :\r\n\r\n$ a x \\plus{} b y \\plus{} c z \\equal{} 1 ,$\r\n$ a x^2 \\plus{} b y^2 \\plus{} c z^2 \\equal{} 2 ,$\r\n$ a x^3 \\plus{} b y^3 \\plus{} c z^3 \\equal{} 6 ,$\r\n$ a x^4 \\plus{} b y^4 \\plus{} c z^4 \\equal{} 24 ,$\r\n$ a x^5 \\plus{} b y^5 \\plus{} c z^5 \\equal{} 120 ,$ and\r\n$ a x^6 \\plus{} b y^6 \\plus{} c z^6 \\equal{} 720.$\r\n\r\nThen find the value of $ a x^7 \\plus{} b y^7 \\plus{} c z^7.$\r\n :idea:", "Solution_1": "the relation is a cyclic relationship\r\n\r\n\r\n\r\nhave you noticed that the eqns follow the following sequence1!,2!,3!,4!,5!,6!\r\ntherefore the answer is 7! or 1540 \r\n{here ! denotes factorial;n!=n(n-1)(n-2)....321}", "Solution_2": "Let $ u_n=ax^n+by^n+cz^n$, then $ u_{n+3}=\\sigma_1 u_{n+2}-\\sigma_2 u_{n+1}+\\sigma_3u_n$,\r\nwere $ \\sigma_1=x+y+z,\\sigma_2=xy+yz+zx,\\sigma_3=xyz$.\r\nIt give system of equations (n=1,2,3) for $ \\sigma s_i$:\r\n$ (n+3)!=(n+2)!\\sigma_1-(n+1)!\\sigma_2+n!\\sigma_3$ or\r\n$ 6\\sigma_1-2\\sigma_2+\\sigma_3=24,$\r\n$ 24\\sigma_1-6\\sigma_2+2\\sigma_3=120,$\r\n$ 120\\sigma_1-24\\sigma_2+6\\sigma_3=720$.\r\nIt give $ \\sigma_3=12,\\sigma_2=36,\\sigma_3=24$ (x,y,z are roots $ t^3-12t+36t-24=0$).\r\nTherefore $ u_7=12u_6-36u_5+24u_4=-2880$.", "Solution_3": "[quote=\"Rust\"]Let $ u_n = ax^n + by^n + cz^n$, then $ u_{n + 3} = \\sigma_1 u_{n + 2} - \\sigma_2 u_{n + 1} + \\sigma_3u_n$,\nwere $ \\sigma_1 = x + y + z,\\sigma_2 = xy + yz + zx,\\sigma_3 = xyz$.\nIt give system of equations (n=1,2,3) for $ \\sigma s_i$:\n$ (n + 3)! = (n + 2)!\\sigma_1 - (n + 1)!\\sigma_2 + n!\\sigma_3$ or\n$ 6\\sigma_1 - 2\\sigma_2 + \\sigma_3 = 24,$\n$ 24\\sigma_1 - 6\\sigma_2 + 2\\sigma_3 = 120,$\n$ 120\\sigma_1 - 24\\sigma_2 + 6\\sigma_3 = 720$.\nIt give $ \\sigma_3 = 12,\\sigma_2 = 36,\\sigma_3 = 24$ (x,y,z are roots $ t^3 - 12t + 36t - 24 = 0$).\nTherefore $ u_7 = 12u_6 - 36u_5 + 24u_4 = - 2880$.[/quote]\r\n\r\nOh! Good idea.\r\n\r\nMy idea is the same as yours.\r\n\r\nBut you have a little mistake , I think. So I would like you to compute again. :wink:" } { "Tag": [ "search", "national olympiad" ], "Problem": "These have been in ML for a while, but just for the sake of completness I will post them here so that they can be added to the resources sections, as well as the other Argentina's TST\r\nDay one\r\n\r\n1- http://www.mathlinks.ro/Forum/viewtopic.php?search_id=129882059&t=35067\r\n2- http://www.mathlinks.ro/Forum/viewtopic.php?search_id=129882059&t=35064\r\n3- http://www.mathlinks.ro/Forum/viewtopic.php?search_id=129882059&t=35065\r\n\r\nDay two\r\n\r\n4-\r\n5- http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1089870445&t=35066\r\n6- http://www.mathlinks.ro/Forum/viewtopic.php?search_id=129882059&t=34564", "Solution_1": "Can you post the problem 4 ?", "Solution_2": "I would like to, but I can't find it, I will ask someone today if they have it", "Solution_3": "Here is problem $ 4$\r\n(I can't edit the original post)\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=1633744#1633744" } { "Tag": [ "geometry", "3D geometry", "pyramid", "ratio" ], "Problem": "A plane parallel to the base of a pyramid seperates the pyramid into two pieces with equal volumes. If the height of the pyramid is 12, find the height of the top piece.(please show your work!!!)\r\n\r\n\r\nI posted an image of what the diagram looks like, but i think you might be able to do it without the diagram.", "Solution_1": "[hide]\nThe total height is $12$. The height of the smaller pyramid is $12-x$\n\nCall the base of the big one $y$\n\nThe ratio of the heights is $\\frac{12-x}{12}$, so the ratio of bases is $\\frac{(12-x)^2}{144}$.\n\nso $(12-x)^2$ is the base of the smaller pyramid and $144$ is the base of the bigger one.\n\nApplying the formula for volume\nSo $(12-x)^2\\cdot \\frac 13 (12-x)=(12-x)^3\\cdot\\frac 13$ is half of $144\\cdot12\\cdot\\frac 13=576$\n\nSet up the equation:\n$\\frac 13 (12-x)^3=\\frac 12 576$\nwe are looking for $12-x$\n$(12-x)^3=864$\n$12-x=6\\sqrt[3]{4}$\n[/hide]", "Solution_2": "[quote=\"biffanddoc\"] \nSo $(12-x)^2\\cdot \\frac 13 12-x=(12-x)^3\\cdot\\frac 13$ is half of $144\\cdot12\\cdot\\frac 13=576$\n [/quote]\r\ni don't get what you did here...", "Solution_3": "The method I use for these type of problems is really weird... but it's effective and fast :D (in my opinion)\r\n\r\nCall the height x. Since all dimensions of the small pyramid is similar to the big pyramid, you can compare x with 12 directly (without any other variables). Since this is 3-dimensional, call the area of the small pyramid $x^3$ and the area of the big pyramid $12^3$. Since you want the area of the small pyramid to be half of the big pyramid, set $x^3$ equal to half of 1728 (864) and simplify, which gives x to be $6\\sqrt[3]{4}$", "Solution_4": "[hide]Nice work Biffanddoc, maybe this will seem a little easier though.\n\nIf the volumes are equal, then the volume of the smaller pyramid is $\\frac 12$ of the whole thing. If the ratio of the Volumes is $\\frac 12$ then the ratio of the lengths, heights, or any linear comparison will be $\\sqrt[3]{\\frac 12}$. That simplifies to $\\frac{\\sqrt[3]{4}}{2}$. If that is the ratio, multiply it by 12, the height of the whole thing, and you get $6\\sqrt[3]{4}$ the height of the smaller pyramid.\n\nHope that helps, \n\nFrost\n\n\n[/hide]", "Solution_5": "[quote=\"frost13\"][hide]Nice work Biffanddoc, maybe this will seem a little easier though.\n\nIf the volumes are equal, then the volume of the smaller pyramid is $\\frac 12$ of the whole thing. If the ratio of the Volumes is $\\frac 12$ then the ratio of the lengths, heights, or any linear comparison will be $\\sqrt[3]{\\frac 12}$. That simplifies to $\\frac{\\sqrt[3]{4}}{2}$. If that is the ratio, multiply it by 12, the height of the whole thing, and you get $6\\sqrt[3]{4}$ the height of the smaller pyramid.\n\nHope that helps, \n\nFrost\n\n\n[/hide][/quote]\r\nNice solution :D ." } { "Tag": [ "trigonometry", "calculus", "integration", "algebra", "function", "domain", "modular arithmetic" ], "Problem": "find all triple positive integers such that:\r\n$ x\r\ny\\plus{}xz\\plus{}yz\\minus{}xyz\\equal{}2$", "Solution_1": "[quote=\"asintota\"]find all triple positive integers such that:\n$ x y \\plus{} xz \\plus{} yz \\minus{} xyz \\equal{} 2$[/quote]\r\n\r\n\r\n[hide=\"approach\"]\ncall xy+xz+yz=a\nthis is like finding the roots of $ x^3\\plus{}ax\\plus{}2\\minus{}a\\equal{}0$\n\n1, 1, 1 obviously works :D \n[/hide]", "Solution_2": "any other approach??", "Solution_3": "[hide=\"Big Hint on Simplification\"]\nThis problem begs for some sort of factoring solution. Since we have xyz terms and -xy aso. terms, our best bet is to use:\n[b]\n(x-1)(y-1)(x-1)[/b]\n= xyz-xy-yz-zx+x+y+z-1\n= -(xy+yz+zx-xyz) + x+y+z-1\n= x+y+z-3\n[b]= (x-1)+(y-1)+(z-1)[/b]\n\nDefine r=x-1, s=y-1, t=z-1, each a non-negative integer. Then:\n[b]\nrst=r+s+t[/b]\n[/hide]", "Solution_4": "[quote=\"asintota\"]find all triple positive integers such that:\n$ x y \\plus{} xz \\plus{} yz \\minus{} xyz \\equal{} 2$[/quote]\r\nDiophantient's gonna try out this Diophantine.\r\n[hide]\n$ xyz\\minus{}xy\\minus{}yz\\minus{}xz\\equal{}\\minus{}2$\n$ (x\\minus{}1)(y\\minus{}1)(z\\minus{}1)\\equal{}x\\plus{}y\\plus{}z\\minus{}3$\nIncrement them all.\n$ abc\\equal{}a\\plus{}b\\plus{}c$\nExcept now they're nonnegatives.\nIf anyone is 0, the only soln is (0,0,0) so now they're positive again.\nLet $ a\\equal{}\\tan A$, $ b\\equal{}\\tan B$, and $ c\\equal{}\\tan C$.\nThis gives us that $ A\\plus{}B\\plus{}C\\equal{}180$.\nSo, the only positive integral values of $ \\tan \\theta$ in the domain given for rational $ \\theta$, which contains all numbers in its range, are when $ \\theta\\equal{}0,45,180$.\nThe only one which works out is 0/0/180, which corresponds with 1,1,1 in the original.[/hide]", "Solution_5": "Um like I'm pretty sure permutations of $ (2,3,4)$ is also a solution.", "Solution_6": "[quote=\"diophantient\"][quote=\"asintota\"]find all triple positive integers such that:\n$ x y \\plus{} xz \\plus{} yz \\minus{} xyz \\equal{} 2$[/quote]\nDiophantient's gonna try out this Diophantine.\n[hide]\n$ xyz \\minus{} xy \\minus{} yz \\minus{} xz \\equal{} \\minus{} 2$\n$ (x \\minus{} 1)(y \\minus{} 1)(z \\minus{} 1) \\equal{} x \\plus{} y \\plus{} z \\minus{} 3$\nIncrement them all.\n$ abc \\equal{} a \\plus{} b \\plus{} c$\nExcept now they're nonnegatives.\nIf anyone is 0, the only soln is (0,0,0) so now they're positive again.\nLet $ a \\equal{} \\tan A$, $ b \\equal{} \\tan B$, and $ c \\equal{} \\tan C$.\n[b]This gives us that $ A \\plus{} B \\plus{} C \\equal{} 180$.[/b]\nSo, the only positive integral values of $ \\tan \\theta$ in the domain given for rational $ \\theta$, which contains all numbers in its range, are when $ \\theta \\equal{} 0,45,180$.\nThe only one which works out is 0/0/180, which corresponds with 1,1,1 in the original.[/hide][/quote]\r\n\r\nexplain the bolded part?", "Solution_7": "Mmmhmm... looks like the trig argument isn't complete :P \r\n\r\nContinued from my previous post:\r\n[hide=\"Done Algebraically\"]\nSo we have $ rst\\equal{}r\\plus{}s\\plus{}t$ for $ r,s,t>0$.\n\nIf any one equals zero, the rest must also, so $ (r,s,t)\\equal{}(0,0,0)$ is a solution.\n\nOtherwise, they are all positive. wlog let $ r \\leq s \\leq t$. Now write:\n\n$ r(st\\minus{}1) \\equal{} s\\plus{}t \\implies st\\minus{}1 \\equal{} \\frac{s\\plus{}t}{r} \\leq s\\plus{}t$ \n$ \\iff st\\minus{}s\\minus{}t\\minus{}1 \\leq 0 \\iff st\\minus{}s\\minus{}t\\plus{}1 \\leq 2 \\iff (s\\minus{}1)(t\\minus{}1) \\leq 2$\n\nIf neither $ s$ nor $ t$ is $ 1$, then we have that $ t\\minus{}1 \\leq \\frac{2}{s\\minus{}1} \\leq 2 \\implies t \\leq 3$.\n\nIf $ s\\equal{}1$, then $ r\\equal{}1$, and we have $ t\\equal{}2\\plus{}t$, a contradiction. $ t\\equal{}1 \\implies r,s\\equal{}1$, which leads to the same contradiction.\n\nThis gives three possible considerations: $ (s,t)\\equal{}(2,2) \\qquad (s,t)\\equal{}(2,3) \\qquad (s,t)\\equal{}3,3$\n\nThe only one that gives an integer solution in $ r$ is $ (s,t)\\equal{}(2,3) \\implies r\\equal{}1$, so $ (r,s,t)\\equal{}(1,2,3)$ is the only other solution.\n\nRemoving ordering and translating to $ x,y,z$, we have that the solutions are $ (x,y,z)\\equal{}(1,1,1)$ and all permutations of $ (x,y,z)\\equal{}(2,3,4)$\n[/hide]", "Solution_8": "[b]diophantient[/b], there's no reason to assume that the degree measure of the angles is rational. The trig substitution can be made to work, but it's less effort just to bound.", "Solution_9": "[quote=\"t0rajir0u\"][b]diophantient[/b], there's no reason to assume that the degree measure of the angles is rational. The trig substitution can be made to work, but it's less effort just to bound.[/quote]\r\nWhoops, I totally missed that :(\r\n\r\nHow would you complete the argument?", "Solution_10": "Another way to finish from $ abc\\equal{}a\\plus{}b\\plus{}c$:\r\n\r\n[hide]Assume WLOG that $ a\\le b\\le c$. Taking the equation $ \\pmod{c}$, we get $ a\\plus{}b\\equiv 0\\pmod{c}$. Therefore , $ a\\plus{}b\\equal{}0,c,2c$. Now we're just a little easy casework away from the answer. :) [/hide]" } { "Tag": [], "Problem": "Okay, the point of the problem is to solve who likes who\r\n\r\nLinda and Louise work at the candy store\r\nMike's sister, Jane, is dating the check-out clerk at McCaffrey's\r\nBruce likes tall girls\r\nBob is going to art school out-of-town, so he only gets to see his girlfriend on weekends\r\nLouise broke up with Richard last year because he didn't want to quit his job to go to college\r\nLinda just got accepted to law school\r\nMike never seems to run out of Candy bars...\r\nBecause of her physique, Jeri may be the first woman basketball player at her school\r\nOn weekdays, sometimes Sue goes out with guys other than her boyfriend!! (Shhhhhhh, he doesn't know about it !!)\r\nOnly one boy has a steady job (he works for a living)\r\nChuck has a sweet tooth, but his girlfriend works in a natural foods store and only brings him tofu!\r\nHenry never like girls who were smart\r\nMary's main hobby is yoga\r\n\r\nI know that I'm supposed to make one of those charts...\r\nHere are the things that I know\r\n\r\nMike likes either Linda or Louise\r\nMike can't like Jane\r\nBob can't like Sue\r\nChuck can't like Linda or Louise\r\nHenry can't like Linda\r\n\r\nThat's as far as I know... So who likes who? (Please show you logic, I'm confused with this problem :blush: And I usually solve them real fast...)", "Solution_1": "The problem is certainly ill-posed for many reasons, but, stretching the logic a bit, we get \r\n\r\nMike-Linda\r\nBob-Sue\r\nBruce-Jeri\r\nHenry-Louise\r\nChuck-Mary\r\nRichard-Jane\r\n\r\nThis requires interpreting \"Bruce likes tall girls\" as \"Bruce likes Jeri\" (which is a probable, but formally incorrect conclusion because we are told absolutely nothing about the height of any other girl) and \"Sue goes with other guys on weekdays\" as \"Bob likes Sue\" (which is also a probable but formally incorrect conclusion). Adding the correct \"Richard likes Jane\" to this list (Jane likes a working boy and Richard has a job and only one boy has a job), we reduce the problem to 3 boys and 3 girls about which we have enough information to fill the table in a unique way.", "Solution_2": "Shouldn't this be in \"Puzzles and Brainteasers\"?", "Solution_3": "Ah... Heh, I didn't know where to put this so... Well anyways, I solved now... :D" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $S$ be a subset of $N$ with the following properties:\r\n\r\na)Among each $2003$ consecutive natural numbers there exists at least one contained in $S$;\r\n\r\nb) If $n\\in S$ and $n>1$ ,then $[\\frac n2]\\in S$. ( where $[x]$ denote the greatest integer less than or equal to $x$).\r\n\r\nprove that $S=N$.", "Solution_1": "supose that $ n\\in N $ but $n \\not\\in S$. \r\nwe will work in binary base and $x_2$ will mean the representation of $x$ in binary base \r\nso $[ \\frac{x}{2}]_2$ is obtained by deleting the last digit of $x_2$\r\nlet $p_2$ be a number formed by adding 11 of 0 at the end of $n_2$and \r\n$q_2$ formed by adding 11 of 1 to the end of $n_2$ between $p_2$ and $q_2$ we have 2048 numbers ($2^{11}$) so at least one of them will be in $S$(because 2048>2003 :D). let it be $a_1$ let $a_{n+1}=[\\frac{a_n}2]$ so $a_{12}$ will be n cause you practicaly delete the last digit from the representation in binary base of $a_n$ every time you apply that.\r\nso $n \\in S$." } { "Tag": [ "inequalities", "calculus", "integration", "algebra", "polynomial", "complex numbers", "inequalities proposed" ], "Problem": "Let $a_1,a_2,\\ldots,a_n$ be complex numbers of unit modulus. Prove that there is a complex number $z_0$ with $|z_0|=1$ such that\r\n\\[\r\n|z_0-a_1|\\cdots|z_0-a_n|\\geq 1.\r\n\\]", "Solution_1": "Hyper-classical problem. Here is a nice short solution: use the modulus inequality in the integral $\\int_{0}^{2\\pi}{p(e^{it})e^{-int}dt}=2\\pi$. Here p is the polynomial having roots $a_i$. We don\"t need the assumption that $a_i$ have modulus 1." } { "Tag": [ "trigonometry", "geometry", "cyclic quadrilateral", "area of a triangle", "Heron\\u0027s formula", "trig identities", "Law of Cosines" ], "Problem": "Inscribed in a circle is a quadrilateral having sides of lengths 25, 36, 52, and 60 taken consecutively. Find the diameter of the circle.", "Solution_1": "[hide]I believe that the answer is 28.5, but I did it in such a weird way I don't remember all the steps. All I know is that I used trig and ratios.[/hide]", "Solution_2": "Since the quadrilateral is inscribed in a circle, it is cyclic.\r\n\r\nHence, we're allowed to use Brahmagupta's Formula which states that for cyclic quadrilateral, its area $K$ is defined as:\r\n\r\n\\[\\sqrt{(s-a)(s-b)(s-c)(s-d)}\\]\r\n\r\nFor more information, go \r\n[url=http://mathworld.wolfram.com/BrahmaguptasFormula.html]Here[/url]", "Solution_3": "Actually, the question asks for the diameter of the circle, not the area. ;)", "Solution_4": "My bad.. Should've read the question more carefully.", "Solution_5": "You can find the length of a diagonal by using law of cosines twice.\r\nThen use the formula $R=\\frac{abc}{4K}$.\r\n$K$ is the area of triangle, which can be calculated by Heron's formula etc." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that$a,b,c>0 and a+b+c=1$\r\n\r\n$a/(b^{2}+c)+b/(c^{2}+a)+c/(a^{2}+b)\\geq 3/(4(ab+bc+ca))$", "Solution_1": "A long time ago, I also conjecture a nice inequality\r\n\r\n[i]Prove or disprove that that if $a,b,c$ are non-negative real numbers and[/i] $a+b+c=3$ then \\[\\frac{a}{b^{2}+c}+\\frac{b}{c^{2}+a}+\\frac{c}{a^{2}+b}\\ge \\frac{3}{2}.\\] I hope to see a nice solution to it.", "Solution_2": "HELP,I need help.", "Solution_3": "[quote=\"qiuxuezhe\"]Prove that$a,b,c>0 and a+b+c=1$\n\n$a/(b^{2}+c)+b/(c^{2}+a)+c/(a^{2}+b)\\geq 3/(4(ab+bc+ca))$[/quote]\nIt's wrong! Try $a=\\frac{1}{3}$, $b=\\frac{2}{3}$ and $c\\rightarrow0^+$.\n[quote=\"Flove\"]A long time ago, I also conjecture a nice inequality\n\n[i]Prove or disprove that that if $a,b,c$ are non-negative real numbers and[/i] $a+b+c=3$ then \\[\\frac{a}{b^{2}+c}+\\frac{b}{c^{2}+a}+\\frac{c}{a^{2}+b}\\ge \\frac{3}{2}.\\] I hope to see a nice solution to it.[/quote]\nToday it's very easy! :lol:", "Solution_4": "[quote=\"Flove\"]A long time ago, I also conjecture a nice inequality\n\n[i]Prove or disprove that that if $a,b,c$ are non-negative real numbers and[/i] $a+b+c=3$ then \\[\\frac{a}{b^{2}+c}+\\frac{b}{c^{2}+a}+\\frac{c}{a^{2}+b}\\ge \\frac{3}{2}.\\] I hope to see a nice solution to it.[/quote]\n\nby Cauchy-Bunjakowski-Schwarz we get\n$\\frac{a^2}{ab^2+ac}+\\frac{b^2}{bc^2+ba}+\\frac{c^2}{ca^2+cb}\\geq \\frac{(a+b+c)^2}{ab^2+bc^2+ca^2+ab+bc+ca}=\\frac{9}{ab^2+bc^2+ca^2+ab+bc+ca}\\geq \\frac{3}{2}$\nthis is equivalent to\n$3(ab^2+bc^2+ca^2)+3(ab+bc+ca)\\le 18$\n$3(ab^2+bc^2+ca^2)=(a+b+c)(ab^2+bc^2+ca^2)=ab^3+bc^3+ca^3+(ab+bc+ca)^2-abc(a+b+c) $\n$ab+bc+ca=p $ and $abc=r $\nBy Vasc's ineq $ 3(ab^3+bc^3+ca^3) \\le (a^2+b^2+c^2)^2 $\nwe need to prove $ 27p+9r-7p^2-27\\ge 0 $\nif $ 18/7\\le p\\le 3 $\nBy Schur's ineq $ 9r \\ge 12p-27 $\nneed to prove $ 27p-7p^2-27+12p-27=39p-7p^2-54\\ge 0 $\nthat's $ (7p-18)(p-3)\\le 0 $ ,it's sure\nif $ p\\le 18/7$, How to solve $ 27p+9r-7p^2-27\\ge 0 $???", "Solution_5": "No, yanzongz. Your way is wrong after first step. :wink:", "Solution_6": "[quote=\"arqady\"]No, yanzongz. Your way is wrong after first step. :wink:[/quote]\n\nWe need to prove \\[ \\frac{a}{b^{2}+c}+\\frac{b}{c^{2}+a}+\\frac{c}{a^{2}+b}\\ge\\frac{3}{2}. \\]\n\nby Cauchy-Bunjakowski-Schwarz\n$ \\frac{a^2}{ab^2+ac}+\\frac{b^2}{bc^2+ba}+\\frac{c^2}{ca^2+cb}\\geq\\frac{(a+b+c)^2}{ab^2+bc^2+ca^2+ab+bc+ca}=\\frac{9}{ab^2+bc^2+ca^2+ab+bc+ca} $\n\nSo we can prove $ \\frac{9}{ab^2+bc^2+ca^2+ab+bc+ca}\\geq\\frac{3}{2} $\nWhat i write is easy , it's right :(", "Solution_7": "[quote=\"yanzongz\"]\nSo we can prove $ \\frac{9}{ab^2+bc^2+ca^2+ab+bc+ca}\\geq\\frac{3}{2} $\nWhat i write is easy , it's right :([/quote]\nThis inequality is wrong. Try $c\\rightarrow0^+$. :wink:" } { "Tag": [ "trigonometry", "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c$ be positive reals s.t. $a^2+b^2+c^2+2abc=1$. Prove that \\[a+b+c\\leq\\frac{3}{2}\\]", "Solution_1": "$a=cosA,b=cosB,c=cosC$ with $A+B+C=\\pi$", "Solution_2": "somebody know where can I find topic with proof of $\\cos ^2A+\\cos ^2B+\\cos ^2C+2\\cos A \\cos B \\cos C =1$ ? I'm sure I've seen it before ... :?", "Solution_3": "But it is very simple... I am sure you can prove it yourself.", "Solution_4": "Actually this inequality comes from this one:\r\nLet $x,y,z$ be positive reals greater than 1 s.t. $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=2$. Prove that \\[\\sqrt{x+y+z}\\geq\\sqrt{x-1}+\\sqrt{y-1}+\\sqrt{z-1}\\]This one is easy from Cauchy.", "Solution_5": "[quote=\"mecrazywong\"]Actually this inequality comes from this one:\nLet $x,y,z$ be positive reals greater than 1 s.t. $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=2$. Prove that \\[\\sqrt{x+y+z}\\geq\\sqrt{x-1}+\\sqrt{y-1}+\\sqrt{z-1}\\][/quote]\r\n\r\nActually, the equivalence of this to the initial inequality $a+b+c\\leq\\frac32$ for positive reals a, b, c satisfying $a^2+b^2+c^2+2abc=1$ is not too obvious: One has to square the inequality $\\sqrt{x+y+z}\\geq\\sqrt{x-1}+\\sqrt{y-1}+\\sqrt{z-1}$, to cancel terms and then to apply the substitution\r\n\r\n$a=\\sqrt{\\left(y-1\\right)\\left(z-1\\right)}$; $b=\\sqrt{\\left(z-1\\right)\\left(x-1\\right)}$; $c=\\sqrt{\\left(x-1\\right)\\left(y-1\\right)}$;\r\n\r\nfor the other direction, one has to substitute\r\n\r\n$x=\\frac{bc+a}{a}$; $y=\\frac{ca+b}{b}$; $z=\\frac{ab+c}{c}$.\r\n\r\nShowing that the conditions $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=2$ and $a^2+b^2+c^2+2abc=1$ are equivalent is an easy calculation.\r\n\r\nNote that the inequality $\\sqrt{x+y+z}\\geq\\sqrt{x-1}+\\sqrt{y-1}+\\sqrt{z-1}$ was also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=25650 and the inequality $a^2+b^2+c^2+2abc=1$ at http://www.mathlinks.ro/Forum/viewtopic.php?t=23888 and (in a slightly modified form) at http://www.mathlinks.ro/Forum/viewtopic.php?t=25977 .\r\n\r\nNow, the actual reason why I am writing this message is to show that the initial inequality $a+b+c\\leq\\frac32$ for positive reals a, b, c satisfying $a^2+b^2+c^2+2abc=1$ can also be proved directly (i. e. without any substitution):\r\n\r\nAssume, for the sake of contradiction, that $a+b+c>\\frac32$. Then, if we succeed to prove that $a^2+b^2+c^2+2abc>1$, then we will get a contradiction with the condition $a^2+b^2+c^2+2abc=1$, so that our assumption must be wrong, and the inequality is proven. So all we have to do is to show that $a^2+b^2+c^2+2abc>1$.\r\n\r\nAt first,\r\n\r\n$2\\left(a+b+c\\right)\\left(a^2+b^2+c^2\\right)+9abc-\\left(a+b+c\\right)^3$\r\n$=a\\left(a-b\\right)\\left(a-c\\right)+b\\left(b-c\\right)\\left(b-a\\right)+c\\left(c-a\\right)\\left(c-b\\right)\\geq 0$\r\n\r\nby Schur's inequality. Thus,\r\n\r\n$2\\left(a+b+c\\right)\\left(a^2+b^2+c^2\\right)+9abc\\geq\\left(a+b+c\\right)^3$.\r\n\r\nThus,\r\n\r\n$2\\left(a+b+c\\right)\\left(a^2+b^2+c^2\\right)+6abc\\left(a+b+c\\right)$\r\n$>2\\left(a+b+c\\right)\\left(a^2+b^2+c^2\\right)+6abc\\cdot\\frac32$ (since $a+b+c>\\frac32$)\r\n$=2\\left(a+b+c\\right)\\left(a^2+b^2+c^2\\right)+9abc\\geq\\left(a+b+c\\right)^3$.\r\n\r\nDividing this inequality by a + b + c, we get\r\n\r\n$2\\left(a^2+b^2+c^2\\right)+6abc>\\left(a+b+c\\right)^2$.\r\n\r\nApplying $a+b+c>\\frac32$ again, we see that\r\n\r\n$2\\left(a^2+b^2+c^2\\right)+6abc>\\left(\\frac32\\right)^2=\\frac94$.\r\n\r\nOn the other hand, by Cauchy-Schwarz, $\\left(1^2+1^2+1^2\\right)\\cdot\\left(a^2+b^2+c^2\\right)\\geq\\left(1\\cdot a+1\\cdot b+1\\cdot c\\right)^2$, so that $3\\cdot\\left(a^2+b^2+c^2\\right)\\geq\\left(a+b+c\\right)^2$, and thus $a^2+b^2+c^2\\geq\\frac{\\left(a+b+c\\right)^2}{3}>\\frac{\\left(\\frac32\\right)^2}{3}=\\frac34$. Adding this to the inequality $2\\left(a^2+b^2+c^2\\right)+6abc>\\frac94$ obtained above, we get\r\n\r\n$3\\left(a^2+b^2+c^2\\right)+6abc>3$.\r\n\r\nDivision by 3 yields $a^2+b^2+c^2+2abc>1$. This yields the required contradiction, and thus our inequality is proven.\r\n\r\n darij" } { "Tag": [ "MATHCOUNTS", "geometry" ], "Problem": "This is my first time coaching a Mathcounts team and naturually, I hope my team makes it to the State level :) \r\n\r\nHow hard it is to make it to the State level must vary depending on what Chapter you are in. I live in Nassau County, NY where the top 3 teams go to the State competition.\r\nI'm giving last year's competition to my students for practice. How many correct on the 2006 Chapter competition would be \"average?\" What would be the minimum score of a student that made it to the State competition?\r\n\r\nI doubt I can get a definitive answer, but I'm just looking for a ballpark figure. For instance, the Sprint round has 30 problems. Does 20 correct put a student in line to possibly make it to States or is 20 average?", "Solution_1": "It's definitely above average, but probably too low to place in the top 10 or so individuals (that depends on your chapter though).", "Solution_2": "On last year's chapter competition, here is some data from the chapter containing Albany, NY. They had 143 students compete. Their average sprint score was 15.7; their average target score was 9.5.\r\n\r\nRegarding making it to states, the team score matters more than the individual score, at least in New York. (In each NY chapter, the top three teams automatically qualify for states; in addition, the top two other students qualify.) In the chapter containing Albany, the third-best team scored a 52.75 last year. In Manhattan, the third-best team (the team I coached) scored a 56.50. Because those two counties are probably the most competitive in New York, I'm guessing that a lower score would qualify in Nassau County.", "Solution_3": "It depends. In my county a 20/7 or better is enough to be almost guaranteed for countdown, and this year I predict it'll only take a 30 to make top 10. Last year the lowest score was $33 \\pm 2$ and the highest was a 45. I know that third written had a 37. Also, the top two teams get to go two state, and usually a score of like 47 or higher is good enough to make it. In my county doing well isn't hard even though last year two people on the nationals team were from it (different schools) and 4 people over all were in countdown at states (1, 3, 5, 6). \r\nHowever in the sugarland county in Texas the competition is fierce. Teams have gotten scores of up to 65 before. Usually there are 3 to four teams in the sixties. Last year a 44 was the minimum to make countdown.", "Solution_4": "I think you made a typo on the 5th line, 4th word.\r\n\r\nanyways, 34 people from my chapters make it to nationals.\r\n-jorian", "Solution_5": "[quote=\"jhredsox\"]anyways, 34 people from my chapters make it to nationals.\n-jorian[/quote]\r\n\r\nI think you mean states. :wink:", "Solution_6": "oh yah, this is like the 3rd time i've done this :P\r\n-jorian", "Solution_7": "I agree that the chapter you live in matters (imagine living in a district where there are only 3 people competing :D) but also the particular test also matters. Some years are definitely harder than others.", "Solution_8": "In my chapter, to go to state in individuals, you needed 38. HOwever 6 teams made States, so all you need to do is make top 6 teams.", "Solution_9": "[quote=\"jhredsox\"]I think you made a typo on the 5th line, 4th word.\n\nanyways, 34 people from my chapters make it to nationals.\n-jorian[/quote]\r\nNo, that isn't a typo, read the sentence carefully.", "Solution_10": "[quote=\"indianamath\"]HOwever 6 teams made States, so all you need to do is make top 6 teams.[/quote]\r\n\r\nDo you mean in your chapter or in general? Because like it varies by chapter, in Massachusetts one chapter was like top 3 teams/individuals, mine was top 7. In Delaware, chapter doesn't even exist.\r\n\r\nLast year I believe the State Individual cutoff was 39 in my chapter (7th place), and the team cutoff was around 51.\r\n\r\nBut it really depends on your state and chapter, some are much stronger than others.", "Solution_11": "In my chapter (somewhere in the bay area) the cutoff for individuals making it to states was 44 (The top 4 teams and next highest 5 individuals made it). I don't know what the team cutoff was. The top team in our chapter won both chapter and state, they got 64.5 at chapters and a 58.25 at states.\r\nI think our chapter is harder than a normal chapter so I have no idea what the cutoff for a normal chapter would be.", "Solution_12": "in our chapter, north florida (big bend), 10th was a 38. (24/7). and 4th was a 42. (28/7).", "Solution_13": "[quote=\"Ubemaya\"]The top team in our chapter won both chapter and state, they got 64.5 at chapters and a 58.25 at states.\n[/quote]\r\nWow 64.5? That's pretty good. That's like 46 45 44 43 + 20 on the team round. That was your team, and don't make sure random comment saying you got a 17 or something at chapter :)", "Solution_14": "Our chapter is very small (60 or so competitors) and there was a kid who had 32 total and still got 10th, but in NY I doubt it would be that easy to get in the top 10. At our chapter the lowest person that actually made it to state as an individual had 34 or so, so 20 on the sprint round probably wouldn't cut it :wink:", "Solution_15": "[quote=\"13375P34K43V312\"][quote=\"Ubemaya\"]The top team in our chapter won both chapter and state, they got 64.5 at chapters and a 58.25 at states.\n[/quote]\nWow 64.5? That's pretty good. That's like 46 45 44 43 + 20 on the team round. That was your team, and don't make sure random comment saying you got a 17 or something at chapter :)[/quote]\r\n\r\nNo the teams score was 46 46 something, they had two out of the three perfects in our chapter.\r\nI got a 26 at chapter, it was above average :) .\r\n\r\nI just realized... I probably have to get like a 43 or 44 to reach my goals." } { "Tag": [ "probability" ], "Problem": "Joe has a fair 6-sided die. He rolls it 4 times. What is the probability of the numbers being 1 ,2, 3, and 4, in that order? What are the odds of the numbers being 1 ,2, 3, and 4, in that order? Explain. :D", "Solution_1": "[quote=\"Arvind_sn\"]Joe has a fair 6-sided die. He rolls it 4 times. What is the probability of the numbers being 1 ,2, 3, and 4, in that order? What are the odds of the numbers being 1 ,2, 3, and 4, in that order? Explain. :D[/quote]\r\n\r\nIsn't the first and second question the same thing? \r\n[hide]$\\displaystyle\\left(1/6\\right)^4=\\frac{1}{1296}$[/hide]", "Solution_2": "[quote=\"ch1n353ch3s54a1l\"][quote=\"Arvind_sn\"]Joe has a fair 6-sided die. He rolls it 4 times. What is the probability of the numbers being 1 ,2, 3, and 4, in that order? What are the odds of the numbers being 1 ,2, 3, and 4, in that order? Explain. :D[/quote]\n\nIsn't the first and second question the same thing? \n[hide]$\\displaystyle\\left(1/6\\right)^4=\\frac{1}{1296}$[/hide][/quote]\r\n\r\nOdds are expresesd in a different way, I think odds would be $\\boxed {1296: 1}$ Not too sure though as I sort of forgot how to express them.", "Solution_3": "Well...that's BASICALLY the same thing. \r\nWouldn't it be $1295: 1$?", "Solution_4": "Probability is the amount of chances a certain outcome wanted over the total number of possiblities.\r\nOdds is the amount of chanes wanted over the amount of chances not wanted. \r\nProbability is 1/1296\r\nOdds are 1/1295", "Solution_5": "[hide]Probability is $\\frac{1}{6}^4=\\frac{1}{1296}$ Odds are $1: 1295$[/hide]", "Solution_6": "[quote][hide]Probability is the amount of chances a certain outcome wanted over the total number of possiblities.\nOdds is the amount of chanes wanted over the amount of chances not wanted.\nProbability is 1/1296\nOdds are 1/1295[/hide][/quote]\r\n\r\nCouldn't have put it better. :D", "Solution_7": "[quote=\"ch1n353ch3s54a1l\"]Well...that's BASICALLY the same thing. \nWouldn't it be $1295: 1$?[/quote]\r\n\r\n>_> he's the only correct guy here.\r\n\r\nbtw the AoPS book has a good explanation on odds :p .", "Solution_8": "Actually, for a more descriptive way to describe odds, you all should visit http://en.wikipedia.org/wiki/Odds...", "Solution_9": "[quote=\"Iron Fist\"][quote=\"ch1n353ch3s54a1l\"]Well...that's BASICALLY the same thing. \nWouldn't it be $1295: 1$?[/quote]\n\n>_> he's the only correct guy here.\n\nbtw the AoPS book has a good explanation on odds :p .[/quote]\r\nIt actually depends, whether it is the odds in favor of that happening or against that happening.", "Solution_10": "[quote=\"ragnarok23\"][quote=\"Iron Fist\"][quote=\"ch1n353ch3s54a1l\"]Well...that's BASICALLY the same thing. \nWouldn't it be $1295: 1$?[/quote]\n\n>_> he's the only correct guy here.\n\nbtw the AoPS book has a good explanation on odds :p .[/quote]\nIt actually depends, whether it is the odds in favor of that happening or against that happening.[/quote]\r\n\r\nOdds against is what's usually assumed." } { "Tag": [ "limit", "integration", "calculus", "calculus computations" ], "Problem": "For $ n$ an integer, evaluate\r\n\\[ \\lim_{n \\rightarrow \\infty} \\left( \\frac{1}{\\sqrt{n^2 \\minus{} 0^2}} \\plus{} \\frac{1}{\\sqrt{n^2 \\minus{} 1^2}} \\plus{} \\cdots \\plus{} \\frac{1}{\\sqrt{n^2 \\minus{} (n \\minus{} 1)^2}} \\right).\r\n\\]", "Solution_1": "[quote=\"aidan\"]For $ n$ an integer, evaluate\n\\[ \\lim_{n \\rightarrow \\infty} \\left( \\frac {1}{\\sqrt {n^2 \\minus{} 0^2}} \\plus{} \\frac {1}{\\sqrt {n^2 \\minus{} 1^2}} \\plus{} \\cdots \\plus{} \\frac {1}{\\sqrt {n^2 \\minus{} (n \\minus{} 1)^2}} \\right).\n\\]\n[/quote]\r\n$ L\\equal{}\\lim_{n\\rightarrow\\infty}\\frac{1}{n}\\sum_{k\\equal{}0}^{n\\minus{}1}\\frac{1}{\\sqrt{1\\minus{}\\left(\\frac{k}{n}\\right)^{2}}}\\equal{}\\int_{0}^{1}\\frac{dx}{\\sqrt{1\\minus{}x^{2}}}\\equal{}\\frac{\\pi}{2}$ :)", "Solution_2": "Could you further elaborate? I'm lost as to how that's converted to an integral. I thought the factor of $ \\frac{1}{n}$ would become zero as $ n$ approached infinity.", "Solution_3": "It's straightforward from these 2 links :\r\nhttp://en.wikipedia.org/wiki/Riemann_sum\r\nhttp://en.wikipedia.org/wiki/Riemann_integral", "Solution_4": "I see.. thanks :)" } { "Tag": [ "calculus", "integration", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "Suppose \r\n$\\int_0^1 |f(x)| dx < \\infty$ and $\\int_1^\\infty \\frac{|f(x)|}{x^2} dx < \\infty$\r\nShow that\r\n$\\int_0^\\infty \\sin ax dx \\int_a^\\infty f(y)e^{-xy}dy = a \\int_a^\\infty \\frac{f(y)}{a^2+y^2}dy.$", "Solution_1": "There's a notational problem involving the variable $x$. If it is a dummy variable inside an integral, then it should not appear elsewhere. If we interpret the left hand side as\r\n\r\n$\\int_0^{\\infty}\\sin ax\\left(\\int_a^{\\infty}f(y)e^{-xy}dy\\right)dx$\r\n\r\nthen the whole thing is just a matter of justifying the use of Fubini's Theorem.", "Solution_2": "yes, all we need to do is to clarify the validity of the exchange of integral signs...\r\ni just met a problem on that..." } { "Tag": [ "inequalities", "function", "logarithms", "trigonometry", "geometry", "circumcircle", "inradius" ], "Problem": "[size=150]prove that\n\n If sin (A/2) sin (B/2) sin (C/2) = 1/8\nthen triangle ABC is equilateral[/size]", "Solution_1": "Well-known. By Jensen on, for example, the function $\\ln{\\sin{x}}$ we get that \\[\\sin{\\frac{A}{2}}\\sin{\\frac{B}{2}}\\sin{\\frac{C}{2}} \\leq \\frac{1}{8}\\] with equality iff $ABC$ is equilateral.", "Solution_2": "yes :10:", "Solution_3": "http://www.mathlinks.ro/Forum/viewtopic.php?t=45005\r\n\r\nCan you see the relations between my problem and yours? :)", "Solution_4": "[quote=\"Arne\"]\\[\\sin{\\frac{A}{2}}\\sin{\\frac{B}{2}}\\sin{\\frac{C}{2}} \\leq \\frac{1}{8}\\][/quote]\r\n\r\nThere are dozens of ways to prove this inequality. For instance, it follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=30785 , applied to $a=\\frac{A}{2}$, $b=\\frac{B}{2}$, $c=\\frac{C}{2}$. Or note that $r=4R\\sin\\frac{A}{2}\\sin\\frac{B}{2}\\sin\\frac{C}{2}$, where R is the circumradius and r is the inradius of triangle ABC, so $\\sin\\frac{A}{2}\\sin\\frac{B}{2}\\sin\\frac{C}{2}\\leq\\frac18$ is equivalent to $r\\leq 4R\\cdot\\frac18$, i. e. to $r\\leq\\frac{R}{2}$, what is Euler's inequality. Another proof was given in http://www.mathlinks.ro/Forum/viewtopic.php?t=42758 post #4 or http://www.mathlinks.ro/Forum/viewtopic.php?t=21188 .\r\n\r\n darij" } { "Tag": [ "modular arithmetic", "number theory", "prime numbers" ], "Problem": "Which prime numbers are 1 greater than a square? Which are 1 less than a square? Which are 1 more than a number to the powe of n and which are 1 less than a number to the power of n?", "Solution_1": "$ p \\plus{} 1 \\equal{} x^2 \\\\\r\np \\equal{} (x \\minus{} 1)(x \\plus{} 1)$\r\ngiven p is prime it has only two divisors so\r\n$ x \\minus{} 1 \\equal{} 1 \\\\\r\nx \\plus{} 1 \\equal{} p \\\\\r\np \\equal{} 3$", "Solution_2": "[quote=\"AndrewTom\"]Which prime numbers are 1 greater than a square? Which are 1 less than a square? Which are 1 more than a number to the powe of n and which are 1 less than a number to the power of n?[/quote]\r\n\r\n$ \\boxed{\\text{Which prime numbers are 1 greater than a square?}}$\r\n\r\nThere are too many primes like that ,e.g.\r\n\r\n$ 2 \\equal{} 1^2 \\plus{} 1 ,5\\equal{}2^2\\plus{}1,17 \\equal{} 4^2 \\plus{} 1,37\\equal{}6^2\\plus{}1,101 \\equal{} 10^2 \\plus{} 1,197 \\equal{} 14^2 \\plus{} 1,257 \\equal{} 16^2 \\plus{} 1,\\cdots$\r\n\r\n:starwars: :D", "Solution_3": "[hide=\"1 less than $ x^n$\"]\n$ p\\equal{}x^n\\minus{}1$\n$ p\\equal{}(x\\minus{}1)(x^{n\\minus{}1}\\plus{}x^{n\\minus{}2}\\plus{}...\\plus{}x\\plus{}1)$\nPrime only has two divisors, so $ (x\\minus{}1)\\equal{}1$, $ p\\equal{}2^{n\\minus{}1}\\plus{}2^{n\\minus{}2}\\plus{}...\\plus{}2\\plus{}1$\n\n[url]http://en.wikipedia.org/wiki/Mersenne_prime[/url]\n[/hide]\n\n[hide=\"hint on 1 more than $ x^2$\"]If $ p\\equal{}x^2\\plus{}1$, then, for any $ k0$ be this minimum : $m=\\min_{t\\in [a,b]}|f(t)|$ .Because of $f$ is continous, there is $c_{n}\\in [a,b]$ such that $|\\int_{a}^{b}f(t) e^{nt}dt|=|f(c_{n})\\int_{a}^{b}e^{nt}dt| \\geq m\\int_{a}^{b}e^{nt}dt$\r\nWe can write $m\\int_{a}^{b}e^{nt}dt \\leq |\\int_{a}^{b}f(t)e^{nt}dt| =|\\int_{0}^{b}f(t) e^{nt}dt-\\int_{0}^{a}f(t) e^{nt}dt|$$\\leq |\\int_{0}^{b}f(t) e^{nt}dt|+|\\int_{0}^{a}f(t) e^{nt}dt| \\leq B(b)+B(a)$\r\nBut $\\lim_{n\\rightarrow \\infty}\\int_{a}^{b}e^{nt}dt = \\infty$ , so $m=0$ (contradiction !) . Finally $f$ has to be identically zero .. q.e.d..\r\n :cool:", "Solution_2": "Here's slightly different proof.\r\n\r\nSuppose the contrary, then since $f \\in C^{0}([0,1])$, WLOG one may assume that $f(t) > 0$ in some $(a,b) \\subset [0,1]$. Since $n \\mapsto f(t)e^{nt}$ is $\\ge0$ and is increasing to $+\\infty$ for $t\\in(a,b)$, by Monotone Convergence Theorem we get that\r\n\\[\\lim_{n\\to\\infty}|f_{n}(b)-f_{n}(a)| = \\lim_{n\\to\\infty}\\int_{a}^{b}f(t)e^{nt}dt =+\\infty, \\]\r\ncontradicting the fact that $\\sup |f_{n}(b)-f_{n}(a)| <+\\infty$." } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "$ ABC$ is an right triangle such that $ AB$ is chord. the length of altitude $ CD$ is $ h$.$ M$ and $ N$ are midpoints of $ AD$ and $ BD$ respectively.find the distance of vertex $ C$ to the orthocenter of $ CMN$ traingle.", "Solution_1": "Nobody want to help me?", "Solution_2": "h/2 - use middle line(s)." } { "Tag": [ "inequalities", "trigonometry", "geometry solved", "geometry" ], "Problem": "Prove that for every acute triangle $\\Delta ABC$, we have:$\\sum_{cyclic}\\frac{\\cos B\\cos C}{\\sin 2A}\\ge \\frac{\\sqrt{3}}{2}$", "Solution_1": "It is pretty similar to the previous inequality you posted.\r\n The point is showing $(\\sum cosA)^2 \\geq \\sum sinBsinC$", "Solution_2": "[color=blue][b]Problem.[/b] If ABC is an acute-angled triangle, prove that $\\sum\\frac{\\cos B\\cos C}{\\sin\\left(2A\\right)}\\geq\\frac{\\sqrt3}{2}$, where the $\\sum$ sign stands for cyclic sums.[/color]\r\n\r\nAnother [i]solution:[/i] Since $\\sin\\left(2A\\right)=2\\sin A\\cos A$, the inequality in question, $\\sum\\frac{\\cos B\\cos C}{\\sin\\left(2A\\right)}\\geq\\frac{\\sqrt3}{2}$, rewrites as $\\sum\\frac{\\cos B\\cos C}{2\\sin A\\cos A}\\geq\\frac{\\sqrt3}{2}$. Upon multiplication by 2, this becomes $\\sum\\frac{\\cos B\\cos C}{\\sin A\\cos A}\\geq\\sqrt3$. Since both sides of this inequality are positive (because ABC is an acute-angled triangle and thus cos A, cos B, cos C are positive), squaring it is an equivalence transformation; thus, it remains to prove that $\\left(\\sum\\frac{\\cos B\\cos C}{\\sin A\\cos A}\\right)^{2}\\geq 3$. Now, for any three reals x, y, z, the inequality $\\left(x+y+z\\right)^{2}\\geq 3\\left(yz+zx+xy\\right)$ holds, because\r\n\r\n$\\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$.\r\n\r\nUsing the $\\sum$ sign, this inequality rewrites as $\\left(\\sum x\\right)^{2}\\geq 3\\cdot\\sum yz$.\r\n\r\nApplying this to $x=\\frac{\\cos B\\cos C}{\\sin A\\cos A}$, $y=\\frac{\\cos C\\cos A}{\\sin B\\cos B}$ and $z=\\frac{\\cos A\\cos B}{\\sin C\\cos C}$, we obtain $\\left(\\sum\\frac{\\cos B\\cos C}{\\sin A\\cos A}\\right)^{2}\\geq 3\\cdot\\sum\\frac{\\cos C\\cos A}{\\sin B\\cos B}\\cdot\\frac{\\cos A\\cos B}{\\sin C\\cos C}$, so that $\\left(\\sum\\frac{\\cos B\\cos C}{\\sin A\\cos A}\\right)^{2}\\geq 3\\cdot\\sum\\frac{\\cos^{2}A}{\\sin B\\sin C}$. But in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=110333#110333]http://www.mathlinks.ro/Forum/viewtopic.php?t=15558 post #4[/url] I showed that $\\sum\\frac{\\cos^{2}A}{\\sin B\\sin C}\\geq 1$, so that\r\n\r\n$\\left(\\sum\\frac{\\cos B\\cos C}{\\sin A\\cos A}\\right)^{2}\\geq 3\\cdot\\sum\\frac{\\cos^{2}A}{\\sin B\\sin C}\\geq 3\\cdot 1=3$,\r\n\r\nqed.\r\n\r\n Darij", "Solution_3": "[quote=\"James Potter\"]Prove that for every acute triangle $\\Delta ABC$, we have:$\\sum_{cyclic}\\frac{\\cos B\\cos C}{\\sin 2A}\\ge \\frac{\\sqrt{3}}{2}$[/quote]\r\n[b] Lemma.[/b] Let be given a triangle $ABC$, we have: \\[\\frac{MA}{a}+\\frac{MB}{b}+\\frac{MC}{c}\\ge\\sqrt 3\\quad \\forall\\; M\\]\r\nAnd then apply this [b]Lemma.[/b] for triangle $A'B'C'$ and the point $H$\r\nwhere: $A'\\in BC: AA'\\perp BC,\\,B'\\in CA: BB'\\perp CA,\\,C'\\in AB: CC'\\perp AB$ and $H=AA'\\cap BB'\\cap CC'$" } { "Tag": [], "Problem": "az doostane aziz mikham ke hamanande maratone hendese dar in maratone tarkibiat ke faghat ekhtesas be soalate dar hade marhale 2 dare hozoore fa'ali dashte bashan.\r\n \r\nSOALE 1; 2n noghte dar safhe darim.va yek noghgteye P darim.sabet konid noghteye P daroone tedade zoji az mosallashaye tashkil shode az in noghat gharar darad.( lotfan rahe hal kamel neveshte she :wink: )", "Solution_1": "soal: $8$adade mosbat darim ke kuchektar mosaviye 2004 hastand, sabet konid $4$az anha mesle $a,b,c,d$ra mitavan tori entekhab kard ke: $4+d= 0$. Prove that:\r\n$ a_1^{a_1}a_2^{a_2}...a_n^{a_n}\\geq (a_1a_2...a_n)^{\\frac {a_1 \\plus{} a_2 \\plus{} ...a_n}{n}}$", "Solution_1": "Is the inequality inverse?\r\n\r\n0=1\r\n\r\n(a1/m)^a1 *(a2/m)^a2\u2026\u2026. *(an/m)^an >=(a1/m)^m *(a2/m)^m\u2026\u2026. *(an/m)^m\r\n\r\na1^a1 *a2^a2\u2026\u2026. *an^an >= (a1*a2*\u2026.*an) *m^((a1+a2\u2026.+an)-n*m)\r\n\r\na1^a1 *a2^a2\u2026\u2026. *an^an >= (a1*a2*\u2026.*an) *m", "Solution_2": "Yes, the inequality is inverse. I have edited. Thanks for appointing my mistake.", "Solution_3": "[quote=\"Inequalities Master\"]Let $ a_1,a_2,...,a_n > 0$. Prove that:\n$ a_1^{a_1}a_2^{a_2}...a_n^{a_n}\\geq (a_1a_2...a_n)^{\\frac {a_1 \\plus{} a_2 \\plus{} ...a_n}{n}}$[/quote]\r\nis it yours\r\n[hide=\"hint\"]first,rewrite it in the form $ ln (RHS)\\le ln(LHS)$Assume $ a_1\\le ..a_n$ and use Chebysev ineq\nIt is quite old,Inequalities Master[/hide]" } { "Tag": [ "HMMT", "MIT", "college", "email", "geometry", "calculus", "Princeton" ], "Problem": "OK,\r\n\r\nI don't think SC will be sending a team to HMMT (we's a po' state). However, I would like to go, and I know we have a few other strong students in the state who would also like to go. If worst comes to worst, we can do some fundraising (aka begging). So what do teams look like? Who is in QUAGGA? I assume that's Florida? Who are the NC people going with? Also, if anyone on a decent team has an open spot for me, PM me and we can discuss it (I can pay my own way).", "Solution_1": "Yeah, I actually looked into that a few days ago and according to the website the deadline's passed already. I'm not all that involved in the contest so you'll have to e-mail hmmt-request at mit dot edu and ask them.", "Solution_2": "Where did you see that? I looked at the website, and you can pay up until the day of the contest...but I can't find any deadlines. I will email them and find out, I guess.\r\n\r\nBut anyway, the website says you can still change your roster, so if a team has an open spot...", "Solution_3": "Yeah, you can still change your roster. I guess I was replying to the \"I don't think SC will be sending a team\". If there are still open spots then you could probably still sign up, but I don't know if the cap was reached or not.", "Solution_4": "Yeah, I emailed him, and December 7th was the deadline. In fact, some teams that registered on time were waitlisted. Oh well, I'll just have to find a team that will let me have a spot. Even if I'm not on a killer team, I can still place individually in Algebra or perhaps Combinatorics.", "Solution_5": "QUAGGA", "Solution_6": "WOOT MINUS QUAGGA", "Solution_7": "Vestavia is going as a school, like it should be", "Solution_8": "[quote=\"Cicatriz\"]Vestavia is going as a school, like it should be[/quote]\r\n\r\nWell the rules say you don't need to be a school...and think about it: Vestavia vs. a normal high school isn't fair. We can't take classes called \"Math Team.\"\r\n\r\nBut I think I may have found a team, we'll have to work out the details though.", "Solution_9": "[quote=\"mysmartmouth\"][quote=\"Cicatriz\"]Vestavia is going as a school, like it should be[/quote]\n\nWell the rules say you don't need to be a school...and think about it: Vestavia vs. a normal high school isn't fair. We can't take classes called \"Math Team.\"\n\nBut I think I may have found a team, we'll have to work out the details though.[/quote]\r\n\r\nthatd suck. at our middle school, math competitions is an elective. its the best class ever.", "Solution_10": "[quote=\"mysmartmouth\"][quote=\"Cicatriz\"]Vestavia is going as a school, like it should be[/quote]\n\nWell the rules say you don't need to be a school...and think about it: Vestavia vs. a normal high school isn't fair. We can't take classes called \"Math Team.\"\n\nBut I think I may have found a team, we'll have to work out the details though.[/quote] \r\n\r\nand Vestavia vs. both states of Carolina or the combined power of Florida and choice picks from Illinois/DC is any more fair?", "Solution_11": "a) Why not?\r\n\r\nThere are advantages to having a strong enough school to field a decent HMMT team. Since you're all living around the same area and go to the same school it is much easier to get together for practice.\r\n\r\nb) SC/NC isn't forming a team for HMMT. Even if they were that wouldn't be any more \"unfair\" than NC by itself.", "Solution_12": "Honestly,\r\n\r\nSC + NC < NC\r\n\r\nbecause teams were picked subjectively. We didn't take the 10 best overall, we took the 7 best SCers, and the 3 best NCers. However, NCer #10 is probably better than SC #1.", "Solution_13": "Yeah, note how Jeremy Hahn was on the central MD team.", "Solution_14": "that was actually something different\r\n\r\nthey passed up noah blach though wtf", "Solution_15": "I just think it's more interesting to compare either schools or states. Obviously one school can have strengths over other school but that kind of variety makes it interesting. If you want something completely fair, we can just have one team per division and they will compete against their own score 10 times, so it will always be a 10-way tie for first. That'd make everyone happy, right?\r\n\r\nI was on a conglomerate team, so I can't say it's completely unforgivable, but I couldn't really feel that much pride about our success as a team. \r\n\r\nWho's on \"Woot minus Quagga\"?", "Solution_16": "[quote=\"Cicatriz\"]Who's on \"Woot minus Quagga\"?[/quote]\r\n\r\n$\\bullet$ Calculus\r\n$\\bullet$ chess64\r\n$\\bullet$ E^(pi*i)=-1\r\n$\\bullet$ K81o7\r\n$\\bullet$ n^4+4\r\n$\\bullet$ NeverOddOrEven\r\n$\\bullet$ pkerichang\r\n$\\bullet$ SplashD\r\n\r\nIt would be so much better if Dnas agreed to come, and also if we had the Quagga WOOTlings...", "Solution_17": "Hmm. I think our team is pretty good.\r\n\r\nWe're only recruiting 2-3 people for our first team. And well, one of them is a time honored tradition. Yay, Arnav! It is only a competition. Its giving an opportunity to people who otherwise wouldn't have it; to go with a team that has a good chance of placing in the top 5. I think you all underestimate the actual strength of the Florida team, when we have our best people. Notice that at Princeton, several of our 2nd team members placed in the top 5 or 10.\r\n\r\nThere are a couple people who have asked to be on a Florida team, and I'm totally cool with that. I'm delegating them to our 2nd and 3rd team. People just like us, I guess. As far as the NC+SC thing goes, more power to them. The WOOT team idea is just a bit ridiculous though. There isn't even like a state behind that. But we'll see how it goes.\r\n\r\n\r\nEDIT: Giving it more thought: I've decided that for all you stragglers without a team, we have 0 spots on our third team. I suggest you all come up with a team name that I approve of.\r\n\r\nEDIT 2: Wow that went fast.", "Solution_18": "[quote=\"Pakman2012\"]The WOOT team idea is just a bit ridiculous though. There isn't even like a state behind that.[/quote]\r\n\r\nWhat... it's clearly MD+NC+NY+NJ+CA.", "Solution_19": "I don't really see what the big deal is with teams consisting of people from all over. Instead of making things less fair I can argue it does the opposite. What if I were from a school that was unable to field an HMMT team and thus would not go unless I joined another team? On the other hand, it gives people or schools that never win a chance to do so. Yeah, we all know TJ, Exeter, AAST, Vestavia, etc are good schools. You've won/placed over and over again. At the least, we don't want you to grow complacent at the top ;)", "Solution_20": "I'd hardly compare Vestavia's ability at HMMT with TJ or Exeter...", "Solution_21": "Vestavia is mostly notable for its performance on Mu Alpha Theta, isn't it?", "Solution_22": "[quote=\"chess64\"][quote=\"Pakman2012\"]The WOOT team idea is just a bit ridiculous though. There isn't even like a state behind that.[/quote]\n\nWhat... it's clearly MD+NC+NY+NJ+SC.[/quote]\r\n\r\nWhoa, who's the WOOT person from SC? If you don't have one, and want one... :wink:", "Solution_23": "[quote=\"mysmartmouth\"][quote=\"chess64\"][quote=\"Pakman2012\"]The WOOT team idea is just a bit ridiculous though. There isn't even like a state behind that.[/quote]\n\nWhat... it's clearly MD+NC+NY+NJ+SC.[/quote]\n\nWhoa, who's the WOOT person from SC? If you don't have one, and want one... :wink:[/quote]\r\n\r\nWoops, that was a typo. No one from SC... actually wait we do have someone from CA though... so yeah, replace the SC with CA.", "Solution_24": "[quote=\"Phelpedo\"]Vestavia is mostly notable for its performance on Mu Alpha Theta, isn't it?[/quote]\r\n\r\nYea but our ability to perform drops quickly when you focus on tournaments like arml/hmmt.", "Solution_25": "dude we should send a team called quagga IMO team" } { "Tag": [ "trigonometry" ], "Problem": "$\\sin x + \\sin y = m, \\cos x +\\cos y = n$\r\nFind $\\tan{(x+y)}$", "Solution_1": "Based on identity $\\frac {\\sin{x}+\\sin{y}}{\\cos{x}+\\cos{y}}=\\tan{(\\frac{x+y}{2})}$ , we get $\\tan({\\frac{1}{2} (x+y)})=\\frac{m}{n}$\r\n\r\nLet $(x+y)=A$ therefore based on identity $\\tan{(2.(\\frac{1}{2}A))}=\\frac{2 \\tan{\\frac{1}{2}A}}{1-\\tan^2{\\frac{1}{2}A}}$\r\n\r\nWe get $\\tan(x+y)=\\frac{2mn}{n^2-m^2}$" } { "Tag": [ "trigonometry", "function", "algebra", "domain" ], "Problem": "Given $\\{ a_{n}\\}$, $a_{0}= \\frac{1}{3}$, $a_{n}= \\sqrt{\\frac{1+a_{n-1}}{2}}$. Show that $\\{ a_{n}\\}$ is monotonically increasing.", "Solution_1": "Assume inductively that $a_{n}< 1$. Then\r\n\r\n$a_{n+1}\\;=\\; \\sqrt{\\textstyle \\frac{1+a_{n}}{2}}\\;<\\; \\sqrt{\\textstyle \\frac{1+1}{2}}\\;=\\; 1,$\r\n\r\ngiving $a_{n}< 1$ since $a_{1}={\\textstyle \\frac{1}{3}}< 1.$ Therefore $a_{n}= 1-b_{n}$ where $0 < b_{n}< 1,$ which implies\r\n\r\n${a_{n+1}\\;=\\; \\sqrt{\\textstyle \\frac{1+a_{n}}{2}}\\;=\\; \\sqrt{\\textstyle 1-\\frac{b_{n}}{2}}\\;>\\; \\sqrt{\\textstyle 1-\\frac{b_{n}}{2}+b_{n}(b_{n}-\\frac{3}{2})}}\\;=\\; \\sqrt{(1-b_{n})^{2}},$\r\n\r\ni.e.\r\n\r\n$a_{n+1}\\;>\\; 1-b_{n}= a_{n}.\\;\\;$ [b]q.e.d[/b]", "Solution_2": "you can show that 0 \\frac{1}{3}= a_{0}$, so we always have $a_{n+1}> a_{n}$, which means $\\{ a_{n}\\}$ is monotonically increasing." } { "Tag": [ "function", "calculus", "derivative", "algebra unsolved", "algebra" ], "Problem": "[b](1) find Min value of (x^2+y^2)^1/2 +(x^2+(y-1))^1/2 = \n(x^2+y^2)^1/2 square root of (x square +y square) [/b]", "Solution_1": "[quote=\"jagdish\"][b](1) find Min value of (x^2+y^2)^1/2 +(x^2+(y-1))^1/2 = \n(x^2+y^2)^1/2 square root of (x square +y square) [/b][/quote]\r\n\r\nWe are looking for the minimum value of $ \\sqrt{x^2\\plus{}y^2}\\plus{}\\sqrt{x^2\\plus{}y\\minus{}1}$\r\n\r\nWlog say $ x\\geq 0$\r\n\r\n1) if $ x\\in[0,1)$\r\nWe need then $ y\\geq 1\\minus{}x^2\\geq 0$ and then ($ y\\geq 0$) $ \\sqrt{x^2\\plus{}y^2}\\plus{}\\sqrt{x^2\\plus{}y\\minus{}1}$ is an increasing function of $ y$ and minimum is reached for $ y\\equal{}1\\minus{}x^2$\r\n\r\nThis minimum is $ \\sqrt{x^2\\plus{}(1\\minus{}x^2)^2}$ $ \\equal{}\\sqrt{(x^2\\minus{}\\frac 12)^2 \\plus{}\\frac 34}$ and this minimum is reached for $ x^2\\equal{}\\frac 12$\r\nSo, if $ x\\in[0,1)$, minimum is reached when $ x^2\\equal{}y\\equal{}\\frac 12$ and the value is $ \\sqrt{\\frac 12 \\plus{}\\frac 14}\\equal{}\\frac{\\sqrt 3}2$\r\n\r\n2) if $ x\\geq 1$ and $ y\\geq 0$\r\nThen $ \\sqrt{x^2\\plus{}y^2}\\plus{}\\sqrt{x^2\\plus{}y\\minus{}1}$ is an increasing function of $ x$ and minimum is reached for $ x\\equal{}1$ and the value is $ \\sqrt{y^2\\plus{}1}\\plus{}\\sqrt{y}$ whose minimum is reached for $ y\\equal{}0$ and value is $ 1$\r\n\r\n3) if $ x\\geq 1$ and $ y\\leq 0$\r\nThen $ \\sqrt{x^2\\plus{}y^2}\\plus{}\\sqrt{x^2\\plus{}y\\minus{}1}$ is an increasing function of $ x$ and minimum is reached for $ x\\equal{}\\sqrt{1\\minus{}y}$ and the value is $ \\sqrt{y^2\\plus{}1\\minus{}y}$ whose minimum is reached for $ y\\equal{}0$ and is $ 1$\r\n\r\n\r\nSo the minimum value is obtained when $ x^2\\equal{}y\\equal{}\\frac 12$ and the value is $ \\boxed{\\frac{\\sqrt 3}2}$", "Solution_2": "[b](1) Thanks PCO for giving a NICE solution.\nbut actually my question is (1) \nfind Min value of (x^2+y^2)^1/2 +(x^2+(y-1)^1/2)^1/2 = \n(x^2+y^2)^1/2 square root of (x square +y square)[/b]", "Solution_3": "So new problem (please try to give the good questions on the first try) :(\r\n\r\n[quote=\"jagdish\"][b](1) but actually my question is (1) \nfind Min value of (x^2+y^2)^1/2 +(x^2+(y-1)^1/2)^1/2 [/quote]\r\n\r\nWe are looking for the minimum value of $ \\sqrt {x^2 \\plus{} y^2} \\plus{} \\sqrt {x^2 \\plus{} \\sqrt{y \\minus{} 1}}$\r\n\r\nWlog say $ x\\geq 0$\r\n$ y\\geq 1$\r\n\r\nFor $ x\\geq 0$ and $ y\\geq 1$, $ \\sqrt {x^2 \\plus{} y^2} \\plus{} \\sqrt {x^2 \\plus{} \\sqrt{y \\minus{} 1}}$ is an increasing function of both variables $ x$ and $ y$ and so minimum is reached when $ x\\equal{}0$ and $ y\\equal{}1$ and is $ 1$", "Solution_4": "If you are allowed to use Calculus, you could use derivitives to find critical values and apply sign chart. Minimum occurs when left side is negative and right side is positive. But taking derivitive is a booger." } { "Tag": [ "calculus", "integration", "ratio", "floor function", "quadratics", "geometric sequence", "algebra" ], "Problem": "For a positive number such as 3.27, 3 is referred to as the integral part of the number and .27 as the decimal part. Find a positive number such that its decimal part, its integral part, and the number itself form a geometric progression.", "Solution_1": "[hide]It's $ \\phi\\equal{}\\frac{1\\plus{}\\sqrt5}{2}$.\n\nIf the integral part is greater than 1, then the common ratio will be too large. So the integral part is one.\nIf $ x$ is the decimal portion, then $ \\frac{1}{x}\\equal{}\\frac{x\\plus{}1}{1}$. Solve for $ x\\plus{}1$ to get $ \\phi$.[/hide]", "Solution_2": "Nice. I actually got that as well except I somehow did algebra wrong and stared at it until I saw your solution. :blush:", "Solution_3": "Here's the same problem with a different wording.\r\n\r\nLet $ x$ be a real number and $ \\lfloor x \\rfloor$ be the floor function. For which values of $ x$ do $ x\\minus{}\\lfloor x \\rfloor, \\lfloor x \\rfloor, x$ form an increasing geometric sequence?", "Solution_4": "Nice, it seems your wording makes the problem appears harder though. :D", "Solution_5": "[hide]Let $ d$ be the decimal part (which is not zero), and $ r$ be the geometric progression common ratio. The integral part is $ dr$ and the number itself is $ dr^2$ (the next term in the geometric sequence). But also $ dr^2\\equal{}dr\\plus{}d$. Solving the quadratic equation $ r^2\\equal{}r\\plus{}1$ gives us the same positive answer.[/hide]" } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "Hi there i am trying to make this equation look exact. \r\n\r\n$(Cos(2y)-Sin (x)) dx-2 Tan (x) Sin (2y) dy = 0$\r\n\r\nWhat I've done so far is take the partial with respect to x and y.\r\n\r\nSo, my \r\n\r\n$M_{y}$ is equal to $-2 Sin (2y)-0$ and, \r\n\r\nmy $N_{x}$ is equal to $-2(Sec^{2}(x)) Sin (2y)$\r\n\r\nWhich makes it not exact. So, then I tried using \r\n$\\frac{M_{y}-N_{x}}{-N}$ and,\r\n\r\nhere is where I have tried so many times to find out a way to find an Integration Factor (I.F.)\r\n\r\nAny suggestions will help, thanks for your time.", "Solution_1": "hello, i have got the following solution\r\n$ y(x)\\equal{}\\frac{1}{2}\\arccos\\left(\\frac{\\sin(x)^2\\plus{}C}{2\\sin(x)}\\right)$\r\nSonnhard." } { "Tag": [ "logarithms", "floor function", "irrational number", "number theory unsolved", "number theory" ], "Problem": "Hey all, \r\nI have some difficulties solving a problem.\r\n\r\nProve that there exist a power of 2 such that its decimal representation starts with 2009.\r\n\r\nLooks very interesting and I believe that it is a popular problem, but I couldn't google it.", "Solution_1": "More generally\r\n\r\n[color=blue]Prove that given any positive integer $ N$, there are infinitely many powers of $ 2$ whose decimal representation start with $ N$.[/color]\r\n\r\nWe look for integers $ n,k$ such that $ N10^n < 2^k < (N\\plus{}1)10^n$; then the decimal representation of $ 2^k$ will start with $ N$. Take decimal logarithms to get $ n\\plus{} \\log N < k\\log 2 < n \\plus{} \\log(N\\plus{}1)$. But $ \\log 2$ is irrational (easy to prove), and some celebrated theorems (Dirichlet, Kronecker, Weyl) state that the fractional parts of the integer multiples of any irrational number are dense in $ (0,1)$. Therefore there exist infinitely many positive integers $ \\ell$ such that $ \\ell\\log2 \\minus{} \\lfloor \\ell\\log2 \\rfloor \\in (0, \\log (N\\plus{}1) \\minus{} \\log N)$. Denoting $ m \\equal{} \\lfloor \\ell\\log 2 \\rfloor$, this is equivalent to $ 0 < \\ell\\log 2 \\minus{} m < \\log (N\\plus{}1) \\minus{} \\log N$. Take now some integer $ M$ large enough so that $ M \\plus{} \\log N > 1$, and take $ \\displaystyle t \\equal{} \\left \\lfloor \\frac {M \\plus{} \\log N} {\\ell\\log 2 \\minus{} m} \\right \\rfloor$. Then $ M \\plus{} \\log N < (t\\plus{}1)(\\ell\\log 2 \\minus{} m) < M \\plus{} \\log (N\\plus{}1)$. All that is left is take $ n \\equal{} M \\plus{} (t\\plus{}1)m$ and $ k \\equal{} (t\\plus{}1)\\ell$.\r\n\r\nMoreover, this proof works for powers of any positive integer $ d$ other than $ 2$, with the exception of $ 10$ and its powers (which obviously start with $ 100\\ldots$). This ensures that $ \\log d$ is irrational, and the proof goes the same.", "Solution_2": "Thanks a lot. I also found a solution for 123454321 instead of 2009, but I wasn't quite convinced by that solution.", "Solution_3": "I followed mavropnevma's steps and it produced n=129932880. The first digits of 2^n are 200913254645...", "Solution_4": "The first is n = 4469 :)" } { "Tag": [ "AMC" ], "Problem": "During which years were calculators allowed on the AMC/AHSME?", "Solution_1": "1994-2007.", "Solution_2": "why did they stop allowing the use of it?", "Solution_3": "It removed the unfair advantage students with high tech calculators had over those with low tech ones. Additionally, it is not possible to just bash numbers for answers anymore." } { "Tag": [], "Problem": "Find $ x$ if $ \\displaystyle \\frac{2}{x} \\minus{} \\frac{3}{5} \\plus{} \\frac{1}{x} \\equal{} \\frac{1}{5}$.", "Solution_1": "First simplify this equation. \r\n\r\n$ \\frac{2}{x} \\plus{} \\frac{1}{x} \\equal{} \\frac{4}{5}$\r\n\r\nWe want to eliminate the denominators, so multiply both sides by $ 5x$.\r\n\r\n$ 10 \\plus{} 5 \\equal{} 4x$\r\n\r\n$ 15 \\equal{} 4x$\r\n\r\n$ \\boxed{\\frac{15}{4}} \\equal{} x$" } { "Tag": [ "conics", "parabola", "analytic geometry", "algebra unsolved", "algebra" ], "Problem": "Let p and q be positive numbers such that y=x^2 -2px + q has no common point with the x-axis. Prove that there exists points A and B on the parabola such that the segment AB is parallel to to the x axis and 0.Prove that\r\n$ \\frac{x(y\\plus{}z)^2}{2x\\plus{}y\\plus{}z}\\plus{}\\frac{y(x\\plus{}z)^2}{x\\plus{}2y\\plus{}z}\\plus{}\\frac{z(x\\plus{}y)^2}{x\\plus{}y\\plus{}2z}$\u2265$ \\sqrt(3xyz(x\\plus{}y\\plus{}z))$", "Solution_1": "[quote=\"SUPERMAN2\"]Given x,y,z>0.Prove that\n$ \\frac {x(y \\plus{} z)^2}{2x \\plus{} y \\plus{} z} \\plus{} \\frac {y(x \\plus{} z)^2}{x \\plus{} 2y \\plus{} z} \\plus{} \\frac {z(x \\plus{} y)^2}{x \\plus{} y \\plus{} 2z}\\geq\\sqrt{3xyz(x \\plus{} y \\plus{} z)}$[/quote]\r\nBut it's obvious by SOS. :wink:", "Solution_2": "Please can you post a solution ?", "Solution_3": "$ \\sum_{cyc}\\frac {x(y \\plus{} z)^2}{2x \\plus{} y \\plus{} z} \\minus{} \\sqrt {3xyz(x \\plus{} y \\plus{} z)} \\equal{}$\r\n$ \\equal{} \\sum_{cyc}\\left(\\frac {x(y \\plus{} z)^2}{2x \\plus{} y \\plus{} z} \\minus{} yz\\right) \\plus{} xy \\plus{} xz \\plus{} yz \\minus{} \\sqrt {3xyz(x \\plus{} y \\plus{} z)} \\equal{}$\r\n$ \\equal{} \\sum_{cyc}\\frac {z^2(x \\minus{} y) \\minus{} y^2(z \\minus{} x)}{2x \\plus{} y \\plus{} z} \\plus{} \\sum_{cyc}\\frac {z^2(x \\minus{} y)^2}{2\\left(xy \\plus{} xz \\plus{} yz \\plus{} \\sqrt {3xyz(x \\plus{} y \\plus{} z)}\\right)} \\equal{}$\r\n$ \\equal{} \\sum_{cyc}(x \\minus{} y)\\left(\\frac {z^2}{2x \\plus{} y \\plus{} z} \\minus{} \\frac {z^2}{2y \\plus{} x \\plus{} z}\\right) \\plus{}$\r\n$ \\plus{} \\sum_{cyc}\\frac {z^2(x \\minus{} y)^2}{2\\left(xy \\plus{} xz \\plus{} yz \\plus{} \\sqrt {3xyz(x \\plus{} y \\plus{} z)}\\right)} \\equal{}$\r\n$ \\equal{} \\sum_{cyc}(x \\minus{} y)^2\\left(\\frac {z^2}{2\\left(xy \\plus{} xz \\plus{} yz \\plus{} \\sqrt {3xyz(x \\plus{} y \\plus{} z)}\\right)} \\minus{} \\frac {z^2}{(2x \\plus{} y \\plus{} z)(2y \\plus{} x \\plus{} z)}\\right).$\r\nThus, it remains to prove that\r\n$ (2x \\plus{} y \\plus{} z)(2y \\plus{} x \\plus{} z)\\geq2\\left(xy \\plus{} xz \\plus{} yz \\plus{} \\sqrt {3xyz(x \\plus{} y \\plus{} z)}\\right).$\r\nBut $ (2x \\plus{} y \\plus{} z)(2y \\plus{} x \\plus{} z)\\geq2\\left(xy \\plus{} xz \\plus{} yz \\plus{} \\sqrt {3xyz(x \\plus{} y \\plus{} z)}\\right)\\Leftrightarrow$\r\n$ \\Leftrightarrow2x^2 \\plus{} 2y^2 \\plus{} z^2 \\plus{} 3xy \\plus{} xz \\plus{} yz\\geq2\\sqrt {3xyz(x \\plus{} y \\plus{} z)},$\r\nwhich is true because $ x^2 \\plus{} y^2 \\plus{} z^2\\geq xy \\plus{} xz \\plus{} yz\\geq\\sqrt {3xyz(x \\plus{} y \\plus{} z)}.$\r\n\r\nIt seems that the following inequality is true too.\r\nLet $ x,$ $ y$ and $ z$ are positive numbers. Prove that:\r\n\\[ \\frac {x(y \\plus{} z)^2}{3x \\plus{} 2y \\plus{} 2z} \\plus{} \\frac {y(x \\plus{} z)^2}{2x \\plus{} 3y \\plus{} 2z} \\plus{} \\frac {z(x \\plus{} y)^2}{2x \\plus{} 2y \\plus{} 3z}\\geq\\frac{4}{7}\\sqrt{3xyz(x \\plus{} y \\plus{} z)}\\]" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "HI, I am a high school student. I did really well on the AP CALC BC exam. But my school doesnt offer any math class to me b/c i finished calc bc. And I studied multi-calc on my own. SO , im just wondering if u are really good with calc, are there any OFFICIAL calculus contests, exams, or competitions? \r\n\r\nALSO, are there any official and outstanding math research competitions for high school juniors?\r\nINTEL is for seniors only. And i dont wanna form a team...", "Solution_1": "The Siemens Westinghouse competition, and you can do Intel on your own if you want." } { "Tag": [ "percent", "AMC" ], "Problem": "At Jefferson Summer Camp, $ 60\\%$ of the children play soccer, $ 30\\%$ of the children swim, and $ 40\\%$ of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer?\r\n\r\n$ \\textbf{(A)}\\ 30\\% \\qquad\r\n\\textbf{(B)}\\ 40\\% \\qquad\r\n\\textbf{(C)}\\ 49\\% \\qquad\r\n\\textbf{(D)}\\ 51\\% \\qquad\r\n\\textbf{(E)}\\ 70\\%$", "Solution_1": "[hide=\"Click for solution\"]\nLet us assume there are $ 50$ children at the camp. We know that:\n\n$ 0.6 \\times 50\\equal{}30$ children play soccer.\n\n$ 0.3 \\times 50\\equal{}15$ children swim.\n\n$ 0.4 \\times 30\\equal{}12$ children play soccer and swim.\n\nThere are $ 50\\minus{}15\\equal{}35$ children who do not swim and $ 35\\minus{}17\\equal{}18$ of them play soccer. The desired percentage is thus $ 100 \\times \\frac{18}{35}$, which is just over $ 50 \\%$, so the answer is $ \\boxed{\\textbf{(D)}}$.\n[/hide]", "Solution_2": "[quote=\"AIME15\"][hide=\"Click for solution\"]\nLet us assume there are $ 50$ children at the camp. We know that:\n\n$ 0.6 \\times 50 \\equal{} 30$ children play soccer.\n\n$ 0.3 \\times 50 \\equal{} 15$ children swim.\n\n$ 0.4 \\times 30 \\equal{} 12$ children play soccer and swim.\n\nThere are $ 50 \\minus{} 15 \\equal{} 35$ children who do not swim and $ 35 \\minus{} 17 \\equal{} 18$ of them play soccer. The desired percentage is thus $ 100 \\times \\frac {18}{35}$, which is just over $ 50 \\%$, so the answer is $ \\boxed{\\textbf{(D)}}$.\n[/hide][/quote]\r\nWhere did u get the 17 to subtract from 35?", "Solution_3": "Assuming $ 50$ kids, you know that there are $ 30$ kids that play soccer, therefore if $ 12$ play both sports, you know $ 18$ swims only. Similarly, you know that there are only $ 3$ kids who only swim. That sums to $ 12 \\plus{} 18 \\plus{} 3 \\equal{} 33 \\rightarrow50 \\minus{} 33 \\equal{} 17$ kids who don't do either sport, which you need to subtract from the total kids after subtracting swimmers.", "Solution_4": "Or you can just stop at saying that there are $ 30$ kids that play soccer and $ 12$ swim, so only $ 30\\minus{}12\\equal{}18$ play soccer and don't swim.", "Solution_5": "I got confused with the wording of \"non-swimmers\". I thought it was anyone who didn't swim, so I got my fraction to be 36/46 because the 36 kids who play soccer and don't swim plus the 10 kids who didn't do anything is 46. I'm basing these numbers off of a camp with 100 children.", "Solution_6": "You are counting wrong. Soccer players can also be swimmers...", "Solution_7": "I still don't get this, can someone please explain this thoroughly? \nI don't get where you are getting all the numbers from.", "Solution_8": "Assume that there are 100 children in the camp. \n$60\\% \\times 100=60$ of them play soccer. \n$30\\% \\times 100=30$ of the children swim. \nOut of the 60 soccer players, $40\\% \\times 60=24$ of them swim.\nSo $60-24=36$ play soccer but do not swim.\nThere are 30 children who swim, so there are $100-30=70$ total non-swimmers.\nWe are looking for the percent of non-swimmers playing soccer, which is $\\frac{36}{70}$, which is slightly over $\\frac{35}{70}=\\frac{1}{2}=50\\%$. So the answer is (D) 51%.\n\nIt is really helpful if you draw a Venn Diagram.", "Solution_9": "[hide=Solution]Assume that there are $1000$ children in the camp. \n$600$ play soccer, $300$ swim, and $40\\%$ of $600=240$ play soccer and swim. So there are $700$ non-swimmers in the camp. \n$600-240=360$ non-swimmers play soccer, so $\\frac{360}{700}$ of the non-swimmers play soccer. This can be simplified to $\\frac{18}{35}$, which is about $\\boxed{\\textbf{(D)}\\ 51\\% \\qquad}$ [/hide]", "Solution_10": "7.5 year bump :rotfl: ", "Solution_11": "[quote=ethanlanc]7.5 year bump :rotfl:[/quote]\n\nmegarnie is probably going to bump every single AMC problem thread ever so expect more of that :P. Although seriously I know I can't make you not post solutions (you're allowed to) but this is getting a little annoying to see what I mean just scroll down C&P and it's all AMCs (mostly).", "Solution_12": "[quote=HrishiP][quote=ethanlanc]7.5 year bump :rotfl:[/quote]\n\nmegarnie is probably going to bump every single AMC problem thread ever so expect more of that :P. Although seriously I know I can't make you not post solutions (you're allowed to) but this is getting a little annoying to see what I mean just scroll down C&P and it's all AMCs (mostly).[/quote]\n\nI won't bump all amc problems only some" } { "Tag": [ "quadratics" ], "Problem": "Find the Roots of $ 8x^{4}\\minus{}8x^{2}\\plus{}1$", "Solution_1": "Rational or other kind as well?", "Solution_2": "other kind as well", "Solution_3": "i will just write the answer down, someone else please show how you work them out.\r\n\r\n$ x\\equal{}\\frac{\\pm\\sqrt{\\sqrt{2}\\plus{}2}}{2}$ or $ \\frac{\\pm{\\sqrt{2\\minus{}\\sqrt{2}}}}{2}$", "Solution_4": "[hide=\"Hint\"]\nLet $ x^{2}\\equal{}a$ and then just use the quadratic formula\n[/hide]" } { "Tag": [ "quadratics" ], "Problem": "Its idea is Intermediate but since it has a lot of square roots, I decide to post on Pre-Olympiad. It's very easy for Pre-Olympiad level so have fun on \"easy solving.\"\r\n\r\nNumber 1\r\n\r\nFor what integers $b$ and $c$ is $x = \\sqrt{19} + \\sqrt{98}$ a root of the equation $x^4+bx^2+c = 0$?\r\n\r\n[hide=\"View\"]\nI got $b = -234, c = 6241$ if anyone wanted the answer.[/hide]", "Solution_1": "The equation is a quadratic in $x^2$ with roots $x^2 = \\frac{-b \\pm \\sqrt{b^2 - 4c}}{2}$. We have that $117 + 14\\sqrt{38}$ is the square of a root. Thus, for some integers b and c, we have $234 + 28\\sqrt{38} = -b \\pm \\sqrt{b^2 - 4c}$.\r\n\r\nNow it is clear $b = -234$ etc.", "Solution_2": "Very nice method, Singular.\r\n\r\nFrom his answer, it is pretty easy to find $c$ because once you found what $b$ equals:\r\n\r\nYou have $x^4-234x^2+c = 0$ and putting $\\sqrt{19} + \\sqrt{98}$ for $x$ gives $-6241+c = 0$ which tells $c = 6241$.\r\n\r\nI got the answer by first setting $x-(\\sqrt{19} + \\sqrt{98})$ and multiplying it by $x+(\\sqrt{19}+\\sqrt{98})$ and then I'll have quadratic equation. Using some playing with positive and negative sign, you can come up with quartic equation, $x^4-234x^2+6241 = 0$, telling $b = -234$ and $c=6241$.\r\n\r\nAgain, nice easy method, Singular. :D" } { "Tag": [ "MATHCOUNTS", "\\/closed" ], "Problem": "For the last couple days my Aops hasn't been updating properly. Basically, even through there have been new posts made in certain sub-forums, I can only view the old posts from 1 -2 days ago. This is happening in getting started, mathcounts and some other topics.. What can I do?\r\n\r\nEDIT- nvm, i see what the problem is- I have to open the topic and the refresh, I use to be able to refresh the homepage and everything else would be updated as well....", "Solution_1": "Deeply werid. Try deleting all your old internet files, and perhaps your cookies." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Does there exist a function $ f\\colon \\mathbb{R}\\to\\mathbb{R}$ such that for all x the following equality holds \r\n$ f(sinx)\\plus{}f(sin^2x)\\plus{}f(sin^3x)\\equal{}f(cosx)\\plus{}f(cos^2x)\\plus{}f(cos^3x)$ \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n_______________________________________ \r\nAzerbaijan Land of the Fire :ninja: :ninja:", "Solution_1": "[quote=\"Pirkuliyev Rovsen\"]Does there exist a function $ f\\colon \\mathbb{R}\\to\\mathbb{R}$ such that for all x the following equality holds \n$ f(sinx) \\plus{} f(sin^2x) \\plus{} f(sin^3x) \\equal{} f(cosx) \\plus{} f(cos^2x) \\plus{} f(cos^3x)$ [/quote]\r\n\r\nYes, it does : $ f(x)\\equal{}a$" } { "Tag": [ "algebra", "polynomial", "calculus", "derivative", "search", "algebra proposed" ], "Problem": "Let $ n>2$ be an integer. Show, without calculus, that the polynomial $ P(x)\\equal{}x^n\\minus{}nx^{n\\minus{}1}\\plus{}1$ has $ n$ distinct zeros.\r\n\r\n[hide]I think it's also true that $ P(x)\\equal{}0$ has $ n$ distinct real roots, but I could not find a way of proving it. Any help on this will be greatly appreciated. :) [/hide]", "Solution_1": "That depends on what you mean by \"calculus.\" I can define the [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=225324]formal derivative[/url] without any knowledge of calculus and show that it has the power to detect repeated roots (and in fact, the only reason we want to define the derivative formally is to detect repeated roots in fields where we do not have a natural notion of limit). \r\n\r\nOne can in fact say something much stronger: $ P(x)$ is [i]irreducible[/i] for $ n > 2$ by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1088214531&t=150]Perron's criterion[/url] and, since $ \\mathbb{Q}$ is a [url=http://en.wikipedia.org/wiki/Separable_extension]perfect field[/url], cannot have repeated roots. (I do not claim to have satisfied your constraint; Perron's criterion requires a fair bit of analysis to prove, and the proof that fields of characteristic zero are perfect requires the use of the formal derivative.)\r\n\r\nEdit: Your conjecture fails for $ n \\equal{} 4$ but suggests an interesting exercise. Let $ Q(x)$ be a polynomial of degree $ n$ with $ n$ distinct real roots. Prove or disprove that $ P(x) \\equal{} Q(x) \\minus{} Q'(x)$ also has $ n$ distinct real roots.", "Solution_2": "Thanks for your comments. By \"without calculus\" here I meant that, \"use only plain and simple algebra, no advanced things i.e. do it with no knowledge of calculus\" (kinda like the rule at IMO as [url=http://en.wikipedia.org/wiki/International_Mathematical_Olympiad]Wikipedia[/url] says:)\r\n\r\n[quote=\"Wikipedia\"]They require no knowledge of higher mathematics such as calculus and analysis[/quote]\r\n\r\nThe $ nx^{n\\minus{}1}$ surely makes one believe that the solution to this problem must have something to do with derivatives. But I just put that expression for fun, it's nothing special really! I can give you examples of hundreds of polynomials which have nothing to do with derivatives. :)", "Solution_3": "That's not what I mean.\r\n\r\n[b]Theorem:[/b] A polynomial $ P(x)$ has no repeated roots if and only if $ \\gcd(P(x), P'(x)) \\equal{} 1$, and this is true over [i]any[/i] field.\r\n\r\nThis has nothing to do with the form of the problem; it's simply both the most basic and the most useful tool we have to deal with repeated roots. The calculation becomes trivial: $ P'(x) \\equal{} nx^{n\\minus{}1} \\minus{} n(n\\minus{}1) x^{n\\minus{}2}$ which has roots $ 0$ and $ n\\minus{}1$, neither of which are roots of the original equation - and voila!", "Solution_4": "I know it isn't very hard if we use derivatives, I also found the same solution as yours. But I have another solution where I don't use any calculus at all. I was looking for such a solution. That's why I said, \"no calculus!\" Hope you understand what I mean. :)", "Solution_5": "I believe that [b] Nayel [/b] is correct. I 'm still working on it.", "Solution_6": "@[b] t0rajir0u [/b] Did you mean that we could define the derivative of a polynomial as:\r\n\r\nDefinition: Let $ f(x) \\equal{} \\sum_{i\\equal{}0}^{n} {a_i x^i} \\in F[x]$. The derivative of $ f$ is defined as:\r\n\\[ f'(x) \\equal{} \\sum_{i\\equal{}0}^{n} {i a_i x^i}\\].\r\n\r\nFrom this definition, we could prove the following properties:\r\ni) $ (f\\plus{}g)' \\equal{} f' \\plus{} g'$\r\nii) $ [f(g)]' \\equal{} f'(g) g'$\r\netc.\r\n\r\nAnd then applying these properties to prove that: \r\n\r\n$ (x\\minus{}\\lambda)^k \\| f(x)$ if and only if $ f^{i} (\\lambda) \\equal{} 0$ for all $ 0 \\leq i \\leq k\\minus{}1$ and $ f^{k} (\\lambda) \\not\\equal{} 0$.\r\n\r\n\r\n@[b] Nayel [/b] I don't think that the terms $ n x^{n\\minus{}1}$ could be replaced arbitrarily.", "Solution_7": "Yes. Usually I believe the defining properties are something along these lines:\r\n\r\n- The derivative operator $ \\frac{d}{dx}$ takes polynomials to polynomials and is linear,\r\n- $ \\frac {d}{dx} x \\equal{} 1$, and\r\n- $ \\frac {d}{dx} fg \\equal{} f \\frac {dg}{dx} \\plus{} g \\frac {df}{dx}$.\r\n\r\nFrom these three properties everything else follows. As for your second comment, $ nx^{n \\minus{} 1}$ can be replaced by $ kx^{n \\minus{} 1}$ for $ |k| > 2$; see my comment." } { "Tag": [ "ARML", "geometry", "calculus", "integration", "trigonometry", "rotation" ], "Problem": "1.\r\nIn right triangle $ABC$, $\\angle C = 90^\\circ$, $c=3$, and $a+b=\\sqrt{17}$. Compute the area of the triangle.\r\n\r\n\r\n2.\r\nIf $(a+b):(b+c):(c+a) = 6:7:8$, and $a+b+c=14$, compute the value of $c$.\r\n\r\n\r\n3.\r\nLet $z$ be a root of $z^5-1=0$, with $z \\not= 1$. Compute the value of\r\n\r\n\\[z^{15} + z^{16} + z^{17} + \\cdots + z^{50}. \\]\r\n\r\n\r\n4.\r\nA regular polygon of $2n$ sides is inscribed in a circle of radius 5. The sum of the squares of the distances from any point on the circle to the vertices of the polygon is 600. Compute the value of the integer $n$.\r\n\r\n\r\n5.\r\nThe circles, each of radius 5, lie in perpendicular planes. They intersect in two points, $X$ and $Y$, thus determining a common chord $\\overline{XY}$. If $XY = 8$, compute the distance between the centers of the circles.\r\n\r\n\r\n6.\r\nHow many positive integers less than 54 are equal to the product of their proper divisors?\r\n\r\n\r\n7.\r\nFind all positive integers $n$, less than 17, for which $n! + (n+1)! + (n+2)!$ is an integral multiple of 49.\r\n\r\n\r\n8.\r\nIn triangle $ABC$, $a \\ge b \\ge c$. If $\\displaystyle \\frac{a^3+b^3+c^3}{\\sin^3{A} + \\sin^3{B} + \\sin^3{C}} = 7$, compute the maximum possible value of $a$.", "Solution_1": "[hide=\"Solution to #3\"]We can write the solutions to $z^5=1$ as $e^{.4k\\pi i}$ for $k\\in \\{1,2,3,4\\}$. WLOG assume k=1. \n\\[z^{15}+z^{16}+z^{17}+\\cdots+z^{50}\\]\n\\[=\\frac{z^{15}(z^{36}-1)}{z-1}\\]\n\\[=\\frac{e^{14.4\\pi i}-1}{e^{.4\\pi i}-1}\\]\nNote that $e^{14.4\\pi i}$ is $e^{.4\\pi i}$ rotated around the origin in the Argand diagram in the Complex Plane 7 times. Therefore we have $\\frac{e^{.4\\pi i}-1}{e^{.4\\pi i}-1}=\\boxed{1}$[/hide]", "Solution_2": "That was one of my current ARML homework problems. Luckily, I had solved it yesterday." } { "Tag": [ "geometry", "3D geometry", "algorithm" ], "Problem": "My best time is 1:01. :first:", "Solution_1": "I am one of those uber-uncreative people who can't solve. :( \r\n\r\nIsn't there another poll like this already?", "Solution_2": "Well, not on the official polls, so I thought it would be ok.\r\n\r\nThere might be one somewhere else though...\r\n\r\nThis is funny $ \\rightarrow$ :stretcher:\r\n\r\nEDIT: Some people have discussed this before, but it wasn't put on a poll, according to my search.", "Solution_3": "There might already be one in Math Strategy Games. Anyway I can solve it in about 6-7 minutes [u][i]with[/u][/i] a Seven-Step guide that came with it. I'd be lost without it. :D", "Solution_4": "2-5 is unrealistic\r\nit should be like sub 20\r\n20-40\r\n40-1 min\r\n1min-2 min\r\n2min+\r\ncan't solve\r\nanyone can do it in 2 min with practice", "Solution_5": "I could never solve the Rubik's Cube, no matter how often I tried, since I was five years old. :( :rotfl:", "Solution_6": "My record is 97.04...sad.", "Solution_7": "Sadly mine is 107.58 I suck", "Solution_8": "My single record is maybe about 8 seconds...(ie a very lucky one)\r\n\r\nMy best average of 12 is about 19 sec.", "Solution_9": "Need to have more poll options for fast people (sub 20 people). My record is 26.5 (actually 26.49) seconds.", "Solution_10": "3:44 is my fastest so far. I used the seven-step solution guide, but I was a little more creative and made it a 6-step solution guide with shortcuts.\r\n\r\nEdit: 3:31.689\r\n\r\nEdit 2: 2:35.99", "Solution_11": "My record is 1:29.", "Solution_12": "My record is 1:21. I got lucky and it skipped a few steps for me.\r\n\r\nHowever my average falls between 1:40 and 2 min.", "Solution_13": "Who here CAN use the solutions guide and do it under three minutes?", "Solution_14": "[quote=\"1=2\"]Who here CAN use the solutions guide and do it under three minutes?[/quote]\r\n\r\nWhen I first solved the Cube, I did it with the solutions (F2L Beginners method) and averaged a little under 3 minutes. Then I started memorizing easy algorithms and my time halved.", "Solution_15": "[quote=\"7h3.D3m0n.117\"][quote=\"1=2\"]Who here CAN use the solutions guide and do it under three minutes?[/quote]\n\nWhen I first solved the Cube, I did it with the solutions (F2L Beginners method) and averaged a little under 3 minutes. Then I started memorizing easy algorithms and my time halved.[/quote]\r\n\r\nHow do you average under three minutes first time around? I have a rubix cube that doesn't turn easily, is that the issue?\r\n\r\nP.S. 1:41.85", "Solution_16": "[quote=\"1=2\"]\nHow do you average under three minutes first time around? I have a rubix cube that doesn't turn easily, is that the issue?[/quote]\r\n\r\nYou should see my cube...it can't turn, the stickers are dying, and it's amazing how I could solve the cube in my lifetime. I'm trying to get another cube but here's my method:\r\n\r\nBefore learning to solve the actual cube I got bored and started this fun competition between me and my sister. The goal was just to solve one side (the edges didn't matter). Eventually we added multiple sides (independant of each other) and we averaged about less than 20 seconds per side (it's not really solved because of the edges). We then made it challenging and solved the side for real (with edges). Our average didn't increase that much because we knew how the algorithms. When it was time to solve the whole cube, the first step (cross + corners) takes no time at all. \r\n\r\nAlso, my moves with the cube have no hesitation to them (other than converting steps and analyzing my states). So to solve for the second layer I hardly do any pausing at all.", "Solution_17": "49 secs. Haven't done it in a while though.", "Solution_18": "[quote=\"1=2\"]Who here CAN use the solutions guide and do it under three minutes?[/quote]\r\n\r\nI got 36 seconds just now using the beginners method. :)", "Solution_19": "I memorized the solution guide and solved in 1:52.\r\nWay better than before. \r\n:D", "Solution_20": "I tried solving my friends' at CTY, but i just couldn't....\r\nand I thought people who SOLVED rubix cubes were amazing..\r\ni didn't know there were people who could cube in like 45 seconds..\r\n\r\ni know how to make a checkerboard pattern though...\r\nabout ALL i know..\r\n\r\nmy friend used a 5x5 cube and made a heart out of it...\r\nshe is skilled..", "Solution_21": "i solved mine in less than one second or as i like to call \"instantly\"\r\n\r\nall you have to do is take it out of the box. :P Just kidding!", "Solution_22": "yay i got lucky and skipped a few steps!\r\nstill got 1min30 tho...", "Solution_23": "[quote=\"moogra\"][quote=\"1=2\"]Who here CAN use the solutions guide and do it under three minutes?[/quote]\n\nI got 36 seconds just now using the beginners method. :)[/quote]\r\n\r\nWow that's impressive with the beginners :) . If I'm unfortunate, I'll average about 36 seconds just to get the first side and end up with a little under two minutes as my time.", "Solution_24": "My best was 56.something secs. Haven't done it in a long time though..." } { "Tag": [ "inequalities" ], "Problem": "The relation $ x^2(x^2 \\minus{} 1)\\ge 0$ is true only for:\r\n\r\n$ \\textbf{(A)}\\ x \\ge 1\\qquad \\textbf{(B)}\\ \\minus{} 1 \\le x \\le 1\\qquad \\textbf{(C)}\\ x \\equal{} 0,\\, x \\equal{} 1,\\, x \\equal{} \\minus{} 1\\qquad \\\\\\textbf{(D)}\\ x \\equal{} 0,\\, x \\le \\minus{} 1,\\, x \\ge 1\\qquad \\textbf{(E)}\\ x \\ge 0$", "Solution_1": "[quote=\"sunehra\"]The relation $ x^2(x^2 \\minus{} 1)\\ge 0$ is true only for:\n\n$ \\textbf{(A)}\\ x \\ge 1\\qquad \\textbf{(B)}\\ \\minus{} 1 \\le x \\le 1\\qquad \\textbf{(C)}\\ x \\equal{} 0,\\, x \\equal{} 1,\\, x \\equal{} \\minus{} 1\\qquad \\\\\n\\textbf{(D)}\\ x \\equal{} 0,\\, x \\le \\minus{} 1,\\, x \\ge 1\\qquad \\textbf{(E)}\\ x \\ge 0$[/quote]\r\n\r\n$ x^2(x^2 \\minus{} 1)\\ge 0 \\Leftrightarrow x\\equal{}0$ or $ x^2\\minus{}1 \\ge 0$ , so the answer is $ D$", "Solution_2": "[hide=\"Algebraic Solution\"]\nWe have $ x^2 \\ge 0$ for all $ x$, so we must have $ x^2 \\minus{} 1 \\ge 0\\implies x \\le \\minus{} 1$ and $ x \\ge 1$ and $ x \\equal{} 0$ (for equality). Therefore, $ \\fbox{(D)}$ is the answer.\n[/hide]\n[hide=\"Graphing Solution\"]\nIf you can get a general idea of how it looks, then this problem is very straight forward. We use the fact that an even power graph does up at the left and up at the right. The graph will intersect for the first time at $ x \\equal{} \\minus{} 1$, then it will be tangent to the origin as it works its way up, and finally it will work its way down and then up again intersecting the $ x$-axis for a final time at $ x \\equal{} 1$. The graph is above the $ x \\minus{} axis$ or touching it at $ x \\equal{} 0$, $ x \\ge 1$, and $ x \\le \\minus{} 1$ making the answer $ \\fbox{(D)}$.\n[/hide]" } { "Tag": [ "inequalities", "induction", "function", "inequalities theorems" ], "Problem": "If $n \\geq 2$ and $a_1, \\ldots, a_n > -1$, and that all the a_i's have the same sign, then \\[\\prod_{i=1}^n (1+a_i) \\geq 1+ \\sum_{i=1}^n a_i\\]", "Solution_1": "what happens if some of ai are 0?", "Solution_2": "[quote]what happens if some of ai are 0?[/quote]\r\nAlright, just to be safe, I've changed it to include equality.", "Solution_3": "extra 0's doesn't matter. \r\n\r\nthe product gets x1 so doesn't change, the sum gets +0 so doesn't change :)", "Solution_4": "hmm. math induction works.", "Solution_5": "Yes, that works. There are a lot of working methods though :)", "Solution_6": "Let $x > -1 \\neq 0$, $a \\in \\mathbb{R}$.\r\nThen:\r\n\r\n$(1+x)^a>1+ax$, if $a>1$ or $a<0$\r\n\r\n$(1+x)^a<1+ax$, if $0 -1 \\neq 0$, $a \\in \\mathbb{R}$.\nThen:\n\n$(1+x)^a>1+ax$, if $a>1$ or $a<0$\n\n$(1+x)^a<1+ax$, if $01$ or $a<0$, $f''(x)>0$, so $x=0$ is the minimum of the function which is 0. Otherwise, $f''(x)<0$, so $x=0$ is the maximum of the function.", "Solution_8": "Is there any non-analysis appraoch. Surely!\r\nThough I like Langrange multipliers...\r\n\r\nBomb", "Solution_9": "Isnt this solvable by AMGM :D :D \r\n\r\nBomb", "Solution_10": "[quote=\"bomb\"]To Billzhao, Isn't your proposed inequality Wierstrass' inequality...\n\nBomb[/quote]No... It's bernouilli's inequality. As the title says. ;)", "Solution_11": "It's not Bernoulli, I'm pretty sure it is Wierstrass when all a's >= 0 but when 0>=a's>=-1, it becomes Boniferri's inequalty....\r\n\r\nBomb" } { "Tag": [], "Problem": "Okay, here's how it works. I'm going to say a phrase, and you guys have to ask yes or no questions to try and solve it. If you think you've solved it, I'll ask you 3 yes or no questions to see if you really have solved. Once someone has solved the puzzle, I or the problem solver will explain the puzzle. I know it sounds vague, but I can't think of a better way to word it.\r\n\r\nThe phrase is:\r\n[size=150][b]\"Jeanette & Giuseppe\"[/b][/size]\r\n\r\n\r\nOh, I'm pretty sure this belongs here since it was in a book called [i]The Art of Problem Solving[/i], but not exactly sure.", "Solution_1": "Does it have something to do with the similarities in their names? Like, \"jea\" and \"gui\" sound the same, then they both have \"ette\" and \"eppe\"?", "Solution_2": "[quote=\"JS1527\"]Does it have something to do with the similarities in their names? Like, \"jea\" and \"gui\" sound the same, then they both have \"ette\" and \"eppe\"?[/quote]\r\n\r\nno", "Solution_3": "Could this be any more vague?", "Solution_4": "[quote=\"jelyman\"]Could this be any more vague?[/quote]\r\n\r\n\r\nyes", "Solution_5": "[quote=\"jelyman\"]Could this be any more vague?[/quote]\r\n\r\nYes", "Solution_6": "Is \"Guiseppe\" spelt right? Its not Giuseppe?", "Solution_7": "[quote=\"TripleM\"]Is \"Guiseppe\" spelt right? Its not Giuseppe?[/quote]\r\n\r\nTo answer your questions realisticly, I must have typed it wrong, I really don't know.\r\n\r\nAnd to answer the questions as it if was in the puzzle, No.", "Solution_8": "Do the answers to questions have something to do with the initial phrase (I mean, if the phrase had changed would you answer some questions in a differentmanner)?" } { "Tag": [ "inequalities", "algebra", "polynomial", "inequalities proposed" ], "Problem": "If $ a,b,c$ are nonnegative real numbers, then\r\n\r\n$ \\sum (a\\minus{}b)(a\\minus{}c)(a\\minus{}2b)(a\\minus{}2c)(a\\minus{}3b)(a\\minus{}3c) \\ge 0$,\r\n\r\nwith equality for $ a\\equal{}b\\equal{}c$, for $ \\frac {a}{2}\\equal{}b\\equal{}c$ or any cyclic permutation, and again for $ \\frac a{3}\\equal{}b\\equal{}c$ or any cyclic permutation.", "Solution_1": "[quote=\"Vasc\"]If $ a,b,c$ are nonnegative real numbers, then\n\n$ \\sum (a \\minus{} b)(a \\minus{} c)(a \\minus{} 2b)(a \\minus{} 2c)(a \\minus{} 3b)(a \\minus{} 3c) \\ge 0$,\n\nwith equality for $ a \\equal{} b \\equal{} c$, for $ \\frac {a}{2} \\equal{} b \\equal{} c$ or any cyclic permutation, and again for $ \\frac a{3} \\equal{} b \\equal{} c$ or any cyclic permutation.[/quote]\r\n\r\nActually, the following stronger and extended inequality holds.\r\n\r\nIf $ a,b,c$ are real numbers, then\r\n\r\n$ \\sum (a \\minus{} b)(a \\minus{} c)(a \\minus{} 2b)(a \\minus{} 2c)(a \\minus{} 3b)(a \\minus{} 3c) \\ge 3(a\\minus{}b)^2(b\\minus{}c)^2(c\\minus{}a)^2$,\r\n\r\nwith equality for $ a \\equal{} b \\equal{} c$, for $ \\frac {a}{2} \\equal{} b \\equal{} c$ or any cyclic permutation, and again for $ \\frac a{3} \\equal{} b \\equal{} c$ or any cyclic permutation", "Solution_2": "[quote=\"Vasc\"]If $ a,b,c$ are nonnegative real numbers, then\n\n$ \\sum (a \\minus{} b)(a \\minus{} c)(a \\minus{} 2b)(a \\minus{} 2c)(a \\minus{} 3b)(a \\minus{} 3c) \\ge 0$,\n\nwith equality for $ a \\equal{} b \\equal{} c$, for $ \\frac {a}{2} \\equal{} b \\equal{} c$ or any cyclic permutation, and again for $ \\frac a{3} \\equal{} b \\equal{} c$ or any cyclic permutation.[/quote]\r\n\r\nit is equivalent to :\r\n\r\n$ 1 \\minus{} 12q \\plus{} 26r \\plus{} 44q^2 \\minus{} 286qr \\plus{} 819r^2 \\minus{} 12q^3 \\geq 0$\r\n\r\nhence if : $ \\delta \\equal{} 13(28q \\minus{} 5)(27q^2 \\minus{} 38q \\plus{} 10) \\leq 0$ ==> $ q\\leq \\frac {5}{28}$ we are done otherwise if $ q \\geq \\frac {5}{28}$ then:\r\n\r\n$ \\displaystyle ( \\sum{a^2} \\minus{} \\sum{ab} ) (LHS) \\equal{} \\frac {13(a \\minus{} b)^2(a \\minus{} c)^2(b \\minus{} c)^2 ( 18\\sum(ab) \\minus{} 5\\sum(a^2))}{4} \\plus{} ( \\sum{a^4} \\minus{} \\frac {7\\sum{ab(a \\plus{} b)}}{2} \\plus{} 11\\sum{ (ab)^2} \\minus{} 5abc\\sum{a})^2$\r\n\r\nthis solution is perhaps wrong, so someone check it please :)", "Solution_3": "I think it is right (for $ p\\equal{}a\\plus{}b\\plus{}c\\equal{}1$). Nice, Anas. :lol:", "Solution_4": "Thank you VasC, I think it works for the stronger one :wink: \r\n\r\nP.S: what is your solution please ? :)", "Solution_5": "[quote=\"anas\"]Thank you VasC, I think it works for the stronger one :wink: \n\nP.S: what is your solution please ? :)[/quote]\r\nI have 4 solutions to this ineguality, based on 4 new theorems concerning the sixth degree symmetric homogeneous polynomial inequality $ f_6(x,y,z)\\ge 0$: for either real or nonnegative $ x,y,z$; with either necessary and sufficient conditions, or only sufficient conditions. Notice that the theorems with sufficient conditions are much more practical. The theorem below is for nonnegative real numbers, with necessary and sufficient conditions.\r\n \r\n[b]Theorem. [/b]Let $ f_6(x,y,z)\\ge 0$ be a sixth degree symmetric homogeneous polynomial inequality, written in the form\r\n\\[ Ar^2 \\plus{} g_1(p,q)r \\plus{} g_2(p,q)\\ge 0,\\]\r\nwhere $ A > 0$, $ p \\equal{} x \\plus{} y \\plus{} z$, $ q \\equal{} xy \\plus{} yz \\plus{} zx$, $ r \\equal{} xyz$. Let\r\n\\[ d(p,q) \\equal{} g_1^2(p,q) \\minus{} 4Ag_2(p,q),\\]\r\n\r\n\\[ \\delta_0 \\equal{} d(t,1),\\ \\ \\ \\delta(t) \\equal{} d(t \\plus{} 2,2t \\plus{} 1),\\ \\ \\ h(t) \\equal{} 2At \\plus{} g_1(t \\plus{} 2,2t \\plus{} 1).\\]\r\nSuppose that $ f_6(x,1,1)\\ge 0$ and $ f_6(0,y,z) \\ge 0$ for any nonnegative real numbers $ x,y,z$. The inequality $ f_6(x,y,z) \\ge 0$ holds for all nonnegative real numbers $ x,y,z$ if and only if one of the following two conditions is fulfilled:\r\n\r\n(a) if $ t\\ge 1$ and $ \\delta(t) > 0$, then $ h(t)\\le 0$; \r\n\r\n(b) if $ 0 < t\\le 1$ and $ \\delta(t) > 0$, then $ h(t)\\ge 0$, and if $ t\\ge 2$ and $ \\delta_0(t) > 0$, then $ g_1(t,1)\\ge 0$." } { "Tag": [ "analytic geometry", "vector", "number theory unsolved", "number theory" ], "Problem": "Solve the equation $ x! \\plus{} y! \\plus{} z! \\equal{} 2^{u!}$", "Solution_1": "[b]Claim:[/b] up to permutation of the first $ 3$ coordinates the solution vectors $ (x,y,z,u)$ are:\r\n$ (0,0,2,2),(0,1,2,2),(1,1,2,2)$\r\n\r\n[i]Proof:[/i] Since $ x!\\plus{}y!\\plus{}z!\\geq 3$ holds, we have $ u>1$. Up to permuation we can assume $ x\\leq y\\leq z$, therefore $ x!|2^{u!}$, i.e. $ x<3$. The case $ x\\equal{}0,1$ requires $ y\\equal{}0,1$ and yields $ z!\\equal{}2^{u!}\\minus{}2$. Since $ 2^{u!}\\minus{}2\\equiv 2\\bmod{4}$ we must have $ z<4$, which gives $ z\\equal{}u\\equal{}2$.\r\n\r\nThe case $ x\\equal{}2$ requires $ y!|2^{u!}\\minus{}2$, i.e. $ y<4$. The sub-case $ y\\equal{}2$ yields $ z!\\equal{}2^{u!}\\minus{}4$, i.e. $ z\\geq 4,u\\geq 3$, hence $ z<6$, since $ 2^{u!}\\minus{}4\\equiv 4\\bmod{8}$. But $ z\\equal{}4,5$ do not permit $ u$.\r\n\r\nThe subcase $ y\\equal{}3$ yields $ z!\\equal{}2^{u!}\\minus{}10$, i.e. $ z<4$. But the only candidate $ z\\equal{}3$ does not permit a valid $ u$." } { "Tag": [ "algebra", "polynomial", "function", "real analysis", "real analysis unsolved" ], "Problem": "1.Does anybody know nice asymptotics for $ F(x) \\equal{} \\sum \\limits_{l \\equal{} 0}^{k} (J_l^{(0,1)}(x))^2$ where $ J_l$ are normalized Jacobi polynomials orthogonal with respect to $ (1 \\plus{} x)$?\r\n2. Prove that $ max_{[ \\minus{} 1,1]} (1 \\plus{} x)\\sum \\limits_{l \\equal{} 0}^{k} (J_l^{(0,1)}(x))^2 \\equal{} 2F(1)$\r\nThanks\r\nEdit. For the second one I've just found nice general Sego theorem. It says that if we have the sequence of orthogonal polynomials on the segment $ [a,b]$ with respect to non-decreasing weight $ \\omega (x)$ then the function $ |p_n (x)\\parallel{}\\omega (x)|^{\\frac{1}{2}}$ attains maximum at the point $ b$", "Solution_1": "I wonder if you have taken a look at the book \"Orthogonal polynomials\" by Gabor Szego. There is a lot of such stuff there :) (not that I remember much of it nowdays :().", "Solution_2": "Yes, I've actually found Szego theorem there. But for the first question I still did not get what I need.", "Solution_3": "OK, I'll ask a few friends of mine who should know such things. Is the normalization condition that the $ L^2((1\\plus{}x)\\,dx)$-norm of each polynomial is $ 1$, or something else?", "Solution_4": "Yes, you are absolutely right (their norm in $ L_2((1\\plus{}x))dx$ is $ 1$).\r\nThanks a lot." } { "Tag": [ "geometry", "rectangle", "ratio" ], "Problem": "When you cut a certain rectangle in half, you obtain two rectangles that are both similar to the original rectangle. Find the ratio of the longer side length to the shorter side length.", "Solution_1": "[hide=\"Solution\"]\nAssume that the rectangle's dimensions are x and y, where $ x > y$.\nThen $ \\frac{\\frac{x}{2}}{y} \\equal{} \\frac{y}{x}$.\nThen $ x^2 \\equal{} 2 y^2$\nso $ x \\equal{} \\sqrt{2} y$\n\n$ \\sqrt{2} : 1$\n[/hide]\r\nwhich happens to be more or less the ratio of the A4 and related sheets. They designed it that way, so you could photocopy two A4's onto one A4 with the right ratio. Um, I got distracted." } { "Tag": [], "Problem": "Asertain the digits $ \\{x,y,z\\}\\subset\\overline {0,9}$ so that $ \\overline {yz}^2 \\equal{} \\overline {xyz}$ , i.e. $ (10y \\plus{} z)^2 \\equal{} 100x \\plus{} 10y \\plus{} z$ .", "Solution_1": "We have $ \\overline{yz}(\\overline{yz}\\minus{}1)\\equal{}100x$ then we have two case:\r\n-) $ \\overline{yz}\\vdots 25,\\overline{yz}\\minus{}1\\vdots 4$: easy to see $ \\overline{yz}\\equal{}25$\r\n-) $ \\overline{yz}\\vdots 4,\\overline{yz}\\minus{}1\\vdots 25$: easy to see no result!" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Why must a periodic function $f: \\mathbb{R}\\rightarrow\\mathbb{R}$ be uniformly continuous in order to be expressed as $f\\left(x\\right)=\\sum_{n\\in\\mathbb{Z}}a_{n}e^{2\\pi inx}$?", "Solution_1": "Depends on what you mean by \"expressed\". An infinite series of functions can converge in a number of ways. If it converges to $f$ uniformly, then the limit must be a uniformly continuous function by a standard theorem of real analysis. Other types of convergence (pointwise, a.e., $L^{2}$, ...) do not require continuity of $f$." } { "Tag": [ "vector", "induction", "linear algebra", "matrix", "inequalities", "linear algebra unsolved" ], "Problem": "Hello. Please try to prove the following:\r\n\r\nLet $ T : V \\rightarrow V$ be a linear operator in a unitary space $ V$.\r\nProve that if\r\n(1) For all $ \\lambda$ eigenvalue of $ T$ $ |\\lambda|=1$.\r\n(2) For each vector $ a$ in $ V$ $ \\parallel T(a)\\parallel \\leq \\parallel a\\parallel $.\r\nthen $ T$ is an isometry.\r\n\r\n(Here $ \\parallel x\\parallel $ is the length of x.)\r\nPlease give a hint for a proof, or a proof.\r\nI know one proof using induction on V's dimention, but I would like to see another proof.\r\nThanks in advance for any ideas/hints.", "Solution_1": "I assume that you mean \"complex inner product space\" when you say \"unitary space\". This also means that all eigenvalues exist, and we don't get anything funny from that.\r\n\r\nFor all $ x$, $ \\langle x, T^{*}Tx\\rangle =\\langle Tx,Tx\\rangle=\\|Tx\\|^{2}\\le \\|x\\|^{2}$. Since $ T^{*}T$ is self-adjoint and positive semidefinite, it is similar to a real diagonal operator; there is an orthonormal basis $ B$ in which the matrix of $ T$ is diagonal with entries in $ [0,\\infty)$. By the norm condition, all diagonal entries in this basis must be $ \\le 1$.\r\n\r\nOn the other hand, the determinant of $ T$ is a unit since all of its eigenvalues have magnitude 1, and this means that the determinant of $ T^{*}T$ is 1. The only way for this to happen with our other inequalities is for $ T^{*}T$ to be the identity, and that makes $ T$ an isometry.", "Solution_2": ":thumbup: Thank you, it's a very nice proof!\r\n\r\n(and yes, you figured out correctly what I meant.)" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "[url=http://bildr.no/view/509704][img]http://bildr.no/thumb/509704.jpeg[/img][/url]\r\n\r\nI can not understand the sentence denoted by red line.\r\n\r\nI feel a little shy to ask so many questions.\r\nI am studying by myself. I need friends and teachers.\r\nSo I think I will ofter ask questions at the beginning of self-studying.\r\n\r\nDon't laugh at me.\r\n\r\nThanks very much.", "Solution_1": "The misunderstanding stems from the vicious way (that I noticed on other occasions in your posts) the authors of those articles use the fraction symbol. So, when they wrote, just before the red-underlined passage,\r\n[quote]... $ y \\geq x \\plus{} 1/f(f(x)) \\minus{} 1$ ...[/quote]\r\nwhat they actually meant was $ y \\geq (x \\plus{} 1)/(f(f(x)) \\minus{} 1)$. \r\nNow the passage you underlined becomes intelligible, since $ f(x\\plus{}y) \\geq f(x) \\plus{} yf(f(x))$ is just the functional relation, to which we apply $ f(x) \\geq 0$, and $ yf(f(x)) \\geq x \\plus{} 1 \\plus{}y$, which is just a rewrite of the relation I edited.\r\n\r\nBy the way, I sent you an e-mail on your personal address, with a pdf document attached, in which there is a lot about this very problem (I am the author of a much more difficult version of this - which was asked at some Italian contest) :)" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Prove or disprove that:for any $ n \\geq 2$ there exist 2n distinct positive integer $ a_1,...,a_{2n}$ such that:$ a_1^2\\plus{}...\\plus{}a_n^2\\equal{}a_{n\\plus{}1}^2\\plus{}...\\plus{}a_{2n}^2$.", "Solution_1": "[quote=\"k.l.l4ever\"]Prove or disprove that:for any $ n \\geq 2$ there exist 2n distinct positive integer $ a_1,...,a_{2n}$ such that:$ a_1^2 \\plus{} ... \\plus{} a_n^2 \\equal{} a_{n \\plus{} 1}^2 \\plus{} ... \\plus{} a_{2n}^2$.[/quote]\r\n\r\nSince $ (a\\plus{}2)^2\\plus{}(2a\\minus{}1)^2\\equal{}(a\\minus{}2)^2\\plus{}(2a\\plus{}1)^2$ $ \\forall a$, it's obvious that if the property is true for $ n$, it's true for $ n\\plus{}2$. So we just have to show for $ n\\equal{}2$ and $ n\\equal{}3$\r\n\r\n$ n\\equal{}2$ : $ 6^2\\plus{}7^2\\equal{}2^2\\plus{}9^2$\r\n$ n\\equal{}3$ : $ 4^2\\plus{}6^2\\plus{}19^2\\equal{}2^2\\plus{}3^2\\plus{}20^2$\r\n\r\nHence the result", "Solution_2": "How about this?\r\nProve or disprove that:for any $ n \\geq 2$ there exist $ 2n\\plus{}1$ distinct positive integer $ a_1,a_2,...,a_{2n\\plus{}1}$ such that:\r\n$ a_{1}^{2}\\plus{}...\\plus{}a_{n}^{2}\\equal{}a_{n\\plus{}1}^{2}\\plus{}...\\plus{}a_{2n}^{2}\\equal{}a_{2n\\plus{}1}^2$", "Solution_3": "[quote=\"k.l.l4ever\"]How about this? [...][/quote]\r\n\r\nYou're welcome\r\nI'm glad to have helped you.\r\n\r\nAnd, btw, for you second problem :\r\nLet $ u\\equal{}\\sum_{k\\equal{}1}^{n\\minus{}1}(15k\\plus{}5)^2$\r\nLet $ v\\equal{}\\sum_{k\\equal{}1}^{n\\minus{}1}(15k\\plus{}3)^2$\r\n\r\n$ \\sum_{k\\equal{}1}^{n\\minus{}1}(2(v\\plus{}1)(15k\\plus{}5))^2\\plus{}((v\\plus{}1)(u\\minus{}1))^2$ $ \\equal{}\\sum_{k\\equal{}1}^{n\\minus{}1}(2(u\\plus{}1)(15k\\plus{}3))^2\\plus{}((u\\plus{}1)(v\\minus{}1))^2$ $ \\equal{}((v\\plus{}1)(u\\plus{}1))^2$", "Solution_4": "[quote=\"pco\"][quote=\"k.l.l4ever\"]Prove or disprove that:for any $ n \\geq 2$ there exist 2n distinct positive integer $ a_1,...,a_{2n}$ such that:$ a_1^2 \\plus{} ... \\plus{} a_n^2 \\equal{} a_{n \\plus{} 1}^2 \\plus{} ... \\plus{} a_{2n}^2$.[/quote]\n\nSince $ (a\\plus{}2)^2\\plus{}(2a\\minus{}1)^2\\equal{}(a\\minus{}2)^2\\plus{}(2a\\plus{}1)^2$ $ \\forall a$, it's obvious that if the property is true for $ n$, it's true for $ n\\plus{}2$. So we just have to show for $ n\\equal{}2$ and $ n\\equal{}3$\n\n$ n\\equal{}2$ : $ 6^2\\plus{}7^2\\equal{}2^2\\plus{}9^2$\n$ n\\equal{}3$ : $ 4^2\\plus{}6^2\\plus{}19^2\\equal{}2^2\\plus{}3^2\\plus{}20^2$\n\nHence the result[/quote]\nHello, what do you mean with showing the cases $n=2$ and $n=3?$ i understood it so that if $n$ is even then we can use the identity which you showed for the pairs, and if $n$ is odd then we can replace six of them so: $4; 6; 19$ -->RHS and $2; 3; 20$-->LHS and then we have remained even amount of numbers on both sides which we have to replace and we do it with the useful identity $ (a\\plus{}2)^2\\plus{}(2a\\minus{}1)^2\\equal{}(a\\minus{}2)^2\\plus{}(2a\\plus{}1)^2$ is it so?" } { "Tag": [ "inequalities", "number theory unsolved", "number theory" ], "Problem": "Let n be a positive integer, and let d(n) denote the sum of the positive divisors of n, including 1 and n itself. Prove that:\r\n d(1)/1+d(2)/2+d(3)/3+......+d(n)/n<=2n", "Solution_1": "let k be a numebr smaller than $n$, then let f(k) be the numebr of multiples of $k$ in the set $(1,2,3....,n)$. since $d(m)/m=\\sum_{d|n}1/d$ we have that k contributes $f(k)/k$ to this sum, but $f(k)$ is at most $n/k$ so we get our sum is at most $n(1+1/2^2+1/3^2....)<2n$", "Solution_2": "[quote=\"Pascual2005\"]since $d(m)/m=\\sum_{d|n}1/d$ [/quote]\r\nwhat is m? :read:", "Solution_3": "...and you may use $\\frac{\\pi^2}{6}n$ instead of $2n$..." } { "Tag": [], "Problem": "A merchant buys a number of oranges at $ 3$ for $ 10$ cents and an equal number at $ 5$ for $ 20$ cents. To \"break even\" he must sell all at:\r\n\r\n$ \\textbf{(A)}\\ \\text{8 for 30 cents} \\qquad\r\n\\textbf{(B)}\\ \\text{3 for 11 cents} \\qquad\r\n\\textbf{(C)}\\ \\text{5 for 18 cents} \\\\\r\n\\textbf{(D)}\\ \\text{11 for 40 cents} \\qquad\r\n\\textbf{(E)}\\ \\text{13 for 50 cents}$", "Solution_1": "[hide=\"Solution\"]He buys $ n$ oranges.\nHe spends $ 10n/3\\plus{}4n\\equal{}22n/3$ cents.\nIf he sells them for $ c$ cents per orange, he makes $ 2cn$ cents.\n$ 2cn\\equal{}22n/3\\implies c\\equal{}11/3$.\nSo the answer is $ \\boxed{\\textbf{(B)}}$.\n[/hide]" } { "Tag": [ "\\/closed" ], "Problem": "When I add a second entry to a category, it doesn't show up when I click on the category. How can I add more entries to categories?\r\n\r\nEDIT: Ah, never mind. It seems to be working now.", "Solution_1": "Hmmm... mine isn't working...", "Solution_2": "Did you remember to click \"Submit\" after selecting the category you wanted? :)" } { "Tag": [ "geometry", "perimeter", "circumcircle", "geometry proposed" ], "Problem": "Given an acute angled triangle with perimeter $L$, if its pedal triangle (the inscribed triangle of smallest perimeter) has perimeter $P$, then $\\frac{P}{L} \\leq \\frac{1}{2} $. The equality holds iff the triangle is equilateral.\r\n\r\nI cannot find a purely geometric solution. :?", "Solution_1": "Let ABC be our triangle and A', B', C' are the projections.\r\nO the center of circumcircle and H the ortocenter.\r\nUsing AO is perpendicular to B'C' then\r\nArea(ABC) = 1/2 AO*B'C' + 1/2 BO*A'C' + 1/2 CO*A'B'.\r\nhense L*r=P*R or P/L=r/R \u2264 1/2", "Solution_2": "Very nice solution!", "Solution_3": "Yes, very nice. \r\n\r\nCongratulations.", "Solution_4": "Regard $A''$, $B''$, $C''$ as the midpoints of sides $BC$, $CA$, $AB$ respectively, then we get the triangle $A''B''C''$ has perimeter $M$.\r\nObviously, $M = \\frac{1}{2} L$, and for $P \\leq M$, we have $\\frac{P}{L} \\leq \\frac12$, the equality holds iff $A'=A''$, $B'=B''$, $C'=C''$, mean the triangle is equilateral.", "Solution_5": ":D" } { "Tag": [ "conics", "parabola", "algebra proposed", "algebra" ], "Problem": "The sequences $\\{a_{n}\\},\\ \\{b_{n}\\},\\{c_{n}\\}$ are defined as follows.\r\n\\[a_{1}=2,\\ \\ a_{n+1}=4a_{n}\\]\r\n\r\n\\[b_{1}=3,\\ \\ b_{n+1}=b_{n}+2a_{n}\\]\r\n\r\n\\[c_{1}=4,\\ \\ c_{n+1}=\\frac{1}{4}c_{n}+a_{n}+b_{n}\\]\r\nLet $P_{n},\\ Q_{n}$ be the intersection points between the parabola $y=a_{n}x^{2}+2b_{n}x+c_{n}$ and the $x$ -axis. Find $\\sum_{k=1}^{n}P_{k}Q_{k}.$", "Solution_1": "$a_{n}=2*4^{n-1},b_{n}=\\frac{4}{3}4^{n-1}+\\frac{8}{3},c_{n}=\\frac{8}{9}4^{n-1}+\\frac{32}{9}-\\frac{4}{9}4^{1-n}$.\r\nIt give $\\sum_{k=1}^{n}P_{k}Q_{k}=2\\sqrt 2 \\frac{4^{n}-1}{3*4^{n-1}}$.", "Solution_2": "Your answer is incorrect. We don't need to find all of $a_{n},\\ b_{n},\\ c_{n}.$", "Solution_3": "Yes \r\n$b_{b+1}^{2}-a_{n+1}c_{n+1}=(b+2a_{n})^{2}-4a_{n}(\\frac{1}{4}c_{n}+a_{n}+b_{n})=b_{n}^{2}-a_{n}c_{n}=b_{1}^{2}-a_{1}c_{1}=1.$\r\nTherefore $\\sum_{k=1}^{n}P_{k}Q_{k}=1+4^{-1}+...+4^{1-n}=\\frac{4^{n}-1}{3*4^{n-1}}$.", "Solution_4": "That's right." } { "Tag": [ "limit", "logarithms", "calculus", "calculus computations" ], "Problem": "$ a_1,a_2,...,a_n>0$,Find:\r\n$ \\lim_{x\\longrightarrow 0}(\\dfrac{\\sum a_n^x}{n})^\\dfrac{1}{x}$", "Solution_1": "For big $ x$ we have:\r\n\r\n$ \\sum a_n^x \\sim \\left(\\max a_n\\right)^x$\r\n\r\nThis means that:\r\n\r\n$ \\left(\\frac{1}{n} \\sum a_n^x\\right)^{\\frac{1}{x}} \\sim \\left(\\frac{1}{n} \\left(\\max a_n\\right)^x \\right)^{\\frac{1}{x}} \\equal{} \\max a_n \\sqrt[x]{n}$\r\n\r\nFinally:\r\n\r\n$ \\lim _{x\\to\\infty}\\left(\\frac{1}{n} \\sum a_n^x\\right)^{\\frac{1}{x}} \\equal{} \\max a_n \\lim _{x\\to\\infty} \\sqrt[x]{n} \\equal{} \\max a_n$", "Solution_2": "[quote=\"milin\"]For big $ x$ we have:\n\n$ \\sum a_n^x \\sim \\left(\\max a_n\\right)^x$\n[/quote]\r\n\r\nHow about when $ a_1\\equal{}a_2\\equal{}...\\equal{}a_n\\equal{}1$,then the equality means $ n\\sim 1$ which i can't understand\u3002\r\nand i think the answer to the problem is $ \\sqrt[n]{a_1a_2..a_n}$by the power ineq:\r\n$ f(2)\\ge f(1)\\ge \\sqrt[n]{a_1a_2..a_n}\\ge f(\\minus{}1)$\r\n\r\ni am sorry if i made any mistake", "Solution_3": "When $ \\forall n \\,a_n \\equal{} 1$ then you have:\r\n\r\n$ \\left(\\frac {1}{n}\\sum_n a_n^x\\right)^{\\frac {1}{x}} \\equal{} \\left(\\frac {1}{n}\\sum_n 1\\right)^{\\frac {1}{x}} \\equal{} \\left(1\\right)^{\\frac {1}{x}} \\equal{} 1 \\equal{} \\max a_n$\r\n\r\nBut you are right.. It should have been only\r\n\r\n$ \\left(\\frac {1}{n}\\sum_n a_n^x\\right)^{\\frac {1}{x}} \\sim \\left(\\frac {1}{n} \\left( \\max a_n\\right)^x \\right)^{\\frac {1}{x}}$\r\n\r\nwithout the remark of the sum being similar to the maximal value.\r\n\r\nAlso, as for the $ \\frac{1}{n}$ factor - you can remove it immediately because $ \\lim _{ x\\to \\infty} \\frac{1}{\\sqrt[x]{n}} \\equal{} 1$.", "Solution_4": "[quote=\"milin\"]For big $ x$ [/quote] We have $ x \\to 0$, not $ x \\to \\infty$ ;)\r\n\r\n[hide=\"answer (no solution)\"]$ \\sqrt[n]{a_1 \\cdot a_2 \\cdots a_n}$[/hide]", "Solution_5": ":blush: Ups..", "Solution_6": "so can anyone give the solution?\r\nthanks so much~~", "Solution_7": "One way to do this, is to take the logarithm of the expression, and then \r\nyou have to deal with a limit of the form \"$ \\frac 0 0$\", which you probably know how do, \r\ne.g. using differentiation." } { "Tag": [ "geometry" ], "Problem": "I'm taking those three SAT II's this saturday, almost without notice. I'm completing AP Calc BC, AP Chem, and AP Physics C. If anyone here has taken them before, can you give me advice as to what I should review/skim; especially physics and chem. thanks", "Solution_1": "Math should be easy as pi for you.\r\n\r\nIf you're ready for AP Chem, you'll be ready for SAT chem. I took SAT II chem half way through the year and got a 770. The AP exam covers so much more material than the AP exam.\r\n\r\nI haven't taken physics so I don't know.\r\n\r\nJust look at sparknotes.com or check some books out at the library and look them over a little bit. But if you're ready for the AP chem and AP physics you shouldn't have much of a problem.", "Solution_2": "Yeah, the SAT II's are way easier than the AP. I think the questions are more general too (at least for the American History one).", "Solution_3": "Chem should be a breeze compared to the AP.\r\nNote that AP Physics C does not cover everything on the SATII, like vibrations, I think. Grab a review book and check.", "Solution_4": "Just make sure you look at the structure of the nasty true/false section. It gets so misleading, because overall it will be a wrong statement, but part i and part ii by themselves might be completely fine.", "Solution_5": "[quote=\"Bictor717\"]Chem should be a breeze compared to the AP.\nNote that AP Physics C does not cover everything on the SATII, like vibrations, I think. Grab a review book and check.[/quote]\r\n\r\nDitto the AP Physics C vs. SAT II stuff. The SAT II mostly covers AP Physics B material and Physics C is MUCH more limited in scope, so don't walk into the test expecting to see all Mechanics and E&M questions. There will be questions on literally EVERYTHING. So do some review.", "Solution_6": "yeah, thanks for the advice. i finished today.\r\n\r\nthe math was a lot easier than i expected. i thought that they would turn it up a notch for the sat ii, level 2. its kind of sad when you compare it to the mathematics entrance exams at japanese universities (like the ones frt has posted).\r\n\r\nthe chemistry wasn't as bad as i thought. u guys were right that the ap test covers so much more. the test was basic chemistry that i could confidently work through.\r\n\r\nunfortunately, you guys were also right about the way the physics compares to physics c. i was aware that there would be some optics and waves problems, but there were a lot more than i expected. i only had like 2 days for any kind of review, since i only knew i was going to take the test last thursday. i went over optics in physics b last year, but it is the area of physics i have most neglected. since i want atleast a 750 on the physics, i'll probably need to take this again.\r\n\r\nonce again, thanks for the advice anyways.", "Solution_7": "Do you guys think that these SAT II's would be too hard for someone who hasn't taken physics yet (not offered except to seniors :mad: ) and whose chemistry class was utterly worthless? (we've learned nothing, at least in terms of chemistry. I have gotten better at sudoku and cribbage...)", "Solution_8": "for physics it would be too hard if you havn't had the class. if you really want to, you could learn it on your own, but you would have to put ample time into reading and understanding concepts from a preparation book or text book.\r\n\r\nfor chemistry its the same kind of thing\r\n\r\noverall the tests aren't that hard if you take time to study for them directly. you can do it if you work at it." } { "Tag": [ "analytic geometry", "graphing lines", "slope", "calculus", "calculus computations" ], "Problem": "Find the necessary and sufficient condition such that 2 curves $ ax^2 \\minus{} by^2 \\equal{} 1, cx^2 \\minus{} dy^2 \\equal{} 1\\ (abcd\\neq 0)$ have a intersecion of point and their tangent line at $ P$ are perpendicular to each other.", "Solution_1": "Slope of the first curve$ \\equal{}m_{1}\\equal{}\\frac{ax}{by}$\r\n\r\nSlope of the second curve$ \\equal{}m_{2}\\equal{}\\frac{cx}{dy}$\r\n\r\nSince the two tangents are perpendicular,we have\r\n$ m_{1} . m_{2}\\equal{}\\minus{}1$\r\n\r\nwhich gives \r\n$ \\frac{acx^2}{bdy^2}\\equal{}\\minus{}1$..................(1)\r\n\r\nAlso,$ ax^2 \\minus{} by^2 \\equal{} cx^2 \\minus{} dy^2$\\\\since (x,y) is the point of intersection\r\n\r\n$ \\frac{x^2}{y^2}\\equal{}\\frac{b\\minus{}d}{a\\minus{}c}$...................(2)\r\n\r\nEliminating x and y from (1) and (2),we get\r\n\r\n$ \\frac{1}{a}\\plus{}\\frac{1}{b}\\equal{}\\frac{1}{c}\\plus{}\\frac{1}{d}$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers. Prove that\r\n$ \\left(\\frac{3a}{a\\plus{}2b\\plus{}2c\\plus{}2d}\\right)^{\\frac{3}{4}}\\plus{}\\left(\\frac{4b}{b\\plus{}3c\\plus{}3d\\plus{}3a}\\right)^{\\frac{2}{3}}\\plus{}$\r\n$ \\plus{}\\left(\\frac{6c}{c\\plus{}5d\\plus{}5a\\plus{}5b}\\right)^{\\frac{3}{5}}\\plus{}\\left(\\frac{d}{a\\plus{}b\\plus{}c}\\right)^{\\frac{1}{2}}\\geq2$", "Solution_1": "[quote=\"arqady\"]Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers. Prove that\n$ \\left(\\frac {3a}{a \\plus{} 2b \\plus{} 2c \\plus{} 2d}\\right)^{\\frac {3}{4}} \\plus{} \\left(\\frac {4b}{b \\plus{} 3c \\plus{} 3d \\plus{} 3a}\\right)^{\\frac {2}{3}} \\plus{}$\n$ \\plus{} \\left(\\frac {6c}{c \\plus{} 5d \\plus{} 5a \\plus{} 5b}\\right)^{\\frac {3}{5}} \\plus{} \\left(\\frac {d}{a \\plus{} b \\plus{} c}\\right)^{\\frac {1}{2}}\\geq2$[/quote]\r\nJust by AM-GM:\r\n$ \\left(\\frac {3a}{a \\plus{} 2b \\plus{} 2c \\plus{} 2d}\\right)^{\\frac {3}{4}} \\ge \\frac{2a}{a\\plus{}b\\plus{}c\\plus{}d}$\r\n$ \\left(\\frac {4b}{b \\plus{} 3c \\plus{} 3d \\plus{} 3a}\\right)^{\\frac {2}{3}} \\ge \\frac{2b}{a\\plus{}b\\plus{}c\\plus{}d}$\r\n$ \\left(\\frac {6c}{c \\plus{} 5d \\plus{} 5a \\plus{} 5b}\\right)^{\\frac {3}{5}} \\ge \\frac{2c}{a\\plus{}b\\plus{}c\\plus{}d}$\r\n$ \\left(\\frac {d}{a \\plus{} b \\plus{} c}\\right)^{\\frac {1}{2}} \\ge \\frac{2d}{a\\plus{}b\\plus{}c\\plus{}d}$", "Solution_2": "Yes, ehku! :lol:", "Solution_3": "[quote=\"ehku\"][quote=\"arqady\"]Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers. Prove that\n$ \\left(\\frac {3a}{a \\plus{} 2b \\plus{} 2c \\plus{} 2d}\\right)^{\\frac {3}{4}} \\plus{} \\left(\\frac {4b}{b \\plus{} 3c \\plus{} 3d \\plus{} 3a}\\right)^{\\frac {2}{3}} \\plus{}$\n$ \\plus{} \\left(\\frac {6c}{c \\plus{} 5d \\plus{} 5a \\plus{} 5b}\\right)^{\\frac {3}{5}} \\plus{} \\left(\\frac {d}{a \\plus{} b \\plus{} c}\\right)^{\\frac {1}{2}}\\geq2$[/quote]\nJust by AM-GM:\n$ \\left(\\frac {3a}{a \\plus{} 2b \\plus{} 2c \\plus{} 2d}\\right)^{\\frac {3}{4}} \\ge \\frac {2a}{a \\plus{} b \\plus{} c \\plus{} d}$\n$ \\left(\\frac {4b}{b \\plus{} 3c \\plus{} 3d \\plus{} 3a}\\right)^{\\frac {2}{3}} \\ge \\frac {2b}{a \\plus{} b \\plus{} c \\plus{} d}$\n$ \\left(\\frac {6c}{c \\plus{} 5d \\plus{} 5a \\plus{} 5b}\\right)^{\\frac {3}{5}} \\ge \\frac {2c}{a \\plus{} b \\plus{} c \\plus{} d}$\n$ \\left(\\frac {d}{a \\plus{} b \\plus{} c}\\right)^{\\frac {1}{2}} \\ge \\frac {2d}{a \\plus{} b \\plus{} c \\plus{} d}$[/quote]\r\n\r\nCan you explain, please? First, for example :)", "Solution_4": "[quote=\"Yuriy Solovyov\"][quote=\"ehku\"][quote=\"arqady\"]Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers. Prove that\n$ \\left(\\frac {3a}{a \\plus{} 2b \\plus{} 2c \\plus{} 2d}\\right)^{\\frac {3}{4}} \\plus{} \\left(\\frac {4b}{b \\plus{} 3c \\plus{} 3d \\plus{} 3a}\\right)^{\\frac {2}{3}} \\plus{}$\n$ \\plus{} \\left(\\frac {6c}{c \\plus{} 5d \\plus{} 5a \\plus{} 5b}\\right)^{\\frac {3}{5}} \\plus{} \\left(\\frac {d}{a \\plus{} b \\plus{} c}\\right)^{\\frac {1}{2}}\\geq2$[/quote]\nJust by AM-GM:\n$ \\left(\\frac {3a}{a \\plus{} 2b \\plus{} 2c \\plus{} 2d}\\right)^{\\frac {3}{4}} \\ge \\frac {2a}{a \\plus{} b \\plus{} c \\plus{} d}$\n$ \\left(\\frac {4b}{b \\plus{} 3c \\plus{} 3d \\plus{} 3a}\\right)^{\\frac {2}{3}} \\ge \\frac {2b}{a \\plus{} b \\plus{} c \\plus{} d}$\n$ \\left(\\frac {6c}{c \\plus{} 5d \\plus{} 5a \\plus{} 5b}\\right)^{\\frac {3}{5}} \\ge \\frac {2c}{a \\plus{} b \\plus{} c \\plus{} d}$\n$ \\left(\\frac {d}{a \\plus{} b \\plus{} c}\\right)^{\\frac {1}{2}} \\ge \\frac {2d}{a \\plus{} b \\plus{} c \\plus{} d}$[/quote]\n\nCan you explain, please? First, for example :)[/quote]\r\nBy AM-GM inequality we have\r\n$ \\sqrt[4]{\\left(\\frac{a\\plus{}2b\\plus{}2c\\plus{}2d}{3a}\\right)^3} \\\\\r\n\\equal{}\\sqrt[4]{1.\\left(\\frac{a\\plus{}2b\\plus{}2c\\plus{}2d}{3a}\\right)\\left(\\frac{a\\plus{}2b\\plus{}2c\\plus{}2d}{3a}\\right)\\left(\\frac{a\\plus{}2b\\plus{}2c\\plus{}2d}{3a}\\right)} \\\\ \\le \\frac{1}{4}\\left(1\\plus{}3.\\left(\\frac{a\\plus{}2b\\plus{}2c\\plus{}2d}{3a}\\right)\\right) \\\\ \\equal{} \\frac{a\\plus{}b\\plus{}c\\plus{}d}{2a}$\r\nHence,\r\n$ \\left(\\frac {3a}{a \\plus{} 2b \\plus{} 2c \\plus{} 2d}\\right)^{\\frac {3}{4}} \\ge \\frac {2a}{a \\plus{} b \\plus{} c \\plus{} d}$" } { "Tag": [ "geometry" ], "Problem": "Hello all first time poster here.\r\n\r\nI'm having a bit of a problem with the question in the pic.\r\n[url=http://img373.imageshack.us/my.php?image=img6716xn0.jpg][img]http://img373.imageshack.us/img373/2046/img6716xn0.th.jpg[/img][/url]\r\nNow I'm 96% sure that Car1 is ahead of car2 I'm just unsure how to work this kind of problem out, any tips? -thanks a bunch.", "Solution_1": "Welcome!\r\n\r\nThe trick you can use is the fact that area under the graph represents the displacement of the car." } { "Tag": [ "integration", "trigonometry", "limit" ], "Problem": "1. Sa se determine aria marginita de graficele functiilor $\\ f_n $ si $\\ f_{n+1} $, unde $\\ f_n $:[0, $\\infty $) $ \\to $ $\\mathbb{R}$ (n $\\in \\mathbb{N}^*$), $\\ f_n(x) $ = $\\ x^n lnx$ (x>0) si $\\ f_n $ (0)=0 .\r\n\r\n2. Fie F(x)=$\\int _{x}^{2x}{\\frac {t^2}{t^2+\\sin^2{t}}dt}$, $x \\in \\mathbb{R}^* $. Atunci F este impara.", "Solution_1": "1. Se arata usor ca $f_{n}$ sunt continue in $0$, $f_{n}$ si $f_{n+1}$ au puncte comune $\\left( 0,0\\right) $ si $\\left( 1,0\\right) $. Deducem ca aria cautata este\r\n\\[\r\n\\underset{a\\searrow0}{\\lim}\\int_{a}^{1}\\left| f_{n+1}\\left( t\\right)\r\n-f_{n}\\left( t\\right) \\right| dt=\\frac{1}{\\left( n+1\\right) ^{2}}%\r\n-\\frac{1}{\\left( n+2\\right) ^{2}}.\r\n\\]\r\n(vezi poza atasata)\r\n2. Daca $f$ e o functie para, avem\r\n\\[\r\n\\int_{-x}^{0}f\\left( t\\right) dt=\\int_{0}^{x}f\\left( t\\right) dt,\r\n\\]\r\npentru orice $x.$ Atunci\r\n\\[\r\nF\\left( -x\\right) =\\int_{-x}^{-2x}f=\\int_{-x}^{0}f+\\int_{0}^{-2x}f=\\int\r\n_{0}^{x}f+\\int_{2x}^{0}f=-\\int_{x}^{2x}f=-F\\left( x\\right) .\r\n\\]" } { "Tag": [ "LaTeX", "superior algebra", "superior algebra unsolved" ], "Problem": "attached the question in a doc.", "Solution_1": "Please write your question here in LaTeX. Not every one has (or wants to use) Word...", "Solution_2": "no Word this time:", "Solution_3": "I think [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=159144]this[/url] is an answer. This is made less clear by the lack of a subscript on your $ \\aleph$.\r\n\r\nIncidentally, we can read plaintext pseudo-markup stuff just fine. It's having to download and open something outside the browser that really drives people away.\r\n\r\nMost $ \\text{\\LaTeX}$ codes are pretty intuitive; \\aleph is $ \\aleph$, \\subset is $ \\subset$ (or \\subseteq is $ \\subseteq$), and so on. The forum interprets the codes, so it's the standard around here. If you feel like learning it, there's a lot of good stuff in [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]this subforum[/url].", "Solution_4": "thanks, I'll look into it..." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let a1 = 1, and an = (n - 1)an-1 + 1, for n > 1. \r\n \r\nFor what values of n is an divisible by n? \r\n\r\nSo far I've found primes 2, 5, 13, 37, 463, 83507 306759, 412021, and 466123 computationally, plus 4. \r\nAny product of the above also works.\r\nHowever, I can't ascertain which primes will work, even though I can see that the general solution is an = SUMk=0n-1 (n-1)!/k!\r\n\r\nAny ideas?", "Solution_1": "Something is wrong. 83507, 306759, 412021, and 466123 are not primes.\n2, 5, 13, 37, 463 are the only currently known primes that satisfy the required property.\nSee http://oeis.org/A064384 for further details and references.\nNot necessary prime $ n$'s are given in http://oeis.org/A064383" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "For what positive integers $ n$ is it true that \"every $ g$ in a finitely generated group $ G$ satisfies $ g^n \\equal{} 1$\" implies $ G$ is finite? \r\n\r\nThis is obvious when $ n \\equal{} 2$ and has been assigned to me when $ n \\equal{} 3$ (although I couldn't prove it), but I don't know the general answer. A proof for the $ n \\equal{} 3$ case would be nice.", "Solution_1": "There are counterexamples for bigger $ n$, see http://en.wikipedia.org/wiki/Burnside%27s_problem ." } { "Tag": [ "function", "number theory proposed", "number theory" ], "Problem": "In the following $ {\\mathbb N} $ denotes the set of positive integers and $ a,b $ are given integers in ${\\mathbb N}. $ \r\nBy $ \\mbox{\\rm Card}(M) $ is denoted the numbers of elements of a certain set $ M $. \r\nLet us consider two functions $ f,g : {\\mathbb N}\\to {\\mathbb N} $ such that $ f(xy)=g(ax+by) $ for all $ x,y \\in {\\mathbb N} .$ \r\n\r\n[b]1). Find an upper-bound for $ \\mbox{\\rm Card}\\left(f({\\mathbb N})\\right)\\; . $\n2) . In case $ a=b=1 $, it's true that $ \\mbox{\\rm Card}\\left(f({\\mathbb N} )\\right)\\le 3 \\;\\; $ ? [/b]", "Solution_1": "$ 2)$\r\n$ f(x) = f(1.x) = g(x + 1)$ so $ g(x) = f(x - 1)$\r\nthen $ f(xy) = g(x + y) = f(x + y - 1)$\r\nlet $ x\\in\\mathbb{N}$\r\n\r\nwe take $ (x_n)$ such that $ x_0 = x$ and \r\n\r\nif $ x_n$ isn't prime $ x_{n + 1} = \\frac {x_n}{p} + p - 1$ where $ p = min\\{q\\ primes: \\ q|x_n\\}$\r\nand we have $ f(x_{n + 1}) = f(\\frac {x_n}{p} + p - 1) = f(p.\\frac {x_n}{p}) = f(x_n)$\r\n\r\nif $ x_n$ is prime>2 $ x_{n + 1} = \\frac {x_n + 5}{2}$\r\nand we have $ f(x_n) = f((x_n - 1) + (2) - 1) = f(2(x_n - 1)) = f(4.\\frac {x_n - 1}{2}) = f(\\frac {x_n - 1}^{2} + 4) = f(x_{n + 1})$\r\n\r\nif $ x_n = 2$ $ x_{n + 1} = x_n$\r\n\r\nwe have $ \\forall n\\in\\mathbb{N}: \\ if x_n\\ge 6: \\ x_{n+1} 0$ these solutions are $ L^1$ and of one sign (moreover for $ \\alpha,\\beta > 1$ they are nice continuous bounded functions).\r\n\r\nEdit: Having given it a second thought, I realize I need to pin down more conditions to specify the time-dependent behavior. I'll go over the derivation and see if I can come up with the correct conditions. Edit 2: Hah looks like you pointed that out a minute earlier, I'll think about it.", "Solution_9": "Great, I see. Now, what dependence of $ t$ are you looking for? (Your functions that do not depend on $ t$ definitely solve the system, so why do you need anything else?)", "Solution_10": "Ok, so it should be something like: $ u(x,0) \\equal{} 0$, $ v(x,0) \\equal{} \\phi(x)$ for $ x\\in (0,\\gamma)$ where $ \\int_0^\\gamma \\phi (x)dx \\equal{} 1$ and $ \\phi(\\gamma) \\equal{} 0$. $ \\phi$ shouldn't matter, but if it makes it any easier it's physically sensible to have it be a $ \\delta$ function centered somewhere in the interval $ (0,\\gamma)$.", "Solution_11": "OK, so do you now give up the conditions $ u(0,t)\\equal{}v(\\gamma,t)\\equal{}0$? (they will be quite hard to maintain in the new version of the problem when both $ u$ and $ v$ are prescribed for all $ x$ at the moment $ 0$)", "Solution_12": "No, those boundary conditions must stay. But why do you think they are necessarily inconsistent?", "Solution_13": "[b]JoeBlow[/b], you tried to solve it in Maple? It can solve it. Answer consist hypergeometric series. And if $ \\alpha\\ne \\minus{} \\beta$, then from condition $ v(\\gamma,t) \\equal{} 0$ we get that $ v$ is constant in time, it is a function only $ x$.\r\n\r\n[hide=\"image\"][img]http://pic.ipicture.ru/uploads/090206/fzVMk8WQml.png[/img][/hide]", "Solution_14": "I solved the thing numerically in Mathematica for $ \\gamma \\equal{} 1$, $ \\alpha \\equal{} 4$, $ \\beta \\equal{} 3$, $ \\phi(x) \\equal{} 6\\left(\\frac {1}{4} \\minus{} \\left(\\frac {1}{2} \\minus{} x\\right)^2\\right)$ and it returns curves that settle into the correct asymptotic behavior, and satisfy the PDE everywhere to within precision except on what seems to be a shock wave.\r\n\r\nI'm guessing the point is that Maple returns classical solutions, whereas solutions to the PDE with those BCs may only exist in a weak sense.", "Solution_15": "Here is one trick that may be useful or useless depending on what you really want to do next. If you set\r\n\\[ \\begin{aligned} u(x,t) & \\equal{} f(e^t)e^{i\\lambda e^t(x \\minus{} \\gamma)} \\\\\r\nv(x,t) & \\equal{} g(e^t)e^{i\\lambda e^tx} \\end{aligned}\r\n\\]\r\nwith any $ \\lambda\\in \\mathbb R$, you get the system\r\n\\[ \\begin{aligned} T f'(T) & \\equal{} (1 \\minus{} \\beta)f \\plus{} \\alpha e^{i\\lambda\\gamma T}g \\\\\r\nT g'(T) & \\equal{} (1 \\minus{} \\alpha)g \\plus{} \\beta e^{ \\minus{} i\\lambda\\gamma T}f \\end{aligned}\r\n\\]\r\nwhere $ T \\equal{} e^t$ is a new variable.\r\n(check it!)\r\n\r\nThus, you have a system of first order ODE's for each frequency (notice that the frequency itself changes with time, so to get the coefficients at frequency $ \\lambda$ at time $ t$, you need to start with the coefficients at frequency $ \\lambda e^{ \\minus{} t}$ at time $ 0$). \r\n\r\nThe system of ODE's can be \"solved\" by modified \"power series technique\" but I'm not sure this is of any value whatsoever.\r\n\r\nSo, in a sense, the \"explicit\" solution is here: take the Fourier transforms of $ u$ and $ v$ at $ t \\equal{} 0$ and solve the above system of ODE's to get the Fourier transforms at time $ t$. The big problem is whether the next question you were going to ask is tractable in these terms.", "Solution_16": "Thanks very much for your ideas, I'll mess around with it.", "Solution_17": "You are welcome. :) Feel free to ask more questions any time. The things that can be easily concluded from the \"Fourier method\" I proposed are that if $ u$ and $ v$ were $ C^\\infty$ on the real line and supported in $ [0,\\gamma]$ at $ t\\equal{}0$, then they will remain $ C^\\infty$ for all times $ t>0$ and will be supported in the same interval (so, there are no \"shock waves\" or other singularities in this case; the solution is classical; also, the boundary conditions you requested will hold automatically). It seems like with some extra effort one can show that the solutions tend to the steady state ones exponentially fast but I'll have to check many details to ensure that I, indeed, have a proof of it. Since I have no idea what you are really after, I'll stop here and forget about the problem for the time being..." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "let a,b,c,d,e,f be six positive integers satisfying abc=def.\r\nprove that : a(b^2+c^2)+d(e^2+f^2) is a composite number", "Solution_1": "Since $abc=def,$ we can write\r\n\\[ \\frac{ab}{de}=\\frac{f}{c}=\\frac{x}{y}, \\]\r\nwith $\\gcd\\left( x,y\\right) =1.$ There exist the positive integers $\\alpha,\\beta$ such that $ab=\\alpha x,$ $de=\\alpha y,$ $f=\\beta x$ and $c=\\beta y.$\r\n\r\nSimilarly, writing\r\n\\[ \\frac{ac}{df}=\\frac{e}{b}=\\frac{z}{t}, \\]\r\nwith $\\gcd\\left( z,t\\right) =1,$ we can find the positive integers $\\gamma,\\delta,$ such that $ac=\\gamma z,$ $df=\\gamma t,$ $e=\\delta z$ and $b=\\delta t.$\r\n\r\nWe have\r\n\\[ a\\left( b^{2}+c^{2}\\right) +d\\left( e^{2}+f^{2}\\right) =ab\\cdot b+ac\\cdot c+de\\cdot e+df\\cdot f= \\]\r\n\\[ =\\alpha x\\cdot\\delta t+\\gamma z\\cdot\\beta y+\\alpha y\\cdot\\delta z+\\gamma t\\cdot\\beta x=\\left( \\alpha\\delta+\\beta\\gamma\\right) \\left( xt+yz\\right) , \\]\r\na composite number.", "Solution_2": "[quote=\"tranthanhnam\"]let a,b,c,d,e,f be six positive integers satisfying abc=def.\nprove that : a(b^2+c^2)+d(e^2+f^2) is a composite number[/quote]\r\ntranthanhnam, why do you often post problems in ussue of recent Math & Magazine ?? :mad: \r\nThis problem isn't hard but you don't permit to do this.\r\nStop it right now", "Solution_3": "He did not claim it of his own. Perhaps he is seeking a different or at least a solution. I don't see a problem in that. If a solution is meant to be submitted and the due date is not reached yet, then there's a problem. I don't know for sure, so I won't judge anyone. Other than that, it's perfectly fine. Besides, I see the problems :D", "Solution_4": "[quote=\"tranthanhnam\"]Let $a,b,c,d,e,f$ be six positive integers satisfying $abc=def$. \n\nProve that : $a(b^2+c^2)+d(e^2+f^2)$ is a composite number.[/quote]$a(b^2+c^2)+d(e^2+f^2)=\\frac{(de^2+ac^2)(de^2+ab^2)+(def+abc)(def-abc)}{de^2}$\n\n$=\\frac{(de^2+ac^2)(de^2+ab^2)}{de^2}.$", "Solution_5": "@Ji Chen: but how do you know that $\\frac{(de^2+ac^2)(de^2+ab^2)}{de^2}$ is not a prime?", "Solution_6": "That is easy, each of the factors $de^2+ac^2$ and $de^2+ab^2$ at the numerator is strictly larger than the denominator $de^2$, so none can disappear when dividing the numerator by the denominator.", "Solution_7": "Oh thanks... Didn't notice that... :blush:" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "The following proposition is true or false:\r\nLet n is given positive integer, we can find 8 positice integer numbers a, b, c, d, e, f, g, h so that: $n=\\frac{2^a-2^b}{2^c-2^d}.\\frac{2^e-2^f}{2^g-2^h}$", "Solution_1": "I don't think it is from the German TST 2005. Check here: http://www.mathlinks.ro/Forum/resources-64-54-2005.html", "Solution_2": "It's German TST 2006 actually :) http://www.mathlinks.ro/Forum/viewtopic.php?p=383965#p383965" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "Prove that there exist [b]ten[/b] triangles in a plane, no two of them having all the vertices same, such that all of them have a common circumcircle.\r\nDeduce the same problem for [b]twenty[/b] triangles also.", "Solution_1": "I'm not sure, but could we construct a circle, and draw 10 distinct triangles in the circle? (30 random points on the circumference connecting groups of 3)", "Solution_2": "Really? Are you absolutely sure.....cause I have a solution. Here it follows:\r\nConsider any circle in the plane. You can take [b]five[/b] distinct points on it. With these point you can construct exactly ten triangles, no two the same(5C3=10), and hence is the problem.\r\nFor twenty triangles, take [b]six [/b]points and continue(6C3=20)\r\nI am not sure if there is any gross mistake in this proof. Please point out if there is.", "Solution_3": "Sunkern_sunflora: From what you say, I think you meant that there should be exactly ten triangles possible through the points, and all of them must have a common circumcircle. Or it might be what most ppl including SimonM thought. In either case the problem looks trivial.", "Solution_4": "I guess I understand. Sorry everyone. Thanx Akashnil." } { "Tag": [ "Olimpiada de matematicas" ], "Problem": "[color=blue]Hay una desigualdad que encontr\u00e9 emp\u00edricamente por casualidad, pero no logro demostrarla.\n\nSean [b][i]a[/i]1, [i]a[/i]2, \u2026 [i]an[/i][/b] reales positivos tales que el producto de todos ellos es igual a [b]1[/b]. Demostrar que su sumatoria es mayor que [b][i]n[/i]-1[/b].[/color]", "Solution_1": "Bueno, una desigualdad facil.\r\nPor $MG-MA$ tenemos que\r\n(a1+a2+...+an)/n >= (a1a2...an)^1/n = 1\r\nentonces a1+a2+...+an >= n >n-1\r\n :D", "Solution_2": "[color=blue]\u00a1Gracias! \u00bfC\u00f3mo no me pude dar cuenta de lo sencilla que era la soluci\u00f3n?\n\nDe paso, te agradezco nuevamente porque tambi\u00e9n conjeturaba que era mayor o igual que [b][i]n[/i][/b] dicha suma.\n\nGracias, nuevamente.[/color]", "Solution_3": "Uno facil.\r\nsean, $a,b,c,d$ numeros reales positivos.\r\ndonde: $a,b,c < d$.\r\ndemostrar que :\r\n $d^{3}> d(ab+bc+ca)-3abc$.", "Solution_4": "[color=blue]Estuve intentando resolver tu problema, aev5peru, pero no llego a nada \u00fatil. Mis acotaciones no me permiten llegar a la soluci\u00f3n. \u00bfMe puedes dar alguna pista?\n\nSaludos.[/color]", "Solution_5": "[quote=\"R.G.A.M.\"][color=blue]Estuve intentando resolver tu problema, aev5peru, pero no llego a nada \u00fatil. Mis acotaciones no me permiten llegar a la soluci\u00f3n. \u00bfMe puedes dar alguna pista?\n\nSaludos.[/color][/quote]\r\nese problema se me ocurrio desps de ver el siguiente.\r\nsi $d>a,b,c>0$ demostrar que $d^{2}>d(a+b+c)-(ab+bc+ca)$,\r\nespero que esta ayuda sea suficiente.\r\nchau.", "Solution_6": "[color=blue]Disc\u00falpame por no haber preguntado antes. Cuando dices que $d>a;b;c>0$, \u00bfest\u00e1s afirmando que es $d$ es estrictamente mayor que los dem\u00e1s t\u00e9rminos? Hasta ahora supuse como cierto lo que acabo de mencionar.\n\nMuchas gracias.[/color]", "Solution_7": "[quote=\"R.G.A.M.\"][color=blue]Disc\u00falpame por no haber preguntado antes. Cuando dices que $d>a;b;c>0$, \u00bfest\u00e1s afirmando que es $d$ es estrictamente mayor que los dem\u00e1s t\u00e9rminos? Hasta ahora supuse como cierto lo que acabo de mencionar.\n\nMuchas gracias.[/color][/quote] SI, eso sirve vastante en mi demostracion." } { "Tag": [], "Problem": "Let $x,y,z$ be positive integers. Consider the value of\r\n\r\n$\\alpha = |\\frac{x-y}{x+y}+\\frac{y-z}{y+z}+\\frac{z-x}{z+x}|$\r\n\r\na) Prove that $\\alpha <1$\r\n\r\nb)Suppose now that $x,y,z$ are the side lengths of a triangle (which may be degenerate).\r\n\r\n(i) Prove that $\\alpha <\\frac{1}{8}$\r\n\r\n(ii) Find the absolute maximum of $\\alpha$.", "Solution_1": "1. you mean prove $0< \\alpha < 1$", "Solution_2": "[quote=\"courtrigrad\"]1. you mean prove $0< \\alpha < 1$[/quote]\r\n\r\nno.....consider $x=y=z$ \r\n\r\nAlso the >0 is implied since it is absolute value....", "Solution_3": "So, any suggestions?" } { "Tag": [ "calculus", "number theory proposed", "number theory" ], "Problem": "Let $ m,n,k$ be positive integers such that\r\n$ mn\\equal{}k^{2}\\plus{}k\\plus{}3$\r\nProve that there exist odd integers $ x,y$ satisfy one of the equations\r\n$ x^{2}\\plus{}11y^{2}\\equal{}4m$ and $ x^{2}\\plus{}11y^{2}\\equal{}4n$", "Solution_1": "[quote=\"ThangTongHopK41\"]Let $ m,n,k$ be positive integers such that\n$ mn = k^{2} + k + 3$\nProve that there exist odd integers $ x,y$ satisfy one of the equations\n$ x^{2} + 11y^{2} = 4m$ and $ x^{2} + 11y^{2} = 4n$[/quote]\r\nIndeed it is a very nice problem . You can see a solution of iura in this links :\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=407830#407830\r\nAlthough I want to post an other solution . \r\n[i]\nNote 1: If $ p$ is a prime satisfy condition :exist two relative prime number $ (a,b)$ such that $ p|a^2+11b^2$ then $ 4p$ can represent in the form $ x^2+11y^2$\n[/i]\r\nProof : Because $ p|a^2+11a^2$ so exist an integer $ k$ such that $ p|k^2+11$\r\nLet $ m=[\\sqrt{p}]$\r\nConsider set : \r\n\\[ A=\\{x+ky|x,y\\in (0,..,m)\\}\r\n\\]\r\nBecause $ |A|>p$ so exist two different elements $ (x_1,y_1),(x_2,y_2)$ in A such that \r\n$ x_1+ay_1\\equiv x_2+ay_2 (\\mod p)$\r\nLet $ u=|x_1-x_2|,v=|y_1-y_2|$ and easy to check that $ u,v$ is not all zero . \r\nThus $ u^2\\equiv k^2v^2(\\mod p)$ \r\nAnd because $ k^2\\equiv -11 (\\mod p)$ so we have $ p|u^2+11v^2$ .And because $ u^2,v^2$ are less than $ p$ so it can take only one follow value $ {p,2p,3p,...,10\\}}$ .By considering modulo $ 4$ and modulo $ 7$ we get $ u^2+11v^2\\in \\{p,3p,4p,5p,9p\\}$ . If $ u^2+11v^2$ is equal $ p,4p$ then statement true. We only need to consider other cases . It can be done by some algebra calculus . For example when $ 3p=u^2+11v^2$\r\nThen we have $ u=3t+1,v=3j+1$ (notice that sign $ +$ can be replace by $ -$ ) \r\nThus $ p=3k^2+2k+11(3j^2+2j)+4$ . It is easy to check that : \r\n\\[ 4p=(k+11j+4)^2+11(k-j)^2\r\n\\]\r\nIn case $ u^2+11v^2=5p$ ,it can be done by same method as above represent . The case $ 9p=u^2+5v^2$ is the longest ,but you can check that it is true . \r\nTherefore notice 1 has been proved . \r\nNote 2 : If $ 4m$ and $ 4n$ can represent in the form $ u^2+11v^2$ then $ 4mn$ can represent in this form . \r\nNow consider our problem . $ 4mn=(2k+1)^2+11$\r\nBy the lemma $ 4m,4n$ can represent in the form $ u^2+11v^2$\r\nSuppose that $ 4m=u_1^2+11v_1^2$ and $ 4n=v_2^2+11u_2^2$\r\nIf at least of two pairs $ (u_1,v_1)$ $ (v_2,u_2)$ has odd elements then statement is true . And because $ u_1,v_1$ and $ u_2,v_2$ has same parity so we only need to consider the case : \r\n$ u_1\\equiv v_1 \\equiv 0 (\\mod 2)$ and $ v_2\\equiv u_2\\equiv 0 (\\mod 2)$\r\nThus $ m,n$ can represent in the form $ u^2+11v^2$ .From condition it is easy to check that $ m,n$ are odd . Therefore exist two odd integer $ x,y$ such that $ 4m=x^2+11y^2$\r\nProblem claim .", "Solution_2": "In principle, this is just an easy question of arithmetic of $ \\mathbb Z[\\frac{1\\plus{}\\sqrt{\\minus{}11}}{2}]$ and it's suborder $ \\mathbb Z[\\sqrt{\\minus{}11}]$.\r\nMuch further ,it is possible to classify all numbers of type $ x^2\\plus{}xy\\plus{}3y^2$ or $ x^2\\plus{}11y^2$. For how to do this, see http://www.mathlinks.ro/viewtopic.php?t=232507 . This directly applies to $ x^2\\plus{}xy\\plus{}3y^2$, and $ x^2\\plus{}11y^2$ is just a little more work.", "Solution_3": "Let $ m,n,k$ be positive integers such that\r\n$ mn \\equal{} k^{2} \\plus{} k \\plus{} 3$\r\nProve that there exist odd integers $ x,y$ satisfy one of the equations\r\n$ x^{2} \\plus{} 11y^{2} \\equal{} 4m$ and $ x^{2} \\plus{} 11y^{2} \\equal{} 4n$\r\n\r\nIndeed it is a very nice problem . You can see a solution of iura in this links :\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=407830#407830\r\nAlthough I want to post an other solution . \r\n[i]\nNote 1: If $ p$ is a prime satisfy condition :exist two relative prime number $ (a,b)$ such that $ p|a^2 \\plus{} 11b^2$ then $ 4p$ can represent in the form $ x^2 \\plus{} 11y^2$\n[/i]\r\nProof : Because $ p|a^2 \\plus{} 11b^2$ so exist an integer $ k$ such that $ p|k^2 \\plus{} 11$\r\nLet $ m \\equal{} [\\sqrt {p}]$\r\nConsider set :\r\n\\[ A \\equal{} \\{x \\plus{} ky|x,y\\in (0,..,m)\\}\\]\r\nBecause $ |A| > p$ so exist two different elements $ (x_1,y_1),(x_2,y_2)$ in A such that \r\n$ x_1 \\plus{} ky_1\\equiv x_2 \\plus{} ky_2 (\\mod p)$\r\nLet $ u \\equal{} |x_1 \\minus{} x_2|,v \\equal{} |y_1 \\minus{} y_2|$ and easy to check that $ u,v$ is not all zero . \r\nThus $ u^2\\equiv k^2v^2(\\mod p)$ \r\nAnd because $ k^2\\equiv \\minus{} 11 (\\mod p)$ so we have $ p|u^2 \\plus{} 11v^2$ .And because $ u^2,v^2$ are less than $ p$ so it can take only one follow value $ \\{p,2p,3p,...,10p \\}$ .By considering modulo $ 4$ and modulo $ 7$ we get $ u^2 \\plus{} 11v^2\\in \\{p,3p,4p,5p,9p\\}$ . If $ u^2 \\plus{} 11v^2$ is equal $ p,4p$ then statement true. We only need to consider other cases . It can be done by some algebra calculus . For example when $ 3p \\equal{} u^2 \\plus{} 11v^2$\r\nThen we have $ u \\equal{} 3t \\plus{} 1,v \\equal{} 3j \\plus{} 1$ (notice that sign $ \\plus{}$ can be replaced by $ \\minus{}$ ) \r\nThus $ p \\equal{} 3k^2 \\plus{} 2k \\plus{} 11(3j^2 \\plus{} 2j) \\plus{} 4$ . It is easy to check that :\r\n\\[ 4p \\equal{} (k \\plus{} 11j \\plus{} 4)^2 \\plus{} 11(k \\minus{} j)^2\\]\r\nIn case $ u^2 \\plus{} 11v^2 \\equal{} 5p$ ,it can be done by same method as above represent . The case $ 9p \\equal{} u^2 \\plus{} 11v^2$ is the longest ,but you can check that it is true . \r\nTherefore notice 1 has been proved . \r\nNote 2 : If $ 4m$ and $ 4n$ can represent in the form $ u^2 \\plus{} 11v^2$ then $ 4mn$ can represent in this form . \r\nNow consider our problem . $ 4mn \\equal{} (2k \\plus{} 1)^2 \\plus{} 11$\r\nBy the lemma $ 4m,4n$ can represent in the form $ u^2 \\plus{} 11v^2$\r\nSuppose that $ 4m \\equal{} u_1^2 \\plus{} 11v_1^2$ and $ 4n \\equal{} v_2^2 \\plus{} 11u_2^2$\r\nIf at least of two pairs $ (u_1,v_1)$ $ (v_2,u_2)$ has odd elements then statement is true . And because $ u_1,v_1$ and $ u_2,v_2$ has same parity so we only need to consider the case : \r\n$ u_1\\equiv v_1 \\equiv 0 (\\mod 2)$ and $ v_2\\equiv u_2\\equiv 0 (\\mod 2)$\r\nThus $ m,n$ can represent in the form $ u^2 \\plus{} 11v^2$ .From condition it is easy to check that $ m,n$ are odd . Therefore exist two odd integer $ x,y$ such that $ 4m \\equal{} x^2 \\plus{} 11y^2$\r\nProblem claim .\r\n\r\n There are too many mistakes in that solution . I have edited some of them" } { "Tag": [ "geometry" ], "Problem": "A right triangle has an inscribed circle cutting the hypotenuse into lengths of 7 and 11. What is the area of the triangle?\r\n\r\nCan someone show me how to do this hard problem?", "Solution_1": "Hint: From a point, the two tangents to the same circle are of equal length", "Solution_2": "From warut suk's hint, I guess then, since the tangents from one point is equal, you can set up the equation as $(7+x)^2+(11+x)^2=18^2$. Then solve for $x$ and then find out each leg right? Thus, you have $x^2+18x-77=0--->x=3.569805...--->\\frac{(14.569805...)(10.56980...)}{2}$, thus the area is $77$.", "Solution_3": "[quote=\"Iversonfan2005\"]From warut suk's hint, I guess then, since the tangents from one point is equal, you can set up the equation as $(7+x)^2+(11+x)^2=18^2$. Then solve for $x$ and then find out each leg right? Thus, you have $x^2+18x-77=0--->x=3.569805...--->\\frac{(14.569805...)(10.56980...)}{2}$, thus the area is $77$.[/quote]\r\n\r\n\r\nI write $(7+x)^2+(11+x)^2=18^2$ as $(11+x-7-x)^2+2(7+x)(11+x)=18^2$ \r\n\r\nHence the area $\\frac{1}{2}(7 + x)(11 + x)=\\frac{{(18 - 4)(18 + 4)}}{4}=77$\r\n\r\nI think this works out to be simpler. :)", "Solution_4": "Ok, i guess if the circle would have cut the hypotenuse in 9:9 the sides would have been equal....and 45 degrees.\r\nthe smaller angle=a\r\nhypotenuse=c\r\na=45-2*45/9=35 deg\r\nThats why i guess that the area is:\r\nc*cos35*c*sin35/2=76,11510235\r\n\r\nIt may be terribly wrong, but its almost close to 77 anyway.", "Solution_5": "I think vidyamanohars's way is the easiest. Same as mine.", "Solution_6": "Yes, Iversonfan2005, you should think carefully before solving for x! :)", "Solution_7": "[quote=\"warut_suk\"]Yes, Iversonfan2005, you should think carefully before solving for x! :)[/quote]\r\n\r\noh well, still got the right answer :P even though it got messy :?", "Solution_8": "Yeah, my method was similar...\r\nLet the length of the shorter leg be $a$. Then we're trying to find the area, which is $\\frac{a(a+4)}{2}$. We know that\r\n$a^2+(a+4)^2=324$\r\n$2a^2+8a+16=324$\r\n$2a(a+4)=308$\r\n$\\frac{a(a+4)}{2}=\\frac{308}{4}=77$" } { "Tag": [ "Support", "geometry" ], "Problem": "assume that $C$ is a convex polygon ,for any edge like $AB$ we consider the vertex which has the most distance to this edge like $D$ ,and then prove that $\\sum_{A,B are vertex} 0.$ How about $ a<0$?\r\n\r\nLet's wait others solution.", "Solution_8": "Can anyone solve this problem?", "Solution_9": "[quote=\"kunny\"][quote=\"i_like_pie\"][hide=\"#1 Solution\"]The area under the parabola is $ \\int_\\alpha^\\beta ax^2 + bx + c\\,\\mathrm{d}x$, and the area under $ y = ux + v$ is $ \\int_\\alpha^\\beta ux + v\\,\\mathrm{d}x$.\n\nAs Carcul has already said, $ u = a\\left(\\alpha + \\beta\\right) + b$ and $ v = c - a\\alpha\\beta$.\n\nThe second integral now becomes $ \\int_\\alpha^\\beta x\\left(a\\left(\\alpha + \\beta\\right) + b\\right) + c - a\\alpha\\beta\\,\\mathrm{d}x$.\n\\begin{align*}\\int_\\alpha^\\beta x\\left(a\\left(\\alpha + \\beta\\right) + b\\right) + c - a\\alpha\\beta\\,\\mathrm{d}x & = \\left[\\frac {x^2\\left(a\\left(\\alpha + \\beta\\right) + b\\right)}{2} + cx - ax\\alpha\\beta\\right]_\\alpha^\\beta \\\\\n \\\\\n& = \\frac {\\beta^2\\left(a\\left(\\alpha + \\beta\\right) + b\\right)}{2} + c\\beta - a\\alpha\\beta^2 - \\frac {\\alpha^2\\left(a\\left(\\alpha + \\beta\\right) + b\\right)}{2} - c\\alpha + a\\alpha^2\\beta \\\\\n \\\\\n& = \\frac {a\\alpha\\beta^2 + a\\alpha\\beta^3 + b\\beta^2}{2} + c\\beta - a\\alpha\\beta^2 - \\frac {a\\alpha^3 + a\\alpha^2\\beta + b\\alpha^2}{2} - c\\alpha + a\\alpha^2\\beta\\qquad (1)\\end{align*}\n\n\\begin{align*}\\int_\\alpha^\\beta ax^2 + bx + c\\,\\mathrm{d}x & = \\left[\\frac {ax^3}{3} + \\frac {bx^2}{2} + cx\\right]_\\alpha^\\beta \\\\\n \\\\\n& = \\frac {a\\beta^3}{3} + \\frac {b\\beta^2}{2} + c\\beta - \\frac {a\\alpha^3}{3} - \\frac {b\\alpha^2}{2} - c\\alpha\\qquad(2)\\end{align*}\n(1) can be re-written as $ \\frac {3a\\alpha\\beta^2 + 3a\\beta^3 + 3b\\beta^2 + 6c\\beta - 6a\\alpha\\beta^2 - 3a\\alpha^3 - 3a\\alpha^2\\beta + 3b\\alpha^2 - 6c\\alpha + 6a\\alpha^2\\beta}{6}$.\n\n(2) can be re-written as $ \\frac {2a\\beta^3 + 3b\\beta^2 + 6c\\beta - 2a\\alpha^3 - 3b\\alpha^2 - 6c\\alpha}{6}$.\n\nSubtracting the second overly complicated expression from the first yields $ \\frac {3a\\alpha\\beta^2 + a\\beta^3 - 6a\\alpha\\beta^2 - a\\alpha^3 - 3a\\alpha^2\\beta + 6a\\alpha^2\\beta}{6}$.\n\nThis can then be further simplified to $ \\frac {a\\left( - 3\\alpha\\beta^2 + \\beta^3 - \\alpha^3 + 3\\alpha^2\\beta\\right)}{6}$.\n\nFinally, factoring the numerator gives $ A = \\frac {a\\left(\\beta - \\alpha\\right)^3}{6}$.[/hide][/quote]\n\nGood job, i_like_pie :) You showed in case of $ a > 0.$ How about $ a < 0$?\n\nLet's wait others solution.[/quote]\nedit-[hide=\"solution\"]\ntake the integral of the parabola minus the integral of the line for $ a<0$\ni don't know how to do integral notation in latex.\ni got $ \\frac {a(\\alpha - \\beta) ^3}{6}$ for the case of $ a < 0$\n[/hide]", "Solution_10": "No one seems to be posted solution any more, so I post the solution of problem 1.\r\n\r\nFrom the condition, we can write $ ax^2 \\plus{} bx \\plus{} c \\minus{} (ux \\plus{} v) \\equal{} a(x \\minus{} \\alpha)(x \\minus{} \\beta)\\ (\\alpha < \\beta)$, thus desired area is $ S \\equal{} \\minus{} a\\int_{\\alpha }^{\\beta} (x \\minus{} \\alpha )(x \\minus{} \\beta)dx \\equal{} \\frac {a}{6}(\\beta \\minus{} \\alpha )^3$ for $ a > 0.$ Similarly we have $ S \\equal{} \\minus{} \\frac {a}{6}(\\beta \\minus{} \\alpha )^3$ for $ a < 0.$ As the result, We can find directly the area as $ S \\equal{} \\left|\\int_{\\alpha}^{\\beta} a(x \\minus{} \\alpha )(x \\minus{} \\beta)\\ dx\\right| \\equal{} \\frac {|a|}{6}(\\beta \\minus{} \\alpha )^3$.\r\n\r\nNote: $ \\boxed{\\int_{\\alpha}^{\\beta} (x \\minus{} \\alpha )(x \\minus{} \\beta)\\ dx \\equal{} \\minus{} \\frac {1}{6}(\\beta \\minus{} \\alpha )^3},$ which is written in Japanese high school mathematics text today.\r\n\r\nSo can we for Problem 2." } { "Tag": [], "Problem": "Fie $\\ a,b,c,d$ naturale consecutive.Sa se arate ca\r\n $\\ [\\frac{\\sqrt a+\\sqrt b+\\sqrt c+\\sqrt d}{2}]$=$\\ [\\sqrt{a+b+c+d}]$, unde $\\ [x]$ este partea intreaga a numarului real $\\ x$.\r\n $\\ Dana\\ Piciu,\\ Craiova$", "Solution_1": "enuntul e lafel cu\r\n$\\left[\\frac{\\sqrt{n} + \\sqrt{n+1}+\\sqrt{n+2}+\\sqrt{n+3}}{2}\\right]=\\left[\\sqrt{4n+6}\\right]$ pentru n natural.\r\ndar oricare n natural el e cuprins intre 2 patrate\r\n$k^2\\leq n \\leq (k+1)^2$ (se observa $[\\sqrt{n}]=k$)\r\nse poate observa ca $LHS \\geq 2k$\r\nse poate observa deasemeni ca $RHS \\geq 2k$\r\nP.R.A. daca $RHS=2k+1 \\ \\Rightarrow (2k+1)^2 \\leq 4k^2 +6 \\ \\Rightarrow$\r\n contradictie pentru k suficient de mare.\r\nP.R.A daca $LHS=2k+1 \\ \\Rightarrow (2k+1)^2 \\leq LHS \\Rightarrow$\r\n -//-\r\n\r\ndeci de la un k suficient de mare(maximul din cele doua parti de mai sus)\r\nrezulta ca $LHS=RHS=2k$.\r\nsper sa fie bine", "Solution_2": "Sorry, dar ce inseamna $\\ LHS$ si $\\ RHS$?", "Solution_3": "LHS=membrul stanga\r\nRHS=membrul drept" } { "Tag": [ "quadratics", "algebra", "quadratic formula" ], "Problem": "Given $ x \\equal{} \\bar{x} \\plus{} \\beta \\sqrt{D^2 \\plus{} \\bar{x}^2}$, how does one obtain $ \\bar{x} \\equal{} \\gamma^2x \\minus{} \\beta\\gamma\\sqrt{D^2\\plus{}(\\gamma x)^2}$? The simplification $ \\gamma \\equal{} \\frac {1}{\\sqrt{1 \\minus{} \\beta^2}}$ may be used.\r\n\r\nThe problem i'm facing is that i don't get the correct required expression..", "Solution_1": "[quote=\"aidan\"]Given $ x \\equal{} \\bar{x} \\plus{} \\beta \\sqrt {D^2 \\plus{} \\bar{x}^2}$, how does one obtain $ \\bar{x} \\equal{} \\gamma^2x \\minus{} \\beta\\gamma\\sqrt {D^2 \\plus{} (\\gamma x)^2}$? The simplification $ \\gamma \\equal{} \\frac {1}{\\sqrt {1 \\minus{} \\beta^2}}$ may be used.\n\nThe problem i'm facing is that i don't get the correct required expression..[/quote]\r\nWriting $ x \\equal{} \\bar{x} \\plus{} \\beta \\sqrt {D^2 \\plus{} \\bar{x}^2}$ as \r\n$ x \\minus{} \\bar{x} \\equal{} \\beta \\sqrt {D^2 \\plus{} \\bar{x}^2}$ and squaring on both sides, we get\r\n[hide]$ (x \\minus{} \\bar{x})^2 \\equal{} \\beta ^2 ({D^2 \\plus{} \\bar{x}^2})$\ncollecting all the like terms, we get\n$ (1\\minus{}\\beta ^2) \\bar{x}^2 \\minus{}2x\\bar{x}\\plus{}x^2\\minus{}\\beta ^2 D^2\\equal{} 0$\nDividing throughout by $ (1\\minus{}\\beta ^2)$ and using $ \\gamma \\equal{} \\frac {1}{\\sqrt {1 \\minus{} \\beta^2}}$, we have\n$ \\bar{x}^2 \\minus{}2x\\gamma ^2\\bar{x}\\plus{}\\gamma ^2(x^2\\minus{}\\beta ^2 D^2)\\equal{} 0$\nCompleting the perfect sqaure, we have \n$ (\\bar{x} \\minus{}x\\gamma ^2)^2\\equal{}\\gamma ^4x^2\\minus{}\\gamma ^2x^2\\plus{}\\gamma ^2\\beta ^2 D^2\\equal{}\\gamma ^2x^2(\\gamma ^2\\minus{}1)\\plus{}\\gamma ^2\\beta ^2 D^2$\nObserving $ \\gamma ^2\\minus{}1\\equal{}\\gamma ^2\\beta ^2$, we can write \n$ (\\bar{x} \\minus{}x\\gamma ^2)^2\\equal{}\\gamma ^4\\beta ^2x^2\\plus{}\\gamma ^2\\beta ^2 D^2$\nFurther simplification gives\n$ (\\bar{x} \\minus{}x\\gamma ^2)^2\\equal{}\\gamma ^2\\beta ^2((\\gamma x)^2\\plus{} D^2)$\nTaking the square roots on both sides, we can get [/hide]the required expression as one of the answers.", "Solution_2": "Oh.. I didn't divide throughout by $ \\frac {1}{\\sqrt{1 \\minus{} \\beta^2}}$ first..I just applied the quadratic formula directly. Thanks :)" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "Giving: $ F(x)\\equal{}\\frac{ln(1\\plus{}x)}{x}$ $ \\forall x\\neq0$ and $ F(0)\\equal{}1$.\r\n\r\nevaluate this limit by using that: $ e^{x}\\equal{}1\\plus{}x\\plus{}\\frac{x^{2}}{2}\\plus{}x^{2}t(x)$ with $ \\lim_{x\\to0}t(x)\\equal{}0$ :\r\n\r\n$ \\lim_{x\\to0}\\frac{ln(x\\plus{}1)\\minus{}1}{x}$\r\n\r\nthanx!.", "Solution_1": "hello, by the rule of L'Hospital we get\r\n$ \\lim_{x \\to 0}\\frac{\\ln(x\\plus{}1)\\minus{}1}{x}\\equal{}\\lim_{x \\to 0}\\frac{1}{1\\plus{}x}\\equal{}1$.\r\nSonnhard.", "Solution_2": "$ \\lim_{x\\to0}\\frac{ln(x\\plus{}1)\\minus{}1}{x} \\equal{} \\left|\\begin{array}{l} y\\equal{}\\ln(1\\plus{}x) \\to 0 \\\\ \\end{array}\\right| \\equal{} \\lim_{y\\to 0} \\frac{y\\minus{}1}{e^y\\minus{}1} \\equal{} \\lim_{y\\to 0} \\frac{y\\minus{}1}{1\\plus{}y\\plus{}y^2/2 \\plus{} y^2o(1)\\minus{}1} \\equal{} \\lim_{y\\to 0} \\frac{y\\minus{}1}{y\\plus{}yo(1)} \\equal{} \\lim_{y\\to 0} \\frac{1\\minus{}\\frac{1}{y}}{1\\plus{}o(1)} \\equal{}1$", "Solution_3": "This limit does not exist, does it? It goes to $ \\plus{}\\infty$ from the left and $ \\minus{}\\infty$ from the right. I thought L'Hopstal's rule was only applicable for limits of the indeterminate forms $ 0/0$ or $ \\infty/\\infty$. This one is not even an indeterminate form. That said, I am quite sure I am missing something important. Could someone point it out?", "Solution_4": "Haaaaaaa)))))))\r\n[b]KBriggs[/b] you are right )))\r\nthis limit is not exists. What a stupid mistake i made ) \r\n\\[ \\lim_{y\\to 0}\\frac{y\\minus{}1}{e^{y}\\minus{}1} \\equal{} \\frac{\\minus{}1}{0} \\equal{}\\infty\\]\r\n\r\ni garble this limit with $ \\lim_{x\\to0}\\frac{ln(1\\plus{}x)}{x}$... and i was inattentive )) :blush: \r\nThanks [b]KBriggs[/b]", "Solution_5": "I don't understand why defining F(x) was necessary." } { "Tag": [ "trigonometry", "geometry", "algebra", "polynomial", "logarithms", "analytic geometry", "graphing lines" ], "Problem": "1) What is the leftmost digit of $2006^{2006}$\r\n\r\n2) In $\\triangle{ABC}$, $AD$ and $AE$ trisect $\\angle{BAC}$. $BD=2, DE=3, EC=6$, what is the length of the shortest side of $\\triangle{ABC}$?\r\n\r\n3) Find the largest value of $\\frac{y}{x}$ if $(x-3)^2+(y-3)^2=6$\r\n\r\n4) $z^4+az^3+bz^2+cz+d$ has only complex roots which lie on a circle in a complex plane centered at $0+0i$ and has radius $1$. What is the sum of the reciprocals of the roots in terms of $a$, $b$, $c$, or $d$?", "Solution_1": "[hide=\"Number 2\"]\n[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=578[/img]\nBy the angle bisector theorem we have \\[ 3x=2e \\ \\text{and} \\ y=2d. \\]\n\nFrom the Law of Cosines and since $\\cos \\angle AEC=-\\cos \\angle AEB$ and $\\cos \\angle ADB=-\\cos \\angle ADC$\n\n\\begin{eqnarray} x^2 &=& d^2+4+4d\\cos\\angle ADC\\\\ e^2 &=& d^2+9-6d\\cos \\angle ADC\\\\ y^2 &=& e^2+36-12e\\cos \\angle AEC\\\\ d^2 &=& e^2+9+6e\\cos \\angle AEC\\\\ \\end{eqnarray}\n\nSo then\n\n\\[ 3(1)+2(2) = 3x^2+3e^2=5d^2+30 \\]\n\\[ (3)+2(4) = y^2 + 2d^2 = 3e^2 + 54. \\]\n\nThen plug $x=\\frac 23e$ in the first and $y=2d$ in the second to get\n\n\\begin{eqnarray*} 2d^2-e^2 &=& 18\\\\ -5d^2 +\\frac{10}3e^2 &=& 30 \\Rightarrow e = 3\\sqrt{10} && d=3\\sqrt{6} \\end{eqnarray*}\n\nSo then $x=\\frac 23 e=2\\sqrt{10}=\\sqrt{40}$ and $y=2d=6\\sqrt{6}=\\sqrt{216}$. Clearly $AB=x=2\\sqrt{10}$ is the smallest.\n[/hide]\r\n\r\nI'll leave the rest for other people to solve...", "Solution_2": "[quote=\"joml88\"][hide=\"Number 2\"]\n[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=578[/img]\nBy the angle bisector theorem we have \\[ 3x=2e \\ \\text{and} \\ y=2d. \\]\n\nFrom the Law of Cosines and since $\\cos \\angle AEC=-\\cos \\angle AEB$ and $\\cos \\angle ADB=-\\cos \\angle ADC$\n\n\\begin{eqnarray} x^2 &=& d^2+4+4d\\cos\\angle ADC\\\\ e^2 &=& d^2+9-6d\\cos \\angle ADC\\\\ y^2 &=& e^2+36-12e\\cos \\angle AEC\\\\ d^2 &=& e^2+9+6e\\cos \\angle AEC\\\\ \\end{eqnarray}\n\nSo then\n\n\\[ 3(1)+2(2) = 3x^2+3e^2=5d^2+30 \\]\n\\[ (3)+2(4) = y^2 + 2d^2 = 3e^2 + 54. \\]\n\nThen plug $x=\\frac 23e$ in the first and $y=2d$ in the second to get\n\n\\begin{eqnarray*} 2d^2-e^2 &=& 18\\\\ -5d^2 +\\frac{10}3e^2 &=& 30 \\Rightarrow e = 3\\sqrt{10} && d=3\\sqrt{6} \\end{eqnarray*}\n\nSo then $x=\\frac 23 e=2\\sqrt{10}=\\sqrt{40}$ and $y=2d=6\\sqrt{6}=\\sqrt{216}$. Clearly $AB=x=2\\sqrt{10}$ is the smallest.\n[/hide]\n\nI'll leave the rest for other people to solve...[/quote]\r\n\r\n :yup: except I didn't use trig to solve it, I simply used angle bisector theorem, and found the lengths of all the bisectors, etc..", "Solution_3": "How'd you find the lengths of the bisectors....Stewart's? If so, I wouldn't realy call that simpler :P Look at where I added up the numbered equations (i.e. $3(1)+2(2)$, etc). Look at what those equal...they look suspiciously like what Stewart's would give you (in fact they are what you would get :D ). Indeed, you can actually prove Stewart's with LoC (just do what I did but with variables instead).", "Solution_4": "[quote=\"joml88\"]How'd you find the lengths of the bisectors....Stewart's? If so, I wouldn't realy call that simpler :P Look at where I added up the numbered equations (i.e. $3(1)+2(2)$, etc). Look at what those equal...they look suspiciously like what Stewart's would give you (in fact they are what you would get :D ). Indeed, you can actually prove Stewart's with LoC (just do what I did but with variables instead).[/quote]\r\n\r\n[hide=\"length of bisectors\"]Let $AB, AD, AE, AC$ be $w,x,y,z$ respectively. Using angle bisector theorem, we know that $2y=3w$, $x^2+6=wy$, $z=2x$, $y^2+18=xz$, so we have a bunch of equations now. We can get the system of equations: $\\frac{2}{3}y^2-x^2+6=0$ and $y^2+18-2x^2=0$. So we can get $x=\\sqrt{54}, y=\\sqrt{90}$, thus $w=2\\sqrt{10}$ and $z=6\\sqrt{6}$, so $\\boxed{2\\sqrt{10}}$ is the answer[/hide]\r\n\r\nI would've used Stewart's, but I almost never use it in geometry problems (hard to memorize anyway)", "Solution_5": "Here's a problem gauss202 posted but noone solved yet..\r\n\r\n$1+11+111+....+11....11111$ (2002 ones) has how many ones?\r\n[hide=\"solution\"]\n\nNotice that it is a finite geometric series for each one of the terms $(1+10+100...)$ etc. So we can put this as $\\frac{1+10}{1+10}+\\frac{1+10+100}{1+10}+....+\\frac{1+10+...1000...}{1+10}=\\frac{2002+11111...11110}{11}$, which contains 2002 ones. So we have $182+\\frac{11111...110}{11}$ and then doing long division, we see that the quotient has 2002 digits and is in the form $101010101...10$ which contains $1001$ ones. and adding 182, it still has $\\boxed{1001} ones$[/hide]", "Solution_6": "Oh, well...I don't consider $x^2+6=wy$ to be the angle bisector theorem. I consider it to be a combination of angle biscector and stewart's/LoC. But, whatever, it doesn't really matter which way you look at it :P", "Solution_7": "If anybody needs hints, here they are:\r\n\r\n[hide=\"hint problem 1\"]logs[/hide]\n\n[hide=\"hint problem 3\"]graph[/hide]\n\n[hide=\"hint problem 4\"]transform the polynomial[/hide]", "Solution_8": "About number 4...does it give you extraneous information. Because I got an answer but I don't think I used the facts that the roots are on the unit circle... (I might've missed something while doing the problem)", "Solution_9": "[quote=\"joml88\"]About number 4...does it give you extraneous information. Because I got an answer but I don't think I used the facts that the roots are on the unit circle... (I might've missed something while doing the problem)[/quote]\r\n\r\nUm..to be honest, it was an old AHSME problem that I remembered but I don't think it gave anything else :? , what exactly did you get because now that I think about it, there can be more than one way to express the answer.\r\n[hide=\"what I got\"]\n$\\frac{-c}{d}$ or since it's a unit circle $-c$ or just $c$[/hide]\r\n\r\nI forgot the answer choices and I don't remember which test or question this was :?", "Solution_10": "Yeah, that's what I got as well.", "Solution_11": "[quote=\"joml88\"]Yeah, that's what I got as well.[/quote]\r\n\r\nshould i/you post the solution for it or should we let others solve it (if anyone else even cares)\r\n\r\nThe thing I wasn't sure about is whether the transformation of the polynomial for the reciprocal roots works for polynomials with only complex roots as well :?:", "Solution_12": "[hide=\"Number 1\"] Well...technically if this were an AMC type problem i suppose it could be done without a calculator, but caclulators are allowed...so.\n\nLet $n$ be the number of digits in $2006^{2006}$. Then $n=[\\log{2006^{2006}}]+1$. Thus $n=6225$. If we divide $2006^{2006}$ by $10^{6224}$, we will be left with the first digit and then the first few numbers following the decimal place after that..\n\nDenote $q$ to be the quotient we are looking for, then $[q]$ will be the leftmost digit of $2006^{2006}$.\n\n$q=2006^{2006} / 10^{6224}\\\\ \\log{q}=2006\\log{2006} -6224\\\\ \\log{q} = 0.47584294...\\\\ q=2.99118...$\n\nThe leftmost digit is thus $\\boxed{2}$.[/hide]", "Solution_13": "[quote=\"arkmastermind\"][hide=\"Number 1\"] Well...technically if this were an AMC type problem i suppose it could be done without a calculator, but caclulators are allowed...so.\n\nLet $n$ be the number of digits in $2006^{2006}$. Then $n=[\\log{2006^{2006}}]+1$. Thus $n=6225$. If we divide $2006^{2006}$ by $10^{6224}$, we will be left with the first digit and then the first few numbers following the decimal place after that..\n\nDenote $q$ to be the quotient we are looking for, then $[q]$ will be the leftmost digit of $2006^{2006}$.\n\n$q=2006^{2006} / 10^{6224}\\\\ \\log{q}=2006\\log{2006} -6224\\\\ \\log{q} = 0.47584294...\\\\ q=2.99118...$\n\nThe leftmost digit is thus $\\boxed{2}$.[/hide][/quote]\r\n\r\n :yup: , nice to see you posting here Asif, and no it wasn't an AMC problem, I just made it up..but I don't think you can do it without a calculator.", "Solution_14": "for number 3, i didnt really work it out, but im guessing $3+\\sqrt{6}$?", "Solution_15": "[quote=\"arkmastermind\"]for number 3, i didnt really work it out, but im guessing $3+\\sqrt{6}$?[/quote]\r\n\r\n[hide=\"mysolution\"]Obviously we have a circle centered at $(3,3)$ with radius $\\sqrt{6}$, the largest solution point is tangent to the circle so by drawing that tangent, we form a right triangle with base $\\sqrt{6}$ and hypotenuse $3\\sqrt{2}$ (from the $45,45,90$ triangle that contains the center of the circle at it's top vertex). Notice that $\\frac{y}{x}$ seems strangely falmiliar, and it involves $\\tan$, so by calling the small angle of the right triangle with sides $3\\sqrt{2}, \\sqrt{6}, 2\\sqrt{2}$ as $\\theta$, we want to know the entire angle, which is $\\theta+45$, so $\\frac{y}{x}=\\tan{(\\theta+45)}=\\frac{\\frac{\\sqrt{2}}{2}+1}{1-\\frac{\\sqrt{2}}{2}}=\\boxed{2\\sqrt{2}+3}$[/hide]\r\n\r\nhmm, i didn't get what you got so one of us is wrong..so how did you solve it? :?\r\n\r\nEDIT: Didn't rationalize the denominator right", "Solution_16": "i dont seem to be able to visualize your solution that well...\r\n\r\ni didnt really work it out, i just took a guess that the largest value for y/x would occur either at either of the circles four extreme points (topmost, bottom most, left most, right most). upon inspection it seemed that the leftmost point on the circle would have the greatest slope from the origin to it, and thus the greatest y/x. so the y value was 3 and the x value was $3-\\sqrt{6}$ so, dividing those i got $3+\\sqrt{6}$. \r\n\r\nBut seeing as you posted the problems...you probably have the right answer :P", "Solution_17": "Hmm, I disagree with both of you! lol\r\n\r\nBut I checked with Geometer's Sketchpad to see if my answer agrees, and it does :D\r\n\r\n\r\n\r\nFirst, let $\\frac yx=k$ so that $y=kx$. This is just a series of lines through the origin with different slopes. We are looking for the one with the largest slope (i.e. $k$). The line has to intersect the circle though. So imagine picking a point on the circle and drawing a line through it and the origin. Then just imagine moving this point around so that the line will be getting either more or less steep. When will it be tilting upwards the most? This will occur when the line is tangent to the circle (actually there are two places where it will be tangent...the other is the minimum). The reason that Asif's reasoning is wrong is that, although we are moving the point on the circle as far to the left as possible (in a sense), we have to be aware of the fact that once we pass the point of tangency we will still be moving left but there will be another point of interesection with the circle to the right of the point of tangency. So instead of moving the point as far left as possible, we have to consider the farthest right intersection point and move it as far left as possible (this occurs when it [i]is[/i] the other intersection point).\r\n\r\nOk, so now we know what we're after. Let $(x,y)$ be the point of tangency. The slope between the line through $(0,0)$ and $(x,y)$ and the line through $(x,y)$ and $(3,3)$ are perpendicular. Thus:\r\n\r\n\\[ \\frac{y-3}{x-3}=-\\frac xy \\\\ \\Rightarrow x^2+y^2 = 3x+3y. \\]\r\n\r\nThen expand $(x-3)^2+(y-3)^2 = 6$ and plug in from above, cancel, simplify, etc. and get $x+y=4\\Leftrightarrow y=4-x$. Plug this back into the original equation for the circle and we find that $x=2\\pm\\sqrt{2}$ which makes $y=2\\mp\\sqrt{2}.$ Therefore, the minimum $\\frac yx$ is $\\frac{2-\\sqrt{2}}{2+\\sqrt{2}}=3-2\\sqrt{2}$ and the max is its reciprocal: $3+2\\sqrt{2}$.\r\n\r\nWhew!\r\n\r\n\r\nOh, and Shobhit, how can you have a 45-45-90 right triangle with legs $\\sqrt{6}$ and hypotenuse $3\\sqrt{2}$ ;)", "Solution_18": "ah duh. of course it had to be the tangent line. *hits self on head* \r\n\r\nlol. maybe i should have actually worked it out instead of guessing around :roll:", "Solution_19": "[quote=\"joml88\"]Hmm, I disagree with both of you! lol\nOh, and Shobhit, how can you have a 45-45-90 right triangle with legs $\\sqrt{6}$ and hypotenuse $3\\sqrt{2}$ ;)[/quote]\r\n\r\nJoe, I didn't mean that was the triangle, i got $3\\sqrt{2}$ FROM the hypotenuse of the 45,45,90 triangle, and I guess I did it right since I got the same answer as you, I just didn't rationalize $\\frac{\\frac{\\sqrt{2}}{2}+1}{1-\\frac{\\sqrt{2}}{2}}$ correctly since it's equal to $2\\sqrt{2}+3$ not $2\\sqrt{2}+6$ :blush: , I made a diagram to explain myself, the $45,45,90$ triangle comes from the $x$ and $y$ values of the center of the circle...", "Solution_20": "Oh gothca. I was just quickly scanning your solution to see where you might have gone wrong and once I got there I stopped reading :oops: .\r\n\r\nBtw, no offense or anything but that diagram is pretty horrendous :D You know that paint has line and ellipse tools, right? You can even hold down shift and it will make the lines at straight angles, i.e. the lines will go in one of the four direction or one of the four intermediate directions. Holding the shift key down for the ellipse tool draws circles. Paint is far from great but you can still make semi-decent diagrams with it.", "Solution_21": "[quote=\"joml88\"]Oh gothca. I was just quickly scanning your solution to see where you might have gone wrong and once I got there I stopped reading :oops: .\n\nBtw, no offense or anything but that diagram is pretty horrendous :D You know that paint has line and ellipse tools, right? You can even hold down shift and it will make the lines at straight angles, i.e. the lines will go in one of the four direction or one of the four intermediate directions. Holding the shift key down for the ellipse tool draws circles. Paint is far from great but you can still make semi-decent diagrams with it.[/quote]\r\n\r\nI know, i was just doing the graph quickly and didn't really think about it's neatness :rotfl: . I cared more about getting the picture to work, since it was the first picture i posted on Aops." } { "Tag": [ "MATHCOUNTS", "analytic geometry" ], "Problem": "Finally I find a problem worth calling \"MATHCOUNTS [b]Challenge[/b]\" :) \r\n\r\nEither prove or disprove that Associative Property is valid under exponent: a^(b^c) equal or not equal to (a^b)^c\r\n\r\nalso, do the same for Commutative Property under Exponent.", "Solution_1": "hmm...its a bit of common sense...\r\n\r\n[quote=\"Tare\"]a^(b^c)[/quote]\nThat equals a^b^c...\n\n[quote=\"Tare\"](a^b)^c [/quote]\r\nequals abc...\r\n\r\nClearly, they're not equal...or am i misunderstanding something?\r\n\r\nI'd put it in spoiler...but its probly not hard enuf for that...", "Solution_2": "Hmm...that's what I thought before my math teacher confuted me :lol: by raising tons of examples...\r\nsorry, just had to say that or else I'll go crazy :lol:\r\n\r\nyou know, maybe, just maybe I used \"confuted\" correctly this time...*Apollo 13 theme song plays* :)\r\nOnto MithsApprentice! :lol: ", "Solution_3": "Are you sure? I think he's right. a^b^c = a^bc if and only if bc = b^c, and that's clearly not the case all the time.", "Solution_4": "Well...I'll leave it up to you guys prove it (or disprove it). Remember, I'm not saying which is right because that's part of the question.", "Solution_5": "remember...to disprove something, you need only offer one counter-example. For example, if we wanted to disprove the conjecture that n:^2:+n+41 outputs a prime for each n. we need only show that it doesn't for n=41.", "Solution_6": "Well, I mean, it's trivial. Here's a case where that doesn't work.\r\n\r\na = 4 b= 2, c = 3\r\n\r\n4^2^3 = 4 ^8\r\n4^2*3 = 4^6\r\n\r\n4^6 /= 4^8.", "Solution_7": "[quote=\"MysticTerminator\"]remember...to disprove something, you need only offer one counter-example. For example, if we wanted to disprove the conjecture that n:^2:+n+41 outputs a prime for each n. we need only show that it doesn't for n=41.[/quote]\r\n\r\nn=40 is actually the smallest positive integer such that it's not a prime.", "Solution_8": "Any proofs/disproofs for the Commutative Property?\r\n\r\n...is \"disproof\" a word? :)\r\nwhew, it would've been embarrasing if I made up a word unintentionally (like the time my teacher misheard me say \"coordinate system\" and thought I said \"para-coordinate system\", which later on was defined as \"coordinate systems in which instead of points, there are poligonal \"boxes\" in which numbers fit, as is the Pascal's Triangle\"...oh well :) )", "Solution_9": "Yes, it is. Often on problems you will be asked to prove or disprove.", "Solution_10": "So whats the answer?", "Solution_11": "[hide]$ a^{b}=b^{a}$\nWell, just take $ a=2$ and $ b=3$.\n$ 2^{3}\\not= 3^{2}$[/hide]", "Solution_12": "wouldnt $ a^{(}b^{c})$ be $ a^{bc}$ and $ (a^{b})^{c}$ be $ a{}^{b}{}^{c}$ which is $ a^{bc}$ so they are equal." } { "Tag": [], "Problem": "What is the 100th digit to the right of the decimal point in the\ndecimal representation of $ \\frac{13}{90}$?", "Solution_1": "After the initial $ 1$, all digits to the right are $ \\fbox{4}$ in the decimal representation of $ \\frac{13}{90}$ due to properties of fractions in the form $ \\frac{n}{9}$ being expressed with a $ 0$ after the $ 9$ in the denominator." } { "Tag": [ "Sequence", "bounded", "algebra", "IMO", "imo 1991", "construction", "Harm Derksen" ], "Problem": "An infinite sequence $ \\,x_{0},x_{1},x_{2},\\ldots \\,$ of real numbers is said to be [b]bounded[/b] if there is a constant $ \\,C\\,$ such that $ \\, \\vert x_{i} \\vert \\leq C\\,$ for every $ \\,i\\geq 0$. Given any real number $ \\,a > 1,\\,$ construct a bounded infinite sequence $ x_{0},x_{1},x_{2},\\ldots \\,$ such that\r\n\\[ \\vert x_{i} \\minus{} x_{j} \\vert \\vert i \\minus{} j \\vert^{a}\\geq 1\r\n\\]\r\nfor every pair of distinct nonnegative integers $ i, j$.", "Solution_1": "I'm not sure if this works:\r\n\r\nAfter we have found $ x_0, x_1, ... x_{n\\minus{}1}$ satisfying the conditions, we now choose an $ x_n$ as follows. Draw open intervals of lengths $ 2/1^a, 2/2^a, 2/3^a, ... , 2/n^a$ around $ x_{n\\minus{}1}, x_{n\\minus{}2}, ... x_1, x_0$ (so that each of these points are the midpoints of the intervals. It is clear that if $ x_n$ does not lie in any of these intervals, it satisfies the conditions of the problem. Thus we choose the $ x_n$ with smallest absolute value that does not lie in these $ n$ open intervals.\r\n\r\nNow, because the series $ 1/1^a \\plus{} 1/2^a \\plus{} ...$ converges for all real $ a > 1$, these intervals cover a total area that approaches a limit, say c. Then, we can always find $ x_n$ with absolute value less than c, so the sequence is bounded.", "Solution_2": "Could you not simply do x_i = -1 ^ i? It doesn't seem like a very interesting problem unless you're looking for a monotonic bounded infinite sequence, in which case x_i = [sum from 1 to i] 2/k^{1 + epsilon} would suffice, where epsilon is small.", "Solution_3": "Yesterday I was solving this problem and I thought about exactly the same solution as alkjash's solution. But I don't know if it's really correct, beacuse we should \"construct\" such a sequence and I don't know if we could say it's construction. What's your opinion?", "Solution_4": "Actually, the minimal $a$ for which such a sequence exists is $a=1$, which follows from the problem [url=http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=A624&l=en]K\u00f6mal A.624[/url]. ([hide=Explanation.]\nOf course, in the IMO problem, the interval $[-C;C]$ can be dilated into interval $[0;1]$, hence such a sequence exists for some $a$ if and only if there is a sequence $x_1,x_2,\\dots\\in[0;1]$ satisfying $|x_i-x_j|\\ge \\frac{c}{|i-j|^a}$ for all $i,j\\ge 1$ ($i\\neq j$) with a suitable constant $c$.\n\nPart b) in the K\u00f6mal (or CIIM) problem is in fact equivalent to giving a sequence satisfying $|x_i-x_j|\\ge \\frac c{|i-j|}$ for some suitable small $c$ - if we give a sequence $x_1,x_2,\\dots\\in[0;1]$ satisfying the conditions in part b), then we can simply take $y_i=x_{ri}$. (The converse is also true, but takes a bit more effort to prove.)\n\nPart a) in the K\u00f6mal problem says that if $x_1,x_2,\\dots\\in[0;1]$, then for suitable $K>0$, and any positive integer $r$, we can find $i,j$ with $|i-j|\\ge r$ and $|x_i-x_j|<\\frac{K}{|i-j|}$. Suppose that we have a sequence $x_1,x_2,\\dots\\in[0;1]$ that satisfies $|x_i-x_j|\\ge \\frac{c}{|i-j|^a}$ $\\forall i,j\\ge 1$ ($i\\neq j$) for some $c>0$ and $a<1$, and let's take the appropriate $K$. Then if $|i-j|\\ge r$ and $|x_i-x_j|<\\frac{K}{|i-j|}$, then from the condition we get\n$$\\frac{K}{|i-j|}\\ge \\frac{c}{|i-j|^a}\\implies \\frac{K}{c}\\ge |i-j|^{1-a}\\ge r^{1-a},$$\nwhich gives a contradiction for $r>(K/c)^{1/(1-a)}$.\n[/hide])\n\nI have posted a solution to the K\u00f6mal/CIIM problem [url=http://artofproblemsolving.com/community/c7h621441]here[/url].\n\nThere's an easy construction for the IMO problem as well, in the case $a=1$ actually. Just let $x_n=\\{n\\sqrt 2\\}$ and use the well-known estimate\n$$|a\\sqrt 2-b|=\\frac{|2a^2-b^2|}{|a\\sqrt 2+b|}\\ge \\max\\left(1,\\frac1{a\\sqrt 2+(a\\sqrt 2+1)}\\right)\\ge \\frac1{4a}$$\nfor positive integer $a,b$. (This is a Liouville-type estimate such as [url=http://www.artofproblemsolving.com/community/c2927h1057645_liouvilles_approximation_theorem]here[/url].)", "Solution_5": "http://artofproblemsolving.com/community/c7h551121" } { "Tag": [ "modular arithmetic" ], "Problem": "Recall that the Fibonacci numbers F(n) are defined by F(0) = 0, F(1) = 1,\r\nand F(n) = F(n\u22121)+F(n\u22122) for n \u2265 2. Determine the last digit of\r\nF(2006)", "Solution_1": "[hide]Consider the sequence mod 10\n1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,1,7,8,5,3,8,1,9,0,9,9,8,7,5,2,7,9,6,5,1,6,7,3,0,3,3,6,9,5,4,9,3,2,5,7,2,9,1,0,1,1 and now we finally have a repetition after 60 terms :D. So we want the 26th term, $ \\boxed{3}$[/hide]", "Solution_2": "[hide=\"Less brute force and more generalizable\"]Find the units digits of each term until one is a multiple of 10. This occurs at $ F_{15}$. This sequence will repeat itself, but each time it is multiplied by $ F_{14}$. In this case the units digit of that is 7.\n\nSo $ F_{n}\\equiv 7F_{n\\minus{}15}\\pmod{10}$, and $ F_{2006}\\equiv 7^{133}F_{11}\\equiv 7 F_{11}\\pmod{10}$ (note $ 7^{4}\\equiv 1\\pmod{10}$). The units digit of $ F_{11}$ is 9, and $ 7\\cdot 9\\equiv\\boxed{3}\\pmod{10}$.[/hide]", "Solution_3": "Excellent observation!" } { "Tag": [], "Problem": "Find X :\r\n- X is a natural number\r\n- X has a most number of divisol\r\n- 9b$", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?p=356406#p356406" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Consider a $m*n$ board. On each box there's a non-negative integrer number assigned. An operation consists on choosing any two boxes with $1$ side in common, and add to this $2$ numbers the same integrer number (it can be negative), so that both results are non-negatives. \r\nWhat conditions must be satisfied initially on the assignment of the boxes, in order to have, after some operations, the number $0$ on every box?.", "Solution_1": "If we color the cells of the board like those of a chessboard (alternatively black and white) then the sum of the numbers in the white cells must be equal to that of the numbers in the black cells. \r\n\r\nThe necessity is clear: in each move, you make equal contributions to both sums (white and black), and if in the end you reach $0$, i.e. both sums equal to $0$, then they must have been equal in the beginning as well.\r\n\r\nIf, conversely, the two sums are equal, we can clear the board one column at a time. When we're left with only one column, we continue clearing up cells one by one until we're left with a $2\\times 1$ board. By the hypothesis (white and black sums are equal) the two numbers we're left with are equal and we can finish.", "Solution_2": "Why the black sum must be equal to the white sum?\n\nThanks" } { "Tag": [ "logarithms", "trigonometry", "calculus", "calculus computations" ], "Problem": "Answer the following question.\r\n\r\n[1] $\\frac{d}{dx} \\frac{x}{\\sqrt{x^2+x+1}}$\r\n\r\n[2] $\\frac{d}{dx} \\ln (x+\\sqrt{x^2+1})$\r\n\r\n[3] $\\frac{d}{dx} \\sin (1+\\ln x)$\r\n\r\n[4] Let $x=\\frac{e^{3t}}{1+t^2},y=\\frac{t}{1+t^2}$, find $\\frac{dy}{dx}$\r\n\r\n[5] Sinsyu University Faculty of Textile Sience and technology\u3000 \r\n\r\n(1) Let $x^2-y^2=a^2$, find $\\frac{d^2y}{dx^2}$\r\n\r\n(2) Let $y=\\ln (1+\\cos x)^2$, find the value of $\\frac{d^2y}{dx^2}+2e^{-\\frac{y}{2}}$", "Solution_1": "hello, to the first we have by the chaine and quotient rule\r\n$ \\frac{d}{dx}\\frac{x}{\\sqrt{x^2\\plus{}x\\plus{}1}}\\equal{}\\\\\r\n\\frac{\\sqrt{x^2\\plus{}x\\plus{}1}\\minus{}x\\frac{1}{2}(x^2\\plus{}x\\plus{}1)^{\\frac{1}{2}}(2x\\plus{}1)}{x^2\\plus{}x\\plus{}1}\\equal{}\\\\\r\n\\frac{x^2\\plus{}x\\plus{}1\\minus{}x\\frac{1}{2}(2x\\plus{}1)}{(x^2\\plus{}x\\plus{}1)\\sqrt{x^2\\plus{}x\\plus{}1}}\\equal{}\\\\\r\n\\frac{x\\plus{}2}{2(x^2\\plus{}x\\plus{}1)^{\\frac{3}{2}}}$\r\nSonnhard.", "Solution_2": "[quote=\"kunny\"]Answer the following question.\n\n[1] $ \\frac {d}{dx} \\frac {x}{\\sqrt {x^2 \\plus{} x \\plus{} 1}}$\n\n[2] $ \\frac {d}{dx} \\ln (x \\plus{} \\sqrt {x^2 \\plus{} 1})$\n\n[3] $ \\frac {d}{dx} \\sin (1 \\plus{} \\ln x)$\n\n[4] Let $ x \\equal{} \\frac {e^{3t}}{1 \\plus{} t^2},y \\equal{} \\frac {t}{1 \\plus{} t^2}$, find $ \\frac {dy}{dx}$\n\n[5] Sinsyu University Faculty of Textile Sience and technology\u3000 \n\n(1) Let $ x^2 \\minus{} y^2 \\equal{} a^2$, find $ \\frac {d^2y}{dx^2}$\n\n(2) Let $ y \\equal{} \\ln (1 \\plus{} \\cos x)^2$, find the value of $ \\frac {d^2y}{dx^2} \\plus{} 2e^{ \\minus{} \\frac {y}{2}}$[/quote]\r\n\r\n[2] $ \\frac {d}{dx} \\ln (x \\plus{} \\sqrt {x^2 \\plus{} 1})\\equal{} \\frac{1}{x\\plus{}\\sqrt{x^2\\plus{}1}} .(1\\plus{}\\frac{x}{\\sqrt{x^2\\plus{}1}})$\r\n3/ $ cos(1\\plus{}lnx) .\\frac{1}{x}$", "Solution_3": "[2] You can express your answer in simple form.", "Solution_4": "[quote=\"kunny\"][2] You can express your answer in simple form.[/quote]\r\nyes, $ (2)\\equal{} \\frac{1}{\\sqrt{x^2\\plus{}1}}$", "Solution_5": "can I do like this::?\r\n[4] $ \\frac{dy}{dx}\\equal{}\\frac{dy}{dt}\\cdot\\frac{dt}{dx}$\r\n\r\nwhere $ \\frac{dy}{dt}\\equal{}\\frac{(1\\plus{}t^2)\\minus{}2t^2}{(1\\plus{}t^2)^2}$\r\nand $ \\frac{dx}{dt}\\equal{}\\frac{3e^{3t}(1\\plus{}t^2)\\minus{}2e^{3t}t}{(1\\plus{}t^2)^2}\\rightarrow\\frac{dt}{dx}\\equal{}\\frac{(1\\plus{}t^2)^2}{3e^{3t}(1\\plus{}t^2)\\minus{}2e^{3t}t}$\r\nso $ \\frac{dy}{dx}\\equal{}\\frac{dy}{dt}\\cdot\\frac{dt}{dx}\\equal{}\\frac{(1\\plus{}t^2)\\minus{}2t^2}{(1\\plus{}t^2)^2}\\cdot\\frac{(1\\plus{}t^2)^2}{3e^{3t}(1\\plus{}t^2)\\minus{}2e^{3t}t}\\equal{}\\frac{1\\minus{}t^2}{e^{3t}(3t^2\\minus{}2t\\plus{}3)}$", "Solution_6": "can I also make like this?\r\n[5]\r\n(1) $ x^2 \\minus{} y^2 \\equal{} a^2$\r\n\r\n$ 2x \\minus{} 2y\\frac {dy}{dx} \\equal{} 2a\\frac {da}{dx}\\rightarrow\\frac {dy}{dx} \\equal{} \\frac {x}{y} \\minus{} \\frac {a}{y}\\frac {da}{dx}$\r\n\r\n$ 2 \\minus{} 2(\\frac {dy}{dx})^2 \\minus{} 2y\\frac {d^2y}{dx^2} \\equal{} 2\\frac {da}{dx} \\plus{} 2a\\frac {d^2a}{dx^2}$\r\n\r\n$ 1 \\minus{} (\\frac {x^2}{y^2} \\plus{} \\frac {a^2}{y^2}(\\frac {da}{dx})^2 \\minus{} \\frac {2xa}{y^2}\\frac {da}{dx}) \\minus{} y\\frac {d^2y}{dx^2} \\equal{} \\frac {da}{dx} \\plus{} a\\frac {d^2a}{dx^2}$\r\n\r\n$ y\\frac {d^2y}{dx^2} \\equal{} 1 \\minus{} \\frac {x^2}{y^2} \\minus{} \\frac {a^2}{y^2}(\\frac {da}{dx})^2 \\plus{} \\frac {2xa}{y^2}\\frac {da}{dx} \\minus{} \\frac {da}{dx} \\minus{} a\\frac {d^2a}{dx^2}$\r\n\r\n$ \\frac {d^2y}{dx^2} \\equal{} \\frac {1}{y} \\minus{} \\frac {x^2}{y^3} \\minus{} \\frac {a^2}{y^3}(\\frac {da}{dx})^2 \\plus{} \\frac {2xa}{y^3}\\frac {da}{dx} \\minus{} \\frac {1}{y}\\frac {da}{dx} \\minus{} \\frac {a}{y}\\frac {d^2a}{dx^2}$\r\n\r\n\r\n(2)$ y \\equal{} \\ln(1 \\plus{} \\cos x)^2 \\equal{} 2\\ln(1 \\plus{} \\cos x)$\r\n$ y' \\equal{} \\minus{} 2(\\frac {\\sin x}{1 \\plus{} \\cos x}$\r\n$ y'' \\equal{} \\minus{} 2(\\frac {\\cos x(1 \\plus{} \\cos x) \\plus{} \\sin^2x}{(1 \\plus{} \\cos x)^2} \\equal{} \\minus{} 2(\\frac {1}{1 \\plus{} \\cos x})$\r\n\r\n$ 2e^{ \\minus{} \\frac {y}{2}} \\equal{} 2\\frac {1}{e^{\\frac {y}{2}}} \\equal{} 2(\\frac {1}{1 \\plus{} \\cos x})$\r\nso $ y'' \\plus{} 2e^{ \\minus{} \\frac {y}{2}} \\equal{} 0$\r\n\r\nsorry if I made a mistake..\r\nthx", "Solution_7": "hello, to [5] 1), here i have $ 2x - 2yy^' = 0 \\Leftrightarrow y^' = \\frac {x}{y}$.\r\nBy differentiating again we get $ y^{''} = \\frac {y - xy^'}{y^2} = \\frac {y^2 - x^2}{y^3}$. By $ y^2 - x^2 = - a^2$ we get $ y^{''} = -\\frac {a^2}{y^3}$.\r\nSonnhard.", "Solution_8": "are you sure $ a$ doesn't contain $ x$\r\nwhat if $ a(x)$?? :maybe:", "Solution_9": "hello, i have assumed that $ a$ is a constant.\r\nSonnhard." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$ a,b,c>0$\r\n\r\n$ \\frac{1}{6a^2\\plus{}bc}\\plus{}\\frac{1}{6b^2\\plus{}ac}\\plus{}\\frac{1}{6c^2\\plus{}ab}\\ge\\frac{9}{7}\\frac{1}{ab\\plus{}bc\\plus{}ca}$", "Solution_1": "I think Iran 1996 might be useful... Have you got a proof of your inequality?", "Solution_2": "I found it on polish mathematical forum (probably You know limes123 matematyka.pl) I can't prove this.", "Solution_3": "[quote=\"matex (L)(L)(L)\"]$ a,b,c > 0$\n\n$ \\frac {1}{6a^2 \\plus{} bc} \\plus{} \\frac {1}{6b^2 \\plus{} ac} \\plus{} \\frac {1}{6c^2 \\plus{} ab}\\ge\\frac {9}{7}\\frac {1}{ab \\plus{} bc \\plus{} ca}$[/quote]\r\nIt is a privet case of the following Vasc's inequality. \r\nLet $ a,$ $ b$ and $ c$ are positive numbers and $ r\\geq3\\plus{}\\sqrt7.$ Prove that\r\n\\[ \\frac{1}{ra^2\\plus{}bc}\\plus{}\\frac{1}{rb^2\\plus{}ac}\\plus{}\\frac{1}{rc^2\\plus{}ab}\\geq\\frac{9}{(r\\plus{}1)(ab\\plus{}ac\\plus{}bc)}\\]", "Solution_4": "Where I can find prove of this generalization?", "Solution_5": "[quote=\"matex (L)(L)(L)\"]Where I can find prove of this generalization?[/quote]\r\nHere:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=100399" } { "Tag": [ "geometry", "3D geometry", "modular arithmetic", "floor function", "ceiling function" ], "Problem": "If a and b are natural numbers and $ 5a^3\\equal{}6b^4$, what is the smallest possible value of $ a \\plus{} b$?", "Solution_1": "[hide=\"My Solution\"]\nRearranging the equation in terms of $ a$, we have $ a^3 \\equal{} \\frac{6b^4}{5}$. \n\nWe have $ 6 \\equal{} 2\\cdot 3$ and the numerator is being divided by $ 5$ so we can conclude that $ b \\equal{} 2^h\\cdot 3^i\\cdot 5^j$. Substituting that, we have \n\n$ a^3 \\equal{} \\frac{6(2^h\\cdot 3^i\\cdot 5^j)^4}{5} \\equal{} 2^{4h\\plus{}1}\\cdot 3^{4i \\plus{} 1}\\cdot 5^{4j \\minus{} 1}$.\n\nIn order to minimize $ a$, we must minimize $ a^3$ and doing that will minimize $ b^4$. Because $ a$ is a perfect cube, we must have the following:\n\n$ 4h \\plus{} 1 \\equiv 0 \\pmod{3} \\implies h \\equiv 2 \\pmod{3}\\\\\n4i \\plus{} 1 \\equiv 0 \\pmod{3} \\implies i \\equiv 2 \\pmod{3}\\\\\n4j \\minus{} 1 \\equiv 0 \\pmod{3} \\implies j \\equiv 1 \\pmod{3}$\n\nIn order to minimize $ a$, we must minimize the powers of $ 2$, $ 3$, and $ 5$. In order to do that, we must have $ h \\equal{} i \\equal{} 2$ and $ j \\equal{} 1$. So we have:\n\n$ 5a^3 \\equal{} 6(2^2\\cdot 3^2\\cdot 5)^4 \\implies a \\equal{} 2^3\\cdot 3^3\\cdot 5 \\equal{} 1080$\n\nIt follows that $ b \\equal{} 180$. Therefore, $ a \\plus{} b \\equal{} 1080 \\plus{} 180 \\equal{} \\boxed{1260}$.\n[/hide]", "Solution_2": "alternatively, proceed using divisibility, through repeated application of the lemma:\r\n\r\nif $ p|a^n$, where $ p$ is prime, then $ p|a$.\r\n\r\n[hide=\"Proof\"] Start from euclid's lemma ($ p|ab$, where $ p$ is prime, implies $ p|a$ or $ p|b$). The statement results from repeated application of this lemma. Let $ b\\equal{}a^{n\\minus{}1}$. Either $ p|a$ or $ p|a^{n\\minus{}1}$. But $ p|a^{n\\minus{}1}$ implies $ p|a$ or $ p|a^{n\\minus{}2}$, and continuing in this way we find that all cases lead to $ p|a$.\n[/hide]\n\nNote that $ p^m|a^n$ only if $ p|a^n$. In fact the more general result that $ p^m | a^n$ implies $ p^{\\lfloor m/n \\rfloor} | a$.\n[hide=\"Proof\"]This can be shown by noting $ p^m|a^n \\implies p|a^n \\implies p|a$. Now let $ a\\equal{}bp$; we have $ p^m|b^np^n$, hence $ p^{m\\minus{}n}|b^n$ if $ m \\ge n$. Hence we can apply this result up to as many times as $ \\lceil m/n \\rceil$ since this is the number of times you have to subtract $ n$ from $ m$ to get a nonpositive which will result in the $ m < n$ case.\n[/hide]\n\nLast comment: for the general composite numbers, we can put together the result for each prime factors to get $ p_1^{e_1}p_2^{e_2} \\cdots | a^n$ implies $ p_1^{\\lceil e_1/n \\rceil} p_2^{\\lceil e_2/n \\rceil} \\cdots | a$.\n\n[hide=\"Solution\"]\n$ 5|6b^4$ implies $ 5|b$ so let $ b\\equal{}5b'$\n$ 6|5a^3$ implies $ 6|a$ (since $ 6$ is squarefree) so let $ a\\equal{}6a'$.\n\nPlugging back in\n$ 5(6a')^3 \\equal{} 6(5b')^4$\n$ 6^2 a'^3 \\equal{} 5^3 b'^4$\n\nNow $ 5^3|36a'^3$ implies $ 5|a'$. Let $ a'\\equal{}5a''$\n$ 6^2|125b'^4$ implies $ 6|b'$. Let b'=$ 6b''$.\n\nPlugging in again\n$ 6^2 5^3 a''^3 \\equal{} 5^3 6^4 b''^4$\n$ a''^3 \\equal{} 6^2 b''^4$\n\nNow $ 6^2|a''^3$ implies $ 6|a''$. Let $ a''\\equal{}6a'''$\nSubstituting,\n$ 6^3 a'''^3 \\equal{} 6^2 b''^4$\n$ 6a'''^3 \\equal{} b''^4$\n\n$ 6|b''^4$ implies $ 6|b''$ so let $ b''\\equal{}6b'''$. Substituting again\n$ 6a'''^3 \\equal{} 6^4 b'''^4$\n$ a'''^3 \\equal{} 6^3 b'''^4$\n\n$ 6^3|a'''^3$ implies $ 6|a'''$. Now letting $ a'''\\equal{}6a''''$ we have\n$ a''''^3 \\equal{} b'''^4$\n\nThe smallest positive integer that is both a cube and a 4th power is $ 1$ (and the sequence of cube 4th powers are given by $ a_n\\equal{}n^{12}$), so our minimum is achieved at $ a''''\\equal{}b'''\\equal{}1$. Working backwards:\n$ a\\equal{}6a'\\equal{}6(5a'')\\equal{}30(6a''')\\equal{}180(6a'''') \\equal{} 1080$\n$ b\\equal{}5b'\\equal{}5(6b'') \\equal{} 30(6b''') \\equal{} 180$\n\nThus the minimum $ a\\plus{}b$ is $ 180\\plus{}1080 \\equal{} 1260$.[/hide]" } { "Tag": [ "Alcumus", "FTW", "rate problems", "Support" ], "Problem": "Hey Everyone,\r\n\r\nI used to see Alcumus, but I didn't pay much attention to it. But now I want to do good so as to get in all-time high scores (inspiration by nikeballa96). Everyone who's really smart already have the position. I'm terrible in FTW so I atleast want to get the high scores in this. Currently, my goal is to get in Daily Top Ten everyday. \r\n\r\nIs this enough or should I set like a certain number of points and karma a day? Can someone in the list give me some advice? I'm a newbie so I need a lot of help.", "Solution_1": "It really depends on how much time you are willing to spend. I used to do a 100 points a day, but after you complete all the problems, you get the ones you got wrong the first time again for only one point. I f you want to be on Top 10 daily, you should aim for about 45 points and 400 Karma points. Also, remember to rate problems!", "Solution_2": "That shouldn't be the point. Who really cares how high in the rankings you are? Do it becuase you have fun, want to get better, and love problems! :)", "Solution_3": "Haha, yeah...\r\nI've only gotten into daily top 10 like 2-3 times and that was when I actually did alcumus for a while, not like the 30-45 minutes I usually do." } { "Tag": [ "MATHCOUNTS", "geometry" ], "Problem": "Oh my goodness, my region's competition is on Feb. 3rd. Exactly two weeks. :( It'll be my first, can anyone from the Rochester, NY area tell me what it's like?", "Solution_1": "have you been there before?\r\n\r\nhow many advance?\r\n\r\ni suppose (i can't think of his name) but the guy who led the team from ny to nationals could help you :)\r\n\r\n-jorian\r\n\r\nmy chapter in 14\r\nmy state in 42", "Solution_2": "Just a reminder to all, once your chapter competition is done, DO NOT discuss the competition. Many chapters don't finish until March. \r\n\r\n\r\nI guess as advice, be careful and don't be nervous. If the competition is hard, it is hard for everyone. If it's easy, just be really careful, because then, the person w/ least dumb mistakes wins.", "Solution_3": "[quote=\"kyyuanmathcount\"]Just a reminder to all, once your chapter competition is done, DO NOT discuss the competition. Many chapters don't finish until March. \n\n\nI guess as advice, be careful and don't be nervous. If the competition is hard, it is hard for everyone. If it's easy, just be really careful, because then, the person w/ least dumb mistakes wins.[/quote]\r\n\r\nThe person with no mistakes also wins. :roll: \r\n\r\nI hate school because that is where I make the mistakes, chapter where I make mistakes.", "Solution_4": "haha, 7 teams from my chapter advance, 12 individuals :)\r\n\r\n-jorian\r\n\r\nand i'm 2nd on my team (behind guess who)\r\n\r\noh, i would like to bring up that dan's actually a lot smarter than i thought", "Solution_5": "bah, smartness is just a delusion, one that people are under when they make it to national mathcounts.", "Solution_6": "[quote=\"egghead91\"]Oh my goodness, my region's competition is on Feb. 3rd. Exactly two weeks. :( It'll be my first, can anyone from the Rochester, NY area tell me what it's like?[/quote]interesting... My chapter competition is on the same day...\r\n\r\n@egghead: Where are you taking it? I'm taking it at Niskayuna...\r\n\r\nWell, my first year was last year, so I only know what an easy competition is like.\r\n\r\nYou get there, I get a little shaky, maybe I'm a sugar low guy...\r\n\r\nThe sprint round is a little boring, the target round is a little better, but the team round is REALLY fun! Also, if you get into the countdown round, it should be really fun.", "Solution_7": "Yeah, chapter cd is fun. Luckilly, my chapter isn't great, b/c it's where I make all of my mistakes. I got 43 last year... :blush: I get really nervous there, though. at states, I'm almost TOO confident. I'd rather not discuss my composure at nats...", "Solution_8": "[quote=\"1=2\"][quote=\"egghead91\"]Oh my goodness, my region's competition is on Feb. 3rd. Exactly two weeks. :( It'll be my first, can anyone from the Rochester, NY area tell me what it's like?[/quote]interesting... My chapter competition is on the same day...\n\n@egghead: Where are you taking it? I'm taking it at Niskayuna...\n\nWell, my first year was last year, so I only know what an easy competition is like.\n\nYou get there, I get a little shaky, maybe I'm a sugar low guy...\n\nThe sprint round is a little boring, the target round is a little better, but the team round is REALLY fun! Also, if you get into the countdown round, it should be really fun.[/quote]\r\nReally, I always thought the opposite. THe sprint round is a otn of fun (the last ten problems are actually good ones), the target is boring (the county target problems have always been much easier than they should, at least in my opinion), and the team round is boring because the questions are too easy considering that you have four people to work on it, the problems are also way too varied in difficulty (four of the first five are usually easy enough to be in the first 10 of the sprint, with one of them being hard enough for 11-20. Then 3 of the last five could be easy enough to be 1-15 and two of them could be 16-25 difficulty wise).", "Solution_9": "[quote=\"bpms\"][quote=\"1=2\"][quote=\"egghead91\"]Oh my goodness, my region's competition is on Feb. 3rd. Exactly two weeks. :( It'll be my first, can anyone from the Rochester, NY area tell me what it's like?[/quote]interesting... My chapter competition is on the same day...\n\n@egghead: Where are you taking it? I'm taking it at Niskayuna...\n\nWell, my first year was last year, so I only know what an easy competition is like.\n\nYou get there, I get a little shaky, maybe I'm a sugar low guy...\n\nThe sprint round is a little boring, the target round is a little better, but the team round is REALLY fun! Also, if you get into the countdown round, it should be really fun.[/quote]\nReally, I always thought the opposite. THe sprint round is a otn of fun (the last ten problems are actually good ones), the target is boring (the county target problems have always been much easier than they should, at least in my opinion), and the team round is boring because the questions are too easy considering that you have four people to work on it, the problems are also way too varied in difficulty (four of the first five are usually easy enough to be in the first 10 of the sprint, with one of them being hard enough for 11-20. Then 3 of the last five could be easy enough to be 1-15 and two of them could be 16-25 difficulty wise).[/quote]\r\n\r\nSame here, I always thought that sprint was the most fun because of the time pressure (although there isn't much at chapters.) Chapter target is boring because you finish really early, because usually the problems are pretty easy (chapter targets could be more fun if a calculator was prohibited). Team is okay..... but this is my opinion", "Solution_10": "[quote=\"jhredsox\"]haha, 7 teams from my chapter advance, 12 individuals :)\n\n-jorian\n\nand i'm 2nd on my team (behind guess who)\n\noh, i would like to bring up that dan's actually a lot smarter than i thought[/quote]\r\n\r\nDUDE, are you serious?!?!? Two teams from our chapter advance and I don't remember how many individuals, but definitely less than ten. Maybe three or four.\r\n\r\nI went the last two years and did really bad. (I got in the top 30 and got scores of around 30-ish :oops: ) I plan on doing a lot better this year (hopefully).\r\n\r\nI personally think team rounds are pretty fun, but there's really no one that smart at my school so they just leave all the problems that look hard to me. :roll: This year though, we might actually have a chance because last year we got third (errrr!!! So close!!) and this year we have a new kid whose pretty smart. :lol:", "Solution_11": "[quote=\"laughinghead505\"][quote=\"jhredsox\"]haha, 7 teams from my chapter advance, 12 individuals :)\n\n-jorian\n\nand i'm 2nd on my team (behind guess who)\n\noh, i would like to bring up that dan's actually a lot smarter than i thought[/quote]\n\nDUDE, are you serious?!?!? Two teams from our chapter advance and I don't remember how many individuals, but definitely less than ten. Maybe three or four.\n\nI went the last two years and did really bad. (I got in the top 30 and got scores of around 30-ish :oops: ) I plan on doing a lot better this year (hopefully).\n\nI personally think team rounds are pretty fun, but there's really no one that smart at my school so they just leave all the problems that look hard to me. :roll: This year though, we might actually have a chance because last year we got third (errrr!!! So close!!) and this year we have a new kid whose pretty smart. :lol:[/quote]\r\nLast year a 32 would've gotten you in countdown in my county.", "Solution_12": "Whoa. our cutoff, in redneck country, was a 38 or 39.\r\n\r\nand on the number of people sent to state, we send 2 teams and 4 individuals. but it really turns out to be like 2 teams and 1 individual because normally at least 3 of the top 4 individuals were on one of the two top teams.", "Solution_13": "[quote=\"laughinghead505\"][quote=\"jhredsox\"]haha, 7 teams from my chapter advance, 12 individuals :)\n\n-jorian\n\nand i'm 2nd on my team (behind guess who)\n\noh, i would like to bring up that dan's actually a lot smarter than i thought[/quote]\n\nDUDE, are you serious?!?!? Two teams from our chapter advance and I don't remember how many individuals, but definitely less than ten. Maybe three or four.\n\nI went the last two years and did really bad. (I got in the top 30 and got scores of around 30-ish :oops: ) I plan on doing a lot better this year (hopefully).\n\nI personally think team rounds are pretty fun, but there's really no one that smart at my school so they just leave all the problems that look hard to me. :roll: This year though, we might actually have a chance because last year we got third (errrr!!! So close!!) and this year we have a new kid whose pretty smart. :lol:[/quote]\r\n\r\nYeah 7 teams is a lot.... it might be because he lives in a small state.", "Solution_14": "Well, my opinion of the sprint being boring was from the 2006 chapter.(that's the only one I've been to)", "Solution_15": "I believe you are allowed to say things about rankings, team or individual, just not anything about scores, problems, solutions, difficulties, etc.", "Solution_16": "second place, ubemaya. oh my god, what a tragedy.", "Solution_17": "u got second, and think ur an idiot...ur NOT!!!", "Solution_18": "[quote=\"perfect628\"]u got second, and think ur an idiot...ur NOT!!![/quote]\r\n\r\nCan a person who got seventh think he is an idiot? :fool:", "Solution_19": "not in ur chapter!!!\r\n\r\njorian", "Solution_20": "yeah. in m ine, though, yes. i can't wait 2 c who that is...", "Solution_21": "[quote=\"star99\"]second place, ubemaya. oh my god, what a tragedy.[/quote]\r\nIt might be a tragedy for ubemaya.", "Solution_22": "not according to his goals 4 this year(in his sig)", "Solution_23": "Okay I changed my goals. Anyway, it's not the place that sucks, it's the score that I got.", "Solution_24": "uhh, just so a mod/admin doesn't see that, ur not supposed to say that ur score (sucked/was great/was average/etc.)\r\n\r\n-jorian", "Solution_25": "[quote=\"jhredsox\"]uhh, just so a mod/admin doesn't see that, ur not supposed to say that ur score (sucked/was great/was average/etc.)\n\n-jorian[/quote]\r\nBut who knows what ubemaya considers to suck? \r\nI HAVEN'T DONE CHAPTERS YET SO THE BELOW IS ALL SPECULATION\r\nLet's say the chapter was hard. I got a 45. I am content. I got a 44. I am content. I got a 43, I am unhappy. I got a 42, I am sad.\r\n\r\nLet's say it was easy. I got a 46. I am happy. I got a 45 I am sad.\r\n\r\n\r\n\r\nBig difference.", "Solution_26": "Please see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=132111]my announcement[/url] on what is and is not allowed to be discussed.\r\n\r\nIn a nutshell: you can say your ranking and your school's ranking. Nothing else. [b]No scores.[/b] This includes whether you thought you did better or worse than your were expecting.\r\n\r\nWe'll tell you when you can talk freely, but it won't be until after all the chapters are completed in late February.", "Solution_27": "T-T. Saturday: I took a Tylenol for my flu and went off to MATHCOUNTS. I'm always sick before/after math competitions / piano recitals. EH, I had to leave at like 1 something. Did not do well at ALL. And I'm still sick, missing a field trip tomorrow... *stabs the flu* :mad: :mad: :mad: :mad:\r\n\r\nAlso, there are like 30 schools in the SCV chapter so if we sent like 7 teams... :/", "Solution_28": "i wish they sent seven....they only sent 5...", "Solution_29": "I thought SCV was like the biggest chapter or something. I could just be making things up, but... SCV is so competitive!" } { "Tag": [ "LaTeX", "Support" ], "Problem": "I don't know if this has been asked before (I didn't find it). \r\n\r\nThe MiKTex download site does not include Ubuntu as one of the compatible systems. Is there another software (comparable to MiKTeX and TeXnic center) for Ubuntu?", "Solution_1": "[code]sudo apt-get install tetex-base tetex-bin tetex-extra[/code]\r\n\r\nalso http://ubuntuforums.org/showpost.php?p=26528&postcount=3", "Solution_2": "teTeX is now obsolete and has been superceded by TeX Live; also Kile uses KDE which may not be suitable if sticking to Gnome.\r\nI would look in the package manager to see if Tex Live and Texmaker (or Kile if you want to use it) are there but, if not, then there are instructions at http://ubuntuforums.org/showthread.php?t=131507.\r\n\r\nTeX Live can be found [url=http://www.tug.org/texlive/]here[/url] with Ubuntu files [url=http://packages.ubuntu.com/search?keywords=texlive]here[/url].", "Solution_3": "Wait wha?\r\n\r\nMy server still uses tetex. What's wrong with it/", "Solution_4": "There's nothing wrong with it and you can continue to use it for as long as you wish. But it's no longer maintained so will never be updated. See the De-support notice at [url=http://www.tug.org/tetex/]The teTeX Homepage[/url], the recommendation to use TeX Live and CTAN's announcement at http://www.mail-archive.com/ctan-ann@dante.de/msg01091.html", "Solution_5": "I just stumbled across this older post while reading through the forum. I've been using LaTeX on Ubuntu Linux since some years, perhaps I could give some advice on it, if you still have questions.\r\nI'm using Kile on GNOME, without any problems, it just installed some needed KDE parts, and I'm very satisfied with it. Further I'm using TeX Live mentioned above.\r\n\r\nConcerning the first question regarding MiKTeX: the MiKTeX package manager has been ported to Unix and can be used on Ubuntu Linux too. I've used it on Ubuntu 7.04, 7.10 and the current 8.04, the mpm has been very useful for me. [url=http://texblog.net/latex-archive/linux/mpm-miktex-package-manager/]Here[/url] I've described the installation of the MiKTeX package manager on Ubuntu 8.04 with the needed packages and commands.\r\n\r\nJust to mention, TeX Live 2008 has been [url=http://texblog.net/latex-archive/latex-general/texlive-2008/]released[/url], and it comes together with it's own package manager [url=http://tug.org/texlive/doc/texlive-en/texlive-en.html#x1-410006]tlmgr[/url].\r\n\r\nStefan" } { "Tag": [], "Problem": "There are $ 24$ four-digit whole numbers that use each of the\nfour digits $ 2$, $ 4$, $ 5$ and $ 7$ exactly once. Only one of these four-digit\nnumbers is a multiple of another one. Which of the following is it? \\\\ \\\\\n$ \\text{(A)}\\ 5724 \\qquad \\text{(B)}\\ 7245 \\qquad \\text{(C)}\\ 7254 \\qquad \\text{(D)}\\ 7425 \\qquad \\text{(E)}\\ 7542$ \\\\ \\\\\nSelect the most appropriate letter.", "Solution_1": "Consider the multiples of some smaller numbers that are made up of $ 2$, $ 4$, $ 5$ and $ 7$.\r\n\r\n$ 2457$: $ 4914$, $ 7371$, ...\r\n\r\n$ 2475$: $ 4950$, $ 7425$, ...\r\n\r\nWe notice that $ 7425$ is also made up of the digits $ 2$, $ 4$, $ 5$ and $ 7$, so the answer is $ \\boxed{D}$" } { "Tag": [ "trigonometry", "limit", "function", "ratio", "calculus", "derivative", "calculus computations" ], "Problem": "Find:\r\n\r\n$ \\lim_{x\\to\\infty}\\frac{x \\minus{} \\cos{x}}{x}$\r\n\r\nintuitively i think since,\r\n \r\n$ \\frac{x \\minus{} \\cos{x}}{x} \\equal{} 1 \\minus{} \\frac{\\cos{x}}{x}$ \r\n\r\nand as x goes to infinity, $ \\cos{x}$ will keep oscillating between $ \\minus{}1$ and $ 1$, but, since $ x$ gets much bigger the second term goes to zero and limit reaches $ 1$. This [u]is[/u] the answer at the back of the book. Is there any other better way of solving this perhaps using l'Hospital or simple limit techniques? Also, when $ x\\to\\infty$ what would one do in general with a trig limit since trig functions don't have any limit at infinity?", "Solution_1": "First of all, your intuitive sense is fine. You don't need any more techniques than that.\r\n\r\nThe issue I hammer on repeatedly with my class is that you have to understand size. You have to know what grows faster than what else. Looking at this, $ x\\to \\infty,$ and in fact $ x$ is a pretty simple size all by itself. Now, how big is $ \\cos x?$ I usually state that as \"one, only wigglier.\" That is, the observation that $ |\\cos x|\\le 1$ is the only thing you need to know. When you add (or subtract) two things of different size, the larger one dominates and the smaller one becomes inconsequential, so in $ x\\minus{}\\cos x,$ it's the $ x$ that really matters (unless you do something that in effect subtracts that $ x$ back off).\r\n\r\nWhat about L'H\u00f4pital? This problem breaks L'H\u00f4pital. To be more specific, the statement of L'H\u00f4pital is this: if the original limit is a $ \\frac00$ or $ \\frac{\\infty}{\\infty}$ indeterminate case, and [b]if the ratio of the derivatives has a limit[/b], then the original limit is the limit of the ratio of the derivatives. But what if the ratio of the derivatives doesn't exist? In this case it doesn't - but the original limit exists anyway.\r\n\r\nI have no doubt that the textbook author intentionally planted this problem to demonstrate that particular shortcoming of L'H\u00f4pital.\r\n\r\nI also spend a lot of effort in my class trying both to talk students out of using L'H\u00f4pital for everything and also talking them out of trying to make everything a simple equality (which is what I think you're uncomfortable with, since you can't write $ \\lim_{x\\to\\infty}\\cos x$ as anything meaningful). The understanding of size is a big part of that.", "Solution_2": "[quote]which is what I think you're uncomfortable with, since you can't write $ \\lim_{x\\to\\infty}\\cos x$ as anything meaningful[/quote]\r\n\r\nYes this was the main reason I asked the question. Thanks a lot for the detailed explanation.", "Solution_3": "hello, it is\r\n$ \\lim_{x \\to \\infty}\\frac{x\\minus{}\\cos(x)}{x}\\equal{}1$\r\nSonnhard." } { "Tag": [ "quadratics" ], "Problem": "The roots of $ (x^2\\minus{}3x\\plus{}2)(x)(x\\minus{}4)\\equal{}0$ are:\n\n$\\textbf{(A)}\\ 4\\qquad\n\\textbf{(B)}\\ 0\\text{ and }4 \\qquad\n\\textbf{(C)}\\ 1\\text{ and }2 \\qquad\n\\textbf{(D)}\\ 0,1,2\\text{ and }4\\qquad\n\\textbf{(E)}\\ 1,2\\text{ and }4$", "Solution_1": "[quote=\"Smartguy\"]The roots of $ (x^2 \\minus{} 3x \\plus{} 2)(x)(x \\minus{} 4) \\equal{} 0$ are:\n(A) 4\n(B) 0 and 4\n(C) 1 and 2\n(D) 0, 1, 2, and 4\n(E) 1, 2, and 4[/quote]\r\n\r\nFactor $ x^2\\minus{}3x\\plus{}2$ to get $ (x\\minus{}1)(x\\minus{}2)$ so its D", "Solution_2": "You can also use process of elimination. Since x is one of the terms, then one of the roots is 0 and so that eliminates all answers except B and D. Clearly, substituting one into the quadratic results in another 0. Therefore, D." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $x, y , z$ three nonnegative real numbers. Prove that \\[1+\\frac{xyz}{(x+y)(y+z)(z+x)}\\geq\\frac{27xyz(x+y+z)}{8(xy+yz+zx)^{2}}.\\]", "Solution_1": "$1+\\frac{xyz}{(x+y)(y+z)(z+x}=\\frac{(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)}$\r\nSo our inequality rewrites as:\r\n$8(xy+yz+zx)^{3}\\geq 27xyz(x+y)(y+z)(z+x)=27(zx+yz)(xy+zx)(yz+xy)$\r\nLet $xy=a, yz=b, zx=c$. We have to prove that:\r\n$8(a+b+c)^{3}\\geq 27(a+b)(b+c)(c+a)$\r\nBut this is just AM-GM for the numbers $a+b, b+c, c+a$ after cubing." } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ R$ be a ring, $ l,n\\in\\mathbb{N}$ and $ \\omega\\in R$ is a primitive nth root of unity. \r\nProve that for any $ 1.<", "Solution_6": "I would make a trigonometric substitution...$ \\cos{\\alpha} \\equal{} \\sqrt {mn}$ or something else...", "Solution_7": "Positive reals $ m,n,p$ satisfy $ mn \\plus{} np \\plus{} pm \\plus{} 2mnp \\equal{} 1$.\r\nProve that $ \\sqrt {mn} \\plus{} \\sqrt {np} \\plus{} \\sqrt {pm} \\leq \\frac {17}{12} \\plus{} mnp$\r\n\r\n[hide=\"Hint\"]\nFrom the given condition, we can deduce that there exist positive reals $ x,y,z$ s.t. $ m \\equal{} \\frac {x}{y \\plus{} z}, n \\equal{} \\frac {y}{z \\plus{} x}, p \\equal{} \\frac {z}{x \\plus{} y}$\n[hide=\"Why?\"]\nThe given condition is equivalent to $ \\frac {1}{m \\plus{} 1} \\plus{} \\frac {1}{n \\plus{} 1} \\plus{} \\frac {1}{p \\plus{} 1} \\equal{} 2$.\n\nLet $ \\frac {1}{m \\plus{} 1} \\equal{} a, \\frac {1}{n \\plus{} 1} \\equal{} b, \\frac {1}{p \\plus{} 1} \\equal{} c$.\nWe have $ m \\equal{} \\frac {1 \\minus{} a}{a} \\equal{} \\frac {b \\plus{} c \\minus{} a}{2a}$ and analogously for $ n,p$.\nSetting $ b \\plus{} c \\minus{} a \\equal{} x, c \\plus{} a \\minus{} b \\equal{} y, a \\plus{} b \\minus{} c \\equal{} z$, we have the desired result. ($ x,y,z$ are positive because $ m,n,p$ are positive)\n[/hide]\n[/hide]", "Solution_8": "[quote=\"Harry Potter\"]$ m,n,p$ are 3 positive real numbers such that : $ mn \\plus{} np \\plus{} mp \\plus{} 2mnp \\equal{} 1$ . Show that : $ \\sqrt {mn} \\plus{} \\sqrt {np} \\plus{} \\sqrt {mp}\\leq \\frac {17}{12} \\plus{} mnp$\n\nI edited Problem >.<[/quote]\r\n\r\n$ k\\equal{}\\frac {17}{12}$ is the best constat?! it such that:\r\n\\[ f(m,n,p)\\equal{}\\sqrt {mn} \\plus{} \\sqrt {np} \\plus{} \\sqrt {pm} \\minus{}mnp \\leq k ?!\\]\r\n\r\nDo you know when does equality hold?", "Solution_9": "[quote=\"zaizai-hoang\"][quote=\"Harry Potter\"]$ m,n,p$ are 3 positive real numbers such that : $ mn \\plus{} np \\plus{} mp \\plus{} 2mnp \\equal{} 1$ . Show that : $ \\sqrt {mn} \\plus{} \\sqrt {np} \\plus{} \\sqrt {mp}\\leq \\frac {17}{12} \\plus{} mnp$\n\nI edited Problem >.<[/quote]\n\n$ k \\equal{} \\frac {17}{12}$ is the best constat?! it such that:\n\\[ f(m,n,p) \\equal{} \\sqrt {mn} \\plus{} \\sqrt {np} \\plus{} \\sqrt {pm} \\minus{} mnp \\leq k ?!\n\\]\nDo you know when does equality hold?[/quote]\r\n\r\nWhen m=n=p :) \r\nWhat Held ? :D \r\n\r\nTo Zaizai : T\u1edb d\u00f9ng \u0111\u1ea1o h\u00e0m \u0111\u1ebf CM b\u00e0i n\u00e0y :lol: Nh\u01b0ng f\u1ea3i \u0111\u1ea1o h\u00e0m m\u1ed9t c\u00e1ch th\u00f4ng minh :blush: N\u1ebfu c\u1eadu c\u00f3 gi\u1ea3i th\u00ec post l\u00ean nh\u00e9 . C\u00e1m \u01a1n nhi\u1ec1u ^^", "Solution_10": "When $ m\\equal{}n\\equal{}p\\equal{}1/2 ?!$ Did you means like that?!\r\nBut when $ m\\equal{}n\\equal{}p\\equal{}1/2$ :\r\n\\[ LHS\\equal{}3/2 \\neq RHS\\equal{}37/24\\] \r\n\r\nI am also misunderstand :maybe:", "Solution_11": "Oh SOrry ^^ \r\nI don't know about equality >.< I created it from Other Problem ^^\r\nIt \u00eds true 100% :D", "Solution_12": "[quote=\"Harry Potter\"]Oh SOrry ^^ \nI don't know about equality >.< I created it from Other Problem ^^\nIt \u00eds true 100% :D[/quote]\r\n\r\nYou don't know this? Really? But you said this one was solved by you by \"special Derivative's method\", totally what do you mean?", "Solution_13": "Oh My GOD :blush: :blush: \r\n\r\n[hide=\"To Zaizai\"]T\u1edb \u0111\u00e3 r\u1ea5t l\u1ecbch s\u1ef1 khi d\u00f9ng Ti\u1ebfng Vi\u1ec7t Zaizai \u1ea1 , thi\u1ebfp ngh\u0129 vi\u1ec7c t\u1edb vi\u1ebft m\u1ea5y c\u00e2u tr\u00ean c\u0169ng kh\u00f4ng t\u1edb m\u1ee9c khi\u1ebfn m\u1ed9t h\u1ecdc sinh su\u1ea5t s\u1eafc v\u1ec1 b\u1ea5t \u0111\u1eb3ng th\u1ee9c nh\u01b0 c\u1eadu ph\u1ea3i t\u00f2 m\u00f2 x\u0103m xoi nh\u01b0 v\u1eady ch\u1ee9 :( hay \u0111\u00f3 ch\u1ec9 l\u00e0 c\u00e1i m\u00e3 c\u1ee7a c\u1eadu th\u00f4i :wink: . N\u1ebfu nh\u01b0 c\u1eadu c\u00f3 l\u1eddi gi\u1ea3i hay nh\u1eefng \u00fd t\u01b0\u1edfng cho b\u00e0i to\u00e1n th\u00ec post l\u00ean c\u00f2n n\u1ebfu kh\u00f4ng th\u00ec kh\u00f4ng n\u00ean post spam nh\u1eefng th\u1ee9 kh\u00f4ng li\u00ean quan , t\u1edb l\u00e0 ch\u1ee7 topic v\u00e0 t\u1edb kh\u00f4ng th\u00edch ng\u01b0\u1eddi ta spam trong Topic c\u1ee7a m\u00ecnh ! [/hide]" } { "Tag": [ "inequalities", "function", "inequalities unsolved", "n-variable inequality" ], "Problem": "let $ x_1$ , $ x_2$ ,.....,$ x_n$ positive reals numbers , $ n > 1$ , and $ \\frac{1}{x_1 \\plus{}1}\\plus{}\\frac{1}{x_2\\plus{}1}\\plus{}.....\\plus{}\\frac{1}{x_n \\plus{}1}\\equal{}1$\r\n\r\nprove that :\r\n\r\n$ \\sqrt[n]{x_1 x_2 ...x_n} \\geq n\\minus{}1$", "Solution_1": "[quote=\"y-a-s-s-i-n-e\"]let $ x_1$ , $ x_2$ ,.....,$ x_n$ positive reals numbers , $ n > 1$ , and $ \\frac {1}{x_1 \\plus{} 1} \\plus{} \\frac {1}{x_2 \\plus{} 1} \\plus{} ..... \\plus{} \\frac {1}{x_n \\plus{} 1} \\equal{} 1$\n\nprove that :\n\n$ \\sqrt [n]{x_1 x_2 ...x_n} \\geq n \\minus{} 1$[/quote]\r\nWe have:$ \\frac{x_1}{x_1\\plus{}1}\\equal{}\\frac{1}{x_2\\plus{}1}\\plus{}....\\plus{}\\frac{1}{x_n\\plus{}1} \\ge (n\\minus{}1) \\sqrt[n\\minus{}1]{\\frac{1}{(x_2\\plus{}1)...(x_n\\plus{}1)}}$\r\nSimilar, we have Q.E.D :)", "Solution_2": "Since the function $ f(x) \\equal{} \\frac {1}{1 \\plus{} e^x}$ is convex, we have from Jensen's Inequality\r\n\r\n$ \\frac {1}{n} \\left( \\frac {1}{1 \\plus{} e^{y_1}} \\plus{} \\frac {1}{1 \\plus{} e^{y_2}} \\plus{} ... \\plus{} \\frac {1}{1 \\plus{} e^{y_n}} \\right) \\ge \\frac {1}{1 \\plus{} e^{\\frac {y_1 \\plus{} y_2 \\plus{} ... \\plus{} y_n}{n}}}$ where $ e^{y_i} \\equal{} x_i$\r\n\r\nHence $ \\frac {1}{1 \\plus{} x_1} \\plus{} \\frac {1}{1 \\plus{} x_2} \\plus{} ... \\plus{} \\frac {1}{1 \\plus{} x_n} \\equal{} 1 \\ge \\frac {n}{1 \\plus{} \\sqrt [n] {x_1x_2x_3...x_n}}$\r\n\r\nFrom which the inequality follows", "Solution_3": "[quote=\"hsbhatt\"]Since the function $ f(x) \\equal{} \\frac {1}{1 \\plus{} e^x}$ is convex,...[/quote]\r\nAre you sure?", "Solution_4": "thank's for the elegant solution", "Solution_5": "[quote=\"quykhtn-qa1\"]We have:$ \\frac{x_1}{x_1\\plus{}1}\\equal{}\\frac{1}{x_2\\plus{}1}\\plus{}....\\plus{}\\frac{1}{x_n\\plus{}1} \\ge (n\\minus{}1) \\sqrt[n\\minus{}1]{\\frac{1}{(x_2\\plus{}1)...(x_n\\plus{}1)}}$\nSimilar, we have Q.E.D :)[/quote]\n\nDon't need to revive.. But, i still didn't understand. What should we do after get \n\n$ \\frac{x_1}{x_1\\plus{}1} \\ge (n\\minus{}1) \\sqrt[n\\minus{}1]{\\frac{1}{(x_2\\plus{}1)...(x_n\\plus{}1)}}$", "Solution_6": "it is just AM-GM :wink:", "Solution_7": "[quote=\"youarebad\"][quote=\"quykhtn-qa1\"]We have:$ \\frac{x_1}{x_1\\plus{}1}\\equal{}\\frac{1}{x_2\\plus{}1}\\plus{}....\\plus{}\\frac{1}{x_n\\plus{}1} \\ge (n\\minus{}1) \\sqrt[n\\minus{}1]{\\frac{1}{(x_2\\plus{}1)...(x_n\\plus{}1)}}$\nSimilar, we have Q.E.D :)[/quote]\n\nDon't need to revive.. But, i still didn't understand. What should we do after get \n\n$ \\frac{x_1}{x_1\\plus{}1} \\ge (n\\minus{}1) \\sqrt[n\\minus{}1]{\\frac{1}{(x_2\\plus{}1)...(x_n\\plus{}1)}}$[/quote]\nSimilar we have\n\\[\\frac{x_{2}}{1+x_{2}}\\ge (n-1)\\sqrt[n-1]{\\frac{1}{(x_{1}+1)(x_{3}+1)...(x_{n}+1)}}\\]\n\\[...\\]\n\\[\\frac{x_{n}}{1+x_{n}}\\ge (n-1)\\sqrt[n-1]{\\frac{1}{(x_{1}+1)(x_{2}+1)...(x_{n-1}+1)}}\\]\nFrom here we have\n\\[\\frac{x_{1}x_{2}...x_{n}}{(1+x_{1})(1+x_{2})...(1+x_{n})}\\geq \\frac{(n-1)^{n}}{(1+x_{1})(1+x_{2})...(1+x_{n})}\\]\nTherefore \n\\[ \\sqrt[n]{x_{1}x_{2}...x_{n}}\\geq n-1 \\]\nComplete prove :lol:", "Solution_8": "[quote=y-a-s-s-i-n-e]let $ x_1$ , $ x_2$ ,.....,$ x_n$ positive reals numbers , $ n > 1$ , and $ \\frac{1}{x_1 \\plus{}1}\\plus{}\\frac{1}{x_2\\plus{}1}\\plus{}.....\\plus{}\\frac{1}{x_n \\plus{}1}\\equal{}1$ prove that $$ \\sqrt[n]{x_1 x_2 ...x_n} \\geq n\\minus{}1$$[/quote]\n[url=https://artofproblemsolving.com/community/c6h2330399p18696367]2001 Kazakhstan[/url]\n" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Prove that the maximal value of $\\prod_{i=1 and a+b+c+d=7, and no circular permutations of a solution, this is the same as w+x+y+z=3, with w,x,y,z>=0\n\n3,0,0,0 [1 way]\n2,1,0,0 [2 ways]\n1,1,1,0 [1 way]\n\nso 4[/hide]", "Solution_10": "[quote=\"vishalarul\"]Seven points are equally distributed on a circle. How many non-congruent quadrilaterals can be drawn w/ vertices chosen from among the seven points?[/quote]\r\njorian is correct\r\n[hide=\"answer\"]\nfirst point has 6 to choose from, then decreases\n(6*5*4)/(3*2) =20\n5*4*3)/(3*2)=10\n4*3*2)/(3*2)=4\n6/6=1\n20+10+4+1=35\neach shape is repeated 7 times.\n35/7 equals 5.\n[/hide]", "Solution_11": "[quote=\"Fanatic\"]Uhh... my answer key says jhredsox's answer was correct.[/quote]\r\n[hide=\"Please show your work.\"]\nIf the answer is 5, as several of you have claimed, please prove you are correct, by listing five non-congruent quadrilaterals? E.g., Label the seven points, A, B, C, D, E, F, and G, in order around the circle. The quadrilaterals are ABCD, ABCE, etc. Answer keys have been known to be incorrect.[/hide]", "Solution_12": "It sounds like many are treating this like a key chain problem, but since some possible quadrilaterals are the same as others that are not only rotations but reflections, while others are not repeated as reflections, I am not sure you can treat it that way. I tried actually making them, and I can't get 5 unique ones no matter how hard I try and it seems clear there aren't.\r\n\r\nHmm....", "Solution_13": "i am normal and i do not am not and will not be good at this kind of quetion because i have not just broken my scooter today!!!!!!! \r\n :arrow: :ninja: :idea:", "Solution_14": "[quote=\"jhredsox\"]is it just[hide](7C4)/7=5\n\nhmm...that doesn't seem quite right for some odd reason[/hide]\n\njorian[/quote]\r\n\r\n7C4 because you are picking 4 points from 7 possible points\r\n\r\ndivided by 7 because you repeat each 7 times \r\n\r\n...like...\r\n\r\n1,2,3,4---2,3,4,5---3,4,5,6---4,5,6,7---5,6,7,1---6,7,1,2---7,1,2,3\r\n\r\n1,3,5,7---2,4,6,1---3,5,7,2---4,6,1,3---5,7,2,4---6,1,3,5---7,2,4,6\r\n\r\n1,4,7,2---2,5,1,4---3,6,2,5---4,7,3,6---5,1,4,7---6,2,5,1---7,3,6,1\r\n\r\n1,5,2,6---2,6,3,7---3,7,4,1---4,1,5,2---5,2,6,3---6,3,7,4---7,4,1,5\r\n\r\n1,6,4,2---2,7,5,3---3,1,6,4---4,2,7,5---5,3,1,6---6,4,2,7---7,5,3,1\r\n\r\n\r\nthose are all of them :)\r\n\r\n-jorian", "Solution_15": "[quote=\"jhredsox\"][quote=\"jhredsox\"]is it just[hide](7C4)/7=5\n\nhmm...that doesn't seem quite right for some odd reason[/hide]\n\njorian[/quote]\n\n7C4 because you are picking 4 points from 7 possible points\n\ndivided by 7 because you repeat each 7 times \n\n...like...\n\n1,2,3,4---2,3,4,5---3,4,5,6---4,5,6,7---5,6,7,1---6,7,1,2---7,1,2,3\n\n1,3,5,7---2,4,6,1---3,5,7,2---4,6,1,3---5,7,2,4---6,1,3,5---7,2,4,6\n\n1,4,7,2---2,5,1,4---3,6,2,5---4,7,3,6---5,1,4,7---6,2,5,1---7,3,6,1\n\n1,5,2,6---2,6,3,7---3,7,4,1---4,1,5,2---5,2,6,3---6,3,7,4---7,4,1,5\n\n1,6,4,2---2,7,5,3---3,1,6,4---4,2,7,5---5,3,1,6---6,4,2,7---7,5,3,1\n\n\nthose are all of them :)\n\n-jorian[/quote]\r\n\r\nlook at your 1,3,5,7 and 1,6,4,2 (or 1,2,4,6 if you want to draw it inorder) case-they are the exact same..the answer i get is 4 \r\nmaybe im wrong, but i think it is 4.", "Solution_16": "oh i'm sorry\r\n\r\nperhapas 1,2,4,7\r\n\r\ni know there's another pattern\r\n\r\nthe point is, you divide by 7 because each can be repeated 7 times\r\n\r\n-jorian", "Solution_17": "Why would you do that-I like ALtheman's way and he got 4, the same answer as mine\r\n\r\nthere are 7 arc lengths and each side of the quadrilateral can contain the arc lengths that add up to 7\r\n\r\nthey can be 2 2 2 1\r\n3 1 1 1\r\n2 1 1 3\r\n2 1 3 1\r\n\r\napart so 4 (for example 2 3 1 1 would not work because you can rotate that-imagine a circle, into 2 1 1 3)\r\n\r\nand fanatic-are you sure that your teacher's answer key correct?", "Solution_18": "Not positive, I guess it COULD be wrong... but I don't think so.\r\n\r\nI did this problem under time constraints, and I know I'm bad at visual stuff. Thus, I just did Jorian's method 7 choose four divided by 7 and it worked... not quite sure why, but intuitively it seemed correct.\r\n\r\nYou're probably missing a quadrilateral somewhere...", "Solution_19": "[hide=\"think i got it if you want\"]\n\nThe $_{7}C_{3}$ divided by 7 idea is a really good one, but there is one problem. You are dividing by 7 because we assume there are 7 of each of the quadrilaterals. However since the Quad created by three points in a row and then skipping 1 can be created 14 ways since it is not symmetric and can be created by taking 3 in a row and then skipping 2.\n\nIn the picture you can see the 5 you get by doing $_{7}C_{3}$ divided by 7, but the top 2 are the same just flipped, so I believe there are only 4.\n\nOh, and I checked the answer key [hide]and 4 is the answer[/hide] \n\n[/hide]", "Solution_20": "don't yu mean 7C4?\r\n\r\nhow would you figure this out on the test?\r\n\r\njorian", "Solution_21": "$_{7}C_{3}$ is the same as $_{7}C_{4}$", "Solution_22": "[quote=\"Altheman\"][hide]the number of arcs that each side subtends uniquely defines a quadrilateral; suppose we have the arcs subtended as a,b,c,d, we have that a,b,c,d>=1 and a+b+c+d=7, and no circular permutations of a solution, this is the same as w+x+y+z=3, with w,x,y,z>=0\n\n3,0,0,0 [1 way]\n2,1,0,0 [2 ways]\n1,1,1,0 [1 way]\n\nso 4[/hide][/quote]\n\nCould you elaborate on this a bit more? It seems interesting but I don't get the concepts behind it.\n\nIs there any other way besides listing and this? I'm very bad at listing >__>" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "How to prove that the group $ G$ defined by the presentation $ $ is reduced to the neutral element ?\r\nI proved that $ a^{2}=b^{-1}a^{3}b$, $ a^{4}=b^{-1}a^{6}b$, etc... but if somebody can give me an idea, thanks for advance! Bob.", "Solution_1": "http://mathlinks.ro/Forum/viewtopic.php?t=151256" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "IMO Shortlist" ], "Problem": "For the USAMO, if you get an answer but you can't prove how you got it, should you bother writing it down? How much is the actual answer to a problem worth, if anything?", "Solution_1": "99% of the time, a USAMO problem starts with \"Prove that...\" so there is no actual answer; it is given to you, and you need to prove it. However, there are some problems where it might say \"find, with proof...\", and I think the answer is worth 0 points there. (The USAMO is very strict, remember).\r\n\r\nBut always write down WHATEVER you have, it doesn't hurt (you can't get a negative score ;))", "Solution_2": "Depends on the problem. On many problems, such as most functional equations, the answer will be obvious. Obvious answers get you no points, but if the answer is not obvious, you might get 1 point. If you can come up with an intuitive argument why it is the answer, you might get 1 or 2 point.", "Solution_3": "Just to throw out an extreme example, the answer for [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1186813#1186813]this IMO shortlist problem[/url] is in Zuming's words \"worth a lot of points\", because half of the problem is solved when you realize what it is. When it was given in the TST 2008 no one got the answer right during the test, and when told what the right answer actually was afterwards several people came up with solutions very quickly.\r\n\r\nBut generally, pythag's description is accurate, though I would caution that the \"intuitive argument\" better have some ideas used in the actual solution to merit more points." } { "Tag": [ "email", "\\/closed" ], "Problem": "Hello. I stoped to receive topics reply notification messages. I remember I didn't visit two topics of which notifications I received. As I understand, no more notifications will be sent until I visit these topics. The trouble is that I don't remember what these topic were since all messages in my mail box are marked as \"read\". How do I continue receiving reply notifications?", "Solution_1": "You can just click \"view your posts\" (you don't have that many) and visit all of them. \r\n\r\nI found email notifications somewhat distracting and turned them off - the \"view your posts\" list serves the same purpose.", "Solution_2": "Yes. But it it also the case that I subscribe to a topic I am interested in, which I didn't create and for which I haven't posted answer yet. Such topics are not in \"view your posts\" section. Without notifications I must visit the forum very often so that be informed about answers for all messages of that sort.\r\n\r\nAnd these two topics I didn't visit was among those I didn't create and gave no answer for.", "Solution_3": "You could add the topic to your favorites, and each time you wished to access it, you could click the favorites button in your userbar.\r\n\r\n(Too add the topic to your favorites: There is a star under the page numbers for the topic. Clicking it will add it. To view your favorites, there's a button saying \"Favorites\" next to \"profile\". At least in MathLinks theme. :maybe:)" } { "Tag": [], "Problem": "How many unique sets of three prime numbers exist for which the sum of the members of the set is 44?", "Solution_1": "$ 44$ is even number.\r\nSo, even+odd+odd i.e. $ 2\\plus{}\\text{odd}\\plus{}\\text{odd}\\equal{}44$.\r\nSo, sum of the two prime number is $ 42$.\r\n\r\n$ 44\\minus{}2\\minus{}3\\equal{}39 \\not\\in \\text{prime}$\r\n$ 44\\minus{}2\\minus{}5\\equal{}37 \\in \\text{prime}$\r\n$ 44\\minus{}2\\minus{}7\\equal{}35 \\not\\in \\text{prime}$\r\n$ 44\\minus{}2\\minus{}11\\equal{}31 \\in \\text{prime}$\r\n$ 44\\minus{}2\\minus{}13\\equal{}29 \\in \\text{prime}$\r\n$ 44\\minus{}2\\minus{}17\\equal{}25 \\not\\in \\text{prime}$\r\n$ 44\\minus{}2\\minus{}19\\equal{}23 \\in \\text{prime}$\r\n\r\nTherefore, $ (2, \\ 5, \\ 37)$, $ (2, \\ 11, \\ 31)$, $ (2, \\ 13, \\ 29)$, $ (2, \\ 19, \\ 23)$.\r\nThus $ \\boxed{4}$.", "Solution_2": "just curious, but could you divide 44 by 11, because 11 is it's biggest prime, and get 4 that way?" } { "Tag": [ "pigeonhole principle" ], "Problem": "Given 16 distinct positive integers not exceeding 100, prove that it is possible to choose 4 distinct integers $ a$, $ b$, $ c$ and $ d$ such that $ a\\plus{}c \\equal{} b \\plus{} d$.\r\n\r\nIs this statement still true for 15 distinct positive integers?", "Solution_1": "[hide]\nSuppose we have $ a,b,c,d$ distinct integers with $ a0$\n\nSo the question becomes: given $ 16$ positive integers less than $ 100$, we can find two pairs who have the same difference.\n\nNow we have $ 1\\le |a_i \\minus{} a_j| \\le 99$. However there are $ \\binom{16}{2} \\equal{} 120$ pairs $ (a_i,a_j)$\n So by the box principle we have two pairs such that; $ a_i \\minus{} a_j \\equal{} a_k \\minus{} a_l$\n\nOr; $ a_i \\plus{} a_l \\equal{} a_k \\plus{} a_j$[/hide]", "Solution_2": "[quote=\"ocha\"][hide]\nSuppose we have $ a,b,c,d$ distinct integers with $ a < b$ and $ c < d$ and $ a \\plus{} b \\equal{} c \\plus{} d$\nThis implies that $ a < c < d < b$. Therefore we can write the expression as $ b \\minus{} c \\equal{} d \\minus{} a$\nWhere $ b \\minus{} c$ and $ d \\minus{} a > 0$\n\nSo the question becomes: given $ 16$ positive integers less than $ 100$, we can find two pairs who have the same difference.\n\nNow we have $ 1\\le |a_i \\minus{} a_j| \\le 99$. However there are $ \\binom{16}{2} \\equal{} 120$ pairs $ (a_i,a_j)$\n So by the box principle we have two pairs such that; $ a_i \\minus{} a_j \\equal{} a_k \\minus{} a_l$\n\nOr; $ a_i \\plus{} a_l \\equal{} a_k \\plus{} a_j$[/hide][/quote]\r\n\r\nhow do u know that $ a_i, a_j, a_k, a_l$ are distinct." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "would it be precal, calc etc?", "Solution_1": "??? What am I missing here?" } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ G$ be a group of order $ p^2$. Prove that $ G$ is abelian.", "Solution_1": "nontrivial $ p$-groups have nontrivial center. so in our case $ G/Z(G)$ has order $ 1$ or $ p$. so $ G/Z(G)$ is cyclic, thus $ G$ is abelian." } { "Tag": [ "algebra", "polynomial", "trigonometry", "ratio", "conics", "parabola", "complex numbers" ], "Problem": "Just a reminder that we will have All-State Math Team practices on Saturday, May 9th, 16th, and 25th. If you haven't sent Mr. R your information sheet and your answers to the problems you were asked to work on, do so as soon as possible.\r\n\r\nBy the way, there are some really good problems floating around in the Intermediate and Pre-Olympiad section lately.", "Solution_1": "Good job at the practices today for those who came. For those in my room I hope you got something out of our talk about Complex Numbers. I know it went by pretty fast. Here are some practice problems to reinforce what we talked about:\r\n\r\n1.) Find all complex numbers $ z$ such that $ z^2 \\equal{} i$\r\n\r\n2.) Find all complex numbers $ z$ such that $ z^3 \\equal{} \\minus{} 8$\r\n\r\n3.) Find all complex numbers $ z$ such that $ z^6 \\plus{} z^3 \\plus{} 1 \\equal{} 0$\r\n\r\n4.) $ P(x)$ is a 6th degree polynomial with integer coefficients such that $ 4$, $ \\sqrt {3}$, and $ 2 \\minus{} i$ are roots. If $ P(2) \\equal{} 17$, find $ P(0)$.\r\n\r\n-----------------------------------------------------------------------------\r\nHere are the solutions to Set 1 of the practice problems given when you were accepted to the team:\r\n\r\n1.) If the number $ A3640548981270644B$ is divisible by 99, compute the ordered pair of digits $ (A,B)$\r\n\r\n[hide=\"Solution:\"] A number is divisible by 9 if and only if the sum of its digits is divisible by 9. So $ A \\plus{} B \\equiv 1 \\bmod 9$\n\nA number is divisible by 11 if and only if the alternating sum of its digits is divisible by 11. So $ A \\minus{} B \\equiv 8 \\bmod 11$\n\n$ A \\equal{} 9$, $ B \\equal{} 1$ satisfies these conditions, so the answer is $ \\boxed{(9,1)}$[/hide]\n\n2.) If $ n$ is the smallest positive integer for which positive $ a$ and $ b$ exist that make $ (a \\plus{} bi)^n \\equal{} (a \\minus{} bi)^n$, compute the numerical value of $ \\frac {b}{a}$\n\n[hide=\"Solution:\"] Rewrite $ a \\plus{} bi$ in polar form as $ re^{i\\theta}$. Then $ a \\minus{} bi$ will be $ re^{ \\minus{} i\\theta}$.\n\nSo $ (a \\plus{} bi)^n \\equal{} (a \\minus{} bi)^n$ means\n$ (re^{i\\theta})^n \\equal{} (re^{ \\minus{} i\\theta})^n$\n$ r^ne^{in\\theta} \\equal{} r^ne^{ \\minus{} in\\theta}$\n\nThis implies either $ r \\equal{} 0$ or $ e^{in\\theta} \\equal{} e^{ \\minus{} in\\theta}$\nSince we know that $ a$ and $ b$ are positive, the former cannot be true (because then both $ a$ and $ b$ would be zero).\n\nSo the latter must be true, and also $ \\theta$ must be between $ 0$ and $ \\frac {\\pi}{2}$ (exclusive).\n\n$ e^{in\\theta} \\equal{} e^{ \\minus{} in\\theta}$ means $ e^{2in\\theta} \\equal{} 1$\nSo $ 2n\\theta \\equal{} 2\\pi k$ for some integer $ k \\implies \\theta \\equal{} \\frac {2\\pi k}{n}$\n\nThe smallest positive integer value of $ n$ that will make $ 0 < \\theta < \\frac {\\pi}{2}$ is $ n \\equal{} 3$ (with $ k \\equal{} 1$).\n\nSo $ \\theta \\equal{} \\frac {\\pi}{3}$ and $ \\tan{\\theta} \\equal{} \\frac {b}{a}$. So $ \\boxed{\\frac {b}{a} \\equal{} \\sqrt {3}}$[/hide]\n\n3.) A mixture of wine and water is made in the ratio of wine:total = k:m. Adding $ x$ units of water or removing $ x$ units of wine $ (x \\neq 0)$ each produces the same new ratio of wine:total. Compute the numerical value of this new ratio.\n\n[hide=\"Solution:\"] Let $ a$ be the amount of wine and $ b$ the amount of water.\nThen $ \\frac {a}{a \\plus{} b \\plus{} x} \\equal{} \\frac {k}{m} \\implies am \\equal{} ka \\plus{} kb \\plus{} kx$,\nand $ \\frac {a \\minus{} x}{a \\plus{} b \\minus{} x} \\equal{} \\frac {k}{m} \\implies am \\minus{} mx \\equal{} ka \\plus{} kb \\minus{} kx$\n\nSubtracting the second equation from the first gives $ mx \\minus{} kx$\nSo $ m \\equal{} k$ and $ \\boxed{\\frac {k}{m} \\equal{} 1}$[/hide]\n\n4.) The length of each side of regular hexagon $ ABCDEF$ is $ 2$. $ A$, $ B$, and $ C$ are the centers of three mutually extermally tangent circles. Compute the radius of the smallest of these three circles.\n\n[hide=\"Solution:\"] First, draw a picture (not shown here). Let $ r_1$ be the radius of the circle centered at $ B$ (which will be the smallest circle), and let $ r_2$ be the radius of the circles centered at $ A$ and $ C$ (notice these circles will have the same radius).\n\nSince $ AB \\equal{} 2$ we have $ r_1 \\plus{} r_2 \\equal{} 2$,\nand because $ AC \\equal{} 2\\sqrt {3}$ (by 30-60-90 triangles, or by law of cosines), we have $ 2r_2 \\equal{} 2\\sqrt {3}$\n\nSolving, we find the $ \\boxed{r_1 \\equal{} 2 \\minus{} \\sqrt {3}}$[/hide]\n\n5.) The parabola $ y \\equal{} ax^2 \\plus{} bx \\plus{} c$ crosses the x-axis at $ (p,0)$ and $ (q,0)$, both to the right of the origin. A circles passes through these two points. Find the length of a tangent from the origin to the circle, in terms of one or more of the coefficients $ a$, $ b$, and $ c$.\n\n[hide=\"Solution:\"] I assumed that the circle was centered on the x-axis (hey, they didn't say that it couldn't be!!). So now draw a picture (not shown).\n\nThe center of the circle will be midway between the roots, at $ \\left(\\frac {p \\plus{} q}{2},0\\right)$, and the radius of the circle will be $ r \\equal{} \\frac {p \\minus{} q}{2}$\n\nThe tangent line from the origin to the circle will make a right angle with the radius of the circle, so by Pythagoras' Theorem the length of the tangent line will be\n$ L \\equal{} \\sqrt {\\left(\\frac {p \\plus{} q}{2}\\right)^2 \\minus{} \\left(\\frac {p \\minus{} q}{2}\\right)^2} \\equal{} \\boxed{\\sqrt {pq}}$ (using difference of squares to simplify)[/hide]\n\n6.) In triangle $ ABC$, $ D$ is the midpoint of $ BC$. If $ \\angle BAD \\equal{} 15^o$ and $ \\angle CAD \\equal{} 30^o$, compute the tangent of angle $ C$.\n\n[hide=\"Solution:\"]Draw a picture!! Let $ BD \\equal{} CD \\equal{} x$ and let $ AD \\equal{} d$. Let angle $ C \\equal{} \\theta$, making angle $ B \\equal{} 135^o \\minus{} \\theta$.\n\nBy the law of sines on triangle $ ABD$,\n$ \\frac {x}{\\sin{15}} \\equal{} \\frac {d}{\\sin{(135 \\minus{} \\theta)}}$\nand by the law of sines on triangle $ ACD$,\n$ \\frac {x}{\\sin{30}} \\equal{} \\frac {d}{\\sin{\\theta}}$\n\nCombining these equations gives\n$ \\frac {\\sin{15}}{\\sin{(135 \\minus{} \\theta)}} \\equal{} \\frac {\\sin{30}}{\\sin{\\theta}}$\n\n$ \\implies \\sin{15} \\cdot \\sin{\\theta} \\equal{} \\sin{30} \\cdot \\sin{(135 \\minus{} \\theta)}$\n\nNote: $ \\sin{15} \\equal{} \\sin{(45 \\minus{} 30)} \\equal{} \\frac {\\sqrt {2}}{2} \\cdot \\frac {\\sqrt {3}}{2} \\minus{} \\frac {\\sqrt {2}}{2} \\cdot \\frac {1}{2} \\equal{} \\frac {\\sqrt {6} \\minus{} \\sqrt {2}}{4}$\n\nSo the equation becomes\n$ \\frac {\\sqrt {6} \\minus{} \\sqrt {2}}{4} \\sin{\\theta} \\equal{} \\frac {1}{2} \\left[\\frac {\\sqrt {2}}{2} \\cdot \\cos{\\theta} \\plus{} \\frac {\\sqrt {2}}{2} \\cdot \\sin{\\theta} \\right]$\n\n$ \\frac {\\sqrt {6} \\minus{} 2\\sqrt {2}}{4} \\sin{\\theta} \\equal{} \\frac {\\sqrt {2}}{4} \\cos{\\theta}$\n$ \\tan{\\theta} \\equal{} \\frac {\\sqrt {2}}{\\sqrt {6} \\minus{} 2\\sqrt {2}} \\equal{} \\frac {1}{\\sqrt {3} \\minus{} 2} \\equal{} \\boxed{ \\minus{} 2 \\minus{} \\sqrt {3}}$[/hide]", "Solution_2": "[quote=\"gauss202\"]\n3.) A mixture of wine and water is made in the ratio of wine:total = k:m. Adding $ x$ units of water or removing $ x$ units of wine $ (x \\neq 0)$ each produces the same new ratio of wine:total. Compute the numerical value of this new ratio.\n\nLet $ a$ be the amount of wine and $ b$ the amount of water.\nThen $ \\frac {a}{a \\plus{} b \\plus{} x} \\equal{} \\frac {k}{m} \\implies am \\equal{} ka \\plus{} kb \\plus{} kx$,\nand $ \\frac {a \\minus{} x}{a \\plus{} b \\minus{} x} \\equal{} \\frac {k}{m} \\implies am \\minus{} mx \\equal{} ka \\plus{} kb \\minus{} kx$\n\nSubtracting the second equation from the first gives $ mx \\minus{} kx$\nSo $ m \\equal{} k$ and $ \\boxed{\\frac {k}{m} \\equal{} 1}$[/quote]\r\n\r\nSubtracting the second from the first gives $ \\minus{} ( \\minus{} mx) \\equal{} kx \\minus{} ( \\minus{} kx) \\equal{} 2kx$\r\nSo $ m \\equal{} 2k$ and $ \\boxed{\\frac {k}{m} \\equal{} \\frac {1}{2}}$\r\n\r\nAfter all, you are adding $ x$ units of water in order to obtain the ratio one way, and as $ x$ cannot be $ 0$ the final ratio cannot indicate that all that is left is wine.\r\n\r\n5.) (Continued)\r\nThe product of the roots of any polynomial is given by the constant term (possibly negated, but not in the case of a polynomial of even degree) divided by the coefficient of the term of greatest degree. This is expressed in this problem as $ \\frac {c}{a}$.\r\n\r\nThus, $ \\displaystyle\\sqrt {pq} \\equal{} \\sqrt {\\frac {c}{a}}$.", "Solution_3": "[hide=\"1.)\"] $ \\displaystyle z^2\\equal{}cis(\\frac{\\pi}{2})$\n$ \\displaystyle z\\equal{}cis(\\frac{\\pi}{4}\\plus{}\\frac{2\\pi k}{2})$ where $ k\\equal{}0,1$\nor\n$ \\displaystyle z\\equal{}\\pm\\frac{\\sqrt{2}}{2}(1\\plus{}i)$[/hide]\n\n[hide=\"2.)\"] $ \\displaystyle z^3 \\equal{} 8cis(\\pi)$\n$ \\displaystyle z\\equal{}2cis(\\frac{\\pi}{3}\\plus{}\\frac{2\\pi k}{3})$ where $ k\\equal{}0,1,2$\nor\n$ \\displaystyle z\\equal{}\\minus{}2, 1\\pm\\sqrt{3}i$[/hide]\n\n[hide=\"3.)\"] $ (z^3\\minus{}1)(z^6\\plus{}z^3\\plus{}1) \\equal{} 0(z^3\\minus{}1)$\n$ z^9 \\equal{} 1$\n$ \\displaystyle z\\equal{}cis(\\frac{2\\pi k}{9})$ where $ k\\equal{}0,1,2,3,4,5,6,7,8$\n\nHowever, we must take away the roots supplied by the extra $ z^3\\minus{}1$, so $ k\\equal{}1,2,4,5,7,8$[/hide]\r\n\r\nOne question on number 4: If a polynomial has integer coefficients and an even root (like 4), can $ P(2)$ ever be odd?", "Solution_4": "Correct. I should have said $ P(2) \\equal{} 8$" } { "Tag": [ "inequalities", "rearrangement inequality", "inequalities unsolved" ], "Problem": "Let $ a,b,c\\geq0$.Prove that:\r\n$ 2(x^5\\plus{}y^5\\plus{}z^5)\\plus{}x^4y\\plus{}y^4z\\plus{}z^4x\\geq3(xy^4\\plus{}yz^4\\plus{}zx^4)$", "Solution_1": "You mean $ x,y,z\\geq0$ right???", "Solution_2": "by rearrangement inequality :\r\n$ 2(x^5 \\plus{} y^5 \\plus{} z^5) \\equal{} 2(x^4x \\plus{} y^4y \\plus{} z^4z)\\geq2(zx^4 \\plus{} xy^4 \\plus{} yz^4)$\r\nand :\r\n$ x^4y \\plus{} y^4z \\plus{} z^4x \\geq x^5 \\plus{} y^5 \\plus{} z^5$\r\n\r\n(not mistake ??)", "Solution_3": "Yes, I meant $ x,y,z\\geq0$.By the way ,mehdi cherif, $ x^4y\\plus{}y^4z\\plus{}z^4x \\geq x^5\\plus{}y^5\\plus{}z^5$ is not a correct inequality.", "Solution_4": "[quote=\"Inequalities Master\"]Let $ a,b,c\\geq0$.Prove that:\n$ 2(x^5 \\plus{} y^5 \\plus{} z^5) \\plus{} x^4y \\plus{} y^4z \\plus{} z^4x\\geq3(xy^4 \\plus{} yz^4 \\plus{} zx^4)$[/quote]\r\nTry the following reasoning: \r\nlet $ x\\equal{}\\min\\{x,y,z\\},$ $ y\\equal{}x\\plus{}u$ and $ z\\equal{}x\\plus{}v.$ :wink:", "Solution_5": "The most dirty solution I think,,\r\n\r\nBy Muirhead inequality,\r\n (5,0,0)>(1,4,0)and (4,1,0)>(1,4,0) so,,\r\n x^5+y^5+z^5>=xy^4+yz^4+zx^4\r\n x^5+y^5+z^5>=xy^4+yz^4+zx^4\r\n x^4y+y^4z+z^4x>=xy^4+yz^4+zx^4\r\n If we sum all of these inequalities,,\r\n We can get 2(x^5+y^5+z^5)+(x^4y+y^4z+z^4x)>=3(xy^4+yz^4+zx^4)\r\n I hate Muirhead inequalities,, but I use it often,,,", "Solution_6": "[quote=\"GoKorea\"]\n x^4y+y^4z+z^4x>=xy^4+yz^4+zx^4\n [/quote]\r\nIt's wrong. Try $ x\\equal{}1,$ $ y\\equal{}2$ and $ z\\equal{}0.$ :wink:", "Solution_7": "I just notice that this problem is pretty hard,,,", "Solution_8": "Sorry,arqady,I still cannot get your method to the end.Can someone help me? :wallbash: :?:", "Solution_9": "For you, Inequalities Master! :lol: \r\n[quote=\"Inequalities Master\"]Let $ x,y,z\\geq0$.Prove that:\n$ 2(x^5 \\plus{} y^5 \\plus{} z^5) \\plus{} x^4y \\plus{} y^4z \\plus{} z^4x\\geq3(xy^4 \\plus{} yz^4 \\plus{} zx^4)$[/quote]\r\nLet $ x \\equal{} \\min\\{x,y,z\\}$ and $ y \\equal{} x \\plus{} u,$ $ z \\equal{} x \\plus{} v.$ Then I have obtained:\r\n$ \\sum_{cyc}(2x^5 \\plus{} x^4y \\minus{} 3xy^4) \\equal{} 8(u^2 \\minus{} uv \\plus{} v^2)x^3 \\plus{} 6(2u^3 \\plus{} u^2v \\minus{} 3uv^2 \\plus{} 2v^3)x^2 \\plus{}$\r\n$ \\plus{} 4(2u^4 \\plus{} u^3v \\minus{} 3uv^3 \\plus{} 2v^4)x \\plus{} 2u^5 \\plus{} u^4v \\minus{} 3uv^4 \\plus{} 2v^5\\geq0.$ :wink:", "Solution_10": "For the collection:\r\nFor all reals $ a,$ $ b$ and $ c$ prove that:\r\n\\[ 2(a^4+b^4+c^4)+abc(a+b+c)\\geq3(a^3b+b^3c+c^3a)\\]", "Solution_11": "Actually,by Schur we know that $ a^4\\plus{}b^4\\plus{}c^4\\plus{}abc(a\\plus{}b\\plus{}c)\\geq 2(a^2b^2\\plus{}b^2c^2\\plus{}c^2a^2)$.\r\nAnd the inequality $ (a^2\\plus{}b^2\\plus{}c^2)^2\\geq 3(a^3b\\plus{}b^3c\\plus{}c^3a)$ is true,but very hard to prove.\r\nHow to prove easier your inequality, arqady? :?:", "Solution_12": "[quote=\"Inequalities Master\"]Actually,by Schur we know that $ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} abc(a \\plus{} b \\plus{} c)\\geq 2(a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2)$\n[/quote]\r\nWhich wrong.\r\nTry $ a\\equal{}\\minus{}1$ and $ b\\equal{}c\\equal{}1.$ :wink:", "Solution_13": "But I meant is true for positive reals.Is your inequality true for all reals? :o", "Solution_14": "[quote=\"Inequalities Master\"]Is your inequality true for all reals? :o[/quote]\r\nYes! :lol:", "Solution_15": "OK, arqady, how to prove it? :?:", "Solution_16": "[quote=\"Inequalities Master\"]OK, arqady, how to prove it? :?:[/quote]\r\nTry the following way:\r\nLet $ a\\equal{}\\min\\{a,b,c\\},$ $ b\\equal{}a\\plus{}x$ and $ c\\equal{}a\\plus{}y...$ :wink:" } { "Tag": [ "search", "\\/closed" ], "Problem": "Why are there two websites for AoPS? Which is the main one? All my posts on this site are on mathlinks. I don't get this, why are there two sites for the same thing?", "Solution_1": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1831775186&t=197230]Searching is your friend.[/url]", "Solution_2": "One additional fact: the merge was in July 2004.", "Solution_3": "What's a portal?", "Solution_4": "[url=http://www.mathlinks.ro/]Going to the Mathlinks site[/url] and seeing the portal on the homepage might help. :) \r\n\r\nAdditionally, [url=http://www.lmgtfy.com/?q=define%3Aportal]Google can also be a great friend.[/url]" } { "Tag": [ "absolute value" ], "Problem": "On the starship Enterprise, Captain Picard continually asks the food replicator to brew him a cup of Earl Grey tea. The proper brewing temperature for Earl Grey tea is 210\u00ba F plus or minus 5 degrees. Write and solve an absolute value equation representing the maximum and minimum brewing temperatures for Earl Grey tea.", "Solution_1": "[quote=\"Interval\"]On the starship Enterprise, Captain Picard continually asks the food replicator to brew him a cup of Earl Grey tea. The proper brewing temperature for Earl Grey tea is 210\u00ba F plus or minus 5 degrees. Write and solve an absolute value equation representing the maximum and minimum brewing temperatures for Earl Grey tea.[/quote]\r\n\r\n[hide]ummm...\n\n$|210\\pm 5|=205---215$.\n\nI dont know what an absolute value equation is.[/hide]" } { "Tag": [ "geometry", "rectangle", "combinatorics solved", "combinatorics" ], "Problem": "More than three quarters of the circumference of a circle is colored black. Prove that there exists a rectangle such that all of its vertices are black.\r\n\r\nActually the result holds if \"three quarters\" is replaced by \"one half\"...", "Solution_1": "Actually, I hope something like this is ok.\r\nSo, consider point $A$ and it's simetric with respect to the center of the circle $A'$. By continuously moving $A$ on the black region, you get a position when $A'$ is also black. Now remove point $A$ and $A'$. The area is still bigger than one half. In the same way, find $B$ and $B'$. Ready :)" } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "This is so stupid question but I keep wondering, what's TST?\r\n\r\n :D \r\n\r\nThank you and Happy Holiday!", "Solution_1": "It stands for Team Selection Test.", "Solution_2": "Hardest test in an national olympiad. They select the team that participates at the International Mathematical Olympiad representing the country. In USA TSTs are only taken by the first 12 USAMOers *with exceptions sometimes* ;)", "Solution_3": "I never even knew they had a TST in the U.S. Does it take place in MOP?", "Solution_4": "[quote=\"h_s_potter2002\"]I never even knew they had a TST in the U.S. Does it take place in MOP?[/quote]US always had a TST since they began participating in IMO (I think). I'm not sure exactly when I takes place, but I think it's at the beginning of MOP. IMO Team members from the site can tell us more (or Steve Dunbar for that matter). \r\n\r\nAt the USAMO official page, on the download part, where you find the USAMOs and the APMOs you will also find the TSTs from the past 4-5 years.", "Solution_5": "The US has only been using a specialized TST to select the team from among the top 12 on the USAMO for the past five years (the first one was in 2000). That is, the TSTs on the AMC website are the only ones that have ever been given (at the top they say \"43rd IMO Team Selection Test\" and such, but don't let that fool you: it's the the test to pick the team for the 43rd IMO, not the 43rd test). It's generally been held on the first Friday and Saturday of MOP. It isn't necessarily true that the USA TST is harder than the corresponding USAMO -- I think that the last couple years the top 12 as a whole have done roughly as well or better on the TST thanoon the USAMO. There have been some quite hard TST's though: I believe that the first TST, in 2000, didn't do as good a job of picking a team as it did of showing how much better Reid Barton was than anyone else. ;-) But they seem to have mostly worked the kinks out since then.", "Solution_6": "[quote=\"Alison\"] I believe that the first TST, in 2000, didn't do as good a job of picking a team as it did of showing how much better Reid Barton was than anyone else. ;-) But they seem to have mostly worked the kinks out since then.[/quote]\r\n\r\nWhere do they publish the individual score of contestants or at least score distributions per problem ? :)", "Solution_7": "[quote]The US has only been using a specialized TST to select the team from among the top 12 on the USAMO for the past five years (the first one was in 2000). That is, the TSTs on the AMC website are the only ones that have ever been given.[/quote]\r\n\r\nThe names of the US IMO team and USAMO top 6 are \r\npublished for the years before 2000, and the two lists\r\ndiffer substantially. TSTs were used, in other words.", "Solution_8": "[quote=\"fleeting_guest\"][quote]The US has only been using a specialized TST to select the team from among the top 12 on the USAMO for the past five years (the first one was in 2000). That is, the TSTs on the AMC website are the only ones that have ever been given.[/quote]\n\nThe names of the US IMO team and USAMO top 6 are \npublished for the years before 2000, and the two lists\ndiffer substantially. TSTs were used, in other words.[/quote]Both Richard and Mathew were involved in the USAMO / TST tesing back then, (Richard was #4 if I remember correctly on USAMO - but he didn't go to the IMO - so tests were issued somehow).", "Solution_9": "[quote=\"orl\"][quote=\"Alison\"] I believe that the first TST, in 2000, didn't do as good a job of picking a team as it did of showing how much better Reid Barton was than anyone else. ;-) But they seem to have mostly worked the kinks out since then.[/quote]\n\nWhere do they publish the individual score of contestants or at least score distributions per problem ? :)[/quote]\r\n\r\nActually I referred to TST scores and not those of the USAMO... ;)", "Solution_10": "Team selection for the US has varied quite a bit over the years. Some years it was just the top 6 (or 8 if you go back far enough) on the USAMO. When Dave & I were at MOP (late 80s), the team was determined by the USAMO plus 3 or 4 more tests in the first week or two of MOP (plus a little discretion of the IMO deputies). I think the process is more highly defined now, and therefore better.", "Solution_11": "[quote=\"Valentin Vornicu\"]Both Richard and Mathew were involved in the USAMO / TST tesing back then, (Richard was #4 if I remember correctly on USAMO - but he didn't go to the IMO - so tests were issued somehow).[/quote]\r\n\r\nMe too -- I was tied for #6 on the USAMO in 1988, but did really badly on the TSTs (there were 3 my year), so I didn't go to the IMO. (Not that I would have necessarily went anyway; they would have had to break the tie for #6 somehow.)\r\n\r\nIn the early days (late 1970s/early 1980s) and also for a while in the 1990s I believe, the US did away with the TSTs and just took the top 6 scorers on the USAMO for the IMO team.", "Solution_12": "TST scores are not made public, usually even the team members don't know their exact scores. But little bits of information get leaked around MOP ;-)...\r\n\r\nHmmm... fascinating. I only knew about the team selection process from when I was at MOP and what I'd heard from veterans and staff members who were there in the late 90's, although I think I was vaguely aware that at one point people's performance on MOP tests fed into the team selection process. (Somebody needs to write a history of MOP. :-)", "Solution_13": "The TSTs were done away with after 1993. In 1994 and 1995 the team was selected solely from USAMO scores. I don't know how long that lasted. They probably thought it was a good idea at first because 1994 was the year the US team had the only team ever to have 6 perfect 42's at the IMO." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "I think this is a pretty hard inequality, shall we discuss its solution", "Solution_1": "This one has been locked.\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=227462", "Solution_2": "Can you show me how to write in latex. I am so sorry", "Solution_3": "I can only tell you this is really an easy problem.I got the solution.", "Solution_4": "can you show your solution so that we can discuss together, please", "Solution_5": "[quote=\"Cai Ni\"]I can only tell you this is really an easy problem.I got the solution.[/quote]\r\nSorry,I don't think it is an easy problem.\r\nCan you send me your proof?", "Solution_6": "[quote=\"shfdfzhjj\"][quote=\"Cai Ni\"]I can only tell you this is really an easy problem.I got the solution.[/quote]\nSorry,I don't think it is an easy problem.\nCan you send me your proof?[/quote]\r\nI think it quite nice and not too easy .\r\n\r\nMy hint is use AM-GM then use a known Vasc 's lemma :wink:\r\n\r\nTo Cai Ni : Sorry but I don't want you post solution because this problem haven't limited on MYM in Vietnam", "Solution_7": "[quote=\"onlylove_math\"]\n\nMy hint is use AM-GM then use a known Vasc 's lemma :wink:[/quote]\r\n\r\nI think we get the same proof. :)", "Solution_8": "[quote=\"onlylove_math\"]\n\nTo Cai Ni : Sorry but I don't want you post solution because this problem haven't limited on MYM in Vietnam[/quote]\r\n\r\nSorry,I'm not good at English.Actually,I don't understand what's the meaning.It seems that in some problem we can't show our answer?\r\n :?: \r\nOK.I use Holder,then AM-GM.\r\nWe may easily remember the similar problem.(The inequality from 42th IMO)\r\nSo I try it and luckily I managed to get the solution.\r\nOh,yes.I think it's not a very easy problem,just I solved it quickly.", "Solution_9": "Interesting", "Solution_10": "[quote=\"Cai Ni\"][quote=\"onlylove_math\"]\n\nTo Cai Ni : Sorry but I don't want you post solution because this problem haven't limited on MYM in Vietnam[/quote]\n\nSorry,I'm not good at English.Actually,I don't understand what's the meaning.It seems that in some problem we can't show our answer?\n :?: \nOK.I use Holder,then AM-GM.\nWe may easily remember the similar problem.(The inequality from 42th IMO)\n[/quote]\r\nPlease check your proof,I think the step you use AM-GM is wrong.", "Solution_11": "Dont discuss more,please\r\n[quote =\"Cai Ni\"\t\r\n]Sorry,I'm not good at English.Actually,I don't understand what's the meaning.It seems that in some problem we can't show our answer? [/quote]\r\nIt means here is the new problem inthis issue of MYM(in our country)which is running\r\nbesides\r\n[quote=\"Cai Ni\"]Please check your proof,I think the step you use AM-GM is wrong[/quote]\r\nSure????,I have a genaral one for it (and it is similar to an ineq that I posted it) and I think i wasn't wrong (I used AMGM):lol:", "Solution_12": "[quote=\"Allnames\"]Please check your proof,I think the step you use AM-GM is wrong[/quote]\nSure????,I have a genaral one for it (and it is similar to an ineq that I posted it) and I think i was wrong (I used AMGM):lol:[/quote]\r\n\r\nI am not sure,I haven't seen the proof yet.Would you like to send your proof or genaral one to me? :maybe: \r\nThank you very much.", "Solution_13": "Mod please lock this topic. :mad: This is Vo Quoc Ba Can 's problem in this month's issue of Mathematics and Youth Magazine!\r\n\r\n@Lamminh: c\u00f2n l\u00e0 h\u1ecdc sinh LHP th\u00ec \u0111\u1eebng bao gi\u1edd l\u00e0m c\u00e1i tr\u00f2 n\u00e0y nh\u00e9.Th\u1eadt x\u1ea5u h\u1ed5!", "Solution_14": "[quote=\"lamminh_LHP_TC\"]I think this is a pretty hard inequality, shall we discuss its solution[/quote]\r\nI hope deadline has past,here is my proof :) :", "Solution_15": "[quote=\"shfdfzhjj\"][quote=\"lamminh_LHP_TC\"]I think this is a pretty hard inequality, shall we discuss its solution[/quote]\nI hope deadline has past,here is my proof :) :[/quote]\r\nYes shfdfzhjj,here is also the proof that I said to you above,of course,the general one is also not hard,You can kill it by the same way\r\n$ \\sum \\frac{a}{\\sqrt [k] {8c^3\\plus{}1}}\\ge 3^{\\frac{k\\minus{}2}{k}}$\r\nCan_hang problem is the case $ k\\equal{}2$ :)b", "Solution_16": "[quote=\"lamminh_LHP_TC\"]Let $ a,b,c$ be positive real numbers such that $ abc \\equal{} 1,$ then we have this inequality:\n\n$ \\frac {a}{\\sqrt {8c^3 \\plus{} 1}} \\plus{} \\frac {b}{\\sqrt {8a^3 \\plus{} 1}} \\plus{} \\frac {c}{\\sqrt {8b^3 \\plus{} 1}}\\geq1.$[/quote]From $ abc \\equal{} 1$, we may substitute $ a \\equal{} \\frac {x}{y},b \\equal{} \\frac {y}{z},c \\equal{} \\frac {z}{x},$ and the switching inequality is\n\n$ 1\\leq\\sum_{cyc}{\\frac {a}{\\sqrt {8c^3 \\plus{} 1}}} \\equal{} \\sum_{cyc}{\\sqrt {\\frac {x^5}{y^2(8z^3 \\plus{} x^3)}}}.$\n\nUsing the H\u00f6lder inequality, \n\n$ \\left(\\sum_{cyc}{\\sqrt {\\frac {x^5}{y^2(8z^3 \\plus{} x^3)}}}\\right)^2\\sum_{cyc}{xy^2(8z^3 \\plus{} x^3)}\\geq\\left(\\sum{x^2}\\right)^3,$\n\nwe have to prove $ \\left(\\sum{x^2}\\right)^3\\geq\\sum_{cyc}{xy^2(8z^3 \\plus{} x^3)}.$\n\nBut $ \\left(\\sum{x^2}\\right)^3 \\minus{} 8xyz\\sum_{cyc}{yz^2} \\minus{} \\sum_{cyc}{y^4z^2}$\n\n$ \\equal{} \\sum_{cyc}{(y \\minus{} z)^2(2x^3y \\plus{} 2x^2y^2 \\plus{} 4xyz^2 \\plus{} 2xz^3 \\plus{} y^4 \\plus{} y^2z^2)}\\geq 0.$\n[quote=\"Cai Ni\"]OK.I use Holder,then AM-GM.\n\nWe may easily remember the similar problem.(The inequality from 42th IMO).[/quote]See : http://www.mathlinks.ro/viewtopic.php?t=15294\n[quote=\"shfdfzhjj\"]$ \\sum_{cyc}{\\frac {a}{\\sqrt {8c^3 \\plus{} 1}}}\\geq\\sum_{cyc}{\\frac {a}{2c^2 \\plus{} 1}} \\equal{} \\sum_{cyc}{\\frac {x^3}{y(2z^2 \\plus{} x^2)}}\\geq\\frac {(\\sum{x^2})^2}{\\sum{xy(2z^2 \\plus{} x^2)}}\\geq 1$[/quote]$ \\Longleftarrow \\left(\\sum{x^2}\\right)^2 \\minus{} 2xyz\\sum{x} \\minus{} \\sum_{cyc}{x^3y} \\equal{} \\sum_{cyc}{(y \\minus{} z)^2\\left(\\frac {x^2}{2} \\plus{} xz \\plus{} y^2\\right)}\\geq0.$" } { "Tag": [ "probability and stats" ], "Problem": "Let X(i), i=1,\u2026,n, be an exchangeable sequence of random variables with P[X(i)>0]=1 for each i. For 1 < k < n fixed, find\r\nE[sum{X(i)(i=1 to k)}/sum{X(i)(i=1 to n)}]\r\nHint: first prove that, conditioned on sum{X(i)} (i=1 to n), the sequence X(i), i=1,\u2026,n is still exchangeable.", "Solution_1": "Honestly I have no idea where to start here." } { "Tag": [ "calculus", "derivative", "algebra", "polynomial", "geometry", "geometric transformation", "combinatorics proposed" ], "Problem": "Let $A=\\{a_1,a_2,...,a_n\\}$ be set of positive integers. Define $A^*=\\{a_i+a_j|1\\leq i 0$, s.t. $ a^2 \\plus{} b^2 \\plus{} c^2\\geq 3$.Prove that:\r\n$ \\frac {a^5 \\minus{} a^2}{a^5 \\plus{} b^2 \\plus{} c^2} \\plus{} \\frac {b^5 \\minus{} b^2}{a^2 \\plus{} b^5 \\plus{} c^2} \\plus{} \\frac {c^5 \\minus{} c^2}{a^2 \\plus{} b^2 \\plus{} c^5}\\geq 0$", "Solution_1": "[quote=\"Inequalities Master\"]Let $ a,b,c > 0$, s.t. $ a^2 \\plus{} b^2 \\plus{} c^2\\geq 3$.Prove that:\n$ \\frac {a^5 \\minus{} a^2}{a^5 \\plus{} b^2 \\plus{} c^2} \\plus{} \\frac {b^5 \\minus{} b^2}{a^2 \\plus{} b^5 \\plus{} c^2} \\plus{} \\frac {c^5 \\minus{} c^2}{a^2 \\plus{} b^2 \\plus{} c^5}\\geq 0$[/quote]\r\nIt seem similar to an IMO problem(i remember it is problem 3 in IMO 2005,sorry if if i have mistakes)" } { "Tag": [ "summer program", "MathPath", "AwesomeMath" ], "Problem": "Why don't we try to get this active again? Do people from Ohio still go on here?", "Solution_1": "there are no contests in the summer... @?!", "Solution_2": "Still, there were no posts since the end of May :o \r\nI guess this is a bad time, especially since people are at Mathpath and other stuff... I'll try in September/October.", "Solution_3": "\"Other stuff\" includes AwesomeMath starting a week from Monday :jump: !", "Solution_4": "yeah let's make this place more active." } { "Tag": [ "IMO", "IMO 2003" ], "Problem": "IMO 2003 Japan is the biggest IMO ever ( at least that's what the Japanese tell us on their official site ). \r\n\r\nAny opinions?", "Solution_1": "it seems that the majority of people really guessed the answer :D" } { "Tag": [ "calculus", "function", "calculus computations" ], "Problem": "What are 2 critical numbers and how do differ?", "Solution_1": "I'm not sure what you're asking, but if this is calculus-related, then try posting this [url=http://www.artofproblemsolving.com/Forum/index.php?f=66]here[/url].", "Solution_2": "He's just asking for a definition.\r\nWell, without going into the calculus of it all, I'll try to define it (this is not rigorous):\r\nGiven a function f(x) (doesn't necessarily have to be continous), the critical numbers of f(x) are either:\r\n\r\nThe X values for which we get out a local maximum or minimum.\r\n\r\nOR\r\n\r\nIf f(x) is non-continous, the point at which the graph shifts from one piece to another (think piece-wise defined).\r\n\r\n\r\n[url]http://mcraefamily.com/MathHelp/CalculusTheorem3aCriticalNumbers.htm[/url]\r\nThat should clarify, if you've had a bit of calc." } { "Tag": [ "number theory", "Diophantine equation", "number theory unsolved" ], "Problem": "Hello.\r\n\"Find all numbers that are simultanously triangular and quadrangular, that is, find all pairs $ (a,b)\\in\\mathbb{N}$ such that $ \\frac{a(a\\plus{}1)}{2}\\equal{}b^2$.\r\n\r\nFor example, $ 36$ and $ 1225$ work. Are the only ones?", "Solution_1": "$ \\sum_{k \\equal{} 1}^n k \\equal{} m^2 \\iff \\frac{n(n \\plus{} 1)}{2} \\equal{} m^2$\r\nNow let $ n \\equal{} \\frac {a \\minus{} 1}{2}, m \\equal{} \\frac {b}{2}$, so\r\n\\[ a^2 \\minus{} 2 b^2 \\equal{} 1\\]\r\nSo this problem is amount to the Pell Equation.\r\n\r\nHense the first few minimal pair of $ (a, b)$ are $ (3, 2)$.\r\nOnce the fundamental solution is found, all remaining solutions may be calculated algebraically as\r\n\\[ a_n \\plus{} b_n\\sqrt {2} \\equal{} (3 \\plus{} 2\\sqrt {2})^n\\]\r\nThus $ (a, b)$ are\r\n$ (3, 2)$\r\n$ (17, 12)$\r\n$ (99, 70)$\r\n$ (577, 408)$\r\n$ (3363, 2378)$\r\n$ (19601, 13860)$\r\n$ (114243, 80782)$\r\n$ (665857, 470832)$\r\n$ (3880899, 2744210)$\r\n$ (22619537, 15994428)$\r\n\r\n\r\nNow $ (n, m) \\equal{} \\left(\\frac {a \\minus{} 1}{2}. \\frac {b}{2}\\right)$, so $ (n, m)$ are\r\n$ (1, 1)$\r\n$ (8, 6)$\r\n$ (49, 35)$\r\n$ (288, 204)$\r\n$ (1681, 1189)$\r\n$ (9800, 6930)$\r\n$ (57121, 40391)$\r\n$ (332928, 235416)$\r\n$ (1940449, 1372105)$\r\n$ (11309768, 7997214)$\r\n\r\ncf.\r\na : [url=http://www2.research.att.com/~njas/sequences/A001108]a(n)-th triangular number is a square: a(n+1) = 6*a(n)-a(n-1)+2, with a(0) = 0, a(1) = 1.[/url]\r\nb : [url=http://www2.research.att.com/~njas/sequences/A001109]a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0, a(1)=1.[/url]\r\n\r\nAnd this Diophantine Equation is [url=http://mathworld.wolfram.com/SquareTriangularNumber.html]Square Triangular Number[/url]" } { "Tag": [ "function" ], "Problem": "[url]http://hubblesite.org/newscenter/archive/releases/2007/17/[/url]\r\n\r\nAnother one for dark matter.\r\nScore\r\n$\\text{Dark Matter}\\ 1$\r\n$\\text{MOND}\\ 0$\r\n :rotfl:", "Solution_1": "Actually...\r\n\r\nDark Matter: 2\r\nMOND: 0\r\nMy Dark Matter Theory: 0 :( \r\n\r\nMy idea doesn't really make much sense anymore, now that we've actually \"seen\" dark matter... :roll: Oh, well. It was fun while it lasted.", "Solution_2": ":lol: Your theory was interesting, the coffee cup analog was a cool way to put it.\r\n\r\nMOND does not seem to be popular, I guess people like contemplating the mysterious connotation associated with \"dark\" matter.", "Solution_3": "Yeah. I don't think people really like messing with Newtonian dynamics, either. MOND makes things so messy...what about Occam's razor? MOND is definitely not the simplest solution.", "Solution_4": "MOND does make things weird, but so does Quantum Mechanics with it own math and that works.\r\nDoes Occom's Razor always yeild the answer?\r\n\r\nThough, I still stick to Classical Newtonian Mechanics. :)", "Solution_5": "I don't think Occam's Razor is always right, but what if it is? Maybe we only THINK classical mechanics and quantum mechanics work. The real answer to all of physics--both macro and micro scale--could be something so obvious and simple that it's been right in front of physicists' noses for decades. I have no idea what it would be, though, and that's assuming Occam's Razor is always correct (which it isn't).\r\n\r\nAnyway, for me MOND is just so weird that it's counterintuitive. Why would gravity not decrease proportionally as one gets far away from the center of it, even for very small amounts? There has to be a better answer, and for now, dark matter is it.", "Solution_6": "Good point. :)\r\nThat description kind of fits the present conflict with String Theory, it is so convulted that many are doubting it validity.", "Solution_7": "Why are there so many extra dimensions in string theory? I never really understood that. They're not where the dark matter is, so what's their function? Is there even a way to prove their existence, besides the fact that they \"fit the equations\"?", "Solution_8": ":rotfl: \r\nThat is why it is now being refuted. Essentially they add dimensions until the math works out. \r\nString theorists also attempt to apply stings to everything, yes including dark matter, black holes, and gravity.", "Solution_9": "What would a black hole string look like? :huh: How can one make a point into a two-dimensional string? And how could strings be applied to dark energy?", "Solution_10": "Well, when I asked a string theorist of ISSYP last year, he did not really give me details about strings and dark matter-energy, he just said that String Theory has an intepretation of it.\r\n\r\nAbout black holes, he said that in String Theory a black hole (or atleast one type) can be described by a string (or loop) being around the event horizon, but still not many details." } { "Tag": [], "Problem": "$ 6)$ $ x$ varies jointly as $ z$ and $ y$ and inversely as the square root of $ w$. If $ x\\equal{}2$ when $ z\\equal{}7$, $ y\\equal{}3$, and $ w\\equal{}9$, what is $ z$ when $ x\\equal{}5$, $ y\\equal{}14$, and $ w\\equal{}16.$ \r\n\r\n$ 13)$ Simplify:$ \\frac{2000^{3}\\plus{}64}{2000^{3}\\minus{}6(2000)^{2}\\plus{}24(2000)\\minus{}32}$", "Solution_1": "[hide=\"1\"]since $ x$ is jointly proportional to $ y$ and $ z$ and inversely proportion to the root of $ w$, we have\n$ \\frac{x\\sqrt{w}}{zy}$ plugging in the values we have $ \\frac{x\\sqrt{w}}{zy}\\equal{}\\frac{6}{21}$\nnow plugging in the new values, we have $ \\frac{5\\sqrt{16}}{14z}\\equal{}\\frac{6}{21}$ solving, we get $ z\\equal{}5$..unless i made a computation error[/hide]\n\n[hide=\"2\"]is there some mistake. i tried factoring it in the form $ (x\\plus{}y)^{3}$ but it's not working, unless i made some mistake[/hide]", "Solution_2": "When it says that $ x$ varies jointly as $ y$ and $ z$ and inversely as the $ \\sqrt{w}$ doesn't that translate to $ \\frac{xyz}{\\sqrt{w}}$\r\n\r\n[hide] $ \\frac{(2)(7)(3)}{\\sqrt{9}}\\equal{} 14$ which gives us $ \\frac{(5)(14)(z)}{\\sqrt{16}}\\equal{}\\frac{70z}{4}$ Then $ z \\equal{}\\frac{4}{5}$ [/hide]", "Solution_3": "[quote=\"SnipedYou\"]When it says that $ x$ varies jointly as $ y$ and $ z$ and inversely as the $ \\sqrt{w}$ doesn't that translate to $ \\frac{xyz}{\\sqrt{w}}$[/quote]\r\nIf $ a$ varies inversely to $ b$, then it's $ ab$. Jointly is $ \\tfrac{a}{b}$.", "Solution_4": "yes, i like pie is right. when $ x$ is jointly proportional with two other variables u can really say that those two other variables are both inversely proportional to each other and x is directly proportional to them. inversely proportional means has a constant product. e.g. $ a$ and $ b$ are inversely proportional, thus their product $ ab$ is constant. \r\n\r\nhas anyone tried the second problem?", "Solution_5": "Oops I forgot :blush: Well here's my attempt at 13 (which is probably wrong too)\r\n\r\n[hide] $ 2000^{3}\\plus{}64$ can be factored into $ (2000\\plus{}4)(2000^{2}\\minus{}8000\\plus{}16)$ Now we can do some canceling. We have $ \\frac{(2000\\plus{}4)(2000^{2}\\minus{}8000\\plus{}16)}{2000^{3}\\minus{}6(2000)^{2}\\plus{}24(2000)\\minus{}32}$ Which can be reduced to $ \\frac{2004}{2000^{3}\\minus{}32}$ There's probably an error in there.\n[/hide]", "Solution_6": "[quote=\"Bicameral\"]$ 13)$ Simplify:$ \\frac{2000^{3}\\plus{}64}{2000^{3}\\minus{}6(2000)^{2}\\plus{}24(2000)\\minus{}32}$[/quote]\r\n\r\n$ k \\equal{} 2000$\r\n\r\n$ \\frac{k^{3}\\plus{}64}{k^{3}\\minus{}6k^{2}\\plus{}24k\\minus{}32}\\equal{}\\frac{(k\\plus{}4)(k^{2}\\minus{}4k\\plus{}16)}{(k\\minus{}2)(k^{2}\\minus{}4k\\plus{}16)}\\equal{}\\frac{2004}{1998}\\equal{}\\frac{334}{333}$" } { "Tag": [], "Problem": "Can some tell me what the symbol [b]>>[/b] means please? (From a physics problem.)", "Solution_1": "Can you use it in an expression? I suspect that it means implies but I am not sure.", "Solution_2": "[url]http://en.wikipedia.org/wiki/Inequality[/url]\r\nIt means much greater than. (There is no exact mathematical criteria for what [i]much greater[/i] than means, but it typically implies that there is a difference of at least several orders of magnitude.)", "Solution_3": "Thanks very much! :)" } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "calculate this \r\n$ \\int \\frac{(1\\plus{}\\sin{x})e^x}{1\\plus{}\\cos{x}} dx$", "Solution_1": "$ \\int \\frac{(1\\plus{}\\sin{x})e^x}{1\\plus{}\\cos{x}} dx\\equal{}e^{x}\\tan\\left(\\frac{x}{2}\\right)\\plus{}C$.", "Solution_2": "hello\r\n[hide=\"solution\"]\nwrite $ sinx\\equal{}2sin\\frac{x}{2}cos\\frac{x}{2}$\n $ 1\\plus{}cosx\\equal{}2cos^2\\frac{x}{2}$\n$ \\int \\frac{(1\\plus{}\\sin{x})e^x}{1\\plus{}\\cos{x}} dx\\equal{}\\int e^{x}(\\frac{sec^2{\\frac{x}{2}}}{2}\\plus{}tan\\frac{x}{2}$\n$ \\equal{}e^{x}tan\\frac{x}{2}$\n[/hide]\r\nthank u" } { "Tag": [ "search", "combinatorics unsolved", "combinatorics" ], "Problem": "Graph Airlines $ (GA)$ operates flights between some of the cities of the Republic of Graphia. There are at least three $ GA$ flights from each city, and it is possible to travel from any city in Graphia to any city in Graphia using $ GA$ flights. $ GA$ decides to discontinue some of its flights. Show that this can be done in such a way that it is still possible to travel between any two cities using $ GA$ flights, yet at least $ 2/9$ of the cities have only one flight.", "Solution_1": "It's from Turkey 2002. Here are two other places (one very recent) where it has appeared on the forum, neither with any solutions:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1842627630&t=316753\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1842627630&t=192924", "Solution_2": "[quote=\"JBL\"]It's from Turkey 2002. Here are two other places (one very recent) where it has appeared on the forum, neither with any solutions:\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1842627630&t=316753\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1842627630&t=192924[/quote]\r\n\r\nthanks for links but there is no solution in this topics" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "Suppose that $ G$ is finite and nilpotent group and $ p$ is a prime number such that $ p\\mid \\| G \\|$.Prove that $ p \\mid \\| Z(G) \\|$.", "Solution_1": "$ G$ is the direct product of it's sylow subgroups. then we may assume that $ G$ is a finite $ p$-group. but then the claim is clear since nontrivial $ p$-groups have nontrivial center (a consequence of the class equation)." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Prove that {1,2,3,...,100} can be partitioned into 5 sets such that no 2 elements in a set differ by a perfect square.", "Solution_1": "My sets are:\r\n[hide]\nThe numbers ending in 1 and 4\nThe numbers ending in 2 and 5\nThe numbers ending in 3 and 6\nThe numbers ending in 7 and 9\nThe numbers ending in 8 and 0\n\nThe difference between any two numbers in the same set ends in 3, 7, 2, 8, or 0. But no square less than 100 ends in any of those digits, so no difference is a square.\n[/hide]" } { "Tag": [ "calculus", "integration", "function", "derivative", "geometry", "trigonometry", "real analysis" ], "Problem": "Can we find a function - or does there exist a function $f$ , which has derivative $\\forall x \\ in I$ , I is an iterval , but $\\int_{a}^{b}f'(x)dx$ does not exist ?($a,b \\ in I$).\r\n\r\n Or the existence of derivative is sufficient for the integral to exist ? I suspect that $f'$ must be something more , for the integral to exist , but is this really true and why?\r\n Thanks - Babis", "Solution_1": "suppose the area of the $f'$ in the interval $(a,b)$ tend to infinity\r\n\r\n\r\nor is undefinied in some point $x$ in the interval\r\n\r\nthe improper integral tends to infinity and a lot of cases", "Solution_2": "It depends on what kind of integral we're looking at. The derivative can easily fail to be absolutely integrable, even if improper integrals exist.\r\nAs for improper Riemann integrals- how many bad points are you willing to avoid?", "Solution_3": "Well , the problem concerns high school analysis.I was wondering , if we must allways consider a function to have continoous derivative , if we want $f'$ to be integratable on [a, b]. The newton theorem states f' must be integratable ,but I thought that $f'$ has this propetry.It sees no, so I was seeking an example. If you have any , please send it.\r\n [u] Babis[/u]", "Solution_4": "The function $f(x)=x^{2}\\sin(\\frac1{x^{2}})$ is differentiable on $[-1,1]$. It's not Riemann integrable because the derivative is unbounded.", "Solution_5": "jmerry didn't explicitly mention that he intends to define $f(0)=0;$ that falls under the heading of \"removable discontinuities deserve to be removed.\" The differentiability at zero is less than obvious, but it is differentiable there.", "Solution_6": "[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=132593]Here[/url] you will find the answer to all of your questions (see fedja's example).", "Solution_7": "Just take any continuous function on $(a,b)$ whose integral fails to exist. It is the derivative of one of its primitives ...", "Solution_8": "Really thanks ! \r\n\r\n It was a little difficult for me to read again all this nice theory about intregrals , after 30 years fom the university years! Your help was great! You saved me much time. To all problems I will give the condition that $f'$ is continous , so we'll have not problems with my students in high school!\r\n Thank you again!\r\n Babis" } { "Tag": [ "quadratics", "induction" ], "Problem": "Let $ f(x)\\equal{} ax^{2}\\plus{}bx\\plus{}c$ be a quadratic such that $ a,b,c\\in\\mathbb{R}^{\\plus{}}$ and $ a\\plus{}b\\plus{}c\\equal{}1$. Prove that if given a sequence of positive real numbers $ x_{1},x_{2},........,x_{n}$ such that their product is $ 1$ then:\r\n\r\n$ f(x_{1})f(x_{2})......f(x_{n})\\geq 1$.", "Solution_1": "Using induction I'll prove that :\r\n$ p(n) : f(x_{1})f(x_{2})...f(x_{n})\\geq\\left (f(P_{n})\\right)^{n}$ ,where $ P_{n}\\equal{}\\sqrt[n]{x_{1}x_{2}...x_{n}}$\r\n$ p(1)$ is obvious , $ p(2)$ is true by Cauchy-Schwartz.\r\nSupose that $ p(n)$ is true we must prove that $ p(n\\plus{}1)$ is true:\r\nFrom $ p(n)$ we get that :\r\n$ f(x_{1})f(x_{2})...f(x_{n})f(x_{n\\plus{}1})\\geq\\left(f(P_{n})\\right)^{n}f(x_{n\\plus{}1})$ so it's enough to prove that $ (f(P_{n}))^{n}f(x_{n\\plus{}1})\\geq\\left(f(P_{n\\plus{}1})\\right)^{n\\plus{}1}$ ,but this is just Holder's inequality.\r\nTherefore $ f(x_{1})f(x_{2})...f(x_{n})\\geq\\left (f(P_{n})\\right)^{n},\\forall n\\in\\mathbb N^{*}$ .In particular for this problem $ P_{n}\\equal{}1 ,\\forall n\\in\\mathbb N^{*}$ and the result follows." } { "Tag": [], "Problem": "A map's key equates $ 1$ inch with $ 30$ miles. If two cities are exactly $ 25$ miles apart, how many inches apart are they on the map? Express your answer as a common fraction.", "Solution_1": "Since we know that 1 inch= \t30 miles, and 5 miles= \t$ \\dfrac{1}{6}$ [size=150], [/size]$ \\dfrac{1}{6} * 5$ equals $ \\dfrac{5}{6}$" } { "Tag": [ "logarithms", "floor function", "geometric series" ], "Problem": "Compute $ \\sum_{n \\equal{} 1}^{6237}\\lfloor \\log_2 n \\rfloor.$\r\n\r\nAn interesting problem: Let $ a_k$ be the coefficient of $ x^k$ in the expansion of $ \\sum_{i\\equal{}1}^{99} (x\\plus{}1)^i$. Find the value of $ \\lfloor a_4/a_3 \\rfloor$.", "Solution_1": "[hide=\"1\"]\nWe know that $ 2^{12}<6237<2^{13}$, so for values from $ 4096$ to $ 6237$ ($ 2142$ numbers) we get $ 12$.\n\nLet's consider $ \\sum_{n\\equal{}1}^{4095}\\lfloor\\log_2 n\\rfloor$. For $ n\\equal{}1$, it's $ 0$. For $ n\\equal{}2,3$, it's $ 1$. For $ n\\equal{}4,5,6,7$, it's $ 2$. For $ n\\equal{}8,9,10,11,12,13,14,15$ it's $ 3$.\n\nNoticing this pattern, we find that that sum is equivalent to $ \\sum_{n\\equal{}1}^{11}n{2^n}$.\n\nLet $ S\\equal{}1\\cdot2^1\\plus{}2\\cdot 2^2\\plus{}3\\cdot2^3\\plus{}\\cdots\\plus{}11\\cdot2^{11}$.\n\n$ 2S\\equal{}1\\cdot 2^2\\plus{}2\\cdot 2^3\\plus{}\\cdots\\plus{}11\\cdot2^{12}$.\n\n(I fudged a little here to get a geometric series, but accounted for it by subtracting off $ 2^0$ and $ 2^1$).\n$ S\\minus{}2S\\equal{}\\minus{}S\\equal{}2\\plus{}(2^0\\plus{}2^1\\plus{}2^2\\plus{}2^3\\plus{}\\cdots\\plus{}2^{11})\\minus{}2^0\\minus{}2^1\\minus{}11\\cdot 2^{12}$.\n\n$ \\minus{}S\\equal{}2\\plus{}2^{12}\\minus{}4\\minus{}11\\cdot2^{12}\\equal{}\\minus{}10\\cdot 2^{12}\\minus{}2$\n\n$ S\\equal{}10\\cdot 4096\\plus{}2\\equal{}40962$.\n\nThen $ 40962\\plus{}2142\\cdot 12\\equal{}\\boxed{66666}$ :o\n[/hide]\n\n[hide=\"2\"]\nThe coefficient of $ x^k$ in $ (x \\plus{} 1)^i$ is $ \\binom{i}{k}$, so what we're dealing with here is $ a_k \\equal{} \\sum_{i \\equal{} 1}^{99}\\binom{i}{k}$. For $ i\\leq k$ this is $ 0$, so what we have is $ \\sum_{i \\equal{} k}^{99}\\binom{i}{k} \\equal{} \\binom{100}{k \\plus{} 1}$ by Hockey Stick.\n\nTherefore, $ \\frac {a_4}{a_3} \\equal{} \\frac {\\binom{100}{5}}{\\binom{100}{4}} \\equal{} \\frac {100!}{95!5!}\\cdot\\frac {96!4!}{100!} \\equal{} \\frac {96}{5}$.\n\nOur answer is, then, $ \\boxed{19}$.\n[/hide]" } { "Tag": [ "algebra", "polynomial", "inequalities", "inequalities proposed" ], "Problem": "For any non-negative real numbers $a,b,c$ the following inequalities holds :\r\n$1. \\ 64(a^{2}+b^{2}+c^{2})(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}) \\ge 9(a+b)^{2}(b+c)^{2}(c+a)^{2}$\r\n$2. \\ (a^{2}+b^{2})(b^{2}+c^{2})(c^{2}+a^{2}) \\ge 8 \\left(\\frac{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}{a+b+c}\\right)^{2}$", "Solution_1": "Problem1:Use two inequalities:\r\n 1.$\\sum a^{2}\\sum a^{2}b^{2}\\geq (\\sum a^{2}b)^{2}$\r\n 2.$\\sum a^{2}\\sum a^{2}c^{2}\\geq (\\sum a^{2}c)^{2}$", "Solution_2": "[color=darkblue][b]Problem 2[/b][/color]\r\n\r\nThis inequality is identical whith:\r\n$\\sum_{sym}a^{6}b^{2}+2\\sum_{sym}a^{5}b^{3}+2\\sum_{sym}a^{5}b^{2}c+2\\sum_{sym}a^{4}b^{3}c+4\\sum_{cyc}a^{3}b^{3}c^{2}\\geq 6\\sum_{cyc}a^{4}b^{4}+12\\sum_{cyc}a^{4}b^{2}c^{2}.$\r\n\r\nNow by $AM\\geq GM$, we have:\r\n$\\sum_{sym}a^{6}b^{2}+2\\sum_{sym}a^{5}b^{3}\\geq 6\\sum_{cyc}a^{4}b^{4}$ \r\n\r\n$\\sum_{sym}a^{5}b^{2}c+\\sum_{sym}a^{4}b^{3}c+2\\sum_{cyc}a^{3}b^{3}c^{2}\\geq 6\\sum_{cyc}a^{4}b^{2}c^{2}.$ \r\n\r\n :wink: \r\n\r\nFitim Dika", "Solution_3": "Thank you, silver1989 and Fitim.\r\nThe proof of silver1989 is similar to mine.\r\nI will re-read the Fitim's (I don't have time now, sorry). But do you have another proof not expanding ?\r\n\r\n[b]Ineq. 3.[/b] $(a^{2}+b^{2})(b^{2}+c^{2})(c^{2}+a^{2}) \\ge 8 \\left(\\frac{abc(a^{2}+b^{2}+c^{2})}{ab+bc+ca}\\right)^{2}$\r\n[b]Ineq. 4.[/b] $3(a^{2}+b^{2})(b^{2}+c^{2})(c^{2}+a^{2}) \\ge (ab+bc)^{3}+(bc+ca)^{3}+(ca+ab)^{3}$" } { "Tag": [ "inequalities", "calculus", "function" ], "Problem": "Let $ a,b,c$ be positive real numbers satisfying $ abc\\equal{}1$. Prove that \\[ 3(a\\plus{}b\\plus{}c)\\ge\\sqrt{4a\\plus{}5}\\plus{}\\sqrt{4b\\plus{}5}\\plus{}\\sqrt{4c\\plus{}5}\\]", "Solution_1": "Let $ f: D\\to\\mathbb R$ , $ f(x) =\\sqrt{4x+5}$ ,$ f$ is concave\r\nFrom Jensen's inequality we have that :\r\n$ \\sum f(a)\\leq 3f(\\frac{a+b+c}{3}) = 3\\sqrt{\\frac{4}{3}(a+b+c)+5}\\leq 3(a+b+c)$ $ \\Leftrightarrow$\r\n$ (a+b+c)^{2}-\\frac{4}{3}(a+b+c)-5\\geq$ $ \\Leftrightarrow$ $ a+b+c\\geq 3$ which it's obvious.\r\n\r\n\r\nAnother solution:\r\nThe given inequality is equivalent to:\r\n$ \\sum(\\sqrt{4a+5}-3a+\\frac{7}{3}\\cdot ln(a))\\leq 0$\r\nLet $ f: D\\to\\mathbb R$ $ f(x)=\\sqrt{4x+5}-3x+\\frac{7}{3}\\cdot ln(x))$ \r\nBecause $ f''(x)<0$ and $ f'(1)=0$ we are done :D", "Solution_2": "Um...okay. Great, thanks. :| But there are solution(s) using elementary and basic inequalities. :maybe: Actually I don't consider Jensen's inequality as elementary because it uses calculus(which I haven't pretty much studied :blush: ). Anyway the following inequality is also true:\r\n\r\nLet $ x,y,z>0$. Prove that \\[ \\sqrt{x^{2}\\plus{}3xy}\\plus{}\\sqrt{y^{2}\\plus{}3yz}\\plus{}\\sqrt{z^{2}\\plus{}3zx}\\le 2(x\\plus{}y\\plus{}z)\\]", "Solution_3": "[hide]$ \\sum x^{2}\\ge\\sum xy$\n$ \\Rightarrow 4\\sum x^{2}\\plus{}8\\sum xy\\ge 3\\sum x^{2}\\plus{}9\\sum xy$\n$ \\Rightarrow 4(x\\plus{}y\\plus{}z)^{2}\\ge 3\\sum (x^{2}\\plus{}3xy)\\ge (\\sum\\sqrt{x^{2}\\plus{}3xy})^{2}$ by Cauchy[/hide]", "Solution_4": "Both are very nice solutions! (Although I'm not sure if I understand Svejk's last one pretty much :blush: . So I'm still looking for another solution of my first inequality...)\r\n\r\nHi SplashD, can you prove this one:\r\n\\[ \\sqrt{3x^{2}\\plus{}xy}\\plus{}\\sqrt{3y^{2}\\plus{}yz}\\plus{}\\sqrt{3z^{2}\\plus{}zx}\\le 2(x\\plus{}y\\plus{}z)\\]", "Solution_5": "[hide]\\[ \\sum\\sqrt{3x^{2}\\plus{}xy}\\equal{}\\sum\\sqrt{(2x)(\\frac{3x}{2}\\plus{}\\frac{y}{2})}\\le\\sum\\frac{2x\\plus{}(\\frac{3x}{2}\\plus{}\\frac{y}{2})}{2}\\equal{}2(x\\plus{}y\\plus{}z)\\][/hide]", "Solution_6": "[quote=\"nayel\"]I don't consider Jensen's inequality as elementary because it uses calculus(which I haven't pretty much studied :blush: ). [/quote]\r\n\r\nIt's not that Jensen's inequality itself is calculus - all it really does is extrapolate from the elementary (not the calculus) definition of convexity. The part that could be considered calculus is knowing when a function is convex, but certain functions can easily be proven convex from the elementary definition.", "Solution_7": "[quote=\"nayel\"]But there are solution(s) using elementary and basic inequalities.[/quote]is Cauchy allowed?\r\nWe know $ a+b+c\\geq 3$ from AM-GM.\r\n\\begin{eqnarray*}\\sqrt{4a+5}+\\sqrt{4b+5}+\\sqrt{4c+5}&\\leq&\\sqrt{3}\\sqrt{4(a+b+c)+15}\\\\ &\\leq&\\sqrt{a+b+c}\\sqrt{4(a+b+c)+5(a+b+c)}\\\\ &=&3(a+b+c)\\end{eqnarray*}\r\nnot so different from Svejk's solution though" } { "Tag": [ "trigonometry" ], "Problem": "I know this is probably the wrong place to post this, but I don't know where else. I have a TI30 scientific calculator. However, whenever I ask for Sin it gives the value of Cosine and vice versa (the same is true for arcsin and arccos). is is possible to reprogram such a calculator? thanks :D", "Solution_1": "well unless you have a program button then you can't \r\n\r\nyou can program calculators using BASIC with TI-83s and up", "Solution_2": "2nd f, u need to turn that button on" } { "Tag": [ "search", "MATHCOUNTS", "factorial", "geometry", "calculus", "AMC 8", "AMC 10" ], "Problem": "Hi, I'm curious, as a grown-up example of a student who skipped a grade (from fifth to seventh) in school, how many of the rest of you have a declared grade skip? In other words, how many of you are in a school grade higher than the grade you would be placed in considering only your age? How many grades have you skipped? \r\n\r\nAnd, most importantly, when you register for an AMC test, or a regional talent search achievement test, or for Mathcounts, or for some other math contest, do you register according to the grade you are in from the point of view of the school's records, or according to the grade you would be in from the point of view of your age? \r\n\r\nI'm wondering how common the phenomenon is of multiple grade-skipping among you math-eager young people. Here in Minnesota, I know of a local example of a young man (that is, a boy) who is no more than eight years old, taking an accelerated Algebra I & II course this year, who is listed in that course's records as an eighth-grader. If I'm doing my math right, that is something like a quintuple grade skip. I don't know his family well enough (yet) to ask how they sign him up for talent search testing, which is something I presume they do for him. I also don't know if he is signed up for this year's AMC-8, but he sure ought to be. And, of course, someone ought to invite him to join this forum . . . \r\n\r\nThanks for any observations any of you can provide,", "Solution_1": "I am currently skipping two grades, [b]only in math[/b]. I started skipping in fifth since California did not allow skipping grades let alone a partial grade-skip. Instead I was assigned to a cource about \"How to be successful in life\" in fifth grade. (I'm sure it's experimental since I've never heard of it before nor after I took the cource; Probably because I've got a 100 on STAR 4 years running :) (I moved to the U.S. in second grade))\r\nAfter I came to Massachusetts, my 6th grade math teacher (no longer a teacher; quit after she had me :twisted:; actually, she had a baby so she quit) placed me in eighth (which might've been a mistake since I skipped a lot of material like factorials and box-and-whisker graph and such). Now I am in eighth and is currently taking tenth Honors over at a high school which is across a huge field. (My middle school, another middle school, and the high school shares the same field)\r\n\r\nI tend to take full advantage of this and register for all that I can. For example, MATHCOUNTS and AMC 10. The odd thing is since I dropped a grade when I first came here (since I didn't know any English), so I qualify for the grade below me, my original grade, and the one above. (because of the grade I'm currently in, my age, and my math class's grade, in that order)\r\n\r\nI tend to see a lot of imaturity among those who skip grades and vice versa. This is obviously because the kids around them have more maturity then him/her, and this difference in mental age can affect the child a lot as they grow older. I'd say that you shouldn't skip grades once you're halfway through Elementary, even before. Or else you'll get this obnoxious selfish impudent kid with no friends [size=59]a.k.a. M*x*m[/size]\r\nMy two cents on skipping grades.", "Solution_2": "Skipping grades in math does not count as \"skipping a grade\". Generally, children may skip a grade in elementary school (and thus skip an entire grade, not just take advanced courses in one subject area), though there was an IMOer at my school a few years ago that graduated one year early, so I guess that counts as skipping a grade (senior year).", "Solution_3": "Well there's a pretty obvious difference between those who partly skip grade and take a couple of subject over at the high school and those who don't because we literary go back-and-forth between our schools. I also see the \"Indulgence\"syndrome even when a child is only skipping one grade.\r\n\r\nI guess this is a special case, but still there are a lot of them so I think this should be into consideration. I know of three people who skips only math in AoPS, and I'm sure there's a lot more people in the world who do \"partial skips.\"", "Solution_4": "Well, if you mean skipping grades in only one subject, then I am sure almost everyone here is skipping grades in math.\r\n\r\nBy the way, what is the \"Indulgence\" syndrome?", "Solution_5": "I named it \"Indulgence\" syndrome since I forgot the scientific name of it, but it's the syndrome when people, usually children, becomes less mature than others because they are spoiled by someone such as their parents, their classmates, etc.", "Solution_6": "[quote=\"Tare\"]I named it \"Indulgence\" syndrome since I forgot the scientific name of it, but it's the syndrome when people, usually children, becomes less mature than others because they are spoiled by someone such as their parents, their classmates, etc.[/quote]\r\n\r\nhmmm....i dunno. they might be immature at the beginning, but i think they usually grow out of it (personal experience, and one rather good friend)", "Solution_7": "[quote=\"Tare\"]I tend to take full advantage of this and register for all that I can. For example, MATHCOUNTS and AMC 10. The odd thing is since I dropped a grade when I first came here (since I didn't know any English), so I qualify for the grade below me, my original grade, and the one above. (because of the grade I'm currently in, my age, and my math class's grade, in that order)[/quote]\r\n\r\nI'm trying to be clear about this. As far as I know, anyone can take a math contest at an unusually LOW age. I know of summer math programs that have minimum age requirements, but I don't know of any contests that do. Thus my oldest son started taking part in a junior high mathematics contest we have locally when he was of fourth-grade age (mostly to fill out the team). He was the youngest kid in the whole state in that program--unless someone equally young was designated as belonging to a higher grade. He is on the league records as a fourth-grader, because I don't declare a grade skip for him (yet). People who homeschool can be vague about grades. \r\n\r\nBut I wonder how many unusually young participants in math contests declare their grade as a grade above the grade placement for their age. For example, there are AMC-8 statistics \r\n\r\nhttp://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2002-8a/02stats8.html \r\n\r\nthat go all the way down to fourth grade, but I wonder if some of the nominal eighth graders who take AMC-8 are actually one or two years younger than most other eighth-graders. What grade would you declare for yourself in signing up for AMC-10? \r\n\r\nHas anyone reading this thread done a talent search test \r\n\r\nhttp://www.ditd.org/Cybersource/record.aspx?sid=12649&scat=902&stype=110 \r\n\r\nfor example an out-of-level SAT I test? If so, was your test result reported as matching the grade you would be in by age, or matching some other grade? \r\n\r\nInquiring minds want to know . . .", "Solution_8": "Well... in math I skipped grades... in orchestra too... but nothing else...\r\nand about skipping whole grades...\r\none of my friend just skipped half of kinderg. and half of 1st grade... which doesn't really say anything about him being smart.", "Solution_9": "I know that you can still qualify even if you're lower than the grade the the competition was meant for (you could even take AMC 12 in first for all I know), but you wouldn't get the [b]opportunity[/b]to do so if you don't go to high school/middle school. Sometimes you have to even [b]make [/b]that opportunity. For example, my school didn't offer MATHCOUNTS so I went and set up a MATHCOUNTS \"meetings\" which doesn't really help us with math (since the only teacher available was a 6th math teacher who majored in \"acting\") but it will at least provide some oppotunity for the more fortunate who will come to my middle school. If I stayed back that fateful year from saying \"yes\", then I probably wouldn't have known about AMC, even this website! It's all about whether you know the competitions exists rather than whether or not you qualify.\r\n...I sure wish I've known of MATHCOUNTS earlier :cry: , and I'm sure there are even less fortunate around me. So when I say \"qualify\" I mean actually signing up and doing it, not just qualify for it and do nothing.\r\n\r\nP.S. My friend (who's about a dozen :wink: ) took SAT and he was in the grade that he was currently in, not by his age.", "Solution_10": "if you start school early. like kindergaten and stuff, it doesnt really matter that much. like you can tell all your friends once youve reached like the 7th grade, that you were super smart and skipped a grade. yup yup", "Solution_11": "[color=cyan]In general, you are supposed to indicate the level of your education, not the level of your age. That is, if you have a six-year-old taking in 12th grade, she should indicate her grade as 12, not as kindergarten.[/color]", "Solution_12": "Never skipped a full grade. I did start a year early compared to most people in my state, though, because I lived in California as a kid.\r\n\r\nI think that by the time you're in high school, \"skipping grades\" in individual subjects is no longer meaningful. That's because in high school, people have a lot more choice over what courses they want to take, and it's less unusual to find kids who don't adhere to a specific guideline. For instance, in my middle school most kids took prealgebra in 7th grade and algebra in 8th grade. In high school, on the other hand, there is a 7th grader and a sophomore in geometry, a freshman and a senior in AP Bio, etc.", "Solution_13": "I skipped a grade; 2nd grade that is. It's pretty cool, but there are drawbacks. I'm almost always 1 to 1.5 years younger than my peers. But it's also advantageous in that I \"started\" life earlier than they did.\r\n\r\n-interesting_move", "Solution_14": "When I was little, I skipped first and third grades. I'm glad I did, or I would have been bored in school (still was, but less so).", "Solution_15": "[quote=\"Ravi B\"]When I was little, I skipped first and third grades. I'm glad I did, or I would have been bored in school (still was, but less so).[/quote]\r\n\r\nThanks to you and to the others who revitalized this thread today. So now you participate in math contests, at a grade level two grades higher than your age placement would be, right? \r\n\r\nJust wondering . . .", "Solution_16": "Well, I'm an adult now. But, yes, when I was a kid, I participated in math contests two years ahead of my age. For example, I made it to the US Olympiad at ages 14 and 15 (11th and 12th grades), but never went beyond that. Perhaps if I had those two extra years, I could have went farther... There's advantages and disadvantages to skipping grades.", "Solution_17": "It is kind of nice to be able to declare a grade fitting for age, but I'm not sure exactly how ethical/permitted that is, especially when most people can not declare a lower grade. It would been nice to have an extra year in MathCounts and AMC/AIME/USAMO/(IMO?). Maybe judging by age would be more fair, but then you have people who are just older or just younger than the threshold, so neither way is quite fair. I suppose grade-skippers just all have that extra challenge to face. \r\n\r\nSadly enough, I don't think I got much out of skipping a grade, since I'm still taking math quite a few grades higher, as most of us are, I would think (who else would be such math geeks? :-P).\r\n\r\nIn a larger sense, though, these contests don't matter that much, because in the end you're going to have basically the same opportunities in research/business/whatever.", "Solution_18": "I skipped first grade. I don't suggest it; I think it was a big mistake. (Although I do have the benefits of not having to take the new SAT's)\r\n\r\nSkipping puts you behind socially. In elementary school, other kids don't like the idea of having someone once a grade lower than them in their class. As a result, people who skip are often ostracized, impeding the development of social skills and potentially making their school lives miserable. I got lucky in that I had and have always grown earlier than those my age, so I could fit in physically with the rest of my grade; however, other people aren't as lucky, so shorter stature and an underdeveloped appearance give classmates something else to pick on them with.\r\n\r\nSkipping did not make me any less bored in school. It is a matter of learning speed; if they raise your grade level so that the rest of the class is \"caught up\" with you, but you learn faster than the rest of the class, you are doomed to be bored within a couple of weeks. \r\n\r\nIt's not really worth it: you only gain one year, which in the long run isn't much, and you lose so much more from your school experience. To add insult to injury, you're still bored too! I don't recommend it at all. Instead, I suggest looking for alternative schooling if possible (schools like AAST if available, which run on an accelerated and more in-depth program). But even if this isn't a possibility, skipping grades is not the solution.", "Solution_19": "[quote=\"Nukular\"]Skipping did not make me any less bored in school. It is a matter of learning speed; if they raise your grade level so that the rest of the class is \"caught up\" with you, but you learn faster than the rest of the class, you are doomed to be bored within a couple of weeks. [/quote]\r\n\r\nRight. I went to quite a good school for lower and middle school, and they were kind enough to skip me through 3rd and 5th grade math, but the fundamental problem wasn't that I knew too much, it was that I learned it too fast. That isn't a problem that skipping grades will really deal with.", "Solution_20": "Skipping four years of college math solved that problem pretty well for me.", "Solution_21": "my teachers told my parents i could have skipped from 4th to 6th, but since i was already the shortest and smallest in my class..... (still am, by the way)", "Solution_22": "[quote=\"yif man12\"]my teachers told my parents i could have skipped from 4th to 6th, but since i was already the shortest and smallest in my class..... (still am, by the way)[/quote]\r\n\r\nlol, I'm the shortest in my grade too.\r\n\r\nI never really skipped a grade, but I started skipping math in 1st and english in 2nd. Now I'm up one grade in english honors. and just two grades up in math. :(", "Solution_23": "I've never skipped a grade, but supossedly I'm \"accelerated\" in math and Spanish, which means that I take math two years ahead of me, and can take Spanish II in 9th grade.\r\n\r\nBut skipping math doesn't really help me. Our Algebra class was soooooooooo stupid this year. I mean our midterm was harder than our final! Even a person just taking 8th grade math just answer some questions. All skipping math does for me is make me look really really smart.\r\n\r\nBut anyone who's bored in math shouldn't worry. There will always be something to challenge us in AoPS, no matter what our age. Unless you've like written 1,000 world reknown papers on math by the time your 18.", "Solution_24": "I skipped in math and got into 'honors' programs otherwise...my mom skipped two grades and said she had too much trouble adjusting and regretted, so I was never encouraged to. But even skipping in math is a problem sometimes. I never took algebra II but I picked it up in my extremely slow-paced precal class. On the other hand, at least 3 people on my school math team never took geometry which is a mistake, especially since one of them never bothered to study it independently either.", "Solution_25": "Um, I guess I'm just a year ahead of my class since I'm in the advance program...but my school won't let me double up on math courses, which sucks. My mom...her parents MADE her skip grades. I think she graduated HS when she was 14/15, and graduated college when she was about 17. My mom hated my parents because she really liked her HS...and she only had a couple of years. I like learning stuff independently, however, that only happens when I'm motivated (which isn't often).", "Solution_26": "I've never skipped a grade, but I am 2-5 grades ahead in math. (depending on where you start from and whether you count not taking AB as a skip: in my area, a considerable percentage of students take Al II in 9th grade, but Al I is 'On-Level.')\r\n\r\nI do have a few friends that skipped a grade, though, but they don't seem to have any problems in school.", "Solution_27": "I skipped a grade in China( from 2nd to 4th), not sure if that counts.", "Solution_28": "[quote=\"beta\"]I skipped a grade in China( from 2nd to 4th), not sure if that counts.[/quote]\r\n\r\nI think that would count as a grade skip, especially if you are younger than most other people in your grade now in the United States. Is grade skipping common in China? One of my best buddies fifteen years ago was a guy from Hunan who had entered university at age fourteen, I think, and was in Minnesota studying for a graduate degree at that time, I think barely twenty-some years old.", "Solution_29": "I agree with Nukular, as my grade-skipping experience sounds similar. Grade skipping sidesteps the issue.", "Solution_30": "They offered me to skip 2 grades, but my father said no cuz he did when he was small and he had been bullied for it so he said I couldn't \r\n\r\nthat's been 10 years of boring school since then :(", "Solution_31": "My school wanted me to skip from 7th to 9th but my dad said absolutly not. His father did and had always told him it was a mistake. So I am still in my \"age grade\" unless you count my going to take algebra 1 as an 8th grader where it is normally a 9th grade course here. I luckly have nice teachers who allow me to work ahead and last year allowed me to do the work that the 8th honors were doing even though I was still in the 7th grade class.", "Solution_32": "I skipped two grades, but in France, it always means for all subjects.\r\nNow, I'm 16 and this year was my first and last IMO, as next year I'll be at college (well... not exactly, but more or less so... something special in France). But this is the only problem... In fact, I think it was a very good thing (except for IMO)... Or I'd still be doing boring math at high school :D", "Solution_33": "I skipped 2nd and 6th grades. I don't think it really mattered so much, and I got exposed to lots more math than if I didn't. For example, right now I have finished all the way up to Multivariable Calculus in math classes, which is pretty good for a 14 yr old (I don't know about the ppl here though). I think skipping younger grades or grades where you lived in a \"podunk\" (thank you Mr. Rucscyzk, you saved my SAT) are perfectly acceptable and actually better for your development. :D", "Solution_34": "This is an interesting issue for us, because my son started kindergarten \"early\" -- that is he barely missed the cutoff, but was allowed to start at \"almost 5yo\" anyhow. \r\n\r\nA few years ago, we moved him to an ungraded private school. That makes declaring grades all the more complicated. :)\r\n\r\nFor math competitions and other academic areas, we declare his grade according to the grade he would be in based on when he started Kindergarten, and when he will presumably move to high school (so the grade that is high for his age). For summer camp where they bunk kids by grade, we also use this grade.\r\n\r\nBut for sports, we declare his grade according to his birth date (the grade he would have been in if he had started kindergarten at \"almost 6yo\" according to our state cutoffs). Since he is not especially strong at sports, this doesn't seem unethical to us. He's also small for both his age and grade, no matter which grade we declare.\r\n\r\nFor kids enrolled in normal (graded) schools I don't think there's really any choice as to which grade to declare. For students in a higher grade for math, though, I think it is usually appropriate to declare the overall grade, and not the math grade. \r\n\r\n(I did not skip any grades in school, but did take math 1 year ahead, starting in 8th grade. In that case I qualified for contests intended for students in my general grade, not my math grade, though in some cases I could do both.)\r\n\r\nFor Mathcounts, you can't participate before 6th grade, and you can only compete for 3 years. I specifically contacted MathCounts about how to declare grades for the kids in our ungraded school, and they said to base it not on age as much as on when they will be leaving our ungraded school and entering 9th grade, as we'd basically want them to be eligible for the last 3 years at our school.\r\n\r\nFor kids who would remain homeschooled or in ungraded situations beyond 8th grade, I suspect they do have rules based on age, but I'm not sure.\r\n\r\nSome contests may have rules pertaining to the declaration of grades for homeschoolers -- it is worth checking with them.", "Solution_35": "Issues of age come up about the IMO, as well.\r\nThere were proposals (not implemented) \r\nto limit students to 3 IMO's.\r\nThere were also some concerns about countries\r\nwhere high school can extend to age 19 or grade 13\r\n(e.g. Germany, Canada). \r\nIt's also not unknown for strong students \r\nto use their final year of high school as a platform \r\nto train for the IMO or enter the Westinghouse,\r\nrather than going directly to university. In a sense\r\nthis is sandbagging, though they probably \r\nlearn more than they would at college during that\r\ntime. \r\n\r\ntokenadult, you asked about starting college \r\nearly in China. Since there are national entrance\r\nexams for college, one can, in principle, \r\nbypass high school entirely. I get the impression \r\nthat it's not uncommon for people in rural areas\r\nwith bad or nonexistent high schools to prepare\r\ndirectly for the college admission exams.\r\nThere's no reason this can't happen at 14-15.", "Solution_36": "I never really skipped a grade... nobody has in our school... since the heads of the lower, middle, and upper (it's a K-12 private school) are pretty oldfashioned... it takes a whole lot of convincing to put the advanced math students from 7th grade into Hon. Geometry in 8th grade... that's what I did... so I did skip a grade in math...", "Solution_37": "[quote=\"tokenadult\"]Has anyone reading this thread done a talent search test \n\nhttp://www.ditd.org/Cybersource/record.aspx?sid=12649&scat=902&stype=110 \n\nfor example an out-of-level SAT I test? If so, was your test result reported as matching the grade you would be in by age, or matching some other grade? [/quote]\r\n\r\nFor the talent search programs I know about, students in school are required to take the appropriate test and have their results interpreted according to the grade in which they are registered (to the potential disadvantage of grade-skipped kids). Again, I'm not sure of the rules for homeschoolers.", "Solution_38": "[quote=\"tokenadult\"]Has anyone reading this thread done a talent search test \n\nhttp://www.ditd.org/Cybersource/record.aspx?sid=12649&scat=902&stype=110 \n\nfor example an out-of-level SAT I test? If so, was your test result reported as matching the grade you would be in by age, or matching some other grade? [/quote]\r\n\r\nI've done the Midwest Talent Search twice, and both times I've gotten two score reports - One was just by everyone who took it, including high-school students (from the test making company), and there was one by the grade that I was in (from the college that ran the talent search).", "Solution_39": "skipped a grade of math, skipped a grade of english, did a course during the summer b/c the school had some rule about it so that put me two yrs ahead in math, and i started spanish a year early. well, skipping thoses classes helped me not be bored. so that gives me only two years in math, one year in english, one year in spanish. haha, i feel stupid :blush:", "Solution_40": "for some reason our school wouldn't let me or my brother go ahead to be taking algebra 1 in 6th grade. We both did perfectly on this algebra prognosis test and they still wouldn't let us take '8th honors' in 6th grade.... But for another guy they did!... It make no sense and makes me kind of sad, mad, angry", "Solution_41": "sorry to rub it in, but i would be mad too. I would ask my parents to talk to the teacher, though.", "Solution_42": "I've skipped 2 grades in math (If you actually count skipping pre-algebra as skipping a grade)(In other words, I was in algebra 1 back in 7th)\r\nBeing at TJ, I am currently very fortunate. However, some people in the area had much more opportunity than me (especially those that attended Longfellow MS). Basically, my 1st and 2nd grade educations were a waste, but then I went to a GT center in 3rd. That went pretty well until 6th grade, at which point both GT 6th grade teachers retired, and we got two new one. I can say, without exagerating, that I, as a 6th grader, was far smarter than my math/science teacher. I then went to Irving MS, which was pretty mediocre compared to other schools in fairfax.\r\n\r\nThere are two things here that annoy me:\r\n1. I was talking to a TJ senior that went to my elementary school, and apparently, up until my year (or possibly the year before that), they let one or two kids per year take algebra 1 in 6th grade. I'm assuming this oppurtunity not being brought to my knowledge had to do with the shifting of teachers.\r\n\r\n2. The people that went to Longfellow, as well as a few other schools in the area, had the oppurtunity to get into calculus in 9th grade. I didn't.\r\n\r\nOkay, I'm done complaining now :D", "Solution_43": "I skipped 5th because when I took the mathcounts test when I was in 4th grade, I could have made the team if I was in 6th grade.", "Solution_44": "Actually, I have never skipped a grade, but, as I am homeschooled, I was able to do lots of extra work and do two extra math grades in my spare time." } { "Tag": [ "conics", "parabola", "symmetry", "analytic geometry", "graphing lines", "slope", "calculus" ], "Problem": "OK here's the situation.. (If I'm not giving enough information, please let me know - I wasn't sure)\r\n\r\nAn object is being projected into the air (ignore air resistance) from height zero at an angle 'theta'. If the object must land 5 metres ahead (horizontal displacement) and can never reach a height of 2.5 metres or more, what is the maximum angle of launch?\r\n\r\nI got an answer, but I don't know if it's right:[hide]tan(theta)<4, so theta<76 degrees?[/hide]", "Solution_1": "Equation of a parabola (with vertical symmetry axis) passing through the points (0, 0), (2.5, 2.5), (5, 0) is $y = -0.4x^2 + 2x$, as can be calculated by substituting these 3 coordinate pairs into $y = ax^2 + bx + c$ and calculating a, b, c. Slope of the parabola tangent at x is the derivative $y' = -0.8x + 2$ and at x = 0, the slope is $\\tan \\theta_{\\text{max}} = y'(0) = 2,\\ \\theta_{\\text{max}} \\doteq 63.5^\\circ$.", "Solution_2": "Another solution using the discriminant of a quadratic polynomial\r\n\r\nRelative to X-Y axes through point of projection the equation of the path of the projectile is \r\n\r\nY = X.tanA - g.X^2/{2.V^2.(CosA)^2}............. (1) where V is speed of projection and A is angle of projection.\r\n\r\nThe range R is given by R=2.V^2.SinA.CosA/g from which we find that g/2.V^2 =SinA.CosA/R .Now writing this expression for g/2.V^2 into equation1\r\n\r\nthe path of the particle becomes Y = X.tanA -X^2.TanA/R and if Y cannot exceed a height h then the quadratic inequality\r\n\r\nX^2.TanA/R -X.TanA +h >=0 must hold for all values of X and A.This happens iff the discriminant is negative that is if\r\n\r\n(TanA)^2 <=4.h.TanA/R or TanA<=4h/R and thus the greatest possible value for TanA is 4h/R.", "Solution_3": "[quote=\"Michael Niland\"]Another solution using the discriminant of a quadratic polynomial\n\nRelative to X-Y axes through point of projection the equation of the path of the projectile is \n\nY = X.tanA - g.X^2/{2.V^2.(CosA)^2}............. (1) where V is speed of projection and A is angle of projection.\n\nThe range R is given by R=2.V^2.SinA.CosA/g from which we find that g/2.V^2 =SinA.CosA/R .Now writing this expression for g/2.V^2 into equation1\n\nthe path of the particle becomes Y = X.tanA -X^2.TanA/R and if Y cannot exceed a height h then the quadratic inequality\n\nX^2.TanA/R -X.TanA +h >=0 must hold for all values of X and A.This happens iff the discriminant is negative that is if\n\n(TanA)^2 <=4.h.TanA/R or TanA<=4h/R and thus the greatest possible value for TanA is 4h/R.[/quote]I'm sorry but I found your solution hard to read. To make it easier for others, I edited your solution with $\\LaTeX$\n\n[quote=\"Michael Niland\"]Another solution using the discriminant of a quadratic polynomial\n\nRelative to $x,y$-axes through point of projection the equation of the path of the projectile is \n\n$y=x\\tan A-\\frac{gx^2}{2v^2\\cos^2A}$............. (1)\n\nwhere $v$ is speed of projection and $A$ is angle of projection.\n\nThe range $R$ is given by $R=\\frac{2v^2\\sin A\\cos A}{g}$ from which we find that $\\frac{g}{2v^2}=\\frac{\\sin A\\cos A}{R}$. Now writing this expression for $\\frac{g}{2v^2}$ into equation (1)\n\nthe path of the particle becomes $y=x\\tan A-x^2\\frac{\\tan A}{R}$ and if $y$ cannot exceed a height $h$ then the quadratic inequality\n\n$x^2\\frac{\\tan A}{R}-x\\tan A+h\\ge0$ must hold for all values of $x$ and $A$.This happens iff the discriminant is negative that is if\n\n$\\tan^2A\\le\\frac{4h\\tan A}{R}$ or $\\tan A\\le\\frac{4h}{R}$ and thus the greatest possible value for $\\tan A$ is $\\frac{4h}{R}$.[/quote]" } { "Tag": [ "probability" ], "Problem": "Consider f(n-1) (denoting the sample space). Its last 2 endings maybe\r\n\r\nHH\r\nHT\r\nTH\r\nTT\r\n\r\nSo we can derive f(n) by adding a T to f(n-1) or a H to f(n-1) except where the last 2 digits of f(n-1) are HH.\r\n\r\nSince the last 2 digits are HH, the previous digit must be T.\r\nSo We must find the number of cases with THH in f(n-1), which turns out to be equal to f(n-4). So converting this to probability...\r\n\r\n$ P(N) \\equal{} P(N \\minus{} 1) \\minus{} (1 \\minus{} p)p^3P(N \\minus{} 4)$\r\n\r\nTo solve this let $ P(N) \\equal{} a^n$\r\n\r\nwe get $ a^4 \\equal{} a^3 \\minus{} (1 \\minus{} p)p^3$\r\n\r\nEdit: Where did you get this sum da?", "Solution_1": "Euclidean: Is this right or wrong da?", "Solution_2": "Illai.. Sorry da its wrong.", "Solution_3": "Whats wrong with the logic? I think I've got all cases??? Give your answer.", "Solution_4": "Raga: I think I've got the mistake.\r\n\r\nAnswer is \r\n\r\n$ F(n) \\equal{} pF(n\\minus{}1) \\plus{} (1\\minus{}p)pF(n\\minus{}2) \\plus{} p^2(1\\minus{}p)F(n\\minus{}3)$\r\n\r\nPut $ F(n) \\equal{} a^n$\r\n\r\n$ a^n \\equal{} pa^{n\\minus{}1} \\plus{} p(1\\minus{}p)a^{n\\minus{}2} \\plus{} p^2(1\\minus{}p)a^{n\\minus{}3}$\r\n$ a^3 \\equal{} pa^2 \\plus{} p(1\\minus{}p)a \\plus{} p^2(1\\minus{}p)$\r\n\r\nwhich I'm sure you can obviously solve. :rotfl: :rotfl:", "Solution_5": "Here s the final answer:\r\n\r\nthe required probability is $ (0.75)^{n\\plus{}1} (2 \\plus{} \\alpha^{n\\plus{}1} \\plus{} \\beta^{n\\plus{}1})$where $ \\alpha$ and $ \\beta$ are the roots of $ 3x^{2} \\plus{} 2x \\plus{} 1 \\equal{}0$." } { "Tag": [ "limit", "trigonometry", "algebra unsolved", "algebra" ], "Problem": "find:\r\n\r\n$\\lim_{n\\rightarrow \\infty}\\sin \\frac 1{2}+\\sin\\frac 1{2^{2}}+\\dots+\\sin\\frac 1{2^{n}}$", "Solution_1": "Nice! Do you know the source?\r\n\r\nAt least we're sure that the limit of $s_{n}=\\sin \\frac 1{2}+\\sin\\frac 1{2^{2}}+\\dots+\\sin\\frac 1{2^{n}}$ exists,\r\nbecause $s_{n}$ is increasing and bounded by the known sequence $t_{n}=\\frac 1{2}+\\frac 1{2^{2}}+\\dots+\\frac 1{2^{n}}<1$,\r\nsince $0<\\sin\\frac 1{2^{n}}<\\frac 1{2^{n}}.$", "Solution_2": "[quote=\"Amir.S\"]find:\n\n$\\lim_{n\\rightarrow \\infty}\\sin \\frac 1{2}+\\sin\\frac 1{2^{2}}+\\dots+\\sin\\frac 1{2^{n}}$[/quote]\r\ncotgx-cotg2x=$\\frac1{sin2x}$", "Solution_3": "[quote=\"ali666\"]cotgx-cotg2x=$\\frac1{sin2x}$[/quote]\r\nNice! This would be useful if we wanted to evaluate\r\n$\\frac 1{\\sin 2}+\\frac 1{\\sin 2^{2}}+\\dots+\\frac 1{\\sin 2^{n}}=\\cot 2-\\cot 2^{n+1}$,\r\nand in this case I'd say that the limit doesn't exist.", "Solution_4": "[quote=\"lordWings\"]Nice! Do you know the source?\n\nAt least we're sure that the limit of $s_{n}=\\sin \\frac 1{2}+\\sin\\frac 1{2^{2}}+\\dots+\\sin\\frac 1{2^{n}}$ exists,\nbecause $s_{n}$ is increasing and bounded by the known sequence $t_{n}=\\frac 1{2}+\\frac 1{2^{2}}+\\dots+\\frac 1{2^{n}}<1$,\nsince $0<\\sin\\frac 1{2^{n}}<\\frac 1{2^{n}}.$[/quote]\r\n\r\nactually I proved the limit exsist as you did.But I couldn't find it.\r\n\r\ncan anyone find it?", "Solution_5": "[quote=\"lordWings\"][quote=\"ali666\"]cotgx-cotg2x=$\\frac1{sin2x}$[/quote]\nNice! This would be useful if we wanted to evaluate\n$\\frac 1{\\sin 2}+\\frac 1{\\sin 2^{2}}+\\dots+\\frac 1{\\sin 2^{n}}=\\cot 2-\\cot 2^{n+1}$,\nand in this case I'd say that the limit doesn't exist.[/quote]\r\n :blush:", "Solution_6": "Following ali666's idea,\r\n$\\sin\\alpha=s(2\\alpha)-s(\\alpha)$,\r\nwhere\r\n$s(\\alpha)=\\lim_{n\\to\\infty}\\sin\\frac\\alpha{2}+\\sin\\frac\\alpha{2^{2}}+\\dots+\\sin\\frac\\alpha{2^{n}}$.\r\nThe limit always exists, for every $\\alpha\\in\\mathbb{R}$, because the sequence of finite sums, at least after a certain term, is monotonic and bounded.\r\nCan we find a closed expression for $s$?", "Solution_7": "$\\sin x= \\frac{e^{ix}-e^{-ix}}{i}$ :)", "Solution_8": "$\\sin x=\\frac{e^{ix}-e^{-ix}}{2i}$. :wink:", "Solution_9": "How does that help?\r\nI'm starting to think that we won't find a closed expression." } { "Tag": [ "geometry", "circumcircle" ], "Problem": "The area of a circle circumscribed about a regular hexagon is $ 2\\pi$. What is the area of the hexagon?", "Solution_1": "[hide=\"Solution\"]\nSince the area of the circumscribed circle is $ 2\\pi$ we notice that the radius of the circle is $ \\sqrt{2}$.\n\nA hexagon is made up by 6 equilateral triangles, whose side lengths are equal to the radius of the circumscribed circle.\n\nThe area of an equilateral triangle with side length $ s$ is $ \\dfrac{s^2\\sqrt{3}}{4}$. Therefore the area of the hexagon is,\n\\[ A\\equal{}6\\cdot\\left(\\dfrac{2\\sqrt{3}}{4}\\right)\\implies A\\equal{}\\boxed{3\\sqrt{3}}\\]\n[/hide]" } { "Tag": [ "geometry", "circumcircle", "algebra unsolved", "algebra" ], "Problem": "let $ z_1,z_2,z_3$ is complex number such that $ |z_1|\\equal{}|z_2|\\equal{}|z_3|$=R\r\nProve that:\r\n$ |z_1 \\minus{} z_2| . |z_2 \\minus{} z_3| \\plus{} |z_2 \\minus{} z_3| . |z_3 \\minus{} z_1| \\plus{}|z_3 \\minus{} z_1| . |z_1 \\minus{} z_2| \\leq 9.R^2$", "Solution_1": "[quote=\"math man\"]let $ z_1,z_2,z_3$ is complex number such that $ |z_1| \\equal{} |z_2| \\equal{} |z_3|$=R\nProve that:\n$ |z_1 \\minus{} z_2| . |z_2 \\minus{} z_3| \\plus{} |z_2 \\minus{} z_3| . |z_3 \\minus{} z_1| \\plus{} |z_3 \\minus{} z_1| . |z_1 \\minus{} z_2| \\leq 9.R^2$[/quote]\r\n\r\nConsider triangle $ ABC\\ : \\ A(z_1)\\ ;\\ B(z_2)\\ ;\\ C(z_3)$\r\n\r\n$ Inq < \\equal{} > ab \\plus{} bc \\plus{} ca \\le 9R^2$\r\n\r\nCall $ O$ is circumcenter of $ \\triangle ABC$ : \r\n\r\n$ 0 \\le 9.\\vec{OG}^2 \\equal{} (\\vec{OA} \\plus{} \\vec{OB} \\plus{} \\vec{OC})^2 \\equal{} 3R^2 \\plus{} 2\\sum \\vec{OA}.\\vec{OB} \\equal{} 3R^2 \\plus{} \\sum (OA^2 \\plus{} OB^2 \\minus{} AB^2) \\equal{} 9R^2 \\minus{} a^2 \\minus{} b^2 \\minus{} c^2$\r\n\r\n$ \\equal{} > ab \\plus{} bc \\plus{} ca \\le a^2 \\plus{} b^2 \\plus{} c^2 \\le 9R^2$" } { "Tag": [ "logarithms", "Vieta", "algebra", "binomial theorem", "algebra proposed" ], "Problem": "If $ a,b,c$ are the roots of $ x^3\\minus{}x\\minus{}1\\equal{}0$, compute:\r\n\r\n$ \\frac{1\\minus{}a}{1\\plus{}a}\\plus{}\\frac{1\\minus{}b}{1\\plus{}b}\\plus{}\\frac{1\\minus{}c}{1\\plus{}c}$", "Solution_1": "[quote=\"moldovan\"]If $ a,b,c$ are the roots of $ x^3 \\minus{} x \\minus{} 1 \\equal{} 0$, compute:\n\n$ \\frac {1 \\minus{} a}{1 \\plus{} a} \\plus{} \\frac {1 \\minus{} b}{1 \\plus{} b} \\plus{} \\frac {1 \\minus{} c}{1 \\plus{} c}$[/quote]\r\n\r\nWe have $ 1\\equal{}a^3\\minus{}a\\equal{}b^3\\minus{}b\\equal{}c^3\\minus{}c$ , than use it repeatedly\r\n\r\nthe sum $ \\equal{}\\frac{a^3\\minus{}2a}{a^3}\\plus{}\\frac{b^3\\minus{}2b}{b^3}\\plus{}\\frac{c^3\\minus{}2c}{c^3}$\r\n$ \\equal{}3\\minus{}2(\\frac{1}{a^2}\\plus{}\\frac{1}{b^2}\\plus{}\\frac{1}{c^2})$\r\n$ \\equal{}3\\minus{}2(\\frac{a^2\\minus{}1}{a}\\plus{}\\frac{b^2\\minus{}1}{b}\\plus{}\\frac{c^2\\minus{}1}{c})$\r\n$ \\equal{}3\\minus{}2(a\\plus{}b\\plus{}c)\\plus{}2(\\frac{ab\\plus{}bc\\plus{}ca}{abc})$\r\n$ \\equal{}3\\minus{}2(0)\\plus{}2(\\frac{\\minus{}1}{1})$\r\n$ \\equal{}1$", "Solution_2": "$ x^3\\minus{}x\\minus{}1\\equal{}(x\\minus{}a)(x\\minus{}b)(x\\minus{}c)\\Longrightarrow \\ln |x^3\\minus{}x\\minus{}1|\\equal{}\\ln |x\\minus{}a|\\plus{}\\ln |x\\minus{}b|\\plus{}\\ln |x\\minus{}c|$\r\n\r\n$ \\Longrightarrow \\frac{1}{x\\minus{}a}\\plus{}\\frac{1}{x\\minus{}b}\\plus{}\\frac{1}{x\\minus{}c}\\equal{}\\frac{3x^2\\minus{}1}{x^3\\minus{}x\\minus{}1}$, yielding $ \\frac{1}{1\\plus{}a}\\plus{}\\frac{1}{1\\plus{}b}\\plus{}\\frac{1}{1\\plus{}c}\\equal{}2.$\r\n\r\n$ \\therefore \\frac{1\\minus{}a}{1\\plus{}a}\\plus{}\\frac{1\\minus{}b}{1\\plus{}b}\\plus{}\\frac{1\\minus{}c}{1\\plus{}c}\\equal{}\\minus{}1\\cdot 3\\plus{}2\\left(\\frac{1}{1\\plus{}a}\\plus{}\\frac{1}{1\\plus{}b}\\plus{}\\frac{1}{1\\plus{}c}\\right)\\equal{}\\minus{}3\\plus{}2\\cdot 2\\equal{}\\boxed{1}.$", "Solution_3": "1996 Canada, #1:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77670[/url]", "Solution_4": "[hide=\"Solution\"]$ \\frac{1\\minus{}a}{1\\plus{}a}\\equal{}\\minus{}1\\plus{}\\frac{2}{1\\plus{}a}$.\n\n$ \\sum_{\\text{cyc}}\\minus{}1\\plus{}\\frac{2}{1\\plus{}a}\\equal{}\\minus{}3\\plus{}2\\sum_{\\text{cyc}}\\frac{1}{1\\plus{}a}$.\n\nLet $ f(x)\\equal{}x^3\\minus{}x\\minus{}1$. If $ x\\equal{}\\frac{1}{1\\plus{}a}$, then $ a\\equal{}\\frac{1}{x}\\minus{}1$, so $ \\frac{1}{1\\plus{}a}$ is a root of $ f\\left(\\frac{1}{x}\\minus{}1\\right)$.\n\nIt's not hard to see that $ x^3\\minus{}2x^2\\plus{}3x\\minus{}1$ has the same roots as $ f\\left(\\frac{1}{x}\\minus{}1\\right)$. (Binomial theorem, then multiply by $ \\minus{}x^3$).\n\nFrom Vieta's, the sum of the roots of $ g$ is $ 2$, so the answer is $ \\minus{}3\\plus{}2\\cdot 2\\equal{}4\\minus{}3\\equal{}\\boxed{1}$.[/hide]" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Prove that every subgroup of a cyclic group is cyclic.", "Solution_1": "Let $ G$ be a cyclic group and $ H$ a subgroup. Pick a generator for $ G$, call it $ g$ then order each element based on powers of multiples (if we write additively) of $ g$. Ie $ g < 2g < 3g...$ etc. This gives an isomorphism to either $ \\mathbb{Z}$ or $ \\mathbb{Z}/n\\mathbb{Z}$ where $ n$ is the order of the group. We will now denote $ g$ as $ 1$. let $ H$ be a subgroup of $ G$, under the ordering, there is a unique smallest element of $ H$, call it $ h$. Suppose now $ h$ does not generate $ H$, then there is an element $ f$ that is not a multiple of $ h$. Now, they must not share a common factor, otherwise it would be in the subgroup $ H$, so $ (h,f) \\equal{} 1$. Then by Bezout's identity, you can find multiples of h and f that have a difference of $ 1$ but because $ H$ a group, $ 1$ therefore is in the group which means $ H \\equal{} G$. Thus $ h$ generates $ H$. q.e.d." } { "Tag": [ "probability" ], "Problem": "(Sprint) A best-of-five series ends when one team wins three games. The probability of team A defeating team B in any game is 1/6 . What is the probability that team A will win the series? Express your answer as a fraction in lowest terms.\r\n\r\n (Sprint) We select 6 numbers at random, with replacement, from the set of integers from 1 to 600 inclusive. What is the probability that the product of the 6 numbers is even ? Express your answer as a commom fracton in lowest terms.\r\n\r\n(Target) Find the probability that four randomly selected points of the geoboard shown will be vertices of a square who sides have integer length. Give your answer as a common fraction. The distance between two adjacent vertical or horizontal dots is 1.\r\n . . . . . . \r\n . . . . . . \r\n . . . . . . \r\n . . . . . . \r\n . . . . . . \r\n . . . . . . \r\n\r\nWhat kind of level are they? I say they are pretty hard, probly national lvl?", "Solution_1": "[quote=\"kidwithshirt\"](Sprint) A best-of-five series ends when one team wins three games. The probability of team A defeating team B in any game is 1/6 . What is the probability that team A will win the series? Express your answer as a fraction in lowest terms.\n\n (Sprint) We select 6 numbers at random, with replacement, from the set of integers from 1 to 600 inclusive. What is the probability that the product of the 6 numbers is even ? Express your answer as a commom fracton in lowest terms.\n\n(Target) Find the probability that four randomly selected points of the geoboard shown will be vertices of a square who sides have integer length. Give your answer as a common fraction. The distance between two adjacent vertical or horizontal dots is 1.\n . . . . . . \n . . . . . . \n . . . . . . \n . . . . . . \n . . . . . . \n . . . . . . \n\nWhat kind of level are they? I say they are pretty hard, probly national lvl?[/quote]\r\n[hide=\"1\"]A's combos: win three games: AAA\nwin four games: AABA, ABAA, BAAA\nwin five games: BBAAA, BABAA, ABBAA, BAABA, ABABA, AABBA\nThere are $\\frac{5!}{3!2!}=10$ ways to get a team A to win, and we have all ten. So adding up the probabilities we get\n$\\frac{1}{216}+\\displaystyle\\left(\\frac{5}{1296}\\right)\\times3+\\displaystyle\\left(\\frac{25}{7776}\\right)\\times6=\\boxed{\\frac{23}{648}}$\nIs that correct?[/hide]\n[hide=\"2\"]\nWe find the probability the product is odd. This is $\\displaystyle\\left(\\frac{1}{2}\\right)^6$ and the probability it is even is $1-$ that answer, $\\boxed{\\frac{63}{64}}$. [/hide]\n[hide=\"3\"]I'm getting kicked off the comp so i'm gonna guess $\\boxed{55}$[/hide]", "Solution_2": "I would say national\r\n[hide=\"number 1\"]\nWays that A wins the series\n3 wins. $\\frac{1}{216}$\n3 wins one loss.\nThe last must be a win, so we can arrange the loss in 3 places.\nThus it is $\\frac{15}{6^4}=\\frac{5}{432}$\n3 wins two losses.\nAgain, a win must be last, so we can arrange the losses in $\\binom 42=6$ ways.\n$\\frac{150}{6^5}=\\frac{25}{6^4}=\\frac{25}{1296}$.\nSumming all of them, $\\frac{46}{1296}=\\frac{23}{648}$\n[/hide]\n[hide=\"number 2\"]\nThe only way for it to be odd is if all of the numbers are odd.\nThere is a $\\frac 12$ chance for each number, since there is replacement.\n$\\frac 12^6=\\frac{1}{64}$\nThus the answer is $\\frac{63}{64}$[/hide]\n[hide=\"number 3\"]\nThere are $25+16+9+4+1=55$ squares.\nThere are $36$ points, and thus $\\binom{36}{4}$ ways to choose the points.\nSo the answer=$\\frac{1}{1171}$ ( I used a calculator to evaluate)[/hide]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ a,b\\ge 0$ and $ a\\plus{}b\\equal{}1$. Prove that:\r\n$ \\sqrt{a^2\\plus{}b}\\plus{}\\sqrt{b^2\\plus{}a}\\plus{}\\sqrt{1\\plus{}ab}\\le 3$.", "Solution_1": "Easy!\r\n\r\n$ \\sqrt {a^2 \\plus{} b} \\plus{} \\sqrt {b^2 \\plus{} a} \\plus{} \\sqrt {1 \\plus{} ab}$\r\n$ \\equal{} \\sqrt {3}\\sqrt {(a \\plus{} b)^2 \\plus{} (a \\plus{} b) \\plus{} 1 \\minus{} ab}\\leq \\sqrt{3}\\sqrt{3\\minus{}ab}\\leq 3$.", "Solution_2": "my solution uses Cauchy-Schwarz:\r\n\r\n$ \\sqrt{a^2\\plus{}b}\\plus{}\\sqrt{b^2\\plus{}a}\\plus{}\\sqrt{1\\plus{}ab}\\leq3$\r\n$ (\\sqrt{a^2\\plus{}b}\\plus{}\\sqrt{b^2\\plus{}a}\\plus{}\\sqrt{1\\plus{}ab})^2\\leq9$\r\nBecause of Cauchy-Schwarz-Inequality\r\n$ LHS\\leq(1\\plus{}1\\plus{}1)(a^2\\plus{}b\\plus{}b^2\\plus{}a\\plus{}1\\plus{}ab)\\leq9$ /: 3\r\n$ a^2\\plus{}b^2\\plus{}a\\plus{}b\\plus{}ab\\plus{}1\\leq3$ \r\nbecause $ a\\plus{}b\\equal{}1$\r\n$ a^2\\plus{}ab\\plus{}b^2\\leq1$ /+ab\r\n$ (a\\plus{}b)^2\\leq1\\plus{}ab$ which is true, as $ a\\plus{}b\\equal{}1$ and $ a,b\\geq0$\r\nEquality holds if $ ab\\equal{}0$ and if $ a^2\\plus{}b\\equal{}b^2\\plus{}a\\equal{}1\\plus{}ab$ , that is if $ a\\equal{}1$ and $ b\\equal{}0$ or rather if $ a\\equal{}0$ and $ b\\equal{}1$" } { "Tag": [ "number theory open", "number theory" ], "Problem": "I have this code that a professor posed to use in class. He found it on the internet and has been unable to solve it. The only clue he gave us was A = 1.\r\n\r\nHere it is. Both parts are decoded seperately:\r\n\r\nPart 1: [66-9-13-18-1-19-95-12-3-14-5-12-15-7]\r\nPart 2: [1-12-14-56-18-21-1-23-15-5-20-18-13-18-22-9-25-15-21-15-15-14-4-20]", "Solution_1": "Use [url=http://www.cryptool.com/]CrypTool[/url]" } { "Tag": [ "ratio", "complex numbers", "geometry proposed", "geometry" ], "Problem": "Let $ABCD$ and $A'B'C'D'$ be two squares in the same plane and they are oriented in the same way. Let $A'',B'',C''$ and $D''$ be the midpoints of $[AA'], [BB'], [CC']$ and $[DD']$. Show that the quadrilateral $A''B''C''D''$ is also a square.", "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": "The small letters denote the complex numbers corresponding to the capital letters.\r\n\r\nAssume we have $\\frac{c-d}{a-d}=i,\\frac{c'-d'}{a'-d'}=i$, so $\\frac{c''-d''}{a''-d''}=\\frac{\\frac{c+c'}2-\\frac{d+d'}2}{\\frac{a+a'}2-\\frac{d+d'}2}=i$, meaning that $A''D''C''$ is a right isosceles triangle. We do the same for the others, and we are done.\r\n\r\nI said \"assume\" in the first sentence because those ratios could also be $-i$. That doesn't matter. What matters is that they are both $i$ or $-i$ at the same time (because the two squares have the same orientation)." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "f(x) is defined in the positive integer, and satisfies (1) and (2)\r\n\r\n(1) if n is a positive integer, than $ f(f(n))\\equal{}4n\\plus{}\\frac{3}{2} (3\\plus{}(\\minus{}1)^n)$\r\n\r\n(2) if n is a positive integer, than $ f(2^n)\\equal{}2^{n\\plus{}1}\\plus{}3$\r\n\r\nfind f(322).\r\n\r\nThanks.", "Solution_1": "Perhaps you wanted to find meant find $ f(319)$?\r\n\r\nBecause we can, with some effort, find that $ f(n) \\equal{} 2n \\plus{} \\frac {3}{2}\\cdot\\left(( \\minus{} 1)^n \\plus{} 1\\right)$ for $ n$ of the form $ n \\equal{} g(k,x) \\equal{} 2^{k \\plus{} x} \\plus{} 2^{k \\plus{} 1} \\minus{} \\frac {( \\minus{} 1)^k \\plus{} 3}{2}$, for $ x,k\\geq 0$, if I am not mistaken.\r\n\r\nFurthermore $ f(g(k,x)) \\equal{} g(k \\plus{} 1,x)$. These relationships can be derived by starting with (2) and applying induction. Applied to finding $ f(319)$, we have $ g(5,3) \\equal{} 319$ and thus $ f(319) \\equal{} g(6,3) \\equal{} 638$.\r\n\r\nThis can be seen much more easily for finding $ f(319)$ by starting from $ f(2^3)$.\r\n\r\nBut there are no $ (k,x)$ yielding $ g(k,x) \\equal{} 322$! Furthermore, $ f(f(n))$ cannot be a power of $ 2$ for any positive $ n$, rendering (2) useless except as a first step. And furthermore we cannot find $ n$ such that $ f(f(n))$ is $ 322$, because we have a contradiction in the parity of $ n$. This seems to imply that it is impossible to directly derive $ f(n)$ for all $ n$, unless we simply use the formula we found for the special cases. I have not yet looked at other methods extending this formula to arbitrary $ n$, because I am hoping that you simply made a typo..." } { "Tag": [ "inequalities", "function", "pigeonhole principle" ], "Problem": "Let f(X,Y) be the function $\\displaystyle f(X,Y) = \\frac{(2a+1)X - Y + aXY + a}{(2+a)Y - aX + XY + 1}$ where $a$ is positive, \r\n\r\nand let $S$ be a set of $n$ positive real numbers, where $n > a + 2$ .\r\n\r\nProve that we can found two distinct numbers $X,Y \\in S$ such that $\\displaystyle 1 < f(X,Y) \\leq \\frac{a.n+1}{n-2-a}$\r\n\r\n :cool: Do you like pigeons ? :cool:", "Solution_1": "Sorry, I made a mistake :blush: !! The linequality has to be :\r\n\r\n$ a < f(X,Y) \\leq \\frac{an+1}{n-2-a}$\r\n\r\n :cool: Sorry again :cool:", "Solution_2": "Well if no one else is going to try it, I'll post a solution, since this thing has been frustrating me for a helluva long time.\r\n\r\nThe first part is straightforward:\r\nThe equation can be factored as follows\r\n$F(X,Y)=\\frac{a(X+1)(Y+1)+(a+1)(X-Y)}{(X+1)(Y+1)-(a+1)(X-Y)}$ and for $x>y$, F(X,Y)>0\r\n\r\nAnd now the part thats been bugging me for a fricken long time... :evil: \r\nI only came to this solution after noticing a pattern using the calculator (was trying to disprove it, I was getting a different result on the RHS b/c of a different pigeonholing)...so someone please explain how the solution can be more easily noticed, or show a simpler one.\r\n\r\nNow we pigeonhole on $(\\frac{0}{n-1},\\frac{1}{n-2}],[\\frac{1}{n-2},\\frac{2}{n-3}]....$ So now there must exist $X,Y \\in s$ such that $X-Y \\le \\frac{k}{n-1-k}-\\frac{k-1}{n-k}=\\frac{n-1}{(n-k-1)(n-k)}$\r\n\r\nLetting $X=\\frac{k}{n-1-k}$ and $Y=\\frac{k-1}{n-k}$, $(Y+1)(X+1)=\\frac{(n-1)^2}{(n-k-1)(n-k)}$\r\n\r\nBTW, well see that the maximum value of F(X,Y) is independent of k, so we wont have to worry about minimizing $(Y+1)(X+1)$\r\nSo now $F(X,Y)\\le\\frac{a(n-1)^2+(a+1)(n-1)}{(n-1)^2-(a+1)(n-1)}=\\frac{an+1}{n-2-a}$\r\n\r\nThere has got to be a nicer method, so someone PLEASE POST if u know it...since F(X,Y) has many nice properties like F(x,y)=F(1/y,1/x), however I kept getting stuck working with that...And if there isn't a nicer solution :( then how do u think of this craziness in the time given for a contest?", "Solution_3": "Sorry GeoMar, your proof does'nt work, because you have $k < n$ so $X-Y \\leq (n-1)/2$, but in fact the minimum of $X-Y$ can be so big I want ..\r\n :cool: Try again :cool:", "Solution_4": "I dont see the problem still... the smallest X-Y can be as large as you like, that is true... but two values of s will fall in the same interval... for example the intervals for n=6 would be as follows...\r\n\r\n$\\displaystyle(0,\\frac{1}{4}],[\\frac{1}{4},\\frac{2}{3}],[\\frac{2}{3},\\frac{3}{2}],[\\frac{3}{2},\\frac{4}{1}],[\\frac{4}{1},\\infty)$ \r\nso there must exist an X,Y\r\nsuch that $X-Y\\le \\frac{k}{n-1-k}-\\frac{k-1}{n-k}$ for $k \\in(1,5) $ This isnt necessarily the smallest X-Y...\r\nbut i guess the RHS can be approach infinity... is that a problem?\r\n\r\ni editted out some mistakes", "Solution_5": "Hi GeoMar, had you got a nice day ? Your method is nice but I'm still unconvinced !\r\nIt's not very clear, but It seems that you use two sentences : \r\n(1)$ 0 < U-V < X-Y ==> f(U,V) < f(X,Y)$ , and , (2) There are $X,Y,k$ such that $X,Y \\in S$ , $1 \\leq k < n-1$ and $X,Y \\in [\\frac {k-1}{n-k}, \\frac {k}{n-k-1}]$\r\n\r\nAm I right ? If yes, you have to prove them ... \r\n\r\n:cool: I will post later my solution .. :cool:", "Solution_6": "Ahh I see...\r\nI'm sorry for being unclear and incomplete, but im a beginner at this...I'll try to explain a bit better\r\nAlso, I said some stuff that didn't make sense in the original post, but I think I got my ideas across\r\n\r\n(2) came from that there are n values in s and there are n-1 intervals (for k=1,2,3...n-1) (although k cant =n-1 it can approach it)\r\n\r\nAnd (1) i didn't explain, and I dont even think that it is correct.\r\nWhat i guess I have to prove is that F(X,Y) is the maximum for any two values in the in the interval for which X,Y were in. But I'm not sure how this can be done, perhaps setting $x=\\frac{k-\\alpha}{n-1-k}$ and $Y=\\frac{k-\\beta}{n-k-1}$ working that out?\r\nbut I'm not sure and if I have to do that...it becomes too messy\r\n\r\nSo I would really like to see another solution, because this just gets yuckier and yuckier.", "Solution_7": "My general problem is : \r\n\r\nLet $f$ be a bijective increasing function $f : ]a,b[ \\rightarrow T$ , where $a < b$ are real numbers and $T$ is sub-set of $R$ (real numbers) .\r\n\r\nLet $g$ be the inverse of $f$ , $g : T \\rightarrow ]a,b[$ and let $S$ a finite sub-set of $T$ , with $n+1$ elements .\r\n\r\nIf $a < u < b$ and $u + \\frac{b-a}{n} < v$ , then there are $X,Y \\in S$ such that $f(u) < f(u + g(X)-g(Y)) < f(u + \\frac{b-a}{n})$ .\r\n\r\nCould you prove it ? :D Take$ f:]-a,1[ \\rightarrow ]0, \\infty [ $such that $ f(x) = \\frac{a+x}{1-x} $ :lol: Here you are ? :lol: \r\n\r\n:cool: You can produce a lot of nice problems from the general form :cool:" } { "Tag": [ "AMC", "AIME", "geometry", "linear algebra", "matrix", "LaTeX", "rotation" ], "Problem": "Hi,\r\n\r\nI'm going to post AIME problems (or problems in that level) in this thread now and then. Any of you who want to practice with me on this thread can post the answers to the problems. Please do not post any problems though. I like to keep track of the number of problems I did. :) \r\n\r\n[color=blue]Problem 1[/color]\r\n\r\nThe triangle $ABC$ has $\\angle A = 60^\\circ$, $\\angle B = 45^\\circ$. The bisector of $\\angle A$ meets $BC$ at $T$ where $AT = 24$. If the area of the triangle $ABC$ is expressed in form $a + b \\sqrt{c}$ where $c$ is not divisible by squares of any prime, compute $a+b+c$.\r\n\r\n[color=blue]Problem 2[/color]\r\n\r\nFind $c$ if $a,b,$ and $c$ are positive integers which satisfy $c = (a+bi)^3 - 107i$, where $i = \\sqrt{-1}$.\r\n\r\n[color=blue]Problem 3[/color]\r\n\r\nFind the value of $a_2+a_4+a_6+ \\cdots + a_{98}$ if $a_1,a_2,a_3 \\cdots$ is an arithmetic progression with common difference 1, and $a_1+a_2+a_3+ \\cdots + a_{98} = 137$.\r\n\r\n[color=blue]Problem 4[/color]\r\n\r\nThe integer $n$ is the smallest positive multiple of 15 such that every digit is either 0 or 8. Compute $\\frac{n}{15}$.\r\n\r\n[color=blue]Problem 5[/color]\r\n\r\nHow many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers?\r\n\r\n[color=blue]Problem 6[/color]\r\n\r\nLet $f(x) = |x-p| + |x-15| + |x-p-15|$, where $0 < p < 15$. Determine the minimum value taken by $f(x)$ for $x$ in the interval $p \\leq x \\leq 15$.\r\n\r\n[color=blue]Problem 7[/color]\r\n\r\nWhat is the largest 2-digit prime factor of the integer $n = \\binom{200}{100}$.\r\n\r\n[color=blue]Problem 8[/color]\r\n\r\nThe numbers 1447, 1005, and 1231 have something in common: each is a 4-digit number beginning with 1 that has exactly two identical digits. How many such numbers are there?\r\n\r\n[color=blue]Problem 9[/color]\r\n\r\nA point $P$ is chosen in the interior of $\\Delta ABC$ so that when lines are drawn through $P$ parallel to the sides of $\\Delta ABC$, the resulting smaller triangles have areas 4,9, and 49, respectively. Find the area of $\\Delta ABC$.\r\n\r\n[color=blue]Problem 10[/color]\r\n\r\nWhen a right triangle is rotated about one leg, the volume of the cone produced is $800 \\pi cm^3$. When the triangle is rotated about the other leg, the volume of the cone produced is $1920 \\pi cm^3$. What is the length (in $cm$) of the hypotenuse of the triangle?\r\n\r\n[color=blue]Problem 11[/color]\r\n\r\nWhat is the sum of the solutions of the equation\r\n\r\n$ \\sqrt[4] x = \\frac{12}{7 - \\sqrt[4] x}$?\r\n\r\n[color=blue]Problem 12[/color]\r\n\r\nEvaluate the product\r\n\r\n$(\\sqrt 5 + \\sqrt 6 + \\sqrt 7)(\\sqrt 5 + \\sqrt 6 - \\sqrt 7)(\\sqrt 5 - \\sqrt 6 + \\sqrt 7)(- \\sqrt 5 + \\sqrt 6 + \\sqrt 7)$\r\n\r\n[color=blue]Problem 13[/color]\r\n\r\nIf $\\tan x + \\tan y = 25$ and $\\cot x + \\cot y = 30$, what is $\\tan(x+y)$?\r\n\r\n[color=blue]Problem 14[/color]\r\n\r\nDetermine $3x_4+2x_5$, if $x_1,x_2,x_3,x_4,$ and $x_5$ satisfy the system of equations given below:\r\n\r\n\\[2x_1+x_2+x_3+x_4+x_5 = 6\\\\\r\nx_1+2x_2+x_3+x_4+x_5 = 12\\\\\r\nx_1+x_2+2x_3+x_4+x_5 = 24\\\\\r\nx_1+x_2+x_3+2x_4+x_5 = 48\\\\\r\nx_1+x_2+x_3+x_4+2x_5 = 96\\]\r\n\r\n[color=blue]Problem 15[/color]\r\n\r\nWhat is the largest positive integer $n$ for which $n^3+100$ is divisible by $n+10$?\r\n\r\n[color=blue]Problem 16[/color]\r\n\r\nThe pages of a book are numbered 1 through $n$. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in the incorrect sum of 1986. What was the number of the page that was added twice?\r\n\r\n[color=blue]Problem 17[/color]\r\n\r\nOne commercially available ten button lock may be opened by depressing - in any order - the correct five buttons. The sample shown at right has {1,2,3,6,9} as its combination. Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations. How many additional combinations would this allow?\r\n\r\n[color=blue]Problem 18[/color]\r\n\r\nIn $\\Delta ABC$, $\\tan(\\angle CAB) = \\frac{22}{7}$ and the altitude from $A$ divides $BC$ into segments of lengths 3 and 17. What is the area of $\\Delta ABC$?\r\n\r\n[color=blue]Problem 19[/color]\r\n\r\nFind $x^2+y^2$ if $x$ and $y$ are positive integers such that\r\n\r\n\\[xy+x+y = 71\\\\\r\nx^2y+xy^2 = 880\\]\r\n\r\n[color=blue]Problem 20[/color]\r\n\r\nFor how many positive integers $n < 1000$ such that $\\lfloor {\\log_2 n} \\rfloor$ is positive and even? As usual, $\\lfloor {z} \\rfloor$ denotes the greatest integer function of $z$.\r\n\r\n[color=blue]Problem 21[/color]\r\n\r\nThe square below is magic. It has a number in each cell. The sums of each row and column and of the two main diagonals are all equal. Compute $x$. ($*$ are places with no number or variable defined in the problem)\r\n\r\n\\[x\\ 19\\ 96\\\\\r\n1\\ *\\ *\\\\\r\n*\\ *\\ \\ *\\]\r\n\r\n[color=blue]Problem 22[/color]\r\n\r\nFind the value of $m$ if $m = \\sqrt{1 + 28 \\cdot 29 \\cdot 30 \\cdot 31}$.\r\n\r\n[color=blue]Problem 23[/color]\r\n\r\n10 points lie on a circumference of the circle. How many distinct convex polygons can be formed by connecting some or all of the points?\r\n\r\n[color=blue]Problem 24[/color]\r\n\r\nFor some digit $d$, we have $0.{\\overline{d25}} = \\frac{n}{810}$, where $n$ is a positive integer. Find $n$.\r\n\r\n[color=blue]Problem 25[/color]\r\n\r\nWe have five consecutive positive integers whose sum is cube and sum of the three middle is a square. What is the smallest possible middle integer?\r\n\r\n[color=blue]Problem 26[/color]\r\n\r\nThe function $f$ holds $f(x) + f(x-1) = x^2$ for all $x$ values. Given that $f(19) = 94$, find the remainder when $f(94)$ is divided by 1000.\r\n\r\n[color=blue]Problem 27[/color]\r\n\r\n$k$ is a unique integer such that:\r\n\r\n$\\lfloor {\\log_2 1} \\rfloor + \\lfloor {\\log_2 2} \\rfloor + \\lfloor {\\log_2 3} \\rfloor + \\cdots \\lfloor {\\log_2 k} \\rfloor = 1994$. Find $k$.\r\n\r\n[color=blue]Problem 28[/color]\r\n\r\nThe roots of $x^3+3x^2+4x-11 = 0$ are $a,b,c$. Let's say there exists another cubic polynomial with roots $a+b, b+c, c+a$. This polynomial is in form $x^3+rx^2+sx+t = 0$. What is $t$?\r\n\r\n[color=blue]Problem 29[/color]\r\n\r\nStarting with a unit square, a sequence of square is generated. Each square in the sequence has half the side-length of its predecessor and two of its sides bisected by its predecessor's sides as shown. Find the total area enclosed by the first five squares in the sequence.\r\n\r\nFor image, please click:\r\n\r\nhttp://www.kalva.demon.co.uk/aime/aime95.html\r\n\r\n(Do not look at the answer!)\r\n\r\n[color=blue]Problem 30[/color]\r\n\r\nAssume that $a,b,c,$ and $d$ are positive integers such that $a^5 = b^4, c^3 = d^2,$ and $c-a = 19$. Determine $d-b$.", "Solution_1": "Cool.\r\n\r\nExtend altitude from C to AB, and a perpendicular from T to E on AB.\r\n\r\nSo, $TE = 12$. $AE = 12\\sqrt{3}$.\r\n\r\nIsosceles triangles, 30-60-90, 45-90-45... WHOA.\r\n\r\nAnyways, $CF = 12\\sqrt{3}$.\r\n\r\nSo, 291, after calculations. Darn if my calculations are wrong...\r\n\r\nThis is an AIME problem?... \r\n\r\nWhoaness...\r\n\r\nThis one's just here for practice. Don't look if you don't want to: \r\n\r\n[hide]Problem:\n\nHere's my problem. (AIME)\n\nIn triangle ABC, a point P exists such that the triangles formed by drawing parallel lines through P to each of the sides are of areas 4, 9, and 49.\n\nWhat is [ABC]?[/hide]", "Solution_2": "[hide=\"Answer\"]Bisecting $\\angle A$ gives two $30^\\circ$ angles. Draw the perpendicular from $T$ to $AB$ to get a 30-60-90 triangle and an isosceles right triangle, labeling the intersection as point $D$, so we have $TD=12$ and $AD=12\\sqrt{3}$, and we see that $AB=12+12\\sqrt{3}$. Now, draw the perpendicular from $C$ to $AB$, labeling the intersection as point $E$, and we have 30-60-90 $\\triangle ACE$, with $CE=12\\sqrt{3}$ and $AE=12$. Therefore, the area of $\\triangle ABC=\\frac{1}{2}(12)(12\\sqrt{3})+\\frac{1}{2}(12\\sqrt{3})^2=216+72\\sqrt{3}=a+b\\sqrt{c}$, so $a+b+c=291$.[/hide]", "Solution_3": "[quote=\"Silverfalcon\"]\n[color=blue]Problem 1[/color]\n\nThe triangle $ABC$ has $\\angle A = 60^\\circ$, $\\angle B = 45^\\circ$. The bisector of $\\angle A$ meets $BC$ at $T$ where $AT = 24$. If the area of the triangle $ABC$ is expressed in form $a + b \\sqrt{c}$ where $c$ is not divisible by squares of any prime, compute $a+b+c$.\n[/quote]\n\nMy solution:\n\n[hide]Note that $\\angle ACB = 180^\\circ - 60^\\circ - 45^\\circ = 75^\\circ$. This means that the triangle $ACT$ has interior angles $\\angle TAC = 60^\\circ$, $\\angle ACT = 75^\\circ$, and $\\angle ATC = 75^\\circ$. $ATC$ is therefore an isosceles triangle, with $AT = AC = 24$.\n\nNow apply the Law of Sines to find\n\n$AB = \\frac{24 \\sin 75^\\circ}{\\sin 45^\\circ}$\n\n$AB = 24\\sqrt{2}\\left(\\frac{\\sqrt{3} + 1}{2\\sqrt{2}}\\right)$\n\n$AB = 12(\\sqrt{3} + 1)$\n\nSo we now have\n\n$[ABC] = \\frac{1}{2} (AB)(AC) (\\sin \\angle BAC)$\n\n$ = \\frac{1}{2}(24)(12)(\\sqrt{3} + 1)(\\sin 60^\\circ)$\n\n$ = 216 + 72\\sqrt{3}$\n\nTherefore,\n\n$a + b + c = 216 + 72 + 3 = \\fbox{291}$\n\n$QED$[/hide]\n\n[quote=\"Silverfalcon\"]\n[color=blue]Problem 2[/color]\n\nFind $c$ if $a,b,$ and $c$ are positive integers which satisfy $c = (a+bi)^3 - 107i$, where $i = \\sqrt{-1}$.[/quote]\n\nMy solution:\n\n[hide]\nExpanding the cube and simplifying gives\n\n$c = a^3 - 3ab^2 + (3a^2b - b^3 - 107)i$\n\nIf $c$ is to be a real number, we must have\n\n$3a^2b - b^3 - 107 = 0$\n\nRearranging and factoring gives\n\n$b(3a^2 - b^2) = 107$\n\nNoting that $a,b$ are positive integers and $107$ is prime, we must have either \n\n$3a^2 - b^2 = 107$\n$b = 1$ \n\nor \n\n$b = 107$\n$3a^2 - b^2 = 1$\n\nThe second set of equations must be discarded, as they yield a non-integer $a$. The first set of equations gives $(a,b) = (\\pm 6, 1)$. $a$ must be positive, so we have $(a,b) = (6,1)$. Remembering that we have\n\n$c = a(a^2 - 3b^2)$\n\nwe can now plug in our solution for $(a,b)$ to obtain\n\n$c = (6)(33) = \\fbox{198}$\n\n$QED$[/hide]", "Solution_4": "shouldnt it be $i = \\sqrt{-1}$, not $i=-1$. That might be a little different after that then.", "Solution_5": "[quote=\"Silverfalcon\"]\nwhere $i = -1$.[/quote]\r\n\r\nOh, and I believe you meant $i = \\sqrt{-1}$ ;)", "Solution_6": "I did problem 2 somewhere before, recently...\r\n\r\n[hide=\"Answer\"]$c=(a+bi)^3-107i \\Rightarrow (a+bi)^3=a^3-3ab^2+3a^2bi-b^3i$\n$3a^2bi-b^3i=107i \\Rightarrow b(3a^2-b^2)=107 \\Rightarrow a=6$, $b=1$\nTherefore, we have $c=216-18+108i-i-107i=216-18=198$.[/hide]", "Solution_7": "[quote=\"zanttrang\"][quote=\"Silverfalcon\"]\nwhere $i = -1$.[/quote]\n\nOh, and I believe you meant $i = \\sqrt{-1}$ ;)[/quote]\r\n\r\nYes. I edited now. I think I made typo when I was writing the original wording of the problem (probably forgot the square root part)", "Solution_8": "3 more problems.. :P \r\n\r\n(I'll try to add new problems every day... Hopefully at that rate, I'll be able to cover most of the AIMEs by this summer. :D )", "Solution_9": "Problem 4:\r\n\r\n[hide]To be divisible by three, the number must have a multiple of three eights. As in, three eights, six eights, and so on. And to be divisible by 5, it must end in a zero. Thus the number is 8880 when divided by 15 yields 592. Yeesh, that was like a MATHCOUNTS countdown round problem.[/hide]", "Solution_10": "On #3, I'm assuming that common difference means that $a_1+1=a_2$, $a_2+1=a_3$, etc.\r\n\r\n[hide=\"3\"]$a_1+(a_1+1)+(a_1+2)+...+(a_1+97)=137 \\Rightarrow 98\\cdot a_1+\\frac{97(98)}{2}=137 \\Rightarrow 98\\cdot a_1=-4616 \\Rightarrow a_1=-\\frac{2308}{49}$\n$a_2+a_4+...+a_{98}=49\\cdot a_1+49^2=-2308+2401=93$[/hide]\n\n[hide=\"4\"]The sum of the digits will be divisible by $3$, and will end in either a $5$ or a $0$. Since all digits are either $0$ or $8$, the number obviously ends in a $0$, and must include 3 $8$'s to sum to a multiple of $3$. Therefore, the smallest possible multiple of $15$ consisting only of $8$'s and $0$'s is $8880$, and $\\frac{8880}{15}=592$.[/hide]\n\n[hide=\"5\"]All odd numbers can be found by taking the difference of consecutive squares, so we have eliminated the numbers $1,3,5,...,999$. Next, if we try taking the difference between $(n+2)^2$ and $n^2$, we get $4n+4$, so this set will include all multiples of $4$. $(n+3)^2-n^2=6n+9$, which consists only of odd numbers, and we have already eliminated all odd numbers. Similarly, $(n+4)^2-n^2=8n+16$, which is multiples of $8$, all of which have been covered by the multiples of $4$. Therefore, the only numbers that cannot be expressed as the difference between two perfect squares are numbers in the form $2k$, where $k$ is any odd integer, so there are $250$ numbers of this kind in the given interval, and $750$ can be expressed in this way.\n\nThis also proves that, with $n$ numbers in a given set, $\\frac{n}{4}$ of those numbers cannot be expressed as the difference between two perfect squares.[/hide]\r\n\r\n[b]EDIT:[/b] [quote=\"sirorange\"]Yeesh, that was like a MATHCOUNTS countdown round problem.[/quote]\r\n\r\nI'd guess it was #1 or #2, possibly on one of the easier AIME's.", "Solution_11": "[hide=\"3\"]\nLet $N=a_2+a_4+...+a_{98}$. Then $2N-49=a_1+a_2+a_3+...a_{98}=137$ $\\Longrightarrow$ $\\boxed{N=93}$\n[/hide]", "Solution_12": "I think you mean #3. ;) That's a nice proof, but I have one question: how do you get that $2N-49=a_1+a_2+...+a_{98}$?", "Solution_13": "[quote=\"JesusFreak197\"]On #3, I'm assuming that common difference means that $a_1+1=a_2$, $a_2+1=a_3$, etc.\n\n[hide=\"3\"]$a_1+(a_1+1)+(a_1+2)+...+(a_1+97)=137 \\Rightarrow 98\\cdot a_1+\\frac{97(98)}{2}=137 \\Rightarrow$\n$98(a_1+\\frac{97}{2})=137 \\Rightarrow a_1=-47$\n\nFrom here, we can see that $a_{48}=0$, so the sum $a_1+a_2+a_3+...+a_{95}=0$, and $a_{96}+a_{98}=48+50=98$[/hide]\n\n[/quote]\n\n[hide=\"problem with your solution\"]You made a calculation error. If you look back over your equations, you'll find that $48a_1 = -4616$. So, your method no longer works out nicely. We can proceed by simple brute force:\n\n$a_2 + a_4 + ... + a_{98} = \\frac{98a_1}{2} + 1 + 3 + ... + 97$\n\n$= \\frac{-4616}{2} + \\frac{(98)(98)}{(2)(2)}$\n\n$=-2308 + 2401$\n\n$= \\fbox{93}$[/hide]", "Solution_14": "[quote=\"themonster\"][hide=\"4\"]\nLet $N=a_2+a_4+...+a_{98}$. Then $2N-49=a_1+a_2+a_3+...a_{98}=137$ $\\Longrightarrow$ $\\boxed{N=98}$\n[/hide][/quote]\n\n[hide=\"problem with your solution\"]Check your math and you'll find that actually, $N = \\fbox{93}$ ;)[/hide]", "Solution_15": "I have a question on 23\r\n[quote]Problem 23 \n\n10 points lie on a circumference of the circle. How many distinct convex polygons can be formed by connecting some or all of the points? \n[/quote]\r\n\r\nI originally thought the answer was (10,3)+....+(10,10) also but if you label the points going in order 1,2,3...10\r\nthen would connecting 1,2,3 be the same polygon as 2,3,4?\r\nIt says how many DISTINCT", "Solution_16": "How would connecting $(1,2,3)$ be the same as connecting $(2,3,4)$ unless they are evenly spaced? It never says that they are evenly spaced.", "Solution_17": "[quote=\"JesusFreak197\"]How would connecting $(1,2,3)$ be the same as connecting $(2,3,4)$ unless they are evenly spaced? It never says that they are evenly spaced.[/quote]It doesn't say they aren't evenly spaced. When I did this question I got it but in a lot of problems, it does assume they are evenly spaced. (Which would make the problem harder, too.)", "Solution_18": "O really? I've always thought that if it doesn't specify, it's NOT evenly spaced. Especially if its in a polygon, cuz regular polygons are so easy to work with.", "Solution_19": "It's actually not too hard if they're all evenly spaced. Once you figure out a pattern, you can apply it to every polygon from 3 sides to 10 sides. (Although, there's only one 10-sided polygon in there anyway...)", "Solution_20": "Since this was #1, I assumed that the points are evenly spaced.", "Solution_21": "Well, what was the answer? I don't think my solution is assuming that the points are evenly spaced, but it seems to be the simplest option...", "Solution_22": "968 was the correct answer.", "Solution_23": "What's the answer to 26? Me and Jesus are disagreeing :D", "Solution_24": "For 26, 561 is the correct answer.", "Solution_25": "Oh, I'm an idiot. I forgot that it's missing every other odd number. :roll:\r\n\r\nSo, mine should have come out the exact same as krusty's. It is now fixed. :)", "Solution_26": "It is evenly spaced \r\nbut \r\nthe real AIME specified that having different vertices constitutes a different polygon\r\n(It was a little NOTE under the problem)", "Solution_27": "[quote=\"amirhtlusa\"]$ \\displaystyle (\\sqrt 5 \\plus{} \\sqrt 6 \\plus{} \\sqrt 7)(\\sqrt 5 \\plus{} \\sqrt 6 \\minus{} \\sqrt 7)(\\sqrt 5 \\minus{} \\sqrt 6 \\plus{} \\sqrt 7)( \\minus{} \\sqrt 5 \\plus{} \\sqrt 6 \\plus{} \\sqrt 7)$\n\nif i recall correctly:\n$ (a \\plus{} b \\plus{} c)(a \\plus{} b \\minus{} c)(a \\minus{} b \\plus{} c)( \\minus{} a \\plus{} b \\plus{} c)$ simplifies to:\n$ \\minus{} a^4 \\plus{} 2a^2b^2 \\minus{} b^4 \\plus{} 2b^2c^2 \\minus{} c^4\\Rightarrow$\nthe answer is :\n$ \\boxed{34}$[/quote]\r\n\r\nThis question seemed odd, to me. I just multiplied out all the brackets and simplified it. The first pair of brackets leads to $ 4\\plus{}2\\sqrt{30}$ while the second pair leads to $ \\minus{}4\\plus{}2\\sqrt{30}$, which means that the final result is $ 4 \\times 30 \\minus{} 16 \\equal{} 104$.", "Solution_28": "don't revive such old threads", "Solution_29": "If you have something new to add, you can." } { "Tag": [ "function", "Euler", "calculus", "calculus computations" ], "Problem": "Does anybody know where I can download a computer program that will repeat Euler's method over and over and over again? It'd be even better if the program produced a graph.\r\n\r\nThanks. :)", "Solution_1": "I believe theres a computer forum somewhere on mathlinks.", "Solution_2": "The TI-86 and TI-89 both have such a program built in. They also produce piecewise-linear graphs of the solution.\r\n\r\nI have a program written for TI-83/84's that does Euler's Method, but it would take me a while to type it up. You may be able to find a similar one on the internet.", "Solution_3": "We're talking about Euler's method for approximating the solution to an ODE, right?\r\n\r\nOut of curiosity, what is this needed for?\r\n\r\nHow long do you think it would take to program it?", "Solution_4": "Maple can do it also, so I assume Mathematica should be able to, and if Matlab can't, you could write a program for that.", "Solution_5": "[quote=\"Aunt Sally\"]How long do you think it would take to program it?[/quote]\r\n\r\nShouldn't take very long; I had to implement this in Maple on one of my finals awhile back.", "Solution_6": "Assume you're solving $y'=g(x,y);\\,y(x_0)=y_0$ on the interval $[x_0,x_1].$ Let's suppose we want to use $n$ steps in the interval.\r\n\r\nHere's a version of the program that's almost (but not quite) in BASIC:\r\n\r\n[code](Do some stuff to set up a plotting window.)\n(Input n, x0, x1, y0.)\n(If appropriate, write a subroutine/external function to compute g(x,y).)\n\nLET dx=(x1-x0)/n\nLET x=x0\nLET y=y0\n\nFOR j=1 to n\n PLOT (x,y); (leave the \"pen\" down)\n LET dy = g(x,y)*dx\n LET x = x + dx\n LET y = y + dy\nNEXT j\n\n(pick up \"pen\")\nEND[/code]\nIf you don't write a subroutine to compute $g(x,y),$ you can just put that formula in line in the appropriate place.\n\nThe plotting details (including the commands for leaving the \"pen\" down and picking it up) will vary from language to language and implementation to implementation; so also the preliminary details of setting up the plotting window, and whether you want additional amenities, like axes and tick marks.\n\nThe thing is, as soon as you have this, you can improve upon Euler's method - after all, it has only first order accuracy, which is very poor for numerical purposes. The \"improved Euler\" or second order Runge-Kutta method requires only one additional line:\n\n[code]FOR j=1 to n\n PLOT (x,y); (leave the \"pen\" down)\n LET k1 = g(x,y)\n LET k2 = g(x+dx,y+k1*dx)\n LET x = x + dx\n LET y = y + (k1+k2)*dx/2\nNEXT j[/code]\nAnd the fourth-order Runge-Kutta method is only slightly more complicated than that:\n\n[code]FOR j=1 to n\n PLOT (x,y); (leave the \"pen\" down)\n LET k1 = g(x,y)\n LET k2 = g(x+dx/2,y+(k1*dx)/2)\n LET k3 = g(x+dx/2,y+(k2*dx)/2)\n LET k4 = g(x+dx,y+k3*dx)\n LET x = x + dx\n LET y = y + (k1+2*k2+2*k3+k4)*dx/6\nNEXT j[/code]", "Solution_7": "MatLab rules, do it there." } { "Tag": [], "Problem": "\u039a\u0391\u039b\u0397\u03a3\u03a0\u0395\u03a1\u0391 \u03a3\u0395 \u039f\u039b\u039f\u03a5\u03a3 \r\n\u039d\u0395\u039f\u03a4\u0391\u03a4\u039f \u039c\u0395\u039b\u039f\u03a3 \u039a\u0391\u0399 \u039c\u0391\u039b\u039b\u039f\u039d \u039b\u0399\u0393\u039f \u03a4\u03a3\u0391\u039a\u03a9\u039c\u0395\u039d\u039f\u03a3 \u039c\u0395 \u03a4\u0391 \u039c\u0391\u0398\u0397\u039c\u0391\u03a4\u0399\u039a\u0391......\r\n\u0391\u039b\u039b\u0391 \u039f\u039b\u039f \u039a\u0391\u0399 \u03a0\u03a1\u039f\u03a3\u03a0\u0391\u0398\u03a9.\u039a\u0391\u03a4\u0399 \u0395\u0399\u039d\u0391\u0399 \u039a\u0391\u0399 \u0391\u03a5\u03a4\u039f!\r\n\u0398\u0391 \u0397\u0398\u0395\u039b\u0391 \u0391\u039d \u039c\u03a0\u039f\u03a1\u0395\u0399\u03a4\u0395 \u039d\u0391 \u039c\u039f\u03a5 \u03a0\u0395\u0399\u03a4\u0395 \u03a4\u0397 \u039b\u03a5\u03a3\u0397 \u03a4\u0397\u03a3 \u03a0\u0391\u03a1\u0391\u039a\u0391\u03a4\u03a9 \u0391\u03a3\u039a\u0397\u03a3\u0397\u03a3.\r\n\r\n10 \u03b5\u03c0\u03b9\u03b2\u03ac\u03c4\u03b5\u03c2 \u03c0\u03bf\u03c5 \u03b8\u03b5\u03c9\u03c1\u03bf\u03cd\u03bd\u03c4\u03b1\u03b9 \u03b4\u03b9\u03b1\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03bf\u03b9, \u03b2\u03c1\u03af\u03c3\u03ba\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bc\u03ad\u03c3\u03b1 \u03c3\u03b5 \u03ad\u03bd\u03b1 \u03bb\u03b5\u03c9\u03c6\u03bf\u03c1\u03b5\u03af\u03bf \u03ba\u03b1\u03b9 \u03b1\u03c0\u03bf\u03bc\u03ad\u03bd\u03bf\u03c5\u03bd 3 \u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 (\u03a3\u03c4\u0391, \u03a3\u03c4\u0392, \u03a3\u03c4\u0393) \r\n\r\n\u0391. \u039c\u03b5 \u03c0\u03cc\u03c3\u03bf\u03c5\u03c2 \u03c4\u03c1\u03cc\u03c0\u03bf\u03c5\u03c2 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bd \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b2\u03b9\u03b2\u03b1\u03c3\u03c4\u03bf\u03cd\u03bd \u03cc\u03bb\u03bf\u03b9 \u03bf\u03b9 \u03b5\u03c0\u03b9\u03b2\u03ac\u03c4\u03b5\u03c2 \u03b1\u03c0\u03cc \u03c4\u03bf \u03bb\u03b5\u03c9\u03c6\u03bf\u03c1\u03b5\u03af\u03bf \u03bc\u03b5 \u03b4\u03b5\u03b4\u03bf\u03bc\u03ad\u03bd\u03bf \u03cc\u03c4\u03b9 \u03c3\u03b5 \u03ba\u03ac\u03b8\u03b5 \u03c3\u03c4\u03ac\u03c3\u03b7 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bd \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b2\u03b9\u03b2\u03ac\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1\u03c2 \u03ad\u03c9\u03c2 \u03ba\u03b1\u03b9 10 \u03b5\u03c0\u03b9\u03b2\u03ac\u03c4\u03b5\u03c2;\r\n\u0392. \u039c\u03b5 \u03c0\u03cc\u03c3\u03bf\u03c5\u03c2 \u03c4\u03c1\u03cc\u03c0\u03bf\u03c5\u03c2 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bd \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b2\u03b9\u03b2\u03b1\u03c3\u03c4\u03bf\u03cd\u03bd \u03cc\u03bb\u03bf\u03b9 \u03b1\u03c0\u03cc \u03c4\u03bf \u03bb\u03b5\u03c9\u03c6\u03bf\u03c1\u03b5\u03af\u03bf \u03b1\u03bd \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03ba\u03b1\u03c4\u03ad\u03b2\u03bf\u03c5\u03bd \u03b1\u03c0\u03cc 4 \u03c3\u03b5 \u03b4\u03cd\u03bf \u03bf\u03c0\u03bf\u03b9\u03b5\u03c3\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 2 \u03c3\u03c4\u03b7\u03bd \u03ac\u03bb\u03bb\u03b7;\r\n\u0393. \u039c\u03b5 \u03c0\u03cc\u03c3\u03bf\u03c5\u03c2 \u03c4\u03c1\u03cc\u03c0\u03bf\u03c5\u03c2 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bd \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b2\u03b9\u03b2\u03b1\u03c3\u03c4\u03bf\u03cd\u03bd \u03cc\u03bb\u03bf\u03b9 \u03b1\u03c0\u03cc \u03c4\u03bf \u03bb\u03b5\u03c9\u03c6\u03bf\u03c1\u03b5\u03af\u03bf \u03b1\u03bd \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03ba\u03b1\u03c4\u03ad\u03b2\u03bf\u03c5\u03bd 4 \u03c3\u03c4\u03b7\u03bd \u03a3\u03c4\u0391 \u03ae 4 \u03c3\u03c4\u03b7\u03bd \u03a3\u03c4\u0392 (\u03cc\u03c7\u03b9 \u03cc\u03bc\u03c9\u03c2 \u03b1\u03c0\u03cc 4 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 \u03b4\u03cd\u03bf \u03c4\u03b1\u03c5\u03c4\u03cc\u03c7\u03c1\u03bf\u03bd\u03b1); \r\n\r\n\u0394\u03b5\u03bd \u03b8\u03b1 \u03bb\u03b7\u03c6\u03b8\u03b5\u03af \u03c5\u03c0\u03cc\u03c8\u03b7 \u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac \u03b1\u03c0\u03bf\u03b2\u03af\u03b2\u03b1\u03c3\u03b7\u03c2 \u03c4\u03c9\u03bd \u03b5\u03c0\u03b9\u03b2\u03b1\u03c4\u03ce\u03bd \u03c3\u03b5 \u03ba\u03ac\u03b8\u03b5 \u03c3\u03c4\u03ac\u03c3\u03b7\r\n\r\n\r\n\u03a3\u0391\u03a3 \u0395\u03a5\u03a7\u0391\u03a1\u0399\u03a3\u03a4\u03a9 :lol:", "Solution_1": "\u039b\u03bf\u03b9\u03c0\u03cc\u03bd, \u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c3\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03b4\u03b9\u03b1\u03af\u03c4\u03b5\u03c1\u03b1 \u03b4\u03cd\u03c3\u03ba\u03bf\u03bb\u03b7, \u03b1\u03bb\u03bb\u03ac \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd \u03b2\u03b1\u03c1\u03b5\u03c4\u03ae (\u03ba\u03b1\u03b9 \u03b9\u03b4\u03b9\u03b1\u03af\u03c4\u03b5\u03c1\u03b1 \u03c0\u03bf\u03bb\u03cd\u03c0\u03bb\u03bf\u03ba\u03b7 \u03b1\u03bd \u03b4\u03b5 \u03be\u03ad\u03c1\u03b5\u03b9\u03c2 \u03c3\u03c5\u03bd\u03b4\u03c5\u03b1\u03c3\u03c4\u03b9\u03ba\u03ae). \u0391\u03c1\u03c7\u03b9\u03ba\u03ac \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c4\u03b7\u03c1\u03ae\u03c3\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 \u03c4\u03c1\u03cc\u03c0\u03bf \u03bc\u03b5 \u03c4\u03bf\u03bd \u03bf\u03c0\u03bf\u03af\u03bf \u03b2\u03b3\u03b1\u03af\u03bd\u03bf\u03c5\u03bd \u03bf\u03b9 \u03b5\u03c0\u03b9\u03b2\u03ac\u03c4\u03b5\u03c2 (\u03cc\u03c0\u03c9\u03c2 4 \u03c3\u03c4\u03b7 \u03a3\u03c4\u0391, 4 \u03c3\u03c4\u03b7 \u03a3\u03c4\u0392, 2 \u03c3\u03c4\u03b7 \u03a3\u03c4\u0393), \u03b1\u03bd \u03b4\u03b7\u03bc\u03b9\u03bf\u03c5\u03c1\u03b3\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c0\u03af\u03bd\u03b1\u03ba\u03b1 \u03ba\u03b1\u03b9 \u03b5\u03c6\u03b1\u03c1\u03bc\u03cc\u03c3\u03b5\u03b9\u03c2 \u03c4\u03b7 \u03b8\u03b5\u03bc\u03b5\u03bb\u03b9\u03ce\u03b4\u03b7 \u03b1\u03c1\u03c7\u03ae \u03c4\u03b7\u03c2 \u03c3\u03c5\u03bd\u03b4\u03c5\u03b1\u03c3\u03c4\u03b9\u03ba\u03ae\u03c2 \u03b1\u03c0\u03b1\u03c1\u03af\u03b8\u03bc\u03b7\u03c3\u03b7\u03c2, \u03b8\u03b1 \u03b2\u03c1\u03b5\u03b9\u03c2 10!\r\n\u0391\u03c5\u03c4\u03cc \u03ae\u03c4\u03b1\u03bd \u03c4\u03bf \u0392 \u03b5\u03c1\u03ce\u03c4\u03b7\u03bc\u03b1.\r\n\u0393\u03b9\u03b1 \u03c4\u03bf \u0391 \u03b5\u03c1\u03ce\u03c4\u03b7\u03bc\u03b1 \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b2\u03c1\u03b5\u03b9\u03c2 \u03c4\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c4\u03c1\u03cc\u03c0\u03c9\u03bd \u03c0\u03bf\u03c5 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bd \u03bd\u03b1 \u03b2\u03b3\u03bf\u03c5\u03bd \u03bf\u03b9 \u03b5\u03c0\u03b9\u03b2\u03ac\u03c4\u03b5\u03c2 \u03c0\u03bf\u03c5 \u03b9\u03c3\u03bf\u03cd\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c4\u03c9\u03bd \u03bb\u03cd\u03c3\u03b5\u03c9\u03bd \u03c4\u03b7\u03c2 \u03b4\u03b9\u03bf\u03c6\u03b1\u03bd\u03c4\u03b9\u03ba\u03ae\u03c2 \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7\u03c2 $ x1 \\plus{} x2 \\plus{} x3 \\equal{} 10$ \u03bc\u03b5 $ x1, x2, x3$ \u03c6\u03c5\u03c3\u03b9\u03ba\u03bf\u03cd\u03c2, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae 10 \u03b1\u03bd\u03ac 2 \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae $ \\frac{10!}{2!(10\\minus{}2)!}$ . \u03a0\u03bf\u03bb\u03bb\u03b1\u03c0\u03bb\u03b1\u03c3\u03b9\u03ac\u03b6\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c4\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c4\u03c9\u03bd \u03c4\u03c1\u03cc\u03c0\u03c9\u03bd \u03bc\u03b5 \u03c4\u03b9\u03c2 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03ce\u03c3\u03b5\u03b9\u03c2 \u03c3\u03b5 \u03ba\u03ac\u03b8\u03b5 \u03c4\u03c1\u03cc\u03c0\u03bf \u03b2\u03c1\u03af\u03c3\u03ba\u03b5\u03b9\u03c2 $ \\frac{(10!)^{2}}{2!(10\\minus{}2)!}$, \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b1\u03c0\u03ac\u03bd\u03c4\u03b7\u03c3\u03b7 \u03c3\u03c4\u03bf \u03b5\u03c1\u03ce\u03c4\u03b7\u03bc\u03b1 \u0391.\r\n\u03a4\u03bf \u03b5\u03c1\u03ce\u03c4\u03b7\u03bc\u03b1 \u0393 \u03bb\u03cd\u03bd\u03b5\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03b1\u03c1\u03c7\u03ae \u03c4\u03b7\u03c2 \u03c3\u03c5\u03bc\u03c0\u03b5\u03c1\u03af\u03bb\u03b7\u03c8\u03b7\u03c2-\u03b5\u03be\u03b1\u03af\u03c1\u03b5\u03c3\u03b7\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd \u03b5\u03c6\u03b1\u03c1\u03bc\u03bf\u03b3\u03ae \u03c4\u03c9\u03bd \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03bc\u03b5 \u03ad\u03bd\u03b1 \u03c0\u03b1\u03c1\u03b1\u03bb\u03b1\u03b3\u03bc\u03ad\u03bd\u03bf \u03c4\u03c1\u03cc\u03c0\u03bf. \u03a4\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c4\u03c9\u03bd \u03c4\u03c1\u03cc\u03c0\u03c9\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 10 \u03b1\u03bd\u03ac 2 \u03c0\u03bb\u03b7\u03bd (10-4) \u03b1\u03bd\u03ac 1 \u03ba\u03b1\u03b9 \u03cc\u03bb\u03bf \u03b1\u03c5\u03c4\u03cc \u03b5\u03c0\u03af 10!, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03bc\u03b5 $ 10!(\\frac{10!}{2!(10\\minus{}2)!}\\minus{}{\\frac{(10\\minus{}4)!}{1!(10\\minus{}4\\minus{}1)!}})$\r\n\r\n\u0391\u03bd \u03ad\u03c7\u03b5\u03b9\u03c2 \u03bf\u03c0\u03bf\u03b9\u03b1\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03b1\u03c0\u03bf\u03c1\u03af\u03b1 \u03c3\u03c4\u03b5\u03af\u03bb\u03b5 \u03bc\u03bf\u03c5 \u03c4\u03b7 \u03c3\u03b5 \u03b1\u03c5\u03c4\u03ae \u03c4\u03b7 \u03c3\u03b5\u03bb\u03af\u03b4\u03b1.", "Solution_2": "\u038c\u03c3\u03bf \u03c0\u03b9\u03bf \u03c0\u03bf\u03bb\u03cd \u03c3\u03ba\u03ad\u03c6\u03c4\u03bf\u03bc\u03b1\u03b9 \u03b1\u03c5\u03c4\u03ae \u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7, \u03c4\u03cc\u03c3\u03bf \u03c0\u03b9\u03bf \u03c0\u03bf\u03bb\u03cd \u03bc\u03bf\u03c5 \u03b1\u03c1\u03ad\u03c3\u03b5\u03b9. \u0394\u03b5 \u03b8\u03ad\u03bb\u03b5\u03b9 \u03b9\u03b4\u03b9\u03b1\u03af\u03c4\u03b5\u03c1\u03b5\u03c2 \u03b3\u03bd\u03ce\u03c3\u03b5\u03b9\u03c2 \u03b1\u03bb\u03bb\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03bf\u03bc\u03c8\u03ae \u03c0\u03b5\u03c1\u03af\u03bb\u03b7\u03c8\u03b7 \u03c4\u03c9\u03bd \u03ba\u03c5\u03c1\u03b9\u03cc\u03c4\u03b5\u03c1\u03c9\u03bd \u03c3\u03c4\u03bf\u03b9\u03c7\u03b5\u03af\u03c9\u03bd \u03c4\u03b7\u03c2 \u03c3\u03c5\u03bd\u03b4\u03c5\u03b1\u03c3\u03c4\u03b9\u03ba\u03ae\u03c2." } { "Tag": [ "linear algebra", "matrix", "number theory unsolved", "number theory" ], "Problem": "Let $ a$ be a fixed integer. Find all integer solutions $ x,y,z$ of the system:\r\n\r\n$ 5x\\plus{}(a\\plus{}2)y\\plus{}(a\\plus{}2)z\\equal{}a,$\r\n$ (2a\\plus{}4)x\\plus{}(a^2\\plus{}3)y\\plus{}(2a\\plus{}2)z\\equal{}3a\\minus{}1,$\r\n$ (2a\\plus{}4)x\\plus{}(2a\\plus{}2)y\\plus{}(a^2\\plus{}3)z\\equal{}a\\plus{}1.$", "Solution_1": "[hide]Take equation #2 and subtract equation #3 to get: \n$ (a^2\\minus{}2a\\plus{}1)(y\\minus{}z) \\equal{} 2a\\minus{}2 \\Rightarrow (a\\minus{}1)^2(y\\minus{}z) \\equal{} 2(a\\minus{}1)$ \n\nThus, either $ a \\equal{} 1$ OR $ y\\minus{}z \\equal{} \\dfrac{2}{a\\minus{}1}$ which must be an integer, \n\nSo, $ a\\minus{}1 \\equal{} \\pm1 , \\pm2 \\Rightarrow a \\equal{} \\minus{}1,0,2,3$. \n\nTherefore, integer solutions $ (x,y,z)$ can only occur when $ a \\in \\{\\minus{}1,0,1,2,3\\}$. \n\n$ a \\equal{} \\minus{}1$ yields: \n$ \\begin{pmatrix} 5 & 1 & 1 \\\\ 2 & 4 & 0 \\\\ 2 & 0 & 4 \\end{pmatrix}\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} \\equal{} \\begin{pmatrix} \\minus{}1 \\\\ \\minus{}4 \\\\ 0 \\end{pmatrix}$ $ \\Longrightarrow (x,y,z) \\equal{} (0,\\minus{}1,0)$ \n\n$ a \\equal{} 0$ yields: \n$ \\begin{pmatrix} 5 & 2 & 2 \\\\ 4 & 3 & 2 \\\\ 4 & 2 & 3 \\end{pmatrix}\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} \\equal{} \\begin{pmatrix} 0 \\\\ \\minus{}1 \\\\ 1 \\end{pmatrix}$ $ \\Longrightarrow (x,y,z) \\equal{} (0,\\minus{}1,1)$ \n\n$ a \\equal{} 1$ yields: \n$ \\begin{pmatrix} 5 & 3 & 3 \\\\ 6 & 4 & 4 \\\\ 6 & 4 & 4 \\end{pmatrix}\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} \\equal{} \\begin{pmatrix} 1 \\\\ 2 \\\\ 2 \\end{pmatrix}$ $ \\Longrightarrow (x,y,z) \\equal{} (\\minus{}1,2\\minus{}t,t)$ (Which is an integer solution iff $ t \\in \\mathbb{Z}$\n\n$ a \\equal{} 2$ yields: \n$ \\begin{pmatrix} 5 & 4 & 4 \\\\ 8 & 7 & 6 \\\\ 8 & 6 & 7 \\end{pmatrix}\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} \\equal{} \\begin{pmatrix} 2 \\\\ 5 \\\\ 3 \\end{pmatrix}$ $ \\Longrightarrow (x,y,z) \\equal{} (\\minus{}6,5,3)$ \n\n$ a \\equal{} 3$ yields: \n$ \\begin{pmatrix} 5 & 5 & 5 \\\\ 10 & 12 & 8 \\\\ 10 & 8 & 12 \\end{pmatrix}\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} \\equal{} \\begin{pmatrix} 3 \\\\ 8 \\\\ 4 \\end{pmatrix}$ $ \\Longrightarrow (x,y,z) \\equal{} (\\minus{}\\frac{2}{5}\\minus{}2t,t\\plus{}1,t)$ (Which is not an integer solution for any $ t \\in \\mathbb{R}$.)\n\nTherefore, the solutions are: \n$ a\\equal{}\\minus{}1$: $ (x,y,z) \\equal{} (0,\\minus{}1,0)$ \n$ a\\equal{}0$: $ (x,y,z) \\equal{} (0,\\minus{}1,1)$ \n$ a\\equal{}1$: $ (x,y,z) \\equal{} (\\minus{}1,2\\minus{}t,t)$ for all $ t \\in \\mathbb{Z}$ \n$ a\\equal{}2$: $ (x,y,z) \\equal{} (\\minus{}6,5,3)$ [/hide]" } { "Tag": [ "ratio", "geometry", "AMC", "AMC 8" ], "Problem": "What is the ratio of the area of the shaded square to the area of\r\nthe large square? (The figure is drawn to scale.)\r\n[asy]/* AMC8 1998 #13P */\nsize(1inch,1inch);\npair r1c1=(0,0), r1c2=(10,0), r1c3=(20,0), r1c4=(30, 0), r1c5=(40, 0);\npair r2c1=(0,10), r2c2=(10,10), r2c3=(20,10), r2c4=(30, 10), r2c5=(40, 10);\npair r3c1=(0,20), r3c2=(10,20), r3c3=(20,20), r3c4=(30, 20), r3c5=(40, 20);\npair r4c1=(0,30), r4c2=(10,30), r4c3=(20,30), r4c4=(30, 30), r4c5=(40, 30);\npair r5c1=(0,40), r5c2=(10,40), r5c3=(20,40), r5c4=(30, 40), r5c5=(40, 40);\ndraw(r1c1--r5c1--r5c5--r1c5--r1c1--r5c5);\ndraw(r5c1--r3c3);\ndraw(r4c4--r2c4--r3c5);\nfill(r2c2--r3c3--r2c4--r1c3--cycle);[/asy]", "Solution_1": "Let the shaded square's side length be $ x$. By isosceles right triangles, the side length of the large square is $ 2\\sqrt{2}x$. Ratio of side length of small square to side length of large square is $ \\frac{1}{2\\sqrt{2}}$, so the ratio of their areas is the square of that, or $ boxed{\\frac{1}{8}}$." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "AIME", "USAJMO", "AMC 10", "AMC 12" ], "Problem": "This message is to announce changes in the structure and selection\nprocess for the 2010 USAMO.\n\nOn May 18, 2009, the MAA Executive COmmittee formally approved the attached proposal from the Committee on the American Mathematics Competitions.\n\nIn short, the changes are:\n\nThe current USA Mathematical Olympiad administered to about 500\nstudents in all grades will be split into two flights:\n\n1. The USA Mathematical Olympiad administered to about 270 students\nwho qualify from the AMC 12 with a top USAMO index based on their AMC\n12 score. (along with a few students who took only the AMC 10\nand scored 11 or more on the AIME.)\n \n2. The USA Junior Mathematical Olympiad administered to about 230\nstudents who qualify from the AMC 10 with a top USAMO index based on\ntheir AMC 10 score.\n\n3. The current \"floor score\" system will no longer be used.\n\n\nIn practical terms this means:\n1. For 11th and 12th graders, there is essentially no change.\n\n2. Students who want to qualify for the USAMO should take the AMC 12,\nsince the AMC 12 is the gateway to the USAMO. (The exception is\nstudents who take the AMC 10 only but score 11 or more on the AIME.\nEvidence from the last few years indicates that this is typically\nabout 5 students. These students later excelled at all levels of the\nAMC program, and so based on this evidence we have built in this\nexception.)\n\n3. Students in 10th grade and below who want a chance at the USAMO\nshould take the AMC 12, but may also want to hedge their chances by\ntaking the AMC 10. Students in 10th grade and below who take the AMC\n12 will have their AMC 12-based USAMO index considered without\nconsideration of age or grade or AIME score. Of course this means\nthey are considered with 11th and 12th graders and compete for the\napproximately 250-270 USAMO spots on AMC 12 index alone. Students in\n10th grade and below who qualify for the USAMO are eliminated from the\npool of AMC 10 takers competing for the 230 invitations to the USA Junior\nMath Olympiad.\n\n4. The gateway for the USA Junior Math Olympiad is the AMC 10. This\nautomatically limits participants in the Junior Math Olympiad to\ngrades 10 and below.\n\n5. The USA Junior Mathematical Olympiad would have 6 problems and be\nadministered over 2 days, the same as the USAMO. Problems J1,J2 on\nDay 1, and Problems J4, J5 would be different from the USAMO, and\nwould be close in level and content to problems 13-15 on the AIME.\nProblem J3 would be the same as Problem 1 (Day 1) on the USAMO,\nProblem J6 would be the same as Problem 4 (Day 2) on the USAMO. All\nsix problems on the USA Junior Mathematical Olympiad would require\nwritten answers, perhaps a detailed algebraic, number theoretic,\ncombinatoric or geometric solution rather than a proof. Problems\nwould be graded on the same 7 point scale (42 points total) as the\nUSAMO, and have the same rigorous grading.\n\n6. Selection to \"Red MOSP\" will work in the following way (assuming\nfunding for Red MOSP continues.) Among all 9th graders taking the\nUSAMO and JMO, the scores on problems J3 = 1 and J6 = 4 will be\ntotalled and ordered. Then we will select the students with the top\n25 scores on these two problems. Ties among the top scores in each\ncategory will be broken by using the scores from the other 4 problems\non each Olympiad.", "Solution_1": "So if someone hypothetically qualified for both JMO and USAMO, can they choose which one to take or is it left to the guidelines?", "Solution_2": "After discussing that issue, we decided that: No, if someone qualifies for both USAMO and USAJMO, they will be invited to the USAMO only. See the last sentence of point 3:\r\n[quote]Students in\n10th grade and below who qualify for the USAMO are eliminated from the\npool of AMC 10 takers competing for the 230 invitations to the USA Junior\nMath Olympiad. [/quote]\r\nWe want students working at the best of their abilities.\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions", "Solution_3": "Out of curiosity (and to get a vague general idea), what would it have take on the AMC 10/12 and AIME to qualify for the USAJMO and USAMO respectively this year (had this new system been in place this year)?\r\n\r\nEDIT: Also, a question.\r\n\r\nClearly, if 1/4 are too hard on USAMO, they will be on USAJMO too. If students can solve problems that are of the approximate difficulty of 13-15 on the AIME, don't you think they will be able to score an 11 or higher? That is, won't the problems on the USAJMO be either too easy or too hard to everyone who qualifies?", "Solution_4": "A quick skim of the data suggests that it would have taken an index of approximately 210 on both the \r\nAMC10+10*AIME -> USAJMO and AMC12+10*AIME -> USAMO\r\nthis 2009 contest year.\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions", "Solution_5": "So a tenth grader who wants to try to make USAMO and MOSP most take the AMC12?", "Solution_6": "[quote]Clearly, if 1/4 are too hard on USAMO, they will be on USAJMO too. If students can solve problems that are of the approximate difficulty of 13-15 on the AIME, don't you think they will be able to score an 11 or higher? That is, won't the problems on the USAJMO be either too easy or too hard to everyone who qualifies?[/quote]\r\n\r\nThe item difficulty on the both the AIME and the USAMO as well as analysis of scores by grade indicates that neither of those concerns is valid. Students solving some or all of problems 13-15 on the AIME should be scoring 11 or higher on the AIME and will be taking the USAMO, in any case. \r\n\r\nFinally, don't you trust the USAMO Committee, all of them highly qualified mathematicians, most of them MO veterans and IMO veterans, and with years of experience in creating problems, tests, and contests including the USAMO and AIME to carefully estimate and calibrate the difficulty of the contests? My intention is to indicate the relative difficulty level in relation to something most readers on this forum will recognize. I am sure that the USAJMO will be a satisfactory experience for those who qualify.\r\n\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions", "Solution_7": "Since I'm going to be a senior next year, the same number of spots are still here, so I guess I'm really not affected by this. Just gotta hit the books now and qualify :]", "Solution_8": "I think this is a very thoughtful and creative approach. \r\n\r\nThe USAMO was originally designed for roughly 100+ students, of whom 8 would be designated \"winners\" and another 20 or so of the most promising students in 11th grade and below would be invited to the Math Olympiad Summer Program. Moreover, all qualifiers were selected from a single qualifying path (originally just the AHSME, later just a single version of the AHSME and a single version of the AIME.)\r\n\r\nExpecting a single test with six questions to be appropriate for 500 students, some as young as 4th grade (!), selected by a combinatorial explosion of paths (10A/12A/10B/12B/USAMTS with multiplicity possible and AIME I or AIME II) is really a lot to ask! The goal of identifying this year's team and very strong prospects for teams in the next two years (i.e., black/blue MOP) require different questions than the goals of encouraging and developing the talents of younger students with much less experience.\r\n\r\nThere are no perfect solutions, but I like this idea a lot, in principle. The junior math olympiad strikes me as having great possibilities as a \"stepping stone\" for younger students with less experience.", "Solution_9": "Stepping stones yes, but if students are able to do that well that young, they shouldn't be worried about having \"too hard\" of a USAMO. The USAMO is the most prominent competition in America (for high schoolers) for a reason, by adding a newer, easier level, I am of the opinion we give some of these younger kids a disservice. Rather than push them to work harder and master more difficult problems, we are giving them easier, AIME level problems, which they could get from elsewhere (USAMTS perhaps?)\r\n\r\nAlso, now qualification for red MOP has been reduced from 6 problems to 2. Is that a good thing?", "Solution_10": "Also, is there still the state best rule? That is, the highest index in the state? So now will there be [i]two[/i] of those? That is, will there be a qualifier for both USAJMO and USAMO solely based on the fact they live in a less competitive state?", "Solution_11": "How will CGMO selection work? Will it be done similarly to the Red cutoff? It doesn't affect me personally, but it hasn't been addressed yet.", "Solution_12": "[quote=\"MellowMelon\"][quote=\"CircleSquared\"]Could you please enlighten me about these \"qualification problems\"? The way I see it, narrowing down the selection criteria from 6 problems to 2 makes the system more based on luck and unstable, and would result in Red MOP qualifiers who potentially are farther from the top 25 9th graders than the qualifiers by the old system.[/quote]\nQualification problems means the issue of taking the AMC10 being beneficial to qualifying for USAMO, which is the topic that has been hotly debated on this forum.\n\n[/quote]\r\n\r\nYes, we have solved the problem of taking the 10 or not if you want to take the Junior, but now 10th graders who want to make MOP, the real goal of the whole process, now must take the AMC 12.", "Solution_13": "this basically makes the amc10 useless", "Solution_14": "I am hijacking D2 (daughter #2)'s account to post here.\r\n\r\n[quote]but now 10th graders who want to make MOP, the real goal of the whole process, now must take the AMC 12.[/quote]\r\n\r\nThat is not true. A 10th grader can still take AMC10 but score 11 and above on AIME will still qualify for USAMO and have a short at MOP.\r\n\r\nNo system is absolutely right for every occasion. If an underclassman wants to qualify for USAMO (which definitely sounds better than qualifying for USAJMO) should choose to take AMC12. \r\n\r\nLike bookaholic, D1 (just finish her freshman year in college), always did AMC12 in high school. She did AMC10 incidentally in 8th grade (first time knew there was such thing like AMC) and never took AMC10 again. D2 is about finishing her 7th grade and took AMC10 this year. This is the first time that her middle school ever participated in AMC10 (at my suggestion and support). I will suggest to her middle school to do both AMC10 and AMC12 next year.\r\n\r\nOne solution is: students should push their schools to participate in both AMC/A and AMC/B, then an underclassman would have opportunity to do both 10 and 12 and have chance to qualify USAMO (if made the index for USAMO) or USAJMO (if made the index for USAJMO).", "Solution_15": "[quote=\"CatalystOfNostalgia\"][quote=\"AMCDirector\"](assuming funding for Red MOSP continues.)[/quote]\n\nOn a tangent, this is pretty ominous. Is this likely not to happen?[/quote]\r\nThis was probably more of a reminder that the funding is a source outside of the MAA, so it's not something they can guarantee. They probably don't know or have an idea if the money is going to get pulled from them or not, but it's probably safe to count on it continuing.", "Solution_16": "As of this writing, there are now 50 posts in this thread filled with conjecture, speculation, questions, opinions and counter-opinions. This is too many, too diverse, and too cacophonous to respond to intelligently in this thread.\r\n\r\nIf you have a question about either a fact or a procedure related to the 2010 USAMO or the 2010 USA Junior Math Olympiad, please begin a new thread and ask your question in that new thread. (One question per thread for simplicity and clarity, please). Either I or AMCFinance (E. Claassen) will answer your question in that thread for all to see. Thank you.\r\n\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions", "Solution_17": "I'm about to do some splitting on this thread.", "Solution_18": "[quote=\"Kent Merryfield\"][quote=\"mathemonster\"]this basically makes the amc10 useless[/quote]\nI can't let this comment pass unchallenged.[/quote]\r\n\r\nLol. Troll fed.\r\n\r\nThis is a very interesting decision. I do not like the possible departure from proof though. I think that all the problems should be proofs, and if you want an easier contest, you should just ask easier proof questions.", "Solution_19": "^ They are proof question, but at 13-15 AIME [i]level[/i] Is this definite?[/i]", "Solution_20": "Actually,\r\n[quote=\"AMCDirector\"]All\nsix problems on the USA Junior Mathematical Olympiad would require\nwritten answers, perhaps a detailed algebraic, number theoretic,\ncombinatoric or geometric solution rather than a proof.\n[/quote]", "Solution_21": "1) Can 10th graders qualify for blue MOP through USAJMO?\r\n2) If so, how will these 10th graders be compared to those (10th graders) who took the USAMO in terms of MOP qualification?", "Solution_22": "Questions about the USAJMO question format have been addressed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=280817]here[/url] by AMC officials.\r\n\r\nmathking: I have not seen this answered for sure. But if USAJMO contestants could qualify for blue MOP the whole idea would be laughably stupid. Based on what the USAJMO difficulty sounds like, most blue MOP qualifiers should be able to get a perfect on it, and people not at blue level would probably be able to as well.", "Solution_23": "[quote]1) Can 10th graders qualify for blue MOP through USAJMO? [/quote]\n\nNo.\n\n[quote] If so, how will these 10th graders be compared to those (10th graders) who took the USAMO in terms of MOP qualification?[/quote]\r\n\r\nBy the answer to 1), not applicable.\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions", "Solution_24": "As someone who essentially qualified for the USAMO in 9th grade [i]because[/i] of the expansion and who met the blue mop cutoff (although was placed in red mop, which I feel was better for me), I am somewhat biased against this change. Even as a ninth grader, I felt like the AMC10 was uninteresting, having gotten a 146.5 on it the year before (that was before the 2.5->1.5 score change). Therefore, I only took the AMC12 my freshman year, even though I realized that it may hurt my chances of making USAMO. I ended up with a 144/6, for a 204 index. The cutoff that year was 197.5 (If I remember correctly), but if the expansion were not in place, I feel like I would not have qualified. What worries me is that I took the AMC12 because I find easy problems uninteresting, not because it would help qualifying for any future exam, so I am not sure that I would have taken the AMC10 at all even with this new rule. I would strongly object to having no way to qualify for USAJMO having only taken the AMC12. Actually, I am worried that you might accidentally miss some 10th graders who are worthy of blue mop (as would have been the case if I had been a year older), whereas with the expansion this is pretty much impossible.\r\n\r\nI do realize that I am by far the exception, and you won't miss more than one or two such people, but I definitely feel that it's doing a disservice to that minority of underclassmen who would want to take the AMC12 rather than the AMC10 for the interesting problems and are worthy of making USAJMO but not necessarily USAMO, or those ninth graders who are ready for red mop but miss the USAMO cutoff and didn't try for the USAJMO cutoff.", "Solution_25": "I took 2 AMC12s both 9th and 10th grade for the same reason. I didn't qualify in 9th though I got an 8 on the AIME.\r\n\r\nI think it should be at least part of the objective of the selection processes to have people taking the most challenging tests they will do reasonably well on. I see no reason for someone who will get a 0 on the USAMO to take it instead of an easier test, but that isn't true on AMC10/12 for anyone with a remote chance of making either Olympiad. We can all do the AMC12, and the selection process shouldn't make us waste our time on the AMC10.\r\n\r\n(For the record, I'll be an 11th grader next year and be mostly unaffected by the exact process.)", "Solution_26": "There is no possible solution that will leave everybody happy. The AMC committee or whatever attempts to find the solution that satisfies the most people. This is what they came up with. Obviously, there will always be some people who don't like their decisions. Personally, I think it is a good move. I will be in 7th grade and will take the AMC 12 next year based on this change. This is what the AMC committee is looking for. They are encouraging taking the AMC 12.", "Solution_27": "Does this mean that AMC 12 is gonna get much harder?(Since we will only see AMC 12 index)", "Solution_28": "Because 10th graders in USAJMO can't qualify for blue MOP, doesn't that mean that USAJMO is essentially a dead end for 10th graders? :|", "Solution_29": "[quote=\"AMCDirector\"]As of this writing, there are now 50 posts in this thread filled with conjecture, speculation, questions, opinions and counter-opinions. This is too many, too diverse, and too cacophonous to respond to intelligently in this thread.\n\nIf you have a question about either a fact or a procedure related to the 2010 USAMO or the 2010 USA Junior Math Olympiad, please begin a new thread and ask your question in that new thread. (One question per thread for simplicity and clarity, please). Either I or AMCFinance (E. Claassen) will answer your question in that thread for all to see. Thank you.\n\n\nSteve Dunbar\nMAA Director, American Mathematics Competitions[/quote]\r\n\r\nAfter Mr. Dunbar made this comment, there have been entirely one another page about it. I think that the discussion will continue, so I'm going to lock the thread and ask everyone to ask questions separately. This is, after all, an announcement thread." } { "Tag": [ "function", "complex analysis", "calculus", "calculus computations" ], "Problem": "How to prove that Taylor's expansion is valid to expand a function?\r\n\r\nThank you for helping. :P", "Solution_1": "You can prove for example that it's Lagrange rest tends to zero.", "Solution_2": "Here is a less elementary way.\r\n\r\nSay you have $ f(x)$ a real function on a set $ (-R,R)$ for $ R>0$.\r\n\r\nIf you can find a holomorphic complex function $ g(z)$ in the open disk $ \\{|z|0$.\n\nIf you can find a holomorphic complex function $ g(z)$ in the open disk $ \\{|z| 3\\sqrt{2}$.\r\nNow here's the rigorous explanation:\r\n[hide=\"My approach\"]As Kouichi Nakagawa noticed, we may generalize this to:\n$ \\sum \\frac{1}{\\sqrt{1 \\plus{} x_k}} \\le \\frac{n\\sqrt{2}}{2}$,\nfor any positive $ x_k$ with product $ 1$ and $ n \\ge 2$ a positive integer.\nConsider the function $ f : \\mathbb{R} \\rightarrow \\mathbb{R}_\\plus{}^*$, given by the law\n$ f(t) \\equal{} \\frac{1}{\\sqrt{1 \\plus{} e^t}}$.\nNow we notice that $ f$ is concave on $ (\\minus{}\\infty, \\mathrm{ln} 2]$ and convex on $ [\\mathrm{ln} 2, \\infty)$. Now let $ a_i \\equal{} \\mathrm{ln}(x_i)$. The sum of all $ a_k$'s is constant (equal to $ 0$). However, by LCRCF theorem we know that the maximum of the left-hand side, which is $ \\sum f(a_k)$, is attained for $ n \\minus{} 1$ equal $ a_k$'s, let them be $ a_1 \\equal{} ... \\equal{} a_{n\\minus{}1} \\equal{} a \\le b \\equal{} a_n$, with $ (n\\minus{}1)a \\plus{} b \\equal{} 0$.\nNow consider $ x \\equal{} e^a$ and $ y \\equal{} e^b \\equal{} \\frac{1}{x^{n\\minus{}1}}$, with $ 0 < x \\le 1 \\le y$. It remains to be shown\n$ \\frac{n \\minus{} 1}{\\sqrt{1 \\plus{} x}} \\plus{} \\sqrt\\frac{x^{n\\minus{}1}}{1 \\plus{} x^{n\\minus{}1}} \\le \\frac{n\\sqrt{2}}{2}$. However this is false because the left-hand side is greater than $ \\frac{n \\minus{} 1}{\\sqrt{1 \\plus{} x}} > \\frac{n}{\\sqrt{2}}$, which for $ n \\equal{} 6$ (the topic problem's case) can be checked as true for $ x < \\frac{7}{18}$.\nConsequently the problem is not correct.\nHowever, there is a related result, namely Vasc's well-known\n$ \\sum \\sqrt{\\frac{2a}{a \\plus{} b}} \\le 3$.[/hide]", "Solution_3": "Thanks for your help :) Here is wrong solution, but I can not find out my mistake. Hope someone will help me :wink: \r\nBy substitution $ b/a\\equal{}x$, $ c/b\\equal{}y$, $ d/c\\equal{}z$, $ e/d\\equal{}t$, $ f/e\\equal{}u$, $ a/f\\equal{}v$, we have $ xyztuv\\equal{}1$ and the inequality takes the form\r\n\\[ \\frac1{\\sqrt{1\\plus{}x}}\\plus{}\\frac1{\\sqrt{1\\plus{}y}}\\plus{}\\frac1{\\sqrt{1\\plus{}z}}\\plus{}\\frac1{\\sqrt{1\\plus{}t}}\\plus{}\\frac1{\\sqrt{1\\plus{}u}}\\plus{}\\frac1{\\sqrt{1\\plus{}v}}\\leq3{\\sqrt2}.\\]\r\nConsider the function \r\n\\[ f(w)\\equal{}\\frac{\\sqrt2}2\\minus{}\\frac1{\\sqrt{1\\plus{}w}}\\minus{}\\frac{\\sqrt2}8\\ln w.\\]\r\nCalculating the derivative of the function, we obtain\r\n\\[ f'(w)\\equal{}\\frac{4w\\minus{}\\sqrt2(w\\plus{}1)^{3/2}}\t{8w(w\\plus{}1)^{3/2}}.\\]\r\nThe equation $ f'(w)\\equal{}0$ has two roots $ w\\equal{}1$, and $ w\\equal{}2\\plus{}\\sqrt5$. Then we have\r\n\\[ f(w)\\geq \\min\\left[f(1), f\\left(\\frac{96}{100}\\right)\\right]\\equal{}0.\\]\r\nIt follows that $ f(w)\\geq 0$ for all $ 0\\leq w\\leq \\frac{96}{100}$. This implies that the desired inequality is true for all $ 0\\leq x,y,z,t,u,v\\leq\\frac{96}{100}$. Otherwise, for $ x,y,z,t,u,v\\geq\\frac{96}{100}$, we have\r\n\\[ f(x)\\plus{}f(y)\\plus{}f(z)\\plus{}f(t)\\plus{}f(u)\\plus{}f(v)\\leq \\frac6{\\sqrt{1\\plus{}\\frac{96}{100}}}\\leq 3\\sqrt2.\\]\r\nThe proof is complete. But I am really strange because of your comment... Where is the mistake in my proof :)", "Solution_4": "[quote=\"zaizai-hoang\"][...] This implies that the desired inequality is true for all $ 0\\leq x,y,z,t,u,v\\leq\\frac {96}{100}$. Otherwise, for $ x,y,z,t,u,v\\geq\\frac {96}{100}$ [...] [/quote]\r\nAnd then, what happens when not all the variables are on one side of $ \\frac{96}{100}$ ?" } { "Tag": [ "geometry", "perimeter" ], "Problem": "An artist has 50 inches of oak trim to frame a painting. The frame is to have a border 3 inches wide surrounding the painting.\r\n\r\nIf the painting is rectangular with a length twice its width, what are the dimensions of the painting? What are the dimensions of the frame?", "Solution_1": "Can you please stop creating so many different topics?\r\n\r\nIf you have several questions, compile them into a list and make one big topic with 5-10 questions.\r\nThat would be much better than having the entire topic list be populated by your questions\r\n\r\nthank you\r\n\r\n[hide]\nthe dimensions are\nW x 2W\n\nthe perimeter is 6W\nbecause the border is 3 inches wide, we have to add 3 inches for each corner, or 12 inches\n\n6W + 12 = 50\n6W = 38\nW = 6.33333\n[/hide]", "Solution_2": "[quote=\"chickendude\"]Can you please stop creating so many different topics?\n\nIf you have several questions, compile them into a list and make one big topic with 5-10 questions.\nThat would be much better than having the entire topic list be populated by your questions\n\nthank you\n\n[hide]\nthe dimensions are\nW x 2W\n\nthe perimeter is 6W\nbecause the border is 3 inches wide, we have to add 3 inches for each corner, or 12 inches\n\n6W + 12 = 50\n6W = 38\nW = 6.33333\n[/hide][/quote]\r\n\r\nWait, wouldn't it be 6x+24 because of the 3 inches on each side? so wouldnt x equal 4 and 1/3?\r\nor did i screw up", "Solution_3": "[quote=\"Concentric\"][quote=\"chickendude\"]Can you please stop creating so many different topics?\n\nIf you have several questions, compile them into a list and make one big topic with 5-10 questions.\nThat would be much better than having the entire topic list be populated by your questions\n\nthank you\n\n[hide]\nthe dimensions are\nW x 2W\n\nthe perimeter is 6W\nbecause the border is 3 inches wide, we have to add 3 inches for each corner, or 12 inches\n\n6W + 12 = 50\n6W = 38\nW = 6.33333\n[/hide][/quote]\n\nWait, wouldn't it be 6x+24 because of the 3 inches on each side? so wouldnt x equal 4 and 1/3?\nor did i screw up[/quote]\r\nI think that you can just put one 2x+6 in. piece of wood on the top and bottom so the two sides would only need x in. wood each (the width of 3 in. on top and bottom takes off 6 in. from left and right sides). 2*(2x+6)+2*x=6x+12" } { "Tag": [ "modular arithmetic", "number theory proposed", "number theory" ], "Problem": "[color=Red]Prove that for no $ a,b,c \\in \\mathbb{N}$ \n\n$ \\frac{a^2\\plus{}b^2\\plus{}c^2}{3(ab\\plus{}bc\\plus{}ca)}$\n\nis an integer.\n(sorry if it has been posted before :oops:)[/color]", "Solution_1": "it is a nice application of jacobi symbol.I will post my solution later because there are some others who are still trying.. :wink:", "Solution_2": "For $ a\\in\\mathbb{N}^*$ and $ p$ is an odd prime, let $ \\left( \\frac{a}{p} \\right)$ denote the Legendre's symbol.\r\n\r\nAssume the contradiction that there exists $ k\\in\\mathbb{Z}$ such that $ \\frac{a^2 \\plus{} b^2 \\plus{} c^2}{3(ab\\plus{}bc\\plus{}ca)} \\equal{} k$ then $ (a\\plus{}b\\plus{}c)^2 \\equal{} (3k\\plus{}2)(ab\\plus{}bc\\plus{}ca)$. Let $ p$ be a prime such that $ p \\equiv 2 \\pmod{3}$ and $ p^{2s\\plus{}1} \\parallel 3k\\plus{}2$, we have $ p^{2s\\plus{}1} | (a\\plus{}b\\plus{}c)^2$ then $ p^{2s\\plus{}2} | (a\\plus{}b\\plus{}c)^2$ and hence $ p | a\\plus{}b\\plus{}c$ and $ p | ab\\plus{}bc\\plus{}ca$. Therefore, $ ab \\plus{} c(a\\plus{}b) \\equiv ab \\minus{} (a\\plus{}b)^2 \\equiv 0 \\pmod{p}$, meanwhile $ p | a^2 \\plus{} ab \\plus{} b^2$. From this we can deduce that $ p | (2a\\plus{}b)^2 \\plus{} 3b^2$ and so $ \\left( \\frac{\\minus{}3}{p} \\right)\\equal{} 1$. Moreover, since $ p \\equiv 2 \\pmod{3}$ then $ \\left( \\frac{p}{3} \\right) \\equal{} \\minus{}1$ and the Gauss's laws yields $ \\left( \\frac{\\minus{}1}{p} \\right) \\cdot \\left( \\frac{3}{p} \\right) \\cdot \\left( \\frac{p}{3} \\right) \\equal{} \\left( \\frac{p}{3} \\right) \\equal{} \\minus{}1$ or $ (\\minus{}1)^{\\frac{p\\minus{}1}{2}\\plus{}\\frac{p\\minus{}1}{2}\\cdot\\frac{3\\minus{}1}{2}} \\equal{} (\\minus{}1)^{p\\minus{}1} \\equal{} \\minus{}1$, contradiction. QED.", "Solution_3": "There is a small mistake in Mashimaru's solution . I really want to complete that nice one :lol: \r\n\r\n Frist , You should assume WLOG that $ (a ; b ;c ) \\equal{} 1 \\ \\ (*)$\r\n\r\n For $ a\\in\\mathbb{N}^*$ and $ p$ is an odd prime, let $ \\left( \\frac {a}{p} \\right)$ denote the Legendre's symbol.\r\n\r\nAssume the contradiction that there exists $ k\\in\\mathbb{Z}^{ \\plus{} }$ such that $ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{3(ab \\plus{} bc \\plus{} ca)} \\equal{} k$ then $ (a \\plus{} b \\plus{} c)^2 \\equal{} (3k \\plus{} 2)(ab \\plus{} bc \\plus{} ca)$. Let $ p$ be a prime such that $ p \\equiv 2 \\pmod{3}$ and $ p^{2s \\plus{} 1} \\parallel 3k \\plus{} 2$, we have $ p^{2s \\plus{} 1} | (a \\plus{} b \\plus{} c)^2$ then $ p^{2s \\plus{} 2} | (a \\plus{} b \\plus{} c)^2$ and hence $ p | a \\plus{} b \\plus{} c$ and $ p | ab \\plus{} bc \\plus{} ca$. \r\n \r\nWe can prove that $ p \\geq 5$ , Indeed , if $ p \\equal{} 2$ . Due to the fact that $ p | a \\plus{} b \\plus{} c$ , we can deduce that in three numbers $ a ; b ; c$ , Exist at least one even number , assume $ 2 |b$ then $ a;c$ have the same parity , but $ 2 | ac$ so $ a ; c$ are also even integer \r\n $ 2 | gcd(a;b;c )$ , Contradiction with $ (*)$ \r\n\r\nTherefore, $ ab \\plus{} c(a \\plus{} b) \\equiv ab \\minus{} (a \\plus{} b)^2 \\equiv 0 \\pmod{p}$, meanwhile $ p | a^2 \\plus{} ab \\plus{} b^2$. From this we can deduce that $ p | (2a \\plus{} b)^2 \\plus{} 3b^2 \\rightarrow \\left( \\frac { \\minus{} 3}{p} \\right) \\ \\equal{} \\ 1$ .\r\n\r\n (We have $ p | (2a \\plus{} b)^2 \\plus{} 3b^2$ .Then if $ p|b \\rightarrow p| (2a \\plus{} b) \\rightarrow p|2a \\rightarrow p|a$ , $ p|b$\r\n\r\n But $ p | a \\plus{} b \\plus{} c \\rightarrow p|c$ , Contracdition with $ (*)$\r\n\r\n So , we must have $ (b ; p) \\equal{} 1$ . On the other hand , $ \\minus{} 3b^2 \\equiv (2a \\plus{} b)^2 \\ \\ (mod \\ \\ p)$ \r\n $ \\minus{} 3 (b .b^{ \\minus{} 1} )^2 \\equiv (b^{ \\minus{} 1}(2a \\plus{} b))^2 \\equiv \\minus{} 3 \\ \\ (mod \\ \\ p) \\rightarrow \\left( \\frac { \\minus{} 3}{p} \\right) \\ \\equal{} \\ 1$ )\r\n\r\n. Moreover, since $ p \\equiv 2 \\pmod{3}$ then $ \\left( \\frac {p}{3} \\right) \\equal{} \\minus{} 1$ and the Gauss's laws yields \r\n\r\n$ \\left( \\frac { \\minus{} 3}{p} \\right) \\equal{} \\left( \\frac { \\minus{} 1}{p} \\right) \\cdot \\left( \\frac {3}{p} \\right)$\r\n\r\n$ \\equal{} \\left( \\frac { \\minus{} 1}{p} \\right) \\cdot \\left( \\frac {p}{3} \\right) ( \\minus{} 1)^{\\frac {p \\minus{} 1}{2} \\cdot \\frac {3 \\minus{} 1}{2} } \\equal{} \\minus{} ( \\minus{} 1)^{\\frac {p \\minus{} 1}{2} \\plus{} \\frac {p \\minus{} 1}{2} \\cdot \\frac {3 \\minus{} 1}{2} } \\ \\equal{} \\ \\minus{} ( \\minus{} 1)^{p \\minus{} 1} \\equal{} \\minus{} 1$, Contradiction", "Solution_4": "Foolish posts!\r\nMashimaru (pham hy hieu) and nguyenvuthanhha shouldn't repeat the solution in Dusan Djukic's article, rigidly. This problem is only an application of the Fermat little theorem. :wink:", "Solution_5": "[quote=\"N.N.Trung\"]This problem is only an application of the Fermat little theorem. :wink:[/quote]can you post your solution?", "Solution_6": "You can see the solution [hide=\"in this file\"]http://www.imomath.com/tekstkut/quadcong_ddj.pdf[/hide]\nThe problem appeared in [hide=\"Mathlinks contest edition 7_round 3\"]http://www.mathlinks.ro/contest.php?edition=7&round=3[/hide]", "Solution_7": "[quote=\"Mashimaru\"]For $ a\\in\\mathbb{N}^*$ and $ p$ is an odd prime, let $ \\left( \\frac {a}{p} \\right)$ denote the Legendre's symbol.\n\nAssume the contradiction that there exists $ k\\in\\mathbb{Z}$ such that $ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{3(ab \\plus{} bc \\plus{} ca)} \\equal{} k$ then $ (a \\plus{} b \\plus{} c)^2 \\equal{} (3k \\plus{} 2)(ab \\plus{} bc \\plus{} ca)$. Let $ p$ be a prime such that $ p \\equiv 2 \\pmod{3}$ and $ p^{2s \\plus{} 1} \\parallel 3k \\plus{} 2$, we have $ p^{2s \\plus{} 1} | (a \\plus{} b \\plus{} c)^2$ then $ p^{2s \\plus{} 2} | (a \\plus{} b \\plus{} c)^2$ and hence $ p | a \\plus{} b \\plus{} c$ and $ p | ab \\plus{} bc \\plus{} ca$. Therefore, $ ab \\plus{} c(a \\plus{} b) \\equiv ab \\minus{} (a \\plus{} b)^2 \\equiv 0 \\pmod{p}$, meanwhile $ p | a^2 \\plus{} ab \\plus{} b^2$. From this we can deduce that $ p | (2a \\plus{} b)^2 \\plus{} 3b^2$ and so $ \\left( \\frac { \\minus{} 3}{p} \\right) \\equal{} 1$. Moreover, since $ p \\equiv 2 \\pmod{3}$ then $ \\left( \\frac {p}{3} \\right) \\equal{} \\minus{} 1$ and the Gauss's laws yields $ \\left( \\frac { \\minus{} 1}{p} \\right) \\cdot \\left( \\frac {3}{p} \\right) \\cdot \\left( \\frac {p}{3} \\right) \\equal{} \\left( \\frac {p}{3} \\right) \\equal{} \\minus{} 1$ or $ ( \\minus{} 1)^{\\frac {p \\minus{} 1}{2} \\plus{} \\frac {p \\minus{} 1}{2}\\cdot\\frac {3 \\minus{} 1}{2}} \\equal{} ( \\minus{} 1)^{p \\minus{} 1} \\equal{} \\minus{} 1$, contradiction. QED.[/quote]\r\nMy solution only have a bit of difference from Mashimaru's :P :\r\n$ p | a^2 \\plus{} ab \\plus{} b^2|a^3 \\minus{} b^3 \\Rightarrow a^3\\equiv b^3 \\pmod p$.\r\nLet $ p \\equal{} 3k \\plus{} 2$. The Little Fermath theorem shows that $ a^{3k \\plus{} 1}\\equiv b^{3k \\plus{} 1}\\pmod p$.\r\nSo, $ a \\equiv b \\pmod p \\Rightarrow p|(3a^2) \\Rightarrow p|a \\Rightarrow p|b, p|c$ (contradiction).\r\nNo Legendre's symbol in solution! :roll:" } { "Tag": [ "real analysis", "inequalities", "triangle inequality", "real analysis unsolved" ], "Problem": "Studying for my analysis qual, came across this problem:\r\n\r\n$ \\{E_n\\}$ Lebesgue measurable subsets of $ [0,1]$, and $ \\sum_{n\\equal{}1}^\\infty m(E_n) < \\infty$. Define $ f \\equal{} \\sum_{n\\equal{}1}^\\infty \\frac{\\chi_{E_n}}{n}$ and show that $ f \\in L^p[0,1]$ for $ 1 \\leq p < \\infty$.\r\n\r\nI got the case $ p\\equal{}1$, which was rather trivial. However when you stick the $ p$ in there, it messes everything up. It seems like you need to say something about the convergence rate of the measures of E_n and compare this to the convergence rate of $ \\frac{1}{n}$ somehow, but I have no idea.", "Solution_1": "wrong and deleted", "Solution_2": "It looks like you're using Holder's to say:\r\n\r\n$ \\sum_{n \\equal{} 1}^m \\left(\\frac {1}{n}\\right)\\left(\\chi_{E_n}(x)\\right) \\leq \\left(\\sum_{n\\geq 1}^{m}\\frac {1}{n^{p}}\\right)^\\frac {1}{p}\\left(\\sum_{n\\geq 1}^{m}[\\chi_{E_{n}}(x)]^{p}\\right)^\\frac {1}{p}$\r\n\r\nIs that correct? But you can only use Holder's when $ \\frac {1}{p} \\plus{} \\frac {1}{q} \\equal{} 1$.\r\n\r\nEdit: I suppose you could just have it\r\n\r\n$ \\left(\\sum_{n\\equal{}1}^m\\frac {1}{n^{q}}\\right)^\\frac {p}{q}$\r\n\r\nAnd it seems like the rest should work... maybe?", "Solution_3": "It's a series problem in disguise. Apply the triangle inequality; if $ f_n$ is the $ n$th term of that series, we want to show that $ \\sum_n \\|f_n\\|_p<\\infty$. What is $ \\|f_n\\|_p$? It's $ \\frac1n(m(E_n))^{1/p}$; let $ a_n\\equal{}m(E_n)$.\r\nNow, Holder's inequality: $ \\sum_n \\frac1n\\cdot a_n^{1/p}\\le \\|\\frac1n\\|_q\\cdot \\|a_n^{1/p}\\|_p$. The latter term, which is $ \\left(\\sum a_n\\right)^{1/p}$, is always finite for $ 1\\le p<\\infty$. The former is finite for $ 1 \\epsilon\\}$, where\r\n\\[\\Psi_\\epsilon (z) : = \\frac{1}{\\epsilon^{2}}\\Psi (\\frac{z}{\\epsilon}) \\]\r\nfor $z\\in \\mathbb{C}, \\epsilon > 0$ and $\\Psi \\in C^\\infty_{0}(\\mathbb{C}, \\mathbb{R}_{+})$ is such that $supp \\Psi = \\overline{E}$ ($E = \\{z\\in \\mathbb{C}: |z|<1\\}$), $\\Psi(z) = \\Psi(|z|), z \\in \\mathbb{C}$ and $\\int_\\mathbb{C}\\Psi dL^{2}= 1$; $d_\\Omega (z) : = dist (z, Bd \\Omega)$. The proof of \"$\\Leftarrow$\" consists of few steps; first it is shown, that $u_\\epsilon\\in SH(\\Omega_\\epsilon)$, then that the sequence $u_\\epsilon$ decreases when $\\epsilon \\to 0$. Finally it is introduced:\r\n\\[v : = \\lim_{\\epsilon \\to 0}u_\\epsilon \\]\r\nwhich is subharmonic, of course, on $\\Omega$. And there's the moment when the author writes: \"On the other hand, as you know, $u_\\epsilon \\to u$ $L^{2}$-almost everywhere in $\\Omega$, whence $u=v$ $L^{2}$-al. e. in $\\Omega$.\" These words end the proof. And here's my question: what is it: \"as you konw\"?? I don't know, why $u_\\epsilon \\to u$ $L^{2}$-al. e. in $\\Omega$. Please, help me to understand, if you can.\r\n\r\nThanks for your help. ;)", "Solution_1": "It is because almost every point is a [url=http://en.wikipedia.org/wiki/Lebesgue_point]Lebesgue point[/url] of $u$ (this is a non-trivial theorem) and the convergence takes place at every Lebesgue point of $u$ (this you, probably, should be able to prove yourself).", "Solution_2": "fedja, THANK YOU :lol: This (theorem) was the one, that made me so tired! Unfortunately, I'm not too good at measure theory (as can be seen) and your advice seems to be some basic in this theory. Am I right? I must say, that while trying to find an argument for the \"phrase\" in the book, I thought it would be fine if such inequality hold, but I wasn't able to explain why it had to work. Now I konw it's non-trivial. Of course the rest if obvious.\r\n\r\nGreetings! ;)" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "For $a,b,c $ sides of a triangle such that \r\n\r\n$2( \\frac{1}{a} +\\frac{1}{b} + \\frac{1}{c} ) = 4 + \\frac{a}{bc} + \\frac{b}{ca} + \\frac{c}{ab} $ \r\n\r\nprove the inequality\r\n\r\n\r\n$ 64(1-a)(1-b)(1-c) \\leq1 $", "Solution_1": "After some computations, the given condition becomes [tex] 4R+r=4Rp[/tex]. Homogenize the inequality as follows [tex]64\\prod [4Rp-a(4R+r)]\\leq (4Rp)^3 \\Leftrightarrow 64\\prod [4Rrp-ar(4R+r)]\\leq (4Rrp)^3 \\Leftrightarrow 64\\prod (bc-4Rr-r^2)\\leq a^2b^2c^2 \\Leftrightarrow 64\\prod (p^2-ab-ca)\\leq a^2b^2c^2 \\Leftrightarrow \\prod (-a+b+c)^2\\leq a^2b^2c^2 [/tex], which is well known, and the proof is complete." } { "Tag": [ "geometry", "3D geometry" ], "Problem": "How would you solve this problem?\r\n\r\n :arrow: A cube's faces are painted green, then divided into unit cubes. If there are 64 unit cubes that are not painted, how many unit cubes are there in total?\r\n\r\n\r\nI don't understand how to solve this. Please help. :maybe: \r\n\r\nThank you. :roll:", "Solution_1": "If 64 cubes are not painted, then on the inside there's a cube with side length $ \\sqrt[3]{64}\\equal{}4$. Thus, the number of unit cubes is $ (4\\plus{}2)^3\\equal{}216$." } { "Tag": [], "Problem": "Prove that there exists an irrational real number $ x > 0$ such that $ x^{\\sqrt{2}}$ is rational.", "Solution_1": "$ x\\equal{}\\sqrt2^{\\sqrt2}$", "Solution_2": "[hide=\"Explanation\"]We don't even have to bother trying to see if the number in the previous post is rational or not.\n\n[b][u]Case 1:[/u][/b] $ \\sqrt {2}^{\\sqrt {2}}$ [i]is rational [/i]\nThen, $ x \\equal{} \\sqrt {2}$ works.\n\n[b][u]Case 2:[/u][/b] $ \\sqrt {2}^{\\sqrt {2}}$[i] is irrational [/i]\nThen, $ x \\equal{} \\sqrt {2}^{\\sqrt {2}}$ works.[/hide]" } { "Tag": [], "Problem": "Here is another sum.I think I did it, so please check the answer:\r\n\r\n Find the value 0f $ x^3 \\minus{} 6x^2 \\plus{} 18x \\plus{} 11$ when $ x \\equal{} 2 \\minus{} 2^{\\frac {1}{3}} \\plus{} 2^{\\frac {2}{3}}$.My answer is 34.Is it correct?The book gives 33.\r\n\r\nThank you in advance.\r\n\r\nOh!I forgot to mention that I brought 2 to the left and cubed the expression.", "Solution_1": "[quote=\"muks\"]Here is another sum.I think I did it, so please check the answer:\n\n Find the value 0f $ x^3 \\minus{} 6x^2 \\plus{} 18x \\plus{} 11$ when $ x \\equal{} 2 \\minus{} 2^{\\frac {1}{3}} \\plus{} 2^{\\frac {2}{3}}$.My answer is 34.Is it correct?The book gives 33.\n\nThank you in advance.[/quote]\r\n\r\nYou must have made some mistakes on calculation.\r\n\r\n$ x^3 \\minus{} 6x^2 \\plus{} 18x \\plus{} 11\\equal{}(x\\minus{}2)^3\\plus{}6x\\plus{}19\\equal{}2(\\sqrt[3]{2}\\minus{}1)^3\\plus{}6x\\plus{}19$\r\n\r\n$ \\equal{}(2\\minus{}6\\sqrt[3]{4}\\plus{}6 \\sqrt[3]{2})\\plus{}6(2\\plus{}\\sqrt[3]{4}\\minus{}\\sqrt[3]{2})\\plus{}19\\equal{}33$", "Solution_2": "Thank you." } { "Tag": [], "Problem": "Let \\[x = .123456789101112\\ldots998999,\\] where the digits are obtained by writing the integers 1 through 999 in order. The 1983rd digit to the right of the decimal point is\n\n$ \\textbf{(A)}\\ 2\\qquad\\textbf{(B)}\\ 3\\qquad\\textbf{(C)}\\ 5\\qquad\\textbf{(D)}\\ 7\\qquad\\textbf{(E)}\\ 8 $", "Solution_1": "[hide]There are $9$ numbers with one digit. They produce 9 digits.\nThere are $90$ numbers with two digits. They produce 180 digits.\nThere are $900$ numbers with three digits. They produce 2700 digits.\n\nThat's probably too much, don't you think? :D We're looking for the 1983rd one.\n\nLet's say there are $x$ three digit numbers. Then\n\n$1983 = 9 + 180 + 3x\\\\ x=598$\n\nThe digit right before that is D: 7.[/hide]", "Solution_2": "[hide=\"Answer\"]We use 9 digits for the 1-digit numbers and 180 digits for the two-digit numbers, so we have $1983-189=1794$ digits left to use. $1794\\equiv 0\\bmod{3}$, so we are looking at the final digit of a number. $\\frac{1794}{3}=598\\Rightarrow 598+99=697$, so our answer is $7\\Rightarrow \\boxed{D}$.[/hide]" } { "Tag": [ "\\/closed" ], "Problem": "I can't jump to another forums... why?", "Solution_1": "[quote=\"Jos\u00e9\"]I can't jump to another forums... why?[/quote]\r\n\r\nWhat do you mean by \"jump\"? Can you not see any part of our forums other than this one? Or do you mean that where you see links, you cannot access other forums than this one? Or do you mean something else?", "Solution_2": "I'm having this problem too.\r\n\r\nNear the bottom right hand corner there is a place where it says \"Jump to\" and \"select a forum\" and it used to also have a list of all the forums under the \"select a forum\". It just says \"none\" now under \"select a forum\".", "Solution_3": "Clear the cookies (there's a clear cookies set by this board link just above the jumbpox in the forum view) and try again.", "Solution_4": "[quote=\"Valentin Vornicu\"]Clear the cookies (there's a clear cookies set by this board link just above the jumbpox in the forum view) and try again.[/quote]\r\n\r\nSorry, but where is it? I can't find it", "Solution_5": "Go to the front Index and look for it right above the \"Jump to...\" pull down. (This is right above where it lists all the people viewing the forums).", "Solution_6": "[quote=\"joml88\"]Go to the front Index and look for it right above the \"Jump to...\" pull down. (This is right above where it lists all the people viewing the forums).[/quote]\r\n\r\nFixed. Thanks.", "Solution_7": "Without touching anything, the same is happening again!!!!!!!!!!!!", "Solution_8": "[quote=\"Jos\u00e9\"]Without touching anything, the same is happening again!!!!!!!!!!!![/quote]\r\n\r\nWhat can I do????", "Solution_9": "Well, this is really strange... without doing anything, it fixed again!!! why is it?" } { "Tag": [ "calculus", "integration", "function", "calculus computations" ], "Problem": "So what stinks about this problem is I have 1 shot to get them all right and no partial credit! I could figure it if it were just converge or diverge but using only the integral test is the trick because I never use the integral test ha...Can anyone help?\r\n\r\nTest each of the following series for convergence by the Integral Test. If the Integral Test can be applied to the series, enter CONV if it converges or DIV if it diverges. If the integral test cannot be applied to the series, enter NA. (Note: this means that even if you know a given series converges by some other test, but the Integral Test cannot be applied to it, then you must enter NA rather than CONV.)\r\n\r\n1.) {summation 1 to infinity} ne^(6n)\r\n2.) {summation 1 to infinity} ne^(-6n)\r\n3.) {summation 1 to infinity} (n+8)/(-6^n)\r\n4.) {summation 1 to infinity} 8/(n*ln(3n)\r\n5.) {summation 1 to infinity} 8/(n(ln(3n))^3)\r\n\r\nI think it's:\r\n1) NA It's easy to see this series diverges, it's term goes to infinity.\r\n\r\n2) CONV. the function is positive and ultimately deacresing. And it's infinite integral is finite.\r\n\r\n3) Same as 2.\r\n\r\n4) CONV. or NA b/c this easy to see comparing with the p series.\r\n\r\n5) Same as 4.", "Solution_1": "Keep in mind that this problem doesn't care whether you can apply another test or not; the only question is whether you can apply the Integral Test, so don't talk about the other tests.\r\n\r\nDoes 3) mean $ \\sum_{k \\equal{} 1}^{\\infty} \\frac {n \\plus{} 8}{( \\minus{} 6)^n}$? In this case, the Integral Test can't be applied because the integrand isn't continuous.\r\n\r\nI will tell you that 4) and 5) do not have the same answers. Keep working :)", "Solution_2": "[quote=\"MegPie\"]So what stinks about this problem is I have 1 shot to get them all right and no partial credit! [/quote]\r\nI think that's exactly what you should expect in every real life situation. It would be nice if a neuro-surgeon would be allowed to try to operate his patient several times and got the partial credit for just holding the scalpel right but, quite surprisingly, the patient will probably run screaming into the night if he is told about such rules :roll:\r\n\r\nOn the other hand, in this particular set I would be quite perplexed about 1): should I write \"diverges\" or \"NA\"? The answer depends heavily on how exactly the theorem was stated in your course.\r\n\r\nAs to the rest, the integral test application is very simple. On a cookbook level, it reduces to checking that $ a_n\\equal{}f(n)$ where $ f(x)$ preserves sign and is monotone eventually (if not, the test does not apply) and then just checking if $ \\int^\\infty f(x)\\,dx$ converges." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$ a,b,c$ are the sides of a triangle. $ a \\plus{} b \\plus{} c \\equal{} 3$, $ x,y > 0$. Prove that\r\n$ \\sum_{cyc}\\sqrt {(x \\plus{} y)ab \\plus{} xbc \\plus{} xca}\\geq3\\sqrt {(x \\plus{} y)(ab \\plus{} bc \\plus{} ca) \\minus{} 2y}$", "Solution_1": "[quote=\"jedaihan\"]$ a,b,c$ are the sides of a triangle. $ a \\plus{} b \\plus{} c \\equal{} 3$, $ x,y > 0$. Prove that\n$ \\sum_{cyc}\\sqrt {(x \\plus{} y)ab \\plus{} xbc \\plus{} xca}\\geq3\\sqrt {(x \\plus{} y)(ab \\plus{} bc \\plus{} ca) \\minus{} 2y}$[/quote]\r\nThe inequality is true for all $ a,b,c \\ge 0$, not necessary to be the side lenghts of a triangle. \r\nBy sqaring both side we got\r\n\\[ \\Leftrightarrow \\left( {3x \\plus{} y} \\right)\\sum {ab} \\plus{} 2\\sum {\\sqrt {\\left( {x\\sum {ab} \\plus{} yab} \\right)\\left( {x\\sum {ab} \\plus{} bc} \\right)} } \\ge 9\\left( {x \\plus{} y} \\right)\\sum {ab} \\minus{} 18y\r\n\\]\r\n\\[ \\Leftrightarrow 9y \\plus{} \\sum {\\sqrt {\\left( {x\\sum {ab} \\plus{} yab} \\right)\\left( {x\\sum {ab} \\plus{} bc} \\right)} } \\ge \\left( {3x \\plus{} 4y} \\right)\\sum {ab} \r\n\\]\r\nBy the Cauchy Schwarz inequality we have\r\n\\[ \\sum {\\sqrt {\\left( {x\\sum {ab} \\plus{} yab} \\right)\\left( {x\\sum {ab} \\plus{} bc} \\right)} } \\ge 3x\\sum {ab} \\plus{} y\\sqrt {abc} \\left( {\\sum {\\sqrt a } } \\right)\r\n\\]\r\nThus it suffices to show that\r\n\\[ 9 \\plus{} \\sqrt {abc} \\left( {\\sum {\\sqrt a } } \\right) \\ge 4\\sum {ab}\r\n\\]\r\nwhich is true by Schur for third degree, :).", "Solution_2": "Nice proof, Honey_S!\r\n[quote=\"Honey_S\"][quote=\"jedaihan\"]$ a,b,c$ are the sides of a triangle. $ a + b + c = 3$, $ x,y > 0$. Prove that\n$ \\sum_{cyc}\\sqrt {(x + y)ab + xbc + xca}\\geq3\\sqrt {(x + y)(ab + bc + ca) - 2y}$[/quote]\nThe inequality is true for all $ a,b,c \\ge 0$, not necessary to be the side lenghts of a triangle. \n[/quote]\r\nIf so, why ${ (x + y)(ab + bc + ca) - 2y}\\geq0$ $ ?$ :wink:" } { "Tag": [ "inequalities", "trigonometry", "inequalities unsolved" ], "Problem": "$a_1, ..., a_n$ are non-negative and $a_p+a_q \\leq \\pi$ for $p \\neq q$\r\nProve that $\\frac{1}{n}\\sum\\cos{a_i} \\leq \\cos(\\frac{1}{n}\\sum a_i)$", "Solution_1": "Use the concavity of cos on [0,Pi/2].", "Solution_2": "Not quite; we can have $a_i>\\pi/2$", "Solution_3": "[quote=\"blahblahblah\"]Not quite; we can have $a_i>\\pi/2$[/quote]\r\nHmm... does the inequality still hold in that case?" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Given a positive prime $p$ and two positive integers $a,b$, prove that\r\n\r\n$\\binom{pa}{pb} \\equiv \\binom{a}{b} \\mod{p^3}$.", "Solution_1": "Please do not post the same problem in two sub-forums, you have posted this problem in:\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=37337", "Solution_2": "Well, in that thread blahblahblah said that the problem is more appropriate for Olympiad NT.....", "Solution_3": "this problem has appeared here before, try a search.\r\n\r\nedit: I think it was solved by Grobber.", "Solution_4": "[quote=\"zscool\"]edit: I think it was solved by Grobber.[/quote]\r\n\r\nThen it shouldn't be hard to find, just look through his posts!\r\nOn second thought... :lol:", "Solution_5": "[quote=\"zscool\"]this problem has appeared here before, try a search.\n\nedit: I think it was solved by Grobber.[/quote]\r\n\r\nI tried and didn't find anything, but I'll try some more.\r\n\r\nEDIT: ok, i found it, but i dont fully understand the solution... so could anyone either clarify or post another one? thanks.\r\n\r\n[url=http://artofproblemsolving.com/Forum/viewtopic.php?highlight=mod+p%5E3&t=21509]http://artofproblemsolving.com/Forum/viewtopic.php?highlight=mod+p%5E3&t=21509[/url]\r\n\r\nI think I understand grobber's part in the link, but I don't quite get pbornsztein's." } { "Tag": [ "logarithms" ], "Problem": "$ \\log_3 5\\plus{}\\log _3 6\\minus{}\\log _3 10$ is equal with:\r\n\r\n$ (A)$ $ 10$\r\n$ (B)$ $ 0$\r\n$ (C)$ $ \\minus{}1$\r\n$ (D)$ $ 1$\r\n$ (E)$ $ 2$", "Solution_1": "$ \\log_3 5 \\plus{} \\log _3 6 \\minus{} \\log _3 10 \\equal{} \\log_3 \\left(\\frac {5 \\times 6}{10}\\right) \\equal{} \\log_3 3 \\equal{} 1$.\r\nThus, $ \\boxed{D}$." } { "Tag": [ "geometry", "\\/closed" ], "Problem": "For example, would taking intro to geometry cover everything you would learn in a school year of geometry?", "Solution_1": "It depends on your school. But in most cases you would learn more than a school class, have a better understanding. Also a great group of classmates in most cases.", "Solution_2": "So would it be possible to skip one grade of math in school? Like taking geometry eighth grade, taking algebra 2 on AOPS during the summer, and skipping to pre-calc in 9th grade?", "Solution_3": "If your school will let you skip yes!\r\nIts totally possible, as long as you put the work in.", "Solution_4": "If you're trying to use the classes as credit, make sure you talk it over with your school counselor or whoever is in charge of scheduling before you sign up.", "Solution_5": "[quote=\"Raynee\"]So would it be possible to skip one grade of math in school? Like taking geometry eighth grade, taking algebra 2 on AOPS during the summer, and skipping to pre-calc in 9th grade?[/quote]\r\n\r\nEh, I tried to do that with geometry, it didn't work; so it depends on your school.\r\n\r\n@Raynee: The AoPS stuff is way harder than regular school math. I'm getting really board in Geometry so I do Precalc aops during that time.", "Solution_6": "[quote=\"Raynee\"]So would it be possible to skip one grade of math in school? Like taking geometry eighth grade, taking algebra 2 on AOPS during the summer, and skipping to pre-calc in 9th grade?[/quote]\r\n\r\nYou would have to make sure your school will allow you to do this; don't simply assume your school will accept our classes as substitutes. \r\n\r\nOur classes are considerably more challenging than most regular school classes.", "Solution_7": "Yeah some schools like mine are really inflexible, so I'm stuck in algebra II for the year. We just got to learning what $ i$ was :roll: and it's the beginning of 2nd semester. You tend to learn much more in an aops class than a typical school class, also the time you need to devote to the class is much more than you'll probably be used to. But it's definitely worth it! :D", "Solution_8": "AIME15, you have quite some lofty goals lol, at least for the average person. But good for you, and good luck achieving those goals :)", "Solution_9": "I do not think AIME15 is an average person. In fact, I don not think anyone on AoPS is an average person.\r\n\r\nI kinda hope there is some sarcasm in AIM15's siggy. Cause I'm annoyed. But in my state, the is an extra-cric option (with a test) that allows you to do 2 years of math in one year, and is generally accepted throughout the state. So I did Algebra 1 and 2 in 6th grade, and geo+PC in 7th grade and am doing Calc I now. Woot. But I don't think I learned it as well as I could have if it was AoPS, so...meh." } { "Tag": [ "induction", "number theory unsolved", "number theory" ], "Problem": "Prove that for that for every positive integer $ n$, the smallest integer that is greater than $ (\\sqrt {3} \\plus{} 1)^{2n}$ is divisible by $ 2^{n \\plus{} 1}$.", "Solution_1": "in fact $ 2^{n\\plus{}1} | [(\\sqrt{3}\\plus{}1)^{2n}]\\plus{}1$ and also $ 2^{n\\plus{}1} | [(\\sqrt{3}\\plus{}1)^{2n\\plus{}1}]$\r\njust use the fact that the sequence :\r\n$ x_1\\equal{}2,x_2\\equal{}8,x_n\\equal{}2(x_{n\\minus{}1}\\plus{}x_{n\\minus{}2})$\r\nsatisfies $ x_n\\equal{}(\\sqrt{3}\\plus{}1)^n\\plus{}(1\\minus{}\\sqrt{3})^n$\r\nnow use induction and the fact that $ x_n\\equal{}[(\\sqrt{3}\\plus{}1)^n]$ for odd $ n$ and $ x_n\\equal{}[(\\sqrt{3}\\plus{}1)^n]\\plus{}1$ for even $ n$.", "Solution_2": "\\text<\u044f\u0441\u043d\u043e \u0447\u0442\u043e > \\{sqrt(3)+1}^n-{sqrt(3)-1}^n\\ \\text<\u0446\u0435\u043b\u043e\u0435 \u0447\u0438\u0441\u043b\u043e>\r\n\\{sqrt(3)-1}^n \\text<\u0434\u0440\u043e\u0431\u043d\u043e\u0435 \u0447\u0438\u0441\u043b\u043e ,\u0442\u043e\u0433\u0434\u0430 \u0446\u0435\u043b\u043e\u0435 \u0447\u0430\u0441\u0442 \u0434\u0430\u043d\u043d\u043e\u0439 \u0432\u044b\u0440\u043e\u0436\u0435\u043d\u0438\u0439 \u0431\u0443\u0434\u0435\u0442 \u0432\u0435\u0440\u0445\u043d\u043e\u0435 \u0432\u044b\u0440\u043e\u0436\u0435\u043d\u0438\u0435 >\r\n\\text<\u044f\u0441\u043d\u043e \u0447\u0442\u043e \u0432\u0435\u0440\u0445\u043d\u043e\u0435 \u0432\u044b\u0440\u043e\u0436\u0435\u043d\u0438\u0435 \u0434\u0435\u043b\u0438\u0442\u0441\u044f \u043d\u0430 > \\2^(n+1)\r\n\\text<\u0442\u043e \u043c\u043e\u0436\u043d\u043e \u0434\u043e\u043a\u0430\u0437\u0430\u0442 \u043f\u043e \u043c\u0435\u0442\u043e\u0434\u043e\u043c \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u0447\u0435\u0441\u043a\u043e\u0439 \u0438\u043d\u0434\u0443\u043a\u0446\u0438\u0439>\r\n :rotfl:", "Solution_3": "[quote=\"PIRISH\"]\\text<\u044f\u0441\u043d\u043e \u0447\u0442\u043e > \\{sqrt(3)+1}^n-{sqrt(3)-1}^n\\ \\text<\u0446\u0435\u043b\u043e\u0435 \u0447\u0438\u0441\u043b\u043e>\n\\{sqrt(3)-1}^n \\text<\u0434\u0440\u043e\u0431\u043d\u043e\u0435 \u0447\u0438\u0441\u043b\u043e ,\u0442\u043e\u0433\u0434\u0430 \u0446\u0435\u043b\u043e\u0435 \u0447\u0430\u0441\u0442 \u0434\u0430\u043d\u043d\u043e\u0439 \u0432\u044b\u0440\u043e\u0436\u0435\u043d\u0438\u0439 \u0431\u0443\u0434\u0435\u0442 \u0432\u0435\u0440\u0445\u043d\u043e\u0435 \u0432\u044b\u0440\u043e\u0436\u0435\u043d\u0438\u0435 >\n\\text<\u044f\u0441\u043d\u043e \u0447\u0442\u043e \u0432\u0435\u0440\u0445\u043d\u043e\u0435 \u0432\u044b\u0440\u043e\u0436\u0435\u043d\u0438\u0435 \u0434\u0435\u043b\u0438\u0442\u0441\u044f \u043d\u0430 > \\2^(n+1)\n\\text<\u0442\u043e \u043c\u043e\u0436\u043d\u043e \u0434\u043e\u043a\u0430\u0437\u0430\u0442 \u043f\u043e \u043c\u0435\u0442\u043e\u0434\u043e\u043c \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u0447\u0435\u0441\u043a\u043e\u0439 \u0438\u043d\u0434\u0443\u043a\u0446\u0438\u0439>\n :rotfl:[/quote][code][/code][code][/code]", "Solution_4": "The smallest integer greater than it is (sqrt(3) +1)^2n + (sqrt(3) -1)^2n = 2^n * [ (2+sqrt(3))^n + (2-sqrt(3))^n ] . Binomial expansion of the expression in the square braces will reveal that it is a multiple of 2. Hence problem solved[/quote][/code]", "Solution_5": "if $ x = (\\sqrt {3} + 1)^2n$ is a fraction then,its fraction which has to summed up to make an integer is $ y = (\\sqrt {3} - 1)^2n )$\r\n\r\nso, we know that x and y are both divisible by $ 2^{n}$[expand the square] .and we know that $ y < 1$.so, x+y is the least integer.so,add x and y.we get it as x+y=$ 2(%Error. \"comb\" is a bad command.\n{2n}c{2n}3^{n} + %Error. \"comb\" is a bad command.\n{2n}c{2n - 2} . ........... + %Error. \"comb\" is a bad command.\n{2n}c{2n} 1)$.but we know that the whole expression is divisible by again $ 2$.so, it is divisible by $ 2^{n + 1}$.", "Solution_6": "In fact, we can also prove that $ (\\sqrt{3}\\plus{}1)^{2n}$ is not divisible by $ 2^{n\\plus{}2}$.", "Solution_7": "[quote=\"Stephen\"]In fact, we can also prove that $ (\\sqrt {3} \\plus{} 1)^{2n}$ is not divisible by $ 2^{n \\plus{} 2}$.[/quote]\r\nAre you sure about that?\r\nLet $ \\alpha \\equal{} 2\\plus{}\\sqrt{3}, \\beta \\equal{} 2\\minus{}\\sqrt{3}$. Then the smallest integer greater than $ (\\sqrt{3}\\plus{}1)^{2n}$ is $ (\\sqrt{3}\\plus{}1)^{2n} \\plus{} (\\sqrt{3}\\minus{}1)^{2n} \\equal{} (2\\alpha)^n \\plus{} (2\\beta)^n \\equal{} 2^n( \\alpha^n \\plus{} \\beta^n)$.\r\nLet $ a_n \\equal{} \\alpha^n \\plus{} \\beta^n$. Then $ a_1 \\equal{} 4, a_2\\equal{}14$ and $ a_{n\\plus{}2} \\equal{} 4a_{n\\plus{}1}\\minus{}a_n$. Hence $ 2 \\mid a_n$ and $ 4 \\mid a_n$ when $ 2 \\nmid n$ and $ 4 \\nmid a_n$ when $ 2 \\mid n$.\r\nSo $ 2^{n\\plus{}2}$ divides the smallest integer greater than $ (\\sqrt{3}\\plus{}1)^{2n}$ iff $ 2 \\nmid n$.", "Solution_8": "[quote]The smallest integer greater than $ (\\sqrt {3} \\plus{} 1)^{2n}$ is $ (\\sqrt {3} \\plus{} 1)^{2n} \\plus{} (\\sqrt {3} \\minus{} 1)^{2n}$ [/quote]\r\nActually I don't understand how you people have gotten this result. What was the motivation of taking conjugate? Can anyone explain?", "Solution_9": "[quote=\"Moonmathpi496\"][quote]The smallest integer greater than $ (\\sqrt {3} \\plus{} 1)^{2n}$ is $ (\\sqrt {3} \\plus{} 1)^{2n} \\plus{} (\\sqrt {3} \\minus{} 1)^{2n}$ [/quote]\nActually I don't understand how you people have gotten this result. What was the motivation of taking conjugate? Can anyone explain?[/quote]\r\nIf $ \\alpha, \\beta$ is roots of $ x^2\\minus{}nx\\minus{}m\\equal{}0$ and we define $ a_n \\equal{} x \\alpha^n \\plus{} y \\beta^n$ for some $ x,y \\in \\mathbb{C}$ then we have $ a_{n\\plus{}2} \\equal{} na_{n\\plus{}1} \\plus{} ma_n$.\r\n\r\nIf $ |\\alpha| < 1$ we can approximate $ a_n$ to be $ y \\beta^n$ for large $ n$. Don't know how to explain the \"motivation\". It's just a standard thing to do :P", "Solution_10": "thanks mathias:dk...nice!!! \r\nbut plz can u say how can i conclude with the part i mean how can i form the recursion??? :blush: \r\ncan u plz say this??? :oops:", "Solution_11": "Induction!!!" } { "Tag": [ "geometry" ], "Problem": "we have a quadrilateral. we can change its angles, without changeing its sides. which quadrilateral will have the maximum area? there is an easy sol with trigo and calculus.", "Solution_1": "[hide=\"The relevant formula is...\"]\n[url=http://mathworld.wolfram.com/BretschneidersFormula.html]Bretschneider's Formula[/url] [/hide]" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "For n natural denote by f(n) the aruthmetic mean of positive divisors of n.\r\nShow that \\sqrt n <=f(n) <=(n+1)/2", "Solution_1": "Observe that if d_1,...,d_k are the divisors then n/d_1,...,n/d_k are excatly the same divisors ( in other order). Thus, n*(1/d_1+...+1/d_k)=d_1+...+d_k. Then d_1+...+d_k \\geq n*k^2/(d_1+...+d_k) from where we get the left inequality. For the second one, all we need to ptove is that if x_1<... pwned this!\n9. (3,18)(18,3)\n10. Use Cyclic Quads, then use common chords, and you're done!\n[/hide]\r\nBy the way, do you think I will get in???\r\nPlease share your answers!\r\n- Nawahd Werdna", "Solution_2": "I got\r\n[hide]\n1. 25\n2. 139 (double checked with my calculator)\n3. 30 (double checked by brute forcing)\n4. 98 (not sure)\n5. 4,030,056\n6. 197 (fun problem!)\n7. 1020 (not sure, seems kind of high)\n8. 360 (fun problem!) (double checked with my calculator)\n9. Didn't try\n10. Didn't try\n[/hide]\r\nEDIT: how did you go from not having applied to having applied in a couple of minutes, Nawahd Werdna?", "Solution_3": "isn't number 10 the outer napolean triangle?\r\nIt's a very very famous problem, I'm surprsed they used it.\r\nIt's mentioned in Intro to Geometry, so it must be provable using extremely basic techniques.", "Solution_4": "[quote=\"Twiz\"]EDIT: how did you go from not having applied to having applied in a couple of minutes, Nawahd Werdna?[/quote]\r\n\r\nSorry, that was a typo... I thought the application wouldn't reach in time, but it did :)... although, I guess that i did make quite a few errors.... by the way, could you please post your solution to numer 3? I want to see where I went wrong...", "Solution_5": "For Number 3, you use the triangle inequality theorem (sum of 2 sides is greater then the third side). Then you list:\r\n1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. \r\nSum of the other sides, other than 13, must be at least 14.\r\nIf one other side is 12, there are 10 choices, \r\nif one side is 11, there are 8 choices.\r\nIf one side is 10, there are 6 choices\r\nIf one side is 9, there are 4 choices,\r\nif one side is 8, there are 2 choices\r\nIf one side is 7, there are no more choices since 6+7=13, and \r\nall the greater choices, B= 7+ x have been counted. Thus, you have \r\n10 + 8 + 6 + 4 + 2 = 30 choices", "Solution_6": "can somebody please post their solution to number 4 and number 9", "Solution_7": "number 10 is EASILY brute forceable using complex numbers\r\nlol", "Solution_8": "Isn't number 10 the Napoleons theorem?", "Solution_9": "Could someone please tell me whether or not my answers to numbers 4 and 7 are correct?", "Solution_10": "[quote=\"Twiz\"]Could someone please tell me whether or not my answers to numbers 4 and 7 are correct?[/quote]\r\n\r\nI can't be very sure, but I think your #7 is very high compared to mine...", "Solution_11": "numba 10\r\nhttp://mathworld.wolfram.com/OuterNapoleonTriangle.html", "Solution_12": "[quote=\"lena_siz\"]Isn't number 10 the Napoleons theorem?[/quote]\r\n\r\nIt is :o \r\n[url]http://www.cut-the-knot.org/proofs/napoleon_intro.shtml[/url]", "Solution_13": "Meh, I already knew the complex number one, so I was like \"guys, sorry, no cheating here\", and had a REALLY BAD one by rotation, which I think reduced to \"here guys, these triangles are similar...AND OH LOOK THESE LOOK EQUILATERAL I BET THEY ARE QED\"", "Solution_14": "I thought that number 10 didn't involve any concepts with complex numbers, I solved it by rotation and similar triangles. What do the complex numbers have to do with it?", "Solution_15": "[quote=\"lena_siz\"]I thought that number 10 didn't involve any concepts with complex numbers, I solved it by rotation and similar triangles. What do the complex numbers have to do with it?[/quote]\r\nyou can brute force it with complex numbers =D", "Solution_16": "[quote=\"kohjhsd\"][quote=\"lena_siz\"]Isn't number 10 the Napoleons theorem?[/quote]\n\nIt is :o \n[url]http://www.cut-the-knot.org/proofs/napoleon_intro.shtml[/url][/quote]\r\n\r\nlol...i used the same exact website\r\n\r\nfor numba nine!\r\n\r\n\r\n\r\n\r\n$xy+\\frac{x^{3}+y^{3}}{3}=2007$\r\n\r\nLet $p=xy$ and $s=x+y$\r\n\r\nWe know $x^{3}+y^{3}=s^{2}-3sp$\r\n\r\nSo $p+\\frac{s^{3}-3sp}{3}=3^{2}\\cdot 223$\r\n$\\Rightarrow 3p+s^{3}-3sp=3^{3}\\cdot 223$\r\n\r\nWe know that $s^{3}$ must be divisible by 3 so $s$ is also divisible by 3. Now let $3a=s$\r\n\r\n$3p+3^{3}a^{3}-3^{2}ap=3^{3}\\cdot 223$\r\n$\\Rightarrow p+3^{2}a^{3}-3ap=3^{2}\\cdot 223$\r\n$\\Rightarrow 9a^{3}-p(3a-1)=9\\cdot 223$\r\n\r\n$3a-1$ is not divisible by 9 so $p$ must be divisible by 9. Let $9b=a$.\r\n\r\n$9a^{3}-9b(3a-1)=9\\cdot 223$\r\n$\\Rightarrow a^{3}-b(3a-1)=223$\r\n$\\Rightarrow b=\\frac{a^{3}-223}{3a-1}$\r\n\r\n---\r\n\r\nBy vieta's theorem, the solutions to $m^{2}-sm+p=m^{2}-3am+9b=0$ are $x$ and $y$. The discriminant is:\r\n\r\n$9a^{2}-4\\cdot 9b=9a^{2}-36b$.\r\n\r\nThis must be greater than 0 so $a^{2}\\ge 4b$\r\n\r\n$\\Rightarrow a^{2}\\ge 4\\frac{a^{3}-223}{3a-1}$Solving for this equation, we see that $a$ is between 0 and 9 inclusive. I used calculator for this by the way. Now test them out and there is only one pair of $(a,b)$ that gives integers. Then we can find $s$ and $p$ thus giving us $x$ and $y$. The answer is $\\boxed{(3,18) (18,3)}$\r\n\r\nI probably did it the worst, most tedious way but whatever. I think I got an 8 or something", "Solution_17": ":o \r\n\r\nI was going to do that\r\n\r\nbut then I was like\r\n\r\nnah its too long\r\n\r\nwill never work\r\n\r\nAMP writers are evil grr", "Solution_18": "[hide=\"I got\"]\n1. 25\n2. 139\n3. 30\n4. 100\n5. 4030056\n6. 197\n7. 215 (rather brute-force-ish :( )\n8. Didn't do--found the answer to be 360 using my calculator :roll: \n9. (18,3) and (3,18)--very brut-force-ish :mad: \n10. Didn't do--Isn't looking it up somewhat cheap?\n[/hide]\r\n\r\nI wouldn't be surprised if I either over- or under-counted in 7, and am not at all proud of 9--I basically limited the range and then checked each possibility individually.", "Solution_19": "Could you post or PM me your solution to number 4? I'm curious to see whether you overcounted by two or I undercounted by two (or we were both wrong). Reviewing my solution, I feel that I may have overcounted by two, but still can't find where I may have undercounted.", "Solution_20": "My solution 4 is attached. If anyone spots a potential error, please feel free to shoot holes in it.", "Solution_21": "you guy all got 139 for number 3. I got 169, and i used the fact that 2*5 = 10 and 10 * 1 = 10 , could someone please post their solution to this question? :huh:", "Solution_22": "guys* , number 2*, sorry", "Solution_23": "Roadnottaken: Okay, your solution is far more convincing (and neat) than mine. Anyways I found a possible error in my solution (which uses a completely different method than yours does) and an error in your solution that lead to the answer being $96$. In case three, you subtract $4$ cases when the squares contain horizantal pieces and $4$ additional cases when the squares contain vertical pieces; however, when the squares are against the corners, both the vertical pieces and the horizantal pieces are contained in square; therefore, I believe that subtracting $4$ only one time would have accounted for the overcounting. This possible mistake took me about ten minutes to notice, and again, I could be wrong, but I guess we'll see when the answers come out.\r\nEDIT: It turns out that your answer was right.\r\n\r\n[quote=\"lena_siz\"]you guy all got 139 for number 3. I got 169, and i used the fact that 2*5 = 10 and 10 * 1 = 10 , could someone please post their solution to this question? :huh:[/quote]\r\nIt looks like you may have forgotton to consider that a factor of 25 contains two fives and thus adds two zeroes to the product and a factor of 125 contains three fives and thus adds three zeroes to the product (I did this the first time I did one of these types of problems as well).", "Solution_24": "They've posted solutions to the admissions test, but unlike test a, none of the solutions were student-submitted. I wonder why?", "Solution_25": "well, I got 6 problems perfect (ladder, factorial, triangles, 2007 rt. triangle, the fraction sum, primes (loved that one)), had a fairly shoddy proof for 10 (used reflections and concurrency, made one HUGE assupmtion that I didn't prove), and got 1944 for the cubes simply because it was 5:30 on the due date, I didnt want to do case work, and so I said there are $6^{6}$ ways to color the cube, $24$ ways to orient the cube, so the answer is $\\frac{6^{6}}{24}=1944$, which seemed decently reasonable from the small amount of casework I had done previous to writing the solution. I totally didn't read the problem right to see that each face had to be adjacent to faces of different colors.\r\n\r\nI got a call today from Dr. A, and I'm in.", "Solution_26": "Sapphyre, what grade are you in??? I just want to know... because I think that I also got 6 or 7 problems perfect.... So, just curious...\r\n\r\n\r\n- Nawahd Werdna", "Solution_27": "9\r\n[I'm typing this because my message is too short]", "Solution_28": "x^3+y^3+(-1)^3-3(x)(y)(-1)=3*2007-1\r\n\r\n=(x+y-1)(...)", "Solution_29": "you can use analytic geometry for #10 if you're inclined to brute force your way through it.", "Solution_30": "[quote=\"Aryth\"]you can use analytic geometry for #10 if you're inclined to brute force your way through it.[/quote]\r\n\r\nUsing Complex Numbers is much cleaner, and essentially the same thing. It's also a well-known theorem, so you can just google outer napoleonic triangle and BAM", "Solution_31": "I came up with this proof:\r\n\r\nhttp://reflections.awesomemath.org/2007_2/Napoleons_Theorem.pdf", "Solution_32": "that's exactly how I did it (with the inner point and everything) but I pretty much assumed everything rather than proving it, which I knew I couldn't do... :roll:" } { "Tag": [ "trigonometry", "calculus", "derivative", "real analysis", "real analysis unsolved" ], "Problem": "Let $F_n(x)=\\sum_{m=1}^n \\frac{1}{2^m} |\\sin (\\pi m!x)|.$ For rational numbers $r=\\frac{q}{p}\\ (p: natural\\ number,\\ q: integer),$ \r\nif $n\\geq p$ then prove that $F_n(x)$ isn't differentiable at $x=r.$", "Solution_1": "Hmmm... I think you mean \"[b]rational[/b] numbers\". But isn't it obvious that the right derivative exceeds the left derivative at each rational point with small denominator? :?", "Solution_2": "Sorry for that, I will edit.\r\n\r\nkunny" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "hello every body :) !!!\r\n\r\nfind all fuctions $ g: \\mathbb{R}_\\plus{}^* \\longrightarrow \\mathbb{R}_\\plus{}^*$ \r\n\r\n$ g \\left( \\frac{a}{x} \\right) \\equal{} g(x)$ with $ a \\in \\mathbb{R}_\\plus{}^*$\r\n\r\n$ good \\ luck$", "Solution_1": "mathema, is $ a$ represent a given positive real? If so, then there are infinitely many $ g(x)$ because $ g(\\dfrac{a}{\\dfrac{a}{x}})\\equal{}g(x)$.\r\n\r\nIf $ a$ represent arbitrary positive real, then put $ a\\equal{}x$ and we get $ g(x)$ is a constant.", "Solution_2": "Hi stephen :) !\r\n\r\nfor exemple the functions that wa have $ g(\\frac{1}{x} ) \\equal{}g(x)$ \r\n\r\nor also find all functions that: $ g( \\frac{\\pi}{x})\\equal{} g(x)$ ....\r\n\r\nPS: that's obvious that may be exists infinity solutin :)", "Solution_3": "Let $ A$ be the set of all real numbers between 0 and $ \\sqrt{a}$, where $ \\sqrt{a}$ belongs to $ A$, and $ B$ be the set of all real numbers between $ \\sqrt{a}, \\inf$, where $ \\sqrt{a}$ does not belong to $ B$.\r\n\r\nDefine G arbitrarily on $ A$, then for every number in $ B$ define $ f(x)\\equal{}f(\\frac{a}{x})$ where $ \\frac{a}{x}$ must obviously belong in $ A$." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "hi,\r\nlet $a,b,c,d \\in \\mathbb{R}$ and $0 < |a| < |c|$ and $f : \\mathbb{R}\\longrightarrow \\mathbb{R}$ with the property $f(ax+b)=f(cx+d)$, $\\forall x \\in \\mathbb{R}$. Prove that there $\\exists x_{0}\\in \\mathbb{R}$ s.t.\r\nif $f$ is continous in $x_{0}$ than $f$ is constant.", "Solution_1": "Let $x_{0},y_{0}$ defined by $ax_{0}+b=cx_{0}+d=y_{0}$ and $s=a/c, \\ g(y)=f(y_{0}+y/c)$.\r\nThen $g(y)=g(sy), \\ |s|<1$, therefore if $g(y)$ is continiosly in 0, then $g(y)=g(0) \\ \\forall y$ or\r\nif $f(x)$ continiosly in $y_{0}$, then $f(x)=f(y_{0})$." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let $ ABC$ be a right triangle with $ \\angle A \\equal{} 90^\\circ$. Let $ M$ and $ H$ be the mid-point of the side $ BC$ and the foot of the perpendicular from $ A$ to $ BC$, respectively. Extend segment $ AM$ through $ A$ to $ P$. The lines through $ H$ and perpendicular to $ AB$, $ AC$, intersect $ PB$, respectively $ PC$ at $ Q$ and $ R$. Prove that $ A$ is the orthocentre of triangle $ PQR$.", "Solution_1": "Let $ V_C^{\\frac{CP}{CR}}: R\\rightarrow P,H\\rightarrow I,A\\rightarrow J$ we have:\r\n$ \\angle{JAP}\\equal{}90^0\\plus{}\\angle{BAP}\\equal{}90^0\\plus{}\\angle{BIP}\\equal{}\\angle{JIP}$\r\nThen $ AJPI$ is inscribe quadrangle, it mean $ \\angle{IAP}\\equal{}\\angle{IJP}$\r\nNow $ \\angle{IBP}\\equal{}\\angle{IAP}$, then $ \\angle{IBP}\\equal{}\\angle{IJP}$.\r\nHence $ \\angle{BPJ}\\equal{}\\angle{BIJ}\\equal{}90^0$, it mean $ BP,PJ$ are perpendicular.\r\nBut $ PJ,AR$ are parallel then $ BP,AR$ are perpendicular.\r\nSimilarly we have solution!\r\n[img]http://i227.photobucket.com/albums/dd72/TPFC/april.png[/img]" } { "Tag": [ "parameterization" ], "Problem": "Given the graph of $ y\\equal{}f(x)$, how would you sketch the graph of $ y\\equal{}f(\\sqrt{x})$ and $ y\\equal{}\\sqrt{f(x)}$?\r\n\r\nAs an example, consider $ f(x)\\equal{}e^x$, $ x \\in \\mathbb{R}$.", "Solution_1": "Hello\r\nFirst of all we need to understand what parameters r going to change and then we can proceed easily.\r\n$ f(x) = e^x$\r\nAll d three func $ y = f(x),\\sqrt {f(x)},f(\\sqrt{x})$ all are going to meet at$ (0,1)$\r\nFor $ y = \\sqrt {f(x)}$\r\nAs for $ x < 0$ $ f(x) < 1$ so sqrt of this func will give larger value than this function.\r\nBut for $ x > 0$ $ f(x) > 1$ so sqrt of this value will give smaller value than this function.\r\nSo the graph $ y = \\sqrt {f(x)}$ will be above the graph $ y = f(x)$ before it reach $ x = 0$.At $ x = 0$, both the graph will meet and after that $ f(x)$ will come above to $ \\sqrt {f(x)}$. \r\nFor $ y = f(\\sqrt x)$\r\nno need to think about $ - ve x - axis$.For $ + ve$ x-axis ,like $ y = \\sqrt {x}$ it opens concave \r\nat x=0 $ f(x)$ and $ f(\\sqrt {x})$ meets .in the interval $ (0,1)$ $ f(\\sqrt {x})$ will be above $ f(x)$ both will meet again at $ x = 1$.And after that $ f(\\sqrt {x})$ will come below the $ f(x)$.\r\nTry with other functions.Have fun with graphs.\r\nThank u.", "Solution_2": "[quote=\"kabi\"]\nAs for $ x < 0$ $ f(x) < 0$ so sqrt of this func will give larger value than this function.\nThank u.[/quote]\r\n\r\nI don't understand your grammar, sorry. It is not possible for any real value of $ x$ for $ e^x<0$.", "Solution_3": "Hello\r\nSorry but by mistake I wrote that.I have edited that." } { "Tag": [ "calculus", "absolute value", "calculus computations" ], "Problem": "(a)\r\nProve that if f'(x)>=M whenever a<=x<=b, then f(b)>=f(a) + M(b-a)\r\n\r\n(b)\r\nProve that if f'(x)<=m whenever a<=x<=b, then f(b)<=f(a) + m(b-a)\r\n\r\n(c)\r\nFormulate a similar theorem when absolute value of f'(x)<=M for all x in [a,b]", "Solution_1": "By mean value theorem we have $f(b)-f(a)=f'(\\zeta)(b-a)$ for some $\\zeta\\in (a,b)$. Since $f'(\\zeta)\\geq M$, then $f(b)-f(a)\\geq M(b-a)$. The rest problems are similar.", "Solution_2": "thanx,\r\n\r\nHow do I do part (c) of the question?", "Solution_3": "I think it the same :) write $|f'(x)| \\leq M$ as $-M \\leq f'(x) \\leq M$ and use similar argument.", "Solution_4": "$|f(b)-f(a)|\\leq M|b-a|$\r\n\r\nThis is what's called a Lipschitz condition.", "Solution_5": "Thank you for your contributions!" } { "Tag": [ "function", "linear algebra", "linear algebra unsolved" ], "Problem": "les E and F two $ K$vectorial space of finit dimension p,n respectively,and let $ f \\in L(E,F)$.\r\nwe set $ G = \\{u \\in L(F,E)\\ fuf = 0_{L(E,F) \\}}$\r\nprove that G is a subspace of $ L(E,F)$ and find his dimension in function of $ p,n, rank(f)$", "Solution_1": "I cannot decipher this statement, nor can I guess what it should have said. $ G$ is the set of $ u$ such that .. what? The condition that follows doesn't mention $ u.$ And $ fgf$ doesn't make sense as written.\r\n\r\nMaybe $ G\\equal{}\\{g\\in L(F,E)\\mid fgf\\equal{}0\\}?$", "Solution_2": "it's ok now :roll:", "Solution_3": "No, it's not OK now. If $ u\\in L(E,F)$ and $ E$ is not the same space as $ F,$ then $ fuf$ is not defined. Perhaps you want $ u\\in L(F,E)?$", "Solution_4": "im sure now , :huh: i think :!:", "Solution_5": "G is a subspace of L(F,E) :).for the dimension i've just find some in\u00e9qualitise beetwen p,n;rg(f).it is also easy to see that G is the Ker of a linear application.\r\nif you have a nice solution dear Driss post it :D", "Solution_6": "$ Im(f)\\oplus{H}=F,dim(Im(f))=r$. $ u|_{Im(f)}$ is any linear function$ : Im(f)\\rightarrow{ker(f)}$. $ u|_{H}$ is any linear function $ : H\\rightarrow{E}$.\r\nThen $ dim(G)=r(p-r)+(n-r)p=np-r^2$." } { "Tag": [ "email" ], "Problem": "toi m\u1eebng Vi\u1ec7t Nam thay cho tuy\u1ec7t IMO.\r\ntoi thich Vi\u1ec7t Nam lam.", "Solution_1": "Cam on ban rat nhieu, dat nuoc cua chung toi dep nhi? :D Ban thay Ha Long the nao? Toi o do day! :P", "Solution_2": "Viet Nam that dep voi nhung con nguoi that tuyet voi.Ha Long cung vay,that tuyet! Toi hi vong se co dip quay tro lai Viet nam !\r\n(sorry for my bad vietnamese)", "Solution_3": "Hen gap ban khi do. :)", "Solution_4": "Oa anh nao ma moi tieng Viet Nam chuan vay . Cho em dia chi email di .Co gi em giup nang cao von tu Viet :D .[hide]Anh Tuan co thi cho em nha[/hide]", "Solution_5": "Cam on ban rat nhieu'day la email cua minh,ban lien lac voi minh nhe. (shayandashmiz@yahoo.com)\r\ntuan toi \u01b0\u1edbc ao gap ban khi do!", "Solution_6": "ne thiet ! ban hoc tieng viet lau chua ma` noi rat chuan. \r\n ban wa VN tuan toi lam j vay ?", "Solution_7": "[quote=\"Hong Quy\"]ne thiet ! ban hoc tieng viet lau chua ma` noi rat chuan. \n ban wa VN tuan toi lam j vay ?[/quote]\r\nC\u1eadu \u1ea5y n\u00f3i: Tu\u00e2n, t\u00f4i \u01b0\u1edbc ao g\u1eb7p b\u1ea1n khi \u0111\u00f3! :D", "Solution_8": "[quote=\"N.T.TUAN\"][quote=\"Hong Quy\"]ne thiet ! ban hoc tieng viet lau chua ma` noi rat chuan. \n ban wa VN tuan toi lam j vay ?[/quote]\nC\u1eadu \u1ea5y n\u00f3i: Tu\u00e2n, t\u00f4i \u01b0\u1edbc ao g\u1eb7p b\u1ea1n khi \u0111\u00f3! :D[/quote]\n anh ay noi :\n[quote=\"dashmiz\"]Cam on ban rat nhieu'day la email cua minh,ban lien lac voi minh nhe. (shayandashmiz@yahoo.com)\ntuan toi \u01b0\u1edbc ao gap ban khi do![/quote]\r\nLA Y NOI GAP EM MA :P \r\n:spider: \r\nAnh ay noi tieng Viet cung tot ma khi chat voi em toan chat bang tieng Anh :starwars: \r\nLam em toan dung tu dien :read: :wallbash:", "Solution_9": "Khong ngo co nguoi nuoc ngoai tren ML noi tieng Viet gioi nhu vay! Khong biet co phai dashmiz di thi IMO dot vua roi khong nhi?", "Solution_10": "C\u00f4ng nh\u1eadn r\u1ea5t chu\u1ea9n.\u0110\u00e1ng n\u1ec3,\u0111\u00e1ng n\u1ec3!!" } { "Tag": [ "national olympiad" ], "Problem": "Please if it possible someone to send me on private message or via e-mail the problems from Australian Math Olympiad 2008.\r\nWithout solutions.", "Solution_1": "Me too please. :lol:" } { "Tag": [], "Problem": "Where can I find The problems of AHSME 1990-1994? I don't find it.\r\nCan anyone help me?", "Solution_1": "Check the AOPS wiki section [url=http://www.artofproblemsolving.com/Wiki/index.php/AHSME_Problems_and_Solutions]here[/url]", "Solution_2": "Some ASHME problems are missing I believe so the only way to obtain them is to buy one of the contest books the MAA sells.", "Solution_3": "All the ones from 1990 to 1994 can also be found [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44]here[/url]." } { "Tag": [ "geometry", "circumcircle", "angle bisector", "geometry proposed" ], "Problem": "In a trinagle with $ h,t,m,R$ as altitude, internal angle bisector, median(all three corresponding tot he same vertex) and circumradius respectively, prove that \r\n$ 4R^2h^2(t^2\\minus{}h^2)\\equal{}t^4(m^2\\minus{}h^2)$.\r\nTry to avoid a lot of computaion while proceeding by algebra or trigonometry.", "Solution_1": "Note that $ t\\equal{}\\frac{2\\sqrt{bc(p\\minus{}a)}}{b\\plus{}c}, m^2\\equal{}\\frac{2(b^2\\plus{}c^2)\\minus{}a^2}{4}$ and $ 4R^2.h^2\\equal{}4\\frac{a^2b^2c^2}{16S^2}.\\frac{4S^2}{a^2}\\equal{}b^2c^2$\r\nIs it enough to calculate? :)" } { "Tag": [ "calculus", "geometry", "probability", "MATHCOUNTS", "AMC", "AIME", "USA(J)MO" ], "Problem": "I thought it would be cool for all of U.S. students to give their math profile (good for future reference). Here is the format, (and my profile) and please, don't me modest with your achievements; be very specific as well.\r\n\r\n[b]Grade:[/b] 10\r\n[b]Age:[/b] 15\r\n[b]Math Class Currently in:[/b] Calculus BC\r\n[b]Weakest Math Area:[/b] Geometry\r\n[b]Favorite Math Area:[/b] Probability\r\n[b]Chapter Mathcounts (tell ranking, score): [/b]too low to be ranked, 30\r\n[b]State Mathcounts(same as above): [/b]didn't participate\r\n[b]National Mathcounts(same as above):[/b] didn't participate\r\n[b]AMC 10 [i]Highest [/i]Score:[/b] 118\r\n[b]AMC 12 [i]Highest [/i]Score:[/b] 105\r\n[b]AIME [i]Highest [/i]Score:[/b] 2\r\n[b]USAMO [i]Highest [/i]Score: [/b]didn't participate\r\n[b]Other Achievements (list): [/b]qualified for MMPC (Michigan Math Prize Competition) Part 2", "Solution_1": "Grade: 9\r\nAge: 14\r\nMath Class Currently in: Trig/Analysis Honors\r\nWeakest Math Area: dunno\r\nFavorite Math Area: Probability or Algebra\r\nChapter Mathcounts (tell ranking, score): 3, don't remember\r\nState Mathcounts(same as above): 2, don't remember\r\nNational Mathcounts(same as above): 3 written, 40 or 41\r\nAMC 10 Highest Score: 13_ don't remember last digit\r\nAMC 12 Highest Score: 11_ same as above\r\nAIME Highest Score: 7\r\nUSAMO Highest Score: 4", "Solution_2": "[b]Grade:[/b] 11\r\n[b]Age:[/b] 15\r\n[b]Math Class Currently in:[/b] Calculus BC and Statistics (in school); Intermediate Independent Study, Intermediate Counting and Intermediate Trig (on AoPS)\r\n[b]Weakest Math Area:[/b] Number Theory \r\n[b]Favorite Math Area:[/b] Geometry (as of now)\r\n[b]Chapter Mathcounts: [/b]NA (never even did MC)\r\n[b]State Mathcounts: [/b]NA\r\n[b]National Mathcounts:[/b] NA\r\n[b]AMC 10 [i]Highest [/i]Score:[/b] can't remember 80-90 something\r\n[b]AMC 12 [i]Highest [/i]Score:[/b] 106.5\r\n[b]AIME [i]Highest [/i]Score:[/b] 1 :blush: \r\n[b]USAMO [i]Highest [/i]Score: [/b]NA\r\n[b]Other Achievements (list): [/b]SC ARML team(All State Team); placed 2nd and 3rd at a bunch of statewide competitions", "Solution_3": "[b]Grade:[/b] 9\r\n[b]Age:[/b] turned 14 today :) \r\n[b]Math Class Currently In:[/b] Pre-AP Alg. II\r\n[b]Weakest Math Area:[/b] don't know\r\n[b]Favorite Math Area:[/b] don't know\r\n[b]Chapter Mathcounts:[/b] 7th Grade: 33, rank: 23\r\n 8th Grade: 31? (guess) rank: don't know\r\n[b]State Mathcounts:[/b] 7th Grade: 37-38, Rank: 31\r\n 8th Grade: 33, Rank: 51 or 55, don't remember\r\n[b]National Mathcounts:[/b]N/A\r\nI have not competed in AMC\r\n[b]Other Achievements:[/b]can't think of any", "Solution_4": "Grade: 7\r\nAge: 12\r\nCurrent math class: Algebra 1\r\nWeakest Math Area: Geometry and Number Theory\r\nFavorite Math Area: I like everything, even my weak spots, I really can't decide\r\nChapter Mathcounts (tell ranking, score): 24th place, 34 points\r\nState Mathcounts(same as above): N/A\r\nNational Mathcounts(same as above): N/A\r\nAMC 10 Highest Score: N/A\r\nAMC 12 Highest Score: N/A\r\nAIME Highest Score: N/A\r\nUSAMO Highest Score: N/A\r\nOther Achievements (list): none that I can think of\r\n[EDIT] Whoops...I didn't see the 'if you are a [b]high school[/b] student...'", "Solution_5": "Grade: 9\r\nAge: 14 \r\nCurrent math class: Algebra II\r\nWeakest Math Area: Number Theory\r\nFavorite Math Area: Counting and Probability\r\nChapter Mathcounts (tell ranking, score): 7th-3rd(42 or 43)\r\nState Mathcounts(same as above): N/A 7th-6th(40 or something)\r\nNational Mathcounts(same as above): N/A \r\nAMC 10 Highest Score: N/A \r\nAMC 12 Highest Score: N/A \r\nAIME Highest Score: N/A \r\nUSAMO Highest Score: N/A", "Solution_6": "[quote=\"LynnelleYe\"][EDIT] Whoops...I didn't see the 'if you are a [b]high school[/b] student...'[/quote]\r\n\r\nOh no! You've violated a huge rule! Now we will have to forever brand you as an outcast on AoPS and make a giant cyber-bird peck your brains out every daaaaaay!!!!! :D ;) \r\n\r\nGrade-9\r\nAge-14 and some even months\r\nMath Class Currently in: Honors Trig\r\nWeakest Math Area: Your guess is as good as mine \r\nFavorite Math Area: Easy problems that I can do in less than a minute\r\nChapter Mathcounts (tell ranking, score): 7th grade- 3, 40 something. 8th grade-1,44.\r\nState Mathcounts(same as above): 7th-3, 39. 8th- 3, 40. \r\nNational Mathcounts(same as above): 7th- 69, 33. 8th- 38, 33?\r\nAMC 10 Highest Score: 7th grade- 120.5 (B)\r\nAMC 12 Highest Score: Nope \r\nAIME Highest Score: 7th grade-3 \r\nUSAMO Highest Score: guess\r\nOther Achievements: I won a no-talking contest with someone on my 2003 Mathcounts team. Ah, good times. :D", "Solution_7": "[b]Grade:[/b] 10\r\n[b]Age:[/b] 15 (16 on the 19th)\r\n[b]Math Class Currently in:[/b] Math Analysis (basically Algebra II/Trig)\r\n[b]Weakest Math Area:[/b] Probably Probability\r\n[b]Favorite Math Area:[/b] Geometry\r\n[b]Chapter Mathcounts (tell ranking, score): [/b]7th Grade: 2nd (forgot my score, I'm afraid; that'll be something of a theme); 8th Grade: 1st\r\n[b]State Mathcounts(same as above): [/b]7th Grade: 18th or something like that; 8th Grade: 2nd, 44\r\n[b]National Mathcounts(same as above):[/b] 8th Grade: 138th\r\n[b]AMC 10 [i]Highest [/i]Score:[/b] N/A\r\n[b]AMC 12 [i]Highest [/i]Score:[/b] N/A\r\n[b]AIME [i]Highest [/i]Score:[/b] N/A\r\n[b]USAMO [i]Highest [/i]Score: [/b]N/A\r\n[b]Other Achievements (list): [/b]5th Place in the 2004 National Mathematics High School Championship", "Solution_8": "Grade: 11\r\nAge: 16\r\nMath Class Currently In: teaching myself mutivariable calc and linear algebra, without much success, though\r\nWeakest Math: proving stuff\r\nFavourite Math Thingy: Probably calculus, for some reason\r\nMathcounts: none\r\nAMC 12: 116\r\nAIME: 5\r\nUSAMO: not yet :-/\r\nOther: Nope.", "Solution_9": "How fun...does that imply my brains grow back? Like, you know, that Greek god person's liver? LOLOLOL\r\n\r\nAnyway, what scores did you guys get for [i]6th grade[/i] MathCounts? Just curious...", "Solution_10": "You mean Prometheus?\r\n\r\nWasn't in MathCounts in 6th Grade, because I moved halfway through the year and my new school had already started the MathCounts season.", "Solution_11": "I think you are talking about Tityus for the greek myth reference...but I m sure that's besides the point...", "Solution_12": "[b]Grade:[/b] 9\r\n[b]Age:[/b] 14\r\n[b]Math Class Currently in:[/b] Algebra II Honors(next semester...due to block scheduling, i have no maths this semester. *tearz*)\r\n[b]Weakest Math Area:[/b] i dunno\r\n[b]Favorite Math Area:[/b] algebra i guess\r\n[b]Chapter Mathcounts (tell ranking, score):[/b] 1st, 37 i believe\r\n[b]State Mathcounts(same as above):[/b] 15th, 36 or 37 ish\r\n[b]National Mathcounts(same as above):[/b] n/a\r\n[b]AMC 10 Highest Score:[/b] n/a\r\n[b]AMC 12 Highest Score:[/b] n/a\r\n[b]AIME Highest Score:[/b] n/a\r\n[b]USAMO Highest Score:[/b] n/a\r\n[b]Other Achievements (list):[/b] can't think of any atm", "Solution_13": "i didn't do mc in 6th grade", "Solution_14": "[b]Grade:[/b] 10 \r\n[b]Age:[/b] 15 \r\n[b]Math Class Currently in:[/b] H Math Analysis \r\n[b]Weakest Math Area: [/b]Geometry \r\n[b]Favorite Math Area:[/b] Number Theory\r\n[b]Chapter Mathcounts (tell ranking, score):[/b] 22nd, 38\r\n[b]State Mathcounts(same as above):[/b] didn't participate \r\n[b]National Mathcounts(same as above): [/b]didn't participate \r\n[b]AMC 10 Highest Score:[/b] N/A\r\n[b]AMC 12 Highest Score:[/b] 103 \r\n[b]AIME Highest Score:[/b] 2 \r\n[b]USAMO Highest Score: [/b]didn't participate [/b]", "Solution_15": "Yeah, I was kinda talking about Prometheus. Probably because my English class read the story and went really in-depth about it.", "Solution_16": "Grade: 11\r\nAge: 16\r\nMath Class Currently in: Calculus BC/IB Calc HL (school), Modern Algebra (local U.)\r\nWeakest Math Area: Geometry\r\nFavorite Math Area: Analytic Number Theory\r\nChapter Mathcounts: 1st place (8th grade)\r\nState Mathcounts: 4th place (8th grade)\r\nNational Mathcounts: 63rd place (8th grade):?\r\nAMC 10 Highest Score: 150 (9th)\r\nAMC 12 Highest Score: 124 (10th):?\r\nAIME Highest Score: 10 (10th)\r\nUSAMO Highest Score: 17 (10th grade)\r\nOther Achievements (list): Made MOP", "Solution_17": "Aha! Another analytic number theorist!", "Solution_18": "Yeh, there are just way too many of em. Half of the Princeton staff does that. Seriously, when I visited and went to the Math Building, every single person I met was an analytic number theorist.", "Solution_19": "Grade: 12\r\nAge: 17\r\nMath Class Currently in: Indpt Study through University of Illinois-Chicago in group theory. \r\nWeakest Math Area: Geometry \r\nFavorite Math Area: Number Theory.\r\nChapter Mathcounts (tell ranking, score): never did\r\nState Mathcounts(same as above): never did\r\nNational Mathcounts(same as above): never did\r\nAMC 10 Highest Score: Don't remember. Not good.\r\nAMC 12 Highest Score: 100.5\r\nAIME Highest Score: 4\r\nUSAMO Highest Score: never did\r\nOther Achievements (list): Solved the Riemann Hypothesis at the age of 6. Ok, maybe not. But it'd be pretty cool if I did.", "Solution_20": "As some of our peers might say:\r\n\r\n\"Analytic number theory is the new black...\"", "Solution_21": "Who does analytic number theory at Princeton? I guess Sarnak does, but who else?", "Solution_22": "The people I spoke to included only two that I actually remember distinctly. One was Jordan Ellenburg, who professed to being one. The other was a graduate student who was about to present his thesis in a week. He said that his area was anayltic number theory and that most of the people there are the same (He showed me on the board with all of the professors what some people did).", "Solution_23": "I see. Around here, practically everyone is either a ring theorist or a topologist. (Well, okay, there are some people at this university who aren't mathematicians, but we'll ignore them.)", "Solution_24": "[b]Grade:[/b] 12 \r\n[b]Age:[/b] 17\r\n[b]Math Class Currently in:[/b] Self learning real analysis\r\n[b]Weakest Math Area:[/b] geometry (I don't think I'm alone here)\r\n[b]Favorite Math Area:[/b] Number Theory\r\n[b]Chapter Mathcounts (tell ranking, score):[/b] 5th, too long ago to remember...\r\n[b]State Mathcounts(same as above):[/b] 13th, ditto\r\n[b]AMC 12 [i]Highest[/i] Score:[/b] 133\r\n[b]AIME [i]Highest[/i] Score:[/b] 5\r\n[b]Other Achievements (list):[/b] MN ARML Gold and Math Bowl! w00t!", "Solution_25": "Grade 9\r\nAge 14 (15 next month)\r\nCurrent Math Course: Algebra II\r\nWeakest Area: Counting/Probablility\r\nFavorite Area: Geometry\r\nHighest AMC 12 score: 97\r\nnever did mathcount, AIME, and USAMO\r\n\r\nNo other achievements", "Solution_26": "Grade: 12\r\nAge: 16\r\nMath Class Currently in: Linear Alg/ Multivar Calc / other topics \r\nWeakest Math Area: hopefully I've eliminated them by now\r\nFavorite Math Area: Geometry\r\nChapter Mathcounts (tell ranking, score): didn't participate\r\nState Mathcounts(same as above): didn't participate\r\nNational Mathcounts(same as above): didn't participate\r\nAMC 10 Highest Score: don't remember haven't taken it since 9th gr\r\nAMC 12 Highest Score: 132.5 (2004)\r\nAIME Highest Score: 8 (2004)\r\nUSAMO Highest Score: 22 (in 2004)", "Solution_27": "Grade: 11 \r\nAge: 16 \r\nMath Class Currently in: Geometry (going through Geometry Revisited by Coxeter and Greitzer)\r\nWeakest Math Area: Geometry \r\nFavorite Math Area: Number Theory, Algebra, Combinatorics, Discrete are all tied \r\nChapter Mathcounts: 7th-1st, 39 pts; 8th-1st, 43\r\nState Mathcounts: 7th-74th, 16 pts; 8th- 24th, 35 pts\r\nNational Mathcounts: didn't participate \r\nAMC 10 Highest Score: 131\r\nAMC 12 Highest Score: 132.5 \r\nAIME Highest Score: 8\r\nUSAMO Highest Score: 9 \r\nOther Achievements (list): MOP '04", "Solution_28": "Grade: 11\r\nAge: 16\r\nMath Class Currently in: College Course combining Linear Algebra, Multi-D Calc and Differential Equations\r\nWeakest Math Area: don't know \r\nFavorite Math Area: Topology/Knot theory\r\nChapter Mathcounts (tell ranking, score): 7th-9th, don't remember the score, 8th-2nd, 43\r\nState Mathcounts(same as above): 7th-too low, 8th-13th, don't remember the score.\r\nNational Mathcounts(same as above): didn't participate\r\nAMC 10 Highest Score: 150\r\nAMC 12 Highest Score: 130\r\nAIME Highest Score: 5\r\nUSAMO Highest Score: didn't participate\r\nOther Achievements (list): Qualified for the APMO", "Solution_29": "Age: 17, almost 18\r\nsenior\r\nmath studies (self studied, and self studying): \r\nself studied calculus, passed with 5 (11th)\r\nmultivariate calculus (11th)\r\nnumber theory, fractal stuff, complex analysis (12th)\r\n\r\nNote these are practice tests, I've never taken them in a testing environment, because my school doesnt offer these (i'll take them somewhere else though, this year)\r\n\r\namc12 highest: around 22 or 23 correct\r\naime highest: 10-11 correct\r\nusamo: never taken this as a practice test", "Solution_30": "Grade: 12\r\nAge: barely 18\r\nMath Class Currently in: none \r\nWeakest Math Area: \r\nFavorite Math Area: Discrete Math \r\nChapter Mathcounts (tell ranking, score): did not participate \r\nState Mathcounts(same as above): did not participate \r\nNational Mathcounts(same as above): did not participate \r\nAMC 10 Highest Score: never taken \r\nAMC 12 Highest Score: 124?\r\nAIME Highest Score: 7?\r\nUSAMO Highest Score: did not participate \r\nOther Achievements (list): learned virtually all I know by 8th grade (intense algebra, trig, geometry, calculus, logic, games, etc... mostly precalculus stuff though.)", "Solution_31": "Grade: 11\r\nAge: 15\r\nMath Class Currently in: AP Calculus AB (BC is not offered at my school, how sad...-_-')\r\nWeakest Math Area: Geometry\r\nFavorite Math Area: interesting problems, weird ones even\r\nChapter Mathcounts: 2\r\nState Mathcounts: 4\r\nNational Mathcounts: 42\r\nAMC 10 Highest Score: 146.5 (guess what I forgot: how to calculate the area of a parallelogram given base and height, yep)\r\nAIME Highest Score: 6 or 7\r\nOther Achievements (list): top assassin in the Verbal Assassin game :cool:", "Solution_32": "Grade: 12 \r\nAge: 17 \r\nMath Class Currently in: None (will take Enriched Math Topics a.k.a. Calc BC next semester, though) \r\nWeakest Math Area: Number Theory \r\nFavorite Math Area: Number Theory \r\nChapter Mathcounts: Didn't Participate \r\nState Mathcounts: didn't participate \r\nNational Mathcounts: Didn't participate \r\nAMC 10 Highest Score: Didn't participate (did the AMC 12 in 10th grade) \r\nAMC 12 Highest Score: 99.5 (Aargh! So close!) \r\nAIME Highest Score: Didn't participate \r\nUSAMO Highest Score: Didn't participate \r\nOther Achievements: 9th in state at 2003-4 Minnesota H.S. Math League Team Competition", "Solution_33": "What if you're not in high school :P ?\r\n\r\nGrade: not telling\r\nAge: not telling\r\nMath Class Currently in: Calculus BC \r\nWeakest Math Area: Vectors \r\nFavorite Math Area: Probably counting, or maybe adding single-digit numbers\r\nChapter Mathcounts: 1st, 45\r\nState Mathcounts: 4th, 41 \r\nNational Mathcounts: 36th, unknown \r\nAMC 10 Highest Score: didn't take it\r\nAMC 12 Highest Score: didn't take it\r\nAIME Highest Score: didn't take it\r\nUSAMO Highest Score: didn't take it \r\nOther Achievements (list): Probably three or four years younger than most of you, placed 1st in NJ math league, placed 2nd in NJ math league, and I can count to ten on my fingers :D !", "Solution_34": "Grade: 10\r\nAge: 16\r\nMath Class Currently in: Alg II/Trig AL\r\nWeakest Math Area: Probability\r\nFavorite Math Area: Number Theory\r\nChapter Mathcounts (tell ranking, score): didn't participate\r\nState Mathcounts(same as above): didn't participate\r\nNational Mathcounts(same as above): didn't participate\r\nAMC 10 Highest Score: 125 \r\nAMC 12 Highest Score: didn't take it in 9th grade\r\nAIME Highest Score: 4 (Totally would have gotten five if I had remembered about that one single cube left to make it a complete cube on AIME 2004)...\r\nUSAMO Highest Score: didn't participate\r\nOther Achievements (list): None that I can think of...I got a 6-7 trophies at MAO as a freshmen, which was pretty cool.", "Solution_35": "Grade: 9\r\nAge: 13\r\nMath Class Currently in: Calculus BC \r\nWeakest Math Area: Dunno\r\nFavorite Math Area: Dunno\r\nChapter Mathcounts: 7th?\r\nState Mathcounts: 4th.\r\nNational Mathcounts: 6th\r\nAMC 10 Highest Score: 146.5\r\nAMC 12 Highest Score: N/A\r\nAIME Highest Score: 11\r\nUSAMO Highest Score: 7\r\nOther Achievements (list): I'll come up with them later.", "Solution_36": "Grade: 10\r\nAge: 15 \r\nMath Class Currently in: Calculus AB\r\nWeakest Math Area: State MathCounts and AMC tests\r\nFavorite Math Area: I like prime numbers and really hard stuff\r\nChapter Mathcounts: 2nd (4 times...)\r\nState Mathcounts: 36th, 40th :twisted: I was angry at myself for a while..\r\nNational Mathcounts: N.A. \r\nAMC 10 Highest Score: 119 (in 8th grade) (so close...)\r\nAMC 12 Highest Score: 99.5 (in 9th grade) (so close...)\r\nAIME Highest Score: N.A.\r\nUSAMO Highest Score: N.A. \r\nOther Achievements (list): \r\n-I have a row of MathCounts trophies in my bedroom\r\n-17th in MN Math League last year \r\n-ARML! :lol: \r\n-25 w/ 1 commended solution in USAMTS Year 16 Round 1 :lol:", "Solution_37": "[quote]\nNational Mathcounts: 6th \n[/quote]\r\n\r\nWas that this year's competition? If it was... *looks you up in my MathCounts paper thingy showing everybody who competed*", "Solution_38": "Grade: 12\r\nAge: 17\r\nMath Class Currently In: Honor's Mathematics I 295, Problem Solving Seminar 285 (both at U of M)\r\nWeakest Math: advanced geometry\r\nFavourite Math Thingy: number theory\r\nMathcounts: 3rd in region, choked at states\r\nAMC 10: 124 freshmen year\r\nAIME: 1 freshmen year \r\nMy school stopped doing AMC stuff after that, hopefully we'll do it for my senior year\r\nOther: MMPC; 9th grade, 94thish. 10th grade, 46thish, 11th grade 2nd. Two year member of Michigan ARML team. Honorable Mention Lawrence Tech High School Math Competition." } { "Tag": [ "quadratics", "complex analysis", "number theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Hi \r\n\r\nis there an elementary proof that -2 is a square mod p\r\nwhen p is a prime 3 mod 8\r\n\r\nI mean with elementary, that stuff like little fermat is allowed, but not \r\nsomething like quadratic reciprocity\r\n\r\n\r\nthx", "Solution_1": "At least, also the things you named are elementary (not using deeper algebra or complex analysis etc.) and why didn't you post it to the Number Theory section when you want such a solution\u00bf\r\nBut the technique used in http://www.mathlinks.ro/Forum/viewtopic.php?t=23681&highlight= should work.", "Solution_2": "Thanks\r\n\r\nsorry for the misplacement\r\nI am also having trouble deciding whether something should go in college playground or advanced \r\n\r\nhowever, I am familiar with that criterion, it is used in leveque to actually prove reciprocity\r\n\r\nBut for p=1 mod 8 there is a trick : make p=8*v+1\r\n\r\nthen (x^(4*v)+1)*(x^(4*v)-1)=0 holds for all nonzero x\r\n\r\nso there are x for which x^(4*v)+1=0 holds\r\n\r\nor with x^(v)=y\r\n\r\n(y+1/y)^2=2\r\n\r\nis something like this possible here?" } { "Tag": [ "algebra", "polynomial", "induction", "number theory proposed", "number theory" ], "Problem": "Let $a_1,...,a_k$ algebraic integers and $S_n=a_1^n+...+a_k^n$. Prove that $n$ divides $\\sum_{d|n}{ S_d f(\\frac{n}{d})}$ where $f$ is the Moebius function. This generalizes many classical theorems, so it is of interest, I think.", "Solution_1": "so your statement means that $a_1 a_2...a_k$ are the roots of a monic integer polynomial? (all of the same polynomial?????)", "Solution_2": "Harazi if i am not wrong $p^t|a(a^{\\phi(p^t}-1)$ also holds for the ring of algebraic integers.....(btw i cant remember the proof of this fact) using this property the problem becomes very easy!", "Solution_3": "Mayn I see how it is so easy, Pascual? :?", "Solution_4": "Honestly, I don't have a clue what you actually mean, Pascual :huh:", "Solution_5": "i will ppost it this afternoon, i promess...but maybe i am a completely fool and it is just nonsense :(", "Solution_6": "Well, after you post a complete solution probably we'll see. ;)", "Solution_7": "well I dont now if this is true but i am almost sure that $p^t|a^{p^t}-a^{p^{t-1}}$ int he ring of algebriac integers... So from here what can we get? just wathc separately the sum as $\\sum_{d|n}a^df(n/d)$ now iff $p^t|n$ and $p^{t-1}|d$ we have that $a^d$ and $a^{dp}$ are both with coefficient $0$ or with different sign. but from the lemma $p^t|a^{dp}-a^{d}$ so n divides our sum in the ring of algebraic integers. Finally if $a_1, a_2...,a_k$ are associates, aour sum is a integer and therefore it is an integer multiple of $p$ since non integer rationals are not algebraic integers. (i dont know if i am just saying true facts or all of them are just an abuse of properties for integers) :blush:", "Solution_8": "I'm afraid my knowledge of Number Theory is not deep enough to judge this proof :blush:", "Solution_9": "mine neither, thats why i am not going to try and convince you it is true...", "Solution_10": "wella ctually now i am starting to believe my result is a fake and that the only useful thing are ideas as in this problem! [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5452[/url]", "Solution_11": "I am sorry, I am dumb, but I really don't understand your solution. :(", "Solution_12": "ok sorry:\r\n\r\n1. Lets assume what i say is in fact true, Namely for an algebraic integer $a$, there exist an algebraic integer $b$ such that $a^{p^t}-a^{p^{t-1}}=p^tb$ (this porperty holds if it holds for $t=1$ by induction on $t$).\r\n\r\nIf this property is true then we can keep going, otherwise what follows only works for integers.... ;) \r\n\r\n2. we have: $\\sum_{d|n}S_df(n/d)=\\sum_{i=1}^{k}\\sum_{d|n}a_i^df(n/d)$ so we will prove $n|\\sum_{d|n}a_i^df(n/d)$ (in the field of algebraic integers, i mean the quotient is an algebraic integer).\r\n\r\n3. The sum we want to analize has many cero terms, so we only deal with noncero terms, if $n=p_1^{a_1}p_2^{a_2}....p_l^{a_l}$ then $f(n/d)$ is different from $0$ only iff $n/d$ is square free... so $d$ must be divisible by $p_i^{a_i-1}$ for every prime divisor of $n$, call those $d$ \"special divisors\"\r\n\r\n4. Now we prove $p_1^{a_1}$ divides our sum. Group the special divisors in the following way: take a special divisor $d$ such that $p_1^{a_1}$ doesnt divide it and group it with $dp$ it is clear that $f(n/d)=-f(n/dp)$. But also all termas are paired and by the property in (1) We have that all those pairs are multiples of $p^t$ since they are of the form $a^{dp}-a^d=a'^{p^t}-a'^{p^{t-1}}$. So their sum is also divisible by $p^t$ (in the RING of algebraic integers).\r\n\r\n5. Finally we also prove $p_i^{a_i}$ divides the sum, and therefore $n$ divides it in the ring of algebraic integers...\r\n\r\n6. Now if all of the algebraic integers given are associates (Same minimal polynomial) then our sum is an integer, and therefore $n$ would divide it in the integers, since none of the quotients is rational! (rationals are not algebraic integers!)\r\n\r\nFinal Note: well if our algebraic integers are asociates, then we can also consider their minimal polynomial $P(x)=x^m-\\sum x^tb_t$ with $b_t$ integers. and we get $S_{m+h}=\\sum S_{t+h}b_t$ so the sequence $S_n$ satisfies a linear recursion. So it is clear that for a prime power not dividing the product of the roots, this sequence is purely periodic mod $p^t$, so we need somehow that $S_{t+n}-S_n$ to be a multiple of $p^t$ for a fixed $t$ and all $n$. We can also reduce $P(x)$ modulo $p$ and work with it or something like that....my guess is that $t=\\phi(p^t)$ wich would also give a similar solution grouping $S_{dp}-S_d$...", "Solution_13": "Yes it is right!!!!! So we get that $a^p-a=pb$. Now suppose $a^{p^t}-a^{p^{t-1}}=p^tb$ then we also have:\r\n$a^{p^t}=a^{p^{t-1}}+p^tb$ and raising both sides to the p-th power we get:\r\n\r\n$a^{p^{t+1}}=a^{p^{t}}+Kp^{t+1}$ because $p|\\binom{p}{k}$. So property $1$ follows!\r\n\r\nEdit: a post by Harazi dissapeared where he sayed that $(1+a-1)^p-(1+a-1)$ is a multiple of $p$ because all multinomial coefficients are, so from here it follows that $p|a^p-a$, Nice!", "Solution_14": "No, I realized immeidately the mistake. It remains the number $\\frac{(1+a-1)^p-(1+a-1)}{p}$. How can we guarantee that it is algebraic integer? Meanwhile, I think your proof is wrong, since the number $\\frac{(1+\\sqrt{2})^3-(1+\\sqrt{2})}{3}$ is not algebraic integer. Or have I completely lost my mind?", "Solution_15": ":(", "Solution_16": "It is not difficult to see that we only need to consider $n$ being a prime power. Suppose for example that we have $n = p^2 q^3$, then we need to show $n | S_{p^2q^3} - S_{p^2q^2} - S_{pq^3} + S_{pq^2}$. But $S_{p^2q^3} - S_{p^2q^2}$ and $S_{pq^3} - S_{pq^2}$ are both divisible by $q^3$ by considering the prime power cases $n' = q^3, a_i' = a_i^{p^2}$ and $a_i'' = a_i^{p}$. Thus $q^3 | \\sum$ and similarly $p^2 | \\sum$ once we prove the prime power cases. \n\nNow let $n = p^m$ be a prime power, the problem becomes $p^m | \\sum_{i = 1}^{k} (a_i^{p^m} - a_i^{p^{m-1}})$. We use induction on $k$ to show something stronger, that for any $k$-variable symmetric integer polynomial $F$, it holds that $p^m | F(a_1^{p^m}, \\dots, a_k^{p^m}) - F(a_1^{p^{m-1}}, \\dots, a_k^{p^{m-1}}).$ The original problem corresponds to $F(x_1, x_2, \\dots, x_k) = x_1 + x_2 + \\dots + x_k$. Now recall a well-known lemma:\n[b]Lemma.[/b] Suppose $F$ is a symmetric integer polynomial, then $F(x_1^p, x_2^p, \\dots, x_k^p) = F(x_1, x_2, \\dots, x_k)^p + p \\cdot G(x_1, x_2, \\dots, x_k)$ with $G$ also a symmetric integer polynomial. \n\nThe proof of this lemma is simple: just expand $F(x_1, x_2, \\dots, x_k)^p$ and notice that all but few multinomial coefficients are divisible by $p$. \n\nGoing back to our claim, the base case $m = 1$ is immediately settled by the above lemma because $F(a_1^p, \\dots, a_k^p) - F(a_1, \\dots, a_k)^p = p \\cdot G(a_1, \\dots, a_k)$. Since $G(a_1, \\dots, a_k)$, $F(a_1, \\dots, a_k)$ and $F(a_1^p, \\dots, a_k^p)$ are integers (by the fundamental theorem of symmetric polynomials), we deduce by Fermat's little theorem that $F(a_1^p, \\dots, a_k^p) \\equiv F(a_1,\\dots, a_k)^p \\equiv F(a_1, \\dots, a_k) ~(mod~p)$. Assume we have proved the result for $m$, then by lemma we can write \n$$F(a_1^{p^{m+1}}, \\dots, a_k^{p^{m+1}}) - F(a_1^{p^{m}}, \\dots, a_k^{p^{m}}) = \\left[F(a_1^{p^m}, \\dots, a_k^{p^m})^p - F(a_1^{p^{m-1}}, \\dots, a_k^{p^{m-1}})^p\\right] + p \\cdot \\left[ G(a_1^{p^m}, \\dots, a_k^{p^m}) - G(a_1^{p^{m-1}}, \\dots, a_k^{p^{m-1}})\\right].$$\nBy induction hypothesis, $F(a_1^{p^m}, \\dots, a_k^{p^m}) \\equiv F(a_1^{p^{m-1}}, \\dots, a_k^{p^{m-1}}) ~(mod~p^m)$. Thus we routinely have $F(a_1^{p^m}, \\dots, a_k^{p^m})^p \\equiv F(a_1^{p^{m-1}}, \\dots, a_k^{p^{m-1}})^p ~(mod~p^{m+1})$. Also by induction hypothesis we have $p^{m+1} | p \\cdot \\left[ G(a_1^{p^m}, \\dots, a_k^{p^m}) - G(a_1^{p^{m-1}}, \\dots, a_k^{p^{m-1}})\\right]$. The induction step is thus completed. \n", "Solution_17": "[hide=Maybe]It might not hurt to consider Dirichlet Convolution with the mobius function and a function giving the $n$th newton sum of a polynomial. [/hide]" } { "Tag": [ "conics", "ellipse", "analytic geometry", "number theory solved", "number theory" ], "Problem": "The problem is: Find all pairs of positive rational numbers such that x^2 + 3y^2 = 1.\r\n\r\nWhat I did was, I put x = a/b and y = c/d, got b = d = 2k and\r\na^2 + 3c^2 = 4k^2 where gcd(a,c) = gcd(k,c) = gcd(a,k) = 1\r\n\r\nMaybe I'm in the wrong way or maybe I'm just missing something after a hard schoolweek, but I have no idea how to go furter, maybe some hints ? (I'm not expecting a full solutions, It would be great if someone could give me just a helping hint)", "Solution_1": "here is what you can find useful!! :D \r\n\r\nhttp://www-gap.dcs.st-and.ac.uk/~history/HistTopics/Pell.html\r\n\r\ncheers!", "Solution_2": "[quote=\"J\"]The problem is: Find all pairs of positive rational numbers such that x^2 + 3y^2 = 1.\n\nWhat I did was, I put x = a/b and y = c/d, got b = d = 2k and\na^2 + 3c^2 = 4k^2 where gcd(a,c) = gcd(k,c) = gcd(a,k) = 1\n\nMaybe I'm in the wrong way or maybe I'm just missing something after a hard schoolweek, but I have no idea how to go furter, maybe some hints ? (I'm not expecting a full solutions, It would be great if someone could give me just a helping hint)[/quote]\r\n=============\r\nWe can use technics of \"cuts\" to solve this problem. First, we reform the problem to: Find all rational points on the ellipse (E): x^2 + 3y^2=1 (1). Obviously, A(1,0) is one solution. Now, we take arbitrary point B (0, r) in Oy and find the second intersection point of (AB) with. By Viet theorem, coordinates of this point will be rational and we get the solution of (1). Inversely, if we have rational point C on (E) then AC meets Oy at rational poin. Answer: x = (3r^2-1)/(3r^2+1)\r\n y = 2r/(3r^2+1) \r\nwhere r is arbitrary rational number.\r\n\r\nHave joy.\r\n\r\nNamdung" } { "Tag": [ "inequalities", "geometry", "circumcircle", "geometric inequality", "inequalities proposed" ], "Problem": "Hello guys I'm stuck in this geometric inequality , I posted this problem in the Highschool section but seems like nobody can prove it geometrically , so I hope that someone here give another type of solution , here it is\r\n\r\n[i]Show that in any acute-angled triangle there are two sides whose sum is always greater than the sum of the diameters of the incircle and circumcircle . [/i]\r\n\r\nThis means that we need to prove that one of these statements is an assertion:\r\n\r\n$ a \\plus{} c > 2(R \\plus{} r) \\ , \\ b \\plus{} c > 2(R \\plus{} r)$ or $ a \\plus{} b > 2(R \\plus{} r)$", "Solution_1": "hello, i think, this means\r\n$ a\\plus{}c>2(R\\plus{}r)$ or $ b\\plus{}c>2(R\\plus{}r)$ or $ a\\plus{}b>2(R\\plus{}r)$.\r\nSonnhard.", "Solution_2": "I am really stuck in this problem :( could someone give a hint or something ?" } { "Tag": [ "inequalities", "geometry", "circumcircle", "incenter", "inequalities proposed" ], "Problem": "Let a,b,c are positive numbers. Prove that\r\n$ 4a^2b^2c^2\\geq (a^3\\plus{}b^3\\plus{}c^3\\plus{}abc)(a\\plus{}b\\minus{}c)(b\\plus{}c\\minus{}a)(c\\plus{}a\\minus{}b)$ :D", "Solution_1": "we can assume that $ a,b,c$ are the side lengths of a triangle (otherwise RHS would be negative and LHS would be positive so the inequality becomes trivial),so let $ a,b,c$ be the sidelength of triangle $ \\triangle ABC$,and let $ p,r,R$ be the semiperimeter,inradus and circumradius of this triangle respectively,now we know that:\r\n\r\n$ abc\\equal{}4Rrp,a^3\\plus{}b^3\\plus{}c^3\\equal{}2p(p^2\\minus{}3r^2\\minus{}6Rr),(p\\minus{}a)(p\\minus{}b)(p\\minus{}c)\\equal{}pr^2$\r\n\r\nso after simplifying,our inequality will be equivalent to:\r\n\r\n$ 4R^2\\plus{}4Rr\\plus{}3r^2\\geq p^2$\r\n\r\nwhich is equivalent to:\r\n\r\n$ \\overline{HI}^2\\geq 0$\r\n\r\nwhich is obviously true...\r\n\r\n(note that $ H,I$ are the orthocenter and incenter of $ \\triangle ABC$ respectively)", "Solution_2": "[quote=\"BaBaK Ghalebi\"]we can assume that $ a,b,c$ are the side lengths of a triangle (otherwise RHS would be negative and LHS would be positive so the inequality becomes trivial),so let $ a,b,c$ be the sidelength of triangle $ \\triangle ABC$,and let $ p,r,R$ be the semiperimeter,inradus and circumradius of this triangle respectively,now we know that:\n\n$ abc \\equal{} 4Rrp,a^3 \\plus{} b^3 \\plus{} c^3 \\equal{} 2p(p^2 \\minus{} 3r^2 \\minus{} 6Rr),(p \\minus{} a)(p \\minus{} b)(p \\minus{} c) \\equal{} pr^2$\n\nso after simplifying,our inequality will be equivalent to:\n\n$ 4R^2 \\plus{} 4Rr \\plus{} 3r^2\\geq p^2$\n\nwhich is equivalent to:\n\n$ \\overline{HI}^2\\geq 0$\n\nwhich is obviously true...\n\n(note that $ H,I$ are the orthocenter and incenter of $ \\triangle ABC$ respectively)[/quote]\r\n\r\n\r\n\r\n$ (a\\plus{}b\\minus{}c)(a\\plus{}c\\minus{}b)(b\\plus{}c\\minus{}a)>0$ <-> $ a\\plus{}b\\minus{}c>0,a\\plus{}c\\minus{}b<0,b\\plus{}c\\minus{}a<0$? --> a,b,c are triangles sides just mustn't be", "Solution_3": "@zaya_yc:\r\nnote that $ a,b,c$ are positive,now if we had $ a \\plus{} c \\minus{} b < 0$ and $ b \\plus{} c \\minus{} a < 0$ then by adding these two inequalities we get that $ c < 0$ which is a contradiction...\r\n\r\nso we must have $ a \\plus{} b \\minus{} c > 0,b \\plus{} c \\minus{} a > 0,c \\plus{} a \\minus{} b > 0$ i.e. $ a,b,c$ are the side-lengths of a triangle", "Solution_4": "yes, you are right. Congratulation! :D" } { "Tag": [ "quadratics", "modular arithmetic" ], "Problem": "Show that for all $p \\in \\mathbb{P}$, there exists $n \\in \\mathbb{N}$ such that : $6n^2 + 5n + 1 \\equiv 0 \\pmod p$.\r\n\r\n:)", "Solution_1": "[hide=\"One-line solution\"]\n$6n^2 + 5n + 1 = (3n + 1)(2n + 1)$ - the latter clearly cycles through all odd primes and $3n+1$ is even at least once. :)\n[/hide]" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Let the operation $ f$ of $ k$ variables defined on the set $ \\{ 1,2,\\ldots,n \\}$ be called $ \\textit{friendly}$ toward the binary relation $ \\rho$ defined on the same set if \\[ f(a_1,a_2,\\ldots,a_k) \\;\\rho\\ \\;f(b_1,b_2,\\ldots,b_k)\\] implies $ a_i \\; \\rho \\ b_i$ for at least one $ i,1\\leq i \\leq k$. Show that if the operation $ f$ is friendly toward the relations \"equal to\" and \"less than,\" then it is friendly toward all binary relations. \r\n\r\n[i]B. Csakany[/i]", "Solution_1": "The case $n = 1$ is trivial.\n\nWe first look at the case $n \\geq 3$. I will show by induction on $k$ that, if $f$ is friendly toward \"equal to\" and \"less than\", then there exists $i \\in \\{1, \\dots, k\\}$ such that $f(a_1, \\dots, a_k) = a_i$ for any $a_1, \\dots, a_k$.\n\nFor $k = 1$, we have a function $f$ from $\\{ 1, \\dots, n\\}$ to itself, which satisfies $f(a) = f(b) \\implies a = b$ (i.e. $f$ is injective) and $f(a) < f(b) \\implies a < b$ (i.e. $a \\geq b \\implies f(a) \\geq f(b)$, that is, $f$ is non-decreasing). It follows that $f(a) = a$ for all $a \\in \\{1, \\dots, n\\}$.\n\nNow suppose that it's true for $k - 1$ and let's prove the result for $k$.\nWe define $g:(a_1, \\dots, a_{k - 1})\\mapsto f(a_1, \\dots, a_{k - 1}, a_{k - 1})$. It is clear from the definitions that $g$ is friendly toward \"equal to\" and \"less than\". Thus by induction hypothesis, there exists $i \\in \\{1, \\dots, k - 1\\}$ such that $g(a_1, \\dots, a_{k - 1}) = a_i$.\n\nIf $i \\neq k - 1$, then I claim that $f(a_1, \\dots, a_k) = a_i$ for any $a_1, \\dots, a_k$.\nIn fact, for any $a_1, \\dots, a_k$ and any $x \\neq a_i$, we may choose $b_1, \\dots, b_{k - 1}$ such that\n- $b_i = x$;\n- $b_j \\neq a_j$ for any $1 \\leq j < k - 1$;\n- $b_{k - 1} \\neq a_{k - 1}$ or $a_k$ (notice that $n \\geq 3$).\nIt follows that $f(a_1, \\dots, a_k) \\neq f(b_1, \\dots, b_{k - 1}, b_{k - 1}) = b_i = x$. Thus the only possibility is $f(a_1, \\dots, a_k) = a_i$.\n\nIf $i = k - 1$, then the same argument as above shows that, for any $a_1, \\dots, a_k$, we have $f(a_1, \\dots, a_k) = a_{k - 1}$ or $a_k$ (for any $x\\neq a_{k - 1}$ or $a_k$, choose $b_1, \\dots, b_{k - 2}$ such that $b_j \\neq a_j$ and consider $f(b_1, \\dots, b_{k - 2}, x, x)$).\nNow consider the two numbers $f(1, \\dots, 1, 1, 2)$ and $f(2, \\dots, 2, 2, 1)$. They must be different and both are equal to $1$ or $2$. Therefore one of them is equal to $1$.\nWithout loss of generality, we assume in the following that $f(1, \\dots, 1, 1, 2) = 1$.\n\nFor any $a_1, \\dots, a_{k - 2}$, we choose $b_1, \\dots, b_{k - 2}$ such that $a_j \\neq b_j$ and $b_j \\neq 1$ (which is possible because $n \\geq 3$).\nIt follows that $f(b_1, \\dots, b_{k - 2}, 2, 1) = 2$ (because it is not equal to $f(1, \\dots, 1, 1, 2) = 1$) and hence $f(a_1, \\dots, a_{k - 2}, 1, 2) = 1$ (because it is not equal to $f(b_1, \\dots, b_{k - 2}, 2, 1) = 2$).\n\nFor any $a_1, \\dots, a_{k - 1}$ such that $a_{k - 1} \\neq 1$, we choose $b_1, \\dots, b_{k - 2}$ such that $a_j \\neq b_j$.\nIt follows that $f(a_1, \\dots, a_{k - 1}, 1) = a_{k - 1}$, because it is not equal to $f(b_1, \\dots, b_{k - 2}, 1, 2) = 1$.\nThis is also true for $a_{k - 1} = 1$, hence we have $f(a_1, \\dots, a_{k - 1}, 1) = a_{k - 1}$ for any $a_1, \\dots, a_{k - 1}$.\n\nFinally, for any $a_1, \\dots, a_k$ such that $a_k \\neq 1$ or $a_{k - 1}$, we choose $b_1, \\dots, b_{k - 2}$ such that $a_j \\neq b_j$.\nIt follows that $f(a_1, \\dots, a_k) = a_{k - 1}$, because it is not equal to $f(b_1, \\dots, b_{k - 2}, a_k, 1) = a_k$.\nThis is also true for $a_k = 1$ or $a_{k - 1}$, hence we have shown that $f(a_1, \\dots, a_k) = a_{k - 1}$ for any $a_1, \\dots, a_k$.\n\n-----\nIt remains to treat the case $n = 2$.\n\nWe view a relation $\\rho$ as a subset of $\\{1, 2\\}^2$.\nNote that, if $f$ is friendly toward two relations $\\rho_1, \\rho_2$, then $f$ is also friendly toward the union $\\rho_1\\cup \\rho_2$.\nSince every subset of $\\{1, 2\\}^2$ is a union of singletons, it suffices to prove that $f$ is friendly toward each singleton relation.\n\nThere are only four singleton relations:\n- $\\rho_1 = \\{(1, 1)\\}$, i.e. $a\\textrm{ } \\rho_1\\textrm{ } b \\iff a = b = 1$;\n- $\\rho_2 = \\{(2, 2)\\}$, i.e. $a\\textrm{ } \\rho_2\\textrm{ } b \\iff a = b = 2$;\n- $\\rho_3 = \\{(1, 2)\\}$, i.e. $a\\textrm{ } \\rho_3\\textrm{ } b \\iff a < b$;\n- $\\rho_4 = \\{(2, 1)\\}$, i.e. $a \\textrm{ }\\rho_4 \\textrm{ }b \\iff a > b$.\nBy assumption, $f$ is friendly toward $\\rho_3$, and this implies that $f$ is also friendly toward $\\rho_4$.\n\nIt remains to show that $f$ is friendly toward $\\rho_1, \\rho_2$. We prove it for $\\rho_1$, the other proof being identical.\n\nSince $f$ is friendly toward \"equal to\", for any $a_1, \\dots, a_k \\in \\{1, 2\\}$, we have $f(3 - a_1, \\dots, 3 - a_k) = 3 - f(a_1, \\dots, a_k)$.\nThus if $f(a_1, \\dots, a_k)\\textrm{ } \\rho_1\\textrm{ } f(b_1, \\dots, b_k)$, then $$f(a_1, \\dots, a_k) = 1 < 2 = f(3 - b_1, \\dots, 3 - b_k)$$ which implies that there exists $i$ such that $a_i < 3 - b_i$.\nThis is only possible when $a_i = 1$ and $3 - b_i = 2$, or in other words, $a_i\\textrm{ } \\rho_1\\textrm{ } b_i$." } { "Tag": [], "Problem": "$\\frac{1}{1.2.3}+\\frac{1}{4.5.6}+\\frac{1}{7.8.9}+...$", "Solution_1": "$\\frac{1}{1.2.3}+\\frac{1}{4.5.6}+\\frac{1}{7.8.9}+...=\\sum \\frac{1}{k(k+1)(k+2)}=\\sum \\left (\\frac{1}{2}\\left (\\frac{1}{k}-\\frac{1}{k+2}\\right )-\\frac{1}{k+1}+\\frac{1}{k+2}\\right )=...=\\frac{1}{4}$", "Solution_2": "no \r\n\r\n1/(3n(3n-1)(3n-2))", "Solution_3": "We have\r\n\\[\\sum_{k=0}^\\infty \\frac{1}{(3k+1)(3k+2)(3k+3)}= \\sum_{k=0}^\\infty \\left( \\frac{1}{6k+2}+\\frac{1}{6k+6}-\\frac{1}{3k+2}\\right) \\]\r\nNow what?\r\n\r\n(edited)", "Solution_4": "[quote=\"cincodemayo5590\"]We have\n\\[\\sum_{k=0}^\\infty \\frac{1}{(3k+1)(3k+2)(3k+3)}= \\sum_{k=0}^\\infty \\left( \\frac{1}{6k+2}+\\frac{1}{6k+6}-\\frac{1}{3k+2}\\right) \\equiv \\sum_{k=0}^\\infty \\left( \\frac{1}{6k+6}-\\frac{1}{6k+5}\\right) \\]\nNow what?[/quote]\r\nNo, this is not correct. Your result says that the original sum is negative. :wink:\r\n\r\nAlso, I think I have already posted a solution to this problem a while ago.", "Solution_5": "[quote=\"boxedexe\"]\nNo, this is not correct. Your result says that the original sum is negative. :wink:[/quote]\r\nOh, alright. I didn't notice that. :)\r\n\r\nBut still, all reciprocals of the numbers that are $\\equiv 2 \\bmod 6$ will eventually be canceled out by the subtracted terms. I guess it matters \"when\" this happens, i.e. at what summation index.\r\n\r\nSo, is there an easy way to compute the value of this summation?", "Solution_6": "You can do it with calculus...it gets kind of messy.", "Solution_7": "[hide=\"calculus...\"]\nLet $f(x)=\\sum_{k=0}^\\infty \\frac{x^{3k+3}}{(3k+1)(3k+2)(3k+3)}$. Observe that\n\\[f^{(3)}(x)=\\sum_{k=0}^\\infty x^{3k}=\\frac{1}{1-x^{3}}\\]\nThis means our sum is equal to\n\\[\\int_{0}^{1}\\int_{0}^{z}\\int_{0}^{y}\\frac{1}{1-x^{3}}\\ dx \\ dy \\ dz\\]\nTo evaluate this triple integral, use partial fractions (as $(1-x)(1+x+x^{2})=1-x^{3}$) to express $\\int_{0}^{y}\\frac{1}{1-x^{3}}\\ dx$ as a linear combination of $\\ln (1-y)$, $\\ln (1+y+y^{2})$, and $\\arctan \\frac{2y+1}{\\sqrt{3}}$. And that's just the first of three integrals! The others require several integrations by parts, and that's if $\\ln (1+y+y^{2})$ even has nice antiderivatives. If it does, expect answers containing $\\pi$ and $\\ln 3$. Good luck... :rotfl: \n[/hide]", "Solution_8": "[quote=\"scorpius119\"]The others require several integrations by parts, and that's if $\\ln (1+y+y^{2})$ even has nice antiderivatives. [/quote]\r\n\r\n[hide=\"Fortunately...\"] $\\ln (y^{2}+y+1) = \\ln (y-\\omega)+\\ln (y-\\omega^{2})$ [/hide]" } { "Tag": [ "inequalities", "trigonometry" ], "Problem": "for a, b integers such that a+b=1, show that (a+1/a)^2+(b+1/b)^2 is > or equal to 25/2.", "Solution_1": "$a=-1$,$b=2$\r\n\r\n$(-1+\\frac{1}{-1})^{2}+(2+\\frac{1}{2})^{2}=(-2)^{2}+(\\frac{5}{2})^{2}=4+\\frac{25}{4}=\\frac{41}{4}<\\frac{25}{2}$\r\n\r\n :huh:", "Solution_2": "Maybe this problem is limited to $a>0,\\ b>0.$\r\n\r\nFrom $a>0,b>0,a+b=1$, using A.M.-G.M. inequality $2\\sqrt{ab}\\leq a+b=1\\Longleftrightarrow \\frac{1}{ab}\\geq 4$\r\n\r\nThus since $y=x^{2}$ is concavedown, using Jensen's inequality \r\n\r\n${\\Longleftrightarrow \\frac{\\left(a+\\frac{1}{a}\\right)^{2}+\\left(b+\\frac{1}{b}\\right)^{2}}{2}\\geq \\left(\\frac{a+\\frac{1}{a}+b+\\frac{1}{b}}{2}\\right)^{2}\\Longleftrightarrow \\left(a+\\frac{1}{a}\\right)^{2}+\\left(b+\\frac{1}{b}\\right)^{2}}\\geq \\frac{1}{2}\\left(a+b+\\frac{a+b}{ab}\\right)^{2}$\r\n\r\n$=\\frac{1}{2}\\left(1+\\frac{1}{ab}\\right)^{2}\\geq \\frac{1}{2}(1+4)^{2}=\\frac{25}{2}$ the equality holds when $a=b\\ and\\ a+b=1\\Longleftrightarrow a=b=\\frac{1}{2}.$", "Solution_3": "[quote]for a, b [b]integers[/b] such that a+b=1[/quote]\r\n\r\nBut then again, your interpretation seems to be better, and at least it gives a correct problem. :)", "Solution_4": "[color=darkblue]If $p=ab\\ne 0$, then the relation $(*)\\ \\left(a+\\frac{1}{a}\\right)^{2}+\\left(b+\\frac{1}{b}\\right)^{2}\\ge \\frac{25}{2}$ $\\Longleftrightarrow$ $a^{2}+b^{2}+\\frac{1}{a^{2}}+\\frac{1}{b^{2}}\\ge \\frac{17}{2}$ $\\Longleftrightarrow$ $(a^{2}+b^{2})\\left(1+\\frac{1}{a^{2}b^{2}}\\right)\\ge \\frac{17}{2}$ $\\Longleftrightarrow$ $2[1-2(ab)]\\left[(ab)^{2}+1\\right ]\\ge 17(ab)^{2}$ $\\Longleftrightarrow$ $2(1-2p)(p^{2}+1)\\ge 17p^{2}$ $\\Longleftrightarrow$ $4p^{3}+15p^{2}+4p-2\\le 0$ $\\Longleftrightarrow$ $(4p-1)(p^{2}+4p+2)\\le 0$ $\\Longleftrightarrow$ $p=ab\\in \\left(-\\infty\\ ,\\--2-\\sqrt 2\\right]\\cup\\left[-2+\\sqrt 2\\ ,\\ 0\\right)\\cup\\left(0\\ ,\\ \\frac{1}{4}\\right]$, i.e. $0\\ne p\\le\\frac{1}{4}$ and $|p+2|\\ge \\sqrt 2\\ .$\nIf $a>0\\ ,\\ b>0$ and $a+b=1$, then $p>0$ and $4p-1=4ab-(a+b)^{2}=-(a-b)^{2}\\le 0$, i.e. $p\\in \\left(0,\\ \\frac{1}{4}\\right]$ and the inequality $(*)$ is truly.[/color]", "Solution_5": "This was on the recent issue of Mathematical Horizons...", "Solution_6": "and what is Mathematical Horizons?", "Solution_7": "Well, specifically, it was prove $\\sqrt{\\sin^{2}{\\theta}+\\frac{1}{\\sin^{2}{\\theta}}}+\\sqrt{\\cos^{2}{\\theta}+\\frac{1}{\\cos^{2}{\\theta}}}\\ge \\sqrt{10}$.\r\n\r\nMath Horizons is a magazine. Sorry, its actual name is Math Horizons, not mathematical." } { "Tag": [], "Problem": "Please note that the Music & Art forum is intended for the musicians and artists among us to discuss techniques and their work. If you want to talk about your favorite band, please do it in Fun & Games.", "Solution_1": "Right! I do so agree to the terms! :)" } { "Tag": [ "geometry", "rectangle", "analytic geometry", "ceiling function", "function", "number theory", "prime factorization" ], "Problem": "Once again, for those who are curious (me), interested (me), or just wondering if they did it right (me) :D \r\n[i][b]\n10. Asaf is investigating rectangles drawn in the plane such that every vertex has integer coordinates. (The sides of the rectangle need not be parallel to the coordinate axes.) He considers two rectangles to be the same if one can be obtained by rotating and/or translating the other.\n\n 1. How many distinct rectangles can Asaf draw with area 100?\n 2. How many distinct rectangles can Asaf draw with a given integer area A? (See Note 3.)[/b][/i]\r\n\r\nThe formula I used is the well-known theory to find the number of factors for any given number $ N$, where the prime factors of $ N$ are $ N \\equal{} A^a*B^b*C^c...$ with A, B, C... being unique prime factors and a, b, c... being the powers that they're raised to, there are $ (a \\plus{} 1)*(b \\plus{} 1)*(c \\plus{} 1)...$ factors of $ N$.\r\n\r\nAfter finding the number of factors, I divided it by two (to make pairs for the sides of the rectangles), and if there was an odd number of factors (if the number is a square), I bumped it up to the nearest integer.\r\nThus I was able to get $ 13$ rectangles for (1) :D", "Solution_1": "Having seen the questions for the first time right now, - Edit - hmm I misread that you had $ N^2$ .. but don't you have to show that the side lengths are achievable, eg sum of two squares, at least for (2)?\r\n\r\nNot returning :(", "Solution_2": "Without loss of generality, let one vertex be at the origin. Let the other points be $ (a,b)$, and $ (c,d)$ where $ a,b,c,d$ are integers and $ a,b\\ge 0$ (we can do this by rotating either $ 0$, $ 90$, $ 180$, or $ 270$ degrees. Clearly these conditions force the last vertex to be on a lattice point as well.\r\n\r\nWe have that $ |ad\\minus{}bc|\\equal{}A$ and $ ac\\plus{}bd\\equal{}0$.\r\n\r\n$ ac\\equal{}\\minus{}bd\\implies \\frac{a}{d}\\equal{}\\minus{}\\frac{b}{c}\\equal{}\\frac{m}{n}$ where $ (m,n)\\equal{}1$. Then $ a\\equal{}km$, $ d\\equal{}kn$, $ b\\equal{}lm$, $ c\\equal{}\\minus{}ln$ where $ k,l$ are integers.\r\n\r\n(Also cover when $ b,c\\equal{}0$, etc)\r\n\r\nThen $ |ad\\minus{}bc|\\equal{}mn|k^2\\plus{}l^2|\\equal{}A$\r\n\r\nThen I don't know about representation in terms of sum of squares. Part a is easy to do from here.", "Solution_3": "I didn't know about representation as sum of squares either. But the description said there was a fact from advanced number theory, so I figured that must be it, so I looked up \"sum of squares\" on http://www.mathworld.com, and found the \"sum of squares function.\"\r\n\r\nI don't want to paste my solution here unless prompted, because it's rather long, and I'll have to retype the bajillions of times I used Equation Editor. Instead, I'll post an image of the final formula, in all its ugly, difficult-or-impossible-to-simplify glory:\r\n\r\n[img]http://img261.imageshack.us/img261/2330/mathcampquiz0810bansho6.jpg[/img]\r\n\r\nEdit: Oh, I got 10 for part a, and I'm very certain I'm right, because I enumerated all the possibilities, along with the representation of the square of the smaller cofactor as a sum of squares (the larger one represented as a sum of squares is just a multiplication of the smaller one):\r\n\r\n[img]http://img152.imageshack.us/img152/2928/mathcampquiz0810aanspd3.jpg[/img]", "Solution_4": "You did not have these cases:\r\n\r\n$ 2\\sqrt2\\times25\\sqrt2$, $ \\left(2\\sqrt2\\right)^2\\equal{}2^2\\plus{}2^2$, $ \\left(25\\sqrt2\\right)^2\\equal{}25^2\\plus{}25^2$\r\n$ 4\\sqrt5\\times5\\sqrt5$, $ \\left(4\\sqrt5\\right)^2\\equal{}4^2\\plus{}8^2$, $ \\left(5\\sqrt5\\right)^2\\equal{}5^2\\plus{}10^2$\r\n$ 5\\sqrt5\\times4\\sqrt5$, $ \\left(5\\sqrt5\\right)^2\\equal{}5^2\\plus{}10^2$, $ \\left(4\\sqrt5\\right)^2\\equal{}4^2\\plus{}8^2$", "Solution_5": "Oops, I forgot to mention that you have to apply the formula to $ 100^2$ (10,000).\r\nI don't know why, but I'm guessing it has to do with your sum of the squares theory. \r\n\r\nFrom mnmath's presentation of the sum of the squares', I now understand the idea behind it. It looks like the cases for each individual rectangle are the different are the different ways you can split the prime factors of 10,000. \r\n\r\nExample: Above 3 cases. For the first, the squares are (2*2) + (2*2) and (25*25) + (25*25). Which makes up the 2^4*5^4 prime factorization of 10,000.\r\n\r\nHmm, it seems I just stumbled on the final formula to find the number of rectangles by chance, while having no idea of the theory behind it :wink: Oh well. Go 13!", "Solution_6": "http://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares\r\n\r\nThe proof of this theorem that uses Gaussian integers is one of my favorite proofs learned at HCSSiM.", "Solution_7": "[quote=\"mnmath\"]\n$ 4\\sqrt5\\times5\\sqrt5$, $ \\left(4\\sqrt5\\right)^2 \\equal{} 4^2 \\plus{} 8^2$, $ \\left(5\\sqrt5\\right)^2 \\equal{} 5^2 \\plus{} 10^2$\n$ 5\\sqrt5\\times4\\sqrt5$, $ \\left(5\\sqrt5\\right)^2 \\equal{} 5^2 \\plus{} 10^2$, $ \\left(4\\sqrt5\\right)^2 \\equal{} 4^2 \\plus{} 8^2$[/quote]\r\n\r\nThose are the same case...", "Solution_8": "Hmm, I didn't notice that! \r\n\r\nThe last one is $ \\sqrt{40}*\\sqrt{250}$, or $ 2\\sqrt{10}*5\\sqrt{10}$.\r\n\r\nGlad that's done with :P", "Solution_9": "My answer is that the number of squares is\r\n\r\n$ \\sum_{x\\in \\mathbf{B}_A}{\\left\\lceil \\frac{\\tau\\left(\\frac{A}{x}\\right)}{2}\\right\\rceil}$\r\n\r\nwhere $ \\tau (n)$ is the divisor function, and \r\n\r\n$ \\mathbf{B}_A\\equal{}\\{x: x\\equal{}1 \\mbox{ or } x \\mbox{ is squarefree and is a }\\\\\\mbox{product of primes 2 and/or of the form } 4k\\plus{}1,\\mbox{ and } x|A, k,x\\in\\mathbf{N}\\}$.\r\n\r\nPlugging in $ A\\equal{}100$, we get 13 for part a)." } { "Tag": [ "trigonometry", "complex numbers" ], "Problem": "Let $z_1,z_2$ be two complex numbers satisifying: $z_1+z_2=2i$, $|z_1||z_2|=3$. Let $u=\\frac{z_1}{z_2}$, $x=\\arg u$. Find $|u|$ when $\\cos x$ get maximum. Besides, what is the maximum value of $\\cos x$?", "Solution_1": "No one want to try?", "Solution_2": "does the x play any role in the problem?\r\n\r\nEDIT: nevermind, i just read to the first part of the question", "Solution_3": "[quote=\"ifai\"]No one want to try?[/quote]\r\n..." } { "Tag": [ "geometry", "similar triangles" ], "Problem": "Medians $BD$ and $CE$ of triangle $ABC$ are perpendicular, $BD=8$, and $CE=12$. The area of triangle $ABC$ is \r\n\r\n[b](A)[/b] $24$ [b](B)[/b] $32$ [b](C)[/b] $48$ [b](D)[/b] $64$ [b](E)[/b] $96$\r\n\r\nATTACHMENT", "Solution_1": "[hide]since the diags are perpendicular, area of BCDE=1/2*8*12, and by similar triangles, BCDE is 3/4 of the whole triangle, so our answer is 4/3*1/2*8*12=64 or D[/hide]" } { "Tag": [ "geometry", "calculus", "trigonometry", "inequalities", "rectangle", "circumcircle", "triangle inequality" ], "Problem": "Solve, without Calculus, for the maximum possible area of a quadrilateral $ABCD$ in which $AB = BC = 3$ and $CD = 7$.", "Solution_1": "Okay, it's clear that we must have\r\n\r\n$AC\\perp CD$. (That hint was free :P )\r\n\r\n[hide]Drop the perpendicular $P$ from $B$ and let $\\angle PBC=t$\n\n$K=\\triangle_{ABC}+\\triangle_{ACD}$\n$K=2\\cdot\\frac{1}{2}(3\\cos t)(3\\sin t)+\\frac {1}{2}\\cdot 2 \\cdot (3\\sin t)\\cdot 7$\n\nSo we need to maximize $9\\cos t\\sin t+21\\sin t$\n\nAfter some work, we get that $t=16\\sqrt 2$, and yes, I may have used calculus to find that[/hide]", "Solution_2": "Pure geometry + AM-GM :)\r\nLet $x=AD$. We know that area of quadrilateral with sidea $a$, $b$, $c$ and $d$ is maximal iff quadrilateral is inscribed (it is always possible to find inscribed quadrilateral with sides $a$, $b$, $c$ and $d$, if they satisfy triangle inequality, of course :) ). \r\nUsing Heron formula for inscribed quadrilateral, we obtain that \r\n\\[S^2=(p-a)(p-b)(p-c)(p-d)=\\frac{1}{16}(x+7)^2(x-1)(13-x)=\\]\r\n\\[=2\\cdot\\frac{x+7}{4}\\cdot\\frac{x+7}{4}\\cdot\\frac{x-1}{2}\\cdot(13-x)\\leq 2\\cdot \\left(\\frac{\\frac{x+7}{4}+\\frac{x+7}{4}+\\frac{x-1}{2}+(13-x)}{4}\\right)^4=2\\cdot 4^4\\]\r\nand equality holds iff $x=9$.\r\nTherefore $S_{\\max}=16\\sqrt{2}$.", "Solution_3": "To complete my solution without calculus, can anyone prove that:\r\n\r\n$9\\sin t\\cos t+21\\sin t\\leq 16\\sqrt {2}$?\r\n\r\nEquality occurs when $\\cos t=\\frac {1}{3}$ and $\\sin t=\\frac {2\\sqrt {2}}{3}$", "Solution_4": "I was first trying to work pure geometrically. I simplify it a lot, but then, when it comes to find maximal area of one rectangle the same formula as blahblahblah wrote appeared to me, but I also can't find it's maximum without calculus.", "Solution_5": "Still no-one has come up with a completely geometric solution! ;) Nevertheless, I have learned a new way to do this - with AM-GM. Albeit well-known, I'm curious: how does one show that a polygon with given side lengths has maximal area iff it has a circumcircle?", "Solution_6": "Let ABCD be any quadrilateral with AB=3,BC=3,CD=7.Now we don't decrease it's area, when we make from D-> D' such that CD'=7 and angle ACD is pi/2. Now we have quadrilatera ABCD'. We don't decrease it's area, when we make A->A' such that A'B=3 and angle A'BD pi/2.\r\n\r\nFrom this we know that we have to just look at such ABCD, where angles ABD=ACD=pi/2 ( From this we know that this ABCD has circumcircle)\r\n\r\nYou can also see that angles CDB=BDA. \r\n\r\nCan you prove it pure geometrically from this point?", "Solution_7": "[quote=\"ondrob\"]Can you prove it pure geometrically from this point?[/quote]\r\n\r\nAbsolutely.", "Solution_8": "[quote=\"ondrob\"]Let ABCD be any quadrilateral with AB=3,BC=3,CD=7.Now we don't decrease it's area, when we make from D-> D' such that CD'=7 and angle ACD is pi/2. Now we have quadrilatera ABCD'. We don't decrease it's area, when we make A->A' such that A'B=3 and angle A'BD pi/2.\n\nFrom this we know that we have to just look at such ABCD, where angles ABD=ACD=pi/2 ( From this we know that this ABCD has circumcircle)\n\nYou can also see that angles CDB=BDA. \n\nCan you prove it pure geometrically from this point?[/quote]\r\n\r\nIt is necessary for the quadrilateral for have right angles ABD and ACD. As there is just one such quadrilateral, this is sufficient. To get its area, take C' on its circumcircle such that BC' = 7 and C'D = 3. Observe that [ABCD] = [ABC'D]. AD = 2r is a diameter, and by Pythagoras and Ptolemy, we find that 4r 2 - 9 = 14r + 9 which is easily solved for the solution r = 9/2, and by a number of methods [ABCD] = 16 :sqrt: 2.", "Solution_9": "For Mildorf: If you want to prove that a given quadrilateral has maximal area if it is cyclic, with given sides,a,b,c,d, consider the area as\r\n\r\n$\\frac{1}{2}ab (Sin{\\alpha})+\\frac{1}{2}cd (Sin{\\beta})$, and prove this is max if $\\alpha+\\beta=\\pi$.\r\n\r\nThen do the general case!", "Solution_10": "If the quadrilaterial is cyclic, then we can use $\\cal{A}$$=\\sqrt{(s-a)(s-b)(s-c)(s-d)}$.", "Solution_11": "To prove it has max area if it is cyclic,\r\n\r\n$Area=F(x)=\\frac{ab}{2}(\\sin{x})+\\frac{cd}{2}(\\sin{y})$\r\n\r\nWhere x and y are the angles between a,b & c,d, respectively.\r\nAlso, going off the diagonal $ac$ and the cosine rule, we get that\r\n\r\n$a^2+b^2-2ab(\\cos{x})=c^2+d^2-2cd(\\cos{y})$\r\n\r\nDiff $F(x)$ with respect to x.\r\n\r\n$F'(x)=\\frac{ab}{2}(\\sin{x})+\\frac{cd}{2}(\\sin{y})(y')$\r\n\r\nAlso Diff the other condition:\r\n\r\n$ab(\\sin{x})=cd(\\sin{y})(y')$\r\n\r\nRe-writing this, we get $y'=\\frac{ab(\\sin{x})}{cd(\\sin{y})}$\r\n\r\nSubbing this into the first equation and simplifing, we get:\r\n\r\n$F'(x)=\\frac{ab}{2} (\\frac{ \\sin{(x+y)}} {\\sin{y}})$\r\n\r\nThis is zero for a max, so ${\\sin({x+y})=0}$\r\n\r\nor\r\n\r\n$x+y=2{\\pi}$", "Solution_12": "[quote=\"ln(dx/dy)\"]If the quadrilaterial is cyclic, then we can use $\\cal{A}$$=\\sqrt{(s-a)(s-b)(s-c)(s-d)}$.[/quote]\r\n\r\nVery good point... we can used the generalized form, called Brahmagupta's formula, to prove what Mildorf wanted.\r\n\r\n\\[A=\\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\\ \\cos^2\\left(\\frac{B+D}{2}\\right)}\\]\r\n\r\nThe area is maximized when $\\textstyle abcd\\ \\cos^2\\left(\\frac{B+D}{2}\\right)$ is minimized. Since this quantity is always nonnegative, we want it to be zero, which happens iff $B+D=180^\\circ$, that is, when $ABCD$ is cyclic." } { "Tag": [ "function", "LaTeX", "\\/closed" ], "Problem": "I got this when trying to post a message:\r\n[code]\nSQL requests not achieved\n\nDEBUG MODE\n\nSQL Error: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' 19209) GROUP BY ug.user_id' at line 7\n\nSQL Request: SELECT MAX(a.auth_value) as allowed, ug.user_id FROM phpbb_auths a, phpbb_user_group ug WHERE a.obj_type = 'f' AND a.obj_id = 132 AND a.auth_name = 'auth_watch_forum' AND a.group_id = ug.group_id AND ug.user_id IN (, 19209) GROUP BY ug.user_id\n\nLine : 1189\nFile : functions_post.php[/code]\r\n\r\n :maybe:\r\n\r\nEDIT: It happens in the MC forum...I know that...is it a bug or something?", "Solution_1": "I sometimes get that, particularly when I'm trying to load something long, complicated, and with lots of $\\LaTeX$ to render. (Often it happens at the preview stage.) Then I click the \"back\" button on my browser, hit \"preview\" or \"submit\" again, and it goes through on the second try. I don't think the second try has failed me yet.\r\n\r\nSomething like that just happened to me when I tried to revive a topic that was over a year old. (I had my reasons - long story, and it involves a hidden forum.) Again, the second try worked.\r\n\r\nOther than to speculate that it involves some kind of buffer overload or internal time-out, I don't think it particularly matters what the reason is as long as the second try works.", "Solution_2": "No. NO. It goes through when trying to post in a topic posted in like 1 minute ago. When I try teh secnod try, I get \"Someone has already posted a reply\" and it appears to be me. Just that that message is quite annoying." } { "Tag": [ "geometry", "analytic geometry", "ratio", "quadratics", "inequalities", "calculus", "calculus computations" ], "Problem": "I know this has problem backround to it, but I dont have anywhere else to go.\r\n\r\nQuestion Details:\r\nI have two circles centered at the origin, one with radius A and the other with radius b.\r\n\r\nLooking at the hemiwasher (area between) the circles form above the x axis, find the values of A and B that place the center of mass within the hemiwasher itself, not in the open middle space.\r\n\r\nWhat i think i has solved so far: not necesseraly accurate:\r\nI solved the Y value of the center of mass in terms of A and B to be:\r\n\r\nY=(4(A^3-B^3))/(3pi(A^2-B^2))\r\n\r\nhow can I use this to find values of a and b that put the y coordinate of the center of mass between a and b?\r\n\r\nB is the smaller radii; the density is constant, so it is irrelvant.", "Solution_1": "Note that only the ratio of the two radii matters- it's probably a good idea to normalize and make one of them $ 1$.\r\nThis will lead to a quadratic equation.", "Solution_2": "But i am not given a ratio, or are you saying i assign one? like:\r\n\r\nA:B becomes BX:B (x is how many times bigger a is than b)\r\n\r\nAnd Im not seeing a quadratic, are you saying one will appear out of the equation I gave or from somewhere else?", "Solution_3": "The expression you gave is not in lowest terms; there's an obvious common factor of $ A \\minus{} B$. (Note- please be consistent with capitalization of variables)\r\n\r\nIf we set $ A \\equal{} rB$, we're trying to solve $ 1\\le \\frac {4}{3\\pi}\\frac {r^2 \\plus{} r \\plus{} 1}{r \\plus{} 1}\\le r$ with $ r\\ge 1$. Clearing the denominator, $ r \\plus{} 1\\le \\frac {4}{3\\pi}(r^2 \\plus{} r \\plus{} 1)\\le r^2 \\plus{} r$.", "Solution_4": "Ok, this makes sense. Sorry, I didn't see the difference of cubics in the numerator. I'll try to use lowercase for variables.\r\n\r\nBut I'm not seeing how I can use this the find values (or range of values) for a and b specifically. Now I have a weird inequality that needs to be true for the ratio between a and b, but how can I convert that back to a and b?\r\n\r\nSorry for not grasping at the hints you're providing, I appreciate the help though, very greatly.", "Solution_5": "Ok, solving all of my equalities and making sure everything that has to be true stays true, and using the quad formular, I get\r\n\r\nr> or = to 2.02\r\n\r\nThus, A>= 2.02 B\r\n\r\nThanks!" } { "Tag": [ "algebra", "polynomial", "inequalities", "trigonometry", "function", "absolute value", "algebra solved" ], "Problem": "Let $P(X)$ be a polynomial of degree $n$ satisfying\r\n$|P(X)|\\leq 1$ whenever $|X|\\leq 1$.\r\n\r\nProve that $|P'(X)|\\leq n^2$ whenever $|X|\\leq 1$.\r\nDoes the equality hold only for Chebyshev polynomials?", "Solution_1": "Is this for X in R or X in C?", "Solution_2": "It was posted by treegoner and solved by Maubinool,the genaral form of Markov inequality also wrote by flip2004 and Maubinool.Here is the link:\r\nhttp://www.mathlinks.ro/Forum/topic-17044.html", "Solution_3": "Here is a magnificient proof that I found in a material from ENS:\r\n It suffices to prove the identity $ xf'(x)=\\frac{n}{2}\\cdot f(x)+\\frac{1}{n}\\cdot\\sum_{k=1}^{n}{f(xz_k)\\cdot\\frac{2z_k}{(z_k-1)^2}}$ and the triangular inequality. Then, a simple identity based on Chebyshev's polynomials will solve the problem. I find it great!", "Solution_4": "I'm sorry I didn't get it. Could you please explain it in details, Harazi? I would be very grateful", "Solution_5": "Sorry, I was a little bit too expeditive with the proof, but here it goes:\r\n Consider $z_i$ the roots of the polynomial $ z^n+1$. If we write the interpolation formula for the polynomial $\\frac{ f(xz)-f(x)}{z-1}$ in the points $z_i$, we will find an identity in z and x. Now, make z very close in the formula you obtain and you will find the beautiful identity I posted. Next, assume that $\u23a2f(x)\u23a5$ is smaller than 1 in the unit circle (or disk) and apply the triangular inequality in the identity. You will find a majoration for the absolute value of $ f'(x)$. You show then (exercise using the Chebyshev's polynomials) that in fact the majoration you find implies that the absolute value of $f'(x)$ is at most n. This is not hard.", "Solution_6": "A detailed explanation for harazi's post. Specially for Iura and for all other mathlinkers, who don't identify himself with Harazi :)\r\n\r\nO'Hara P. J., Another proof of Bernstein's theorem, AMM 80 (1973), 673-674\r\n\r\n[b]Theorem.[/b]\r\nLet $\\deg f=n$. Then $||f'||\\leq n||f||$, where $||x||=\\max_{|z|=1}||x(z)||$.\r\n\r\n[b]Lemma.[/b]\r\nLet $\\deg f\\leq n$ and $z_1,z_2,...,z_n$ are roots of $z^n+1=0$. Then\r\n\\[tf'(t)=\\frac{n}{2}f(t)+\\frac{1}{n}\\sum_{k=1}^{n}f(tz_k)\\frac{2z_k}{(z_k-1)^2}.\\]\r\n\r\n[b]Proof of lemma.[/b]\r\nLet $g_t(z)=\\frac{f(tz)-f(t)}{z-1}$. We see that $g_t(z)$ is a polynomial of degree at most $n-1$ and $g_t(1)=tf'(t)$. Using a Lagrange's interpolation formulae with points $z_1,z_2,...,z_n$ shows that\r\n\\[g_t(z)=\\sum_{k=1}^n g_t(z_k)\\frac{z^n+1}{(z-z_k)nz_k^{n-1}}=\\frac{1}{n}\\sum_{k=1}^ng_t(z_k)\\frac{z_n^1}{z_k-z}z_k\\]\r\n(since $z_k^{n-1}=-1/z_k$).\r\n\r\nPutting $z=1$ we have\r\n\\[tf'(t)=\\frac{1}{n}\\sum_{k=1}^ng_t(z_k)\\frac{2z_k}{(z_k-1)}=\\frac{1}{n}\\sum_{k=1}^n\\frac{f(tz_k)-f(t)}{(z_k-1)^2}=\\] \\[=\\frac{1}{n}\\sum_{k=1}^n f(tz_k)\\frac{2z_k}{(z_k-1)^2}-\\frac{f(t)}{n}\\sum_{k=1}^n \\frac{2z_k}{(z_k-1)^2}.\\]\r\n\r\nIn order to calculate sum $\\sum_{k=1}^n \\frac{2z_k}{(z_k-1)^2}$, put $f(t)=t^n$. Then $f(tz_k)=-t^n$, so \r\n\\[nt^n=-\\frac{2t^n}{n}\\sum_{k=1}^n\\frac{2z_k}{(z_k-1)^2},\\]\r\ni.e. [b](1)[/b]\r\n\\[\\sum_{k=1}^n \\frac{2z_k}{(z_k-1)^2}=-\\frac{n^2}{2}.\\]\r\n\r\n[b]Lemma is proved.[/b]\r\n\r\nSuppose $|t|=1$. We have \r\n\\[|f'(t)|\\leq \\left(\\frac{n}{2}+\\frac{1}{n}\\sum_{k=1}^n\\left|\\frac{2z_k}{(z_k-1)^2}\\right|\\right)||f||.\\]\r\nLet's show that $2z_k/(z_k-1)^2$ is a real negative number. Indeed, $z_k=e^{is}\\neq 1$, thus \r\n\\[\\frac{2z_k}{(z_k-1)^2}=\\frac{2eI{is}}{(e^{is}-1)^2}=\\frac{2}{e^{is}-2+e^{-is}}=\\frac{1}{\\cos s-1}<0.\\]\r\nIt follows that from [b](1)[/b] we obtain that\r\n\\[\\frac{1}{n}\\sum_{k=1}^n\\left|\\frac{2z_k}{(z_k-1)^2}\\right|=-\\frac{1}{n}\\sum_{k=1}^n \\frac{2z_k}{(z_k-1)^2}=\\frac{n}{2}.\\]\r\n\r\n[b]Theorem is proved.[/b]\r\n\r\n[b]Remark 1.[/b]\r\nFor analytic functions the following inequality holds\r\n\\[\\max_{|z|\\leq 1}|f(z)|=\\max_{|z|= 1}|f(z)|.\\]\r\n\r\n[b]Remark 2.[/b]\r\nHarazi, obviously, didn't solved Iura's problem. ;)\r\n\r\n[b]Remark 3.[/b]\r\nThe mentioned link doesn't lead to a proof, just to hints ;)", "Solution_7": "Myth, these are just your opinions. Do you really think it is difficult to obtain a proof from these hints? Anyway,..." } { "Tag": [], "Problem": "Fie un tetraedru in care AB=1 , AC=4 si BD=7.Stiind ca exista un punct egal departat de cele sase muchii ale tetraedrului sa se calculeze lungimea segmentului CD.\r\n\r\n\r\n :) ASTEPT REZOLVARI :)", "Solution_1": "Daca punctul P este egal departat de laturile triunghiului ABC, el se afla pe perpendiculara pe plan dusa prin centrul cercului inscris acestuia. Analog pt a se demonstra ca P se afla pe perpendiculara dusa pe planul ABD prin centrul cercului inscris triunghiului ABD. Vom nota cu V pr lui P pe AD, cu R pr lui P pe CD, cu S pr lui P pe AC, cu T pr lui P pe AB si cu Q pr lui P pe BD. Se demonstreaza usor ca:\r\n1. DR=DV=DQ=a (DO lat com ,OR si OV raze din T3P si ORD=OVD=90) unde o este centru ci al lui ABC\r\n2. AS=AV=AT=b\r\n3. BT=BQ=c\r\n4. CS=CR=d\r\nDin cele 4 relatii rezulta ca\r\n1. AC=b+d=4\r\n2. AB=b+c=1\r\n3. BD=c+a=7\r\nSi noi vrem sa calculam CD, adica d+a\r\ncare este BD-AB+AC=c+a-b-c+b+d=a+d\r\nadica CD=7-1+4=10\r\nSperca nu am gresit prea tare :P" } { "Tag": [ "trigonometry" ], "Problem": "\u5df2\u77e5x,y,z\u4e3a\u6b63\u5b9e\u6570\u4e14xyz(x+y+z)=1,\u6c42\uff08x+y\uff09\uff08y+z\uff09\u6700\u5c0f\u503c\u3002\r\n\u6211\u6709\u4e00\u4e2a\u5f88abnormal\u7684\u89e3\u6cd5\u3002", "Solution_1": "\\[ (x + y)(y + z) = y(x + y + z) + xz = \\frac{1}{xz} + xz \\geq 2 \\]", "Solution_2": "Thank you,Arne.This is what I want to find.\r\nMy solution is a bit more complex than yours.I used Heron's formula.So I think it abnormal.But in short time I can't make a normal solution which my classmate can accept.Maybe my brain itself is abnormal.", "Solution_3": "\u539f\u662f\u7528\u4e09\u89d2\u5f62\u5916\u5207\u5706\u6765\u505a\u7684\u3002", "Solution_4": "\u8fd9\u4e2a\u9898\u540e\u6765\u4f5c\u4e3a\u8fc7\u9ad8\u8003\u9898\r\n\u524d\u4e24\u5929\u8fd8\u6709\u540c\u5b66\u6765\u95ee\u6211\u7684\uff0c\u662f\u5728\u9ad8\u8003\u590d\u4e60\u6750\u6599\u91cc\u7684\u2026\u2026", "Solution_5": "[quote=\"dingdongdog\"]\u5df2\u77e5x,y,z\u4e3a\u6b63\u5b9e\u6570\u4e14xyz(x+y+z)=1,\u6c42\uff08x+y\uff09\uff08y+z\uff09\u6700\u5c0f\u503c\u3002\n\u6211\u6709\u4e00\u4e2a\u5f88abnormal\u7684\u89e3\u6cd5\u3002[/quote]\r\n\r\n$ (x\\plus{}y)(y\\plus{}z)\\equal{}y(x\\plus{}y\\plus{}z)\\plus{}zx \\ge 2\\sqrt{y(x\\plus{}y\\plus{}z)zx}\\equal{}2$.", "Solution_6": "[quote=\"Fourier\"]\u539f\u662f\u7528\u4e09\u89d2\u5f62\u5916\u5207\u5706\u6765\u505a\u7684\u3002[/quote]\r\n\r\n\u662f\u5185\u5207\u5706\u5427?\r\n\r\n\u5475\u5475, \u90a3\u4f60\u7684\u610f\u601d\u5c31\u662f\u5185\u5207\u5706\u4ee3\u6362\u6cd5\u54af, \u8fd9\u6837\u5417?:\r\n\r\n\u4ee4$ a\\equal{}x\\plus{}y,b\\equal{}y\\plus{}z,c\\equal{}z\\plus{}x$, \u5219$ a,b,c$\u6784\u6210\u4e09\u89d2\u5f62, \u5e76\u4e14\u7531\u6d77\u4f26\u516c\u5f0f\u6613\u77e5\u5176\u9762\u79ef$ S\\equal{}1$, \u4e8e\u662f\u5c31\u6709\r\n\r\n$ (x\\plus{}y)(y\\plus{}z)\\equal{}ab\\equal{}\\frac{2S}{\\sin{C}}\\equal{}\\frac{2}{\\sin{C}} \\ge 2,$\r\n\r\n\u5f53$ C\\equal{}\\frac{\\pi}{2}$\u65f6\u53d6\u7b49\u53f7." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "(a) P1 and P2 write down a number on a piece of paper, and only know the numbers they write down. They give their pieces of paper to a person P3. P3 writes on a piece of paper the sum of the numbers, and any other random number. Now P3 asks P1 if he knows P2's number; and we assume both P1 and P2 have complete, impeccable reasoning and neither lie; and P1 replies says 'yes' or 'no'; then P3 asks P2 if he knows P1's number who says \"yes' or 'no'; and the process keeps going. Prove in finitely many steps, both will end up saying 'yes'. \r\n(b) A more general version: exactly the same but instead of writing the sum and 1 random number, P3 writes down the sum and k other random numbers. Find all k for which the problem works (i think it's all values of k, but i have no idea). \r\n(c) Do parts (a) and (b) but with the product of the two numbers instead of the sum (i think this is considerably easier).", "Solution_1": "do P1,P2 have to make guesses abou that number ?", "Solution_2": "i think the statement of the question is quite clear. \r\np1 and p2 try to work out each other person's number, and they're both super-intelligent. prove at some point each one manages to work out the other person's number." } { "Tag": [ "algebra", "polynomial" ], "Problem": "consider a 1xn board. suppose you color every square with two colors, white and black. Let $ c_n$ be the number of colorings in which no two squares that are colored white are adjacent. find a recurrence relation for $ c_n$ and then derive an explicite formula for $ c_n$.", "Solution_1": "[quote=\"math227\"]consider a 1xn board. suppose you color every square with two colors, white and black. Let $ c_n$ be the number of colorings in which no two squares that are colored white are adjacent. find a recurrence relation for $ c_n$ and then derive an explicite formula for $ c_n$.[/quote]\r\n\r\n[hide]\nLet $ a_{n}$ be the number of valid relations ending with a white one.\nLet $ b_{n}$ be the number of valid relations ending with a back one.\n$ c_{n}\\equal{}a_{n}\\plus{}b_{n}$\n$ b_{n}\\equal{}a_{n\\minus{}1}\\plus{}b_{n\\minus{}1} \\Rightarrow b_{n}\\equal{}c_{n\\minus{}1}$\n$ a_{n}\\equal{}b_{n\\minus{}1}$ since the number of valid relations ending with white that is N blocks long is exactly equal to the number of varid ones, since obviously n-1 needs to be valid, and if n-1 ends with white it's no good.\n$ c_{n}\\equal{}b_{n\\minus{}1}\\plus{}b_{n}$\n$ c_{n}\\equal{}c_{n\\minus{}1}\\plus{}c_{n\\minus{}2}$\nOur first values are \n1-2\n2-3\nThe characteristic porynomiar is\n$ x^{2}\\equal{}x\\plus{}1$\nThis is the fibonacci sequence, with the nth number being $ F_{n\\plus{}2}$\nYou can use Binet's formula if you want to find the explicit value, I don't really remember it to be honest, and latexing it takes a while anyways.\nOr, if you'd never seen the fibonacci before, you could solve the characteristic polynomial, and then that would give you the values needed to find the formula.[/hide]" } { "Tag": [ "ratio", "arithmetic sequence", "geometric sequence" ], "Problem": "Find 3 Arithmetic means between -25 and 5 _______, ______, ________\r\nFind 2 Geometric means between 6 and 20.25. _____,_______\r\n\r\nI could use some help. OK, I think what this means is find the nth term for the numbers in between But how do I get d? Also, what is the difference between Arithmetic and Geometric?", "Solution_1": "[quote=\"nemomath\"]Find 3 Arithmetic means between -25 and 5 _______, ______, ________\nFind 2 Geometric means between 6 and 20.25. _____,_______\n\nI could use some help. OK, I think what this means is find the nth term for the numbers in between But how do I get d? Also, what is the difference between Arithmetic and Geometric?[/quote]\r\n\r\nArithmetic mean of $a_{1},a_{2},\\ldots,a_{n}$ is\r\n\\[\\frac{a_{1}+a_{2}+\\cdots+a_{n}}{n}\\]\r\nwhile the geometric mean is\r\n\\[\\sqrt[n]{a_{1}a_{2}\\cdots a_{n}}. \\]\r\n[hide=\"Generalization\"]Let's say you are given numbers $a_{0}$ and $a_{n+1}$ and told to find the $n$ arithmetic means, $a_{1}$ through $a_{n}$ between them (in your context). Then we need the common difference $d$... And since $a_{1}=a_{0}+d, a_{2}=a_{0}+2d$, etc., it is true that $a_{n+1}=a_{0}+(n+1)d$. Therefore $d=\\frac{a_{n+1}-a_{0}}{n+1}$. Then you can just add them... giving\n\\[a_{k}= a_{0}+kd = \\boxed{a_{0}+\\frac{(a_{n+1}-a_{0})k}{n+1}}. \\]\nIf we were to find the $n$ geometric means between them, then we need common ratio $r$, and since $a_{1}=a_{0}\\cdot r, a_{2}=a_{0}\\cdot r^{2}$, etc., it is true that $a_{n+1}= a_{0}\\cdot r^{n+1}$. Therefore $r=\\sqrt[n+1]{a_{n+1}/a_{0}}$. Then\n\\[a_{k}= a_{0}\\cdot r^{k}= \\boxed{a_{0}\\cdot (a_{n+1}/a_{0})^{\\frac{k}{n+1}}}. \\]\nWorthless in plugging in numbers, but it is more interesting to do things in general.\n[/hide]", "Solution_2": "[quote=\"nemomath\"]Find 3 Arithmetic means between -25 and 5 _______, ______, ________\nFind 2 Geometric means between 6 and 20.25. _____,_______\n\nI could use some help. OK, I think what this means is find the nth term for the numbers in between But how do I get d? Also, what is the difference between Arithmetic and Geometric?[/quote]\r\n$3$ Arithmetic means between $-25$ and $5$: $-25$, ___, ___, ___, $5$\r\n\r\n[hide]So the formula for any term of an arithmetic sequence is $a_{n}=a+(n-1)d$ where $a$ is the first term, $n$ is the $n^{\\text{th}}$ term, and $d$ is the distance between each term. I'm sure you already know this though.\n\nNote that $a_{5}=-25+4d$ and we already know that $a_{5}=5$ so set up and solve:\n\n$a_{5}=-25+4d$\n$5=-25+4d$\n$4d=30$\n$d=\\frac{15}{2}=7.5$\n\nSince $d=7.5$, our series goes like:\n$-25,-25+7.5,-25+2 \\cdot 7.5,-25+3 \\cdot 7.5, 5$\n$\\equiv \\boxed{-25,-17.5,-10,-2.5, 5}$[/hide]\n\n$2$ Geometric Means between $6$ and $20.25$: $6$, ___, ___, $20.25$\n\n[hide]Note that the the formula for geometric progression is $a_{n}=a \\cdot r^{n-1}$ where $a$ ($a=6$) is the first term and $r$ is the common ratio between each term.\n\n$a_{4}=a \\cdot r^{3}$ and we know $a_{4}=20.25$\n\n$a_{4}=6 \\cdot r^{3}$\n$20.25=6 \\cdot r^{3}$\n$r^{3}=\\frac{27}{8}$\n$r=3/2$\n\nSo the sequence is: $\\boxed{6, 9, 13.5, 20.25}$.[/hide]" } { "Tag": [ "geometry", "trigonometry", "inequalities", "rectangle", "triangle inequality" ], "Problem": "To tile a region $R$ means to find a set $S$ of non-overlapping regions such that their union is $R.$ Suppose $R$ is a square with side length $1.$\r\n\r\n$1.$ Is it possible to tile $R$ with finitely many circles?\r\n$2.$ Is it possible to tile $R$ with finitely many $30-60-90$ triangles such that their longest side has a rational length?\r\n\r\n[hide=\"Part 1\"]\n[b]\nClaim: It is not possible to tile $R$ with finitely many circles.[/b]\n[b]Proof:[/b]\nAssume that we can tile $R$ in such a way.\n\nLet the radius of the smallest circle, $\\omega_{0},$ be $r.$ Consider circles $\\omega_{1}, \\omega_{2},$ that are externally tangent to $\\omega_{0}$ and each other. In other words, $\\omega_{0}, \\omega_{1}, \\omega_{2}$ are all externally tangent to each other. However, there is a region between these circles that hasn't been covered. In order to cover this area, we must use a circle of smaller radius than $\\omega_{0}.$ Contradiction! \n\nNow, assume that we have only two circles, $\\omega_{0}$ and $\\omega_{1}.$ Let these circles be externally tangent at point $X.$ Now, choose a point outside these circles and place it arbitrarily close to $X.$ This point must be covered, so, use a circle to do so. But there will be distance between this circle and $\\omega_{0}$ and $\\omega_{1}$ if the circles are not all externally tangent to each other, so let this circle be externally tangent to $\\omega_{0}$ and $\\omega{1}.$ From here, we proceed by following the earlier paragraph to reach a contradiction.\n\nThus, we cannot tile $R$ with finitely many circles.\n\n$QED$\n\n[/hide]\n[hide=\"Part 2\"]\n[b]Claim: It is not possible to tile $R$ in such a way.[/b]\n[b]Proof:[/b]\nWe will proceed by contradiction. Assume that $R$ can be tiled in such a way.\nLet there be $n$ triangles whose hypoteneuse lengths are $s_{1}, s_{2}, \\cdots, s_{n}$ for rational $s_{i}.$ For each $s_{i},$ we know that the area of the corresponding triangle is $\\frac{s_{i}\\sin 30 \\cdot s_{i}\\cos 30}{2}= \\frac{s_{i}^{2}\\sqrt{3}}{8}.$ If we add up the area of these triangles and equate the area to that of the unit square, we get:\n\n$\\frac{\\sqrt{3}}{8}(s_{1}^{2}+s_{2}^{2}+\\cdots+s_{n}^{2}) = 1$\n\nThis implies that at least one $s_{i}$ is irrational. Contradiction! Thus, $R$ cannot be tiled with finitely many $30-60-90$ triangles with rational hypoteneuse length.\n\n$QED$\n\n[/hide]", "Solution_1": "Your part 2 is good. Here's the solution I used for part one when we did this at school.\r\n\r\n[hide]\n, so there must be a circle tangent to the square at every point. Then, since a circle can only be tangent to a square at at most one point, there are infinitely many circles in any such tiling.\n[/hide]", "Solution_2": "these are very nice problems!!!\r\n\r\n@karth, how do you know that the circles don't make a ring or something to that effect...like w_1 touches w_2, w_2 touches w_3, but you have not shown that w_3 touches w_1...like what if there was a ring of circles or something", "Solution_3": "[hide=\"1 with perhaps too much rigor\"]\nIt's not possible. Suppose it is. Then there is a circle with least radius, let $m$ be this radius. Now consider the upper left corner (well, one of the corners) of the square, call it $P$. For any positive $\\epsilon$, all points a distance of $\\epsilon$ away from $P$ and inside the square better be covered. Therefore, by the Triangle Inequality, there exists the center of a circle $C$ with radius $r\\geq m$ and distance at most $r+\\epsilon$ from $P$.\n\nHmm... looks like this circle goes outside the square! Indeed, if we drop a perpendicular from $C$ to the left side of the square and let the foot be $H$, $CH$ will be at least $r$, but also $PC\\leq r+\\epsilon$, so that\n\\[PH=\\sqrt{PC^{2}-CH^{2}}\\leq \\sqrt{(r+\\epsilon )^{2}-r^{2}}\\]\nSimilarly, if we drop a perpendicular to the top of the square and call it $K$, then $PK\\leq \\sqrt{(r+\\epsilon )^{2}-r^{2}}$.\n\nWait! Quadrilateral $PHCK$ is actually a rectangle! This means $PK=CH$, so $\\sqrt{(r+\\epsilon)^{2}-r^{2}}\\geq r$. Um, yea, we can just choose $\\epsilon=\\frac{m}{9^{9}}$ and get a really bad contradiction...\n[/hide]", "Solution_4": "[quote=\"Altheman\"]these are very nice problems!!!\n\n@karth, how do you know that the circles don't make a ring or something to that effect...like w_1 touches w_2, w_2 touches w_3, but you have not shown that w_3 touches w_1...like what if there was a ring of circles or something[/quote]\r\n\r\nHmm... how did I forget to include that part in my solution? I had it somewhere here... :(\r\n\r\nEDIT: Aha! Found it :D I'll edit my earlier solution.", "Solution_5": "[quote=\"Hamster1800\"]\n[hide]\n, so there must be a circle tangent to the square at every point. Then, since a circle can only be tangent to a square at at most one point, there are infinitely many circles in any such tiling.\n[/hide][/quote]\r\n\r\n=Any point on the square must be contained on or within a circle. The circle cannot be outside the square, so no part of it can be on the other side of the line. Only a tangent circle would include a point on a line but be on one side of it.\r\n\r\nOh, it is possible for a circle to \"be tangent to a square at [more than one] one point.\" What is important is that the circle can only be tangent a finite number of points." } { "Tag": [ "calculus", "rotation", "geometry", "3D geometry", "sphere", "integration" ], "Problem": "The results you get from this problem are fairly interesting.\r\n\r\nThe circle with equation x :^2: + y:^2: = 25 is rotated around the x-axis to form a sphere.\r\n\r\na) Write the differential of surface area, dS, in terms of x.\r\n\r\nb) Numerically find the area of the zone between\r\n i) x=0 and x=1\r\n ii) x=1 and x=2\r\niii) x=2 and x=3\r\n iv) x=3 and x=4\r\n v) x=4 and x=5\r\n\r\nc) Prove that the surface area of a sphere of radius r is given by S=4 :pi:r:^2:.\r\n\r\nd) Derive the formula for the volume of a sphere V= 4:pi:r:^3: /3 using spherical shells and the fact that the surface area is equal to 4:pi:r:^2:.\r\n\r\nI'm not necessarily looking for a solution to be posted. But for those of you who know how to use calculus to do this, if you haven't done it before, have a look at it.", "Solution_1": "sigh long time since i've done this. hope this is right.\n\n\n\na. [hide]dS=:sqrt:(1+(dy/dx):^2:)dx, (dy/dx):^2:=x:^2:/(25-x:^2:), dS=5/:sqrt:(25-x:^2:)dx.[/hide]\n\n\n\nb. [hide]A=2:pi:int(y*dS)=2:pi:int(:sqrt:(25-x:^2:)*5/:sqrt:(25-x:^2:)dx)=10:pi: (x from a to b). for all, it is 10:pi:[/hide]\n\n\n\nc. [hide]we use the same method as before except 5 is replaces by r. be 2r:pi: (x from -r to r) = 4:pi:r:^2:.[/hide]\n\n\n\nd. [hide]letting S(r) be the surface area of a sphere with radius r, and using concentric spheres, we see that integral (S(r))=V(r), where V(r) is the volume, because the volume is pretty much all those little areas added up. So, V(r)=4/3:pi:r:^3:+c. since V(0)=0, c=0.[/hide]", "Solution_2": "Your answer for a) is dL, not dS, but you used the correct formula in part b), so it's okay." } { "Tag": [ "geometry", "3D geometry" ], "Problem": "Eight cubes form the figure shown. If the side length of each cube is 3 cm, how many square centimeters are in the surface area of the figure?\n\n[asy]unitsize(1cm);\nfor(int i=0; i<4; ++i) {\ndraw((i,0)--(i,3)--(0.3+i,3.3),linewidth(0.7));\ndraw((0,i)--(3,i)--(3.3,i+0.3),linewidth(0.7));\n}\ndraw((0.3,3.3)--(3.3,3.3)--(3.3,0.3),linewidth(0.7));\ndraw((1,1)--(1.3,1.3),linewidth(0.7));\ndraw((1.3,2)--(1.3,1.3)--(2,1.3),linewidth(0.7));[/asy]", "Solution_1": "This is (12+16+4)(9)=288, unless I miscounted.", "Solution_2": "If the eight cubes were separate, their total surface area would be $ 8\\cdot 6\\cdot 3^2\\equal{}432$. Wherever two faces are glued, $ 2(3^2)\\equal{}18$ is taken from the surface area. There are $ 8$ places where this occurs, so the surface area is $ 432\\minus{}8\\cdot 18\\equal{}\\boxed{288}$.", "Solution_3": "First, I counted the S.A. of the figure, disregarding the fact that each side length of a cube is 3 cm. I used the S.A. formula 2(lw+wh+hl). Notice that this is a 3 by 3 by 1 figure. Then, I subtracted 2 for the two squares missing (front & back) and added 4 because there are four faces in place. Last, I multiplied the expression by 9 because S.A. is 2-dimensional, so 3^2=9. Therefore, (2(9+3+3)-2+4)9=288, Q.E.D." } { "Tag": [], "Problem": "Find the sum of the series\r\n$ 1-\\frac{1}{2}-\\frac{1}{6}-\\frac{1}{8}+\\frac{1}{3}-\\frac{1}{10}-\\frac{1}{12}-\\frac{1}{14}-\\frac{1}{16}+\\frac{1}{5}-...$", "Solution_1": "I'm not sure where $ \\frac{1}{4}$ is, so I'm not sure what kind of pattern you're trying to form. Could you clarify?" } { "Tag": [ "geometry", "trapezoid", "geometry proposed" ], "Problem": "Let $ABCD$ be a trapezoid (without parallel sides). Let $a, \\ b, \\ c, \\ d$ trapezoid sides and $d_{1}, \\ d_{2}$ its diagonals respectively.\r\n\r\nIs it true that $A_{ABCD}= \\frac{1}{2}\\sqrt{4(d_{1}\\cdot d_{2})^{2}-(a^{2}-b^{2}+c^{2}-d^{2})^{2}}$??", "Solution_1": "That formula works for any quadrilateral. :)", "Solution_2": "Cool.\r\n\r\nI didn't know that :o \r\n\r\nDoes anyone have a proof of this one?" } { "Tag": [], "Problem": "The phases of the moon alternate in the following sequence: new (N), waxing crescent (WxC), waxing half (WxH), waxing gibbous (WxG), full (F), waning gibbous (WaG), waning half (WaH), waning crescent (WaC), and back to new (N). This cycle can also be written: N, WxC, WxH, WxG, F, WaG, WaH. WaC, N, \u2026. If the current phase is WxC, what phase will it be in if the phase sequence continues for 12 more phases??", "Solution_1": "[hide]\nthe sequence is\nN, WxC, WxH, WxG, F, WaG, WaH. WaC, N, WxC, WxH, WxG, F, WaG, WaH. WaC,\nso if it is WxC then 12 more phases makes it WaG :) [/hide]", "Solution_2": "[hide]colts18 made it easy wiht the list of phases he provided. Just trace it with your finger, to obtain WaG[/hide]", "Solution_3": "[quote=\"236factorial\"]colts18 made it easy wiht the list of phases he provided.[/quote]\r\ncolts18 is a girl, by the way\r\n :D", "Solution_4": "[quote=\"math92\"][quote=\"236factorial\"]colts18 made it easy wiht the list of phases he provided.[/quote]\ncolts18 is a girl, by the way\n :D[/quote]\r\n\r\nWhoops, there are too many boys on this site :D :P", "Solution_5": "[quote=\"236factorial\"][quote=\"math92\"][quote=\"236factorial\"]colts18 made it easy wiht the list of phases he provided.[/quote]\ncolts18 is a girl, by the way\n :D[/quote]\n\nWhoops, there are too many boys on this site :D :P[/quote]\r\ni agree that there are too many boys on this site. \r\nI think that 96.141592653% of the people on AoPS are boys.\r\n[size=75]that number is just a guess[/size]", "Solution_6": "[quote=\"colts18\"]The phases of the moon alternate in the following sequence: new (N), waxing crescent (WxC), waxing half (WxH), waxing gibbous (WxG), full (F), waning gibbous (WaG), waning half (WaH), waning crescent (WaC), and back to new (N). This cycle can also be written: N, WxC, WxH, WxG, F, WaG, WaH. WaC, N, \u2026. If the current phase is WxC, what phase will it be in if the phase sequence continues for 12 more phases??[/quote]\r\n\r\nThe answer is WaG by just counting.", "Solution_7": "[quote=\"math92\"]\nI think that 96.141592653% of the people on AoPS are boys.\n[size=75]that number is just a guess[/size][/quote]\r\n :rotfl: yeah, just a guess... (a f**king precise guess :P )" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c,d,k$ be positive real numbers. Prove that\r\n\r\n$ \\left.\\frac{a\\plus{}b\\plus{}c\\plus{}d}{2}\\cdot\\left[\\frac{a}{(b\\plus{}c)^k}\\plus{}\\frac{b}{(c\\plus{}d)^k}\\plus{}\\frac{c}{(d\\plus{}a)^k}\\plus{}\\frac{d}{(a\\plus{}b)^k}\\right]\\geq \\frac{a}{(b\\plus{}c)^{k\\minus{}1}}\\plus{}\\frac{b}{(c\\plus{}d)^{k\\minus{}1}}\\plus{}\\frac{c}{(d\\plus{}a)^{k\\minus{}1}}\\plus{}\\frac{d}{(a\\plus{}b)^{k\\minus{}1}}\\right.$", "Solution_1": "[quote=\"marin.bancos\"]Let $ a,b,c,d,k$ be positive real numbers. Prove that\n\n$ \\left.\\frac {a \\plus{} b \\plus{} c \\plus{} d}{2}\\cdot\\left[\\frac {a}{(b \\plus{} c)^k} \\plus{} \\frac {b}{(c \\plus{} d)^k} \\plus{} \\frac {c}{(d \\plus{} a)^k} \\plus{} \\frac {d}{(a \\plus{} b)^k}\\right]\\geq \\frac {a}{(b \\plus{} c)^{k \\minus{} 1}} \\plus{} \\frac {b}{(c \\plus{} d)^{k \\minus{} 1}} \\plus{} \\frac {c}{(d \\plus{} a)^{k \\minus{} 1}} \\plus{} \\frac {d}{(a \\plus{} b)^{k \\minus{} 1}}\\right.$[/quote]\n\nThis follows immediately from your lemma [url=http://www.mathlinks.ro/viewtopic.php?p=1671199]here[/url] with $ k\\equal{}4$\n[quote=\"marin.bancos\"][b]MY LEMMA:[/b]\n\nIf: $ a_i\\geq0,b_i > 0,i \\equal{} 1,2,\\dots,k$ and $ k,n\\in N,k\\geq1$, then:\n$ \\frac {{a_1}^{n \\plus{} 1}}{b_1^n} \\plus{} \\frac {a_2^{n \\plus{} 1}}{b_2^n} \\plus{} \\cdots \\plus{} \\frac {{a_k}^{n \\plus{} 1}}{b_k^n}\\geq\\frac {a_1 \\plus{} a_2 \\plus{} \\cdots \\plus{} a_k}{b_1 \\plus{} b_2 \\plus{} \\cdots \\plus{} b_k}\\cdot\\left(\\frac {{a_1}^n}{b_1^{n \\minus{} 1}} \\plus{} \\frac {{a_2}^n}{b_2^{n \\minus{} 1}} \\plus{} \\cdots \\plus{} \\frac {{a_k}^n}{b_k^{n \\minus{} 1}}\\right)$\n[/quote]\r\n\r\n(putting cyclically $ a_1\\equal{}\\frac {a}{b \\plus{} c}$ etc., $ b_1\\equal{}a$ etc. and $ n\\equal{}k\\minus{}1$), \r\nfollowed by Shapiro for $ 4$ variables :lol:", "Solution_2": "Taking into account that my lemma was improved for $ k>0$, we get the following form of it:\r\n\r\nFor $ a_i\\geq0,b_i>0,i\\equal{}1,2,\\dots,n,\\ n\\in N,n\\geq1$ and $ k>0,$ then:\r\n$ \\left.\\frac{{a_1}^{k\\plus{}1}}{b_1^k}\\plus{}\\frac{a_2^{k\\plus{}1}}{b_2^k}\\plus{}\\cdots\\plus{}\\frac{{a_n}^{k\\plus{}1}}{b_n^k}\\geq\\frac{a_1\\plus{}a_2\\plus{}\\cdots\\plus{}a_n}{b_1\\plus{}b_2\\plus{}\\cdots\\plus{}b_n}\\cdot\\left(\\frac{{a_1}^k}{b_1^{k\\minus{}1}}\\plus{}\\frac{{a_2}^k}{b_2^{k\\minus{}1}}\\plus{}\\cdots\\plus{}\\frac{{a_n}^k}{b_n^{k\\minus{}1}}\\right)\\right.$\r\n\r\nThe inequality could be obtained for $ n\\equal{}4$, $ a_1\\equal{}a, a_2\\equal{}b, a_3\\equal{}c, a_4\\equal{}d$ and $ b_1\\equal{}a(b\\plus{}c), b_2\\equal{}b(c\\plus{}d), b_3\\equal{}c(d\\plus{}a), b_4\\equal{}d(a\\plus{}b)$.\r\nIn fact, this is a new generalization for Shapiro's inequality ($ n\\equal{}4$)." } { "Tag": [ "logarithms" ], "Problem": "$ 3^{2457}\\plus{}1\\equal{}x$\r\nwithout working out the answer , say how many numbers , and which number, how many times used in the answer $ (x)$.", "Solution_1": "[quote=\"binomial_4eva\"]$ 3^{2457} \\plus{} 1 \\equal{} x$\nwithout working out the answer , say how many numbers , and which number, how many times used in the answer $ (x)$.[/quote]\r\n\r\nSimply put, WHAT!?\r\n\r\nAre you looking for a table along the lines of\r\n\r\n1 : 1123\r\n2 : 12\r\n3 : 123\r\n4 : ....\r\n\r\nOr what?", "Solution_2": "this is what i mean, i will give an example here\r\n\r\nwhen $ 100$ is divided by $ 5$ the answer is $ 20$\r\nthere is one $ 2$ and one $ 0$ in $ 20$ .\r\n\r\nhow many of each numbers from the set $ {1,2,3,4...9}$ are used in the answer for this , $ 3^{2457}\\plus{}1\\equal{}?$ i hope its clear now.", "Solution_3": "Good luck with that, I have calculated it and its fairly extreme (there doesn't seem to be an obvious way to do it otherwise)", "Solution_4": "You want the number of digits in $ 3^{2457}\\plus{}1$?\r\n\r\n[hide=\"solution\"]Well, since the units digit of $ 3^{2457}$ is not 9, we just need to find the number of digits in $ 3^{2457}$. We take the logarithm base 10:\n\n$ \\log_{10} 3^{2457}\\equal{}2457\\log_{10} 3$\n\nNow you run for your calculator.[/hide]", "Solution_5": "[quote=\"1=2\"]You want the number of digits in $ 3^{2457} \\plus{} 1$?\n[/quote]\r\n\r\nHe also wants this\r\n\r\n[hide]\n$ {132, 109, 123, 117, 111, 108, 113, 111, 126, 123}$\n[/hide]\r\n\r\nBut I fail to see an elementary way of obtaining that", "Solution_6": "He's asking how many of each digit there are.", "Solution_7": "[quote=\"xpmath\"]He's asking how many of each digit there are.[/quote]\r\n\r\nI cheated and used Maple to get that\r\n\r\n\r\n[hide=\"in here\"]$ 3^{2457}\\plus{}1$\n\n=193607798301087061431583144629329214508737133042108005358818293959351413848579400527015793279969882281884356020894409145212864135998556816842566331095607078650385559292181945035499328241768753060146904329351320388912443191251507252624765009165597363043514732799372576425212921359895791760004246711770732571379038647670332038734854078104264596116768436003570491538730897781713543181289565170181991061594063941081426799389938134125993339341535241610822303654061326578587839902022947739187987221211462138391126919615812857320948025429485836670104664182146679105139921625679465998786170218908742166411834195388405755375661393299315666296665265805234346096462806656269737392537002483513378697189820794421975473420652985535357099489202204950401867071749069193914930675491887606241533434504857738825465090097201796639161408286119392743358706021871174773610403472079491128480628446770638160946727043074441114918267733536174551156877346511943390051908666674282501744003036044614149109709884253208033283550058065252958043809432814044770021395727636507258389606139410944571135838107597919228936011421720714984693698605181870575257822353840898230459634239792074723175097281490221798564[/hide]\r\n\r\n\r\n@SimonM: Are those the numbers of the digits?", "Solution_8": "[quote=\"1=2\"]\n@SimonM: Are those the numbers of the digits?[/quote]\r\n\r\nIndeed they are, yay mathematica!\r\n\r\n[hide]DigitCount[3^{2457} + 1][/hide]", "Solution_9": "@1=2 ya i mean how do you find how many for example $ 2$ is in that long number. without using maple.", "Solution_10": "You use Mathematica instead.\r\n\r\nSeriously, I really doubt there is a good way, known or unknown." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Let regular polygon has $1999$ vertexs .$41$ of them are painted blue and remain are painted yellow.Find number of isoceles triangle which has tthree vertexs same colour.", "Solution_1": "Do you mean that this number is independant of the positions of the blue vertices? :huh: \r\n\r\nPierre.", "Solution_2": "YES>>>>>>>", "Solution_3": "I think that there were many person in VN solved it" } { "Tag": [ "integration", "real analysis", "real analysis unsolved" ], "Problem": "If $f(x) \\geq 0$ and $\\int_0^{\\infty} f(x) dx=0$, does this imply, $\\int_0^{\\infty} f(x) x dx=0$ ? or there is a counter example?\r\n\r\nThanks.", "Solution_1": "Yes, it does.", "Solution_2": "If $f\\geq 0$ and $\\int_{\\mathbb{R}} f(x)dx=0$, then $f(x)=0$ a.e on $\\mathbb{R}$. It follows that $xf(x)=0$ a.e on $\\mathbb{R}$, which in turn implies that\r\n\\[ \\int_{\\mathbb{R}} xf(x)dx=0 \\]" } { "Tag": [ "induction", "function", "calculus", "derivative", "inequalities", "inequalities proposed" ], "Problem": "a) For each $n \\ge 2$, find the maximum constant $c_{n}$ such that\r\n$\\frac 1{a_{1}+1}+\\frac 1{a_{2}+1}+\\ldots+\\frac 1{a_{n}+1}\\ge c_{n}$\r\nfor all positive reals $a_{1},a_{2},\\ldots,a_{n}$ such that $a_{1}a_{2}\\cdots a_{n}= 1$.\r\n\r\nb) For each $n \\ge 2$, find the maximum constant $d_{n}$ such that\r\n$\\frac 1{2a_{1}+1}+\\frac 1{2a_{2}+1}+\\ldots+\\frac 1{2a_{n}+1}\\ge d_{n}$\r\nfor all positive reals $a_{1},a_{2},\\ldots,a_{n}$ such that $a_{1}a_{2}\\cdots a_{n}= 1$.", "Solution_1": "The following general problem holds\r\n\r\nFor all positive real numbers $a_{1},a_{2},...,a_{n}$ with product $n$ then\r\n\\[\\sum \\frac{1}{ka_{i}+1}\\ge \\min\\left\\{1,\\frac{n}{k+1}\\right\\}\\]\r\n\\[\\sum \\frac{1}{ka_{i}+1}\\le \\max\\left\\{n-1,\\frac{n}{k+1}\\right\\}\\]\r\n\r\nI solved it by \"induction\", \"UMV\" or LCF-RCF theorem. I think induction is most elementary way.", "Solution_2": "Edriv could you post your solution ?Thank you", "Solution_3": "An idea that works is studying the function $f : \\mathbb R \\to \\mathbb R$, $f(x) = \\frac1{k e^{x}+1}$.\r\nIf you look at the second derivative, then you'll see that it's concave on one interval and convex on the rest.\r\nOf course, some nasty computations remain, but they can be handled.\r\n\r\nMildorf solved a Romanian TST problem with this idea. I think you know which one I'm talking about.", "Solution_4": "[quote=\"edriv\"][b]Problem 6[/b]\na) For each $n \\ge 2$, find the maximum constant $c_{n}$ such that\n$\\frac 1{a_{1}+1}+\\frac 1{a_{2}+1}+\\ldots+\\frac 1{a_{n}+1}\\ge c_{n}$\nfor all positive reals $a_{1},a_{2},\\ldots,a_{n}$ such that $a_{1}a_{2}\\cdots a_{n}= 1$.[/quote]\r\n\r\nFor $a,\\ b>0$, we have $\\frac{a}{a+1}+\\frac{b}{b+1}\\geq \\frac{a+b}{a+b+1}$, thus $\\\u3000a_{1}+a_{2}+\\cdots+a_{n}\\geq n\\Longrightarrow \\frac{a_{1}}{a_{1}+1}+\\frac{a_{2}}{a_{2}+1}+\\cdots+\\frac{a_{n}}{a_{n}+1}\\geq \\frac{n}{n+1}$ holds inductively.\r\n\r\nRewriting this inequality gives \r\n\r\n\\[n-\\sum_{k=1}^{n}\\frac{1}{a_{k}+1}\\geq 1-\\frac{1}{n+1}\\Longleftrightarrow \\sum_{k=1}^{n}\\frac{1}{a_{k}+1}\\leq \\frac{n^{2}}{n+1}\\].", "Solution_5": "[quote=\"silouan\"]Edriv could you post your solution ?Thank you[/quote]\r\nSame idea http://www.mathlinks.ro/Forum/viewtopic.php?t=143678", "Solution_6": "[hide=\"a\"]\nFor all $n\\geq 2$, we have $c_{n}=1$. To see this, choose $a_{1}=\\epsilon^{n-1}$ and $a_{k}=\\frac{1}{\\epsilon}$ for $k\\neq 1$ and send $\\epsilon \\rightarrow 0$ to have the left hand side approach 1. Therefore, there cannot be a better constant than 1.\n\nTo actually prove the lower bound, WLOG let $a_{1}\\leq a_{2}\\leq \\cdots \\leq a_{n}$. Then $a_{1}a_{2}\\leq 1$, so that\n\\[\\sum_{k=1}^{n}\\frac{1}{a_{k}+1}\\geq \\frac{1}{a_{1}+1}+\\frac{1}{a_{2}+1}\\]\n\\[=\\frac{1}{a_{1}+1}+\\frac{a_{1}}{a_{2}+a_{1}a_{2}}\\geq \\frac{1}{a_{1}+1}+\\frac{a_{1}}{a_{1}+1}=1\\]\nas desired.\n[/hide]\n[hide=\"b\"]\nFirst, $d_{2}=\\frac{2}{3}$, attainable when $a_{1}=a_{2}=1$. To prove the lower bound, let $a=a_{1}$, then $a_{2}=\\frac{1}{a}$. The two-variable inequality is equivalent to\n\\[\\frac{1}{2a+1}+\\frac{a}{a+2}\\geq \\frac{2}{3}\\iff 3(a+2)+3a(2a+1)\\geq 2(2a+1)(a+2)\\]\n\\[\\iff (a-1)^{2}\\geq 0\\]\nFor $n\\geq 3$, we have $d_{n}=1$. This is very similar to part (a): we can let $a_{1}=\\epsilon^{n-1}$ and $a_{k}=\\frac{1}{\\epsilon}$ for $k\\neq 1$ and send $\\epsilon \\rightarrow 0$ to show that there is no better constant than 1.\n\nTo prove the lower bound, we can WLOG let $a_{1}\\leq a_{2}\\leq \\cdots \\leq a_{n}$ and then forget about all terms beyond the third. We also have $a_{1}a_{2}a_{3}\\leq 1$. Now let\n\\[x=\\sqrt[9]{\\frac{a_{2}a_{3}}{a_{1}^{2}}},\\ y=\\sqrt[9]{\\frac{a_{1}a_{3}}{a_{2}^{2}}},\\ z=\\sqrt[9]{\\frac{a_{1}a_{2}}{a_{3}^{2}}}\\]\nThen $a_{1}\\leq \\frac{1}{x^{3}},\\ a_{2}\\leq \\frac{1}{y^{3}},\\ a_{3}\\leq \\frac{1}{z^{3}}$, and $xyz=1$. Then\n\\[\\sum_{k=1}^{n}\\frac{1}{a_{k}+1}\\geq \\sum_{k=1}^{3}\\frac{1}{a_{k}+1}\\geq \\frac{x^{3}}{x^{3}+2}+\\frac{y^{3}}{y^{3}+2}+\\frac{z^{3}}{z^{3}+2}\\]\n\\[=\\frac{x^{2}}{x^{2}+2yz}+\\frac{y^{2}}{y^{2}+2xz}+\\frac{z^{2}}{z^{2}+2xy}\\]\n\\[\\geq \\frac{x^{2}}{x^{2}+y^{2}+z^{2}}+\\frac{y^{2}}{x^{2}+y^{2}+z^{2}}+\\frac{z^{2}}{x^{2}+y^{2}+z^{2}}=1\\]\nas desired.\n[/hide]", "Solution_7": "[quote=\"kunny\"]For $ a,\\ b > 0$, we have $ \\frac {a}{a + 1} + \\frac {b}{b + 1}\\geq \\frac {a + b}{a + b + 1}$\n[/quote]\nHow to prove this inequality? :?: :maybe: \n\n[quote=\"kunny\"]$ \\\u3000a_{1} + a_{2} + \\cdots + a_{n}\\geq n\\Longrightarrow \\frac {a_{1}}{a_{1} + 1} + \\frac {a_{2}}{a_{2} + 1} + \\cdots + \\frac {a_{n}}{a_{n} + 1}\\geq \\frac {n}{n + 1}$ holds inductively.\n\nRewriting this inequality gives\n\\[ n - \\sum_{k = 1}^{n}\\frac {1}{a_{k} + 1}\\geq 1 - \\frac {1}{n + 1}\\Longleftrightarrow \\sum_{k = 1}^{n}\\frac {1}{a_{k} + 1}\\leq \\frac {n^{2}}{n + 1}\n\\]\n.[/quote]\r\nSo...what's the correct value for $ c_n$? I think that, it is $ \\frac{n}{n+1}$ But what about the proof of scorpioa119?\r\n\r\nI am really sorry for bumping this old thread after years...But I really want to know the answer. Please help me. :(\r\nThanks in advance.", "Solution_8": "[quote=edriv]a) For each $n \\ge 2$, find the maximum constant $c_{n}$ such that\n$\\frac 1{a_{1}+1}+\\frac 1{a_{2}+1}+\\ldots+\\frac 1{a_{n}+1}\\ge c_{n}$\nfor all positive reals $a_{1},a_{2},\\ldots,a_{n}$ such that $a_{1}a_{2}\\cdots a_{n}= 1$.\n\nb) For each $n \\ge 2$, find the maximum constant $d_{n}$ such that\n$\\frac 1{2a_{1}+1}+\\frac 1{2a_{2}+1}+\\ldots+\\frac 1{2a_{n}+1}\\ge d_{n}$\nfor all positive reals $a_{1},a_{2},\\ldots,a_{n}$ such that $a_{1}a_{2}\\cdots a_{n}= 1$.[/quote]\n[b][u][color=#f00]Solutions :[/color][/u][/b]\na) I wanna prove that $c_n=1$, without loss of generality, assume that $a_1 \\ge a_2 \\ge ... \\ge a_n$. \n\nIf $a_1 \\ge a_2 \\ge ... \\ge a_{n-1} \\ge 1 \\ge a_n$ we have $\\frac{1}{x+1}+ \\frac{1}{x+1} \\ge \\frac{2}{\\sqrt{xy}+1} \\Leftrightarrow \\frac{(\\sqrt{x}-\\sqrt{y})^2(\\sqrt{xy}-1)}{(x+1)(y+1)} \\ge 0$ which is obvious for $x,y \\ge 1$.\n\nThen by using a simple induction we can prove $\\sum_{k=1}^{n-1} \\frac 1{a_{k}+1} \\ge \\frac{n-1}{\\sqrt[n-1]{a_1a_2....a_{n-1}}+1}$ therefore $\\sum_{k=1}^{n} \\frac 1{a_{k}+1} \\ge \\frac{n-1}{\\sqrt[n-1]{a_1a_2....a_{n-1}}+1}+\\frac 1{a_{n}+1}$\n\nLet $b_i=\\sqrt[n-1]{a_i}$ then $b_1b_2...b_n=1$, so, it remains to prove that $\\frac{n-1}{b_1b_2....b_{n-1}+1}+\\frac 1{b_{n}^{n-1}+1}>1$ \n\nOr, $\\frac{(n-1)b_n}{1+b_n}+\\frac 1{b_{n}^{n-1}+1}>1$, or $\\frac{a_n\\left((n-2)b_n^{n-1}+n-1-b_n^{n-2}\\right)}{\\left(1+b_n\\right)\\left(b_{n}^{n-1}+1\\right)} \\ge 0$ which is obvious for $n \\ge 2$ and $02$ then $\\exists k \\le n-1$, such that $a_1 \\ge a_2 \\ge ... \\ge a_k \\ge 1 \\ge ... \\ge a_n$\n\nTherefore $\\sum_{k=1}^{n} \\frac 1{a_{k}+1} > \\frac 1{a_{n-1}+1}+\\frac 1{a_{n}+1} \\ge \\frac 1{2}+\\frac 1{2}=1$\n\nb) We can do it with the same method using the cases :\nFor $n=2$ we get $\\frac 1{2a_{1}+1}+\\frac 1{2a_{2}+1}=\\frac 23$, \n\n[color=#00f][b]1) [/b][/color]If $a_1 \\ge a_2 \\ge ... \\ge a_{n-1} \\ge 1 \\ge a_n$ for $n \\ge 3$ we get $\\sum_{k=1}^{n} \\frac 1{2a_{k}+1} \\ge \\frac{n-1}{2\\sqrt[n-1]{a_1a_2....a_{n-1}}+1}+\\frac 1{2a_{n}+1} \\ge 1$\n\n[color=#00f][b]2) [/b][/color] If $ a_1 \\ge a_2 \\ge ... \\ge a_{n-2} \\ge 1 \\ge a_{n-1} \\ge a_n$ for $n \\ge 3$ we get $\\sum_{k=1}^{n} \\frac 1{2a_{k}+1} \\ge \\frac{n-1}{2\\sqrt[n-1]{a_1a_2....a_{n-2}}+1}+\\frac 1{2a_{n-1}+1}+\\frac 1{2a_{n}+1} \\ge 1$\n\nThe last inequality is equivalente to $\\frac{1}{2x+1}+\\frac{1}{2y+1}+\\frac{1}{2z+1} \\ge 1$ for $xyz=1$ which is easy!!\n[color=#00f][b]3) [/b][/color] If $a_1 \\ge a_2 \\ge ... \\ge a_k \\ge 1 \\ge ... \\ge a_n$ for $n \\ge 4$ and $k \\le n-2$ \n\nWe get $\\sum_{k=1}^{n} \\frac 1{2a_{k}+1} > \\frac 1{2a_{n-2}+1}+\\frac 1{2a_{n-1}+1}+\\frac 1{2a_{n}+1} \\ge \\frac13+ \\frac13+ \\frac13=1$.\nThen $d_n= \\frac23$ and equality holds if and only if $n=2$.." } { "Tag": [ "logarithms" ], "Problem": "The value of $ \\log_5 \\frac {(125)(625)}{25}$ is equal to:\n\n$\\textbf{(A)}\\ 725 \\qquad\n\\textbf{(B)}\\ 6 \\qquad\n\\textbf{(C)}\\ 3125 \\qquad\n\\textbf{(D)}\\ 5 \\qquad\n\\textbf{(E)}\\ \\text{None of these}$", "Solution_1": "Note that $ \\frac {125 \\cdot 625}{25} \\equal{} \\frac {5^3 \\cdot 5^4}{5^2} \\equal{} 5^5$\r\n\r\nSo $ \\frac {125 \\cdot 625}{25}$ equals $ log_55^5 \\equal{} \\boxed{5}$ (D)", "Solution_2": "Note that $125\\cdot625=78125$ and $\\frac{78125}{25}=3125=5^5$. \n\nSo $\\log_5(5^5)=\\boxed{\\textbf{(D)}\\ 5}$. ", "Solution_3": "[quote=megarnie]Note that $125\\cdot625=78125$ and $\\frac{78125}{25}=3125=5^5$. \n\nSo $\\log_5(5^5)=\\boxed{\\textbf{(D)}\\ 5}$.[/quote]\n\nBIG BUMP!\n\n$$\\frac{125\\cdot625}{5} = 125 \\cdot \\frac{625}{25} = 125\\cdot \\frac{25^2}{25} = 125 \\cdot 25 = 5^3 \\cdot 5^2 = 5^5.$$ So then $$\\log_5 5^5 = 5 \\implies \\mathbf{D}.$$", "Solution_4": "Notice that 125, 625, and 25 are all powers of 5! So, we can rewrite it to solve the problem. Answer: D." } { "Tag": [ "geometry", "ratio", "trigonometry", "geometric transformation", "rotation", "geometry proposed" ], "Problem": "You're given an acute triangle ABC. AD is height. There is a line x that passes through point D (different from BC). On the line x points E and F are taken such that the angles AEB and AFC are 90. L is midpoint of the segment EF and M is midpoint of BC. Find =xyz.\r\n=>(xyz)^2<=xyz(x+y+z).\r\nWe also have:xyz(x+y+z)<=[(xy+yz+zx)^2]/3<=[(x^2+y^2+z^2)^2]/3\r\n=>(xyz)^2<=[(x^2+y^2+z^2)^2]/3.\r\n=>xyz<=(x^2+y^2+z^2)/ :sqrt: 3\r\nThus:(x^2+y^2+z^2)/(xyz)>= :sqrt: 3.\r\n=>min[(x^2+y^2+z^2)/(xyz)]= :sqrt: 3.\r\n\"=\" happens when:x=y=z= :sqrt: 3.\r\nThat is \"dpcm\".", "Solution_2": "[quote=\"From Neptune to Mars\"]I think it isn't difficult problem.\nWe have: $ x \\plus{} y \\plus{} z \\ge xyz.$\n:arrow:$ (xyz)^2 \\le xyz(x \\plus{} y \\plus{} z).$\nWe also have\n:$ xyz(x \\plus{} y \\plus{} z)\\le[(xy \\plus{} yz \\plus{} zx)^2]/3\\le\\frac {(x^2 \\plus{} y^2 \\plus{} z^2)^2}{3}$\n:arrow:$ (xyz)^2\\le\\frac {[(x^2 \\plus{} y^2 \\plus{} z^2)^2]}{3}.$\n:arrow:$ xyz < \\equal{} (x^2 \\plus{} y^2 \\plus{} z^2)/ \\sqrt 3$\nThus:\n$ (x^2 \\plus{} y^2 \\plus{} z^2)/(xyz)\\ge \\sqrt 3.$\n:arrow: $ min[(x^2 \\plus{} y^2 \\plus{} z^2)/(xyz)] \\equal{} \\sqrt 3.$\n\"=\" happens when:x=y=z= \\sqrt 3.\nThat is \"dpcm\".[/quote]" } { "Tag": [ "quadratics", "AMC", "AIME", "geometry", "FTW", "probability", "calculus" ], "Problem": "Kinda like Jeff Foxworthy's you know your a redneck if....\r\nReply with additions.\r\n1.You know your a Math geek if you wanted an N-spire for christmas.\r\n2.You Know your a Math Geek if you prove your math teacher wrong on a daily basis.\r\n3.You Know your a Math Geek if your favorite website is AoPS.\r\n4. You Know your a Math Geek if one of your hobby is pwning people at mental math.", "Solution_1": "[quote=\"monkeygirl13\"]3.You know you're a Math Geek if your home page is AoPS.[/quote]\r\n\r\nFixed. :P", "Solution_2": "1.You know your a Math geek if you wanted an N-spire for christmas. \r\n2.You Know your a Math Geek if you prove your math teacher wrong on a daily basis. \r\n3.You Know your a Math Geek if your homepage is AoPS. \r\n4. You Know your a Math Geek if one of your hobby is pwning people at mental math.\r\n5. You debate with your stuffed animals on how to REALLY extract a quaratic equation.", "Solution_3": "6. You know your a math geek if your new years resolution is to read to 3 AoPS books.\r\n7. You know your a math geek if you read an AoPS book in class, while teh teacher is helping the others factor quads.", "Solution_4": "5. Change quadratic to quartic please.\r\n6. Just no.", "Solution_5": "You know you're not the antithesis of a math geek if you spell \"quadratic\" correctly. \r\n\r\nAnd will you guys actually post attributes of math geeks instead of attributes of normal people?\r\n\r\nEDIT: Wait, nvrm. Didn't see the stuffed animals part.", "Solution_6": "you play with your stuffed animals???\r\n :P", "Solution_7": "You know if you are a math geek if...\r\n1. Your homepage is Wolfram|Alpha\r\n2. You have 50 dollars worth of AMC and AIME tests\r\n3. You have about 10 geometry books, 5 of which are Chinese\r\n\r\nThose are all true for me.", "Solution_8": "Stuffed Animals are beast.\r\n\r\n11) You did all your addition in kindergarten without using counting blocks like the other kids.\r\n12) You laugh when you see someone using their fingers to count how many hours are from 9 am to 5 pm.\r\n13) You ask your mom if she can make pumpkin 3.14159... on the sqrt400 + 2^2 day of November, 2005 + 1\r\n14) You're awesome.", "Solution_9": "12. You know your a math geek if your math teacher searches your backpack for a calc cuz you finished 100 problems in 10 minutes or less.", "Solution_10": "13. you know ur a math geek if you did all the questions in the lesson because you forgot to write down the homework and you're too lazy to call ur friends.\r\n\r\n[size=134]ARE YOU KIDDING ME, STUFFED ANIMALS [i]ARE[/i] AWESOME![/size]", "Solution_11": "[quote=\"piegraph\"]13. you know ur a math geek if you did all the questions in the lesson just for the fun of it.\n[size=134]ARE YOU KIDDING ME, STUFFED ANIMALS [i]ARE[/i] [color=red]NOT[/color] AWESOME![/size][/quote]\r\n\r\nfixed.", "Solution_12": "I like stuffed animals :maybe:\r\n\r\n- ...when you ask your engineering teacher for $ 12\\pi$ square inches of wood for your project.\r\n\r\nYeah, that happened to me. It took me a while to realize.", "Solution_13": "Stuffed Animals FTW!\r\n15. You know your a math geek if you named your stuffed cat pi.", "Solution_14": "16. If you have memorized 134digits of pi, and organized a pi group with 5 followers.\r\nThats what happened to me.", "Solution_15": "[quote=\"eagle2\"]mathlete. to tell you the truth you are sickkk. That girl is UGLY too.\nfor my it would\nbe nice + smart + not tempermental = good. so basically piegraphs things. sick minded creature.[/quote]\nYou are just saying that to piss me off...and yes she is nice. So that DOES work out.", "Solution_16": "ok the truth she is not the best i've seen. She is pretty ugly for a cheerleader. even my sister is better then her.", "Solution_17": "I don't know if you need to, but did you get her permission to put the photo on the forum? Just in case she sues or something.....but since you said she was nice, it should be fine....\n\nSorry for going even more off topic.....", "Solution_18": "Oh goodness.\nThis is getting off topic.\nYou know you're a math geek if you had to prove to your math teacher that 'x cubed' means 'x to the third power', not 'x to the fourth power', and proved it to her by showing her why, by showing how it got it's name by the volume of a cube, and in this process, you were in mental facepalm mode.", "Solution_19": "You know your a math geek if you sign up for all of the AoPS classes at once.", "Solution_20": "If this gets really off topic again, can we please have it locked?\n\n\"......\" you do math at lunch.", "Solution_21": "You know you are a Math Geek if....\n\n1. People don't understand you.\n2. You raise your hand in your math class in order to correct the teacher for about 10 times a day.\n3. People do not get your joke.\n4. You get about 10 percent higher in your calculus or similarly advanced class (for a middle school student) compared to somewhat nerdy student in your middle school gets in Honors Algebra class.\n5. Your life is AoPS.\n6. You would rather buy a full licence for Mathematica instead of WOW CD.", "Solution_22": "[quote=\"lightning\"]I don't know if you need to, but did you get her permission to put the photo on the forum? Just in case she sues or something.....but since you said she was nice, it should be fine....\n\nSorry for going even more off topic.....[/quote]\nShe has already posted the photo publicly on the internet. And it is not like she is going to come here. She may be a math girl, but I doubt she has ever heard of aops, considering she is not even on my school's math team (she thought cheerleading was a bigger priority)", "Solution_23": "Can someone please lock this? It is getting extremely off-topic.\n\nIronically, this is a \"math geek\" thread and we're talking about girls.", "Solution_24": "I do advise that pictures are not posted on the forum without explicit permission. Also, posts like \n\n[quote=\"eagle2\"]mathlete. to tell you the truth you are sickkk. That girl is UGLY too.[/quote]\n\nare completely unnecessary to make. Insulting others' opinions is untasteful and spammy.", "Solution_25": "haha i got pwned.", "Solution_26": "You know you're a math geek if when you saw \"10!,\" you thought \"What were the last two non-zero digits?,\" because AMC 2010 #24 is still in the back of you mind.\n\nBTW, the answer is.. Math geek if you found this out or knew it already...[hide]88[/hide]", "Solution_27": "If you know random math facts and show them off at school.\nIf you get bullied every time after math.", "Solution_28": "You know you are a math geek if you are on Lehigh Valley Fire\n\nNOTE THAT I JUST SAID \"IF\", NOT \"IF AND ONLY IF\"", "Solution_29": "You are a math geek if you corrected your teacher by pointing out 1/2>1/2 is a false statement in third grade." } { "Tag": [ "geometry", "analytic geometry", "absolute value" ], "Problem": "i need help. what is\r\n1) the hockey stick identity\r\n2)the shoelace theorem. i saw it explained once but i didnt really get it\r\n\r\nalso, a complementary problem would be nice\r\n\r\nthanks!", "Solution_1": "um i cant help you with the first one but the way i have understood the shoelace theorum is that you take the verticies of a polygon (not sure if it works with triangles) and list them. and multiply diagonally. e.g. If a polygon has verticies of $ (a,b) (c,d) (e,f) (g,h) (i,j) (k,l)$ and $ (m,n)$ what is the area of this polygon?\r\n\r\nWe list them like this:\r\n\r\na,b\r\nc,d\r\ne,f\r\ng,h\r\ni,j\r\nk,l\r\nm,n\r\n\r\nWe then multiply diagonally from top right to bottom left, then subtract the sums of multiplying bottom right to top left yeilding --- (($ a\\times{d}$) + ($ c\\times{f}$) + ($ e\\times{f}$) + ($ g\\times{j}$) + ($ i\\times{l}$) + ($ k\\times{n}$)) - (($ m\\times{l}$) + ($ k\\times{j}$) + ($ i\\times{h}$) + ($ g\\times{f}$) + ($ e\\times{d}$) + ($ c\\times{b}$))", "Solution_2": "try looking up the hockey stick identity on the internet figure out what it is how to apply it and its origins in pascals triangle.", "Solution_3": "PASCALS TRIANGLE. Draw a hockey stick. and add up the ones on the longer side. O.O its amazing", "Solution_4": "find the area of a quadrilateral with coordinates (0,0), (0,4), (4,4), (4,0),\r\n\r\nwell we know this is a square with side length 4 so the area is 16. so if we do that using shoelace we get\r\n\r\n0 0\r\n0 4\r\n4 4\r\n4 0\r\n(remember you must put the points in order either clockwise or counterclockswise otherwise it won't work)\r\n\r\nit is (0*4 + 0*0 + 4*0) - (0*0 + 4*4 + 4*4) = -32 now you take the absolute value of that and you get 32, now you divide that be 2 and VOILA 16.", "Solution_5": "I think you forgot something, cognos, although it didn't really matter since the first point was (0,0)", "Solution_6": "You have to...... write the first coordinate again.......", "Solution_7": "but you can't go like \r\n\r\n0 0\r\n4 4\r\n4 0\r\n0 4\r\n\r\ntry it. ITs messed up\r\n\r\n\r\nOOPS THE FIRST COORDINATE I TOTALLY FORGOT. SORRY ABACADAE", "Solution_8": "HOCKEY STICK IDENTITY- It should discussed thoroughly in the Intro to Counting/Probability book (in the later chapters). You also might want to try looking it up in the AopsWiki.....", "Solution_9": "[quote=\"mathcrazed\"]PASCALS TRIANGLE. Draw a hockey stick. and add up the ones on the longer side. O.O its amazing[/quote]\n\nhow do you draw a hockey stick?\n\n\n[quote]\t\nHOCKEY STICK IDENTITY- It should discussed thoroughly in the Intro to Counting/Probability book (in the later chapters). You also might want to try looking it up in the AopsWiki.....[/quote]\r\n[url=http://www.artofproblemsolving.com/Wiki/index.php/Combinatorial_identity]wiki doesnt really help...[/url]", "Solution_10": "yes, i would like to know the hockey stick identity as well. first of all i get a headache from looking at the wiki page and second i dont even understand what it does \"summation\" whats that???\r\n\r\nPlease any and all help for me and abacadaea would be greatly apreciated!!!", "Solution_11": "Ahem...Google does better\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=515429#515429", "Solution_12": "i under stand it now :lol: .\r\n\r\nhow do i used it??..... :huh:", "Solution_13": "Say they asked you what $ \\binom{1}{n}\\plus{}\\binom{2}{n}\\plus{}\\binom{3}{n}\\plus{}\\binom{4}{n}\\plus{}\\binom{5}{n}\\plus{}\\binom{6}{n}$ equals.\r\nYou would use the hockey stick identity then.", "Solution_14": "for some reason it seems to me that from what i have heard the hockey stick identity is fairly rare in problems. Could some one give me a problem that involves the use of the hockey stick identity but is not like the one posted before me.", "Solution_15": "[quote=\"ZhangPeijin\"]Say they asked you what $ \\binom{1}{n} \\plus{} \\binom{2}{n} \\plus{} \\binom{3}{n} \\plus{} \\binom{4}{n} \\plus{} \\binom{5}{n} \\plus{} \\binom{6}{n}$ equals.\nYou would use the hockey stick identity then.[/quote]\r\n\r\n\r\nerm... zolojetto.....right there\r\n\r\nand btw, could someone post a solution to that. sorry, im still confused[size=59]only a bit[/size]", "Solution_16": "Peijin, that's not Hockey stick unless $ n \\equal{} 0$ or $ n\\equal{}1$, which makes the sum easy to compute anyways" } { "Tag": [], "Problem": "Which of the following correctly represents $ \\text{CO}_{2}$, and why? Choices $ (2)$ and $ (3)$ both look reasonable to me. Thanks in advance!", "Solution_1": "1. octet rule\r\n2. maximize # of bonds", "Solution_2": "the correct answer would be (3).\r\n(2) is wrong because oxygen have only 6 electrons.\r\nEvery element must have at least eight electrons surround it ( with the exception if boron and beryllium)", "Solution_3": "[quote=\"I_thirst_for_knowledge\"](with the exception if boron and beryllium)[/quote]\r\n\r\nAnd hydrogen. :)", "Solution_4": "[quote=\"Carcul\"][quote=\"I_thirst_for_knowledge\"](with the exception if boron and beryllium)[/quote]\n\nAnd hydrogen. :)[/quote]\r\n\r\nThanks a lot! - I got it now. :)", "Solution_5": "Note also that an element can have more than eight electrons \"around it\", provided it doesn't belong to the second row of the Periodic Table." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "For each positive integer $ n\\leq 49$ we define the numbers $ a_{n}\\equal{}3n\\plus{}\\sqrt{n^{2}\\minus{}1}$ and $ b_{n}\\equal{}2(\\sqrt{n^{2}\\plus{}n}\\plus{}\\sqrt{n^{2}\\minus{}n})$.Prove that there exist two integers A,B such that \r\n$ \\sqrt{a_{1}\\minus{}b_{1}}\\plus{}\\sqrt{a_{2}\\minus{}b_{2}}\\plus{}...\\plus{}\\sqrt{a_{49}\\minus{}b_{49}}\\equal{}A\\plus{}B\\sqrt{2}$", "Solution_1": "$ a_{n}-b_{n}=(2\\sqrt{n/2}-\\sqrt{(n+1)/2}-\\sqrt{(n-1)/2})^{2}$, therefore \r\n${ \\sum_{k=1}^{49}\\sqrt{a_{k}-b_{k}}=\\sum_{k=1}^{49}(\\sqrt{k/2}-\\sqrt{(k-1)/2}})-\\sum_{k=1}^{49}(\\sqrt{(k+1)/2}-\\sqrt{k/2})=7\\sqrt{1/2}-5+\\sqrt{1/2}=4\\sqrt{2}-5$." } { "Tag": [], "Problem": "1) Provide a synthesis of 1-Phenyl-3-bromopropene.\r\n\r\n2) Starting from the compound prepared in part 1), outline reasonable synthesis of 2-Phenylcyclopropanecarboxilic acid.", "Solution_1": "1. benzene to toluene by Friedel and crafts\r\n etards to convert into benzaldehyde\r\n convert into cinnamic acid using perkins\r\n P/HI to convert COOH to CH3 ( i still dunno whether this rn can be used on paper )\r\n NBS to insert Br\r\n\r\nAlternate\r\n \r\n benzene to propyl benzene using $ \\ce{C2H5COCl and AlCl3}$ and then$ \\ce{N2H4}$/$ \\ce{HO\\minus{}}$\r\n brominate using Br2 only to get 1 substituted propyl benzene \r\n alc. KOH\r\n NBS", "Solution_2": "First ,convert the benzene to benzaldehyde by the Gatterman-Koch reaction(treat it with CO in the presence of AlCl3).Then perform aldol condensation with Ethanal.Then reduce the aldehydic group by using LiAlH4.Then Use PBr5 for bromination to get the required product.\r\n\r\nIs it correct :wink: ??????????????????", "Solution_3": "problems i think\r\n1. gatterman koch rn requires HCl also\r\n2. wouldnt you be reducing the double bond also in the process\r\nAnd why do you want to use PBr5 ?\r\nP.S. Chill on the question marks", "Solution_4": "No Only H2/Ni catalysis will reduce all the double bonds in the compound. Use LiAlH4 quantitatively i.e. 1 mole to get my reaction to happen", "Solution_5": "why shouldnt LAH reduce the double bond?", "Solution_6": "When you have a functional aldehyde group having greater nucleophilic charcter(since it is attached to a highly electro negative compound),LAH prefers to attack the aldehyde rather than the double bond. \r\n\r\nIs my explanation satisfactory :wink: ????????????????", "Solution_7": "No i dont think what you say is true.\r\nIf you could accomplish things by mole concept or stoichiometry then why cant you monobrominate aniline?", "Solution_8": "That is because of the extensive electron density at the ortho and para positions due to the electron pair of the NH2 group involving in resonance.....Look at the acetanilide......SInce the lonepair on nitrogen is involved in resonance with the COCH3 group...We get only the paraproduct.........\r\nDo you understand :wink: heyy!!!!!!!!!", "Solution_9": "Instead of just arguing over a never ending matter i accept to agree to Carcul's view on the topic\r\nP.S. Carcul can you please recommend some books for the chem olympiad?", "Solution_10": "OK Let Carcul post the correct answer..................\r\np.s. Carcul--please recommend some chem olympiad books to me also", "Solution_11": "[quote=\"Madness\"]1. benzene to toluene by Friedel and crafts \netards to convert into benzaldehyde \nconvert into cinnamic acid using perkins \nP/HI to convert COOH to CH3 ( i still dunno whether this rn can be used on paper ) \nNBS to insert Br[/quote]\n\nThe problem is in P/HI.\n\n[quote=\"Madness\"]Alternate \n\nbenzene to propyl benzene using and then/ \nbrominate using Br2 only to get 1 substituted propyl benzene \nalc. KOH \nNBS[/quote]\n\nVery nice.\n\n[quote=\"Valeriummaximum\"]First ,convert the benzene to benzaldehyde by the Gatterman-Koch reaction(treat it with CO in the presence of AlCl3).Then perform aldol condensation with Ethanal.Then reduce the aldehydic group by using LiAlH4.Then Use PBr5 for bromination to get the required product.[/quote]\n\nVery nice also. Just a remark: as Madness pointed out, the Gatterman formylation requires the presence of HCl (and also CuCl).\n\n[quote=\"Madness\"]wouldnt you be reducing the double bond also in the process[/quote]\n\nNo: for alpha-beta unsaturated aldehydes and ketones LiAlH4 reduces only the carbonyl.\n\n[quote=\"Madness\"]Carcul can you please recommend some books for the chem olympiad?[/quote]\r\n\r\nThe only books that I can recommend are the ones I use: \"Chemistry: Molecules, Matter, and Change\" from Atkins; for Organic Chemistry I have a lot of books, perhaps Solomons and Carey are my favourite for an introductory course.\r\n\r\n\r\n[b]How about Part 2?[/b]", "Solution_12": "treat it with carbene, it adds to double bond.\r\naq KOH to substitute the bromo group for a OH group.\r\noxidise to acid by adding potassium dichromate in presence of sulphuric acid.\r\n\r\nRight? :D", "Solution_13": "[quote=\"valeriummaximum\"]When you have a functional aldehyde group having greater nucleophilic charcter(since it is attached to a highly electro negative compound),LAH prefers to attack the aldehyde rather than the double bond. \n\nIs my explanation satisfactory :wink: ????????????????[/quote]\r\nactually the reducing nature of $ \\ce{LiAlH4}$ is due to the $ \\ce{H}^{\\minus{}}$ it gives and that attacks a positive centre so there doesn't seem to be a reason of it attacking the already electron sufficient double bond\r\nif u r so worried what about Meerwin-Verley-Pondorf reduction", "Solution_14": "[quote=\"Zimrock\"]treat it with carbene, it adds to double bond. \naq KOH to substitute the bromo group for a OH group. \noxidise to acid by adding potassium dichromate in presence of sulphuric acid.[/quote]\r\n\r\nUnfortunately, carbenes might also react with benzene rings. Also, is a cyclopropane ring stable under acidic/oxidative conditions?", "Solution_15": "But Chemrock told me that Cinnamaldehyde was a special case where Micheal addn takes place ( because of the nucleophilicity of the double bond ) and hence the double bond gets reduced :maybe:", "Solution_16": "Michael addition occurs with weak nucleophiles, like cyanide, enolates, amines, enamines, and Gilman reagents; with strong nucleophiles (like hydride ion, organo lithiums and Grignard reagents) carbonyl addition is the main path.", "Solution_17": "So isnt Cinnamaldehyde a special case then? :maybe:", "Solution_18": "What I said in my post above is the general case. Although, I was not aware of exceptions: so, I consulted one of my books and it turns out that cinnamyl systems, like cinnamaldehyde and cynnamic acid, are indeed special cases - with LiAlH4 [i]both[/i] the C-C double bond and the carbonyl are reduced." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "For any three positive reals a, b, c, we have\r\n\r\n$\\frac{2a^3+abc}{b^2+c^2}+\\frac{2b^3+abc}{c^2+a^2}+\\frac{2c^3+abc}{a^2+b^2} \\ge \\frac{3}{2}(a+b+c)$\r\n.", "Solution_1": "Another vornicu-schur type inequality. :D \r\n\\[\r\n\\begin{gathered}\r\n \\sum {\\frac{{2a^3 + abc}}\r\n{{b^2 + c^2 }}} \\hfill \\\\\r\n \\geqslant \\sum {\\frac{{a^3 + a^2 b + a^2 c}}\r\n{{b^2 + c^2 }}} \\hfill \\\\\r\n = \\left( {a + b + c} \\right)\\sum {\\frac{{a^2 }}\r\n{{b^2 + c^2 }}} \\hfill \\\\\r\n \\geqslant \\frac{3}\r\n{2}(a + b + c) \\hfill \\\\ \r\n\\end{gathered} \r\n\\]\r\nby vornicu-schur and nesbitt", "Solution_2": "[quote=\"siuhochung\"]\n\n $ \\sum {\\frac{{2a^3 + abc}}\n{{b^2 + c^2 }}}\n \\geq \\sum {\\frac{{a^3 + a^2 b + a^2 c}}\n{{b^2 + c^2 }}}$[/quote]\r\n\r\nCould you please explain this step ?", "Solution_3": "[quote=\"Andrei\"][quote=\"siuhochung\"]\n\n $ \\sum {\\frac{{2a^3 + abc}}\n{{b^2 + c^2 }}}\n \\geq \\sum {\\frac{{a^3 + a^2 b + a^2 c}}\n{{b^2 + c^2 }}}$[/quote]\n\nCould you please explain this step ?[/quote]\r\nIt is the vornicu-schur inequality as\r\n\\[\r\n\\begin{gathered}\r\n \\sum {\\frac{{a^3 + abc - a^2 b - a^2 c}}\r\n{{b^2 + c^2 }}} \\hfill \\\\\r\n = \\sum {\\frac{{a(a - b)(a - c)}}\r\n{{b^2 + c^2 }}} \\hfill \\\\\r\n \\geqslant 0 \\hfill \\\\ \r\n\\end{gathered} \r\n\\]\r\nYou can read about vornicu-schur here.\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=156212#p156212", "Solution_4": "The following more general inequality holds:\r\n\r\nIf $a_i$ are positive real numbers then \r\n\r\n$\\frac{(n-1)a_1^n+g}{b_1}+\\frac{(n-1)a_2^n+g }{b_2}+...+\\frac{(n-1)a_n^n+g}{b_n} \\ge \\frac{n}{n-1}(a_1+a_2+...+a_n)$\r\n\r\nwhere $ g=a_1a_2...a_n$ and $b_i=\\sum_{k<>i}{a_k^{n-1}}$ for all $i$.\r\n." } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "rotation", "trigonometry", "angle bisector", "geometry proposed" ], "Problem": "The ray $ [OA$ , the ray $ [OB$, and a positive number $ k$ are given. Let $ P$ be a point on $ [OA$ and $ Q$ be a point on $ [OB$ such that $ OP\\plus{}OQ \\equal{} k$. Prove that the circumcircles of $ \\triangle OPQ$ pass through a fixed point.", "Solution_1": "Let $ M,N$ be two points on the rays $OA,OB$ such that $ OM \\equal{} ON \\equal{} \\frac {_1}{^2}k.$ Then\n\n$ \\frac {_1}{^2}k \\minus{} MP \\plus{} \\frac {_1}{^2}k \\plus{} NQ \\equal{} k \\ \\Longrightarrow MP \\equal{} NQ.$\n\nThus, circles $ \\odot (OMN)$ and $ \\odot (OPQ)$ meet at the center $ K$ of the rotation that takes the oriented segments ${MP}$ and $ NQ$ into each other. But $ K$ is the antipode of $ O$ WRT the circle $ \\odot (OMN)$ because of $ KM \\equal{} KN.$ Then $K$ is fixed.", "Solution_2": "Let $ OM'\\equal{}ON'\\equal{}k/2$ and $ OM'' \\plus{} ON'' \\equal{} k$. Let $ P'$ be the intersection of $ (OM'N')$ and $ (OM''N'')$. $ \\angle M''P'M' \\equal{} N''P'N'$ and $ \\angle OM''P' \\plus{} \\angle ON''P' \\equal{} 180^\\circ$. Also we know $ M'M''\\equal{}NN''$. Use sinus theorem, then $ M'P' \\equal{} P'N'$. $ OM'N'$ is fixed, so its angle bisector and its circumcircle, so their intersection point is fixed. So all $ (OMN)$s pass through a point on angle bisector of $ \\angle MON$.", "Solution_3": "We will use ptolemy. Let $ P$ be the intersection point on angle bisector of $ \\angle MON$ and $ (OMN)$. Use ptolemy. $ (OM \\plus{} ON).MP \\equal{} MN.OP$. Use sine law at $ \\triangle MPN$, then $ \\frac {MN}{sin(180^\\circ \\minus{} \\alpha)} \\equal{} \\frac {MP}{sin(\\alpha/2)}$. Combine two equations. $ OP \\equal{} \\frac {k}{2cos(\\alpha/2)}$ which is fixed. The line $ OP$ is fixed, $ |OP|$ is fixed, so $ P$ is fixed. All $ (OMN)$s pass through the fixed point $ P$." } { "Tag": [ "articles" ], "Problem": "Are low carbs diet healthy? Why dont people just go out and excercise and eat healthy and well balanced diets? and by healthy i dont mean a big mac w/ a DIET coke", "Solution_1": "Moved to the Round Table.", "Solution_2": "Not good at all, just look at the data.\r\n\r\nP.S. Did you know you can now use TeX? Thanks Mr. Rusczyk!", "Solution_3": "Low carb diets aren't healthy at all. When you eat carbs, they are broken down into glucose monomers, which provide the main source of energy in aerobic respiration. When you don't eat carbs, the body must break down fats and stuff. This is the main idea behind the diet.\r\n\r\nProblem is, the body also breaks down proteins, which are critical to virtually every process in the body. Not only are proteins important in muscle structure and movement, but all enzymes are proteins, and as any biology student knows, enzymes facilitate almost all reactions.\r\n\r\nI strongly recommend working out (particularly running or walking) and maintaining a healthy, but carb-containing diet. You may not lose weight, but that's because you're gaining muscle mass at the same time as you're losing fat.", "Solution_4": "Fat is ESSENTIAL - to all those people who have no clue.\r\nIt is important in the cell membrane, it is energy storage, it is lots of vital things your body would die without.\r\nFAT is like ALCOHOL.\r\nBad in excess - good in moderation.", "Solution_5": "the dude that invented low-carb diets died fat. end of story.\r\ni think my point is made.", "Solution_6": "[quote=\"white_horse_king88\"]Fat is ESSENTIAL - to all those people who have no clue.\nIt is important in the cell membrane, it is energy storage, it is lots of vital things your body would die without.\nFAT is like ALCOHOL.\nBad in excess - good in moderation.[/quote]\r\n\r\nNo. I don't agree. I will claim that any alcohol is bad. No one ever listens to me though.", "Solution_7": "[quote=\"ComplexZeta\"]I will claim that any alcohol is bad. No one ever listens to me though.[/quote]\r\n\r\nI will listen to you long enough for you to lay out your evidence for your claim.", "Solution_8": "Well, first of all, studies tend to show that grape juice (which is nonalcoholic) helps the heart as much as wine does, so the alcohol gives no health benefit. But what concerns me more is the bias in published articles. People regularly see articles in, for example, [i]Scientific American[/i] claiming that alcohol in moderation is a health benefit. Well, ok, so maybe a few studies show that something in wine (most likely grapes, but we'll ignore that from now on in this post) helps stop heart attacks. But I worry about which articles get published. If someone does a study that shows that alcohol is harmful, which scientific magazine would publish the result? Everyone knows that alcohol is bad for people, so what's the points in publishing a result that everyone knows? But if a study (that possibly isn't performed with ``scientific rigor,'' whatever that means) shows that alcohol is [i]good[/i] for people, that sounds like a really amazing result because it contradicts the intuitions of all. Therefore, the result will be published all over the place, which makes it seem that loads of studies have verified the results because readers can read the results in so many different sources.", "Solution_9": "my grandfather drank one beer a day for a very good portion of his older years. he lived to be very very old. drinking a beer a day was a backwoods prescription. and like many backwoods prescriptions and wive's tales, people are finding out that there is some truth in it.", "Solution_10": "But I'll claim that he lived for a long time despite drinking the beer, not because of it. And there's nothing at all you can do to demonstrate that I'm wrong.", "Solution_11": "I'm pretty sure that the epidemiological evidence for a reduced mortality associated with MODERATE consumption of alcohol has already been shown to be independent of the source of the alcohol (that is, whether it is grape wine, rice wine, beer, distilled spirits, whatever). There are a lot of differing ethnic groups and religious groups with varied habits in this regard to provide natural experiments for testing hypotheses, and some very reasonable proposed mechanisms for why low-dose occasional ethanonl consumption is good for health. For all that, I often forget to take a dose of the wine that I began keeping around the house after I turned forty. (I also drink beer from time to time, but I don't have any in the house today.) I eat lots of fresh grapes, raisins, and grape juice too, so I ought to live to be 100 :wink: My maternal grandmother made it to age 99.", "Solution_12": "someone told me that drinking grape juice makes you jump higher. is this true?", "Solution_13": "[quote=\"SirErnest\"]someone told me that drinking grape juice makes you jump higher. is this true?[/quote]\r\nOf course not.", "Solution_14": "How much higher? It might be possible to jump a few microns higher after drinking grape juice, but I doubt you'd be able to jump an extra meter.", "Solution_15": "they said it had to do with the glucose in the grapes", "Solution_16": "[quote=\"SirErnest\"]they said it had to do with the glucose in the grapes[/quote]\r\nYou mean fructose right? :)" } { "Tag": [ "conics", "parabola", "ellipse", "inequalities unsolved", "inequalities" ], "Problem": "Hello everyone\r\n\r\nHow do I identify the curves such as this one?\r\n\r\n4x^2 + 9y^2 - 16x +54y +61= 0\r\n\r\nis a parabola formula like that? circle? elipse? I AM LOST! How do I find center or vertex, and radius if applicable...\r\n\r\n\r\nPLEASE HELP! it's urgent!", "Solution_1": "$ 4x^2 \\plus{} 9y^2 \\minus{} 16x \\plus{} 54y \\plus{} 61 \\equal{} 0\\Longleftrightarrow \\frac {(x \\minus{} 2)^2}{9} \\plus{} \\frac {(y \\plus{} 3)^2}{4} \\equal{} 1$ which is an ellipse.", "Solution_2": "How did you do it though? :maybe:", "Solution_3": "Completing square for $ x$ and $ y$ at the same time.", "Solution_4": "Ok bear with me please. I am doing it this way...\r\n\r\n\r\n4x^2 - 16 x + 9y^2 =54y = -61\r\n\r\n4(x^2 - 16x) + 9(y^2 + 54y) = -61\r\n\r\ncompleting square\r\n\r\n4(x^2 - 16x +64) = 9( y^2+ 54y +729) = -61 + 4(64) + 9(729)\r\n\r\n\r\nwhen i divide each side by 6756, it doesn't come out right!!!!!!!!!!!!!", "Solution_5": "In the second line, you made mistakes. :(", "Solution_6": "ohhhhhhhhhh silly me!\r\n\r\n\r\n4(x^2 - 4x) + 9(y^2 + 6y) = -61 \r\n\r\nNow it works perfect!\r\n\r\nThanks a lot! :lol:", "Solution_7": "Congratulations, Narjis! :lol:" } { "Tag": [ "ratio", "geometry", "incenter", "circumcircle", "geometry proposed" ], "Problem": "For the triangle ABC ($b\\ne c$) specify the position of the (remarkable) point\r\n$P\\in IG\\cap BC$ if there is one from the following relations:\r\n\r\n $1).\\ \\ \\frac{2}{a}=\\frac{1}{b}+\\frac{1}{c}\\ \\ \\ \\ 2).\\ \\ \\frac{a}{b+c}=\\frac{bc}{b^2+c^2}.$", "Solution_1": "Denote a = BC, b = CA, c = AB the sides of the triangle $\\triangle ABC$. Let D be the midpoint of the side a = BC, K and K' the intersection of the internal and external bisectors of the angle $\\angle A$ with the line BC, and P the midpoint of the segment KK' (the center of A-Apollonius circle of the triangle $\\triangle ABC$). Consider the triangle $\\triangle ADK$. The centroid G divides the median AD in the ratio\r\n\r\n$\\frac{GD}{GA} = \\frac 1 2$\r\n\r\nThe incenter I divides the internal bisector AK of the angle $\\angle A$ in the ratio\r\n\r\n$\\frac{IA}{IK} = \\frac{c + b}{a}$\r\n\r\nThe internal bisector AK of the angle $\\angle A$ divides the opposite side BC internally in the ratio of the adjacent sides, while the external bisector AK' of the angle $\\angle A$ divides the opposite side BC externally in the ratio of the adjacent sides:\r\n\r\n$\\frac{KB}{KC} = \\frac{K'B}{K'C} = \\frac {AB}{CA} = \\frac c b$\r\n\r\nWLOG, assume that $\\angle C > \\angle B$, hence, c > b. The points B, K, C, K' then follow on the line BC in this order.\r\n\r\n[color=blue][b](1)[/b][/color] Case [color=white].[/color] $\\frac 2 a = \\frac 1 b + \\frac 1 c$.\r\n\r\n$KC = \\frac{ab}{c + b},\\ \\ K'C = \\frac{ab}{c - b}$\r\n\r\n$K'K = K'C + KC = \\frac{ab}{c - b} + \\frac{ab}{c + b} = \\frac{2abc}{c^2 - b^2}$\r\n\r\n$K'D = K'C + CD = \\frac{ab}{c - b} + \\frac a 2 = \\frac{a(2b + c - b)}{2(c - b)} = \\frac{a(c + b}{2(c - b)}$\r\n\r\n$\\frac{K'K}{K'D} = \\frac{2abc}{c^2 - b^2} \\cdot \\frac{2(c - b)}{a(c + b)} = \\frac{4bc}{(c + b)^2}$\r\n\r\n$\\frac{K'K}{K'D} \\cdot \\frac{GD}{GA} \\cdot \\frac{IA}{IK}= \\frac{4bc}{(c + b)^2} \\cdot \\frac 1 2 \\cdot \\frac{c + b}{a} = \\frac{2bc}{a(c + b)}$\r\n\r\nBy Menelaus' theorem for the triangle $\\triangle ADK$, the points I, G, K' are collinear, i.e., the line IG passes through the point K', iff\r\n\r\n$\\frac{2bc}{a(c + b)} = 1,\\ \\ \\frac 2 a = \\frac{c + b}{bc} = \\frac 1 b + \\frac 1 c$\r\n\r\n\r\n[color=blue][b](2)[/b][/color] Case [color=white].[/color] $\\frac{a}{b + c} = \\frac{bc}{b^2 + c^2}$.\r\n\r\n$PK = \\frac{K'K}{2} = \\frac{abc}{c^2 - b^2}$\r\n\r\n$PD = K'D - \\frac{K'K}{2} = \\frac{a(c + b}{2(c - b)} - \\frac{abc}{c^2 - b^2} = \\frac{a[(c + b)^2 - 2bc]}{2(c^2 - b^2)} = \\frac{a(c^2 + b^2}{2(c^2 - b^2)}$\r\n\r\n$\\frac{PK}{PD} = \\frac{abc}{c^2 - b^2} \\cdot \\frac{2(c^2 - b^2)}{a(c^2 + b^2)} = \\frac{2bc}{c^2 + b^2}$\r\n\r\n$\\frac{PK}{PD} \\cdot \\frac{GD}{GA} \\cdot \\frac{IA}{IK}= \\frac{2bc}{c^2 + b^2} \\cdot \\frac 1 2 \\cdot \\frac{c + b}{a} = \\frac{bc (c + b)}{a(c^2 + b^2)}$\r\n\r\nBy Menelaus' theorem for the triangle $\\triangle ADK$, the points I, G, P are collinear, i.e., the line IG passes through the point P, iff\r\n\r\n$\\frac{2bc (c + b)}{a(c^2 + b^2)} = 1,\\ \\ \\frac{a}{c + b} = \\frac{bc}{c^2 + b^2}$", "Solution_2": "Therefore, $P\\in BC\\cap AA$, i.e. $\\frac{PB}{PC}=\\frac{AB^2}{AC^2}$ (AA is the tangent in A at the circumcircle of ABC).\r\n****************************************************************************\r\n[b]The one's solution.[/b]\r\n\r\nLet $X\\in BC$ for which $\\overline {XB}=m\\cdot\\overline{XC},m\\in R,m\\ne 1$. Thus,\r\n\r\n$X\\in GI\\Longleftrightarrow\\left|\\begin{array}{ccc}1&1&1\\\\\\\\a&b&c\\\\\\\\0&1&{-m}\\end{array}\\right|=0\\Longleftrightarrow m=\\frac{c-a}{a-b}.$\r\n\r\n[u]Case 1.[/u].....$a=\\frac{2bc}{b+c}\\Longrightarrow m=\\frac{c-\\frac{2bc}{b+c}}{\\frac{2bc}{b+c}-b}\\Longrightarrow m=\\frac{c}{b}.$\r\n\r\n[u]Case 2.[/u].....$a=\\frac{bc(b+c)}{b^2+c^2}\\Longrightarrow m=\\frac{c-\\frac{bc(b+c)}{b^2+c^2}}{\\frac{bc(b+c)}{b^2+c^2}-b}\\Longrightarrow m={\\left(\\frac{c}{b}\\right)^2.}$" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Prove that $\\mathcal{Q}$ is not isomorphic to a subgroup of $S_{n}$ for $n \\leq 7$.", "Solution_1": "The quaternions $Q$ are defined by $i^{2}=j^{2}=k^{2}=ijk=-1$. If $Q$ is isomorphic to a subgroup of $S_{n}$ ($n \\leq 7$), then $i$, having order $4$, must be identified either with a $4$-cycle $(a,b,c,d)$ or the disjoint composition of a $4$-cycle and a transposition. Either way, the square of such an element $u$ will be $(ac)(bd)$, where $(abcd)$ is the cycle in its decomposition. Moreover, $-i$ will be identified with $u^{3}$ (*). The key is that $(abcd)$ and $(adcb)$ are the two only $4$-cycles whose squares are $(ac)(bd)$. Therefore, among $i,j,k$, two elements at least are identified with the disjoint composition of the $4$-cycle $(a,b,c,d)$ (or $(a,d,b,c)$) and a transposition. The transpositions of those compositions must be pairwise distinct by (*). But this contradicts the fact that $ijk=-1$, because the composition of $3$ transpositions cannot be the identity (since the identity has signature $1$)." } { "Tag": [], "Problem": "There are two movie theaters positioned adjacent to each other, both theaters sell tickets to a two-hour movie. One theater starts selling tickets at 15.00 and the other at 15.15 both theaters play the movie the same amount of times. Which one sells more tickets? Why?", "Solution_1": "[quote=\"aleCTY\"]There are two movie theaters positioned adjacent to each other, both theaters sell tickets to a two-hour movie. One theater starts selling tickets at 15.00 and the other at 15.15 both theaters play the movie the same amount of times. Which one sells more tickets? Why?[/quote]\r\n\r\nHm. Is this really a math problem?\r\nThe cheaper one could sell more tickets because everyone is trying to save money.\r\nThe more expensive one might have a better movie, though.\r\nIs that all the information there is?", "Solution_2": "This is an economics problem. Assuming demand is not perfectly inelastic (i.e. a change in price induces no change in the amount demanded), and watching movies is not a Giffen good (when the price increases, you buy more of it), we should expect the theater that has a lower cost for its movies will sell more tickets.", "Solution_3": "The tickets cost the same. :D \r\nThis has to do with the times.It is a little hard to see how to map this into an equation at first, but it can solved many different ways.", "Solution_4": "This problem also assumes a knowledge of human behavior in this case: 1. people will come to the two theaters at random times. 2. They will buy a ticket to a time that is closest to their time of arrival (without traveling backwards in time.) This problem also assumes a steady flow of people.", "Solution_5": "So 15.00 and 15.15 are times, not prices?", "Solution_6": "Well, the theater that opens at 15:15 should sell more tickets as people that arrive late to the other theater will go like \"Oh no! The movie started. Wait a second, the other theater's movie hasn't started yet. Let's go there!\" Also, the time 15:15 is more likeable since its easy for people to remember.\r\n\r\nHow would you make this into a mathematical equation? Is it even possible? My answer is not very realistic :D", "Solution_7": "[quote=\"aleCTY\"]The tickets cost the same. :D[/quote]\r\n\r\n :blush: I read royalflower's post and assumed the costs were different.\r\n\r\nIn theory, if more people come before 15:00 than between 15:00 and 15:15, then the theater that shows the movie earlier sells more tickets, because (from personal experience), we humans are very impatient and we want to see our movies now.", "Solution_8": "From my personal experiences, humans are always late :D", "Solution_9": "True, but does that mean the theater will bar them for entering? (Then again, I've never gone to a theater, so I don't know the rules.)\r\n\r\nBesides, if you've paid for something, although it would be economically beneficial to buy the later ticket to go see the movie, most people wouldn't buy another ticket.", "Solution_10": "Generally, people buy the ticket when they come to the theater.\r\n :o You've never gone to a movie theater :o You should go see Avatar in IMAX 3-D as your first movie :o", "Solution_11": "[quote=\"darkdieuguerre\"]In theory, [b]if more people come before 15:00 than between 15:00 and 15:15[/b], then the theater that shows the movie earlier sells more tickets, because (from personal experience), we humans are very impatient and we want to see our movies now.[/quote]\r\n\r\nI purposefully assume something about the actual (not expected) distribution of number of arrivals with respect to time, so your statement doesn't make sense then.\r\n\r\n(Of course, prior knowledge could have something to do with it, but I think, at that point, we're delving a bit too much into specifics.)", "Solution_12": "[quote=\"aleCTY\"]One theater starts selling tickets at 15.00 and the other at 15.15 both theaters play the movie the same amount of times. Which one sells more tickets? Why?[/quote]\r\n\r\nYou've told us what time the theaters start selling tickets and that they play the movie (I'm assuming the same movie) the same amount of times (per day), but you haven't told us what times the movies start. Are the movies at each theater also starting 15 minutes apart, or are they starting at the same times?", "Solution_13": "\"selling tickets\" simply refers to the show times. 007math, I think your solution has more to do with human behavior than with math. Think of two waves with the same frequency, but out of phase with each other by 15 minutes. darkdieuguerre got the abstract answer correct.", "Solution_14": "The first one will sell more tickets. Taking into account human behavior -- most people will try to make the earlier movie, using the later movie as a backup in case the first is sold out or they are late.", "Solution_15": "Sorry, I was misunderstanding the problem. I thought is said that the movie theaters [i]open[/i] at the given times, whereas it actually says that start selling tickets at those times.", "Solution_16": "Lets see. People buy tickets around 1 hour before the movie, with maximum occuring in the middle. So let wave 1 be \r\n$ sin(x)$\r\nwave 2 will be \r\n$ sin(x\\minus{} \\frac {\\pi}4)$\r\nwith $ pi$ as defined as one hour\r\nHmm, now to analyze it.\r\n[hide=\"My perspective\"]\nCan't find a difference\n[/hide]", "Solution_17": "It seems like everybody is misinterpreting the problem (which is no wonder given how the OP made ambiguous [b]everything[/b] he could plus a few things he couldn't when posting it). So, here is what (to the best of my mind-reading abilities) was meant.\r\n\r\nThere are two theaters showing the same 2-hour movie. The first theater starts the shows at 3PM, 5PM, 7PM, and 9PM sharp. The second one starts its shows at 3:15, 5:15, 7:15, and 9:15. A few people who have nothing better to do in the afternoon arrive to the street at random times and buy tickets to the earliest show they can go to. Which theater will sell more tickets?", "Solution_18": "I don't think price is really an issue here.\r\n\r\nFor every 2 hour period, the later theater will only sell in a particular 15 minute interval, while the first one will sell in a 105 minute interval. So that makes the first theater sell more?\r\n\r\nOr am I not getting something here?", "Solution_19": "I believe this is pretty much all that was meant but we'd better wait until the OP confirms or denies it. :)", "Solution_20": "[quote=\"fedja\"]It seems like everybody is misinterpreting the problem (which is no wonder given how the OP made ambiguous [b]everything[/b] he could plus a few things he couldn't when posting it). So, here is what (to the best of my mind-reading abilities) was meant.\n\nThere are two theaters showing the same 2-hour movie. The first theater starts the shows at 3PM, 5PM, 7PM, and 9PM sharp. The second one starts its shows at 3:15, 5:15, 7:15, and 9:15. A few people who have nothing better to do in the afternoon arrive to the street at random times and buy tickets to the earliest show they can go to. Which theater will sell more tickets?[/quote]\r\n\r\nWell if it is like that, then if a person arrives at X:00-X:15 will go to the second one, and X:15-(X+2):00 will go the the first one.\r\n(where X is an odd number) \r\nThe first interval is much greater, so there will be more people there", "Solution_21": "Yes, that is the answer and the question. Sorry for the misunderstanding. :oops:" } { "Tag": [ "geometry", "circumcircle", "analytic geometry", "Euler", "Vietnam" ], "Problem": "Let $ k$ is a given positive number. Let $ ABC$ be an acute triangle, $ AB\\neq BC\\neq CA$, let $ O$ be its circumcenter, and $ AD$, $ BE$, $ CF$ be its internal angle bisectors. Points $ L$, $ M$, $ N$ lie on rays $ AD$, $ BE$, $ CF$, respectively such that $ \\frac {AL}{AD} \\equal{} \\frac {BM}{BE} \\equal{} \\frac {CN}{CF} \\equal{} k$. Denote by $ (O_1)$, $ (O_2)$, $ (O_3)$, respectively the circle passes through $ L$ and tangent to $ OA$ at $ A$, passes through $ M$ and tangent to $ OB$ at $ B$, passes through $ N$ and tangent to $ OC$ at $ C$. \r\n$ 1.$ When $ k \\equal{} \\frac {1}{2}$, prove that circles $ (O_1)$, $ (O_2)$, $ (O_3)$ have two common points and the centroid $ G$ of triangle $ ABC$ lies on line connecting these common points.\r\n$ 2.$ Find all $ k$, for which circles $ (O_1)$, $ (O_2)$, $ (O_3)$ are coaxial.", "Solution_1": "Tangents to the circumcircle $(O)$ through $A,B,C$ cut $BC,CA,AB$ at the centers of the the A-,B-,C- Apollonian circles $O_A,$ $O_B,$ $O_C.$ Hence, perpendicular bisectors of $AL,BM,CN$ cut $AO_A,BO_B,CO_C$ at $O_1,O_2,O_3,$ respectively. Isosceles $\\triangle ALO_1$ and $\\triangle ADO_A$ are similar with similarity coefficient $k,$ a.s.o. Thus, $\\frac{_{AO_1}}{^{AO_A}}=\\frac{_{BO_2}}{^{BO_B}}=\\frac{_{CO_3}}{^{CO_C}}=k.$ Now, using barycentric coordinates WRT $\\triangle ABC,$ we have\n\n$O_A(0:b^2:-c^2) \\ , \\ O_1((b^2-c^2)(1-k):b^2k:-c^2k)$ \n\n$O_B(-a^2:0:c^2) \\ , \\ O_2(-a^2k:(c^2-a^2)(1-k):c^2k)$ \n\n$O_C(a^2:-b^2:0) \\ , \\ O_3(a^2k:-b^2k:(a^2-b^2)(1-k))$\n\n$OA=OB=OC$ implies that $O$ has equal power to $(O_1),(O_2),(O_3).$ Thus, $(O_1),$ $(O_2),$ $(O_3)$ are coaxal $\\Longleftrightarrow$ $O_1,O_2,O_3$ are collinear $\\Longleftrightarrow$\n\n$\\left [\\begin {array}{ccc} (b^2-c^2)(1-k) & b^2k& -c^2k\\\\ -a^2k & (c^2-a^2)(1-k)& c^2k\\\\ a^2k& -b^2k& (a^2-b^2)(1-k)\\end{array}\\right]=0$ \n\n$ \\Longleftrightarrow (k-1)(2k-1)(b^2-c^2)(c^2-a^2)(a^2-b^2)=0$\n\nSince $\\triangle ABC$ is not isosceles, $(O_1),$ $(O_2),$ $(O_3)$ are coaxal $ \\Longleftrightarrow$ either $k=1$ or $k=\\frac{_1}{^2}.$ If $k=1,$ then $(O_1),$ $(O_2),$ $(O_3)$ coincide with the Apollonian circles $(O_A),$ $(O_B),$ $(O_C),$ whose common radical axis is the Brocard axis of $\\triangle ABC.$ If $k=\\frac{_1}{^2},$ then $O_1,$ $O_2,$ $O_3$ are the midpoints of $AO_A,$ $BO_B,$ $CO_C$ $\\Longrightarrow$ $O_1,$ $O_2,$ $O_3$ lie on the orthic axis of $\\triangle ABC$ $\\Longrightarrow$ common radical axis of $(O_1),(O_2),$ $(O_3)$ is the Euler line of $\\triangle ABC.$", "Solution_2": "You can draw pictures of items for everyone! :D." } { "Tag": [ "linear algebra", "matrix", "search", "linear algebra unsolved" ], "Problem": "Let $A=[a_{i,j}]\\in{GL}_{n}(\\mathbb{C})$ be a tridiagonal matrix ($a_{i,j}=0$ if $|i-j|>1)$ and $b\\in\\mathbb{C}^{n}$; here $n\\approx{10^{6}}$. $A,b$ are known and you search $x\\in\\mathbb{C}^{n}$ s.t. $Ax=b$.\r\n If I give you $A^{-1}$ (explicitly) what do you do ?", "Solution_1": "I ignore $A^{-1}$ and solve the system by Gaussian elimination. (I am hoping that it's not ill-conditioned.) How many multiplications/divisions do I need for that? It's something like $3\\times 10^{6},$ I think. The problem with $A^{-1}$ is that I'm pretty sure that it's not tridiagonal - that it has nonzero entries in most places. If that's so, then to execute the product $A^{-1}b$ takes about $10^{12}$ multiplications.\r\n\r\nNext time, don't give me $A^{-1}.$ Give me $A=LU$ instead.", "Solution_2": "Well done Kent. \r\nThe key is here:\r\n\"I'm pretty sure that it's not tridiagonal - that it has nonzero entries in most places. \"", "Solution_3": "Oh, and when I said \"ignore $A^{-1},$\" that includes that I don't even download the file that contains it. That's a terabyte file - it's going to take up an awful lot of storage space and take an awfully long time to transmit. We can store $A$ in a few megabytes because we don't need to store the zeros." } { "Tag": [ "geometry", "geometric transformation", "geometry proposed" ], "Problem": "Let P1,P2,Q three points from the plane s.t. 1P2= \\pi/4. Now let consider the points P1' and P2' s.t. QPiPi' =\\pi /2 and PiQ=PiPi',i=1,2.\r\nProove that QP1, P1'P2' and the mediator of P1P2 are concurent.\r\nHope that is correct :)\r\n\r\n[i]This doesn't have a source... i was inspired from a problem from Gamow wich has an interesting story... i will post it after u'll give some solutions(hope that are more than one...i found only one).[/i]", "Solution_1": "don't u have any idea?\r\ni could solve it only with complex numbers\r\nif u want i can post here a solution", "Solution_2": "This may be stupid but...what is a mediator? :(", "Solution_3": "it's a direct (mot-a-mot) translation of \"segment bisector line\" from Romanian :D.", "Solution_4": "perpendicular bisector... thx Valentin" } { "Tag": [], "Problem": "Hi, everybody! I'm new here, and this is my first post.\r\n\r\nGetting to the point, I'm having trouble with one of this year's Warm-Up problems. I can't figure out how to get the answer that they got. Is there something I missed? I'd like it if somebody could help me out. Thanks!\r\n\r\nThe problem is:\r\nA deck of 52 playing cards is printed on a 4-by-13 rectangular grid. If stacking the pieces is allowed after each cut, but no folding is allowed, what is the least number of cuts required to seperate all the cards? (Cuts can be any length, but may go in only one direction per cut.)\r\n\r\nThe answer is 6, but no matter how many different ways I tried it I always ended up with 7. Thanks in advance!", "Solution_1": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75092" } { "Tag": [ "algebra", "function", "domain" ], "Problem": "Hey guys I have a math question.\r\n\r\nIf, f(x)= 11/ (x-10) and g(x)= 13/ (x+11)\r\nWhat is the domain of (f/g) (x) ?\r\n\r\nHere's what I found:\r\nFirst, f/g is (11x + 121) / (13x - 130) and its domain is (-infinity, 10) U (10,infinity).\r\n\r\nBut my book sais the domain is wrong. Can anyone help me out?", "Solution_1": "What does your book say, out of curiosity? That does seem right; I can't see why they would claim that $ x \\equal{} \\minus{}11$ is not in the domain because the quotient of the two functions permits $ x\\equal{}\\minus{}11$ as a valid input... unless there's some convention I'm not aware of which states that the domain of $ (f/g)(x)$ is the intersection of the domains of $ f(x)$ and $ g(x)$.", "Solution_2": "Nope. You are perfectly correct. The domain is $ ( \\minus{} \\infty, 10) \\cup (10, \\infty)$. You can also write it as $ \\{ x | x \\in R$ \\ $ 10 \\}$" } { "Tag": [], "Problem": "Find the least positive integer with $ 96$ divisors.", "Solution_1": "[hide=\"Use this formula\"]If $ x\\equal{}p_1^{a_1}\\cdot p_2^{a_2}\\cdots p_n^{a_n}$ where $ p_i$ are primes, then the no. of divisors $ x$ has is $ (a_1\\plus{}1)(a_2\\plus{}1)\\cdots(a_n\\plus{}1)$[/hide]", "Solution_2": "That is what I did and got $ 27720$.", "Solution_3": "[hide=\"never mind\"]I got something else. $ 96 \\equal{} 2^5\\cdot3$\n\\[ 2^2\\cdot3\\cdot5\\cdot7\\cdot11\\cdot13 \\equal{} \\boxed{60060}\\][/hide]", "Solution_4": "I agree with AndrewTom. \r\n\r\n$ 27720 \\equal{} 2^3 \\times 3^2 \\times 5 \\times 7 \\times 11$.\r\n\r\n$ 2^3 < 13$, so we are allowed to make that substitution." } { "Tag": [ "geometry", "perimeter", "MATHCOUNTS" ], "Problem": "5. The perimeter of a sector of a circle is the sum of the two sides\r\nformed by the radii and the length of the included arc. A sector\r\nof a particular circle has a perimeter of 28 cm and an area of\r\n49 sq cm. What is the length of the arc of this sector?\r\n\r\n6. John noticed that the angle formed by the minute hand and \r\nhour hand on a standard 12-hour clock was 110 degrees when\r\nhe left home after 6 p.m.; it was also 110 degrees when he\r\nreturned after 6 p.m. but before 7 p.m. that same night. If he\r\nleft home for more than five minutes, for how many minutes\r\nwas he away?\r\n\r\n7. Let T be a positive integer whose only digits are 0s and 1s. If \r\nX = T/12 and X is an integer, what is the smallest possible\r\nvalue of X?\r\n\r\n8. Using each of the digits {1,2,3,4} exactly once and using\r\nzero or more + signs, how many different totals can be\r\nobtained? No digit can be used as a power, root, etc. Only\r\naddition can be done. When n digits are written next to each\r\nother, they represent an n-digit number and not the product of\r\nn factors. Four of the totals to be included are [u]1234,[/u]\r\n[u]55[/u]= 12+43, [u]127[/u] = 123+4 and [u]37[/u]=31+2+4.", "Solution_1": "I am at library right now, so I'll do these as soon as I get back. :lol:\r\n\r\nEDIT: Mr.dan80, what level is this? Is this NATS level?\r\nThis is challenging!", "Solution_2": "Before I head home, I looked at #7.\r\nI think I have the answer. I'll only post the answer now and when I get home, I'll post the solution.\r\nEDIT: I am back, so will edit the solution\r\n[hide=\"#7\"]\nI tried to solve T first.\nT has to be factor of 12, which also has to be factor of 2 and 3.\nNote that T only has 1's and 0's for its digits.\n0 can cover up 2, so all we have to do is make the sum of the digits multiples of 3.\nSo my first guess was 1110, but it gave me 92.5 and 92.5 does not qualify as an answer since X= integer.\nSo $ \\ 11100$ should be T, so $ \\ 11100\\div12 \\equal{} \\boxed{925}$\nWhat about the other ones?\nThey wouldn't work because you'll need one or more 0 in the number.\n\n[/hide]\n\nI think I also have the solution for #8 (Is there easier way to do this?)\nI am not sure if I got this right.... :( \nP:S I had to count them and tried them and find a pattern....\n\n[hide=\"#8\"]It's really hard to explain....so bear with me.. :wink: \nFirst thing I want to tell you is that there is a pattern to this problem.\n\nI divided into 3 sections by amount of usages of +'s\n\n[u]No +'s :[/u] $ 4!\\plus{}(4\\cdot3\\cdot2)\\plus{}(4\\cdot3)\\plus{}4\\equal{}64$\n[u]1 + :[/u] You have to be picky about this one. They can NOT be these(took me awhile :lol: ):\n[hide=\"Numbers\"]1,2,3,4,12,13,14,21,23,24,31,32,34,41,42,43,123,124,132,134,142,143,213,214,231,234,241,243,312,314,321,324,341,342,412,413,421,423,431,432,1234,1243,1324,1342,1423,1432,2134,2143,2314,2341,2413,2431,3124,3142,3214,3241,3412,3421,4123,4132,4213,4231,4312,4321.[/hide]\n\nSo either you can count or realize there are 3 numbers in every set.\nSo $ \\ 17\\times3\\equal{}51$ numbers + $ 5$ numbers$ \\rightarrow(46,64,73,37,55) \\equal{} 56$\n\n[u]2 +'s : [/u] You can only achieve $ 2 \\rightarrow (19,28)$\n\n3 +'s : Only $ 1 \\rightarrow (10)$\n\nNow you add them up to get $ \\boxed{123}$ sums.[/hide]\r\n\r\nPheww....... Now I think I can sleep and do the rest two. :lol:", "Solution_3": "I got a solution for #6...I am not sure though..\r\n\r\n[hide=\"#6\"]Since hour-hand moves $ \\frac {3}{5}^{\\circ}$ in a minute and minute hand moves $ 6^{\\circ}$ in a minute, we can make an equation like this:\n\nLet $ m$ be minutes after 6'O clock.\n$ \\frac {3}{5}m \\plus{} 6m \\equal{} 110$ and $ m$ equals out to be $ 16\\frac {2}{3}$ so when he leaves it's [u]6:17 PM.(rounded up)[/u]\n\nYou have to get the time the hour-hand and minute hand elapses together,(synchronizes) and the formula for it is $ x \\equal{} \\text {head start} \\plus{} \\frac {x}{12}$ and you get [u]6:33 PM(rounded up)[/u]\n\nThen you add 17 minutes again to 6:33 PM to get [u]6:50 PM.[/u] \nSince the problem asks you how many minutes he was gone, the answer is $ \\boxed{33}$minutes\n\nEDIT: On second thought on this, can't you just get the minutes after 6'O clock when it synchorizes(before 7'O clock)?[/hide]\r\n\r\nNow I have 1 more question to go!! :lol:", "Solution_4": "DK, the only reason why 1110 did not work is because 12 is divisible by not only 2 and 3, but also 4 because 12 = 3*4\r\n\r\nFor #8, you have to use each number once, so you can't have 123. But you're on the right track of dividing it all into different cases.", "Solution_5": "[quote=\"Chinaboy\"]DK, the only reason why 1110 did not work is because 12 is divisible by not only 2 and 3, but also 4 because 12 = 3*4\n\nFor #8, you have to use each number once, so you can't have 123. But you're on the right track of dividing it all into different cases.[/quote]\r\n\r\nOh... okay. Thanks!\r\nBut on #7, did I get it right?\r\n\r\nOn #8, are you saying that you have to use all four once?(i didn't do that..)", "Solution_6": "[quote=\"DonkeyKong\"][quote=\"Chinaboy\"]DK, the only reason why 1110 did not work is because 12 is divisible by not only 2 and 3, but also 4 because 12 = 3*4\n\nFor #8, you have to use each number once, so you can't have 123. But you're on the right track of dividing it all into different cases.[/quote]\n\nOh... okay. Thanks!\nBut on #7, did I get it right?\n\nOn #8, are you saying that you have to use all four once?(i didn't do that..)[/quote]\r\n\r\nYep, you got 7!\r\n\r\nAnd you do have to use all four at once (or \"exactly once\")", "Solution_7": "[quote=\"chinaboy\"]And you do have to use all four at once (or \"exactly once\")\n[/quote]\r\n\r\nThat makes it clearer.. Thank you!\r\n\r\nOh.. Do you know if I got #6 right?", "Solution_8": "Can anyone give me some hints on #5?\r\n\r\n[hide=\" Where I am so far\"]Perimeter=28\nArea=49\n\n[u]Perimeter:[/u]\n$ \\ 2r\\plus{}(\\cfrac{\\text{central angle}}{180^{\\circ}}\\cdot \\pi \\cdot r) \\equal{} 28$\n\n[u]Area:[/u]\n$ \\frac12r^{2}\\cdot \\theta\\equal{}49$ OR $ \\frac12r^{2}\\cdot 2\\pi \\equal{}49$[/hide]\r\n\r\nAm I on the right track?", "Solution_9": "[quote=\"DonkeyKong\"]Can anyone give me some hints on #5?\n\n[hide=\" Where I am so far\"]Perimeter=28\nArea=49\n\n[u]Perimeter:[/u]\n$ \\ 2r \\plus{} (\\cfrac{\\text{central angle}}{180^{\\circ}}\\cdot \\pi \\cdot r) \\equal{} 28$\n\n[u]Area:[/u]\n$ \\frac12r^{2}\\cdot \\theta \\equal{} 49$ OR $ \\frac12r^{2}\\cdot 2\\pi \\equal{} 49$[/hide]\n\nAm I on the right track?[/quote]\r\n\r\nFor #6, the hour hand does not move .6 degrees. Check you work again.\r\n\r\nFor #5, I think you're using degrees for the first equation and radians for the second one. Choose one of the two. I think you should go ahead with degrees on this one just because it's more Mathcounts-friendly.", "Solution_10": "[quote=\"chinaboy\"]For #6, the hour hand does not move .6 degrees. Check you work again.[/quote]\r\n\r\nWait....would it be $ {\\frac12}^\\circ?$\r\n\r\n[hide=\"If it is..\"]I tried it with $ {\\frac12}^\\circ$ for hour-hand, but still gave me same answer... :wink: [/hide]", "Solution_11": "Talk to me next week about this. I like to draw the picture when I explain clock problems.\r\n\r\nI really admire your work ethic, DK! Keep on trucking!", "Solution_12": "I misunderstood #8....I tried to edit on my previous post, but it won't let me..\r\nSo I'll post another one.\r\n\r\nP:S=Is there easier way than counting??(except dividing into sections..)\r\n\r\n[hide=\"New Solution for #8\"]\nI started from taking the obvious ones out and divided into sections by the # of usages of +'s.\n\nWe can easily perceive that with NO +'s, we get [u]24 possible sum.[/u]\nAlso on using ALL the +'s would get you 10 no matter what.. So [u]1 possible sum. .[/u]\n\nThen I divided into sections.\n\nOne +: [u]16 possible sums.[/u]\n[hide=\" Possible Sums\"]46,55,64,37,73,127,136,325,235,217,316,145,244,415,343,433[/hide]\n\nTwo +'s: [u]2 possible sums.[/u]\n[hide=\" Possible Sums\"]19,28[/hide]\n\nSo we add the possible sums to get $ \\boxed{43}$ possible sums.[/hide]", "Solution_13": "[quote=\"Chinaboy\"]Talk to me next week about this. I like to draw the picture when I explain clock problems.\n\nI really admire your work ethic, DK! Keep on trucking![/quote]\r\n\r\nOkay!\r\n\r\nwait..that means I didn't get it right... :(", "Solution_14": "So far only #7 has been answered correctly. Maybe our Chinaboy can give some more hints at practice!", "Solution_15": "Hey, you weren't there at school today! So I couldn't use your microwave to heat my soup. :( \r\n\r\nHint for 5.\r\n[hide]Come up with two equations (one for perimeter and one for area), each using r for radius and theta for angle (in degrees!).[/hide]\n\nHint for 6\n[hide]Before you do anything, you have to find out how many degrees the hour hand and minute hand move in one minute. (I'll have to go over this with you guys during class because this is VERY important)[/hide]\n\nHint for 8\n[hide]Break it into cases. \nCase 1. No + sign\nCase 2. 1 + sign\n...[/hide]", "Solution_16": "REVISED:\r\n[hide=\"#8\"]\nI started from taking the obvious ones out and divided into sections by the # of usages of +'s.\n\nWe can easily perceive that with NO +'s, we get [u]24 possible sum.[/u]\nAlso on using ALL the +'s would get you 10 no matter what.. So [u]1 possible sum. .[/u]\n\nThen I divided into sections.\n\nOne +: [u]17 possible sums.[/u]\n[hide=\" Possible Sums\"]235,244,325,343,424,433,136,145,316,415,127,217,46,55,37,64,73[/hide]\n\nTwo +'s: [u]2 possible sums.[/u]\n[hide=\" Possible Sums\"]19,28[/hide]\n\nSo we add the possible sums to get $ \\boxed{44}$ possible sums.[/hide]\r\n\r\nIs it right?? I actually tried all of them..\r\nIf it's not right, I might cry... :(", "Solution_17": "YOU GOT IT!", "Solution_18": "YAY! I got it right!", "Solution_19": "[hide]number 6# is 220/5.5=40 so the answer is [size=200][b][u]40![/u][/b][/size][/hide]", "Solution_20": "The circumference of a particular circle is 18 cm. In square centimeters, what is the area of the circle? Express your answer as a common fraction in terms of $\\pi$.\n" } { "Tag": [ "inequalities", "conics", "ellipse" ], "Problem": "Can anybody suggest a proof or web link about well known AM-GM inequality $(a+b+c)^{3}\\geq27abc$ in format without words . I interested proofs without words. Thanks.", "Solution_1": "[quote=\"mayhem\"]Can anybody suggest a proof or web link about well known AM-GM inequality $(a+b+c)^{3}\\geq27abc$ in format without words . I interested proofs without words. Thanks.[/quote]\r\n\r\nWikipedia is your friend.", "Solution_2": "Okay. Proofs aren't till like Geometry/PreCalc, so um...I believe you should post in Intermediate.\r\n\r\nhttp://en.wikipedia.org/wiki/AM-GM", "Solution_3": "I think by proofs without words, he means the little picture things (like the ones in the left menu bar on the AoPS website).\r\n\r\nThe Wikipedia for AM-GM does not have these.\r\n\r\nI haven't seen any myself. However,\r\n\r\none possible idea stems from the AM-GM demonstration that uses a semicircle (attached). However, that one contains only two variables - if you use a semi-elipse, you may be able to extend it into three variables somehow, though it would be difficult to get it into the form you want visually.\r\n\r\nThe other possibility is to use 3-D graphics. However, not only would this be hard to make, it would be difficult to demonstrate that your graphic is a representative case (unlike in the semicircle demonstration of the case of 2, in which case it is \"obvious\" that it holds true for other inputs).\r\n\r\nI donno... you'll have a tough time, but you might be able to make something out of it.", "Solution_4": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23921" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "can you prove :maybe: \r\n\r\n$ \\prod_{n\\equal{}1}^{\\infty}\\;(1\\minus{}x^{2n})\\cdot(1\\plus{}x^{2n\\minus{}1}\\cdot y^{2})\\cdot\\left(1\\plus{}\\frac{x^{2n\\minus{}1}}{y^{2}}\\right)\\;\\;\\equal{}\\;\\;\\boxed{\\sum_{p\\equal{}\\minus{}\\infty}^{\\infty}\\;x^{p^{2}}\\cdot y^{2p}}$", "Solution_1": "This identity is called Jacobi Triple Product.\r\n\r\nYou can find a proof at [url=http://mathworld.wolfram.com/JacobiTripleProduct.html]Mathworld[/url] and at [url=http://en.wikipedia.org/wiki/Jacobi_triple_product]Wikipedia[/url]." } { "Tag": [ "topology" ], "Problem": "This is an Exercises 3.3.4 of Hatcher's Algebraic Topology (p257).\r\n\r\n\"Given a covering space action of a group G on an orientable manifold M by orientation-preserving homeomorphisms, show that M/G is also orientable.\"\r\n\r\nSome attempts so far,\r\n\r\nSince M is an orientable manifold, there exists an orientation $ x \\mapsto \\mu_x$ such that, for each $ x \\in M$, we can assign a local orientatation $ \\mu_x \\in H_n(M, M \\minus{}\\{x\\})$ satisfying a \"local consistency property\" with respect to a neighborhood $ U_x \\subset R^n$ containing x in M.\r\n\r\nBy the definition of a covering space action, for a non-identity element g in G, we choose $ U_x$ containing x such that $ gU_x \\cap U_x \\equal{} \\emptyset$.\r\n\r\nThen, $ H_n(M, M\\minus{}\\{x\\})\\equal{}H_n(U_x, U_x\\minus{}\\{x\\}) \\equal{} H_n(gU_x, gU_x\\minus{}\\{gx\\})$.\r\n\r\nIs this correct? Then, how do I proceed from here?\r\n\r\nThanks.", "Solution_1": "Here something in another language which should easily translate to your reference's:\r\n\r\nSince $ M$ is orientable, there is a global continuous section $ s: M\\to Or(M)$ of the two-fold orientation covering $ \\pi_M: Or(M)\\to M$. The quotient map $ \\psi: M\\to M/G$ induces the natural map $ Or(\\psi): Or(M)\\to Or(M/G)$ with $ \\pi_{M/G}\\circ Or(\\psi)=\\psi\\circ \\pi_M$ with the two-fold covering $ \\pi_{M/G}: Or(M/G)\\to M/G$. Since $ G$ is orientation preserving, the mapping $ Or(\\psi)\\circ s: M\\to Or(M/G)$ descends to a continuous section $ \\overline{s}: M/G\\to Or(M/G)$ of the $ \\pi_{M/G}$. This is equivalent to that $ M/G$ has an orientation (such that $ \\psi$ is orientation preserving)." } { "Tag": [ "geometry", "ratio" ], "Problem": "What is the ratio of the area of a circle and the area of a square inscribed inside that circle?", "Solution_1": "[quote=\"IntrepidMath\"]What is the ratio of the area of a circle and the area of a square inscribed inside that circle?[/quote]\r\n\r\n\r\n[hide] lets say the side of the square is x:\n\nThen the area of the square would be $ x^2 $\n\nAnd the area of the circle would be $\\frac{x\\sqrt2}{2} $ * $\\frac{x\\sqrt2}{2} $ * $\\pi$= $\\frac{x^2\\pi}{2}$\n\n Thus, there ratios are $x^2$ : $\\frac{x^2\\pi}{2}$, siplified = $ 2 : \\pi $[/hide]\r\n\r\nEDIT: $ 2 : \\pi$ is reffering to the \"square : circle\".......", "Solution_2": "Let's say that the radius of the circle is $2$, then the area of that circle is $4pi$, then the diagonal of the square could be $4$ and the side length would be $4/(2^{1/2})$, so the area of the square would be $8$, so the ratio would be $pi/2$.", "Solution_3": "[hide]If each side of the square is $x$, then the area of the square would be $x^2$\n\nSo, the diameter of the circle would be $x\\sqrt{2}$ and the radius would be$ \\frac{x\\sqrt{2}}{2}$\n\nSo the area of the circle would be $\\pi \\left (\\frac{x\\sqrt{2}}{2}\\right)^2$\n\nSimplified:\n\n$\\pi\\left(\\frac{2{x^2}}{4} \\right)$\n\n$\\frac{\\pi{x^2}}{2}$\n\nThe ratio would be $\\frac{\\frac{\\pi{x^2}}{2}}{x^2}$\n\nor\n\n$\\left(\\frac{\\pi{x^2}}{2}\\right)*\\left(\\frac{1}{x^2}\\right)$\n\n$\\frac{\\pi}{2}$\n\n$\\pi:2$[/hide]", "Solution_4": "Area of circle : pi*r^2\r\nDiameter of circle: 2r\r\n\r\nSo diagonal of square = 2r\r\n\r\nUsing 1,1, radical2 relationship, side length of square = 2r/radical2 =\r\n\r\n2r*radical2\r\n-------------- = r*radical2\r\n2\r\n\r\n(r*radical2)^2= area of square = 2r^2\r\n\r\npi*r^2\r\n-------- simplifies to: pi/2 (Answer)\r\n2r^2", "Solution_5": "it is a generally known fact that the area of a square inscribed in a circle is $2r^2$ just as it is a generally known fact that an inscribed dodecagon has an area of $3r^2$ \r\n\r\n$2r^2:\\pi r^2 = 2:\\pi$" } { "Tag": [ "geometry", "trigonometry", "trig identities", "Law of Cosines", "similar triangles", "area of a triangle", "Heron\\u0027s formula" ], "Problem": "In triangle $ABC$, $a=11$, $b=13$, and the median to side $c$ is 10. Compute the area of the triangle.", "Solution_1": "[hide]Stewart's Theorem on ABC\n\n\n\n121(c/2)+169(c/2)=100c+2(c/2)^3\n\n\n\nsolve for c/2=3 :rt5: \n\n\n\nLaw of Cosines on BCM, where M is midpt. AB\n\n\n\n121+100-220cos :theta: =(3 :rt5:)^2=45\n\n\n\ncos :theta:=4/5, sin :theta: =3/5\n\n\n\n2[BCM]=[ABC]=11(10)(3/5)=66[/hide]", "Solution_2": "Here's what I have so far:\n\n[hide]Drawing the three medians, we have the medium to c is split into 10/3 and 20/3. However, this is all I've got, and I'm not sure where to go from here. I'm trying similar triangles but not getting anywhere, and I don't think this problem requires the use of trig (which I don't know a whole lot about). I thought of using Heron's formula, but then I'd need to know side c. Can anyone offer me small hints? Thanks.[/hide]" } { "Tag": [], "Problem": "Three Students were asked to find the identity of the metal in a particular sulfate salt. They dissolved a 0.1472-g sample of the salt in water adn treated it with excess barium chloride, resulting in the precipitation of barium sulfate. After teh precipitate had been filtered and dried, it weighed 0.2327 g. \r\nEach student analyzed the data independently and came to different conclusions. \r\nThe conclusions were that the metal is sodium, titanium or gallium. What further tests you suggest to determine which student is most likely correct.\r\n\r\n\r\nI have worked it out, but I find the same solution... it could be either sodium, titanium and gallium.\r\n\r\ni have tried using a CRC, but I to no avail.\r\n\r\ndoes anyone know how to approach this problem? Thanks\r\n\r\n\r\nI've a strong background in chemistry, but can't figure out this college problem.... it's not for a grade, but I'm obsessed to find the answer.\r\n\r\nthanks!", "Solution_1": "What is your opinion about this problem?", "Solution_2": "What about Atomic Absorption Spectroscopy (A.A.S) or Flame Test?", "Solution_3": "Why in the first place the metal should be sodium, titanium, or galium?" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "calculus", "integration" ], "Problem": "Find all integral solutions to a^2 + b^2 + c^2 = a^2*b^2.", "Solution_1": "tetrahedr0n wrote:Find all integral solutions to a^2 + b^2 + c^2 = a^2*b^2.\n\n\n\n[hide](a-1)(a+1)(b-1)(b+1)=c:^2:+1. Either the LHS is odd, or 4|LHS, as if 2 divides one factor, it will divide the one that is 2 away from it. However, 4 can never divide the RHS, as c:^2: is never 3 mod 4. So, a, b, and c must all be even. a=2x, b=2y, c=2z => (2x-1)(2x+1)(2y-1)(2y+1)=4z:^2:+1 => 4x:^2:y:^2:=x:^2:+y:^2:+z:^2:. Considering this mod 4, we have 0:equiv:sum of three squares. all of them must be 0, or multiples of 4. Thus, x=4l, y=4m, z=4n. So 64l:^2:m:^2:=l:^2:+m:^2:+n:^2:. We can continue to apply mod 4 analysis, so we have that l, m, and n must all be integers that continually yield integers upon divison by 4. The only such integers are 0. Thus, l=m=n=0, and the only solution is a=b=c=0.[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c,d\\ge \\frac {1}{7}$ such that $ a \\plus{} b \\plus{} c \\plus{} d \\equal{} 1$. Show that:\r\n\r\n$ {\\frac {7203}{16}}\\ge \\frac {1}{abcd} \\minus{} \\left(\\frac {1}{a^2} \\plus{} \\frac {1}{b^2} \\plus{} \\frac {1}{c^2} \\plus{} \\frac {1}{d^2}\\right)\\ge 192;$\r\n\r\n$ {\\frac {16807}{16}}\\ge \\frac {2}{abcd} \\minus{} \\left(\\frac {1}{a^2} \\plus{} \\frac {1}{b^2} \\plus{} \\frac {1}{c^2} \\plus{} \\frac {1}{d^2}\\right)\\ge 448.$\r\n\r\nGenerally,if $ k\\ge{\\frac{174}{353}},$then\r\n\r\n$ {\\frac {2401}{16}}(4k\\minus{}1)\\ge \\frac {k}{abcd} \\minus{} \\left(\\frac {1}{a^2} \\plus{} \\frac {1}{b^2} \\plus{} \\frac {1}{c^2} \\plus{} \\frac {1}{d^2}\\right)\\ge {256k\\minus{}64}.$", "Solution_1": "[quote=\"fjwxcsl\"]Let $ a,b,c,d\\ge \\frac {1}{7}$ such that $ a \\plus{} b \\plus{} c \\plus{} d \\equal{} 1$. Show that:\n\n$ {\\frac {7203}{16}}\\ge \\frac {1}{abcd} \\minus{} \\left(\\frac {1}{a^2} \\plus{} \\frac {1}{b^2} \\plus{} \\frac {1}{c^2} \\plus{} \\frac {1}{d^2}\\right)\\ge 192;$\n\n$ {\\frac {16807}{16}}\\ge \\frac {2}{abcd} \\minus{} \\left(\\frac {1}{a^2} \\plus{} \\frac {1}{b^2} \\plus{} \\frac {1}{c^2} \\plus{} \\frac {1}{d^2}\\right)\\ge 448.$\n\nGenerally,if $ k\\ge{\\frac {174}{353}},$then\n\n$ {\\frac {2401}{16}}(4k \\minus{} 1)\\ge \\frac {k}{abcd} \\minus{} \\left(\\frac {1}{a^2} \\plus{} \\frac {1}{b^2} \\plus{} \\frac {1}{c^2} \\plus{} \\frac {1}{d^2}\\right)\\ge {256k \\minus{} 64}.$[/quote]\r\n\r\n\r\n>f:=subs({a=p+1/7,b=1/7+q,c=1/7+r,d=1/7+s},{p=3/7*x/(x+y+z+w),q=3/7*y/(x+y+z+w),r=3/7*z/(x+y+z+w),s=3/7*w/(x+y+z+w)},k/a/b/c/d-1/a^2-1/b^2-1/c^2-1/d^2-m):\r\n>f:=numer(f):\r\n\r\n>factor(subs(x=1,y=1,z=1,w=1,f)):\r\n>MIN:=solve(%,m);\r\n\r\nMIN := 256*k-64\r\n\r\n>factor(subs(x=0,y=0,z=0,w=1,f)):\r\n>MAX:=solve(%,m);\r\n\r\nMAX := 2401/4*k-2401/16\r\n\r\n\r\n>symprove(numer(subs(w=(x+y+z)*t,m=MIN,k=u+174/353,f)));\r\n\r\n\r\n \r\n it is symmetric for . {x, z, y}\r\n\r\n ITS = [u, t]\r\n\r\n\r\n IS = 1, N = 2\r\n\r\n\r\n The inequality holds.\r\n\r\n\r\n ITS = [u, t]\r\n\r\n\r\n IS = 1, N = 2\r\n\r\n\r\n The inequality holds.\r\n\r\n\r\n ITS = [u, t]\r\n\r\n\r\n IS = 1, N = 1\r\n\r\n\r\n The inequality holds.\r\n\r\n\r\n DEGU = 2\r\n\r\n\r\n DEG = 2\r\n\r\n\r\n ITS = [u, ee, t]\r\n\r\n\r\n IS = 1, N = 5\r\n\r\n\r\n The inequality holds.\r\n\r\n\r\n ITS = [u, ee, t]\r\n\r\n\r\n IS = 1, N = 4\r\n\r\n\r\n The inequality holds.\r\n\r\n\r\n ITS = [u, ee, t]\r\n\r\n\r\n IS = 1, N = 3\r\n\r\n\r\n The inequality holds.\r\n\r\n\r\n The original form is psd!(SYM3(1))\r\n\r\n\r\n \u5b83\u662f\u5e73\u51e1\u975e\u8d1f\u7684!\r\n\r\n\r\n TIME = 2.921\r\n\r\n \r\n\r\n \r\n\r\n \r\n>symprove(numer(subs(w=(x+y+z)*t,m=MAX,k=u+174/353,-f)));\r\n\r\n\r\n it is symmetric for . {x, z, y}\r\n\r\n ITS = [u, t]\r\n\r\n\r\n IS = 1, N = 1\r\n\r\n\r\n The inequality holds.\r\n\r\n\r\n ITS = [u, t]\r\n\r\n\r\n IS = 1, N = 1\r\n\r\n\r\n The inequality holds.\r\n\r\n\r\n ITS = [u, t]\r\n\r\n\r\n IS = 1, N = 1\r\n\r\n\r\n The inequality holds.\r\n\r\n\r\n DEGU = 2\r\n\r\n\r\n DEG = 0\r\n\r\n\r\n The original form is psd!(SYM3(1))\r\n\r\n\r\n \u5b83\u662f\u5e73\u51e1\u975e\u8d1f\u7684!\r\n\r\n\r\n TIME = 2.094" } { "Tag": [ "national olympiad" ], "Problem": "Hi all and especially romanian people!!\r\n\r\nDoes anyone have the problems of the current romanian olympiad(2006),\r\nall rounds (or those that have already taken place) as well as the dates of the contests??\r\n\r\nMaybe this can be the thread for the RMO 2006!!\r\n\r\nThanks and good luck to the participants!\r\n\r\n :) :)", "Solution_1": "The romanian olympiad has not yet begun. The first round takes place somewhere in February, in the first week after the vacantion. (I think)\r\n\r\nAlso, I think Valentin will post all the problems.", "Solution_2": "[quote=\"perfect_radio\"]The romanian olympiad has not yet begun. The first round takes place somewhere in February, in the first week after the vacantion. (I think)\n\nAlso, I think Valentin will post all the problems.[/quote]\r\n\r\n\r\nThanks, perfect_radio. :) \r\nThe first round is the local one??", "Solution_3": "[quote=\"socrates\"]The first round is the local one??[/quote]\r\n\r\nYes. Here are the subjects for the eleventh grade.\r\n\r\nMoubinool asked me to post them.", "Solution_4": "[quote=\"perfect_radio\"][quote=\"socrates\"]The first round is the local one??[/quote]\n\nYes. Here are the subjects for the eleventh grade.\n\nMoubinool asked me to post them.[/quote]\r\n\r\n\u039cany thanks perfect radio!\r\nCould you post problems for the other forms, too?/\r\n\r\n :)" } { "Tag": [ "inequalities", "geometry unsolved", "geometry" ], "Problem": "Let $ ABCD$ be a convex quadrilateral such that $ AB$ extended and $ CD$ extended cut at a right angle. Prove that $ AC\\cdot BD>AD\\cdot BC$.", "Solution_1": "i'm assuming that $ B$ lies between $ A$ and $ E$, where $ E$ is the intersection of $ AB$ and $ CD$... suppose wlog that $ AC\\geq BD$... note that angles $ BCD$ and $ ACD$ are obtuse, so $ AD>AC$ and $ BD>BC$, so $ AD\\minus{}BC>AC\\minus{}BD\\geq 0$... since $ AB\\perp CD$ we have that $ AD^2\\plus{}BC^2\\equal{}AC^2\\plus{}BD^2$, so the inequality follows after squaring the last inequality obtained and using this equality... :D" } { "Tag": [ "trigonometry" ], "Problem": "\u03a0\u03b1\u03b9\u03b4\u03b9\u03ac \u03b2\u03ac\u03b6\u03c9 \u03bc\u03b9\u03b1 \u03c0\u03ac\u03c1\u03b1 \u03c0\u03bf\u03bb\u03cd \u03b4\u03cd\u03c3\u03ba\u03bf\u03bb\u03b7 \u03ba\u03b1\u03c4\u03ac \u03c4\u03b7 \u03b3\u03bd\u03ce\u03bc\u03b7 \u03bc\u03bf\u03c5 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c4\u03b7\u03bd \u03bf\u03c0\u03bf\u03af\u03b1 \u03b4\u03b5\u03bd \u03bc\u03c0\u03cc\u03c1\u03b5\u03c3\u03b1 \u03bf\u03cd\u03c4\u03b5 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c3\u03b5\u03b3\u03b3\u03af\u03c3\u03c9. \u038c\u03c0\u03bf\u03b9\u03bf\u03c2 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b5\u03b9 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac \u03bc\u03c0\u03c1\u03ac\u03b2\u03bf.\r\n\r\n\u0398\u03b5\u03c9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03ad\u03bd\u03b1 \u03b8\u03b5\u03c4\u03b9\u03ba\u03cc \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03bf \u03ba. \u0391\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 sqrt(k+1)-sqrt(k) \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03cc \u03bc\u03ad\u03c1\u03bf\u03c2 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03bf\u03cd \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd \u03b6 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf\u03c5 \u03ce\u03c3\u03c4\u03b5 \u03b6^\u03bd=1 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03b8\u03b5\u03c4\u03b9\u03ba\u03cc \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03bf \u03bd.\r\n\r\n\u0392\u03c1\u03ae\u03ba\u03b1 \u03bc\u03b9\u03b1 \u03c0\u03ac\u03c1\u03b1 \u03c0\u03bf\u03bb\u03cd \u03c9\u03c1\u03b1\u03af\u03b1 \u03ba\u03b1\u03b9 \u03ad\u03be\u03c5\u03c0\u03bd\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03bf\u03c0\u03bf\u03af\u03b1 \u03b8\u03b1 \u03b4\u03ce\u03c3\u03c9 \u03bc\u03b9\u03b1 \u03c5\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03b1\u03cd\u03c1\u03b9\u03bf \u03b1\u03bd \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9 \u03b2\u03c1\u03b5\u03b8\u03b5\u03af \u03bb\u03cd\u03c3\u03b7.", "Solution_1": "\u039a\u03bf\u03af\u03c4\u03b1\u03be\u03b5 . \u0397 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03ba\u03b1\u03bb\u03ae \u03ba\u03b1\u03b9 \u03b7 \u03b5\u03ba\u03c6\u03ce\u03bd\u03b7\u03c3\u03ae \u03c4\u03b7\u03c2 \u03b4\u03af\u03bd\u03b5\u03b9 \u03ba\u03ac\u03c4\u03b9 \u03ba\u03b1\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03bb\u03ba\u03c5\u03c3\u03c4\u03b9\u03ba\u03ae .\u0391\u03bb\u03bb\u03ac \u03b3\u03b5\u03bd\u03b9\u03ba\u03ac \u03cc\u03c4\u03b1\u03bd \u03ad\u03c7\u03b5\u03b9 \u03c3\u03c5\u03bd\u03b7\u03bc\u03af\u03c4\u03bf\u03bd\u03b1 \u03c4\u03b1 \u03c0\u03bf\u03bb\u03c5\u03ce\u03bd\u03c5\u03bc\u03b1 Chebychev \u03c3\u03ba\u03bf\u03c4\u03ce\u03bd\u03bf\u03c5\u03bd \u03c3\u03c7\u03b5\u03b4\u03cc\u03bd \u03c4\u03b1 \u03c0\u03ac\u03bd\u03c4\u03b1 . \u039d\u03b1 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5 \r\n\r\n\u0397 \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b7 \u03bc\u03b5 \u03c4\u03bf \u03bd\u03b4\u03bf \u03b7 \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7 \r\n$\\cos{m}=\\sqrt{k+1}-\\sqrt{k}$, \u03cc\u03c0\u03bf\u03c5 $z=\\cos{m}+i\\sin{m}$ \u03ba\u03b1\u03b9 \r\n$z^{n}=1$ \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 . \u03a4\u03cc\u03c4\u03b5 \u03b1\u03c0\u03cc DeMoivre \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \r\n $\\cos{nm}=1$ \u03ac\u03c1\u03b1 \u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2 $\\sqrt{k+1}-\\sqrt{k}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c1\u03af\u03b6\u03b1 \u03c4\u03bf\u03c5 $g(x)-1$, \u03cc\u03c0\u03bf\u03c5 $g(\\cos m)=\\cos{nm}$ (\u0398\u03c5\u03bc\u03af\u03b6\u03b5\u03b9 \u03c0\u03bf\u03bb\u03c5\u03ce\u03bd\u03c5\u03bc\u03bf Chebychev :wink: ) .\r\n\u03a0\u03b1\u03c1\u03b1\u03c4\u03b7\u03c1\u03bf\u03cd\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2 $A=\\sqrt{k+1}-\\sqrt{k}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03bb\u03b3\u03b5\u03b2\u03c1\u03b9\u03ba\u03cc\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03b5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03cc \u03c4\u03bf\u03c5 \u03c0\u03bf\u03bb\u03c5\u03ce\u03bd\u03c5\u03bc\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \r\n$f(x)=x^{4}-(4k+2)x^{2}+1$, \u03ac\u03c1\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03c4\u03bf $f(x)$ \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03af \u03c4\u03bf $g(x)-1$ \r\n\u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03cc\u03bc\u03c9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c4\u03bf\u03c0\u03bf \u03b1\u03c6\u03bf\u03cd \u03c4\u03bf \u03b3\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03bf \u03c4\u03c9\u03bd \u03bc\u03ad\u03c1\u03c4\u03c9\u03bd \u03c4\u03c9\u03bd \u03c1\u03b9\u03b6\u03ce\u03bd \u03c4\u03bf\u03c5 $f(x)$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03bf \u03bc\u03b5 $1$ \u03b5\u03bd\u03ce \u03c4\u03bf\u03c5 $g$ \u03cc\u03c7\u03b9 , \u03ac\u03c4\u03bf\u03c0\u03bf \r\n\r\nQED :wink:", "Solution_2": "\u0388\u03b4\u03c9\u03c3\u03b5\u03c2 \u03bc\u03b9\u03b1 \u03c0\u03b9\u03bf \u03c3\u03cd\u03bd\u03c4\u03bf\u03bc\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03ae \u03c0\u03bf\u03c5 \u03b5\u03af\u03c7\u03b1 \u03b2\u03c1\u03b5\u03b9 \u03c0\u03bf\u03c5 \u03cc\u03bc\u03c9\u03c2 \u03c3\u03c4\u03b7\u03c1\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03af\u03b4\u03b9\u03b1 \u03bb\u03bf\u03b3\u03b9\u03ba\u03ae. \u0393\u03b9\u03b1 \u03bd\u03b1 \u03c3\u03bf\u03c5 \u03c0\u03c9 \u03c4\u03b7\u03bd \u03b1\u03bb\u03ae\u03b8\u03b5\u03b9\u03b1 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03ce \u03bd\u03b1 \u03c4\u03b9\u03c2 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ac\u03b2\u03c9 \u03b1\u03c0\u03cc\u03bb\u03c5\u03c4\u03b1. \u039a\u03ac\u03c0\u03bf\u03c5 \u03c7\u03ac\u03bd\u03bf\u03bc\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 \u03b4\u03cd\u03bf (\u03c3\u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf). \u039c\u03ac\u03bb\u03bb\u03bf\u03bd \u03bc\u03bf\u03c5 \u03b4\u03b9\u03b1\u03c6\u03ad\u03c5\u03b3\u03b5\u03b9 \u03ba\u03ac\u03c4\u03b9 \u03b1\u03c0\u03cc \u03b8\u03b5\u03c9\u03c1\u03af\u03b1. \u03a3\u03af\u03b3\u03bf\u03c5\u03c1\u03b1 \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae... \u039f\u03c0\u03cc\u03c4\u03b5 \u03b5\u03bc\u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03bf\u03bc\u03b1\u03b9 \u03b1\u03c0\u03cc\u03bb\u03c5\u03c4\u03b1 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03bf\u03c5.\r\n\r\n\u039c\u03c0\u03c1\u03ac\u03b2\u03bf!\r\n\r\n\u03a4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c0\u03bf\u03c5 \u03c7\u03ac\u03bd\u03bf\u03bc\u03b1\u03b9 \u03b1\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03bc\u03bf\u03c5 \u03b5\u03be\u03b7\u03b3\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03ba\u03b5\u03af \u03c0\u03bf\u03c5 \u03bb\u03b5\u03c2 \u03cc\u03c4\u03b9 \u03c4\u03bf g(x)-1 \u03ad\u03c7\u03b5\u03b9 \u03b3\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03bf \u03c1\u03b9\u03b6\u03ce\u03bd \u03b4\u03b9\u03ac\u03c6\u03bf\u03c1\u03bf \u03c4\u03bf\u03c5 1. \u0388\u03c7\u03c9 \u03b4\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03c5 \u03bc\u03b9\u03b1 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03b1 \u03c0\u03bf\u03bb\u03c5\u03ce\u03bd\u03c5\u03bc\u03b1 \u03a4\u03c3\u03b5\u03bc\u03c0\u03af\u03c4\u03c3\u03b5\u03c6 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03a4n(x)-1 \u03ad\u03c7\u03b5\u03b9 \u03c1\u03af\u03b6\u03b5\u03c2 \u03c3\u03c4\u03bf {-1, 1}\u03ba\u03b1\u03b9 \u03bc\u03ac\u03bb\u03b9\u03c3\u03c4\u03b1 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 cosk\u03c0/n \u03ba\u03b1\u03b9 \u03c4\u03bf k \u03b1\u03bd\u03ac\u03bb\u03bf\u03b3\u03b1 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03ba\u03bb\u03ac\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03bd \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf \u03b4\u03cd\u03bf (\u03ac\u03c1\u03c4\u03b9\u03bf\u03c2, \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03cc\u03c2). \u0391\u03c5\u03c4\u03cc \u03b5\u03af\u03bd\u03b1\u03b9? \u0395\u03bb\u03c0\u03af\u03b6\u03c9 \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03b3\u03b9\u03b1\u03c4\u03af \u03b1\u03bb\u03bb\u03b9\u03ce\u03c2 \u03ba\u03ac\u03c4\u03b9 \u03b4\u03b5\u03bd \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03af\u03bd\u03c9 \u03ba\u03b1\u03b8\u03cc\u03bb\u03bf\u03c5.", "Solution_3": "\u03a4\u03b9\u03c2 \u03b1\u03c0\u03bf\u03c1\u03af\u03b5\u03c2 \u03c3\u03bf\u03c5 \u03c3\u03c4\u03b9\u03c2 \u03b1\u03c0\u03ac\u03bd\u03c4\u03b7\u03c3\u03b1 \u03c3\u03b5 PM . \r\n\u039c\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03b3\u03c1\u03ac\u03c8\u03b5\u03b9\u03c2 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03ad\u03c7\u03b5\u03b9\u03c2 \u03c5\u03c0\u03cc\u03c8\u03b7\u03bd \u03c3\u03bf\u03c5 ? Thanks", "Solution_4": "\u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03ad \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03c3\u03bf\u03c5 \u03ad\u03c3\u03c4\u03b5\u03b9\u03bb\u03b1 \u03ad\u03bd\u03b1 pm. \u0394\u03b5 \u03b2\u03ac\u03b6\u03c9 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03b4\u03b9\u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c7\u03b5\u03b4\u03cc\u03bd \u03b7 \u03af\u03b4\u03b9\u03b1. \u03a4\u03b7 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03ac \u03c3\u03c4\u03b7\u03bd \u03ad\u03b3\u03c1\u03b1\u03c8\u03b1 \u03c3\u03c4\u03bf pm. :wink: \r\n\r\n\u0395\u03c0\u03af\u03c3\u03b7\u03c2 \u03b6\u03b7\u03c4\u03ce \u03c3\u03c5\u03b3\u03b3\u03bd\u03ce\u03bc\u03b7 \u03c3\u03b5 \u03cc\u03c3\u03bf\u03c5\u03c2 \u03b5\u03af\u03c7\u03b1\u03bd \u03c3\u03c4\u03b5\u03af\u03bb\u03b5\u03b9 pm \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03b5\u03af\u03c7\u03b1 \u03b1\u03c0\u03b1\u03bd\u03c4\u03ae\u03c3\u03b5\u03b9 \u03b3\u03b9\u03b1\u03c4\u03af \u03bd\u03cc\u03bc\u03b9\u03b6\u03b1 \u03cc\u03c4\u03b9 \u03c3\u03b5 \u03b5\u03bd\u03b7\u03bc\u03ad\u03c1\u03c9\u03bd\u03b5 \u03ac\u03bc\u03b1 \u03ad\u03c0\u03b1\u03b9\u03c1\u03bd\u03b5\u03c2 pm :blush: \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03c4\u03b1 \u03b5\u03af\u03c7\u03b1 \u03b4\u03b5\u03b9. :oops:\r\n\r\nA \u03ba\u03b1\u03b9 \u03ba\u03b1\u03bb\u03cc \u03ba\u03b1\u03bb\u03bf\u03ba\u03b1\u03af\u03c1\u03b9 \u03c3\u03b5 \u03cc\u03bb\u03bf\u03c5\u03c2 \u03b4\u03b9\u03cc\u03c4\u03b9 \u03c6\u03b5\u03cd\u03b3\u03c9 \u03b4\u03b9\u03b1\u03ba\u03bf\u03c0\u03ad\u03c2. \u0395\u03bb\u03c0\u03af\u03b6\u03c9 \u03bd\u03b1 \u03ad\u03c7\u03c9 \u03c3\u03cd\u03bd\u03b4\u03b5\u03c3\u03b7 internet \u03b5\u03ba\u03b5\u03af \u03c0\u03bf\u03c5 \u03b8\u03b1 \u03c0\u03ac\u03c9 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03c0\u03b1\u03af\u03bd\u03c9. \u03a3\u03b5 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ae \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03bd\u03b1 \u03bc\u03bf\u03c5 \u03c3\u03c5\u03b3\u03c7\u03c9\u03c1\u03ad\u03c3\u03b5\u03c4\u03b5 \u03bc\u03b9\u03b1 \u03b2\u03b4\u03bf\u03bc\u03ac\u03b4\u03b1 \u03b1\u03c0\u03bf\u03c7\u03ae\u03c2." } { "Tag": [], "Problem": "How did Stanley Miller's classic experiment apply mechanism to the evolution of life?\r\n\r\n--\r\n\r\nThis question is from a workbook related to the Campbell AP Biology Textbook, but I can't seem to find a direct concise answer. Could someone please help? Thanks in advance!", "Solution_1": "actually it didnt provide the mechanism--it was just a proof for the mechanism proposed by oparin and haldane\r\n\r\nother scientists added various mix of chemicals to find out how the present organic chems evolved :)" } { "Tag": [ "AMC" ], "Problem": "Find the largest integer that is always a divisor of\r\n 13^n+5^n-10^n-8^n", "Solution_1": "darkquantum wrote:Find the largest integer that is always a divisor of\n 13^n+5^n-10^n-8^n\n\n[hide]30. n=0, 1 gives 0, n=2 gives 30, n=3 gives 810, and n=4 gives 503*30. The largest integer possible can be 30. Let us show that 30 does indeed divide 13^n+5^n-10^n-8^n for all positive integer n.\n\n\n\n30=2*3*5. 2 obviously divides that (there are 2 odd and 2 even terms). We can write that expression as (13^n-10^n)-(8^n-5^n)=\n\n(13-10)(13^(n-1)+13^(n-2)*10+13^(n-3)*10^2)+...+10^(n-1))-(8-5)(8^(n-1)+8^(n-2)*5+...+5^(n-1).\n\nWe can see that both terms are divisible by 3. Finally we write that expression again as (13^n-8^n)-(10^n-5^n). A similar analysis shows that it is divisible by 5. Therefore, 30 divides it for all n.[/hide]", "Solution_2": "[color=cyan]nicely done. Darkquantum, do you know where Linda got this problem from? It's very nice. An old Mandelbrot, perhps? Or is it her own creation?[/color]", "Solution_3": "um, im not sure. She gave a problem similar to it as a lesson problem, so, im not sure.\r\n\r\nAnd mystic, ur correct!\r\nthe key to this problem is to kno that (a-b) l (a^n-b^n)\r\n\r\nShow the expression divides 5, 3, and then obviously it divides 2 by a simple parity check", "Solution_4": "What you mean is that it is divisible by 5, 3, and then obviously divisible by 2 by a simple parity check.", "Solution_5": "I don't know what you're asking, teratomato, but it's divisible by a judicious use of (a-b) | (a^n - b^n) for n >= 1, and it's divisible by 2 because there are four things being added and two are odd and two are even which makes for an even number.", "Solution_6": "I don't know what you think I'm asking or why you think I'm asking anything.\r\n\r\nI'm not asking. I'm telling.\r\n\r\n[quote=\"darkquantum\"]Show the expression divides 5, 3, and then obviously it divides 2 by a simple parity check.[/quote]\nThe expression doesn't divide 5, 3, or 2. It is divisible by those numbers. That is what I meant. That's why I said:\n[quote=\"teratomato\"]What you mean is that it is divisible by 5, 3, and then obviously divisible by 2 by a simple parity check.[/quote]", "Solution_7": "Sorry. I thought you wrote, \"What do you mean it's divisible . . .\"\r\n\r\nMy fault.", "Solution_8": "oops\r\nbad diction on my part\r\nthx hooyoung" } { "Tag": [], "Problem": "Can't wait to get my score :( ... You guys are really testing my patience lol :D", "Solution_1": "I generally takes up to a month to grade solutions, and though it certainly seems that it's been that long, we still have over a week left to go before we hit that mark, so they might be testing your patience just a little bit longer :lol:", "Solution_2": "We hope to have the scores up this week.", "Solution_3": "Yes! That's great news. I've been waiting for what seems to be an eternity!", "Solution_4": "be grateful -- last year, it was way slower than this, but that was probably result of not as much people getting to work on grading and stuff.", "Solution_5": "^5 for all the grading dudes :lol:", "Solution_6": "Ahhh, that was good. It took me 5 or 10 seconds to catch the pun. Sorry, man. If I was a grader, I would have left you hanging." } { "Tag": [ "analytic geometry", "graphing lines", "slope", "calculus", "derivative", "function", "calculus computations" ], "Problem": ":D :D :ddr: Find the $ x$ coordinate of the point that the slope of the tangent line is maximum. The equation is:\r\n\r\n$ y \\equal{} \\minus{} x^4 \\plus{} 12x^3 \\minus{} 54x^2 \\plus{} 21341234220445723893240575x \\plus{} 13936048234505238907$\r\n\r\n\r\n\r\nWhat do you notice about the last two coefficients (other than their size)? There is something interesting about them that affects the problem. Just think carefully, and you will notice a pattern.", "Solution_1": "What's your answer to the first part? The \"interesting\" thing is that the answer doesn't depend on either of the last two coefficients.", "Solution_2": "[quote=\"t0rajir0u\"]What's your answer to the first part? The \"interesting\" thing is that the answer doesn't depend on either of the last two coefficients.[/quote]\r\nExactly, so what is the answer?", "Solution_3": "Since we know those two coefficients don't matter, how about replacing them with $ 108$ and $ \\minus{}81$?", "Solution_4": "[quote=\"jmerry\"]Since we know those two coefficients don't matter, how about replacing them with $ 108$ and $ \\minus{} 81$?[/quote]\r\nIn your mind you can, but in the problem, the coefficients stay. :evilgrin: Uh oh! The return of the evil smiley!", "Solution_5": "That was meant to be a hint to the solution. In that form, I don't even need to do any calculus to see the answer.", "Solution_6": "then what is the answer? you still have to figure out the answer!", "Solution_7": "jmerry isn't providing the answer because he's leaving the problem as an exercise for others, especially for you since this is (presumably) your homework. \r\n\r\nIn any case, here's what he is saying: if you replace the last two coefficients with the coefficients he gave, then $ y \\equal{} \\minus{}(x \\minus{} 3)^4$ by the binomial theorem. Now it's obvious what the answer has to be.", "Solution_8": "[quote=\"t0rajir0u\"]jmerry isn't providing the answer because he's leaving the problem as an exercise for others, especially for you since this is (presumably) your homework. \n\nIn any case, here's what he is saying: if you replace the last two coefficients with the coefficients he gave, then $ y \\equal{} \\minus{} (x \\minus{} 3)^4$ by the binomial theorem. Now it's obvious what the answer has to be.[/quote]\r\n\r\nIf this is my homework, then that would be fun! I wish this was actually a pop quiz. I would love that. But I cannot stay in my world of fantasies.", "Solution_9": "Okay, there is no maximum. This was a trick. However, there is a relative maximum. Do not change the equation now.\r\n\r\n$ y \\equal{} \\minus{} x^4 \\plus{} 12x^3 \\minus{} 54x^2 \\plus{} 21341234220445723893240575x \\plus{} 13936048234505238907$\r\n\r\nAt what $ x$ coordinate is the slope of the tangent line at a relative maximum?", "Solution_10": "None. The first derivative is strictly decreasing on the whole real line.", "Solution_11": "[quote=\"jmerry\"]None. The first derivative is strictly decreasing on the whole real line.[/quote]\r\n\r\nI guess you did not get the \"relative part\".", "Solution_12": "No, I'm saying that the critical point is not a relative maximum or minimum. It's just a flat spot for the derivative, which decreases on both sides.", "Solution_13": "Please, please, [i]please[/i] do not assume that the people who are answering your questions know less than you do. $ f'(x) \\equal{} 0$ is not a sufficient condition for either a relative maximum or a relative minimum; for a twice differentiable function, a relative minimum also requires $ f''(x) > 0$, whereas here we have $ f''(x) \\equal{} 0$. Graphically, we do not have either a relative maximum or a relative minimum.", "Solution_14": "Okay, I was just trying to throw trick questions. Now let us change the problem a little:\r\n\r\n$ y \\equal{} \\minus{}x^4 \\plus{} 10x^3 \\minus{} 36x^2 \\plus{} 21341234220445723893240575x \\plus{} 13936048234505238907$\r\n\r\nAt what x coordinate is the slope of the tangent line at a relative maximum?" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Find all continuous functions $ f: \\mathbb R\\rightarrow\\mathbb R$ satisfying $ f(f(x))\\equal{}f(x)\\plus{}2x$", "Solution_1": "Please check if the answer below makes sense.\r\nhttp://www.mathlinks.ro/viewtopic.php?p=1328709#1328709" } { "Tag": [], "Problem": "Compute the value of the infinite product $ \\frac{9}{10} \\times \\frac{99}{100} \\times \\frac{999}{1000} \\times \\cdots$.", "Solution_1": "You won't get anything simpler than what the [url=http://en.wikipedia.org/wiki/Pentagonal_number_theorem]pentagonal number theorem[/url] tells you.", "Solution_2": "It converges to something close to $ 0.8900100999989988$. I don't think you'll get a nicer answer than that.", "Solution_3": "Yep; that's what the first few terms $ 1 \\minus{} \\frac {1}{10} \\minus{} \\frac {1}{10^2} \\plus{} \\frac {1}{10^5} \\plus{} \\frac {1}{10^7} \\minus{} \\frac {1}{10^{12}} \\minus{} \\frac {1}{10^{15}}$ get you. Examining the decimal representation carefully here is actually an interesting way to [b]discover[/b] the pentagonal number theorem, although I don't know if it would be easy for anyone to discover any of the proofs independently, none of which (that I know of) are trivial." } { "Tag": [ "trigonometry", "inequalities" ], "Problem": "Let $ x,y,z$ be positive real numbers satisfying $ x\\plus{}y\\plus{}z\\equal{}xyz$. Prove that:\n\\[\\frac1{\\sqrt{1+x^2}}+\\frac1{\\sqrt{1+y^2}}+\\frac1{\\sqrt{1+z^2}}\\leq\\frac{3}{2}\\]", "Solution_1": "Very easy:\r\n$ x \\equal{}\\tan{\\alpha},y\\equal{}\\tan{\\beta},z\\equal{}\\tan{\\gamma}\\Rightarrow\\sum\\alpha \\equal{}\\pi$,because if $ \\tan{\\alpha}\\plus{}\\tan{\\beta}\\plus{}\\tan{\\gamma}\\equal{}\\tan{\\alpha}\\tan{\\beta}\\tan{\\gamma}$,then $ \\alpha\\plus{}\\beta\\plus{}\\gamma\\equal{}\\pi$, so we have to prove that $ \\sum\\cos{\\alpha}\\leq\\frac{3}{2}$,which is true by Jensen :)", "Solution_2": "There is another proof of the last inequality:\r\n$ \\sum\\cos{\\alpha}\\leq\\frac{3}{2}$,for all $ \\alpha,\\beta,\\gamma\\geq 0,\\sum\\alpha\\equal{}\\pi$.\r\nProof(if you don't want to use Jensen):\r\n$ \\sum\\cos{\\alpha}\\equal{}\\frac{R\\plus{}r}{R}\\leq\\frac{3}{2}$,because $ 2r\\leq R$. :wink:", "Solution_3": "[quote=\"Erken\"]$ \\sum\\cos{\\alpha}\\equal{}\\frac{R\\plus{}r}{R}$[/quote]\r\n\r\nHello :) \r\n\r\nCould you please write the proof of this or give me a link", "Solution_4": "[quote=\"Erken\"]$ x \\equal{}\\tan{\\alpha}$[/quote]\r\n\r\nYou supposed that $ \\alpha$\u00a3$ [0,\\pi/2]$\r\n\r\nI'm I right?", "Solution_5": "[quote=\"FOURRIER\"][quote=\"Erken\"]$ x \\equal{}\\tan{\\alpha}$[/quote]\n\nYou supposed that $ \\alpha$\u00a3$ [0,\\pi/2]$\n\nI'm I right?[/quote]\r\nYes,you are right $ \\alpha\\in\\[ 0,\\frac{\\pi}{2}\\]$", "Solution_6": "[quote=\"FOURRIER\"][quote=\"Erken\"]$ \\sum\\cos{\\alpha}\\equal{}\\frac{R\\plus{}r}{R}$[/quote]\n\nHello :) \n\nCould you please write the proof of this or give me a link[/quote]\r\nUse $ r\\equal{}4R\\prod\\sin\\frac{A}{2}$.", "Solution_7": "[quote=chien than]Let $ x,y,z$ be positive real numbers satisfying $ x\\plus{}y\\plus{}z\\equal{}xyz$. Prove that:\n\\[\\frac1{\\sqrt{1+x^2}}+\\frac1{\\sqrt{1+y^2}}+\\frac1{\\sqrt{1+z^2}}\\leq\\frac{3}{2}\\][/quote]\nLet $ x,y,z$ be positive real numbers satisfying $ x\\plus{}y\\plus{}z\\equal{}xyz$. Prove that:\n\\[\\frac1{\\sqrt{1+x^2}}+\\frac1{\\sqrt{1+y^2}}+\\frac4{\\sqrt{1+z^2}}\\leq\\frac{33}{8}\\]", "Solution_8": "Let $ x,y,z$ be positive real numbers satisfying $ x+y+z=xyz$. Prove that\n$$\\frac1{\\sqrt{1+x^2}}+\\frac1{\\sqrt{1+y^2}}+\\frac k{\\sqrt{1+z^2}}\\leq\\frac{2k^2+1}{2k}$$\nWhere $k\\in N^+.$", "Solution_9": " Let $a,b,c$ be positive real numbers satisfying $ ab+bc+ca=1$. Prove that\n$$\\frac1{1+a^2}+\\frac1{1+b^2}+\\frac{k}{1+c^2} \\leq\\frac{(2k+1)^2}{4k}$$\nWhere $k\\geq 1.$ ", "Solution_10": "[quote=sqing]Let $a,b,c$ be positive real numbers satisfying $ ab+bc+ca=1$. Prove that\n$$\\frac1{1+a^2}+\\frac1{1+b^2}+\\frac{k}{1+c^2} \\leq\\frac{(2k+1)^2}{4k}$$\nWhere $k\\geq 1.$[/quote]\n\nThank you for your hard-working, update old problem. Respects!", "Solution_11": "[quote=chien than]Let $ x,y,z$ be positive real numbers satisfying $ x\\plus{}y\\plus{}z\\equal{}xyz$. Prove that:\n\\[\\frac1{\\sqrt{1+x^2}}+\\frac1{\\sqrt{1+y^2}}+\\frac1{\\sqrt{1+z^2}}\\leq\\frac{3}{2}\\][/quote]\n[url]https://artofproblemsolving.com/community/c4h2866653p25472821:[/url]\nLet $a=\\frac{1}{x}, b=\\frac{1}{y}, c=\\frac{1}{z} \\Rightarrow ab+bc+ca=1$. \nWe have $$\\frac{1}{\\sqrt{1+x^2}} =\\frac{a}{\\sqrt{a^2+1}}=\\frac{a}{\\sqrt{(a+b)(a+c)}} = \\sqrt{\\frac{a^2}{(a+b)(a+c)}} \\leq \\frac{1}{2}(\\frac{a}{a+b} + \\frac{a}{a+c})$$\n$$\\Rightarrow \\frac{1}{\\sqrt{1+x^2}} + \\frac{1}{\\sqrt{1+y^2}} + \\frac{1}{\\sqrt{1+z^2}} \\leq \\frac{1}{2}(\\frac{a}{a+b} + \\frac{a}{a+c}+\\frac{b}{b+c} + \\frac{b}{b+a} + \\frac{c}{c+a} + \\frac{c}{c+b}) = \\frac{3}{2}$$\n", "Solution_12": "[quote=sqing][quote=chien than]Let $ x,y,z$ be positive real numbers satisfying $ x\\plus{}y\\plus{}z\\equal{}xyz$. Prove that:\n\\[\\frac1{\\sqrt{1+x^2}}+\\frac1{\\sqrt{1+y^2}}+\\frac1{\\sqrt{1+z^2}}\\leq\\frac{3}{2}\\][/quote]\n[url]https://artofproblemsolving.com/community/c4h2866653p25472821:[/url]\nLet $a=\\frac{1}{x}, b=\\frac{1}{y}, c=\\frac{1}{z} \\Rightarrow ab+bc+ca=1$. \nWe have $$\\frac{1}{\\sqrt{1+x^2}} =\\frac{a}{\\sqrt{a^2+1}}=\\frac{a}{\\sqrt{(a+b)(a+c)}} = \\sqrt{\\frac{a^2}{(a+b)(a+c)}} \\leq \\frac{1}{2}(\\frac{a}{a+b} + \\frac{a}{a+c})$$\n$$\\Rightarrow \\frac{1}{\\sqrt{1+x^2}} + \\frac{1}{\\sqrt{1+y^2}} + \\frac{1}{\\sqrt{1+z^2}} \\leq \\frac{1}{2}(\\frac{a}{a+b} + \\frac{a}{a+c}+\\frac{b}{b+c} + \\frac{b}{b+a} + \\frac{c}{c+a} + \\frac{c}{c+b}) = \\frac{3}{2}$$[/quote]\n\nhow we get $\\sum{\\frac{1}{\\sqrt{x^2+1}}}=\\sum{\\frac{a}{\\sqrt{a^2+1}}}$", "Solution_13": "\na = 1/x, so x = 1/a. Substituting this and factorize 1/a^2 from the square root, then put it on the numerator.\n\n\n", "Solution_14": "okay i get it thanks \n" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "A topological space $ X$ is called pathwise connected if for every $ x,y\\in X$, there exists a continuous map $ \\gamma: [0,1]\\to X$ with $ \\gamma(0)\\equal{}x$ and $ \\gamma(1)\\equal{}y$. Show that every pathwise connected space is connected and show that the converse is false.", "Solution_1": "Suppose $ X$ pathwise connected and not connected. Then let $ U\\cup V\\equal{}X$ be a separation of $ X$. Choose $ x\\in U$ and $ y\\in V$. Let $ \\gamma: [0,1]\\to X$ be the path between $ x$ and $ y$. Then $ \\gamma^{\\minus{}1}(X)\\equal{}\\gamma^{\\minus{}1}(U\\cup V)\\equal{}\\gamma^{\\minus{}1}(U)\\cup\\gamma^{\\minus{}1}(V)\\equal{}[0,1]$, but also $ \\gamma^{\\minus{}1}(U)\\cap\\gamma^{\\minus{}1}(V)\\equal{}\\emptyset$. Thus we have a separation of $ [0,1]$ which is a contradiction since $ [0,1]$ is clearly connected.\r\n\r\nTake $ \\{(x,y): y\\equal{}\\sin(1/x) \\ 04Rr \r\n \r\n (a+b+c)/3>(abc)^(1/3) so p^3>(27/8).abc\r\nR=abc/4s r=s/p\r\n4Rr=4.abc/4s.s/p=abc/p so p^2>abc/p=4Rr", "Solution_2": "I think you are wrong , please be careful", "Solution_3": "i wrote the answer in a clear way plz download the attachment", "Solution_4": "Sorry for my terrible mistake\r\n It shoud be\r\n $ 27r^{2} \\plus{} 7p^{2} \\equal{} 108 Rr$\r\n :oops: \r\n Please try agian\r\n Sorry", "Solution_5": "[size=117][color=darkblue][b]This problem is easily and nicely ![/b][/color][/size]\r\n\r\n[hide=\"Only for nkht-tk14.\"]\n[color=darkblue]With the greatest pleasure !\n\nBy the way, my name is [b]Virgil Nicula[/b] and not [u]Virgil Cula[/u], [b]nkht-tk14[/b] ...[/color][/hide]\r\n[quote=\"nkht-tk14\"][color=darkred]Prove that $ 7p^{2} + 27r^2 = 108Rr$ $ \\Longleftrightarrow$ $ a = b = c$.[/color] [/quote]\r\n[color=darkblue][b][u]Proof.[/u][/b] \n\nWe observe that if $ a = b = c$ , then $ \\{\\begin{array}{c} p = 3r\\sqrt 3 \\\\\n\\ R = 2r\\end{array}\\|$ and $ 27r^2 + 7p^2 = 108Rr$ $ \\Longleftrightarrow$\n\n$ 27r^2 + 7\\cdot 27r^2 = 108\\cdot 2r^2$ $ \\Longleftrightarrow$ $ 27\\cdot 8 = 108\\cdot 2$ , what is truly.\n\nThe well-known inequalities $ \\|\\begin{array}{ccc} \\boxed {p^2 + 5r^2\\ge 16Rr} & \\bigodot & 7 \\\\\n \\\\\n\\boxed {R\\ge 2r} & \\bigodot & 4r\\end{array}\\ \\bigoplus$ $ \\implies$\n\n$ 7\\cdot(p^2 + 5r^2) + 4r\\cdot R\\ge 7\\cdot 16Rr + 4r\\cdot 2r$ $ \\implies$ $ 7p^2 + 27r^2\\ge 108Rr$ .\n\nTherefore, $ 7p^2 + 27r^2 = 108Rr$ $ \\Longleftrightarrow$ $ \\|\\begin{array}{ccc} p^2 + 5r^2 = 16Rr \\\\\n \\\\\nR = 2r\\end{array}\\|$ $ \\Longleftrightarrow$ $ a = b = c$ .\n\n[b][u]An easy extension.[/u][/b] For any $ m\\ge 0$ and $ n\\ge 0$ we have\n\n$ mp^2 + (5m - 2n)r^2 = (16m - n)Rr$ $ \\Longleftrightarrow$ $ a = b = c$ .\n\nFor example, for $ m = 2$ and $ n = 5$ we obtain $ 2p^2 = 27Rr$ $ \\Longleftrightarrow$ $ a = b = c$ .[/color]", "Solution_6": "My proof to the ineq: $ p^{2} \\plus{} 5r^{2} \\geq 16 Rr$ is very long\r\n Can you please show me a good proof and another ineq in triangle you usually use , Virgil Nicila?", "Solution_7": "[quote=\"nkht-tk14\"][color=darkred]Prove that $ p^{2} \\plus{} 5r^{2} \\ge 16 Rr$ ... , [b]Virgil Nicila[/b] ?[/color] [/quote]\r\n[hide=\"Again, only for nkht-tk14\"]\n[color=darkred]By the way, my name is [b]Virgil Nicula[/b] and not [b]Virgil Nicila[/b] .[/color].. [/hide]\r\n[color=darkblue][b][u]Proof.[/u][/b] From the [b]Lagrange's identity[/b] $ \\boxed {\\sum XA^2 \\equal{} 3\\cdot XG^2 \\plus{} \\frac 13\\cdot\\sum a^2}$ obtain\n\nthe inequality $ 3\\cdot\\sum XA^2\\ge \\sum a^2$ for any point $ X$, with equality iff $ X: \\equal{} G$ .\n\nUsing $ IA^2 \\equal{} \\frac {bc(p \\minus{} a)}{p}$ a.s.o., the upper inequality for $ X: \\equal{} I$ becomes :\n\n$ 3\\sum \\frac {bc(p \\minus{} a)}{p}\\ge \\sum a^2$ $ \\Longleftrightarrow$ $ 3p\\cdot\\sum bc\\ge 9abc \\plus{} p\\cdot\\sum a^2$ $ \\Longleftrightarrow$\n\n$ 3(p^2 \\plus{} r^2 \\plus{} 4Rr)\\ge 36Rr \\plus{} 2(p^2 \\minus{} r^2 \\minus{} 4Rr)$ $ \\Longleftrightarrow$ $ \\boxed {p^2 \\plus{} 5r^2\\ge 16Rr}$ ,\n\nwith equality if and only if $ I\\equiv G$ , i.e. the triangle $ ABC$ is equilateral.\n\n[b][u]Remark.[/u][/b] $ IN^2 \\equal{} 9\\cdot IG^2 \\equal{} p^2 \\plus{} 5r^2 \\minus{} 16Rr\\ge 0$ .[/color]\r\n\r\n[size=117][color=red][b]Only so much ![/b][/color][/size]" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "How many real numbers $ a$ are there such that, for these values of $ a$, the equation $ x^{3}= ax+a+1$ has at least one root $ x_{0}$ satisfying\r\n$ (i)$ $ x_{0}$ is an integer and is even.\r\n$ (ii)$ $ |x_{0}| < 1000.$", "Solution_1": "$ a=\\frac{x_{0}^{3}-1}{x_{0}+1}=x_{0}^{2}-x_{0}+1-\\frac{2}{x_{0}+1}$, therefore $ x_{0}=-998,-996,...,998$.\r\nThere are 999 means." } { "Tag": [ "trigonometry" ], "Problem": "\\[ {\\rm{Prove: }} \\& {\\rm{cosec}}^6 \\theta - {\\rm{cot}}^6 \\theta = 3{\\rm{cot}}^2 \\theta {\\rm{ cosec}}^2 \\theta + 1\r\n\\]", "Solution_1": "$ \\Leftrightarrow \\frac{1\\minus{}cos^6\\theta}{sin^6\\theta}\\equal{}\\frac{3cos^2\\theta\\plus{}sin^4\\theta}{sin^4\\theta}$\r\n$ \\Leftrightarrow 1\\minus{}cos^6\\theta\\equal{}3cos^2\\theta sin^2\\theta\\plus{}sin^6\\theta$\r\n$ \\Leftrightarrow 1\\equal{}(sin^2\\theta\\plus{}cos^2\\theta)^3$" } { "Tag": [ "inequalities", "geometry", "circumcircle", "trigonometry", "Euler", "blogs", "geometry proposed" ], "Problem": "Given are $\\triangle ABC$ area $S$ and circumcenter $O$ with any $M$ on plane prove that\r\n$a^{2}MA^{2}+b^{2}MB^{2}+c^{2}MC^{2}\\ge 4\\sqrt{3}S(R^{2}-OM^{2})$\r\nequal when $ABC$ is euqaliteral and $M\\equiv O$\r\nan especial case $M\\equiv O\\Rightarrow a^{2}+b^{2}+c^{2}\\ge 4\\sqrt{3}S$\r\n$M\\equiv I\\Rightarrow \\cos^{2}\\frac{A}{2}+\\cos^{2}\\frac{B}{2}+\\cos^{2}\\frac{C}{2}\\ge\\frac{\\sqrt{3}}{2}(\\sin A+\\sin B+\\sin C)$ :lol:", "Solution_1": "[color=purple]Let D, E,F are on BC, CA, AB such that $MD\\perp BC;ME\\perp CA; MF\\perp AB$. By Euler theorem we have$S_{DEF}=\\frac{S(R^{2}-OM^{2})}{4R^{2}}$ thus$4\\sqrt{3}S(R^{2}-OM^{2}=\\frac{4S_{DEF}}{R^{2}}$\n So your inequality is equipvalent to\n $MA^{2}sin^{2}A+MB^{2}sin^{2}B+MC^{2}sin^{2}C\\geq 4\\sqrt{3}S_{DEF}\\leftrightarrow EF^{2}+DE^{2}+FD^{2}\\geq 4\\sqrt{3}S_{DEF}$\n The last inequlity is well- known.And with my solution we can solve the genenral :D \n$MA^{2}xa^{2}+MB^{2}yb^{2}+MC^{2}zc^{2}\\geq 4\\sqrt{xy+yz+zx}S(R^{2}-OM^{2})$\n Do you have more gemath? :D [/color]", "Solution_2": "Now, [b]gemath[/b] don't take part in Mathlinks Forum but you can find more Geometric Inequalities unsloved of [b]gemath[/b] at [url=http://www.mathlinks.ro/Forum/weblog.php?w=659]his blog[/url]", "Solution_3": "[quote=\"April\"]Now, [b]gemath[/b] don't take part in Mathlinks Forum but you can find more Geometric Inequalities unsloved of [b]gemath[/b] at [url=http://www.mathlinks.ro/Forum/weblog.php?w=659]his blog[/url][/quote]\r\n [color=purple] What a pity! :( Thank April I saw them.Can you send a message to gemath for me? I have many problem and i want to discuss with gemath [/color]", "Solution_4": "Why you want to discuss with [b]gemath[/b]? You can post your problems in this Forum and many Mathlinkers can help you :wink:", "Solution_5": "That's right you should post your problem in ML and many people better than me can help you.\r\nAnd \r\n[quote=\"evarist\"]Let D, E,F are on BC, CA, AB such that $MD\\perp BC;ME\\perp CA; MF\\perp AB$. By Euler theorem we have$S_{DEF}=\\frac{S(R^{2}-OM^{2})}{4R^{2}}$ thus$4\\sqrt{3}S(R^{2}-OM^{2}=\\frac{4S_{DEF}}{R^{2}}$\n So your inequality is equipvalent to\n $MA^{2}sin^{2}A+MB^{2}sin^{2}B+MC^{2}sin^{2}C\\geq 4\\sqrt{3}S_{DEF}\\leftrightarrow EF^{2}+DE^{2}+FD^{2}\\geq 4\\sqrt{3}S_{DEF}$\n The last inequlity is well- known.And with my solution we can solve the genenral :D \n$MA^{2}xa^{2}+MB^{2}yb^{2}+MC^{2}zc^{2}\\geq 4\\sqrt{xy+yz+zx}S(R^{2}-OM^{2})$\n Do you have more gemath? :D [/quote]\r\nYou have true solution but don't think I don't know \"the general problem\" :rotfl: , I don't post it because it don't have any sense. \r\nLet me show a general problem like your \"general problem\" in thinking of children: They think $4$ is divisilble by $2$, $6$ is divisible by $2$ and $8,10..$ is too... hmm there is a problem here ah I think I can generalize it ok we will have $2n$ is divisible by $2$ :rotfl:", "Solution_6": "[color=purple]You sneer me ? Please post your solution for this problem and do you get the stronger as me ? It's very funny gemath! :rotfl: [/color]", "Solution_7": "Ok the following is the stronger \r\ngiven are $\\triangle ABC$ area $S$ and circumcenter $O$ with any $M$ on plane prove that\r\n$a^{2}MA^{2}+b^{2}MB^{2}+c^{2}MC^{2}\\ge 4\\sqrt{3}S|R^{2}-OM^{2}|$\r\nequal when $ABC$ is equilateral and $M\\equiv O$ hope you like :rotfl:", "Solution_8": "Consider the pedal triangle and the inequality $a'^{2}+b'^{2}+c'^{2}\\geq 4\\sqrt{3}S'$." } { "Tag": [ "calculus", "integration", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "show :D \r\n\r\n$ \\int_{0}^{1}\\;\\;\\cot^{-1}\\;\\left(1-x+x^{2}\\right)\\;\\;\\textbf dx\\;=\\;\\boxed{\\frac{\\pi}{2}-\\ln 2}$\r\n\r\nhow about $ \\int_{0}^{1}\\;\\;\\cot^{-1}\\;\\left(1+x+x^{2}\\right)\\;\\;\\textbf dx\\;\\;\\;?$ :|", "Solution_1": "$ \\int_{0}^{1}\\text{arccot}\\left( 1-x+x^{2}\\right) \\, dx = x \\, \\cdot \\, \\text{arccot}\\left( 1-x+x^{2}\\right) \\Big|_{0}^{1}+\\int_{0}^{1}x \\cdot \\frac{2x-1}{1+\\left( 1-x+x^{2}\\right)^{2}}\\, dx$\r\n$ = \\frac{\\pi}4+\\int_{0}^{1}x \\cdot \\left( \\frac1{x^{2}-2x+2}-\\frac1{x^{2}+1}\\right) \\, dx$\r\n$ = \\frac{\\pi}4+\\left[ \\frac12 \\ln \\left( x^{2}-2x+2 \\right)-\\frac12 \\ln \\left( x^{2}+1 \\right)+\\arctan \\left( x-1 \\right) \\right]_{0}^{1}$\r\n$ = \\frac{\\pi}2-\\ln 2$.\r\n\r\nThe other one can be solved in the same way.\r\n\r\nBy the way, isn't this better fitted in the [i]Computations[/i] section?", "Solution_2": "$ I = \\int^{1}_{0}arccot(1-x+x^{2})\\,dx = \\int^{1}_{0}\\arctan\\left(\\frac{1}{1-x+x^{2}}\\right)\\,dx = \\int^{1}_{0}\\arctan\\left(\\frac{1}{1+x(x-1)}\\right)\\,dx$\r\n\r\n$ = \\int^{1}_{0}[\\arctan x-\\arctan(x-1)]\\,dx$.\r\n\r\n$ \\int\\arctan x\\,dx = x \\arctan x-\\int\\frac{x}{1+x^{2}}\\,dx = x \\arctan x-\\frac{1}{2}\\ln(1+x^{2})+C$.\r\n\r\n$ \\int \\arctan(x-1)\\,dx = x \\arctan(x-1)-\\int\\frac{x}{1+(x-1)^{2}}\\,dx$\r\n$ = x \\arctan(x-1)-\\int\\frac{x}{x^{2}-2x+2}\\,dx$\r\n$ = x \\arctan(x-1)-\\frac{1}{2}\\int\\frac{2(x-1)+2}{x^{2}-2x+2}\\,dx$\r\n$ = x \\arctan(x-1)-\\frac{1}{2}\\ln(x^{2}-2x+2)-\\int\\frac{1}{1+(x-1)^{2}}\\,dx$\r\n$ = x \\arctan(x-1)-\\frac{1}{2}\\ln(x^{2}-2x+2)-\\arctan(x-1)$\r\n$ = (x-1) \\arctan(x-1)-\\frac{1}{2}\\ln(x^{2}-2x+2)+C$.\r\n\r\n$ \\Rightarrow I = \\left[x \\arctan x-(x-1) \\arctan(x-1)+\\frac{1}{2}\\ln\\left(\\frac{x^{2}-2x+2}{x^{2}+1}\\right)\\right]^{1}_{0}$\r\n$ = \\frac{\\pi}{2}-\\ln 2$.", "Solution_3": "Similarly, $ \\int^{1}_{0}arccot(1+x+x^{2})\\,dx = \\int^{1}_{0}[\\arctan(x+1)-\\arctan x]\\,dx$\r\n\r\n$ = \\left[(x+1) \\arctan(x+1)-x \\arctan x+\\frac{1}{2}\\ln\\left(\\frac{x^{2}+1}{x^{2}+2x+2}\\right)\\right]^{1}_{0}$\r\n$ = 2 \\arctan(2)-\\frac{\\pi}{2}+\\ln\\left(\\frac{2}{\\sqrt{5}}\\right)$.", "Solution_4": "[quote=\"Carcul\"]$ I = \\int^{1}_{0}arccot(1-x+x^{2})\\,dx = \\int^{1}_{0}\\arctan\\left(\\frac{1}{1-x+x^{2}}\\right)\\,dx = \\int^{1}_{0}\\arctan\\left(\\frac{1}{1+x(x-1)}\\right)\\,dx$\n\n$ = \\int^{1}_{0}[\\arctan x-\\arctan(x-1)]\\,dx$.\n\n$ \\int\\arctan x\\,dx = x \\arctan x-\\int\\frac{x}{1+x^{2}}\\,dx = x \\arctan x-\\frac{1}{2}\\ln(1+x^{2})+C$.\n\n$ \\int \\arctan(x-1)\\,dx = x \\arctan(x-1)-\\int\\frac{x}{1+(x-1)^{2}}\\,dx$\n$ = x \\arctan(x-1)-\\int\\frac{x}{x^{2}-2x+2}\\,dx$\n$ = x \\arctan(x-1)-\\frac{1}{2}\\int\\frac{2(x-1)+2}{x^{2}-2x+2}\\,dx$\n$ = x \\arctan(x-1)-\\frac{1}{2}\\ln(x^{2}-2x+2)-\\int\\frac{1}{1+(x-1)^{2}}\\,dx$\n$ = x \\arctan(x-1)-\\frac{1}{2}\\ln(x^{2}-2x+2)-\\arctan(x-1)$\n$ = (x-1) \\arctan(x-1)-\\frac{1}{2}\\ln(x^{2}-2x+2)+C$.\n\n$ \\Rightarrow I = \\left[x \\arctan x-(x-1) \\arctan(x-1)+\\frac{1}{2}\\ln\\left(\\frac{x^{2}-2x+2}{x^{2}+1}\\right)\\right]^{1}_{0}$\n$ = \\frac{\\pi}{2}-\\ln 2$.[/quote]\r\n\r\nCarcul, if you know that\r\n$ \\int\\arctan x\\,dx = x \\arctan x-\\int\\frac{x}{1+x^{2}}\\,dx = x \\arctan x-\\frac{1}{2}\\ln(1+x^{2})+C$.\r\n\r\nit`s easy to deduce that\r\n$ \\int\\arctan x\\,dx = (x-1) \\arctan (x-1)-\\frac{1}{2}\\ln(1+(x-1)^{2})+C$." } { "Tag": [], "Problem": "If you write the six digits $ 1$,$ 2$,$ 3$,$ 4$,$ 5$, and $ 6$ in all possible orders you will find that there are $ 720$ ways to do this. If the numbers are written in order of least to greatest, the first number on the list is $ 123456$ and the last number is $ 654321$. What number will be the $ 417$th number on the list?", "Solution_1": "#1 to #120 on list have a hundred-thousands digit of 1\r\n\r\n#121 to #240 on list have a hundred-thousands digit of 2\r\n\r\n#241 to #360 on list have a hundred-thousands digit of 3\r\n\r\n#361 to #480 on list have a hundred-thousands digit of 4\r\n\r\nSo we know that the number begins with a $ 4$\r\n\r\nUsing same logic,\r\n\r\n#361 to #384 begin with $ 41$\r\n\r\n#365 to #408 begin with $ 42$\r\n\r\n#409 to #432 begin with $ 43$\r\n\r\nOur number begins with $ 43$\r\n\r\nContinuing this, we find that the $ 417$th number is $ \\boxed{432516}$", "Solution_2": "120 begin with \"1\", another 120 begin with \"2\", another 120 begin with \"3\", and another 120 begin with \"4\". 417 is less than 120*4, so the first digit of 417 is 4. \r\n\r\nOf the numbers beginning with 4, 24 have a second digit of 1, another 24 have a second digit of 2, and another 24 have a second digit of 3. We find that the 417th number has a second digit of 3.\r\n\r\nOf the numbers beginning with 43, 6 have a third digit of 1, and another 6 have a third digit of 2. This means the 417th number has a third digit of 2. \r\n\r\nOf the numbers beginning with 432, 2 have a fourth digit of 1, and another 2 have a fourth digit of 5. \r\n\r\nOur number is then $ \\boxed{432516}$. Hopefully I didn't fail.\r\n\r\nEDIT: tinytim submitted that while I was writing this." } { "Tag": [ "Math Kangaroo" ], "Problem": "This is for anyone is interested in [url=http://www.mathkangaroo.com/2010page/kangur/index.htm]Math Kangaroo[/url]. There is center in the [url=http://www.mathkangaroo.com/MathKangaroo/LocationDetails.jsp?locationID=55]Twin Cities[/url] as well as a new one in [url=http://www.mathkangaroo.com/MathKangaroo/LocationDetails.jsp?locationID=91]Grand Forks, ND[/url] for anyone who lives in the western part of the state.\r\n\r\nThis year, it will be on Thursday, March 18, 2010. Registration must be done by January 20, 2010.\r\n\r\nMath Kangaroo is a math contest for grades 1 through 12. It is a 75-minute multiple choice test with 24 or 30 questions, depending on grade level. Prizes are given to the top scorers.", "Solution_1": "You from Minnesota?", "Solution_2": "Hey Kev, you from minnesota yourself? It says antarctica...", "Solution_3": "Whose kev?\n\nAnd no I am not from Minnesota, I am from Antarctica.", "Solution_4": "No, but Minnesota is within walking distance from my house. (and @abcak I've met you once before)", "Solution_5": "I have met both powerofpi and abcak. Seems interesting (math kangaroo)", "Solution_6": "This is creepy.\n\nUNFORTUNATELY (for you), you guys don't know wizeng :P\n\nI think.", "Solution_7": "I really want to meet Abcak.\nI haven't seen him before.", "Solution_8": "*Bump... Similar thing this year, but the registration deadline is January 10, 2011 and the competition itself is on March 17, 2011.", "Solution_9": "Hah.\n\nSo is this only people in Minnesota and North Dakota.\n\nI really want to meet PowerOfPi; I haven't seen him before :wink:", "Solution_10": "I really want to meet PowerOfPi, AwesomeToad, and abcak, I haven't seen any of them before...", "Solution_11": "[quote=\"ksun48\"]I really want to meet PowerOfPi, AwesomeToad, and abcak, I haven't seen any of them before...[/quote]\n\nLOLWUT WHO ARE YOU???", "Solution_12": "http://m.youtube.com/#/watch?v=jDGnfT_OaCM\n\nThe final summary of this topic", "Solution_13": "[quote=\"PowerOfPi\"]*Bump... Similar thing this year, but the registration deadline is January 10, 2011 and the competition itself is on March 17, 2011.[/quote]\n\nDeadline has passed, but maybe registration will still be possible...", "Solution_14": "This year's registration deadline is the 16th of January. Hurry up and register if you haven't already." } { "Tag": [], "Problem": "There are two samples of gas in a thermally insulated cylinder. One of them is Argon at a temperature of 300 K, and the other is Nitrogen at a temperature of 400 K. There is a thermally insulated wall between them. The two gases have the same pressure and the same volume. Then the wall between the gases is taken away, and the two gases mix. \r\n\r\na) Does the energy of the system change? \r\n\r\nb) What will be the common temperature? \r\n\r\nc) How does the pressure change? \r\n\r\nd) Does the entropy of the system change?", "Solution_1": "a) Considering the cylinder as an isolated system and both gases as ideal, and as the gases don't react with each other, then $\\Delta U = 0$.", "Solution_2": "b) about $355$K\r\nd)The entropy of the system will grow.", "Solution_3": "[quote=\"Thaakisfox\"]b) about $355$K\n[/quote]\r\n\r\nCan you show the work?", "Solution_4": "You mean how to calculate the temperature?", "Solution_5": "[quote=\"Thaakisfox\"]You mean how to calculate the temperature?[/quote]\r\n\r\nYeah, I don't see how to do it.", "Solution_6": "its just conservation of energy:\r\n$\\frac32 pV+\\frac{5}{2}pV = \\frac{3}{2}\\frac{pV}{T_{1}}T+\\frac{5}{2}\\frac{pV}{T_{2}}T$\r\nform here the common temperature:\r\n$T=\\frac{8T_{1}T_{2}}{3T_{2}+5T_{1}}$\r\n$T_{1}$ is the begining temp. of Argon, $T_{2}$ is the begining temp of nytrogen, $p$ and $V$ are the begining volume, and pressure.", "Solution_7": "[quote=\"Thaakisfox\"] $\\frac{3}{2}\\frac{pV}{T_{1}}T$[/quote]\r\n\r\nWhy is this the new energy?", "Solution_8": "[quote=\"cancer\"][quote=\"Thaakisfox\"] $\\frac{3}{2}\\frac{pV}{T_{1}}T$[/quote]\n\nWhy is this the new energy?[/quote]\r\n\r\nCan someone explain? Why divide by $T_{1}$ and multiply by $T$?", "Solution_9": "according to the gas law:\r\n\r\n$pV=n_{1}RT_{1}$ so from here $n_{1}R=\\frac{pV}{T_{1}}$\r\nso if we want to find the energy of the Argon after mixing: (since $n_{1}$ is constant, non of the argon leaves the system)\r\n$E_{1}= \\frac32 n_{1}R T = \\frac32 \\frac{pV}{T_{1}}T$" } { "Tag": [ "Princeton", "college", "FTW", "function", "logarithms", "MATHCOUNTS", "irrational number" ], "Problem": "Currently i am planning a huge celebration of pi day at my school. It will probably involve me saying ~100 digits of pi over the announcements. I have committed only 83 digits to memory, and plan on memorizing 17 more by March.\r\n\r\nWhat about you? All pi fans are encouraged to respond.", "Solution_1": "Princeton has a pi recitation contest on Pi Day. I had just under 100 digits memorized then, but now I am not confident of more than 20 or so. The person who won had nearly 200 I think. \r\n\r\nYou should have put a 100+ option or something :P.", "Solution_2": "I had this one on the shelf, because I knew this day would come :\r\n\r\n[img]http://users.telenet.be/frederic_bel6/Fun/refusepi.jpg[/img]", "Solution_3": "All I know is 3.141592...", "Solution_4": "3.14159265358979323", "Solution_5": "3.1415926\r\n\r\nWhat is the point of memorizing pi? Is it just how nerds show their masculinity?", "Solution_6": "lets see:\r\n3.141592653589793238462643383279502884 and umm... umm..\r\nyeah.\r\n\r\nIf i had any reason to memorize, i could probably get to 150", "Solution_7": "3.141-why?", "Solution_8": "Memorizing pi is a great way to attract girls.\r\n\r\nPerson: Hey baby, your curves are like a circle with a radius of pi. Pi is approximately 3.1415...(353784973489)\r\n\r\nor perhaps, more direct\r\n\r\nPerson: Hey baby, wanna get with someone who knows the first 3489305834890 digits of pi?", "Solution_9": "LOL, you got that totally from the MI forum. \r\n\r\nHey baby, wanna go out w/ a guy who got a 34 on the MMPC? \r\n\r\nI kinda tried that...not like...directly. \r\n\r\nDon't try it in your life, it's not pleasant.\r\n\r\nAnyways...I don't feel like memorizing it, cuase it has no poitn:\r\n\r\n3.141592653589793238462643383279?", "Solution_10": "(20:52:33) Atheist Temple: Hey baby, wanna get with someone who knows the first 3489305834890 digits of pi?\r\n(20:52:34) : math homework\r\n\r\n\r\ndarn i thought it would work better", "Solution_11": "3.14159265358979323844338327950288416...\r\n\r\nhuh\r\n\r\nwhy don't people memorize e?\r\n\r\nI have 2.7182818284530459...\r\n\r\nnot cool", "Solution_12": "I know 50, but feel there is no point in memorizing any more.\r\n\r\n3.14159265358979323846264338327950288419716939937510...", "Solution_13": "Mostly I only remember $3.14$\r\n\r\nNo point, lah :D", "Solution_14": "Hey! I'm a girl here, and memorization of pi is a popular nerdy pastime. :mad:", "Solution_15": "correction: kstan013, there must be a picture of me in the song in order for it to be classified as nerdy -_-. :D \r\n\r\n\r\n\r\n[quote=\"funcia\"] [color=blue]kyyuanmathcount: clip, i'd rather not clog up thread, just look posts above[/color][/quote]\n\n\"obvious\" is my word, I think you are offending him, captain obvious should not have to come to rescue too many times in one thread.\n\nAlso, [quote=\"I\"]I don't care[/quote]", "Solution_16": "[quote=\"kyyuanmathcount\"]correction: kstan013, there must be a picture of me in the song in order for it to be classified as nerdy -_-. :D [/quote]Awww... don't be so hard on yourself :)", "Solution_17": "LOL \r\n\r\nOBVIOUS SHOULDN'T BE PUT HERE SO I SAY OBLIVIOUS THEN. \r\nJK I was just upset after a relative died.", "Solution_18": "[quote=\"kstan013\"][quote=\"kyyuanmathcount\"]correction: kstan013, there must be a picture of me in the song in order for it to be classified as nerdy -_-. :D [/quote]Awww... don't be so hard on yourself :)[/quote]\r\n\r\n :D it's for the good of the rest of the world... and the asians in it :P. \r\n\r\n$\\pi = 4(\\frac{1}{1}-\\frac{1}{3}+\\frac{1}{5}-\\frac{1}{7}+\\frac{1}{9}+-\\cdots )$", "Solution_19": "yeah Asians are so pretty and smart\r\nwhen i told my Chinese friend i wanted to be Asian she \r\npractically threw up on me", "Solution_20": "No need to brownnose now is there?\r\n\r\nAsians include indian people. When we want to refer to chinese, japanese, korean, malaysian etc., we say oriental :).", "Solution_21": "but i meant Asian", "Solution_22": "The only reason I know anything much of pi is because my math teacher in 4th grade had a poster on her wall that had like 100 digits of pi on it. 3.14159265358979323846.", "Solution_23": "I have...\r\n3.1415926535897932384626433832795028841971693993751058209749445923078164062862089.\r\n\r\nKinda a waste of time now... :huh:", "Solution_24": "i simply memorized the first 10 digits\r\n3.14159265\r\ni think it's 10...i never counted it\r\nEdit-OMG ITS ONLY 9 digits! actually without the 3 it's 8...i think i might've forgot 1 or 2 digits", "Solution_25": "$\\pi$ is DA BOMB\r\n\r\nI've memorized about 180 digits, but it has been a long time since I've recited them and I can't say them as well as I used to be able to.", "Solution_26": "this is the key that makes us all wind and the beat comes up on it", "Solution_27": "[quote=\"modeler\"][hide=\"I know a guy who can memorize pi better than anyone of you.\"]\nMy Computer :D \n[/hide][/quote]\r\nyou silly :roll: , computers don't have genders", "Solution_28": "computers have genders.... just like they have generations... vacuum tubes, transistors, newly artificial intelligence.\r\n\r\n\r\nso why not gender? \r\n\r\n\r\nI'm sure they do.\r\n\r\n\r\n\r\nOh yeah, I know $\\pi$\r\n\r\n\r\nIt's approximately $\\frac{22}{7}$!\r\n\r\nwell or $3.14159$", "Solution_29": "My computer is a male. I call him Mac :D" } { "Tag": [], "Problem": "Compute: $4 + 5 - 7 \\times 9 \\div 3$.", "Solution_1": "[hide]9+63/3\n9+21\n30[/hide]", "Solution_2": "[quote=\"LuCy4EvA\"][hide]9+63/3\n9+21\n30[/hide][/quote]\r\n\r\nHow did that minus between the 5 and the 7 change to a plus? I got \r\n-12.", "Solution_3": "oops, wasn't really paying attention... sry!!!\r\n\r\nso it's um...\r\n\r\n[hide]\n9-21 = -12[/hide]\r\n\r\nthanx for finding my mistake, that's why i need more practice with arithmetic! :blush:", "Solution_4": "[hide=\"hinty hint\"]ALWAYS!!! do muliplication and divisions frist , and do addition and subtraction next[/hide]\r\n\r\n :D \r\n\r\nI hate arithmeatic problems, so I won't solve it...", "Solution_5": "[hide]$90-\\frac{63}{3}$ \n$=9-21$ \n$=-12$[/hide]", "Solution_6": "why is it 90 - (63/3)?", "Solution_7": "[hide]\n-12\n[/hide]", "Solution_8": "[quote=\"LuCy4EvA\"]why is it 90 - (63/3)?[/quote]\r\n\r\nUmm... It's not 90- (63/3)\r\n\r\n[hide=\"to LuCy4EvA\"]\n4+5-7*9/3\n=4+5-63/3\n=4+5-21\n=9-21\n=-12\n[/hide]", "Solution_9": "[quote=\"MCrawford\"]Compute: $4 + 5 - 7 \\times 9 \\div 3$.[/quote]\r\n\r\n[hide]\n=4+5-(63/3)\n=4+5-21\n=9-21\n=-12\n[/hide]", "Solution_10": "[quote=\"MCrawford\"]Compute: $4 + 5 - 7 \\times 9 \\div 3$.[/quote]\r\n\r\n[hide=\"solution\"]\n4 + 5 - 7 x 9 / 3\n4 + 5 - (7 x 9 / 3)\n9 - (7 x 3)\n9 - 21\n-12\n[/hide]", "Solution_11": "The trick of solving this problems is to know the order of operation. PEMDAS. \r\n1. parenthesis\r\n2. exponent\r\n3.Multiplication and Division\r\n4. Addition and Subtraction", "Solution_12": "[hide]-12[/hide]", "Solution_13": "A stupid acronym for PEMDAS:\r\n\r\nPlease\r\nExcuse\r\nMy\r\nDear\r\nAunt\r\nSally\r\n\r\n....", "Solution_14": "[quote=\"mathgeek2006\"]A stupid acronym for PEMDAS:\n\nPlease\nExcuse\nMy\nDear\nAunt\nSally\n\n....[/quote]\r\n\r\nWhats stupid about that??", "Solution_15": "4+5-7*9/3\r\n4+5-7*3\r\n4+5-21\r\n9-21\r\n-12", "Solution_16": "[quote=\"GoBraves\"][quote=\"mathgeek2006\"]A stupid acronym for PEMDAS:\n\nPlease\nExcuse\nMy\nDear\nAunt\nSally\n\n....[/quote]\n\nWhats stupid about that??[/quote]\r\nI learned it in school, for one. Also, it's really kinda weird...", "Solution_17": "[quote=\"mathgeek2006\"][quote=\"GoBraves\"][quote=\"mathgeek2006\"]A stupid acronym for PEMDAS:\n\nPlease\nExcuse\nMy\nDear\nAunt\nSally\n\n....[/quote]\n\nWhats stupid about that??[/quote]\nI learned it in school, for one. Also, it's really kinda weird...[/quote]\r\n\r\nI learn tons of stuff in school..........i learn all english, and social studies in school...and I dont think its that weird", "Solution_18": "[quote=\"GoBraves\"][/quote][quote=\"mathgeek2006\"][quote=\"GoBraves\"][quote=\"mathgeek2006\"]A stupid acronym for PEMDAS:\n\nPlease\nExcuse\nMy\nDear\nAunt\nSally\n\n....[/quote]\n\nWhats stupid about that??[/quote]\nI learned it in school, for one. Also, it's really kinda weird...[/quote][quote=\"GoBraves\"]\n\nI learn tons of stuff in school..........i learn all english, and social studies in school...and I dont think its that weird[/quote]\r\nThis is something I learned in 4th grade, though. Now that I'm in 8th, I think that it's stupid.", "Solution_19": "[hide]4+5-7*9/3 \n\n[b]-12. There's your answer.[/b][/hide]", "Solution_20": "[hide]4+5-7*9/3 = 4+5-[(7*9)/3]\n = 4+5-(63/3)\n = 4+5-21\n = -12[/hide]", "Solution_21": "all these quotes are getting confusing\r\n\r\ncan someone tell me how to do them?", "Solution_22": "[hide]Very Easy just do Bedmass\n 9x7=63\n 63 Divided 3 = 21\n 9-21= -12 = The Answer [/hide]", "Solution_23": "[hide]$4+5-7\\cdot 9/3=4+5-7\\cdot 3=4+5-21=9-21=\\boxed{-12}$[/hide]", "Solution_24": "[quote=\"math92\"]all these quotes are getting confusing\n\ncan someone tell me how to do them?[/quote]\r\n\r\nto quote someone's post, there is a \"quote\" button on the bottom right of the post", "Solution_25": "oh i see now, thanks", "Solution_26": "but if u remembr the \"pemdas\" and forget the left to right rule than there is no point of remembering the pemdas rule because u r gonna get it wrong! RITE?", "Solution_27": "yeah, they should make a new acronym: DPEDMASLTR which means do PEDMAS left to right. [/hide]", "Solution_28": "yeah, they should make a new acronym: DPEDMASLTR which means do PEDMAS left to right." } { "Tag": [ "Olimpiada de matematicas" ], "Problem": "Primer nivel, problema 1:\r\n\r\nEn un a\u00f1o que tiene 53 s\u00e1bados, \u00bfqu\u00e9 d\u00eda de la semana es el 12 de mayo? Dar todas las posibilidades.", "Solution_1": "Supongo que el problema se resuelve as\u00ed:\n\nHay dos casos:\n\n-A\u00f1o normal (1)\n\n-A\u00f1o bisiesto (2)\n\nEn (1) tenemos que el a\u00f1o tiene 365 d\u00edas. Dado que $365=52\\cdot 7+1$, entonces al haber 52 semanas es obvio que hay 52 s\u00e1bados, el problema pide un a\u00f1o con 53 s\u00e1bados, por lo tanto ese $+1$ que aparece debe ser s\u00e1bado. Como el a\u00f1o est\u00e1 compuesto por 52 semanas enteras y un s\u00f3lo d\u00eda de otra semana, entonces se concluye que la semana nuestra empieza con el d\u00eda s\u00e1bado. El d\u00eda 365 (31 de diciembre) es s\u00e1bado seg\u00fan lo dicho m\u00e1s arriba. El d\u00eda 1 (1 de enero) es s\u00e1bado tambi\u00e9n, ya que es congruente a 1 m\u00f3dulo 7, al igual que 365. El 12 de mayo es el d\u00eda n\u00famero 132 del a\u00f1o, como $132\\equiv 6 (mod 7)$, se concluye que es jueves, ya que si el s\u00e1bado es congruente a 1, entonces el jueves es congruente a 6, porque la secuencia es s\u00e1bado, domingo, lunes, ..., viernes. Como verificaci\u00f3n tomen el a\u00f1o 2011, es decir este a\u00f1o, y ver\u00e1n que se cumple todo (fue un problema futurista este).\n\nEn (2) tenemos que el a\u00f1o tiene 366 d\u00edas, es decir $366=52\\cdot 7+2$, el s\u00e1bado n\u00famero 53 puede ser $+1$ o el $+2$. Si es el $+1$, entonces se tiene que el d\u00eda 365 es s\u00e1bado, es decir el 30 de diciembre, entonces el 1 de enero tambi\u00e9n es s\u00e1bado (d\u00eda 1). El 12 de mayo en esta oportunidad es el d\u00eda 133 del a\u00f1o, como $133\\equiv 0(mod 7)$, entonces el 12 de mayo caer\u00e1 viernes. Si el s\u00e1bado cae $+2$, entonces el 31 de diciembre es s\u00e1bado, y se tiene que para ser s\u00e1bado debe ser congruente a 2 m\u00f3dulo 7. Como 133 es congruente a 0, entonces debe caer dos d\u00edas antes que el s\u00e1bado, es decir el jueves.\nPara verificar pueden ver el calendario del a\u00f1o 1988\n\nConclusi\u00f3n: el 12 de mayo puede caer jueves o viernes en un a\u00f1o con 53 s\u00e1bados." } { "Tag": [ "FTW", "AMC", "AIME", "geometry", "geometric transformation", "reflection", "analytic geometry" ], "Problem": "Join today! And if you so desire, you may refer to the rules [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=250456]here...[/url] :D", "Solution_1": "I say everything I said in the other thread :D", "Solution_2": "okay I sign up I guess. I'm really only interested in the chapter one though since my goal is to make state this year.", "Solution_3": "I sign up.", "Solution_4": "I sign up on BOGTRO and FanFan's team.", "Solution_5": "i am on fanfans team, but not bogtros. \r\n\r\n\r\n....\r\n\r\n\r\nHA!", "Solution_6": "I don't care. Sign me up on any team. I just wanna focus on chapter. If you don't want to be brought down don;t put me on your team. I seriously am bad.(despite my 1721 FTW rating :P )", "Solution_7": "i join...anyone wanna join me?", "Solution_8": "[u]No Team:[/u]\r\n\r\nAIME15\r\nBOGTRO\r\nDiscrete_Math\r\nFantasyLover\r\nisabella2296\r\njjx1\r\nnikeballa96\r\npythag011", "Solution_9": "aime, pythag, izzy, team?", "Solution_10": "dangit I posted right after aops was down. i said nike want to be a team. that is if its okay if i bring you down :blush:", "Solution_11": "[quote=\"mz94\"]i am on fanfans team \n[/quote]\r\n\r\nlol ernie", "Solution_12": "[quote=\"FantasyLover\"][quote=\"mz94\"]i am on fanfans team \n[/quote]\n\nlol ernie[/quote]\r\n\r\nonly you mean lol mz", "Solution_13": "[quote=\"ernie\"][quote=\"FantasyLover\"][quote=\"mz94\"]i am on fanfans team \n[/quote]lol ernie[/quote]only you mean lol mz[/quote]\n\nLol, I did mean you because you said:\n\n[quote=\"ernie\"]No Team: \nFantasyLover \n[/quote]", "Solution_14": "[quote=\"nikeballa96\"]i join...anyone wanna join me?[/quote]\r\n\r\nill join ur team,,,if it is okay with you. PAY NO ATTENTION TO MY FTW RATING, IT DOES NOT REFLECT MY ABILITIES \r\n\r\nI MEAN IT", "Solution_15": "Um, guys, suma_milli hasn't voted for team captain yet. (for team :D )", "Solution_16": "why does team captain matter again???", "Solution_17": "I thought the captains were supposed to submit the team rounds?", "Solution_18": "my assumption is that ernie wants to make this as close to the real competition as possible, which means any submissions for the team round not from the captain probably won't be graded at all", "Solution_19": "Nope.\r\n\r\nOnly team captain sends answers. Anyone who doesn't abide to that rule will disqualify their whole team. :P", "Solution_20": "I VOTE FOR \"SUPERMATHBOY\"\r\nas teh team captain", "Solution_21": "[quote=\"gauss1181\"]Um, guys, suma_milli hasn't voted for team captain yet. (for team :D )[/quote]\r\n\r\nOh sorry it was because I didn't really have an opinion.. So no vote from me. I guess AIME_is_hard is captain then.", "Solution_22": "I hope this means the tournament can finally start now...", "Solution_23": "Check the other thread, it already has :wink:", "Solution_24": "Are these problems from past mc chapter? did you make them up yourself. They seem awfully hard for chapter :maybe:", "Solution_25": "[quote=\"jjx1\"]Are these problems from past mc chapter? did you make them up yourself. They seem awfully hard for chapter :maybe:[/quote]\r\nI guess he made them up.\r\n\r\nRemember...these tests were made by fellow AoPSers...so they'd definitely be harder than normal.", "Solution_26": "Me sign up plz. :lol:", "Solution_27": "RTFT...you can't sign up anymore.", "Solution_28": "of course, that doesn't bar you from taking the test altogether\r\n\r\nyou just won't be able to participate in the competition aspect", "Solution_29": "Hmmm... since I'm really fair, I'll let you participate as an individual." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "$f$ is continius function.\r\n1. $f(1000)=999$\r\n2.$f(x).f(f(x))$=1\r\nfind $f(500)$", "Solution_1": "Hehehe :).\r\n\r\n$f(999)=\\frac 1{999}, f(1000)=999$, so the image of $f$ includes $500\\in [\\frac 1{999},999]$. Since $f(x)=\\frac 1x$ for all $x$ included in the image of $f$, we get $f(500)=\\frac 1{500}$." } { "Tag": [ "algebra", "polynomial", "Rational Root Theorem" ], "Problem": "$ P(x)\\equal{}ax^{3}\\plus{}bx^{2}\\plus{}cx\\plus{}d$\r\na,b,c,d are integers.\r\nad is odd,bc is even.\r\nProve that not all roots are rational.", "Solution_1": "Over $ \\mathbb{F}_2$ the only polynomial of degree $ 3$ with nonzero constant term and linear factors is $ (x \\plus{} 1)^3$.", "Solution_2": "Sorry,but what is $ F_2$? :blush:", "Solution_3": "I think t0rajir0u is talking about fields and groups.", "Solution_4": "I don't know anything in fields and groups.\r\n\r\nBut,this was how BJV solved it.\r\n\r\n[hide=\"solution\"]Assume that all roots are rational.\nLet them be $ x_1,x_2,x_3$\nNow,\n$ ax_1^3 \\plus{} bx_1^2 \\plus{} cx_1 \\plus{}d\\equal{}0$\nMultiplying the above equation by $ a^2$,\n$ {(ax_1)}^3 \\plus{} b{(ax_1)}^2 \\plus{} ca(ax_1) \\plus{} da^2 \\equal{}0$\n\nTherefore the equation,\n$ y^3 \\plus{} by^2 \\plus{} cay \\plus{} da^2 \\equal{}0$ -------I\nhas roots $ y_1,y_2,y_3$,\n$ y_1\\equal{}ax_1$\n$ y_2\\equal{}ax_2$\n$ y_3\\equal{}ax_3$\n\nI ia a monic polynomial with integer coefficients.\nTherefore, $ y_1,y_2,y_3$ are integers.\n\nNow,\n $ y_1 \\plus{} y_2 \\plus{} y_3\\equal{} \\minus{}b$ -----II\n\n $ \\sum{y_1y_2} \\equal{} ca$ -----III\n\n$ \\prod{y_1} \\equal{} \\minus{}a^2d$ ----IV\n\n\nNow,ad is odd.Therefore $ a^2d$ is odd.\nSo,from IV\n $ y_1,y_2,y_3$ are each odd.\n\nThen from II b is odd.\nFrom III,c is odd.\nTHerefore bc is odd.\nBut this is a contradiction,as bc is even.\nTherefore our assumption that all roots are rational is wrong.\n\nThus proved.\n\n\n[/hide]", "Solution_5": "That's essentially the same solution as mine. $ \\mathbb{F}_2$ is the mathematical structure you get when you ignore everything about the integers except their parity; that is, it has two elements $ \\{ \\text{Even}, \\text{Odd} \\}$ satisfying the usual rules for adding and multiplying even and odd numbers. Over $ \\mathbb{F}_2$, the problem condition reduces to\r\n\r\n$ P(x) \\equal{} x^3 \\plus{} x^2 \\plus{} 1 \\text{ or } x^3 \\plus{} x \\plus{} 1$\r\n\r\nwhere $ 1 \\equal{} \\text{Odd}, 0 \\equal{} \\text{Even}$. On the other hand, by the rational root theorem (which holds in this setting!) the only possible rational root is $ 1$, so\r\n\r\n$ P(x) \\equal{} (x \\plus{} 1)^3 \\equal{} x^3 \\plus{} x^2 \\plus{} x \\plus{} 1$;\r\n\r\ncontradiction.", "Solution_6": "Wow!\r\nThanks a lot t0rajir0u. :)" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "[u][b]The author of this posting is : pvthuan[/b][/u]\r\n____________________________________________________________________\r\n\r\nIf $a, b, c \\geq 0$, then\r\n\r\n\\[\\sqrt{a^4+b^4+c^4}+\\sqrt{abc\\left(a+b+c\\right)} \\geq 2\\sqrt{a^2b^2+b^2c^2+c^2a^2}\\].\r\n\r\nThis was solved by Pham Van Thuan, Hanoi University of Science, and here is my conjecture:\r\n\r\nIf $0 :?", "Solution_8": "The inequality and the conjecture are not true for $b=c\\rightarrow\\infty$ .", "Solution_9": "Can anyone ,please, tell me if this ineq\r\n\r\n\r\n$(a^4+b^4+c^4)(a+b+c)abc \\geq (a^2b^2+b^2c^2+c^2a^2)^2$\r\n\r\nis true?\r\n\r\nThanks a lot.", "Solution_10": "Of course, it is not. Take $c=0$!", "Solution_11": ":blush: :blush: :blush: :blush: :blush: :blush: :blush:", "Solution_12": "Don't worry. We all make small mistakes from time to time." } { "Tag": [ "inequalities", "LaTeX" ], "Problem": "Hi, i am having trouble to solve this inequality when i pass 2 to other side of the equation. i know how to solve if multiple the sides.\r\n\r\n [(3-x)^(1/2)] /x <= 2\r\n\r\nsolve doing it please: \r\n\r\n [(3-x)^(1/2)] /x -2 <= 0\r\n\r\n\r\n\r\n2. Equation\r\n\r\nfind X. [9(7x-3)^2]/(2x+1) = [125(14x^2 +x -3)^1/2]/3\r\n\r\n\r\nThanx! :)", "Solution_1": "hello, writing your inequality in $ \\text{\\LaTeX}$ we have\r\n$ \\frac{\\sqrt{3-x}}{x}\\le 2$, because of the range of definition of the square root we have $ -\\infty0$ we get $ \\sqrt{3-x}\\le 2x$ squaring this we have\r\n$ 0 \\le 4x^2+x-3$ and this is fulfilled for all $ x$ with $ \\frac{3}{4}\\le x\\le3$\r\nCase $ 2$ multiplying by $ x$ with $ x<0$ we get $ \\sqrt{3-x}\\geq 2x$, this is true because we have $ -\\infty2(existence condition)\r\n\r\nanswer: (2,4)\r\n\r\nand i cant see the difference between this question and the first\r\nIn this case, i didnt split in 2 cases as you. and got the correct answer.\r\n\r\nplease, clarify this for me.", "Solution_7": "Well, you can solve this sort of inequation by other methods, since it is does not involve any kind of fallacy. This shape of inequation permitted a more simple form that leads to a easier solution. I don't know what you mean clearly, but I hope your doubt was answered.", "Solution_8": "hello, do you mean $ \\frac{2x\\sqrt{x\\minus{}2}\\minus{}4\\sqrt{x\\minus{}2}}{(x\\minus{}4)\\sqrt{x\\minus{}2}}\\le1$?\r\nSonnhard." } { "Tag": [], "Problem": "For what value of $x$ does\r\n\\[ \\frac{0.\\overline{09}}{x} = 11^{-1}? \\]", "Solution_1": "[hide]When $x=1$.[/hide]", "Solution_2": "[hide]11^-1 is 1/11. .0999999 is 1/11 so 1.[/hide]", "Solution_3": "[quote=\"MCrawford\"]For what value of $x$ does\n\\[ \\frac{0.\\overline{09}}{x} = 11^{-1}? \\][/quote]\r\n\r\n[hide]$\\frac{\\frac{1}{11}}{x}=\\frac{1}{11}$\n\n$x=1$[/hide]", "Solution_4": "[quote=\"A+MATH\"][hide]11^-1 is 1/11. .0999999 is 1/11 so 1.[/hide][/quote]\n\n\n[hide]A+MATH...\n\n0.099999999 is not 1/11.\n\nyou misread the question, its .0909090909...[/hide]", "Solution_5": "[hide=\"However...\"]1/11=0.09090909...[/hide]", "Solution_6": "[quote=\"jli\"][quote=\"A+MATH\"][hide]11^-1 is 1/11. .0999999 is 1/11 so 1.[/hide][/quote]\n\n\n[hide]A+MATH...\n\n0.99999999 is not 1/11.\n\nyou misread the question, its .0909090909...[/hide][/quote]\r\n\r\nHe read the question right, he just didn't write it down right." } { "Tag": [ "algebra", "polynomial", "calculus", "integration", "number theory", "relatively prime" ], "Problem": "Find the least positive integer n satisfying the equation n3+2n=b, where b is an odd perfect square.", "Solution_1": "n will have to be odd\r\nb\u00b2 will have to be 0 mod n\r\na\u00b2=b and thus a is odd\r\n\r\nso a= \\sqrt(n\u00b3+2n)\r\n\r\na is an integer, that means n has common divisors with n\u00b2+2\r\nit's not possible that n divides n\u00b2+2, so at least one of these must contain a perfect square in its factors\r\n\r\nthat cannot be n\u00b2+2 so it must be n\r\nso n can be written as c.k\u00b2 (both being odd, k being the largest possible)\r\n\r\na = \\sqrt(c\u00b3.k^6+2c.k\u00b2)\r\nso k divides a\r\na/k=d is integer\r\n\r\nd = \\sqrt (c\u00b3.k^4 + 2c)\r\nas k was largest possible, c cannot be a square, thus c divides c\u00b2.k^4+2 so 2 divides c\r\n\r\nso either c=1\r\nthen k^4+2 is a perfect square, which is not possible\r\n\r\nso c must be 2\r\nbut it had to be odd\r\n\r\nso this is the third time I tried this problem, I wrote is as long and as clear as possible, always the same result that there are none, but you're asking for the lowest one? damn. :?", "Solution_2": "[quote=\"Peter VDD\"]a is an integer, that means n has common divisors with n2+2\nit's not possible that n divides n2+2, so at least one of these must contain a perfect square in its factors\nthat cannot be n2+2 so it must be n\nso n can be written as c.k2 (both being odd, k being the largest possible)\n[/quote]\r\n\r\nWell Peter, in fact n does not have common divisors with n2 + 2. Because if d is a common divisor of n and n2 + 2, then d divides also (n2 + 2) - n x n = 2, so d|2. But d must be odd (as n is odd) so n and n2 + 2 are coprime. So now you know that the product of two relatively prime integers (i.e. n and n2 + 2) is a perfect square. What does that mean? :) Try to go from there.", "Solution_3": "I don't know\r\n\r\nI thought exactly that way the first time, but scratched it away because I concluded from it there were no solutions\r\n\r\nI know it looks very long and stupid like that, my previous solutions were much shorter and more powerful, but I kept getting no results, so I just threw them away. And I didn't feel like retyping the whole stuff.", "Solution_4": "Are you sure about the problem statement, 3X.lich? :?\r\n\r\nIsn't it n3 + 2n2 = b, where b is an odd perfect square?", "Solution_5": "did you look the source up in your private collection, Arne? ;)", "Solution_6": "I didn't find it till now but I'm sure I saw it before :D\r\nWell, my collection is becoming huge so some things are difficult to find :D", "Solution_7": "well it'd surprise me if you'd have Sergipanas OM 2nd round lv2/q6 (Brazil)\r\n\r\nunless it's one of the famous things I never heard of again :D", "Solution_8": "No, I think I saw it on some British IMO mentoring scheme a long time ago, but I don't find it now.", "Solution_9": "well, we'll hear from lich where he found it ;)", "Solution_10": "Yeah maybe I posted this in the wrong forum, should be at the unsolved problems section... Anyways, that's how I was communicated. I believe it's a typo from that site's part because I found out last night that the equation cannot be satisfied by any positive integer (no solution!!). I thought the problem was interesting but here is the link:\r\n\r\nhttp://www.obm.org.br/regionais/se99.htm\r\n\r\nAnd again, I apologize for not double checking the problem first before posting it. Sorry.", "Solution_11": "oh well no problem\r\n\r\nI thought I became stupid, solving it 3 times, and getting the same wrong solution 3 times :P", "Solution_12": "[quote=\"Arne\"]Isn't it n3 + 2n2 = b, where b is an odd perfect square?[/quote]\r\nThen answer is n=7", "Solution_13": "that would be too easy then\r\n\r\nbut I just got this nice idea: :D\r\n\r\ni think it's n\u00b3 + 2^n = b with b a perfect odd square\r\n\r\nnow let's try to solve th\u00e0t :P", "Solution_14": "I'm quite sure it is n3 + 2n2 = b, but anyway, I'll think of your proposal Peter :D", "Solution_15": "The number must be large if there is one, I'm believing more that it has no solution :( But then, I need to prove that there are no solutions. \r\nn3+2n=b\r\nI'll check back later, gotta run to class.", "Solution_16": "Well I think it's pretty obvious the number needs to be large :D\r\n\r\nIt's not a fill-in excercice you'll get after trying 7 times...", "Solution_17": "Let's just stop trying ... the equation n3+2n=m2 have no integral solutions (rank = 0). I had to consult some literatures on elliptic curves before getting this answer. Seriously, we should forget about this version and think that they had n3+2n2=b as the problem.", "Solution_18": "Not anonymous, I posted the above reply.", "Solution_19": "strange excercice then\r\n\r\nwell it's beginner's level :D", "Solution_20": "[quote=\"Peter VDD\"]strange excercice then\n\nwell it's beginner's level :D[/quote]\r\nHehe yeah I agree Peter. Thx for the suggestion though, at least I got to learn something new out of this ;)" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "Let $ A$ and $ B$ be two fixed points in the circle $ (O,R).$ Determine the position of point $ M$ $ (M \\in (O))$ such that the sum $ MA\\plus{}MB$ attains the maximum value.", "Solution_1": "It is easy to see that $ M$ must be the midpoint of the bigger arc $ AB$ of the circle $ (O).$ Take $ B'$ on the ray $ AM$ such that $ MB \\equal{} MB'.$ The triangle $ \\triangle MBB'$ is isosceles with base $ BB'$ $ \\Longrightarrow$ $ \\angle MB'B \\equal{} \\frac {_1}{^2}\\angle AMB,$ therefore $ B'$ moves on the circumference $(N)$ centered at the midpoint $N$ of the arc $ AB$ and radius $ NA \\equal{} NB.$ \n\nThe length of the chord $AB',$ $ AB' \\equal{} MA \\plus{} MB$ obviously attains its maximum when it is identical to the diameter of $ (N)$ passing through $ A.$ Hence, it follows that $ M \\equiv N.$", "Solution_2": "Sorry, Why do you know B' moves on (N;NA)", "Solution_3": "Dear [b]Mr.luis[/b], the problem first stated that $ A,B$ are two points [b]in[/b] circle $ (O)$, not [b]on[/b] circle $ (O)$" } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "I tried do integration with x, and the result came out as a number which is 81/8, while i havenot enter any interval. I expect it would come out as x^2/2 ... What mode should i change in my calculator ? Thanks", "Solution_1": "Wrong forum :wink:", "Solution_2": "Not the right forum (not even sure what it would go in), but you probably have a variable stored into x, making the calculator treat it as a number rather than a variable. Look in MEM-LINK and delete your x variable, and you'll be okay." } { "Tag": [ "limit", "function", "real analysis", "real analysis unsolved" ], "Problem": "let $A_{n}=\\{n,n+1,n+2,...\\}$\r\nwhy is \r\n$\\cap A_{n}= \\phi$ ?\r\ni mean even at infinity $A_{n}$ and $A_{n+1}$ still have something\r\ncommon.\r\nor better asked\r\nhow much is $\\lim_{n \\to \\infty}\\left( |A_{n}\\cap A_{n+1}| \\right)$ ?", "Solution_1": "[quote]i mean even at infinity$A_{n}$ and $A_{n+1}$ still have something[/quote]\r\n\r\nwe arent concerned about just $A_{n}$ and $A_{n+1}$, when we talk abt intersection we want it to be in [b]ALL[/b] the $A_{n}$. \r\n\r\n suppose $x\\in\\cap A_{n}$ ,$x\\geq n$ for all $n$. From here conclude that the intersection is empty", "Solution_2": "no i dont understand.\r\nfor a finite intersection $x\\in I=\\cap_{i=1}^{k}A_{i}$ it is sufficient\r\nthat $x\\in A_{k}$ so that $x\\in I$.\r\n\r\nfor an infinite intersection $I'=\\cap A_{i}$ what happens ?\r\ni think $|I'|>0$.\r\ni am a little confused,i mean ,how can $|I'|=0$ ?\r\neven at infinity there still remain natural numbers in that\r\nintersection.\r\n\r\nkouboy do you have a proof ?", "Solution_3": "i thought of something\r\nconsider $B_{n}=\\{1,2,...,n-1\\}$\r\nfirst of all $\\mathbb{N}-A_{n}= B_{n}$\r\nnow as $n\\to \\infty$ , $B_{n}\\to \\mathbb{N}$\r\nso if $B_{n}$ will at infinity equal $\\mathbb{N}$ \r\nso that means that by the definition of $B_{n}$ that\r\nat infinity $A_{n}=\\phi$ because $\\mathbb{N}-\\mathbb{N}= \\phi$\r\nbut if one of the $A_{i}$ in the intersection is $\\phi$ then the\r\nwhole intersection is $\\phi$\r\nany suggestions are welcome", "Solution_4": "I think your confusion arises from treating sets as real-valued functions; you seem to approach the question about $\\bigcap_{n=1}^{\\infty}A_{n}$ as some sort of a question about $\\lim_{x\\to\\infty}f(x)$. But these are different games. In particular, your statement $B_{n}\\to\\mathbb N$ is meaningless until you define what convergence ($\\to$) means for sets.", "Solution_5": "its rather intuitive notation\r\nbut $B_{n}\\to \\mathbb{N}$ obviously means that\r\nas $n\\to \\infty$ ,$B_{n}$ is more and more closer to beein $\\mathbb{N}$.\r\n\r\ni tried to define it like so:\r\nlet us define a function $c(A,B)=1-\\frac{1}{|A-B|}$ and call it similarity\r\nof two sets(it can be easily seen that as two sets A,B have less difference\r\none from another the similarity between them is greater).\r\nnow we can make an analogous defintion as for the convergent\r\nsequence( http://mathworld.wolfram.com/ConvergentSequence.html )\r\nonly in place of the modulus we use the $c$ function defined earlier\r\nas follows so,we have $B_{n}\\to \\mathbb{N}$ if\r\nfor any $\\epsilon > 0$ there exists $N$ so that for $n>N$ we have $1-c(B_{n},\\mathbb{N})<\\epsilon$\r\n\r\n\r\nbut then i realised that we are dealing with ordinals and i dont know\r\nhow much sense it would make to have a difference of ordinals...\r\ni guess my conclusion is that i cant define it properly,ill have to think\r\nmore about it\r\nbut maybe it is already defined", "Solution_6": "Measuring similarity of sets is an interesting (and possibly difficult) problem. But I'd like to point out that the original question is not nearly as difficult. Intersections and unions are conceptually simpler than limits. You can see it from comparing their formal definitions: \r\n\\[x\\in\\bigcap_{n=1}^{\\infty}A_{n}\\Leftrightarrow \\forall n\\ x\\in A_{n}\\]\r\nversus\r\n\\[L=\\lim_{x\\to a}f(x) \\Leftrightarrow \\forall \\epsilon>0\\ \\exists \\delta>0\\ \\forall x\\in (a-\\delta,a+\\delta)\\setminus\\{a\\}\\ |f(x)-L|<\\epsilon\\]\r\nInterpreting intersections as limits actually makes a simple concept harder to grasp.", "Solution_7": "so , mlok , how do you suggest i modify the \"attempt at a proof\" above\r\nso that it does not imply limits of sets,and only use intersections ?", "Solution_8": "Suppose that $x$ is an element of $\\bigcap_{n=1}^{\\infty}A_{n}$. By definition of intersection, this means that $x\\in A_{n}$ for all $n=1,2,3,\\dots$. The definition of $A_{n}$ implies that $x$ is an integer such that $x\\ge n$ for any $n=1,2,3,\\dots$. Such $x$ does not exist; therefore, the set $\\bigcap_{n=1}^{\\infty}A_{n}$ has no elements.", "Solution_9": "so it is empty,because there is no $x>\\infty$...\r\nat least thats what i figgure" } { "Tag": [ "inequalities" ], "Problem": "If $ |\\minus{}2a\\plus{}1|<13$, what is the sum of the distinct possible integer values of $ a$?", "Solution_1": "We have the two inequalities:\r\n\r\n$ \\minus{}2a\\plus{}1<13$\r\n\r\n$ \\minus{}2a\\plus{}1>\\minus{}13$\r\n\r\nSolving and combining, $ \\minus{}6 13$\r\n\r\n$ a > \\minus{}6$\r\n\r\n2. $ \\minus{}2a \\plus{} 1 < \\minus{}13$\r\n\r\n$ a < 7$\r\n\r\n$ \\minus{}5 \\plus{} \\minus{}4 \\plus{} \\minus{}3 \\plus{} \\minus{}2 \\plus{} \\minus{}1 \\plus{} 0 \\plus{} 1 \\plus{} 2 \\plus{} 3 \\plus{} 4 \\plus{} 5 \\plus{} 6 \\equal{} \\boxed{6}$\r\n\r\nEDIT: Beaten to it." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Describe a 2-Sylow subgroup of $ S_8$. How many are there?", "Solution_1": "Consider the set X = { {1,2,3,4,5,6,7,8}, {1,2,3,4}, { 5,6,7,8 }, {1,2}, {3,4}, {5,6}, {7,8}, {1}, {2}, {3}, {4}, {5}, {6}, {7}, {8} }. If you imagine its elements (which are 8-, 4-, 2-, and 1-element sets) as nodes, and draw edges between minimal containments, then you get a binary tree.\r\n\r\nThe group S8 acts on sets like this, where g in S8 takes X to { { z^g : z in Y } : Y in X }.\r\n\r\nA sylow 2-subgroup of S8 is the stabilizer of X. It is a self-normalizing subgroup; there are 315 conjugates.\r\n\r\nYou can see this by checking that there are (8 choose 4)/2 ways of choosing the 4-element sets, and for each one there are (4 choose 2)/2 * (4 choose 2)/2 ways of choosing the 2-element sets, and of course just one way to choose the 8-element and the 1-element sets.\r\n\r\nSimilarly, looking at stabilizers of levels of the tree gives a nice normal series for the Sylow and allows you to see its structure and order." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "We consider the triangle ABC such that AB=c, BC=a and AC=b.Prove that\r\n $\\frac{a+b+c}{2}$ $\\sqrt 3$>$\\ l_a +\\ l_b+\\ l_c$, where the notation are that used in an acute triangle.", "Solution_1": "See\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=14431\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=805\r\nThe > has to be replaced by $\\geq$, and the triangle needs not be acute.\r\n\r\n darij", "Solution_2": "Thank you, Darij, but I think it was hard to guess that anywhay...", "Solution_3": "Which book is written by Valentin Vornicu?I'm eager to have a look.", "Solution_4": "It's a book from GIL.If you want it, you can contact Mircea Lascu on this forum by PM.It's a very good book.Normally, it is written by V.V.", "Solution_5": "Thank you very much,stancioiu.But what is the meaning of GIL?And I wonder that what the book's name is.", "Solution_6": "[quote=\"Hawk Tiger\"]Thank you very much,stancioiu.But what is the meaning of GIL?And I wonder that what the book's name is.[/quote]\r\nhttp://www.gil.ro", "Solution_7": "$\\frac{a}{\\sqrt [b+c]}+\\frac{b}{\\sqrt (a+c)} +\\frac{c}{\\sqrt (a+b)}\\geq \\sqrt \\frac{3}{2}$" } { "Tag": [], "Problem": "Using U.S currency, in how many different ways can you pay 25 cents? In how many different ways can you pay 50 cents? (The way depends on how many coins of each kind you use) You must answer both parts correctly!", "Solution_1": "i got an idea. Number of ways to make 25cents squared is 50 cents arrangements\r\n\r\nqdnp\r\n1000\r\n0210\r\n0205\r\n0130\r\n0125\r\n01110\r\n01015\r\n005\r\n004*\r\n003*\r\n002*\r\n001*\r\n00025\r\n\r\n13 ways to make 25\r\nso making 50 is like making 25 twice\r\n169 for 50 except if you use half dollars 170." } { "Tag": [ "vector", "group theory", "superior algebra", "superior algebra theorems" ], "Problem": "I would appreciate some help with the decomposition of $ Sym^3 V$ representation of the symmetric group $ S_3$, where $ V$ is its standard 2-dim representation.\r\n\r\nI think I got it right for $ Sym^2 V$. If $ \\alpha$ and $ \\beta$ are the basis vectors of $ V$, then $ \\alpha \\otimes \\alpha$, $ \\beta \\otimes \\beta$, and $ \\alpha \\otimes \\beta \\plus{} \\beta \\otimes \\alpha$ are basis vectors for $ Sym^2 V$. Then, if $ \\alpha$ and $ \\beta$ are eigenvectors of 3-cycles in $ V$, $ \\alpha \\otimes \\alpha$ and $ \\beta \\otimes \\beta$ are eigenvectors of 3-cycles in $ Sym^2 V$, and $ \\alpha \\otimes \\beta \\plus{} \\beta \\otimes \\alpha$ spans the trivial representation, so $ Sym^2 V \\cong V \\oplus U$, where $ U$ is the trivial representation.\r\n\r\nNow, how to work it out in $ Sym^3 V$?\r\n\r\nThanks in advance!", "Solution_1": "Please disregard, already got help on another forum:\r\nhttps://nrich.maths.org/discus/messages/7601/147966.html?1261852868" } { "Tag": [ "geometry", "rectangle", "vector", "geometric transformation", "geometry proposed" ], "Problem": "Let a rectangle $ABCD$ and let a point $K$ on $AC$ such that $BK$ is perpendicular to $AC$ .Also let $M$ the midpoint of $AK$ and $P$ the midpoint of $CD$ . Find the angle \r\n$\\widehat{PMB}$", "Solution_1": "Perform a spiral similarity $L$ with center $B$, angle $\\angle ABD$ and factor $\\frac{BD}{BA}$\r\n\r\nClearly, $L(A)=D, L(K)=C$. For this reason, we deduce that $L(M)=P$.\r\n\r\nSo $\\Delta MBP$ and $\\Delta ABD$ are similar. So $\\angle PMB = 90^{\\circ}$.", "Solution_2": "Nice solution. :) \r\n\r\nBTW, my solution uses vectors .", "Solution_3": "Thanks silouan :) \r\n\r\nCan I see your solution please?", "Solution_4": "[quote=\"silouan\"]Let a rectangle $ABCD$ and let a point $K$ on $AC$ such that $BK$ is perpendicular to $AC$ .Also let $M$ the midpoint of $AK$ and $P$ the midpoint of $CD$ . Find the angle \n$\\widehat{PMB}$[/quote]\r\n\r\nRight triangles $ABK , BCD$ are similar, since $\\angle BAK = \\angle BDC = x$.But $BM, BR$ are medians , so triangles $BAM, BPD$ are similar. Hence $\\angle ABM = \\angle DBP = y$.But then \r\n $\\angle BMC = \\angle BPC = x+y$, which means that $BMPC$ is incribed.Thus $\\angle PMB = 180-\\angle PCB = 90$!\r\n\r\n[u] Babis[/u]", "Solution_5": "Mr Stergiu this is also my synthetic proof(in fact it same as Jan's one but you don't mention the spiral similarity) .\r\nHere is the solution with vectors \r\nLet $\\vec{BA}=\\vec{a}$ , $\\vec{CB}=\\vec{b}$ and $\\vec{BK}=\\vec{c}$\r\nWe have $\\vec{BM}=\\frac{\\vec{a}+\\vec{c}}{2}$ \r\nand $\\vec{PM}=\\vec{PC}+\\vec{CB}+\\vec{BM}=\\vec{b}+\\frac{\\vec{c}}{2}$ \r\nSo $\\vec{PM}\\cdot\\vec{BM}=\\frac{1}{4}(\\vec{a}+\\vec{c})(2\\vec{b}+\\vec{c})=\\frac{1}{4}(2\\vec{a}\\cdot \\vec{b}+2\\vec{c}\\cdot \\vec{b}+\\vec{a}\\cdot \\vec{c}+(\\vec{c})^{2})=0$\r\n\r\nI think it is not ungly at all .What do you think ?", "Solution_6": "I think it is a very nice proof dear silouan. :)" } { "Tag": [ "integration", "limit" ], "Problem": "Fiind data functia continua$f: [0,1]\\rightarrow R$, sa se calculeze limita sirului $(a_{n}) ,a_{n}=\\int_{0}^{1}x^{n}f(x)dx$", "Solution_1": "Fie $M=\\max\\{f(x)|x\\in [0,1]\\}$. Atunci $|a_{n}|=|\\int_{0}^{1}x^{n}f(x) dx|\\leq |M|\\cdot |\\int_{0}^{1}x^{n}dx|\\leq |M|\\cdot \\frac{1}{n+1}\\to 0$ ,deci $\\lim_{n\\to\\infty}a_{n}=0$.", "Solution_2": "Merit\u0103 pomenit\u0103 \u00een acest context \u015fi problema clasic\u0103:\r\nFiind dat\u0103 func\u0163ia continu\u0103$f: [0,1]\\rightarrow R$, s\u0103 se calculeze limita \u015firului $(b_{n}) , b_{n}=n\\int_{0}^{1}x^{n}f(x)dx$", "Solution_3": "$b_{n}$ nu are limita pentru orice functie continua $f\\: ,\\: f: [0,1]\\rightarrow R$ .\r\nExemplu:$f(x)=1,si \\: f(x)=-1$", "Solution_4": "[quote=\"radac_mail\"]$b_{n}$ nu are limita pentru orice functie continua $f\\: ,\\: f: [0,1]\\rightarrow R$ .\nExemplu:$f(x)=1,si \\: f(x)=-1$[/quote]\r\nCred ca glumesti.\r\nEu cred ca $b_{n}\\to f(1)$ dar nu am reusit sa demonstrez decat cu \"Dominant Convergence Theorem\" ( care nu sunt sigur ca e foarte usor de demonstrat).", "Solution_5": "Cred c\u0103 s-a mai discutat... Oricum, presupunem \"WLOG\" c\u0103 $f(1) = 0$ (altfel consider\u0103m func\u0163ia $f(x)-f(1)$).\r\n\r\nIn continuare, folosesc metoda lui spix:\r\n\r\nFie $\\epsilon > 0$ \u015fi $t \\in (0,1)$ astfel \u00eenc\u00e2t $\\left| f(x) \\right| < \\epsilon, \\, \\forall x \\in [t,1]$. De asemenea, $M = \\sup_{x \\in [0,1]}\\left| f(x) \\right| <+\\infty$.\r\nAvem $\\left| b_{n}\\right| \\leq \\left| \\int_{0}^{t}n x^{n}f(x) \\, dx \\right|+\\left| \\int_{t}^{1}n x^{n}f(x) \\, dx \\right| \\leq M t^{n+1}+\\left( 1-t^{n+1}\\right) \\cdot \\epsilon$.\r\nLu\u00e2nd $n \\to \\infty$, ob\u0163inem $\\limsup \\left| b_{n}\\right| \\leq \\epsilon$.\r\nCum $\\epsilon$ a fost ales arbitrar, deducem c\u0103 $\\lim b_{n}= 0$." } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "$ (I) $ is the incircle of the triangle $ ABC $ . $ (O) $ is the circumcircle of the triangle $ ABC $ . $ (O_1) $ internally tangents $ (O) $ at $ D $ and tangents $ AB,AC $ . $ (O_2) $ internally tangents $ (O) $ at $ E $ and tangents $ AB,BC $ . $ (O_3) $ internally tangents $ (O) $ at $ F $ and tangents $ CA,CB $ . Prove that : $ AD , BE , CF $ are concurrent .", "Solution_1": "Isn't it obvious :?:" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Given an integer N.\r\nWhich pairs of integers such that their sum is equal to N and the second number results from the first one by striking out one of its digits.", "Solution_1": "Hello 7erry,\r\n\r\nAssume the first integer is AdB and the second is AB where B have n digits and d is 0 .. 9.\r\nThen : N = A*10^(n+1) + d*10^n + B + A*10^n + B = 10^n(11A + d) + 2B and B < 10^n\r\nI'll qualify a solution by the set (n,A,d,B) where B < 10^n and d is 0..9\r\n\r\n1) if N is odd, we need n=0, hence B=0, and N = 11A+d\r\n1.1) if Mod(N,11) = 10 ==> no solution\r\n1.2) if Mod(N,11) < 10 ==> exactly one solution (0,[N/11],Mod(N,11),0)\r\n\r\n2) if N is even :\r\n2.1) for n = 0 and Mod(N,11) = 10 ==> no solution\r\n2.2) for n = 0 and Mod(N,11) < 10 ==> one solution (0,[N/11],Mod(N,11),0)\r\n2.3) For any n > 0 such that N >= 10^n, we have exactly two possibilities to write N as 10^n x + 2B with B < 10^n :\r\nN = 10^n [N/10^n] + 2*(Mod(N,10^n)/2) \r\nN = 10^n ([N/10^n]-1) + 2*((10^n + Mod(N,10^n))/2) \r\nHence :\r\nif Mod([N/10^n],11) is 0, we have one solution (n,[[N/10^n]/11],0,Mod(N,10^n)/2)\r\nif Mod([N/10^n],11) is 1..9, we have two solutions (n,[[N/10^n]/11],Mod([N/10^n],11) ,Mod(N,10^n)/2) and (n,[[N/10^n]/11],Mod([N/10^n],11)-1 ,(10^n + Mod(N,10^n))/2) \r\nif Mod([N/10^n],11) is 10, we have one solution (n,[[N/10^n]/11],9 ,(10^n + Mod(N,10^n))/2)\r\n\r\n\r\nExamples :\r\n\r\nN = 1583 : N is odd and mod(N,11) = 10 ==> no solution\r\nN = 1585 : N is odd and mod(N,11) = 1 ==> one solution : (0, 144,1, 0) : 1585 = 1441 + 144\r\nN = 1584 : N is even \r\nn = 0 ==> (0,144, 0, 0) : 1584 = 1440 + 144\r\nn = 1 ==> (1, 14, 4, 2) : 1584 =1442 + 142 and (1, 14, 3, 7) : 1584 = 1437 + 147 \r\nn = 2 ==> (2, 1, 4, 42) : 1584 = 1442 + 142 and (2, 1, 3, 92) : 1584 = 1392 + 192\r\nn = 3 ==> (3, 0, 1, 292) : 1584 = 1292 + 292 and (3, 0, 0, 792) : 1584 = 0792 + 792 :blush: \r\n\r\n-- \r\nPatrick", "Solution_2": "[quote]2.3) For any n > 0 such that N >= 10^n, we have exactly two possibilities to write N as 10^n x + 2B with B < 10^n :\nN = 10^n [N/10^n] + 2*(Mod(N,10^n)/2)\nN = 10^n ([N/10^n]-1) + 2*((10^n + Mod(N,10^n))/2) [/quote]\r\nHow can you claim that there is not any possibility else? :)", "Solution_3": "Hello 7erry,\r\n\r\n[quote=\"7erry\"][quote]2.3) For any n > 0 such that N >= 10^n, we have exactly two possibilities to write N as 10^n x + 2B with B < 10^n :\nN = 10^n [N/10^n] + 2*(Mod(N,10^n)/2)\nN = 10^n ([N/10^n]-1) + 2*((10^n + Mod(N,10^n))/2) [/quote]\nHow can you claim that there is not any possibility else? :)[/quote]\r\n\r\nI don't understand your question.\r\nIf I want to write an even positive integer N as 10^n x + 2B with B < 10^n, n given in N*, I have only two cases :\r\n\r\n1) B < 10^n /2 ==> 2B < 10^n ==> 2B = mod(N, 10^n) and x = [N/10^n]\r\n2) 10^n/2 <= B < 10^n ==> 10^n <= 2B < 2*10^n ==> 0 <= 2B - 10^n < 10^n\r\nSo 10^n x + 2B = 10^n(x+1) + 2B - 10^n and I have just one other solution : 2B - 10^n = mod(N,10^n) and x = [N/10^n]-1\r\n\r\nIs this the answer you're asking for ?\r\n\r\n-- \r\nPatrick" } { "Tag": [ "modular arithmetic" ], "Problem": "I have the problem\r\n\r\nFind all primes $ p$ and $ q$ such that $ p \\plus{} q \\equal{} (p \\minus{} q)^3$.\r\n\r\nThe solution manual says $ p \\minus{} q\\equiv 2p\\mod (p \\plus{} q)$.\r\nThis is like step 1. How'd they get there", "Solution_1": "$ p\\plus{}q\\equiv0\\pmod{p\\plus{}q}$\r\n\r\n$ \\Rightarrow \\minus{}q\\equiv p\\pmod{p\\plus{}q}$\r\n\r\n$ \\Rightarrow p\\minus{}q\\equiv 2p\\pmod{p\\plus{}q}$" } { "Tag": [], "Problem": "I recently came to know about MOEMS and would like to approach my child's elementary school to start a MOEMS groups. Any pointers on how to approach school?\r\nDo people who become PICO ask for some re-embursement from school or is it done totally voluntarily?\r\nThanks\r\nGupta", "Solution_1": "Best way is to sit down with the principal and/or the head of the Parents Association. Bring with you an Informational Packet (available free from MOEMS at info@moems.org/866-781-2411). Most schools run and fund their teams, but in some cases, parents do it. The Parents Association might help. Reimbursement varies from school to school.", "Solution_2": "thankyou. I am going to contact the PTA president.", "Solution_3": "i have had this same problem...\r\nthanks so much for going ahead and asking it! i almost forgot to :lol:" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "Prove that any elementary row [column] operation of type 1 can be obtained by a succession of three elementulary row [column] operations of type 3 followed by one elementary row [column] operation of type 2.\r\n\r\nType 1: Switch any two rows/columns\r\nType 2: Multiply any row/column by a non-zero scalar\r\nType 3: Add a scalar multiple of one row/column to another", "Solution_1": "The single type 2 operation must be multiplication of a row by the scalar $ \\minus{}1.$ How do we know this? Just track the effect of elementary row operations on determinants. Type 3 operations leave determinants unchanged. Type 1 operations multiply determinants by $ \\minus{}1.$ Type 2 operations multiply determinants by the scalar in question.", "Solution_2": "To prove that this combination of 4 operations yields the elementary matrix in question, would it suffice to find 4 such operations that yield a type 1 matrix? If so, how could I go about finding them? If not, what would suffice?", "Solution_3": "Here's the process at work in a $ 2\\times 2$ matrix. (1) Subtract the first row from the second; (2) add the second row to the first; (3) subtract the first row from the second; (4) multiply the second row by $ \\minus{} 1.$\r\n\r\n$ \\begin{bmatrix}1 & 0 \\\\\r\n0 & 1\\end{bmatrix}\\to \\begin{bmatrix}1 & 0 \\\\\r\n\\minus{} 1 & 1\\end{bmatrix}\\to \\begin{bmatrix}0 & 1 \\\\\r\n\\minus{} 1 & 1\\end{bmatrix}\\to \\begin{bmatrix}0 & 1 \\\\\r\n\\minus{} 1 & 0\\end{bmatrix}\\to \\begin{bmatrix}0 & 1 \\\\\r\n1 & 0\\end{bmatrix}.$\r\n\r\nBut the process is completely generic. In any matrix with two or more rows, pick two of those rows, and act as I did above on the first and second rows.\r\n\r\nWhat we learn from this is that we don't need to have three types of elementary row operations; we can omit type 1 and have the other two types suffice for all uses in proofs. (Logically sufficient, but inefficient.)", "Solution_4": "That's actually the same method I came up with and I generalized it for any $ m\\times n$. Thanks." } { "Tag": [ "probability" ], "Problem": "A typical ID number in a college class consists of a four-digit number, such as 0119. When an ID number is randomly selected, what is the probability that no two of its digits are the same? Express your answer as a decimal to the nearest thousandths.", "Solution_1": "We calculate the number of ID's in which no two digits are the same. There are $ 10$ choices for the first digit, $ 9$ for the second, $ 8$ for the third, and $ 7$ for the fourth (for those familiar with this notation, we could just do $ _{10}P_4$). Thus, there are $ 10 \\times 9 \\times 8 \\times 7\\equal{}5040$ ways to pick an ID with no two digits that are the same. There are $ 10^4\\equal{}10000$ total possible ID's, so the probability that no two digits are the same is $ \\frac{5040}{10000}\\equal{}\\boxed{0.504}$." } { "Tag": [ "email" ], "Problem": "Has anyone going to CA received any emails about going (besides the acceptance letter)? I'm worried that I'm missing something...", "Solution_1": "Well, I haven't gotten anything.", "Solution_2": "yeah same, i haven't gotten anything. i was sort of hoping for maybe course info, or confirmation that they received my info forms, but none so far.", "Solution_3": "I'm also worried by the lack of receipt of course choices. Hm...AND I'm going to be at another camp, meaning if AMSP mails the course stuff, my parents will have to somehow get them to me. Hrm. \r\n\r\nI've also noticed a lack of information concerning where in UCSC to go to check in, get stuff, live for three weeks, etc. If that also comes in the mail, that would also pose...um...problems.", "Solution_4": "Received an email 2 minutes ago." } { "Tag": [ "calculus", "integration", "quadratics", "symmetry" ], "Problem": "Find all integral solutions to: $ x^2\\plus{}y^2\\plus{}z^2\\equal{}xyz.$\r\n\r\nWhat is the method for tackling these kids of problems (because obviously you can complete the square or anything) and what is the answer to this specific question?", "Solution_1": "I got a 2.\r\nx=y=z=3\r\nx=y=z=0\r\nfor some reason I feel that there are no more solutions.", "Solution_2": "Is there any actual proof that there are no more solutions?", "Solution_3": "well it is not that hard if you think about it.\r\nyou know it is imposible to solve this equation if the values are different for $ x,y,z$ with one solution by using algebra so we suppose $ x \\equal{} y \\equal{} z$\r\n\r\nso we can turn the equation to $ x^2$+$ x^2$+$ x^2$=$ x^3$\r\nwell because there are three terms \r\n\r\nso that would be $ 3x^2$=$ x^3$\r\n\r\nthen you divide both sidesby $ x^2$\r\n\r\nwell since this is a quadratic we have two solutions.now we factor it. $ x^2(3 \\minus{} x)$\r\n\r\nso one solution is $ 0$ and the other solution is $ 3$ because $ x \\minus{} 3$=$ 0$\r\n\r\n$ x \\equal{} (3,0)$\r\n\r\nanother way you can get three is by dividing \r\n\r\n$ \\frac {3x^2}{x^2}$=$ \\frac {x^3}{x^2}$\r\n\r\n$ 3 \\equal{} x$\r\n\r\nso we know that $ 3$=$ x$", "Solution_4": "[quote=\"calculatorwiz\"]well it is not that hard if you think about it.\nyou know it is impossible to solve this equation if the values are different for $ x,y,z$ with one solution by using algebra so we suppose $ x \\equal{} y \\equal{} z$\n[/quote]\r\n\r\nBut thats pretty intuitive isnt it? Is there like any proof that $ x\\equal{}y\\equal{}z$?", "Solution_5": "Also, calculatorwiz, by dividing by $ x^{2}$, you leave out solutions.\r\nWhen you get to $ 3x^{2} \\equal{} x^{3}$, you have to bring everything equal to 0 and then solve.\r\n$ x^{3} \\minus{} 3x^{2} \\equal{} 0$ \r\n$ x^{2}(x \\minus{} 3) \\equal{} 0$\r\n$ x \\equal{} 0,3$\r\n\r\nHowever, I'm also wondering how you can assume that $ x \\equal{} y \\equal{} z$. How do you find an answer if this is not true?", "Solution_6": "well you can asume $ x\\equal{}y\\equal{}z$\r\n\r\nbecause there are no systems of equations given\r\n\r\nyou can not find all three values from one equation", "Solution_7": "[quote=\"calculatorwiz\"]\nyou can not find all three values from one equation[/quote]\r\n\r\nThats a good point. However, say we were trying to find all integral solutions to $ 3x^{2} \\plus{} y^{2} \\minus{} 4y \\minus{} 17 \\equal{} 0$. Well we could turn that into $ 3x^{2} \\plus{} (y \\minus{} 2)^{2} \\equal{} 21$, and from there its very easy to check all the cases and find all the solutions even if there is no system.\r\n\r\nHowever, I do see your point. Since there is no constant, we cant solve this problem without assuming either $ x \\equal{} y \\equal{} z$ or $ x \\equal{} y \\neq z$.\r\nsince we already considered the former, I will do the latter.\r\n\r\n$ x \\equal{} y \\neq z$. Then $ 2x^2 \\plus{} z^2 \\equal{} x^2z$. \r\n$ z^2 \\equal{} x^2(z \\minus{} 2)$\r\n$ \\frac {z^2}{z \\minus{} 2} \\equal{} x^2$\r\nwe easily see that $ z \\equal{} 6$, $ x \\equal{} y \\equal{} 3$ is a solution, as are its permutations because of symmetry-(3,3,6),(3,6,3),(6,3,3). So now I will kind of guess that the only solution to $ \\frac {z^2}{z \\minus{} 2} \\equal{} x^2$ is (3,6). \r\n\r\nSo if anyone can prove that the only solution to $ \\frac {z^2}{z \\minus{} 2} \\equal{} x^2$ is (3,6) we will be done, right?\r\n\r\nIf that is true, the only solutions are $ \\boxed{(3,3,3), (0,0,0), (3,3,6), (3,6,3), (6,3,3)}$", "Solution_8": "Don't forget about $ (\\minus{}3,\\minus{}3,3), (\\minus{}3,\\minus{}3,6), (\\minus{}3,3,\\minus{}6)$, and all of the permutations of those as well...", "Solution_9": "[quote=\"calculatorwiz\"]well you can asume $ x \\equal{} y \\equal{} z$\n\nbecause there are no systems of equations given\n\nyou can not find all three values from one equation[/quote]\r\nThat is not true... since it is a Diophantine Equations it has many different values", "Solution_10": "well how do you find all the values did you just guess because you are not showing haw you gat the answers" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "solve in nonnegative integers the equation \r\n$ 2^x\\minus{}1\\equal{}xy$", "Solution_1": "It 's similar to the problem\r\nFind x such that $ 2^x \\minus{} 1$ divided by x.\r\nLet p is the smallest prime divisors of x.(1)\r\nBy little Fermat theorem we have:\r\n $ 2^{p\\minus{}1} \\minus{}1$ divided by p\r\nand $ 2^x \\minus{}1$ divided by p\r\nbecause of (1),we have gcd(x,p-1)=1\r\nso $ 2^{1} \\minus{} 1$ divided by p. \r\nContraction.\r\nSo x don't have any prime divisors.\r\nx = 1.\r\nand y=1 ,too.\r\nthe root of the equation is (1,1)", "Solution_2": "PEN A14\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=150382[/url]" } { "Tag": [ "geometry", "geometric transformation", "reflection", "circumcircle", "geometry proposed" ], "Problem": "Let $A',B',C'$ be the reflections of the vertices $A,B,C$ in the sides $BC,CA,AB$ respectively. Let $O$ be the circumcenter of $ABC$. Show that the circles $(AOA'),(BOB'),(COC')$ concur again in a point $P$, which is the inverse in the circumcircle of the isogonal conjugate of the nine-point center.\r\n\r\nIt shouldn't be difficult, but I found it cute :).", "Solution_1": "May be $A', B', C'$ --- reflections of point $O$? Then the solution: \r\nLet $O_A$ --- the circumcenter of triangle $BOC$, $H_A$ --- projection the point $A$ on $BC$, $N$ --- the nine-point center, $N'$ --- isogonal conjugate of this and $M_C, M_B$ --- midpoints of segments $AB, AC$ respectively. Made the composition of inversion in point $A$ with radius is $\\sqrt{{AB\\cdot AC}\\over 2}$ and reflection of the $\\angle BAC$ bisector. Then $M_C\\rightarrow C, M_B\\rightarrow B, O\\rightarrow H_A$ (since $2Rh_a=AB\\cdot AC$). Then line $AN\\rightarrow AO_A$. Made the inversion with center $O$ and radius $R$. Then $O_A\\rightarrow A'$, line $AO_A\\rightarrow$ circle$(AOA')$, $N'\\rightarrow P$. \r\nElse this problem is wrong.", "Solution_2": "I assure you, the problem is not wrong. In fact, [url=http://mathworld.wolfram.com/MusselmansTheorem.html]here[/url]'s a link. $A',B',C'$ are, as stated, the reflections of $A,B,C$ in $BC,CA,AB$ respectively.", "Solution_3": "I just downloaded the article. And it is by treegoner!", "Solution_4": "Sorry, the problem is true. Really, the points $A$, $O$, $A'$, reflection $O$ of $BC$ concircle, hence these facts are equivalent.", "Solution_5": "Dear Mathlinkers,\r\nyou will see on my website (http:/perso.orange/jl.ayme) article 10 Le point de Kostina... a synthetic proof of the Musselman's circle and of the isogonal of the nine point center.\r\nSincerely\r\nJean-Louis" } { "Tag": [], "Problem": "5a-12 3a-2\r\n_________ - ___________\r\na^2-8a+5 a^2-8a+15\r\n\r\nDoes anyone have method for solving this?[/u]", "Solution_1": "please do not post the same question twice.", "Solution_2": "jamesstine, Someone have solved it for you! [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=149331[/url]", "Solution_3": "5a-12 3a-2 \r\n_________ - ___________ \r\na^2-8a+15 a^2-8a+15 \r\n\r\nI made a mistake on the first denominator.", "Solution_4": "[quote=\"jamesstine\"]5a-12 3a-2 \n_________ - ___________ \na^2-8a+15 a^2-8a+15 \n\nI made a mistake on the first denominator.[/quote]\r\n\r\nSorry we should have looked into your problem.\r\n\r\nYour problem is $\\frac{5a-12}{a^{2}-8a+15}-\\frac{3a-2}{a^{2}-8a+15}$, isn't it? :) \r\n\r\nYou can correct by pushing Edit button.", "Solution_5": "Yes that is correct!", "Solution_6": "Since there are common denominators it should just be:\r\n\\begin{eqnarray*}\\frac{(5a-12)-(3a-2)}{a^{2}-8a+15}&=& \\\\ &=& \\boxed{\\frac{2a-10}{(a-5)(a-3)}}\\end{eqnarray*}\r\n\r\nEdit: (2a-1)=(2a-10) i can't type well =P", "Solution_7": "I see how you got the denominator but I don't understand how you got (2a-1)", "Solution_8": "[quote=\"jamesstine\"]I see how you got the denominator but I don't understand how you got (2a-1)[/quote]\r\nShould be $2a-10$.", "Solution_9": "[quote=\"hunter34\"]Since there are common denominators it should just be:\n\\begin{eqnarray*}\\frac{(5a-12)-(3a-2)}{a^{2}-8a+15}&=& \\\\ &=& \\boxed{\\frac{2a-10}{(a-5)(a-3)}}\\end{eqnarray*}\n[/quote]\r\n\r\nBut $\\frac{2a-10}{(a-5)(a-3)}= \\frac{2(a-5)}{(a-5)(a-3)}= \\frac{2}{a-3}$ so this is the simplest form." } { "Tag": [ "integration", "calculus", "derivative", "algebra", "polynomial", "symmetry", "real analysis" ], "Problem": "Let $f\\in C^{4}[a, b]$ function. Prove that there is $c\\in (a, b)$ such that \r\n\r\n \\[ \\int_{a}^{b} f(x)dx=(b-a)f(\\frac{a+b}{2})+\\frac{(b-a)^{2}}{12}[f'(\\frac{3b+a}{4})-f'(\\frac{3a+b}{4})]+\\frac{(b-a)^{5}}{11520}\\cdot f^{(4)}{(c). }\\]", "Solution_1": "I'm only going to work part of it, but enough to know that the formula is correct.\r\n\r\nAssume $a=-h,\\,b=h.$ The general case can be turned into this by translation. Assume that $f$ has as many continuous derivatives as we're going to use below in applying Taylor's theorem. The formula becomes:\r\n\r\n$\\int_{-h}^hf(x)\\,dx=2hf(0)+\\frac{h^2}3\\left[f'\\left(\\frac h2\\right)-f'\\left(-\\frac h2\\right) \\right] +\\frac{h^5}{360}f^{[4]}(c)$\r\n\r\nReplace $f(x)$ by its Taylor polynomial centered at zero. By the symmetry of the interval of integration, all of the odd terms of the Taylor polynomial drop out of the result. We get:\r\n\r\n$\\int_{-h}^hf(x)\\,dx=2hf(0)+\\frac{h^3}3f''(0)+\\frac{h^5}{60}f^{[4]}(0)+O(h^7)$\r\n\r\nNow look at the derivative:\r\n\r\n$f'\\left(\\frac h2\\right)=f'(0)+\\frac h2f''(0)+\\frac{h^2}8f'''(0)+\\frac{h^3}{8\\cdot 3!}f^{[4]} (0)+ch^4+O(h^5)$\r\n\r\n$f'\\left(-\\frac h2\\right)=f'(0)-\\frac h2f''(0)+\\frac{h^2}8f'''(0)-\\frac{h^3}{8\\cdot 3!} f^{[4]} (0)+ch^4+O(h^5)$\r\n\r\nSubtract to get\r\n\r\n$f'\\left(\\frac h2\\right)-f'\\left(-\\frac h2\\right)= hf''(0)+\\frac{h^3}{24}f^{[4]}(0)+O(h^5),$ or\r\n\r\n$\\frac{h^2}3\\left[f'\\left(\\frac h2\\right)-f'\\left(-\\frac h2\\right)\\right]= \\frac{h^3}3f''(0)+\\frac{h^5}{72}f^{[4]}(0)+O(h^7)$\r\n\r\nSubtracting what we have, we get\r\n\r\n$\\int_{-h}^hf(x)\\,dx-\\left[2hf(0)+\\frac{h^2}3\\left[f'\\left(\\frac h2\\right)-f'\\left(-\\frac h2\\right) \\right]\\right]$\r\n\r\n$=\\left(\\frac1{60}-\\frac1{72}\\right)h^5f^{[4]}(0)+O(h^7)=\\frac{h^5}{360}f^{[4]}(0) +O(h^7).$\r\n\r\nTechnically, this doesn't quite prove the mean-value form that Cezar Lupu wrote it in, but once we're this far, we know it's going to be true.\r\n\r\nI do have doubts about the utility of a quadrature formula that involves evaluating a derivative.", "Solution_2": "The formula\r\n\r\n$\\int_{-h}^hf(x)\\,dx\\approx 2hf(0)+\\frac{h^2}6\\sqrt{\\frac{10}3}\\left[f'\\left(\\sqrt{\\frac3{10}}\\,h\\right)- f'\\left(-\\sqrt{\\frac3{10}}\\,h\\right) \\right]$\r\n\r\nshould have even higher precision, assuming I did the algebra correctly.", "Solution_3": "[quote=\"Kent Merryfield\"]The formula\n\n$\\int_{-h}^hf(x)\\,dx\\approx 2hf(0)+\\frac{h^2}6\\sqrt{\\frac{10}3}\\left[f'\\left(\\sqrt{\\frac3{10}}\\,h\\right)- f'\\left(-\\sqrt{\\frac3{10}}\\,h\\right) \\right]$\n\nshould have even higher precision, assuming I did the algebra correctly.[/quote]\r\n\r\nI like this formula, but I would like to see the remainder term \"clean\" and its derivative. ;)", "Solution_4": "The remainder term in the formula in #3 should be $Ch^7f^{[6]}(c)$ for some constant $C$ and some point $c\\in(-h,h).$\r\n\r\nI don't have the time or energy to figure out what the constant $C$ should be." } { "Tag": [ "LaTeX" ], "Problem": "How do you install packages into TexShop? :help:\r\nAlso, where can you obtain packages? :help: \r\n\r\nPlease reply.", "Solution_1": "TeXShop is just an editor so you don't install packages into it. You must first install LaTeX then you can add more packages to your LaTeX installation if you need them. You need to tell TeXShop where to find your latex files.\r\n\r\nYou install LaTeX on the Mac by using [url=http://www.tug.org/mactex/]MacTeX[/url]. There are full instructions on how to download and install there. The large version is an enormous 1.2GB download which means you will probably never have to look for extra packages, though there is a facility to do so built into the system.", "Solution_2": "[quote=\"stevem\"]TeXShop is just an editor so you don't install packages into it. You must first install LaTeX then you can add more packages to your LaTeX installation if you need them. You need to tell TeXShop where to find your latex files.\n\nYou install LaTeX on the Mac by using [url=http://www.tug.org/mactex/]MacTeX[/url]. There are full instructions on how to download and install there. The large version is an enormous 1.2GB download which means you will probably never have to look for extra packages, though there is a facility to do so built into the system.[/quote]\r\nI already have TexShop installed, but I need to use the fancyhdr package and I don't know how. I think I explained myself wrong. I installed basictex instead of the full Mactex package if that makes a difference. (BasicTex is a smaller version of MacTex.)", "Solution_3": "BasicTeX includes the TeX Live Manager which allows you to download and install packages. If you don't already have it, download [url=http://code.google.com/p/mactlmgr/]TeX Live Utility[/url] which is a \"Mac OS X graphical interface for TeX Live Manager\" that will help you to install packages without having to use the command line.", "Solution_4": "This application only works with Leopard (10.5.5). I only have 10.4.11, so is there a similar application that works on my computer? If not where can I find the code to run in Terminal to load packages?\r\n :help:", "Solution_5": "[url=http://www.tug.org/texlive/tlmgr.html]tlmgr - TeX Live package manager[/url]", "Solution_6": "I don't understand. Is this an application, or instructions for loading packages via command line? If it's the latter, is this done in Terminal? How? If it's the former, how do you obtain it?", "Solution_7": "You need to type commands in Terminal. There are better instructions written especially for the Mac at [url=http://mactex-wiki.tug.org/wiki/index.php?title=TeX_Live_Manager]TeX Live Manager[/url].\r\nThere are 3 steps you need to follow there:\r\n1. Update the manager [b]First: Update tlmgr[/b]\r\n2. Tell it where to get the packages [b]Second: Set Your Server[/b]\r\nYou only need to do these 2 steps once. After that it is easy to install packages.\r\n3. Install fancyhdr by typing into the Terminal [code]sudo tlmgr install fancyhdr[/code]" } { "Tag": [], "Problem": "edit edit or delite", "Solution_1": "Have some clue? :?", "Solution_2": "ok just found this nearly 16 year old thread lol\nbumping this for no reason\nalso adventure10 upvoted the posts meaning that they saw it\n" } { "Tag": [ "geometry", "rectangle", "geometry solved" ], "Problem": "Triangle ABC is acute-angled.Angle B is less than angle C.O is circumcentre.AO(extended) cuts BC at D.\r\n\r\nThe circumcentres of the triangles ABD,ACD are E and F respectively.\r\n\r\nThe sides BA, CA are extended beyond A to points G and H respectively such that AG=AC, and AH=AB.\r\n\r\nProve that quadrilateral EFGH is a rectangle iff C-B =60 degrees", "Solution_1": "Is this problem from the shortlist of IMO2004?", "Solution_2": "You mean this problem? [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=205737#p205737]www.mathlinks.ro/Forum/viewtopic.php?p=205737#p205737[/url]" } { "Tag": [ "LaTeX" ], "Problem": "I was doing a math homework in LaTex, but when I type spanish with the acent mark like \"factorizaci\u00f3n\" the letter \"\u00f3\" is omitted after compiled into pdf.\r\n How can I solve this problem? Do I need to include extra package?", "Solution_1": "Put \\usepackage[latin1]{inputenc} in your preamble (ie after \\documentclass and before \\begin{document}). This tells LaTeX that you want to use special characters from Latin languages ie languages in Western Europe." } { "Tag": [ "trigonometry", "inequalities", "function", "geometry proposed", "geometry" ], "Problem": "[u][b]The author of this posting is : darkmaster[/b][/u]\r\n____________________________________________________________________\r\n\r\nA triangle ABC has the property that $l_a$, $m_b$, $h_c$ are concurrent; $l_b$, $m_c$, $h_a$ are concurrent; $l_c$, $m_a$, $h_b$ are concurrent.\r\nProve that $\\frac{\\sin C}{\\cos B} + \\frac{\\sin A}{\\cos C} + \\frac{\\sin B}{\\cos A} \\geq 3\\sqrt3$.", "Solution_1": "[u][b]The author of this posting is : Myth[/b][/u]\r\n____________________________________________________________________\r\n\r\nIt is clear that $ABC$ is an acute triangle (since its altitudes lie inside $ABC$).\r\nFrom Ceva theorem for $l_a$, $m_b$ and $h_c$ we obtain $\\frac{\\sin C}{\\cos B} = \\tan A$, and so on.\r\nTherefore, LHS = $\\tan A + \\tan B + \\tan C \\geq 3\\sqrt 3$ due to Jensen (the function $\\tan x$ is convex on $\\left[0,\\pi/2\\right)$).", "Solution_2": "[quote=\"darkmaster\"]A triangle ABC has the property that $l_a$, $m_b$, $h_c$ are concurrent; $l_b$, $m_c$, $h_a$ are concurrent; $l_c$, $m_a$, $h_b$ are concurrent.\nProve that $\\frac{\\sin C}{\\cos B} + \\frac{\\sin A}{\\cos C} + \\frac{\\sin B}{\\cos A} \\geq 3\\sqrt3$.[/quote]\r\n\r\nI have a stronger assertion: Prove that $\\frac{\\sin C}{\\cos B} + \\frac{\\sin A}{\\cos C} + \\frac{\\sin B}{\\cos A} = 3\\sqrt3$.\r\n\r\nIn other words, prove that if we are given a triangle ABC such that $l_a$, $m_b$, $h_c$ are concurrent, that $l_b$, $m_c$, $h_a$ are concurrent, and that $l_c$, $m_a$, $h_b$ are concurrent, then this triangle ABC is equilateral.\r\n\r\nBTW, Myth, I expected your answer to be \"This is trivial :D \"...\r\n\r\n Darij", "Solution_3": "[u][b]The author of this posting is : Myth[/b][/u]\r\n____________________________________________________________________\r\n\r\n[quote=\"darij grinberg\"]In other words, prove that if we are given a triangle ABC such that $l_a$, $m_b$, $h_c$ are concurrent, that $l_b$, $m_c$, $h_a$ are concurrent, and that $l_c$, $m_a$, $h_b$ are concurrent, then this triangle ABC is equilateral.[/quote]\n\nThis is trivial.\n\n[quote=\"darij grinberg\"]BTW, Myth, I expected your answer to be \"This is trivial :D \"...[/quote]\r\n\r\nI don't like such jokes.", "Solution_4": "[u][b]The author of this posting is : darkmaster[/b][/u]\r\n____________________________________________________________________\r\n\r\n[quote=\"darij grinberg\"]In other words, prove that if we are given a triangle ABC such that $l_a$, $m_b$, $h_c$ are concurrent, that $l_b$, $m_c$, $h_a$ are concurrent, and that $l_c$, $m_a$, $h_b$ are concurrent, then this triangle ABC is equilateral.\n\nBTW, Myth, I expected your answer to be \"This is trivial :D \"...\n\n Darij[/quote]\r\n\r\nThis \"trivival\" problem seems to be too touch for me. Can you give me a hint in proving this triangle ABC is equilateral. I have tried to use Ceva and the only thing I get is that \" if $l_a$, $m_b$, $h_c$ are concurrent, $l_b$, $m_c$, $h_a$ are concurrent, then $l_c$, $m_a$, $h_b$ are also concurrent, and get the inequality\"", "Solution_5": "[u][b]The author of this posting is : Myth[/b][/u]\r\n____________________________________________________________________\r\n\r\nSuppose contrary, and let $b$ be a medium side. Then $l_a$, $m_b$ and $h_c$ can't intersect in the same point.", "Solution_6": "[quote=\"darkmaster\"]This \"trivival\" problem seems to be too touch for me. Can you give me a hint in proving this triangle ABC is equilateral. I have tried to use Ceva and the only thing I get is that \" if $l_a$, $m_b$, $h_c$ are concurrent, $l_b$, $m_c$, $h_a$ are concurrent, then $l_c$, $m_a$, $h_b$ are also concurrent, and get the inequality\"[/quote]\r\n\r\nWell, it isn't trivial... but it isn't hard either.\r\n\r\nAt first, let's prove your assertion that\r\n\r\n[b](1)[/b] If the lines $l_a$, $m_b$, $h_c$ are concurrent, and the lines $l_b$, $m_c$, $h_a$ are concurrent, then the lines $l_c$, $m_a$, $h_b$ are also concurrent.\r\n\r\nIn fact, by Ceva, the lines $l_a$, $m_b$, $h_c$ are concurrent if and only if $\\frac{\\sin C}{\\sin A} = \\frac{\\cos B}{\\cos A}$. Similarly, the lines $l_b$, $m_c$, $h_a$ are concurrent if and only if $\\frac{\\sin A}{\\sin B} = \\frac{\\cos C}{\\cos B}$, and the lines $l_c$, $m_a$, $h_b$ are concurrent if and only if $\\frac{\\sin B}{\\sin C} = \\frac{\\cos A}{\\cos C}$. Now, in order to prove [b](1)[/b], it is enough to show that the equations $\\frac{\\sin C}{\\sin A} = \\frac{\\cos B}{\\cos A}$ and $\\frac{\\sin A}{\\sin B} = \\frac{\\cos C}{\\cos B}$ imply $\\frac{\\sin B}{\\sin C} = \\frac{\\cos A}{\\cos C}$. But this is clear, since $\\frac{\\sin B}{\\sin C} = \\left(\\frac{\\sin C}{\\sin A} \\cdot \\frac{\\sin A}{\\sin B}\\right)^{-1}$ and $\\frac{\\cos A}{\\cos C} = \\left(\\frac{\\cos B}{\\cos A} \\cdot \\frac{\\cos C}{\\cos B}\\right)^{-1}$. Thus, [b](1)[/b] is proven.\r\n\r\nNow, we have to show that under the conditions of [b](1)[/b], the triangle ABC is equilateral. Well, since these conditions are cyclic (although not symmetric), we can assume WLOG that the angle A is the smallest among the angles A, B, C. This means $A\\leq B$ and $A\\leq C$. Now, as Myth showed, the triangle ABC must be acute-angled, and since the function sin is increasing and the function cos is decreasing on [0; 90], it follows that $\\sin A\\leq \\sin C$, while $\\cos A\\geq \\cos B$.\r\n\r\nBut we know that $\\frac{\\sin C}{\\sin A} = \\frac{\\cos B}{\\cos A}$. This immediately contradicts $\\sin A\\leq \\sin C$ and $\\cos A\\geq \\cos B$ except for the case when sin A = sin C and cos A = cos B. Thus, A = B = C, and triangle ABC is equilateral.\r\n\r\n Darij", "Solution_7": "[u][b]The author of this posting is : Myth[/b][/u]\r\n____________________________________________________________________\r\n\r\nHmm... Actually, solution is really trivial and doesn't use any calculations or trigonometry.", "Solution_8": "You're right. It's even easier than I thought...\r\n\r\n Darij", "Solution_9": "[u][b]The author of this posting is : darkmaster[/b][/u]\r\n____________________________________________________________________\r\n\r\nThanks, Darij and Myth. Now it's trivival." } { "Tag": [ "inequalities", "absolute value" ], "Problem": "Quick Question:\r\n\r\nWhen you have the $ |3x\\plus{}3|\\ge0$. Then would it be that $ 3x\\plus{}3\\ge0$ and $ \\minus{}(3x\\plus{}3)\\ge0$ \r\n\r\nor would it be $ 3x\\plus{}3\\ge0$ and $ \\minus{}(3x\\plus{}3)\\le0$\r\n\r\n..or something else entirely (i think its the second one i mentioned", "Solution_1": "Well, it's actually both...all absolute values of real numbers are nonnegative. :P \r\n\r\nBut if you have something like $ |3x \\plus{} 3|\\ge5$, then the two relevant inequalities are $ 3x \\plus{} 3\\ge5$ and $ \\minus{} (3x \\plus{} 3)\\le5$ (think \"it has to be outside the range $ ( \\minus{} 5,5)$\")\r\n\r\nIf you had $ |3x \\plus{} 3|\\le5$, then you would just get $ \\minus{} 5\\le3x \\plus{} 3\\le5$ (think \"it has to be in the range $ [ \\minus{} 5,5]$\").\r\n\r\n@Z-coli: Thanks for expanding. But you end up memorizing anyway. :P", "Solution_2": "Well, for things like this I don't bother memorizing. You just have to understand why it works. \r\nLet's say you have $ |x| \\equal{} 5$. You know that $ x \\equal{} 5, \\minus{} 5$. Now, let's say you have $ |x| < 5$. \r\nIf $ x$ is positive, it has to be less than $ 5$, or if it's negative, \r\nit has to be greater than $ \\minus{} 5$ (If it is less than $ \\minus{} 5$, it's absolute value will be greater\r\nthan $ 5$). So, you know that either $ \\minus{} 5 < x < 5$. The same reasoning goes for this problem.", "Solution_3": "Thanks for the reply you guys" } { "Tag": [ "floor function", "combinatorics proposed", "combinatorics" ], "Problem": "A rectangular table $ 9$ rows $ \\times$ $ 2008$ columns is fulfilled with numbers $ 1$, $ 2$, ...,$ 2008$ in a such way that each number appears exactly $ 9$ times in table and difference between any two numbers from same column is not greater than $ 3$. What is maximum value of minimum sum in column (with minimal sum)?", "Solution_1": "Could someone help me with this problem ? I don`t know how start solving !!!\r\n(sorry for my bad English)", "Solution_2": "Do you have any idea? :huh:", "Solution_3": ":wink: I know that Dirichlet Principle hepls but How ???", "Solution_4": "We will denote by $ \\sigma(c)$ the sum of the entries of column $ c$. We claim $ \\max \\min \\sigma(c) \\equal{} 24$, realized for example by the table with three columns equal to $ (1,1,1,3,3,3,4,4,4)$ (having $ \\sigma(c) \\equal{} 24$), one $ (2,2,2,2,2,5,5,5,5)$, one $ (2,2,2,2,5,5,5,5,5)$, and the rest $ (n,n,n,n,n,n,n,n,n)$, for $ 6\\leq n\\leq N\\equal{}2008$.\r\n\r\nThe $ k$ columns contaiming an entry equal to $ 1$ can only have their entries from the set $ \\{1,2,3,4\\}$. Since there are only $ 9\\cdot 4$ such values available, it means that $ 9k \\leq 9\\cdot 4$, hence $ k\\leq 4$.\r\n\r\nIf $ k\\equal{}1$, that column has $ \\sigma(c_1) \\equal{} 9\\cdot 1 < 24$;\r\n\r\nIf $ k\\equal{}2$, those columns have $ \\sigma(c_1) \\plus{} \\sigma(c_2) \\leq 9\\cdot 1 \\plus{} 9\\cdot 4 \\equal{} 45$, hence one of them has $ \\sigma(c) \\leq \\lfloor 45/2 \\rfloor \\equal{} 22 < 24$;\r\n\r\nIf $ k\\equal{}4$, those columns have $ \\sigma(c_1) \\plus{} \\sigma(c_2) \\plus{} \\sigma(c_3) \\plus{} \\sigma(c_4) \\equal{} 9\\cdot 1 \\plus{} 9\\cdot 2 \\plus{} 9\\cdot 3 \\plus{} 9\\cdot 4 \\equal{} 90$, hence one of them has $ \\sigma(c) \\leq \\lfloor 90/4 \\rfloor \\equal{} 22 < 24$;\r\n\r\nIf $ k\\equal{}3$, finally, those columns have $ \\sigma(c_1) \\plus{} \\sigma(c_2) \\plus{} \\sigma(c_3) \\leq 9\\cdot 1 \\plus{} 9\\cdot 3 \\plus{} 9\\cdot 4 \\equal{} 72$, hence one of them has $ \\sigma(c) \\leq \\lfloor 72/3 \\rfloor \\equal{} 24$. A model has been presented with the claim above.", "Solution_5": "Nice !! thanks . :D" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra theorems" ], "Problem": "Prove this facts about supersolvable groups: The commutator subgroup of a supersolvable group is nilpotent.", "Solution_1": "I'm surprised no one answered this. It is a pretty neat result.\r\n\r\n[hide=\"hint\"]\nTry to prove $ [G', G_k]\\subset G_{k\\plus{}1}$ where $ G\\equal{}G_0\\triangleright \\cdots \\triangleright G_N\\equal{}1$ is the normal series with cyclic quotients, by examining the action of $ G$ on $ G_k/G_{k\\plus{}1}$.\n[/hide]" } { "Tag": [ "induction", "inequalities", "function" ], "Problem": "find the value of :\r\n\r\n$\\sqrt{1+1\\cdot\\sqrt{1+2\\cdot \\sqrt{1+3\\cdot\\sqrt{1+4\\cdot\\sqrt{1+5\\cdot\\sqrt{1+\\cdots}}}}}}$", "Solution_1": "I'm pretty sure that's infinite. Don't know how to prove it though.", "Solution_2": "i don't think so....\r\ni have seen a similar problem before, pretty sure it converges...", "Solution_3": "[quote=\"amirhtlusa\"]i don't think so....\ni have seen a similar problem before, pretty sure it converges...[/quote]\r\n\r\nYes, surely...", "Solution_4": "A first step:\r\n\r\n[hide]Let $a_n = \\sqrt{1 + \\sqrt{1 + 2\\sqrt{1 + 3\\sqrt{\\ldots \\sqrt{1+n\\sqrt{1}}}}}}$. Suppose $a_n > 2$. Square this out and subtract one; we get \n$\\sqrt{1 + 2\\sqrt{1 + 3\\sqrt{\\ldots \\sqrt{1+n\\sqrt{1}}}}} > 3$. Now, square this out, subtract one and divide by 2:\n$\\sqrt{1 + 3\\sqrt{\\ldots \\sqrt{1+n\\sqrt{1}}}} > 4$.\n\nAt the $i$th step, I claim we have as a consequence of our assumption that\n$\\sqrt{1 + i\\sqrt{\\ldots \\sqrt{1+n\\sqrt{1}}}} > i + 1$. Then squaring, subtracting 1 and dividing gives us at the $i+1$st step that\n$\\sqrt{1 + (i+1)\\sqrt{\\ldots \\sqrt{1+n\\sqrt{1}}}} > i + 2$.\nThus, by induction the claim is true. However, taking this to the $n-1$st step results in the claim that $\\sqrt n > n$ (or something like that -- I don't think my indices work out exactly, but it's close enough), clearly absurd for integers. Thus our sequence is certainly bounded above by 2. As it is also certainly increasing, we know that it must have some real limit value less than or equal to 2.\n[/hide]", "Solution_5": "AArg, i was trying to majorizate or minorizate noticing some inequalities of each radicals with $2<1+a_i$ since every radical $a_i$ is more than 1, and evalutating the approssimative expression of $S-\\sqrt{S^2-1-S}$...but i arrived nowhere...so tryed with continueed function, because it seems me that could have some joining points with $e^{\\frac{1}{e}}$" } { "Tag": [], "Problem": "A printer can print 10 pages of text per minute or 4 pages of graphics per minute. How many minutes will it take to print 31 pages of text and 7 pages of graphics?", "Solution_1": "You first divide 31 by 10 so you'll need 3.1 minutes to print those pages.\r\nNext you divide 7 by 4 so you'll need 1.75 minutes to print the graphics.\r\n\r\n3.1+1.75= 4.85", "Solution_2": "I reached the same conclusion yet the review says the answer is $\\frac{417}{20}$." } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "Let $\\ ABC$ be an acute triangle, where we take $\\ M$ on $\\ (AB)$ and $\\ N$ on $\\ (AC)$ such that we have\r\n $\\ 0$ and $ x,y,z$ are the positive constants. :)\r\nLet try with it!\r\nRegards", "Solution_2": "I think r is root of an equation. Reads this : http://www.sosmath.com/algebra/facto...11/fac111.html :D Good luck, my friend", "Solution_3": "[quote=\"B_bnguyen\"]I think r is root of an equation. Reads this : http://www.sosmath.com/algebra/facto...11/fac111.html :D Good luck, my friend[/quote]\r\nDead Link!!!\r\nI want to know how does it get that equation. :)", "Solution_4": "[quote=\"HelloWorld\"]How does it get the value $ \\lambda$[/quote]by the hammer :ninja:" } { "Tag": [ "geometry", "3D geometry" ], "Problem": "i have 3 identical cylinder and 18 identical cubes, how many ways can i stack them(1 on top of each other) to make a tower?\r\n\r\npermutation? combination? it has been too long since i did these.", "Solution_1": "If you have $n_{1}$ objects of type $1, n_{2}$ objects of type $2,\\dots,n_{k}$ objects of type $k$, then you can order them in $\\frac{(n_{1}+n_{2}+\\dots+n_{k})!}{n_{1}!n_{2}!\\dots n_{k}!}$ ways.\r\n\r\nIn the given problem, the answer is $\\frac{21!}{18!3!}=\\frac{21\\cdot 20\\cdot 19}{6}=1330$", "Solution_2": "[hide=\"solution\"] you have 21 objects, therefore you have 21!/(18!*3!) ways to place cubes in the tower. each way to place 18 cubes defines the places of the cylinders. since the cylinders are indentical as are the cubes, I believe that should be the answer.[/hide]", "Solution_3": "Another way you can look at the problem, although the previous two solutions given are better and the way I would have done them, is to view them as cubes and cylinders in a bag that you are picking out of. Then, multiplying fractions, it is easy to see the same answer of:\r\n\r\n1330\r\n\r\nThis should work. I tried it out and it worked for me.\r\n\r\nBut I would reccomend sticking to the original solutions provided by sapphyre571 and Farenhajt.", "Solution_4": "[hide]$\\frac{21!}{2!\\cdot3!}=\\boxed{1330}$[/hide]" } { "Tag": [ "inequalities", "inequalities solved" ], "Problem": "Prove the inequality\r\n\r\n \\sum (1+a 2 ) 2 (1+b 2 ) 2 (a-c) 2 (b-c) 2 \\geq (1+a 2 )(1+b 2)(1+c 2 )((a-b)(b-c)(c-a)) 2 \r\n\r\nwhere the initial sum is over cyclic permutation of a,b,c.", "Solution_1": "This is so monstruous and at the same time so nice.\r\n Of course, let a<>b<>c. It becomes sum [(1+a^2)(1+b^2)]/[(1+c^2)(a-b)^2]>=1. Now, by AM-GM we have [(1+a^2)(1+b^2)]/[(1+c^2)(A-B)^2]+[(1+b^2)(1+c^2)]/[(1+a^2)(b-c)^2]>=2(1+b^2)/|(b-a)(b-c)| and three other inequalities. Sum up and you reduced the problem to sum (1+b^2)/|(b-a)(b-c)|>=1. This is true, since |sum (1+b^2)/(b-a)(b-a)|=1." } { "Tag": [ "limit", "logarithms", "function", "inequalities", "integration", "geometry", "rectangle" ], "Problem": "If $ x_n\\equal{}\\frac{1}{1\\cdot n}\\plus{}\\frac{1}{2\\cdot (n\\minus{}1)}\\plus{}\\frac{1}{3\\cdot (n\\minus{}2)}\\plus{}\\cdots\\plus{}\\frac{1}{(n\\minus{}2)\\cdot 3}\\plus{}\\frac{1}{(n\\minus{}1)\\cdot 2}\\plus{}\\frac{1}{n\\cdot 1}$\r\nFind$ \\displaystyle\\lim_{n\\to\\infty}{x_n}$ :blush:", "Solution_1": "$ \\frac{1}{k}\\plus{}\\frac{1}{(n\\plus{}1)\\minus{}k}\\equal{}\\frac{n\\plus{}1}{k((n\\plus{}1)\\minus{}k)}$\r\n\r\nAs a result, $ x_{n}\\equal{}\\sum_{k\\equal{}1}^{n}\\frac{1}{k((n\\plus{}1)\\minus{}k)}\\equal{}\\frac{1}{n\\plus{}1}\\sum_{k\\equal{}1}^{n}\\left[\\frac{1}{k}\\plus{}\\frac{1}{(n\\plus{}1)\\minus{}k}\\right]\\equal{}\\frac{2H(n)}{n\\plus{}1}$.\r\n\r\nTherefore, $ \\lim_{n\\rightarrow\\infty}x_{n}\\equal{}\\lim_{n\\rightarrow\\infty}\\frac{2H(n)}{n\\plus{}1}\\equal{}\\lim_{n\\rightarrow\\infty}\\frac{2\\ln n}{n\\plus{}1}\\equal{}0$\r\n\r\n(Note that in here $ H(n)$ refers to a harmonic function and $ \\lim_{n\\rightarrow\\infty}\\frac{H(n)}{\\ln n}\\equal{}1$)", "Solution_2": "[quote=\"Timestopper_STG\"]$ \\frac {1}{k} \\plus{} \\frac {1}{(n \\plus{} 1) \\minus{} k} \\equal{} \\frac {n \\plus{} 1}{k((n \\plus{} 1) \\minus{} k)}$\n\nAs a result, $ x_{n} \\equal{} \\sum_{k \\equal{} 1}^{n}\\frac {1}{k((n \\plus{} 1) \\minus{} k)} \\equal{} \\frac {1}{n \\plus{} 1}\\sum_{k \\equal{} 1}^{n}\\left[\\frac {1}{k} \\plus{} \\frac {1}{(n \\plus{} 1) \\minus{} k}\\right] \\equal{} \\frac {2H(n)}{n \\plus{} 1}$.\n\nTherefore, $ \\lim_{n\\rightarrow\\infty}x_{n} \\equal{} \\lim_{n\\rightarrow\\infty}\\frac {2H(n)}{n \\plus{} 1} \\equal{} \\lim_{n\\rightarrow\\infty}\\frac {2\\ln n}{n \\plus{} 1} \\equal{} 0$\n\n(Note that in here $ H(n)$ refers to a harmonic function and $ \\lim_{n\\rightarrow\\infty}\\frac {H(n)}{\\ln n} \\equal{} 1$)[/quote]\r\n\r\nYes, the proof is nice, but the problem appears before the Harmonic Series in my book. I've got another proof.\r\n$ \\frac{2}{n\\plus{}1}\\displaystyle\\sum_{i\\equal{}1}^n{\\displaystyle\\frac{1}{i}}<\\frac{2}{n}\\displaystyle\\sum_{i\\equal{}1}^n{\\displaystyle\\frac{1}{i}}<2\\sqrt{\\displaystyle\\frac{\\displaystyle\\sum_{i\\equal{}1}^{n}{\\frac{1}{i^2}}}{n}}$ < $ 2\\sqrt{\\frac{2}{n}}$\r\nwhich uses the Cauchy Inequality or $ A0$\r\n2)$(\\frac{1}{n}\\sum_{k=1}^n a_k^{\\frac{1}{n}})^n$,where $a_n$is a positive increasing sequence.\r\n3)$\\sum_{k=1}^n ln(1+\\frac{(-1)^k}{k^a(ln(k))^b})$,where $a,b>0$\r\n4)$f(x)=\\sum_{n=1}^{\\infty} (-1)^{n-1}ln(1+\\frac{x}{n})$(Equivalent when $x$ goes to $\\infty$)\r\n5)$f(x)=1+\\sum_{n=1}^{\\infty} (1+\\frac{1}{n})^{n^2}\\frac{x^n}{n!}$(Equivalent when $x$ goes to $\\infty$)\r\n6)$\\int_0^{\\infty}(1+\\frac{x}{n})^{-n}x^{-\\frac{1}{n}}dx$\r\n7)$\\int_0^{\\sqrt(n)}(1-\\frac{t^2}{n})^n dt$", "Solution_1": "Killer, Killer, I know one thing: you\"ll never like computing equivalents... ;) ;)", "Solution_2": "Hey guys...I know you all \"love\" these kind of exercises.Why dont you give me a hand to solve them?", "Solution_3": "Killer, please post one exercise [b]per topic[/b], or else all your topics will be closed. \r\nAnd oh, for easy problems use the [url=http://www.mathlinks.ro/Forum/index.php?f=296]Calculus Computations and Tutorials[/url] forums from now on! Topic closed." } { "Tag": [ "geometry", "3D geometry", "algebra", "polynomial", "vector", "geometric series" ], "Problem": "Given $ a_1\\equal{}1, a_2\\equal{}2, a_3\\equal{}2, a_4\\equal{}1$ and $ a_{k}\\equal{}\\frac{a_{k\\minus{}3}\\plus{}a_{k\\minus{}4}}{2}$\r\n\r\nFind the explicit formula.", "Solution_1": "Totally not pretty -- cubic formula and all that jazz. Lots of cube roots of things involving the square root of 33, etc. The limiting value works out to a nice rational number that you should be able to calculate by looking at, say, the 100th term ;). Besides, you know how to approach these in general, don't you? (Nice work on Mandelbrot rd 1, btw. Tech did pretty well overall, too.)", "Solution_2": "We make the guess that it can be modeled as the sum of several geometric series.\r\n\r\nThe characteristic polynomial is \r\n$ 2a_k \\equal{} a_{k \\minus{} 3} \\plus{} a_{k \\minus{} 4}$\r\n\r\n$ 2a_{k \\plus{} 4} \\equal{} a_{k \\plus{} 1} \\plus{} a_k$\r\n$ 2x^4 \\equal{} x \\plus{} 1$\r\n\r\nJBL, doesn't that mean we need to use the QUARTIC formula to get roots?\r\nEDIT: I noticed that 1 is a root, so you can factor that out and use the cubic. NVM above comment. You're right about the ugly cube roots of 33.\r\nSo if you take those roots, you can express the general form like \r\n$ \\alpha \\cdot R_1 ^ n \\plus{} \\beta \\cdot R_2 ^ n \\plus{} \\gamma \\cdot R_3 ^ n \\plus{} \\mu \\cdot R_4 ^ n$\r\n\r\nYou can use the given values and set up a system of four equations to solve for the constants, and then you should be set.\r\n\r\nI'm not sure how well this would work on a contest, but I THINK it kills the problem off. (note this only works on homogenous linear recursions)", "Solution_3": "Where is the problem from? I thought initially that it would become periodic, but after calculating a few terms that's clearly not the case. JBL and DFT are correct about the form the solution should take, as ugly as it might be.\r\n\r\n[quote=\"DiscreetFourierTransform\"]We make the guess that it can be modeled as the sum of several geometric series.[/quote]\r\n\r\n\"Modeled\" implies a degree of inaccuracy. Inspection tells us that homogeneous linear recurrences must have geometric series in their solution sets, and moreover that their solution sets are vector spaces. The result follows." } { "Tag": [ "calculus", "integration", "inequalities", "derivative", "geometry", "trapezoid", "real analysis" ], "Problem": "Show that $\\int_{0}^{1}e^{x^{2}}\\leq \\frac{3e}{5}$.", "Solution_1": "[quote=\"Slizzel\"]Show that $\\int_{0}^{1}e^{x^{2}}\\leq \\frac{3e}{5}$.[/quote]\r\n\r\n$\\int_{0}^{1}\\int_{0}^{1}e^{x^{2}}e^{y^{2}}dxdy=\\left(\\int_{0}^{1}{e^{x^{2}}}dx\\right)^{2}$.\r\n\r\nBe $x=rcost,y=rsint$ then $\\int_{0}^{1}\\int_{0}^{1}e^{x^{2}}e^{y^{2}}dxdy=\\int_{0}^{1}\\int_{0}^{\\frac{\\pi}{2}}re^{r^{2}}drdt=\\frac{e \\pi}{4}$. So $\\int_{0}^{1}e^{x^{2}}=\\sqrt{\\frac{e \\pi}{4}}\\le \\frac{3e}{5}$", "Solution_2": "[quote=\"cckek\"]\n$\\int_{0}^{1}\\int_{0}^{1}e^{x^{2}}e^{y^{2}}dxdy=\\int_{0}^{1}\\int_{0}^{\\frac{\\pi}{2}}re^{r^{2}}drdt$[/quote]\r\ni'm not sure about this ..", "Solution_3": "It's an inequality, and the inequality's pointing in the wrong direction.\r\n\r\nA simple estimate: Since $e^{x^{2}}$ is convex (second derivative $(2+4x^{2})e^{x^{2}}$), the two-interval trapezoid estimate $\\frac14(1+2e^{\\frac14}+e^{1})$ is greater than the integral. This estimate is approximately $1.5716$, which is less than $\\frac35e\\approx 1.6310$.", "Solution_4": "Since $\\int_{0}^{1}e^{x^{2}}dx\\le \\int_{0}^{1}e^{x}dx=e-1$, the estimate would follow from $e-1\\le 3e/5$, which is the same as $e\\le 5/2$... oops. :oops:", "Solution_5": "It is better to use the inequality $e^{y}\\le 1+y+\\frac{y^{2}}2$ for $-1\\le y\\le 0$ and write the integral as $\\int_{0}^{1}e\\cdot e^{x^{2}-1}\\,dx$ ;)", "Solution_6": "How about $e^{x^{2}}\\leq (e-1)x^{2}+1$ on $[0,1]$? Is this okay?", "Solution_7": "Let's see. It gives \r\n\\[\\int_{0}^{1}e^{x^{2}}\\,dx\\le 1+\\frac{e-1}3=\\frac{e+2}3\\]\r\nSo, we need to prove that $e+2<\\frac{9}{5}e$ or $e=1+1+\\frac12+\\dots>\\frac{5}{2}$, which is true. So, indeed, it works perfectly well too :)." } { "Tag": [], "Problem": "Given that $ XY \\equal{} 21$ and $ 1\\leq x\\leq 2$, find the sum of the upper and lower bounds of $ Y$. Express your answer as a decimal.", "Solution_1": "Upperbound: 21/1=21\r\nLowerbound: 21/2=10.5\r\n\r\nSum=31.5" } { "Tag": [ "percent" ], "Problem": "Ok pick a number 10-99\r\n\r\nGot it. Good.\r\n\r\nNow multiply that number by 2\r\n\r\nadd 12 to that number\r\n\r\nnow subtract 10\r\n\r\nadd 6\r\n\r\nnow divide that number by 2\r\n\r\nFinally subtract that number by your original number\r\n\r\nYour answer should be (spoilers)\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n4.", "Solution_1": "How is that reading minds? You still don't know what number I picked. This is more like mind reading:\r\n\r\nPick a prime number less than 50 with two odd digits. The two digits must also be different, so for example 11 won't work.\r\n\r\nI predict that your number is\r\n[hide]37[/hide]", "Solution_2": "oh yeah...pick a positive integer less than or equal to one\r\nmultiply it by 3.12435845247540548\r\nthen divide by 247824973051830413\\\r\n\r\nlemme guess the answer is 1 :o \r\n\r\nobviously this is not mind reading\r\n\r\nsorry :D", "Solution_3": "...............\r\n :|", "Solution_4": "Take a number. Subtract the number from the number itself. result should be 0. :D", "Solution_5": "[quote=\"236factorial\"]Take a number. Subtract the number from the number itself. result should be 0. :D[/quote]\r\n\r\nWHOA THAT'S WHAT I GOT OMG YOU'RE A PSYCHIC", "Solution_6": "whats with the restriction on only numbers 10-99...this works with any number, right?", "Solution_7": "i guess it would take a long time to do it with 5462462 as your number.", "Solution_8": "Pick a number. I predict that your number is...\r\n[hide]not 84756729738572876589278.128735882873875209[/hide]", "Solution_9": "[quote=\"jmadsen\"]Pick a number. I predict that your number is...\n[hide]not 84756729738572876589278.128735882873875209[/hide][/quote]\r\napparently that is exactly what i picked your mind reading skills are horrible", "Solution_10": "[quote=\"jmadsen\"]How is that reading minds? You still don't know what number I picked. This is more like mind reading:\n\nPick a prime number less than 50 with two odd digits. The two digits must also be different, so for example 11 won't work.\n\nI predict that your number is\n[hide]37[/hide][/quote]\r\n\r\nWRONG\r\n\r\ni picked 17", "Solution_11": "[quote=\"jmadsen\"]How is that reading minds? You still don't know what number I picked. This is more like mind reading:\n\nPick a prime number less than 50 with two odd digits. The two digits must also be different, so for example 11 won't work.\n\nI predict that your number is\n[hide]37[/hide][/quote]\r\n\r\nI pick 19..not 37", "Solution_12": "- 13 -\r\n\r\nOh well.", "Solution_13": "eh, I picked 37\r\n\r\nwell I picked 43 at first and then I was like wait, no", "Solution_14": "Pick a number between 1 and 3 [b][i]inclusive[/i][/b]. \r\n\r\nIs it...\r\n[hide] 2! [/hide]\r\n\r\n\r\n :D :D :D :D", "Solution_15": "[quote=\"Hexaditidom\"]Red screwdriver\n\n\nwow :D[/quote]\r\n\r\nwaoh we picked the same thing", "Solution_16": "i got 62 :?", "Solution_17": "order of operations is AWESOME", "Solution_18": "[quote=\"math92\"]i got 62 :?[/quote]You all can't count :D", "Solution_19": "Huh, I got cabbage for the vegetable one. \r\n\r\nOrange Screwdriver? :huh: \r\n\r\nGuess I'm not normal", "Solution_20": "[quote=\"Derek\"]How much is:\n\n[hide=\"1 + 5?\"] [hide=\"3 + 3?\"] [hide=\"4 + 2?\"] [hide=\"Repeat the number 6 in your head as fast as possible\nfor 15 seconds.\"] [hide=\"You were thinking about a CARROT, weren't you!?!\nReally weird, right!?!?!\"][/hide][/hide][/hide][/hide][/hide]\n\nEDIT: added screwed up hide tags[/quote]\r\n\r\ni couldn't come up w/ a vegetable", "Solution_21": "[quote=\"sirkuku\"]Pick a number 10-99\n\nmultiply by 2\n\nadd 47\n\nsubtract 16\n\nadd 10\n\nadd 5\n\nsubtract 8\n\n\ndivide by 2\n\nsubtract by your original number\nYour answer should be \n(Spoilers)!!!!!!!!!\n\n\n\n\n\n\n\n\n19[/quote]\r\n\r\nI'm gonna remove the mystery around these things... :) \r\n\r\nMy number is $x$\r\n\r\n$x \\Rightarrow 2x \\Rightarrow 2x + 47 \\Rightarrow 2x + 31 \\Rightarrow 2x + 41 \\Rightarrow 2x + 46 \\Rightarrow 2x + 38 \\Rightarrow x + 19 \\Rightarrow 19$", "Solution_22": "[quote=\"Hexaditidom\"]Yeah, I heard the 37 problem a while ago. Real interesting, just like this one:\n\n PICK A NUMBER\n1 2 3 4\n\n[hide]Why 3?[/hide][/quote]\n\nI pick 3 because it's between $e$ and $\\pi$ my two favorite numbers :D.\n\nI'm going to post a particularly exhaustive number trick.\n\nPick a number above 10.\nsquare it\nsubtract 4\nmultiply by 2 less than your number\ndivide by 2 more than your number\nsubtract 1\nmultiply by 111,111\ndivide by 1 less than your number\nadd 999,999\ndivide by 1001\nsubtract 111 times your number\nand you get:\n\n[hide=\"spoiler\"]\n\n\nAN EVIL NUMBER BWAHAHAHAHAHA\n[/hide]\r\n\r\nfigure that one out :P", "Solution_23": "[quote=\"Hamster1800\"][quote=\"Hexaditidom\"]Yeah, I heard the 37 problem a while ago. Real interesting, just like this one:\n\n PICK A NUMBER\n1 2 3 4\n\n[hide]Why 3?[/hide][/quote]\n\nI pick 3 because it's between $e$ and $\\pi$ my two favorite numbers :D.\n\nI'm going to post a particularly exhaustive number trick.\n\nPick a number above 10.\nsquare it\nsubtract 4\nmultiply by 2 less than your number\ndivide by 2 more than your number\nsubtract 1\nmultiply by 111,111\ndivide by 1 less than your number\nadd 999,999\ndivide by 1001\nsubtract 111 times your number\nand you get:\n\n[hide=\"spoiler\"]\n\n\nAN EVIL NUMBER BWAHAHAHAHAHA\n[/hide]\n\nfigure that one out :P[/quote]\r\nAt least we don't have to square root it...", "Solution_24": "[quote=\"Hamster1800\"][quote=\"Hexaditidom\"]Yeah, I heard the 37 problem a while ago. Real interesting, just like this one:\n\n PICK A NUMBER\n1 2 3 4\n\n[hide]Why 3?[/hide][/quote]\n\nI pick 3 because it's between $e$ and $\\pi$ my two favorite numbers :D.\n\nI'm going to post a particularly exhaustive number trick.\n\nPick a number above 10.\nsquare it\nsubtract 4\nmultiply by 2 less than your number\ndivide by 2 more than your number\nsubtract 1\nmultiply by 111,111\ndivide by 1 less than your number\nadd 999,999\ndivide by 1001\nsubtract 111 times your number\nand you get:\n\n[hide=\"spoiler\"]\n\n\nAN EVIL NUMBER BWAHAHAHAHAHA\n[/hide]\n\nfigure that one out :P[/quote]\r\n\r\nPick $x$.\r\n\r\n$x \\Rightarrow x^2 \\Rightarrow x^2 - 4 \\Rightarrow (x - 2)^2(x + 2) \\Rightarrow (x-2)^2 \\Rightarrow (x - 3)(x - 1) \\Rightarrow 111111(x-3)(x-1) \\Rightarrow 111111(x - 3) \\Rightarrow 111111(x + 6) \\Rightarrow 111x + 666 \\Rightarrow 666$\r\n\r\nWow. It was actually thought out and not a random list of commands. That's good. :)", "Solution_25": "[quote=\"chesspro\"]I'm gonna remove the mystery around these things... :) \n\nMy number is $x$\n\n$x \\Rightarrow 2x \\Rightarrow 2x + 47 \\Rightarrow 2x + 31 \\Rightarrow 2x + 41 \\Rightarrow 2x + 46 \\Rightarrow 2x + 38 \\Rightarrow x + 19 \\Rightarrow 19$[/quote]\n[quote=\"chesspro\"]Pick $x$.\n\n$x \\Rightarrow x^2 \\Rightarrow x^2 - 4 \\Rightarrow (x - 2)^2(x + 2) \\Rightarrow (x-2)^2 \\Rightarrow (x - 3)(x - 1) \\Rightarrow 111111(x-3)(x-1) \\Rightarrow 111111(x - 3) \\Rightarrow 111111(x + 6) \\Rightarrow 111x + 666 \\Rightarrow 666$\n\nWow. It was actually thought out and not a random list of commands. That's good. :)[/quote]\r\n\r\nyes...we actually can do algebra :o", "Solution_26": "[quote=\"MysticTerminator\"]yes...we actually can do algebra :o[/quote] :help: :surrender: you must be kidding! :)", "Solution_27": "This is a better one:\r\nPick a number of three digits lets say\r\nabc whith a and c not the same digit\r\nThen write the number\r\n|abc-cba|=def\r\nnow add\r\ndef+fed\r\nWhich is your final number?\r\n[hide] Isn\u00b4t it 1089?[/hide]", "Solution_28": "[quote=\"jmadsen\"]Pick a number. I predict that your number is...\n[hide]not 84756729738572876589278.128735882873875209[/hide][/quote]\r\nand if mine was this would this causes problems to you???\r\n\r\n\r\n :rotfl: i'm obiuovsly joking :rotfl:", "Solution_29": "I know where you live.\r\n\r\n[hide]In time. Unless you're a [hide]fairy[/hide], in which case, I'm sorry.[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "For $ 1\\leq |a|\\leq 3,\\ 1\\leq |b|\\leq 3,\\ 1\\leq |c|\\leq 3$, find the maximum of the following expressions.\r\n\r\n(1) $ \\frac {\\sqrt { \\minus{} a^2 \\plus{} 4|a| \\minus{} 3}\\sqrt { \\minus{} b^2 \\plus{} 4|b| \\minus{} 3}\\sqrt { \\minus{} c^2 \\plus{} 4|c| \\minus{} 3}}{|abc|}$\r\n\r\n(2) $ (\\sqrt { \\minus{} a^2 \\plus{} 4|a| \\minus{} 3} \\plus{} \\sqrt {9 \\minus{} a^2})(\\sqrt { \\minus{} b^2 \\plus{} 4|b| \\minus{} 3} \\plus{} \\sqrt {9 \\minus{} b^2})(\\sqrt { \\minus{} c^2 \\plus{} 4|c| \\minus{} 3} \\plus{} \\sqrt {9 \\minus{} c^2})$", "Solution_1": "[quote=\"kunny\"]For $ 1\\leq |a|\\leq 3,\\ 1\\leq |b|\\leq 3,\\ 1\\leq |c|\\leq 3$, find the maximum of the following expressions.\n\n(1) $ \\frac {\\sqrt { \\minus{} a^2 \\plus{} 4|a| \\minus{} 3}\\sqrt { \\minus{} b^2 \\plus{} 4|b| \\minus{} 3}\\sqrt { \\minus{} c^2 \\plus{} 4|c| \\minus{} 3}}{|abc|}$\n\n(2) $ (\\sqrt { \\minus{} a^2 \\plus{} 4|a| \\minus{} 3} \\plus{} \\sqrt {9 \\minus{} a^2})(\\sqrt { \\minus{} b^2 \\plus{} 4|b| \\minus{} 3} \\plus{} \\sqrt {9 \\minus{} b^2})(\\sqrt { \\minus{} c^2 \\plus{} 4|c| \\minus{} 3} \\plus{} \\sqrt {9 \\minus{} c^2})$[/quote]\r\nFor (1) $ \\sqrt { \\minus{} 1 \\plus{} \\frac {4}{|a|} \\minus{} \\frac {3}{a^2}}\\le\\frac {1}{\\sqrt{3}}$ so $ max \\equal{} \\frac {1}{3\\sqrt {3}}$", "Solution_2": "You seem to have a typo.", "Solution_3": "[quote=\"kunny\"]You seem to have a typo.[/quote]\r\nAm I right? (now)", "Solution_4": "Yes! :) \r\nWhen does the equality hold?", "Solution_5": "[quote=\"kunny\"]For $ 1\\leq |a|\\leq 3,\\ 1\\leq |b|\\leq 3,\\ 1\\leq |c|\\leq 3$, find the maximum of the following expressions.\n\n(1) $ \\frac {\\sqrt { \\minus{} a^2 \\plus{} 4|a| \\minus{} 3}\\sqrt { \\minus{} b^2 \\plus{} 4|b| \\minus{} 3}\\sqrt { \\minus{} c^2 \\plus{} 4|c| \\minus{} 3}}{|abc|}$\n\n(2) $ (\\sqrt { \\minus{} a^2 \\plus{} 4|a| \\minus{} 3} \\plus{} \\sqrt {9 \\minus{} a^2})(\\sqrt { \\minus{} b^2 \\plus{} 4|b| \\minus{} 3} \\plus{} \\sqrt {9 \\minus{} b^2})(\\sqrt { \\minus{} c^2 \\plus{} 4|c| \\minus{} 3} \\plus{} \\sqrt {9 \\minus{} c^2})$[/quote]\r\n(2) if I didnt miss in calculations $ \\sqrt { \\minus{} a^2 \\plus{} 4|a| \\minus{} 3} \\plus{} \\sqrt {9 \\minus{} a^2} \\equal{} \\sqrt {3 \\minus{} |a|}(\\sqrt {3 \\plus{} |a|} \\plus{} \\sqrt {|a| \\minus{} 1})\\le \\frac {11\\sqrt {3}}{6}$ so $ max \\equal{} \\frac {11^3\\sqrt {3}}{72}$", "Solution_6": "[quote=\"dima ukraine\"]$ \\sqrt { \\minus{} a^2 \\plus{} 4|a| \\minus{} 3} \\plus{} \\sqrt {9 \\minus{} a^2} \\equal{} \\sqrt {3 \\minus{} |a|}(\\sqrt {3 \\plus{} a} \\plus{} \\sqrt {a \\minus{} 1})$[/quote]\r\n\r\nWhy the equality hold?", "Solution_7": "[quote=\"kunny\"][quote=\"dima ukraine\"]$ \\sqrt { \\minus{} a^2 \\plus{} 4|a| \\minus{} 3} \\plus{} \\sqrt {9 \\minus{} a^2} \\equal{} \\sqrt {3 \\minus{} |a|}(\\sqrt {3 \\plus{} a} \\plus{} \\sqrt {a \\minus{} 1})$[/quote]\n\nWhy the equality hold?[/quote]\r\nHmm I mean it" } { "Tag": [], "Problem": "How do you know how to a attack a certain problem when it is not too easy?", "Solution_1": "Mathematicians intuition, a sixth sense if you will, will develop as you learn more and more problem solving techniques.\r\nYou can guess how to roughly solve the problem using this.", "Solution_2": "Sometimes I have no idea how to approach a problem, but after playing around with it, I seem to find a little possibility. Then I pursue it until I solve the problem. Of course, if the problem is way out my league, then that enlightenment will never come.\r\n\r\n-interesting_move", "Solution_3": "also, asking ppl who kno the question for hints also works!! :D" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Well this is ,a very ordinary problem ,hope it insprises you :) :\r\n\r\nLet $C$ be a circle and $P,Q$ two fixed points (Q on the circle and P outside( in the yard ) :) ).\r\n\r\nand $L$ is fixed line which passes through $P$ and doesnt intersect the circle.\r\n\r\n$X,Y$ are two points on $L$, that varies such that $XP.YP=K$ (a constant ) .let $QX$ and $QY$ meet the circle in $R$ and $S$ .prove that $RS$ passes through a fixed point.", "Solution_1": "[u]Size:[/u] constant or variable (measure).\r\n[u]Pozition:[/u] fixed or mobile (mathematical object - matter, thing).\r\n\r\n[i]Examples:[/i] \"fixed point\" and not \"constant point\"; \"mobile ray\" and not \"variable ray\"; a.s.o.", "Solution_2": "Well thank you very much levi for your lessons but i think the problem is clear enough to be understand . :)", "Solution_3": "Evidently, I understood all ! It is O.K. Sorry, for the my profesional deformation (I am a retired teacher)! I suppose that now you teach, but soon you will teach others. \r\n[i]With friendship and congratulations[/i] for your nice share on this site, [i]Ph-An[/i] (alias Levi).\r\n\r\n[u]Answer.[/u] [b]Let[/b] the fixed circle $w$, the fixed line $d$ such that $d\\cap w =\\emptyset$ and the fixed points $Q\\in w$, $P\\in d$. For the mobile points $\\{X,Y\\}\\subset d$ for which $PX\\cdot PY=k^2\\ (k>0)$ I define the points: $\\{Q,\\underline {R}\\}\\subset QX\\cap w$, $\\{Q,\\underline{S}\\}\\subset QY\\cap w$; $\\{A,B\\}\\subset d$, $P\\in (A,B)$, $PA=PB=k$;$\\{Q,\\underline{U}\\}\\subset QA\\cap w$, $\\{Q,\\underline {V}\\}\\subset QB\\cap w$.[b]Then[/b] the line $RS$ passes through the fixed point $F$, where $F\\in UU\\cap VV$ (XX - the tangent in the point $X\\in w$ to the circle $w$).\r\n\r\n[u]A minute solution.[/u] I note the intersections $M\\in RS\\cap QA$, $N\\in RS\\cap QB$\r\nand the middlepoint $L$ of the segment $[MN]$.\r\nThen ($h.d.$- [b]h[/b]armonic [b]d[/b]ivision; $h.p.$-[b]h[/b]armonic [b]p[/b]encil) \r\n$(X,A,Y,B)-h.d.\\Longleftrightarrow Q(X,A,Y,B)-h.p.\\Longleftrightarrow (R,M,S,N)-h.d.$\r\n$\\Longleftrightarrow LM^2=LN^2=LR\\cdot LS$ and the circle with the diameter $[MN]$\r\nis orthogonally to the circle $w\\Longrightarrow L\\in UU\\cap VV\\Longrightarrow L\\equiv F$.\r\n\r\n[u]A (\"malicious and pedantic\") remark.[/u] The circle $w$, the line $d$ and the points $P,Q,A,B,U,V$ are fixed and the points $X,Y,R,S$ are mobilly. The length of the ray of the circle $w$ and the distances $[AB]$, $[PQ]$ a.s.o. are constant, but the distances $[XY]$, $[RS]$ a.s.o. are variably.", "Solution_4": "I like this problem. My dear Lomos Lupin, are you an other solution ? I offer soon some problems from this family (with harmonical pencils and harmonical divisions)." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $h_{n}=1+\\frac{1}{3}+\\frac{1}{5}+...+\\frac{1}{2n-1}$ with $n\\in N$\r\nProve that:\r\n$\\frac{1}{h^{2}_{1}}+\\frac{1}{3h^{2}_{2}}+...+\\frac{1}{(2n-1)h^{2}_{n}}<2$", "Solution_1": "We have $h_{n}-h_{n-1}=\\frac{1}{2n-1}$, therefore $\\frac{1}{(2i-1)h_{i}^{2}}=\\frac{h_{i}-h_{i-1}}{h_{i}^{2}}<\\frac{1}{h_{i-1}}-\\frac{1}{h_{i}}$. Now use telescopic sum!" } { "Tag": [ "geometry", "circumcircle", "parallelogram", "rectangle", "rhombus", "symmetry", "geometric transformation" ], "Problem": "Let $ABC$ 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 $ABC$. Prove that $CC_1$ bisects $\\angle QC1P$.", "Solution_1": "[color=blue][b]Lemma 1:[/b][/color] A tangential quadrilateral is also cyclic (i.e., bicentric) iff its contact quadrilateral has perpendicular diagonals.\r\n\r\nLet $ABCD$ be the contact quadrilateral of a tangential quadrilateral $PQRS$ with the incircle $(I)$, $A \\in SP, B \\in PQ, \\C \\in QR, D \\in RS$. Assume that a tangential quadrilateral is also cyclic with the circumcircle $(O)$. Invert it in its incircle $(I)$. The vertices $P, Q, R, S$ are carried into to the midpoints $P', Q', R', S'$ of of the chords $AB, BC, CD, DA$, i.e., they form the Varignon parallelogram $P'Q'R'S'$ of the contact quadrilateral $ABCD$ with the sides parallel to the diagonals $AC, BD$: $P'Q' \\parallel S'R' \\parallel AC$, $Q'R' \\parallel S'P' \\parallel BD$. At the same time, the parallelogram $P'Q'R'S'$ is inscribed in the inverted circumcircle $(O')$, hence, it is a rectangle. Thus $P'Q' \\perp Q'R'$, which implies $AC \\perp BD$. Conversely, if the tangential quadrilateral $PQRS$ is not cyclic, the inverted vertices still form a parallelogram $P'Q'R'S'$, which cannot lie on a circle (because the original vertices $P, Q, R, S$ do not), hence, it cannot be a rectangle. Consequently its sides $P'Q', Q'R'$ are not perpendicular, which implies that the diagonals $AC \\parallel P'Q'$, $BD \\parallel Q'R'$ of the contact quadrilateral $ABCD$ are not perpendicular either.\r\n\r\n[color=blue][b]Lemma 2:[/b][/color] The diagonals of a tangential quadrilateral have the same diagonal intersection as the diagonals of its contact quadrilateral. The opposite sides of the contact quadrilateral intersect on the diagonals of the tangential quadrilateral.\r\n\r\nLet $ABCD$ be the contact quadrilateral of a tangential quadrilateral $PQRS$ with the incircle $(I)$, $A \\in SP, B \\in PQ, \\C \\in QR, D \\in RS$, and let the diagonals $AC, BD$ meet at a point $X$. Consider the cyclic contact quadrilateral as a degenerate cyclic hexagon $ABBCCD$ and denote $BB \\equiv PQ, CC \\equiv QR$ the tangents of its circumcircle $(I)$ at the points $B, C$. According to Pascal's theorem, the intersections $X \\equiv AC \\cap BD$, $K \\equiv AB \\cap CD$, $Q \\equiv BB \\cap CC$ are collinear. Likewise, consider the cyclic contact quadrilateral as a degenerate cyclic hexagon $AABCDD$ and denote $AA \\equiv SP, DD \\equiv RS$ the tangents of its circumcircle $(I)$ at the points $A, D$. According to Pascal's theorem, the intersections $X \\equiv AC \\cap BD$, $K \\equiv AB \\cap CD$, $S \\equiv AA \\cap DD$ are collinear. Consequently, the 4 points $X, K, Q, S$ are all collinear, i.e., the diagonal $QS$ of the tangential quadrilateral PQRS passes through the point $X$ and the opposite sides $AB, CD$ meet on this diagonal. In the same way, we can show that the other diagonal $PR$ also passes through the point $X$ and that the opposite sides $BC, DA$ meet on this diagonal. Thus the 2 diagonals intersect at the point $X$.\r\n\r\n[color=blue][b]Lemma 3:[/b][/color] Let the opposite sides $AB, CD$ and $BC, DA$ of any quadrilateral ABCD with perpendicular diagonals (not necessarily cyclic) meet at points $K, L$, respectively. Then the diagonals $AC, BD$ intersecting at a point $X$ bisect the angles formed by the lines $KX, LX$.\r\n\r\nLet $E, G$ be the intersections of the line $LX$ with the opposite sides $AB, CD$ and $F, H$ the intersections of the line $KX$ with the opposite sides $BC, DA$, forming another quadrilateral $EFGH$. In a polarity transformation with respect to an arbitrary circle $(X)$ centered at their common diagonal intersection $X$, the quadrilaterals $ABCD, EFGH$ are transformed into parallelograms $A'B'C'D', E'F'G'H'$. Since the diagonals $AC, BD$ are perpendicular to each other, $A'B'C'D'$ is a rectangle. The quadrilateral $EFGH$ is inscribed in the quadrilateral $ABCD$ and consequently, the rectangle $A'B'C'D'$ is inscribed in the parallelogram $E'F'G'H'$. Since the points $E, G, L$ and $F, H, K$ are collinear with the center $X$ of the polar circle, their polars $e = E'F', g = G'H', l = A'C'$ and $f = F'G', h = H'E', k = B'D'$ are parallel. The rectangle diagonals $l = A'C', k = B'D'$ are thus the middle lines of the parallelogram $E'F'G'H'$. Conversely, the parallelogram diagonals $n = E'G', m = F'H'$ are the middle lines of the rectangle $A'B'C'D'$. Consequently, the parallelogram $E'F'G'H'$ has perpendicular diagonals and it is a rhombus. The opposite sides $EF, GH$ and $FG, HE$ meet at the poles $M, N$ of the rhombus diagonals $m = F'H', n = E'G'$. Since these rhombus diagonals are parallel to the rectangle opposite sides $a = A'B', c = C'D'$ and $b = B'C', d = D'A'$, the poles $M, A, C$ and $N, B, D$ of these parallel lines are collinear with the center $X$ of the polar circle. Hence, the diagonals $AC, BD$ pass through the points $M, N$. The rectangle diagonals $l = A'C', k = B'D'$ are perpendicular to the lines $LX, KX$, identical with the diagonal lines $EG, FH$, connecting their poles $L, K$ with the center of the polar circle. Similarly, the rhombus diagonals $n = E'G', m = F'H'$ are perpendicular to the lines $NX, MX$, identical with the diagonal lines $BD, AC$, connecting their poles $N, M$ with the center of the polar circle. Because of the symmetry of the transformed configuration, the rhombus diagonals are bisectors of the angles between the rectangle diagonals. But this also means that the diagonals of the quadrilateral $ABCD$ are bisectors of the angles between the diagonals of the quadrilateral $EFGH$.\r\n\r\n[color=red][b]Problem:[/b][/color] Let the altitude $CC_1$ of the acute angle triangle $\\triangle ABC$ meet its circumcircle $(O)$ at another point $D$ different from the vertex $C$. Let the tangent to the circumcircle at the point $D$ meet the sides $QR, RP$ of the tangential triangle $\\triangle PQR$ at points $S, T$, respectively. The quadrilateral $PQST$ is tangential and $ADBC$ is its contact cyclic quadrilateral with perpendicular diagonals $CC_1 \\equiv CD \\perp AB$ intersecting at the point $C_1$. This means that the quadrilateral $PQST$ is bicentric (see lemma 1), but we do not need this fact at all. Since the quadrilateral $PQST$ is tangential, its diagonals $PS \\equiv PC_1, QT \\equiv QC_1$ have the same intersection $C_1$ as the diagonals $AB, CD$ of its contact quadrilateral $ADBC$. Moreover, the opposite sides $BC, AD$ and $CA, DB$ intersect at points $K, L$ on the diagonals $QT, PS$ of the tangential quadrilateral $PQST$ (see lemma 2). Since the quadrilateral $ADBC$ has perpendicular diagonals, they bisect the angles formed by the lines $KC_1 \\equiv QC_1, LC_1 \\equiv PC_1$ (see lemma 3), i.e., $CC_1 \\equiv CD$ bisects the angle $\\angle QC_1P$ .", "Solution_2": "[b]Yetti, you are a great !.[/b] I always admired your solutions. You are unique on this site.\r\nThank you for your replies, for your effort in a clear and distinct get up.\r\n\r\nHere is my solution. It isn't necessary as $ABC$ - acute triangle.\r\n\r\nI note $T\\in AB\\cap CC,\\ S\\in AB\\cap CR$. Because $CC_1\\perp AB$, then the ray \r\n$[CC_1$ bisects the angle $\\angle PC_1Q$ iff the division $(TQCP)$ is harmonically.\r\nIt is wellknown that the ray $[CS$ is the simedian from the vertex $C$ of the triangle\r\n $ABC$ and $\\frac{TA}{TB}=\\frac{SA}{SB}=\\left( \\frac ba \\right)^2$. Thus, $R\\ (T,A,S,B)$ - harmonic pencil $\\Longrightarrow$\r\n$R\\ (T,Q,C,P)$ - harmonic pencil $\\Longrightarrow (T,Q,C,P)$ - harmonic division.\r\n\r\n[b]Remark.[/b] Here is a wellknown property of the [i]harmonic division[/i]:\r\nLet $A,B,C,D\\in d$ be four points which belong to the same line $d$\r\n(in the mentioned order) and the point $P\\not\\in d$. [u]Then if two from the following[/u] \r\n[u]affirmations are truly, results that and the other affirmation is truly[/u]:\r\n\r\n$\\blacksquare \\ 1.\\ (A,B,C,D)$ - harmonic division $\\left(\\ \\frac{\\overline {BA}}{\\overline {BC}}=-\\frac{\\overline {DA}}{\\overline {DC}}\\ \\right);$\r\n\r\n$\\blacksquare \\ 2.\\ PB\\perp PD;$\r\n\r\n$\\blacksquare \\ 3.$ The ray $[PB$ bisects the angle $\\angle APC$.\r\n\r\n[b]A similar problem.[/b] Let $ABC$ be a triangle, the its circumcircle $w=C(O,R)$, the reflection $A'$ of the point $A$ w.r.t. the center $O$, the intersection $T\\in BC\\cap A'A'$ and the proiection $P$ of the point $A'$ on the line $OT$. Prove that the ray $[PA'$ bisects the angle $\\angle BPC$.\r\n[u]Indication.[/u] $I\\in BC\\cap PA'\\Longrightarrow (B,I,C,T)$ is a harmonic division.", "Solution_3": "Hi\r\n\r\nYou have written fantastic solutions...\r\n\r\nI have another solution:\r\n\r\nLet Call C1 as H.\r\n\r\nLet M be the foot of the altitude from H on BR\r\n(I mean HM L BR)\r\nand let N be the foot of the altitude from H on AR\r\n\r\nAs shown in the picture:\r\n\r\ntanH2 = SinH2/SinH1 = a/b * x/h\r\n\r\n= SinQ2/SinQ1 * x/h\r\n\r\ntanH3 = inP2/SinP1 * y/h\r\n\r\ntanH2/tanH3 = SinQ2/SinP2 * SinP1/SinQ1 * x/y\r\n\r\n= PH*SinP1 / (QH*sinQ1) * x/y = HM/HN * x/y\r\n\r\nBut we know \r\n\r\nBMH ~ ANH so :\r\n\r\nHM/y = HN/x\r\n\r\nand it means tan H2 = tan H3 \r\n\r\nSo : H2 = H3", "Solution_4": "Well, the circumcircle of triangle ABC is the incircle of triangle PQR (since it touches the sides QR, RP and PQ of this triangle PQR); thus, the points A, B, C are the points of tangency of the incircle of triangle PQR with its sides QR, RP and PQ, so that the lines PA, QB and RC concur at one point, namely at the Gergonne point of the triangle PQR. Hence, if we denote this Gergonne point by K, then the points A, B, C are the points of intersection of the lines PK, QK and RK with the lines QR, RP and PQ. Now, the problem follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=6088 post #2 Theorem 2, applied to the triangle PQR and the point K.\r\n\r\n darij", "Solution_5": "Sorry I coudnt send the picture\r\n\r\nLet:\r\n\r\nx=Ah y=BH\r\n\r\nU = (AC,QH) a=CU b=AU\r\n\r\nV = (BC,PH) c=CV d=BV\r\n\r\nh=AH\r\n\r\nH1 to H4 = AHQ, QHC, CHP, PHB\r\n\r\nQ1,Q2 = AQH, HQP\r\n\r\nP_1,P2 = BPH, HPQ\r\n\r\nThanks", "Solution_6": "Sorry I couldnt send the picture\r\n\r\nLet:\r\n\r\nx=Ah y=BH\r\n\r\nU = (AC,QH) a=CU b=AU\r\n\r\nV = (BC,PH) c=CV d=BV\r\n\r\nh=AH\r\n\r\nH1 to H4 = AHQ, QHC, CHP, PHB\r\n\r\nQ1,Q2 = AQH, HQP\r\n\r\nP_1,P2 = BPH, HPQ\r\n\r\nThanks", "Solution_7": "Sorry I couldnt send the picture\r\n\r\nLet:\r\n\r\nx=Ah y=BH\r\n\r\nU = (AC,QH) a=CU b=AU\r\n\r\nV = (BC,PH) c=CV d=BV\r\n\r\nh=AH\r\n\r\nH1 to H4 = AHQ, QHC, CHP, PHB\r\n\r\nQ1,Q2 = AQH, HQP\r\n\r\nP1,P2 = BPH, HPQ\r\n\r\nThanks", "Solution_8": "If $ CC_1$, $ BN$ ,$ AM$ concur, then the problem will be true by Blanchett's theorem. They concur by Ceva iff $ \\frac{CN}{NA} \\cdot \\frac{AC_1}{C_1B} \\cdot \\frac{BM}{MC}\\equal{}1$. But $ \\frac{CN}{NA}\\equal{}\\frac{[CQC_1]}{[AQC_1]}\\equal{}\\frac{CQ \\cdot CC_1 \\cdot sin QCC_1}{QA \\cdot AC_1 \\cdot sin QAC_1}$. We find the product we want is equal to $ \\frac{CQ}{CP} \\cdot \\frac{BP}{AQ}$, and this is $ 1$ since $ CQ\\equal{}AQ$ and $ BP\\equal{}CP$, so we are done.", "Solution_9": "Er, what's Blanchett's theorem?", "Solution_10": "Apparently [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1422783#1422783]this...[/url]", "Solution_11": "I really doubt that this belongs in the pre-Olympiad section.", "Solution_12": "As a comment, we can prove yetti's first and third lemmas in a much simpler fashion. Using his notation,\r\n\r\nLemma 1: Observe that $ \\angle ASD\\equal{}180\\minus{}\\text{arc}\\, AD$, so $ \\angle ASD\\plus{}\\angle BQC\\equal{}360\\minus{}\\text{arc}\\, AD\\minus{}\\text{arc}\\, BC\\equal{}360\\minus{}2\\angle (AC,BC)$. The equivalence follows easily.\r\n\r\n\r\nLemma 3: Let $ U\\equal{}AC\\cap LK$ and $ V\\equal{}BC\\cap LK$. Then looking at $ \\triangle ALK$, concurrent cevians $ AU$, $ LB$, and $ KD$, and the collinear points $ B$, $ D$, and $ V$, we get that $ (K,L;U,V)\\equal{}\\minus{}1$. Then using the well-known lemma that Virgil mentions finishes the lemma.", "Solution_13": "sorry for spam but I think those kinds problem use complex number is more beutiful than a solution with many lemma or some strange geometry (at me, that is harmonic for example, sorry for my bad knowledge about geometry :( ). I admire darij so much because in his solution often use many lemma :) \r\n if we call the orgin of the circle is O. we have: $ q = \\frac {2ac}{a + c}$; $ p = \\frac {2bc}{b + c}$. $ r = \\frac{2ab}{a + b}$ ${ c_1= \\frac {1}{2}(a + b + c - ab/c}$.\r\nso, the proof is complete. (please notice that $ \\angle BC_1C$ congruent with $ \\angle QC_1C(mod \\pi)$ :wink:", "Solution_14": "[hide = 2-line Projective Solution]\n\nFirst note that $AP$, $BQ$, and $CR$ concur at the Gergonne Point of $\\triangle PQR$. Then, let $X = AB \\cap PQ$ -> a well-known lemma implies that $(X , C ; P , Q) = -1$, and since $\\angle XC_1C = 90$ (by definition), we have $CC_1$ bisects $\\angle AC_1P$ by another well-known lemma.\n\n[/hide]", "Solution_15": "Dear Mathlinkers,\n\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=503523\n\nSincerely\nJean-Louis" } { "Tag": [ "geometry", "parallelogram", "conics", "parabola", "search", "rotation", "geometry unsolved" ], "Problem": "Let $ ABCD$ be a parallelogram with no angle equal to $ 60^{\\circ}$. Find all pairs of points $ E, F$ that lie in the plane of $ ABCD$ and are such that triangles $ AEB$ and $ BFC$ are isosceles, with bases $ AB,BC$, and triangle $ DEF$ is equilateral.", "Solution_1": "It will something like the intersection points of two parabolas, but I'm not sure when is easy to solve for that fourth degree equation...\r\n\r\nMay be there is another aproach...\r\n\r\n\r\n :)", "Solution_2": "It is equivalent to construct an equilateral triangle $ \\bigtriangleup DEF,$ whose the vertices $ E,$ $ F,$ lie on two constant lines, here the midperpendiculars of the side-segments $ AB,$ $ BC$ respectively, of the given parallelogram $ ABCD.$\r\n\r\nIn my drawing, the vertex $ D,$ lies inwardly to the angle $ \\angle MPN,$ where $ M,$ $ N,$ are the midpoints of $ AB,$ $ BC$ and $ P$ is the intersection point, of their midperpendiculars. In this case, I remember that we can construct, by an elementary way, two triangles as the problem states.\r\n\r\nI will search my notes and if I am not mistaken, I will post here later more details.\r\n\r\nKostas Vittas.", "Solution_3": "rotate 60 degrees about D and intersect the image of one perpendicular bisctor with the original version of hte other", "Solution_4": "[b][size=100][color=DarkBlue]GENERAL PROBLEM. \u2013 An angle $ \\angle XOY$ is given and let $ P$ be, an arbitrary point in the plane. Construct an isosceles triangle $ \\bigtriangleup ABC$ with $ PA \\equal{} PB$ and $ \\angle APB \\equal{} \\angle \\omega,$ whose the vertices $ A,$ $ B,$ lie on the lines $ OX,$ $ OY,$ respectively[/color][/size][/b].\r\n\r\nFor the solution, I post here only the schema and I think it is enough.\r\n\r\nKostas Vittas.", "Solution_5": "Well, vittasko, is this really the general problem of the original one?", "Solution_6": "[quote=\"easternlatincup\"]Well, vittasko, is this really the general problem of the original one?[/quote]\r\nI think yes. We can consider the lines $ OX,$ $ OY,$ as the midperpendiculars of the side-segments $ AB,$ $ BC$ and the point $ P,$ as the vertex $ D,$ of the parallelogram $ ABCD,$ in the original problem and $ \\angle \\omega \\equal{} 60^{o}.$\r\n\r\n$ ($ then the labels $ A,$ $ B,$ of the general problem, are corresponded with the $ E,$ $ F$ ones, in your problem $ ).$ \r\n\r\nNever mind, if the vertex $ D$ of $ ABCD,$ lies inwardly to the angle formed by the midperpendiculars of $ AB,$ $ BC.$\r\n\r\nKostas Vittas.", "Solution_7": "[quote=\"vittasko\"]...Never mind, if the vertex $ D$ of $ ABCD,$ lies inwardly to the angle formed by the midperpendiculars of $ AB,$ $ BC.$[/quote]\r\nIn the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=163344]Find a locus[/url], it has been mentioned a construction by the same way, when the point $ P,$ lies inwardly to the angle $ \\angle XOY.$\r\n\r\nAn interesting remark about this simple construction, based on the [b][size=100]Morley\u2019s theorem[/size][/b], is that we can construct infinity geometrically trisectable angles.\r\n\r\nKostas Vittas.", "Solution_8": "I wonder, Kostas, how you know the two triangles you construct are all there are.", "Solution_9": "[quote=\"easternlatincup\"]I wonder, Kostas, how you know the two triangles you construct are all there are.[/quote]\r\nI don't know how to construct other equilateral triangles, as your problem states.\r\n\r\nSo, I didn't search completely, if the two triangles constructed by the way I mentioned, are all there are.\r\n\r\nI also wonder if may be constructed more than two equilateral triangles, as your problem states.\r\n\r\nKostas Vittas.", "Solution_10": "Yes, there are two equilateral triangles, but how do you prove there are no more than these two?" } { "Tag": [], "Problem": "1. How does the mass percentage of carbon in a polyamide, for example (-HNCH2CO^-)x, compare with the mass percentage in a typical monomer, for example H2NCH2CO2H, from which it is formed?\r\n*The numbers are subscripts.\r\nA.The percentage of carbon in the polymer is greater.\r\nB. The percentage of the carbon in the polymer is less.\r\nC. The percentages are equal.\r\nD. There is not enough information to make this comparison.\r\nI really have no clue about this problem. Please help.\r\n\r\n\r\n2. Which one of the following compounds would be expected to show optical activity?\r\nA.CH3CH=CHCOOH\r\nB.HOOCCH2CH(CH3)CH2COOH\r\nC.CH3CH(OH)COOH\r\nD. HOCH2C(CH3)2CH2OH\r\n*The numbers are subscripts.\r\nI'm really confused about optical activity. I'd appreciate any help.[/code]", "Solution_1": "I don't know about the first one.\r\n\r\n[quote=\"RHSLilSweetie07\"]2. Which one of the following compounds would be expected to show optical activity?\nA.CH3CH=CHCOOH\nB.HOOCCH2CH(CH3)CH2COOH\nC.CH3CH(OH)COOH\nD. HOCH2C(CH3)2CH2OH\n*The numbers are subscripts.[/quote]\r\n\r\nI think an organic substance has optical isomers when a carbon is attached to groups that are not the same as any other. I think C would be the correct answer.", "Solution_2": "For number 1, note that when you form the polymer, you remove two hydrogen atoms, and leave all the other atoms unchanged (except for the end units, where only one hydrogen is removed). See if you can figure out the answer given that hint.\r\n\r\nFor 2, I would say it is a bad question. Optical activity means that you have a mixture with an excess of one enantiomer. One way of thinking about an object having an enantiomer is if you hold it up to a mirror, you do not get the same image back--for example, your hand. Zakary is right in that a compound with a carbon attached to four different groups usually has an optical isomer (a synonym for enantiomer). This isn't actually true in general, because there could be a carbon somewhere else in the molecule that is the other one's mirror image.\r\n\r\nIn short, optical activity is a property of a mixture, while existence of an enantiomer is a property of a compound. The answer that is being looked for is indeed C, however.", "Solution_3": "Huh, I didn't realize that, gregx007. Thanks!", "Solution_4": "how do you find the number of isomers of a hydrocarbon?", "Solution_5": "Try every appropriate combination of the atoms involved in the structure? How difficult that would be depends on the complexity of the structure...", "Solution_6": "Yep, there's no easy formula. Doing it systematically (least branched to most branched) and if you have time, writing out the IUPAC name to make sure you didn't accidentally duplicate one are usually the best things you can do to help yourself." } { "Tag": [ "quadratics", "calculus", "integration", "modular arithmetic", "number theory" ], "Problem": "Can you raise a number to a fractional power in mods, i.e. take the square root? I don't think you can, but can anyone prove it?", "Solution_1": "Anyone?", "Solution_2": "Try looking up [url=http://en.wikipedia.org/wiki/Quadratic_residue]Quadratic Residues[/url].", "Solution_3": "Uhhh.. does that mean that you can take square roots of quadratic residues mod p? :?", "Solution_4": "I don't get it.. can someone explain, please? :?", "Solution_5": "Yes, because one of the six rules of mods is that you can raise both sides to any power, and fractions count as powers, therefore you can take roots.", "Solution_6": "The rule is [b]integral[/b] powers.\r\n\r\nIf we have $36 \\equiv 25 \\mod {11}$, then raising both sides to $\\frac{1}{2}$ power, we have $6 \\equiv 5 \\mod {11}$, which is certainly not true. ;)", "Solution_7": "[quote=\"chess64\"]The rule is [b]integral[/b] powers.\n\nIf we have $36 \\equiv 25 \\mod {11}$, then raising both sides to $\\frac{1}{2}$ power, we have $6 \\equiv 5 \\mod {11}$, which is certainly not true. ;)[/quote]\r\n\r\nahh...I see, yes, I thought that the rule applies to $all$ powers.", "Solution_8": "[quote=\"chess64\"]The rule is [b]integral[/b] powers.\n\nIf we have $36 \\equiv 25 \\mod {11}$, then raising both sides to $\\frac{1}{2}$ power, we have $6 \\equiv 5 \\mod {11}$, which is certainly not true. ;)[/quote]\r\n\r\nActually if A^2 = B^2 then A can be either B or -B, here too, $6 \\equiv -5 \\bmod 11 $", "Solution_9": "[quote=\"chess64\"]Can you raise a number to a fractional power in mods, i.e. take the square root? I don't think you can, but can anyone prove it?[/quote]Well I can't answer that question, but I just learned from mathworld that you can have rational numbers in congruences. For example, $2\\cdot4\\equiv1\\pmod{7}\\Longrightarrow2\\equiv\\frac{1}{4}\\pmod{7}$. I didn't know that. If you want to find out about rational roots in congruences, maybe you should take an algebraic number theory course. :D (I actually have no idea what algebraic number theory is, but it could have to do with that.)" } { "Tag": [ "college", "inequalities proposed", "inequalities" ], "Problem": "Let$a_1,a_2,...,a_{2000}$ be real numbers in the interval$[0,1]$.\r\n\r\nFind the Maximum possible value of \r\n\r\n$\\sum_{1\\leq i$ AD \\equal{} \\frac {ac}{b \\plus{} c}.\\frac {sinB}{sin(\\frac {A}{2})}$\r\n=>$ l_a\\equal{}\\frac{2ac}{b\\plus{}c}.cos(\\frac{A}{2}).\\frac{sinB}{sinA}$\r\n=>$ l_a \\equal{} \\frac {2bc}{c \\plus{} b}.cos(\\frac {A}{2})$", "Solution_2": "[hide=\"Hint\"]Extend $ AD$ until it hits the circumcircle of $ \\triangle ABC$.[/hide]", "Solution_3": "[hide=\"Another way\"]\n\nStewart's theorem [/hide]", "Solution_4": "I feel stewart's theorem may be useful here. But i'm probably wrong...", "Solution_5": "[quote=\"Bugi\"][hide=\"Another way\"]\n\nStewart's theorem [/hide][/quote]\r\n\r\nEh? :P", "Solution_6": "OK, full solution\r\n\r\n[hide=\"Solution\"]\n\nAD=l - bisector\nBD=m, CD=n\n\n$ m\\equal{}\\frac{ac}{b\\plus{}c},n\\equal{}\\frac{ab}{b\\plus{}c}$\n\nRecall Stewart's theorem: $ b^2m\\plus{}c^2n\\equal{}a(l^2\\plus{}mn)$, rearrange and express $ l$. Satisfied? :P \n\n[/hide]", "Solution_7": "Oops, it was directed at Poincare, not you. I didn't realize the ambiguity until now. :blush:", "Solution_8": "Lol...That was funny :D \r\n\r\nThis was covered in volume 2 i think..." } { "Tag": [ "topology", "superior algebra", "superior algebra unsolved" ], "Problem": "hi can you help me?", "Solution_1": "Either you are equal to sepidetaghizade or you should ask him/her to build a group on this...\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=119096" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ n\\geq2$ be a positive integer and for real numbers $ a,b$ we have:\r\n\r\n$ a^{n}=a+1,\\;\\; b^{2n}=b+3a$\r\n\r\nwhich one is biGger: $ a$ or $ b$?", "Solution_1": "This Problem purposed in American Mathematical Olympiads and the answer:", "Solution_2": "tha was really nice!Beautiful solution Navid :lol:" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Find all permutations $ (\\mathop x\\nolimits_1,\\mathop x\\nolimits_2,\\mathop x\\nolimits_3,\\mathop x\\nolimits_4,\\mathop x\\nolimits_5,\\mathop x\\nolimits_6,\\mathop x\\nolimits_7,\\mathop x\\nolimits_8,\\mathop x\\nolimits_9)$ of $ (1,2,3,4,5,6,7,8,9)$ such that for any $ i>j,\\left| {\\left. {\\mathop x\\nolimits_i \\minus{} \\mathop x\\nolimits_j} \\right|} \\right.\\ne i\\minus{}j$", "Solution_1": "by using the help of mathematica\r\n\r\nlista[n_] := lista[n] = Flatten[Table[{i, j}, {\r\n i, 1, n - 1}, {j, i + 1, n}], 1]\r\nperm[n_] := perm[n] = Permutations[Range[n]]\r\nchequeo[per_, {j_, i_}] := Abs[per[[i]] - per[[j]]] == i - j\r\ncheq[per_, lista_] := {Or @@ ((chequeo[per, #]) & /@ lista ), per}\r\nf[n_] := Table[cheq[Part[perm[n], a], lista[n]], {a, 1, n!}] \r\nCases[ f[9], {False, b_}] // ColumnForm\r\n\r\nThere are $ 352$ cases...however it is very interesting that there only $ 4$ permutations of $ \\{1,2,3,4,5,6\\}$ with that property.", "Solution_2": "How would one go about solving this for smaller sets? Such as the 6 element set.", "Solution_3": "please show me one permutation please mszew\r\n\r\nthank you", "Solution_4": "[quote=\"Fermat's Little Turtle\"]How would one go about solving this for smaller sets? Such as the 6 element set.[/quote] He showed precisely the computer code he used to do it :wink: \r\n\r\nEquivalently, the question asks for the number of ways of placing 9 nonattacking queens on a 9 by 9 chessboard. The corresponding sequence is [url=http://www.research.att.com/~njas/sequences/A000170]A000170[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].\r\n\r\nOne example of such a permutation is 136824975.", "Solution_5": "@ricardo:\r\n\r\nhere comes all $ 4$ six element permutations:\r\n{2, 4, 6, 1, 3, 5}\r\n{3, 6, 2, 5, 1, 4}\r\n{4, 1, 5, 2, 6, 3}\r\n{5, 3, 1, 6, 4, 2}\r\n\r\nand some of the nine element permutations:\r\n{1, 3, 6, 8, 2, 4, 9, 7, 5}\r\n{1, 3, 7, 2, 8, 5, 9, 4, 6}\r\n{1, 3, 8, 6, 9, 2, 5, 7, 4}\r\n{1, 4, 2, 8, 6, 9, 3, 5, 7}", "Solution_6": "Another problem\r\n\r\nFind all permutations $ (\\mathop x\\nolimits_1,\\mathop x\\nolimits_2,\\mathop x\\nolimits_3,\\mathop x\\nolimits_4,\\mathop x\\nolimits_5,\\mathop x\\nolimits_6,\\mathop x\\nolimits_7,\\mathop x\\nolimits_8,\\mathop x\\nolimits_9)$ of $ (1,2,3,4,5,6,7,8,9)$ such that for any $ i > j,\\left| {\\left. {\\mathop x\\nolimits_i \\minus{} \\mathop x\\nolimits_j} \\right|} \\right.\\ne i \\minus{} j$ or $ 9 \\minus{} (i \\minus{} j)$\r\n\r\nthe answer is no such the permutation .\r\n\r\nCan one show me any solutions please ?", "Solution_7": "Change the code for the aditional condition and there no such permutations, however it remains interesting that the same $ 4$ six number permutation fullfils that aditional criteria.\r\n\r\nThere are 2,10,4,4,0,0 for the 5,6,7,8 and 9 permutations.", "Solution_8": "Any solutions for problem 2 please", "Solution_9": "somebody helps me please" } { "Tag": [ "geometry", "geometric transformation", "dilation" ], "Problem": "Hey guys,\r\nDoes anyone know in what way i should organize my CV / resume from high-school?\r\nCan you give me some templates/examples?\r\nThanks in advance, \r\n:)", "Solution_1": "Here are some things to include.\r\n\r\nAfter reading them, note that you should not go over 2 pages -- if you do, you are almost certainly diluting the good stuff. In the resume, it is [b]your[/b] job (not admissions') to determine the highlights of your experiences and accomplishments. Also, make your margins narrower if you must, and use size-10 font for the body.\r\n\r\n[b]Header:[/b]\r\n- Name\r\n- Date of Birth\r\n- SSN (for identification purposes -- don't leave temporary copies with your SSN lying around)\r\n> Header should be grey or dark grey, but not black.\r\n\r\n[b]Top:[/b]\r\n- Title (Resume, Curriculum Vita/Vitae/Vit\u00e6)\r\n- Name\r\n- website url (if you have one that you want to post)\r\n- optionally, contact information (if you are attaching your resume on Common App, as additional information, it is not very useful to post all your contact info. Otherwise, it can make it look more professional).\r\n\r\n[b]Body:[/b] I would suggest the following sections (if applicable), in the order listed:\r\nI. Education\r\n- List your high schools and the dates.\r\n- This is important if you have more than one school; that way, admissions officers won't superficially look and note that you weren't in any particular school organization for 4 years, and will (hopefully) take more time to absorb the things they see.\r\n- If you only went to one high school, but have a highly competative GPA and/or class rank, you can still put an education section, include the one high school, and your rank and GPA. Coherence is a virtue.\r\n\r\nII. Employment / Internships / Employment and Internships / Work Experience (if this section is substantial)\r\n- Pick the title that is most relevant.\r\n- Order in reverse chronological order, with the most recent jobs first\r\n- For each, include the title of your position (if applicable), a short description of responsibilities, achievements (NOT how much money you made), contributions, skills learned, or anything else that adds dimension to your resume\r\nIMPORTANT CAVEAT - If you started your own business for fun, don't be like, \"Blah blah blah, Inc: Founder and CEO\" \"I oversee all operations and make executive business decisions\", unless there is substance behind the words, because that sounds stupid and childish. Don't say you were in charge of client relations or something if you had 3 clients. Don't say you were in charge of advertising if \"advertising\" means posting it on eBay.\r\n\r\nIII. Community Service\r\n- Include only if you have a significant amount of this. Don't put too much fluff here. If you have only one or two things here, place them under Activities instead.\r\n- Organize in reverse chronological order. Include what you did/achieved/gained/contributed and approximate time investment.\r\n\r\nIV. Activities\r\n- This should be organized in the most relevant way\r\n- If you have SIGNIFICANT athletic experience, you can create a section \"Athletics\" preceeding this section.\r\n- If you want to focus on a few major ones, than write down the names of the major clubs first, along with a description of what you did/accomplished/gained/contributed, much like for the jobs, as well as the dates. Then, list less significant ones without detail.\r\n- Alternately, list all your activities temporally, and include .5 - 2 lines briefly describing the above for each club.\r\nIMPORTANT NOTE: Don't put down stupid things. Some people think that adding a minor activity will either help them or just be overlooked, and cannot hurt. This is wrong. If something contributes to your resume, put it in. If it does not, then it is just diluting the rest of your resume!\r\n\r\nV. Honors and Awards\r\n- Important awards/distinctions you have achieved. Be very discriminating in what you put!\r\nIMPORTANT NOTE: Same as for Activities, but even more important because there is little to describe for awards, usually. It is better to list 2 or 3 important awards alone, than to list 2 or 3 important awards scatterd among 7 unimportant awards.\r\n\r\nVI. Other Skills / {any skills you want to include}:\r\n- Use own discretion on whether or not to include it\r\n- Language proficiencies\r\n- Computer skills (mostly for programming languages. Software familiarity is probably irrelevant for applying to college, though very important for applying for jobs)\r\nNOTE: Do NOT include sports, art, or hobbies here.\r\n\r\nVII. Hobbies\r\n- Hobbies go here.\r\n- Use own discretion on whether or not to include it\r\n- Things that you chose not to put into activities because they would dilute your good stuff might go here.\r\nEXAMPLE: Someone who is heavily into math and science and has a random snowboarding club experience might rather put down snowboarding as a hobby instead of putting it in with their list of activities, which can dilute the list and/or look like fluff.\r\n- Include (if it helps you) how long you've been involved", "Solution_2": "High School Resume Template Tips\r\n\u00b7 If you have work experience, then highlight your responsibilities and achievements relevant to the job you are applying for. For example if you are going for a customer service job, highlight your customer service skills.\r\n\u00b7 If you do not have any work experience simply remove this heading from your resume or CV and focus more on Relevant Skills. List your attributes and capabilities in order of what you know the employer wants. \r\n\u00b7 Limit your resume or CV to maximum 2 sides of paper. \r\n\u00b7 Finally proofread your resume." } { "Tag": [ "algebra", "system of equations" ], "Problem": "If $ a \\plus{} b \\plus{} c \\equal{} 2$, and $ a \\times b \\plus{} 2 \\times c \\equal{} 0$\r\n\r\nthen $ a^3 \\plus{} b^3 \\plus{} c^3 \\equal{} ?$", "Solution_1": "hello, by solving the system $ a\\plus{}b\\plus{}c\\equal{}2,ab\\plus{}2c\\equal{}0$ we get $ a\\equal{}\\minus{}c,b\\equal{}2,c\\equal{}c$ or $ a\\equal{}2,b\\equal{}\\minus{}c,c\\equal{}c$ so we have \r\n$ a^3\\plus{}b^3\\plus{}c^3\\equal{}\\minus{}c^3\\plus{}8\\plus{}c^3\\equal{}8$.\r\nSonnhard.", "Solution_2": "$ \\left\\{\\begin{array}{l}\r\na\\plus{}b\\plus{}c \\equal{} 2 \\\\\r\na b \\plus{} 2 c \\equal{} 0\r\n\\end{array}\\right.\r\n\\Rightarrow\r\n\\left\\{\\begin{array}{l}\r\n(a, b, c) \\equal{} (\\minus{}c, 2, c) \\\\\r\n\\text{or} \\\\\r\n(a, b, c) \\equal{} (2, \\minus{}c, c)\r\n\\end{array}\\right.$\r\n\r\nSo, $ a^3\\plus{}b^3\\plus{}c^3 \\equal{} 8$", "Solution_3": "[hide]\n\nI basically solved the system of equations to get:\nb = 2\na = -c\nc = c\n\nFrom there you do:\n\n$ \\minus{}c \\plus{} 2 \\plus{} c \\equal{} 2$\nand \n$ c \\equal{} 0$\n\n\nso:\n\n$ \\minus{}0^3 \\plus{} 8 \\plus{} 0^3 \\equal{} \\boxed{8}$\n\n[/hide]", "Solution_4": "Substitute for c in the second equation, you get\r\n$ ab \\minus{} 2a \\minus{} 2b \\plus{} 4 \\equal{} 0$\r\n$ (a \\minus{} 2)(b \\minus{} 2) \\equal{} 0$\r\n$ a \\equal{} 2$ or $ b \\equal{} 2$\r\nThis is how you solve the system." } { "Tag": [ "MATHCOUNTS" ], "Problem": "Add seven different fractions with numerator 1 to get 1.", "Solution_1": "[hide]1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/48 + 1/96 = 1[/hide]\r\nThere are actually many possibilities...:D", "Solution_2": "does[hide=\"this work?\"]1/7+1/7+1/7+1/7+1/7+1/7+1/7=1[/hide]", "Solution_3": "oh, it doesn't work because all the fractions have to be different. Well, lightspeed's solution works.", "Solution_4": "Lightspeed's solution works. Math92's was very creative. :lol:", "Solution_5": "Do they all have to be different? I'd enjoy doing this:\r\n[hide]$\\frac{1}{6}+\\frac{1}{6}+\\frac{1}{6}+\\frac{1}{6}+\\frac{1}{6}+\\frac{1}{12}+\\frac{1}{12}$[/hide]", "Solution_6": "[quote=\"math92\"]oh, it doesn't work because all the fractions have to be different. Well, lightspeed's solution works.[/quote]\r\n\r\nOtherwise, math92's would work", "Solution_7": "If math92's solution was a valid answer to this problem, I wouldn't have posted it in the mathcounts forum.", "Solution_8": "Yeah it would be a little too easy then.", "Solution_9": "Yeah, that's true.", "Solution_10": "So, lightspeed's solution is one of the answers.", "Solution_11": "[hide]1/2+1/3+1/12+1/24+1/48+1/72+1/144[/hide]\r\nyour right, there ARE a lot of solutions. In fact, there's an infinite # of them.", "Solution_12": "can u prove it?" } { "Tag": [ "geometry", "function", "integration", "real analysis", "real analysis theorems" ], "Problem": "Prove that the area under the curve from a to b is equal to the antiderivative of b minus the antiderivative of a please.", "Solution_1": "You need to define what you call area and you need to give hypotheses on $f$.", "Solution_2": "for what type of function?\r\n\r\ni think it'd be too hard to prove it for all functions.", "Solution_3": "$F(b) - F(a) = \\int^b_a f(x) dx$\r\n\r\nwhere $f(x) = \\frac{F(x+dx)-F(x)}{dx}$\r\n\r\nthat's the best i can do in terms of a proof, but what that means is f(x) how much F(x) changes. so F(b) minus F(a) is the cumulitive change from a to of f(x), which can be reguarded as an area." } { "Tag": [ "inequalities", "floor function", "function", "integration", "logarithms", "ceiling function", "calculus" ], "Problem": "For a natural number $ n$ and $ n>1$ show that $ \\frac{1}{n}\\plus{}\\frac{1}{n\\plus{}1}\\plus{}\\frac{1}{n\\plus{}2}\\plus{}....\\plus{}\\frac{1}{n^{2}}>1$.\r\n\r\nI have solved thsi but I don't like my proof. So can anyone out her give me a nice and ingenious one?", "Solution_1": "[hide=\"Hint\"]$ AM>HM$[/hide]", "Solution_2": "I tried but what I got was very clumsy! Any other approach?", "Solution_3": "[hide=\"Solution\"]By $ \\text{AM} > \\text{HM}/\\text{Cauchy Schwarz}$\n$ \\left(\\sum_{i\\equal{}n}^{n^2} i\\right)\\left(\\sum_{i\\equal{}n}^{n^2} \\frac{1}{i}\\right)>\\left(\\sum_{i\\equal{}n}^{n^2} 1\\right)^2$\n(Equality doesn't occur since all terms are unequal.)\nSo we need to prove that\n$ \\left(\\sum_{i\\equal{}n}^{n^2} 1\\right)^2\\geq \\left(\\sum_{i\\equal{}n}^{n^2} i\\right) \\iff (n^2\\minus{}n\\plus{}1)^2\\geq \\left(\\sum_{i\\equal{}1}^{n^2} i\\minus{}\\sum_{i\\equal{}1}^{n\\minus{}1} i\\right)$\n$ \\iff (n^2\\minus{}n\\plus{}1)^2\\geq \\left(\\frac{n^2(n^2\\plus{}1)}{2}\\minus{}\\frac{(n\\minus{}1)n}{2}\\right)$\n$ \\iff 2(n^2\\minus{}n\\plus{}1)^2\\geq (n^4\\plus{}n^2\\minus{}n^2\\plus{}n)$\n$ \\iff 2(n^2\\minus{}n\\plus{}1)^2\\geq n(n\\plus{}1)(n^2\\minus{}n\\plus{}1)$\n$ \\iff 2(n^2\\minus{}n\\plus{}1)\\geq n(n\\plus{}1)$\n(Note that $ n^2\\minus{}n\\plus{}1>0$)\n$ \\iff n^2\\minus{}3n\\plus{}2\\geq 0\\iff (n\\minus{}1)(n\\minus{}2)\\geq 0$\nThis is obvious for $ n>1$[/hide]", "Solution_4": "Thanks!Its better than mine!", "Solution_5": "We have from Cauchy-Schwarz\r\n\\begin{align*}\\sum_{i=n}^{n^2}\\frac 1i&=\\frac 12\\sum_{k=0}^{n^2-n}\\left(\\frac{1}{n+k}+\\frac{1}{n^2-k}\\right)\\\\\r\n&\\ge\\frac 12\\sum_{k=0}^{n^2-n}\\frac{4}{n^2+n}\\\\\r\n&=\\frac{2(n^2-n+1)}{n^2+n}\\ge 1\\end{align*}\r\nwhere the last inequality follows from $ 2n^2-2n+2\\ge n^2+n\\iff (n-1)(n-2)\\ge 0$, which is true. And since equality does not hold in the Cauchy-Schwarz for $ n=2$, equality never holds.", "Solution_6": "Can we relplace $ n^{2}$ by a $ k$ and which is the smallest value of $ k$ such that the inequality holds.", "Solution_7": "I don't understand what u said. Are u keeping the ineq intact and Substituting $ n^2$ with a smaller $ k$? :maybe:", "Solution_8": "Yes, I am and I want to know which is the smallest such $ k$?", "Solution_9": "[quote=\"manjil\"]Can we relplace $ n^{2}$ by a $ k$ and which is the smallest value of $ k$ such that the inequality holds.[/quote]\r\n[hide=\"Conjecture\"]$ \\lfloor en \\minus{} 1\\rfloor$\nWhere $ e$ is the exponential constant[/hide]", "Solution_10": "That is very interesting indeed!\r\nWhen we take $ n\\equal{}2$ it shows that $ \\frac{1}{2}\\plus{}\\frac{1}{3}\\plus{}\\frac{1}{4}>1$ i.e., the $ k\\equal{}n\\plus{}2$.\r\n\r\nAgain when we take $ n\\equal{}3$ it shows that $ \\frac{1}{3}\\plus{}\\frac{1}{4}\\plus{}\\frac{1}{5}\\plus{}\\frac{1}{6}\\plus{}\\frac{1}{7}>1$ i.e. the $ k\\equal{}n\\plus{}4$ now!\r\n\r\nWhen we take $ n\\equal{}4$, it shows that $ \\frac{1}{4}\\plus{}\\frac{1}{5}\\plus{}\\frac{1}{6}\\plus{}\\frac{1}{7}\\plus{}\\frac{1}{8}\\plus{}\\frac{1}{9}\\plus{}\\frac{1}{10}>1$\r\nSo lastly our $ k\\equal{}n\\plus{}6$!\r\n\r\nI think the generalisation follows now, doesn't it! :lol: \r\nFor any $ m>1$ the desired $ k$ is expected to be $ k\\equal{}m\\plus{}2(m\\minus{}1)$, I hope! :) \r\n\r\nBut I haven't proved it yet. :(", "Solution_11": "[quote=\"Sunkern_sunflora\"]I think the generalisation follows now, doesn't it! :lol: [/quote]\nUnfortunately it doesn't :!: \n[quote=\"Sunkern_sunflora\"]For any $ m > 1$ the desired $ k$ is expected to be $ k \\equal{} m \\plus{} 2(m \\minus{} 1)$, I hope! :) [/quote]\r\n$ \\sum _{i \\equal{} 6}^{15}\\frac {1}{i}\\approx 1.034896 > 1$\r\nSo for $ n\\equal{}6$, $ 15$ is the minimum $ k$\r\nBut your formula says that the minimum is $ 16$", "Solution_12": "$ k\\leq(3n\\plus{}1)$. Now guys you can do better!", "Solution_13": "[quote=\"manjil\"]$ k\\leq(3n \\plus{} 1)$. Now guys you can do better![/quote]\r\nMy formula is true even at $ 10000$. And I think it is always true. We just need to prove it.", "Solution_14": "[hide=\"Narrow range of k proved\"]Since $ \\frac{1}{x}$ is a decreasing function\n$ \\int_{m}^{m\\plus{}1} \\frac{1}{x} dx\\leq \\frac{1}{m}\\leq \\int_{m\\minus{}1}^{m} \\frac{1}{x} dx$ for $ m>1$\n$ \\int_{a}^{b\\plus{}1}\\frac{1}{x} dx\\leq \\sum _{i\\equal{}a}^{b} \\frac{1}{i}\\leq \\int_{a\\minus{}1}^{b}\\frac{1}{x} dx$\nfor integers $ b>a>1$\nSuppose $ k$ is the required minimum.\nWe need to have \n$ \\sum _{i\\equal{}n}^{k} \\frac{1}{i}\\geq 1$\n[b]and[/b] \n$ \\sum _{i\\equal{}n}^{k\\minus{}1} \\frac{1}{i}\\leq 1$\nWe know that \n$ \\sum _{i\\equal{}n}^{k} \\frac{1}{i}\\leq \\int_{n\\minus{}1}^{k}\\frac{1}{x} dx\\equal{}\\ln k\\minus{}\\ln (n\\minus{}1)\\equal{}\\ln (\\frac{k}{n\\minus{}1})$\nand\n$ \\sum _{i\\equal{}n}^{k\\minus{}1} \\frac{1}{i}\\geq \\int_{n}^{k}\\frac{1}{x} dx\\equal{}\\ln k\\minus{}\\ln n\\equal{}\\ln (\\frac{k}{n})$\nSo \n$ \\ln (\\frac{k}{n\\minus{}1})\\geq 1\\implies \\frac{k}{n\\minus{}1}\\geq e\\implies k\\geq en\\minus{}e$\n[b]and[/b] \n$ \\ln (\\frac{k}{n})\\leq 1\\implies \\frac{k}{n}\\leq e\\implies k\\leq en$\nSo $ en\\geq k\\geq en\\minus{}e$\nThis narrows down our value of $ k$ [b]very much[/b].[/hide]", "Solution_15": "For n = 2, 3 adn 4 it is right. For n>4 we have 1/n+...+1/n^2 > 1/n+...+1/(4n+2) = (1/n+...+1/2n) + (1/(2n+1)+...+1/(4n+2)) > n/2n + (2n+1)/(4n+2) = 1. Q.E.D.", "Solution_16": "Brilliant Akashnil! I thought it would follow generalisation, unfortunately it doesn't! :( \r\nAnyway, your proof really narrows down $ k$ a lot. I think combining your proof and Manjil's conjecture, the problem gets solved as well.\r\n :)", "Solution_17": "[quote=\"Sunkern_sunflora\"]I think combining your proof and Manjil's conjecture, the problem gets solved as well.\n :)[/quote]\r\nIt was my conjecture. However my proof allows $ 2$ or $ 3$ values for $ k$ for all $ n$. Hopefully the interval doesn't get larger as $ n$ gets larger. However I still couldn't pinpoint it :(\r\nI have also proved that a stronger range $ en > k > en \\minus{} \\frac {e \\plus{} 1}{2}$ is possible using the convexity of $ \\frac{1}{x}$. But still there are $ 2$ possibility for $ k$ for some $ n$.\r\nHowever we have got a very nice and strong inequality $ \\sum_{i \\equal{} n}^{\\lceil en \\rceil} \\frac {1}{i} > 1$\r\nLets wait for someone like [t0rajir0u] to solve the problem of $ \\min k$ completely.", "Solution_18": "I thought I was a duffer that I could not solve the $ mink$ but it turns out that its really much more complex than I thought!\r\n\r\nAnyway I was trying to work out on a proof of the inequality without the use of any standard inequality and I have got something which I will post soon enough!\r\n\r\nAbout the $ min k$ problem. I have not been able to solve it completely and each time I think I have done it something occurs. Anyway I sincerely hope someone will be able to solve that!", "Solution_19": "Maybe we can solve the problem for minimum $ k$ by taking the rleft hand side as an integral of some function? It is just an idea, can anyone see if it will work?", "Solution_20": "Or else we can see if we can get some lower bounds for the left hand side expression taking that as a function of $ n$ and finding the minimum $ k$ which satisfies it that is find $ f(k)$!", "Solution_21": "[quote=\"manjil\"]Maybe we can solve the problem for minimum $ k$ by taking the rleft hand side as an integral of some function? It is just an idea, can anyone see if it will work?[/quote]\r\nHave you read post #15 in the thread?", "Solution_22": "Sorry Akashnil I missed that post!" } { "Tag": [ "articles", "LaTeX", "integration" ], "Problem": "Hi, \r\n\r\nOn most math books and articles theres always a label on the extreme right of every equations, like (1), [17]...\r\n\r\nHow can you do that in \\LaTeX\\ ?\r\n\r\nThx", "Solution_1": "Use [code]\\begin{equation} \\int_{-\\infty}^\\infty e^{-x^2}\\; dx = \\sqrt{\\pi} \\end{equation}[/code] to get automatic equation numbers. If you want to number them on your own, do it with [code]\\begin{equation*} \\Gamma(s)=\\int_0^\\infty e^{-t} t^{s-1} \\; dt \\tag{1}\\end{equation*}.[/code]", "Solution_2": "Thank you !\r\n\r\nIt's quite simple...\r\n\r\nHowever...\r\n\r\n[quote]\\begin{equation*} \\Gamma(s)=\\int_0^\\infty e^{-t} t^{s-1} \\; dt \\tag{1}\\end{equation*}.[/quote]\r\n\r\nThat does'nt work, really don't know why...", "Solution_3": "Sorry I wasn't thinking. You still use equation rather than equation* even when you want to choose the equation numbers. But then you also use tag. If you don't want the parentheses around the equation number, you can use \\tag*." } { "Tag": [ "geometry" ], "Problem": "Polygon $ ABCDEF$ is a regular hexagon. What is the measure in\ndegrees of angle $ ABF$?", "Solution_1": "This problem had an intended diagram, but it's missing. Unless someone finds the picture, it's impossible to solve.", "Solution_2": "Actually you don't need the picture, since its a regular hexagon.\r\n\r\nI know the answer is 30, don't know how to explain it.\r\n\r\nHint: The vertices are labeled in order (A then B then C, etc.)", "Solution_3": "Yeah, like ernie said, no diagram necessary.\r\n\r\nThe basic fact from geometry:\r\n\r\nFor a regular polygon of n sides, using degrees:\r\n$ \\text{\"sum of interior angles\"}\\equal{}(n\\minus{}2)\\times 180$\r\n\r\nF.E. triangle gets you 180, square gets you 360, hexagon 720 etc.\r\n\r\nThen each of the 6 interior angles of a regular hexagon is 120. the angle ABF is part of the isocoles triangle ABF. Of course the angle FAB is 120 degrees (because it is an interior angle), and since it is isocoles, the other two angles would have to be 30. (ABF is one of those angles :) )", "Solution_4": "Diagram:\r\n\r\n[asy]pair A=(0,0),\nB=A+4*dir(0),\nC=B+4*dir(60),\nD=C+4*dir(120),\nE=D+4*dir(180),\nF=E+4*dir(240);\nlabel(\"A\", E, N);\nlabel(\"B\", F, W);\nlabel(\"C\", A, S);\nlabel(\"D\", B, S);\nlabel(\"E\", C, W);\nlabel(\"F\", D, N);\ndraw (A--B--C--D--E--F--cycle);\ndraw(F--D);[/asy]\r\n\r\nTriangle ABF is isosceles, and the sum of x of f(y)", "Solution_1": "This is a fairly famous function. Does it have a name?", "Solution_2": "I'm pretty sure g(x) = 0 but I'm thinking how to make it rigorous.", "Solution_3": "I can tell you that you are [hide]correct[/hide], and I can even give you a suggestion for proving the answer:\n\n\n\n[hide]take the cases of irrational and rational numbers seperately. Irrational should be easy. Rational, a little trickier.[/hide]", "Solution_4": "OK here's how we find our delta (see http://mathworld.wolfram.com/Limit.html).\r\nfor an arbitrarily small epsilon. First we find a large enough r such that 1/r < epsilon. Let n = 1*2*3*...*r. Find the largest k such that k/n < x. Let delta = (x+k/n)/2. It is impossible to find an integer q0$, then $lim=0$;\r\n-If $\\beta=0$, $\\alpha>0$, then $lim=0$;\r\n-If $\\beta=0$, $\\alpha=0$, then $lim=\\frac{1}{2}$;\r\n-If $\\beta=0$, $\\alpha<0$, then $lim=1$;\r\n-If $\\beta<0$, $\\alpha\\geq 0$, then $lim=0$;\r\n-If $\\beta<0$, $\\alpha<0$, then let us take two sequences ${x_{n}}$, ${y_{n}}$ which tend to zero and satisfy ${x_{n}}^{2\\alpha}={y_{n}}^{3\\beta}$.\r\nThen $\\frac{{x_{n}}^{2\\alpha}*{y_{n}}^{3\\beta}}{{x_{n}}^{2\\alpha}+{y_{n}}^{3\\beta}}$ $\\rightarrow$ $+\\infty$ as $n\\rightarrow+\\infty$." } { "Tag": [ "geometry", "Euler", "geometric transformation", "reflection", "circumcircle", "angle bisector", "geometry proposed" ], "Problem": "A triangle $ABC$ is given. Let $L$ be its Lemoine point and $F$ its Fermat (Torricelli) point. Also, let $H$ be its orthocenter and $O$ its circumcenter. Let $l$ be its Euler line and $l'$ be a reflection of $l$ with respect to the line $AB$. Call $D$ the intersection of $l'$ with the circumcircle different from $H'$ (where $H'$ is the reflection of $H$ with respect to the line $AB$), and $E$ the intersection of the line $FL$ with $OD$. Now, let $G$ be a point different from $H$ such that the pedal triangle of $G$ is similar to the cevian triangle of $G$ (with respect to triangle $ABC$). Prove that angles $ACB$ and $GCE$ have either common or perpendicular bisectors.", "Solution_1": "Here is something similar to a solution; it would be a solution if it wouldn't use lots and lots of facts not all of which have been synthetically proven.\r\n\r\nFirst let's identify the points we are talking about, with their numbers in the ETC ([3]).\r\n\r\n- The point L is the Lemoine point, also known as symmedian point of triangle ABC. It is $X_6$ in the ETC.\r\n\r\n- The point F is the Fermat point, or, more precisely, the first Fermat point of triangle ABC. It is $X_{13}$ in the ETC.\r\n\r\n- The orthocenter H of triangle ABC is $X_4$, and the circumcenter O is $X_3$ in the ETC.\r\n\r\n- The point D was defined as the point where the reflection $l^{\\prime}$ of the Euler line $l$ of triangle ABC in the line AB meets the circumcircle of triangle ABC (apart from the point H'). Actually, it is a well-known fact ([4]) that the reflections of the Euler line $l$ of triangle ABC in the lines BC, CA and AB concur at one point, which lies on the circumcircle of triangle ABC; this is the so-called [b]Euler reflection point[/b] of triangle ABC. This point can also be described as the point where the reflection $l^{\\prime}$ of the Euler line $l$ of triangle ABC in the line AB meets the circumcircle of triangle ABC (apart from the point H'). Hence, the point D coincides with the Euler reflection point of triangle ABC. This point is $X_{110}$ in the ETC.\r\n\r\n- The point G was defined as the point, different from the orthocenter H of triangle ABC, such that the pedal triangle of G is similar to the cevian triangle of G. According to [1], this yields that the point G is the isogonal conjugate of the Parry reflection point of triangle ABC. Actually, the Parry reflection point is $X_{399}$ in the ETC, and its isogonal conjugate G is $X_{1138}$.\r\n\r\nNote that here we have made the first gap in our synthetic proof: in fact, we have used the result of [1], which was proven using barycentric coordinates.\r\n\r\n- According to [2], the Parry reflection point of triangle ABC is the reflection of the circumcenter O of triangle ABC in the Euler reflection point D. Hence, this Parry reflection point lies on the line OD. On the other hand, according to the ETC ([3]), the Parry reflection point lies on the line FL. Hence, the Parry reflection point coincides with the point of intersection of the lines FL and OD. But this point of intersection is the point E; hence, we see that the point E is the Parry reflection point of triangle ABC. As the point G is the isogonal conjugate of the Parry reflection point of triangle ABC, it thus follows that the points E and G are isogonal conjugates of each other with respect to triangle ABC. Hence, the lines CG and CE are symmetric to each other with respect to the angle bisector of the angle ACB; in other words, the angle bisector of the angle ACB is also an angle bisector of the angle GCE (whether it's the internal or the external angle bisector, we don't know). And the problem is solved.\r\n\r\nHere we have made another gap in our synthetic reasoning, namely we have quoted from the ETC the fact that the Parry reflection point of triangle ABC lies on the line FL.\r\n\r\nAnybody to close the two gaps? I am tired now...\r\n\r\n[b]References[/b]\r\n\r\n[b][1][/b] [url=http://forumgeom.fau.edu/FG2003volume3/FG200310index.html]Jean-Pierre Ehrmann, [i]Similar pedal and cevian triangles[/i], Forum Geometricorum 3 (2003) p. 101-104[/url].\r\n\r\n[b][2][/b] [url=http://groups.yahoo.com/group/Hyacinthos/messages/6200?viscount=1]Darij Grinberg, [i]Hyacinthos message #6200[/i][/url] ([url=http://groups.google.de/group/geometry.college/browse_thread/thread/8260a080b44efaba/6899f6d12e2809c1]mirror at Googlegroups[/url]).\r\n\r\n[b][3][/b] [url=http://faculty.evansville.edu/ck6/encyclopedia/ETC.html]Clark Kimberling, [i]ETC - Encyclopedia of Triangle Centers[/i][/url].\r\n\r\n[b][4][/b] [url=http://de.geocities.com/darij_grinberg/]Darij Grinberg, [i]Anti-Steiner points with respect to a triangle[/i][/url].\r\n\r\n Darij" } { "Tag": [], "Problem": "If the reciprocal of (x - 2) is (x + 2), what is the greatest possible value of x? Express your answer in simplest radical form.", "Solution_1": "[quote=\"sugarwergoingdown\"]If the reciprocal of (x - 2) is (x + 2), what is the greatest possible value of x? Express your answer in simplest radical form.[/quote]\r\n\r\n[hide]$\\frac{1}{x-2}=x+2$\n$x^2-4=1$\n$x^2-5=0$\n$(x-\\sqrt{5})(x+\\sqrt{5})=0$\nMaximum value: $x=\\sqrt{5}$[/hide]", "Solution_2": "yah thats right. :lol:" } { "Tag": [ "analytic geometry", "geometry" ], "Problem": "1.Tangents are drawn to the circle x^2+y^2=12 at the points where it is met by the circle x^2+y^2-5x+3y-2=0 ; find the point of intersection of these tangents.\r\n\r\n2.Prove that along with (1,-2) there is another point the polars of which with respect to the below mentioned circles are the same and find its coordinates .\r\nS1=x^2+y^2+6y+5=0\r\nand\r\nS2=x^2+y^2+2x+8y+5=0\r\n\r\n\r\n3.Prove that the distance of two points P and Q each from the polar of the other with respect to a circle are to one another as the distances of the points from the centre of the circle.\r\n\r\n\r\n4.Prove that the polar of a given point with respect to any one of the circles x^2+y^2-2kx+c^2=0'where k is a variable ,always passes through a fixed point , whatever be the value of k.", "Solution_1": "First question:\r\n\r\nGiven the circle $ x^2 \\plus{}y^2 \\equal{}a^2$ and an external point T (h ,k). Let the tangents from T touch the circle at P and Q.\r\n\r\nThen the chord PQ has equation $ xh\\plus{}yk \\equal{}a^2$. PQ is called the chord of contact of tangents. \r\n\r\nDo you know this result and how to derive it? We will use it to solve your question.\r\n\r\nThe common chord of the two circles is $ (x^2 \\plus{}y^2 \\minus{}12)\\minus{}(x^2 \\plus{}y^2 \\minus{}5x\\plus{}3y\\minus{}2) \\equal{}0$ that is $ 5x\\minus{}3y\\minus{}10 \\equal{}0$. Let the two circles cut at P ,Q and let the tangents to $ x^2\\plus{}y^2\\equal{}12$ at these two points meet at T ( h, k ). Then by the piece of theory stated above PQ will have equation\r\n\r\n$ xh \\plus{}yk \\minus{}12\\equal{}0$ and because this is the line $ 5x\\minus{}3y\\minus{}10 \\equal{}0$ we have $ \\frac{h}{5}$ =$ \\frac{k}{\\minus{}3}$ =$ \\frac{\\minus{}12}{\\minus{}15}$ from which you can solve for h and k.", "Solution_2": "Question four\r\n\r\nWe use this result from coordinate geometry: The polar of a point (p ,q ) with respect to the circle $ x^2 \\plus{}y^2 \\plus{}2gx \\plus{}2fy \\plus{}a\\equal{}0$ is the line\r\n\r\n$ xp\\plus{}yq \\plus{}g(x\\plus{}p) \\plus{}f(y\\plus{}q) \\plus{}a\\equal{}0$ \r\n\r\nIn this question g =-k, f=0 and a=$ c^2$ so for the given circle the polar of the point (p,q ) will be \r\n\r\n$ xp \\plus{}yq \\minus{}k(x\\plus{}p) \\plus{}c^2 \\equal{}0$, that is $ xp\\plus{}yq\\plus{}c^2 \\minus{}k(x\\plus{}p) \\equal{}0$......(*) (p ,q and $ c^2$ are fixed and k is variable)\r\n\r\nThe lines $ xp\\plus{}yq\\plus{}c^2\\equal{}0$ and x+p=0 meet in the (fixed) point ( -p , $ \\frac{\\minus{}p^2\\minus{}c^2}{q}$) and regardless of k this point always lies on the line(*) .Thus we have shown that the polar always passes through a fixed point." } { "Tag": [ "function", "trigonometry", "geometry", "inequalities proposed", "inequalities" ], "Problem": "Show that [tex]|tan x|\\geq |x|, \\forall x: -\\frac{\\pi}{2} < x < \\frac {\\pi}{2}[/tex].", "Solution_1": "Since both $x$ and $\\tan(x)$ are odd functions of $x$, we need only show that $0 \\le x \\le \\tan(x)$ for $x \\in [0, \\frac{\\pi}{2})$. Since $\\tan(0) = 0$, $\\frac{d}{dx}\\tan(x) = \\frac{1}{\\cos^2(x)}$ and $\\frac{d^2}{dx^2}\\tan(x) = \\frac{2\\sin(x)}{\\cos^3(x)} \\ge 0$, we have that $\\frac{d}{dx}\\tan(x) \\ge 1$ for all $x$ in $[0,\\frac{\\pi}{2})$, the result follows immediately.", "Solution_2": "an elementary proof for $x< \\tan x$, $x \\in (0, \\frac{\\pi}{2} )$.\r\nTake $M$ the point corresponding to $x$ on the unit circle.\r\nlet $O$ be the origin and $A$ the point corresponding to $0$ on the unit circle.\r\nlet $T$ be the intersection point of $OM$ with $d$, where $d$ is a line passing through $A$ and paralel to $Oy$.\r\nBy definition, $AT=\\tan x$.\r\nWe have Area of sector $(OAM)$ < area of triangle $(OAT)$ and hence\r\n$\\frac{1^2 \\cdot x}{2}<\\frac{1 \\cdot \\tan x}{2}$\r\nSo $x< \\tan x$, $x \\in (0, \\frac{\\pi}{2} )$. q.e.d.", "Solution_3": "By Taylor's mean value theorem,\r\n\r\n$\\tan x>x+\\frac {1}{3}x^3$ for $\\vert x\\vert <\\pi/2$, which is enough." } { "Tag": [ "probability" ], "Problem": "A few probability exercises:\r\n\r\n1. Bag A contains 5 red and 4 white counters. Bag B contains 6 red and 3 white counters. A counter is picked at random from bag A and placed in bag B. A counter is now picked from bag B. Find the probability that this counter is white.\r\n\r\n2. The two events A and B are such that $P(A)=0.6$ $P(B)=0.2$ $P(A|B)=0.1$. Calculuate the probability that at least one of the events occur. Calculate the probability that exactly one of the events occur. \r\n\r\nI got\r\n\r\n\r\n[hide]\n1. $\\frac{8}{45}$\n\n2. $0.8$, $0.78$ respectively.\n[/hide]\r\n\r\nHowever, according to the answer sheet, those are wrong. Can someone explain why?", "Solution_1": "1\r\n[hide]If you choose red from Bag A\n\n$\\frac 59 \\cdot \\frac {3}{10} = \\frac 16$\n\nIf you choose white from Bag A\n\n$\\frac 49 \\cdot \\frac 25 = \\frac 8{45}$\n\n$\\frac 8{45} + \\frac 16 = \\frac {31}{90}$[/hide]", "Solution_2": "[hide=\"Problem 2 part 1\"]$P(A\\cap B)=P(B)P(A|B)=0.02$\n$P(A)+P(B)-P(A\\cap B)=0.6+0.2-0.02=\\boxed{0.78}$[/hide]", "Solution_3": "i agree with tarquin's answer", "Solution_4": "[quote=\"agolsme\"]A few probability exercises:\n\n1. Bag A contains 5 red and 4 white counters. Bag B contains 6 red and 3 white counters. A counter is picked at random from bag A and placed in bag B. A counter is now picked from bag B. Find the probability that this counter is white.\n\n2. The two events A and B are such that $P(A)=0.6$ $P(B)=0.2$ $P(A|B)=0.1$. Calculuate the probability that at least one of the events occur. Calculate the probability that exactly one of the events occur. \n\nI got\n\n\n[hide]\n1. $\\frac{8}{45}$\n\n2. $0.8$, $0.78$ respectively.\n[/hide]\n\nHowever, according to the answer sheet, those are wrong. Can someone explain why?[/quote]\n\nI believe I can.\n\n[hide]\n\n1. To begin with in bag A you have 5 red, 4 white; and likewise in bag B you have 6 red, and 3 white.\n\nYou move a counter (marble, whatever...) from A to B. This changes the distribution in B. There are now 10 marbles in B, and 8 in A, though the number left in A is irrelevant to the rest of the problem.\n\nIf you chose a red marble, there are now 7 red and 3 white in B: $P_B(\\text{white}|\\text{chose red}) =\\frac{3}{10}$\n\nElse you chose white and there are now 6 red and 4 white in B: $P_B(\\text{white}|\\text{chose white}) =\\frac{4}{10}$\n\nThe marginal probability of choosing a white marble from bag B is given by:\n\n$P_B(\\text{white}) =P_B(\\text{white}|\\text{chose red})\\cdot P_B(\\text{chose red})+P_B(\\text{white}|\\text{chose white})\\cdot P_B(\\text{chose white}) \\\\\n\\\\\n= \\frac{3}{10}\\cdot \\frac{5}{9} + \\frac{4}{10}\\cdot\\frac{4}{9}\\\\\n\\\\\n=\\frac{31}{90}$\n\n2. Given P(A) = 0.6, P(B) = 0.2, and P(A|B) = 0.1\n\na) Calculate the probability that at least one of the events occurs\n\nThe probability that at least one of the events occurs is the probability of their union occuring; i.e., \"A or B or both A and B.\"\n\n$P(A\\cup B) =P(A) + P(B) - P(A\\cap B)\\\\\n\\\\\n= P(A) + P(B)-P(B)P(A|B)\\\\\n\\\\\n= 0.6 +0.2-(0.2)(0.1)\\\\\n\\\\\n= 0.8 -0.02\\\\\n\\\\\n= 0.78\\\\\n$\n\nb) Calculate the probability that exactly one of the events occurs\n\nThis is the probability of the union minus the probability of the intersection; \"A or B, but not both A and B.\"\n\n$P(A\\cup B) - P(A\\cap B) = 0.78 -(0.2)(0.1)\\\\\n\\\\\n= 0.78 - 0.02\\\\\n\\\\\n\\&= 0.76\\\\\n\\\\\n\\blacksquare$[/hide]" } { "Tag": [ "search", "algebra", "polynomial", "quadratics", "function", "domain", "real analysis" ], "Problem": "I want to find a numerical solution to the following non-concave optimization problem $\\max b_{1}\\log(1+\\frac{a_{1}p_{1}}{a_{2}p_{2}+1})+b_{2}\\log(1+\\frac{a_{3}p_{2}}{a_{4}p_{1}+1})-p_{1}-p_{2}$ subject to $p_{1},p_{2}\\geq 0$. where $b_{1},b_{2},a_{1},a_{2},a_{3},a_{4}$ are given positive constants.\r\n\r\nThe problelm seems hard due to lack of concavity. Is there any way to find a global maximum numerially?", "Solution_1": "How large can the constants $a_{1},\\dots,a_{4}$ and $b_{1},b_{2}$ get in the applications you have in mind? I'm especially interested in $b$'s because at the point of maxima we must have $p_{1}< b_{1}$ and $p_{2} 1) = 1/36 ( 0 + 1 + 2 + 3 + 4 + 5) = 5/12.\n\n[/hide]", "Solution_2": "Good, except you should've just noticed that 1, 2, 3, 4, and 5 pattern in the middle and know that sum is 15.", "Solution_3": "Here's an even faster approach, without casework.\r\n\r\n[hide=\"Hint\"]\nWhat's the probability that $a = b$?\n\nThus, what's the probability that $a \\ne b$?\n\nThus, what's the probability that $a > b$?\n[/hide]", "Solution_4": "I did it Barnacle's way.... \r\nit took slightly longer than i wanted it to", "Solution_5": "Haha I wrote it out in a case by case fashion so that others could follow... it probably took me about 15 seconds to get the answer otherwise ;) ...\r\n\r\nRavi your solution is brilliant ... I was looking for a simplification like that but could one...\r\n\r\nFor those who didn't catch on:\r\n\r\nP(a equals b) = 1/6, so P(a doesn't equal b) = 5/6, and half the time a is going to be greater, so [b]5/12[/b]\r\n\r\n:)\r\n\r\nnice work.", "Solution_6": "[hide]$\\frac{5}{12}$[/hide]", "Solution_7": "All of you are correct....\r\n\r\ngo figure", "Solution_8": "whats $a \\ne b$?", "Solution_9": "[hide]5/12, i used barnacle's way[/hide]" } { "Tag": [ "Support" ], "Problem": "Sorry to be a complainer, but it deleted my stats (badges,lessons,current topics) [b]<== Big Annoyance[/b]\r\n\r\nIs there anyway to recover the lessons, topics, and badges I lost?", "Solution_1": "You're not complaining. You should report that. There should be a report button somewhere around in Alcumus. You get Karma!!! :) I don't know but that is a biggie glitchie.", "Solution_2": "When your diagnostic state is done, it deletes all stats, to start over with the normal course of matters.", "Solution_3": "Hello,\r\n\r\nI have just checked your account and all appears normal. You have passed three subjects and have three badges. If you are not seeing as many badges as you use to see, you may wish to check the [url=http://www.artofproblemsolving.com/Alcumus/FAQ.php]FAQ[/url] and read the sections entitles [b]How come I can no longer see subjects on my report page that I use to be able to see?[/b] and [b]How come I can no longer see badges I use to be able to see?[/b]" } { "Tag": [ "summer program", "Mathcamp", "AwesomeMath", "PROMYS", "email" ], "Problem": "is Canada/USA MATHCAMP better \r\n\r\nor \r\n\r\nAwesomeMath Summer Program(AMSP) at Texas better?", "Solution_1": "[quote=\"climskywell\"]is Canada/USA MATHCAMP better \n\nor \n\nAwesomeMath Summer Program(AMSP) at Texas better?[/quote]\r\n\r\nA) This has utterly nothing to do with the American Mathematics Competitions.\r\n\r\nB) That is a question of opinion ;) I would personally recommend Canada/USA MathCamp.", "Solution_2": "So... Motion to move to Other Contests and Programs?\r\n\r\nAnd, why is it better, I've been wondering myself.", "Solution_3": "[quote=\"pieterminate\"]So... Motion to move to Other Contests and Programs?[/quote]\r\n\r\nAgreed... done.", "Solution_4": "This will be AwesomeMath's first year so noone can really give you a first hand opinion of it. However, gathering from the description of the program it will be a lot like MOP. Both camps offer a really solid program so you really can't go wrong. However, it seems to me that AwesomeMath will be more geared toward competition math whereas mathcamp will have more recreational math. I really encourage you to go through both of their websites and formulate your own opinion. From there you should be able to figure out which one better suits your tastes. And I really think it will come down to a matter of taste.", "Solution_5": "Has any one been at AwesomeMath? How will you rate it? :)", "Solution_6": "This is AwesomeMath's first year. It's impossible for anyone to have been there before.", "Solution_7": "Duh... You can tell I read the postings in a rush. Thank you for the info! :D", "Solution_8": "[quote=\"joml88\"]However, it seems to me that AwesomeMath will be more geared toward competition math whereas mathcamp will have more recreational math. I really encourage you to go through both of their websites and formulate your own opinion. From there you should be able to figure out which one better suits your tastes. And I really think it will come down to a matter of taste.[/quote]\r\n\r\nI would reformulate the first quoted sentence just a bit. I would say MathCamp has more upper-division university-level math and more theoretical math. (The term \"recreational math\" has a specific meaning. There definitely is recreational math at MathCamp, but it is possible to be involved in other aspects of math entirely while there.) \r\n\r\nI agree with the rest of the advice quoted above. Visit both Web sites, and THINK about what kind of program is more appealing to you. I note that the MathCamp math jam said that MathCamp is always scheduled for a city that is cool in the summer, and Awesome Math is in Dallas, TX, which is not known for being cool in the summer. If you have a preference for HOT weather in the summer time, you know which program to choose. \r\n\r\nHas everyone finished their application to MathCamp now? The regular application deadline is past, although there may be some chance to get in late.", "Solution_9": "What about Mathcamp vs. PROMYS?\r\n\r\nI just received the admission email from Mathcamp, but I also applied to PROMYS (and I feel like I did better on their qualifying problems :p) so I'm wondering what PROMYS is like.", "Solution_10": "read all about it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=83857]here[/url] or [url=http://www.promys.org]here[/url]" } { "Tag": [], "Problem": "Fie $n$ natural si $x_{k}$ real astfel incat: $3\\leq x\\leq 6$. Sa se arate ca $n$ este multiplu de $18$ daca:\r\n$(x_{1}+x_{2}+...+x_{n})/n= 19/6$\r\n$(x_{1}^{2}+...+x_{n}^{2})/n=21/2$.", "Solution_1": "E clar ca $(x_{k}-3)(x_{k}-6) \\leq 0$; daca aduni expresiile astea si folosesti cele 2 relatii din enunt, iti da ca suma lor e $0$, ceea ce inseamna ca $x_{k}\\in \\{3,6\\}$, deci $3|x_{k}$. Folosind asta si prima relatie iti da imediat ca $18|n$." } { "Tag": [ "AMC", "AIME" ], "Problem": "Is it normal to have an AIME score higher than your AMC score?", "Solution_1": "Ummm.... how is that possible?", "Solution_2": "If you get a 2.5 on the AMC and somehow its IMPOSSIBLY super duper hard but everyone thinks its easy so they answer all of the problems wrong. Except you, you leave one blank. THen you would qualify for AIME and it could be normal and you would get 3 or bettter", "Solution_3": "[quote=\"biffanddoc\"]If you get a 2.5 on the AMC and somehow its IMPOSSIBLY super duper hard but everyone thinks its easy so they answer all of the problems wrong. Except you, you leave one blank. THen you would qualify for AIME and it could be normal and you would get 3 or bettter[/quote]\r\n\r\nWell, you do multiply your aime score by 10 for index, but still................", "Solution_4": "[quote=\"tiredepartment\"]Is it normal to have an AIME score higher than your AMC score?[/quote]\r\n\r\nNo. Even with the 10x factor on the AIME score it's not normal. Possible, but doesn't happen often.", "Solution_5": "138/14 is not that unreasonable.", "Solution_6": "Didn't Tony get 132, 15?", "Solution_7": "I know it happens... it's just not normal. And yes ThAzN1 would be an example. He's special like that :P." } { "Tag": [ "calculus", "integration", "algebra", "polynomial", "function", "real analysis", "real analysis unsolved" ], "Problem": "If $ f(x)$ is continuous on $ [a,b]$ and for any integer $ n\\geq 0$ ,we have\r\n$ \\int_{a}^{b}x^{n}f(x)dx \\equal{} 0$\r\n, show that: $ f(x)\\equiv0$\r\n[color=red]Furthermore[/color], If $ f(x)$ is only Riemann Integrable, can we get the conclusion:\r\n$ f(x) \\equal{} 0 a.e.$\r\na.e. stands for almost everywhere.", "Solution_1": "1. Use the fact that the set of Polynomials are dense in C[a,b]\r\n2. Then if $ \\parallel{}f\\parallel{}^{2}_{2} \\equal{} \\int_{[a,b]}|f|^2dx \\equal{} 0$ then $ f \\equal{} 0$ a.e", "Solution_2": "Thanks 1234567a !\r\n\r\nThe first problem we assume $ f(x) \\in C[a,b]$, and use the fact that the set of Polynomials are dense in $ C[a,b]$, so $ \\exists P(x) \\in C[a,b]$ such that $ |P(x) \\minus{} f(x)| < \\epsilon$. and we have\r\n$ \\int_{a}^{b}|f|^2dx \\equal{} \\int_{a}^{b}P(x)f(x)dx \\minus{} \\int_{a}^{b}(P(x) \\minus{} f(x))f(x)dx$.\r\nbecause:\r\n$ |\\int_{a}^{b}(P(x) \\minus{} f(x))f(x)dx|\\leq\\int_{a}^{b}|P(x) \\minus{} f(x)\\parallel{}f(x)|dx\\leq\\epsilon(b \\minus{} a)\\sup_{x\\in[a,b]}f(x)$\r\nand $ \\int_{a}^{b}P(x)f(x)dx \\equal{} 0$. so we have:$ \\int_{a}^{b}|f|^2dx\\rightarrow0$\r\nthen we get the conclusion:$ f(x) \\equal{} 0$ a.e. and $ f(x)$ is a continuous function ,so we have$ f(x)\\equiv0$\r\n\r\n[color=red]But if $ f(x)$ is only Riemann Integrable. Can we find the Polynomials satifies the equation$ |P(x) \\minus{} f(x)| < \\epsilon$ ?[/color]\r\nso if $ f(x)$ is only Riemann Integrable, can we have the same conclusion?\r\nIn other word: If$ f(x)$is Riemann Integrable in $ [a,b]$ ,and for any integer $ n\\geq0$,we have$ \\int_{a}^{b}x^{n}f(x)dx \\equal{} 0$\r\ncan we get the conclusion:$ f(x) \\equal{} 0$a.e.?\r\n(a.e. stands for almost everywhere)" } { "Tag": [], "Problem": "Buond\u00ec, visto che il forum ufficiale italiano \u00e8 down, un po' di gente si \u00e8 sentita su msn, e dal confronto i risultati dovrebbero essere questi\r\n\r\nABAAC\r\nDEBBE\r\nBADDC\r\nECEAD\r\nEDBCA", "Solution_1": "well, if they are true I do $ 5 \\plus{} 5!$ :lol:", "Solution_2": "solo io e gabriel frequentiamo ML? c'\u00e8 qualche altro italiano? come \u00e8 andata?", "Solution_3": "Da oggi ci sono anch'io! :D \r\nIo ho fatto uguale, mi sono confrontato anche con Giove... perci\u00f2 anch'io dovrei aver fatto $ 5^3$ punti... :lol:", "Solution_4": "sbagliata solo la 9 stupidamente :( \r\n\r\n\r\n120 punti....dovrebbero comunque bastare per le provinciali :P" } { "Tag": [], "Problem": "[url=http://www.crazymonkeygames.com/Endless-Zombie-Rampage.html]Click to play[/url]\r\n\r\nPost your high scores :D and what you bought with your 9000 EXP\r\n\r\nWe're talking survival mode.", "Solution_1": "i hate that annoying game" } { "Tag": [ "trigonometry", "inequalities unsolved", "inequalities" ], "Problem": "$a,b,c>0$. Prove that:\r\n$(a^2+2)(b^2+2)(c^2+2)\\ge 9(bc+ca+ab)$", "Solution_1": "We can set $a=\\sqrt{2}\\tan A, b=\\sqrt{2}\\tan B,\\ c=\\sqrt{2}\\tan C,\\ 00$. Prove that:\n$(a^2+2)(b^2+2)(c^2+2)\\ge 9(bc+ca+ab)$[/quote]\r\n\r\nThis APM0 2OO4 problem 5.\r\nSee [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=APMO+2004&t=19555]post #5[/url].", "Solution_3": "oh , very easy \r\n we have : \r\n ah, APMO have a solution and it like my solution", "Solution_4": "hi\r\n\r\ni hv a dif proof\r\n\r\nout of a,b,c 2 must be say a,b s.t. \\[ (a-1)(b-1)\\geq0 \\]\r\n\r\nusing this fact apply chebbychev's ineq and we get the result easily :!:" } { "Tag": [ "induction", "AMC", "USA(J)MO", "USAMO", "algebra proposed", "algebra" ], "Problem": "Prove that for every $n\\in\\mathbb{N}, \\ n\\geq 3$, there are $n$ natural different numbers $a_1,a_2,...a_n$ such that\r\n $\\frac 1{a_1}+\\frac 1{a_2}+...\\frac 1{a_n}=1$.\r\n\r\n[Well, if $n\\geq 4$, one can prove there are at least two different ways of writing number 1 in the form above.]", "Solution_1": "Another problem related to this problem:\r\nProve that every natural number $n\\ge32$ can be written as sum of \r\nnatural numbers whose sum of inverses(i mean the inverse of $a$ is $\\frac{1}{a}$)\r\nis $1$.", "Solution_2": "This is a well known result. Take, for instance $u_1=2,u_{n+1}=u_1u_2\\ldots u_n+1$.\r\nThen\\[ \\frac{1}{u_1}+\\frac{1}{u_2}+\\ldots +\\frac{1}{u_n-1}=1, \\]and it is easy to see that denominators are distinct.\r\n(Approximation of unity with Egyptian fractions.)", "Solution_3": "[quote=\"spider_boy\"]Another problem related to this problem:\nProve that every natural number $n\\ge32$ can be written as sum of \nnatural numbers whose sum of inverses(i mean the inverse of $a$ is $\\frac{1}{a}$)\nis $1$.[/quote]\r\nActually, the statement is:\r\n(USAMO 1978) An integer will be called good if it can be written as the sum of positive integers (not necessarily distinct) the sum of whose reciprocals is $1$. Given that the integers $33$ through $73$ are good, prove every integer greater than or equal to $33$ is good.", "Solution_4": "[quote=\"enescu\"]This is a well known result. Take, for instance $u_1=2,u_{n+1}=u_1u_2\\ldots u_n+1$\nThen\n$\\frac{1}{u_1}+\\frac{1}{u_2}+\\ldots +\\frac{1}{u_n-1}=1$, [/quote]\r\nMore precisely, \r\n$\\frac{1}{u_1}+\\frac{1}{u_2}+\\ldots +\\frac{1}{u_{n-1}}+\\frac{1}{u_n-1}=1$.\r\nI solved it in the same way.\r\n\r\nHere is the author's solution: \r\n[hide]Suppose we already found $a_1,a_2,...a_n$ s.t. $\\frac 1{a_1}+\\frac 1{a_2}+...\\frac 1{a_n}=1$\nThen $\\frac 1{2}+\\frac 1{2a_1}+\\frac 1{2a_2}+...\\frac 1{2a_n}=1$, so the statement is also true for $n+1$.[/hide]\r\n\r\nI had no idea about the Egyptian fractions. :oops: A very interesting subject, anyway! \r\n\r\nIf you want to amuse yourself converting ordinary fractions into Egyptian ones, see http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fractions/egyptian.html\r\nThere is a \"calculator\" on that page (the second one) which converts a given ordinary fraction into Egyptian sums, finding all the Egyptian sums of shortest length for it." } { "Tag": [ "function", "inequalities", "inequalities unsolved" ], "Problem": "Find the minimum value of this function:\r\nA/(B+2*C+3*D) + B/(C+2*D+3*A)+ C / (D+2*A+3*B) + D/(A+2*B+3*C)\r\n\r\nOr it will be more apparent.\r\n \u3000\u3000A\u3000\u3000\u3000\u3000\u3000\u3000\u3000B\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000C\u3000\u3000\u3000\u3000\u3000\u3000\u3000 D\r\n__________ + ___________+ ___________ + _____________\r\nB+2*C+3*D\u3000\u3000C+2*D+3*A \u3000\u3000\u3000D+2*A+3*B \u3000\u3000\u3000A+2*B+3*C\r\n\r\nA,B,C,D are all postive real.", "Solution_1": "[quote=\"anthony19891019\"]Find the minimum value of this function:\nA/(B+2*C+3*D) + B/(C+2*D+3*A)+ C / (D+2*A+3*B) + D/(A+2*B+3*C)\n\nOr it will be more apparent.\n \u3000\u3000A\u3000\u3000\u3000\u3000\u3000\u3000\u3000B\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000C\u3000\u3000\u3000\u3000\u3000\u3000\u3000 D\n__________ + ___________+ ___________ + _____________\nB+2*C+3*D\u3000\u3000C+2*D+3*A \u3000\u3000\u3000D+2*A+3*B \u3000\u3000\u3000A+2*B+3*C\n\nA,B,C,D are all postive .[/quote]\r\n[b]Problem[/b]: Let $a, b, c, d$ be positive real numbers. Find the minimum value of\r\n$\\dfrac{a}{b+2c+3d}+\\dfrac{b}{c+2d+3a}+\\dfrac{c}{d+2a+3b}+\\dfrac{d}{a+2b+3c}$\r\n[hide=\"solution\"]By letting $a=b=c=d$, we guess the minimum value is $\\dfrac{2}{3}$.\nNow we are going to prove the inequality $\\dfrac{a}{b+2c+3d}+\\dfrac{b}{c+2d+3a}+\\dfrac{c}{d+2a+3b}+\\dfrac{d}{a+2b+3c}\\geq\\dfrac{2}{3}2$\nBy Cauchy-Schwarz Inequality,\n$[(a)(b+2c+3d)+(b)(c+2d+3a)+(c)(d+2a+3b)+(d)(a+2b+3c)](\\dfrac{a}{b+2c+3d}+\\dfrac{b}{c+2d+3a}+\\dfrac{c}{d+2a+3b}+\\dfrac{d}{a+2b+3c})\\geq (a+b+c+d)^2$\nSo it remains to prove\n$\\dfrac{(a+b+c+d)^2}{(a)(b+2c+3d)+(b)(c+2d+3a)+(c)(d+2a+3b)+(d)(a+2b+3c)}\\geq\\dfrac{2}{3}$\n$\\Leftrightarrow 3(a+b+c+d)^2\\geq 2(4ab+4bc+4cd+4da)$\nSince $3(a+b+c+d)^2-8(ab+bc+cd+ca)=(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2\\geq 0$,\nthe original inequality is proved.[/hide]\r\nPS This is a shortlist problem in IMO 1993. Another solution is provided [url=http://www.kalva.demon.co.uk/short/soln/sh9324.html]here[/url]." }