{"year": "1989", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "APMO", "problem": "Let $x_{1}, x_{2}, \\ldots, x_{n}$ be positive real numbers, and let\n\n$$\nS=x_{1}+x_{2}+\\cdots+x_{n}\n$$\n\nProve that\n\n$$\n\\left(1+x_{1}\\right)\\left(1+x_{2}\\right) \\cdots\\left(1+x_{n}\\right) \\leq 1+S+\\frac{S^{2}}{2!}+\\frac{S^{3}}{3!}+\\cdots+\\frac{S^{n}}{n!}\n$$", "solution": "Let $\\sigma_{k}$ be the $k$ th symmetric polynomial, namely\n\n$$\n\\sigma_{k}=\\sum_{\\substack{|S|=k \\\\ S \\subseteq\\{1,2, \\ldots, n\\}}} \\prod_{i \\in S} x_{i},\n$$\n\nand more explicitly\n\n$$\n\\sigma_{1}=S, \\quad \\sigma_{2}=x_{1} x_{2}+x_{1} x_{3}+\\cdots+x_{n-1} x_{n}, \\quad \\text { and so on. }\n$$\n\nThen\n\n$$\n\\left(1+x_{1}\\right)\\left(1+x_{2}\\right) \\cdots\\left(1+x_{n}\\right)=1+\\sigma_{1}+\\sigma_{2}+\\cdots+\\sigma_{n}\n$$\n\nThe expansion of\n\n$$\nS^{k}=\\left(x_{1}+x_{2}+\\cdots+x_{n}\\right)^{k}=\\underbrace{\\left(x_{1}+x_{2}+\\cdots+x_{n}\\right)\\left(x_{1}+x_{2}+\\cdots+x_{n}\\right) \\cdots\\left(x_{1}+x_{2}+\\cdots+x_{n}\\right)}_{k \\text { times }}\n$$\n\nhas at least $k$ ! occurrences of $\\prod_{i \\in S} x_{i}$ for each subset $S$ with $k$ indices from $\\{1,2, \\ldots, n\\}$. In fact, if $\\pi$ is a permutation of $S$, we can choose each $x_{\\pi(i)}$ from the $i$ th factor of $\\left(x_{1}+x_{2}+\\cdots+x_{n}\\right)^{k}$. Then each term appears at least $k$ ! times, and\n\n$$\nS^{k} \\geq k!\\sigma_{k} \\Longleftrightarrow \\sigma_{k} \\leq \\frac{S^{k}}{k!}\n$$\n\nSumming the obtained inequalities for $k=1,2, \\ldots, n$ yields the result.", "metadata": {"resource_path": "APMO/segmented/en-apmo1989_sol.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution 1"}}
{"year": "1989", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "APMO", "problem": "Let $x_{1}, x_{2}, \\ldots, x_{n}$ be positive real numbers, and let\n\n$$\nS=x_{1}+x_{2}+\\cdots+x_{n}\n$$\n\nProve that\n\n$$\n\\left(1+x_{1}\\right)\\left(1+x_{2}\\right) \\cdots\\left(1+x_{n}\\right) \\leq 1+S+\\frac{S^{2}}{2!}+\\frac{S^{3}}{3!}+\\cdots+\\frac{S^{n}}{n!}\n$$", "solution": "By AM-GM,\n\n$$\n\\left(1+x_{1}\\right)\\left(1+x_{2}\\right) \\cdots\\left(1+x_{n}\\right) \\leq\\left(\\frac{\\left(1+x_{1}\\right)+\\left(1+x_{2}\\right)+\\cdots+\\left(1+x_{n}\\right)}{n}\\right)^{n}=\\left(1+\\frac{S}{n}\\right)^{n}\n$$\n\nBy the binomial theorem,\n\n$$\n\\left(1+\\frac{S}{n}\\right)^{n}=\\sum_{k=0}^{n}\\binom{n}{k}\\left(\\frac{S}{n}\\right)^{k}=\\sum_{k=0}^{n} \\frac{1}{k!} \\frac{n(n-1) \\ldots(n-k+1)}{n^{k}} S^{k} \\leq \\sum_{k=0}^{n} \\frac{S^{k}}{k!}\n$$\n\nand the result follows.\nComment: Maclaurin's inequality states that\n\n$$\n\\frac{\\sigma_{1}}{n} \\geq \\sqrt{\\frac{\\sigma_{2}}{\\binom{n}{2}}} \\geq \\cdots \\geq \\sqrt[k]{\\frac{\\sigma_{k}}{\\binom{n}{k}}} \\geq \\cdots \\geq \\sqrt[n]{\\frac{\\sigma_{n}}{\\binom{n}{n}}}\n$$\n\nThen $\\sigma_{k} \\leq\\binom{ n}{k} \\frac{S^{k}}{n^{k}}=\\frac{1}{k!} \\frac{n(n-1) \\ldots(n-k+1)}{n^{k}} S^{k} \\leq \\frac{S^{k}}{k!}$.", "metadata": {"resource_path": "APMO/segmented/en-apmo1989_sol.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution 2"}}
{"year": "1989", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "APMO", "problem": "Prove that the equation\n\n$$\n6\\left(6 a^{2}+3 b^{2}+c^{2}\\right)=5 n^{2}\n$$\n\nhas no solutions in integers except $a=b=c=n=0$.", "solution": "We can suppose without loss of generality that $a, b, c, n \\geq 0$. Let $(a, b, c, n)$ be a solution with minimum sum $a+b+c+n$. Suppose, for the sake of contradiction, that $a+b+c+n>0$. Since 6 divides $5 n^{2}, n$ is a multiple of 6 . Let $n=6 n_{0}$. Then the equation reduces to\n\n$$\n6 a^{2}+3 b^{2}+c^{2}=30 n_{0}^{2}\n$$\n\nThe number $c$ is a multiple of 3 , so let $c=3 c_{0}$. The equation now reduces to\n\n$$\n2 a^{2}+b^{2}+3 c_{0}^{2}=10 n_{0}^{2}\n$$\n\nNow look at the equation modulo 8:\n\n$$\nb^{2}+3 c_{0}^{2} \\equiv 2\\left(n_{0}^{2}-a^{2}\\right) \\quad(\\bmod 8)\n$$\n\nIntegers $b$ and $c_{0}$ have the same parity. Either way, since $x^{2}$ is congruent to 0 or 1 modulo 4 , $b^{2}+3 c_{0}^{2}$ is a multiple of 4 , so $n_{0}^{2}-a^{2}=\\left(n_{0}-a\\right)\\left(n_{0}+a\\right)$ is even, and therefore also a multiple of 4 , since $n_{0}-a$ and $n_{0}+a$ have the same parity. Hence $2\\left(n_{0}^{2}-a^{2}\\right)$ is a multiple of 8 , and\n\n$$\nb^{2}+3 c_{0}^{2} \\equiv 0 \\quad(\\bmod 8)\n$$\n\nIf $b$ and $c_{0}$ are both odd, $b^{2}+3 c_{0}^{2} \\equiv 4(\\bmod 8)$, which is impossible. Then $b$ and $c_{0}$ are both even. Let $b=2 b_{0}$ and $c_{0}=2 c_{1}$, and we find\n\n$$\na^{2}+2 b_{0}^{2}+6 c_{1}^{2}=5 n_{0}^{2}\n$$\n\nLook at the last equation modulo 8:\n\n$$\na^{2}+3 n_{0}^{2} \\equiv 2\\left(c_{1}^{2}-b_{0}^{2}\\right) \\quad(\\bmod 8)\n$$\n\nA similar argument shows that $a$ and $n_{0}$ are both even.\nWe have proven that $a, b, c, n$ are all even. Then, dividing the original equation by 4 we find\n\n$$\n6\\left(6(a / 2)^{2}+3(b / 2)^{2}+(c / 2)^{2}\\right)=5(n / 2)^{2}\n$$\n\nand we find that $(a / 2, b / 2, c / 2, n / 2)$ is a new solution with smaller sum. This is a contradiction, and the only solution is $(a, b, c, n)=(0,0,0,0)$.", "metadata": {"resource_path": "APMO/segmented/en-apmo1989_sol.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution\n\n"}}
{"year": "1989", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "APMO", "problem": "Let $A_{1}, A_{2}, A_{3}$ be three points in the plane, and for convenience,let $A_{4}=A_{1}, A_{5}=A_{2}$. For $n=1,2$, and 3 , suppose that $B_{n}$ is the midpoint of $A_{n} A_{n+1}$, and suppose that $C_{n}$ is the midpoint of $A_{n} B_{n}$. Suppose that $A_{n} C_{n+1}$ and $B_{n} A_{n+2}$ meet at $D_{n}$, and that $A_{n} B_{n+1}$ and $C_{n} A_{n+2}$ meet at $E_{n}$. Calculate the ratio of the area of triangle $D_{1} D_{2} D_{3}$ to the area of triangle $E_{1} E_{2} E_{3}$.\nAnswer: $\\frac{25}{49}$.", "solution": "Let $G$ be the centroid of triangle $A B C$, and also the intersection point of $A_{1} B_{2}, A_{2} B_{3}$, and $A_{3} B_{1}$ 。\n![](https://cdn.mathpix.com/cropped/2024_11_22_831b46bbbb7a3a492868g-3.jpg?height=441&width=449&top_left_y=750&top_left_x=769)\n\nBy Menelao's theorem on triangle $B_{1} A_{2} A_{3}$ and line $A_{1} D_{1} C_{2}$,\n\n$$\n\\frac{A_{1} B_{1}}{A_{1} A_{2}} \\cdot \\frac{D_{1} A_{3}}{D_{1} B_{1}} \\cdot \\frac{C_{2} A_{2}}{C_{2} A_{3}}=1 \\Longleftrightarrow \\frac{D_{1} A_{3}}{D_{1} B_{1}}=2 \\cdot 3=6 \\Longleftrightarrow \\frac{D_{1} B_{1}}{A_{3} B_{1}}=\\frac{1}{7}\n$$\n\nSince $A_{3} G=\\frac{2}{3} A_{3} B_{1}$, if $A_{3} B_{1}=21 t$ then $G A_{3}=14 t, D_{1} B_{1}=\\frac{21 t}{7}=3 t, A_{3} D_{1}=18 t$, and $G D_{1}=A_{3} D_{1}-A_{3} G=18 t-14 t=4 t$, and\n\n$$\n\\frac{G D_{1}}{G A_{3}}=\\frac{4}{14}=\\frac{2}{7}\n$$\n\nSimilar results hold for the other medians, therefore $D_{1} D_{2} D_{3}$ and $A_{1} A_{2} A_{3}$ are homothetic with center $G$ and ratio $-\\frac{2}{7}$.\nBy Menelao's theorem on triangle $A_{1} A_{2} B_{2}$ and line $C_{1} E_{1} A_{3}$,\n\n$$\n\\frac{C_{1} A_{1}}{C_{1} A_{2}} \\cdot \\frac{E_{1} B_{2}}{E_{1} A_{1}} \\cdot \\frac{A_{3} A_{2}}{A_{3} B_{2}}=1 \\Longleftrightarrow \\frac{E_{1} B_{2}}{E_{1} A_{1}}=3 \\cdot \\frac{1}{2}=\\frac{3}{2} \\Longleftrightarrow \\frac{A_{1} E_{1}}{A_{1} B_{2}}=\\frac{2}{5}\n$$\n\nIf $A_{1} B_{2}=15 u$, then $A_{1} G=\\frac{2}{3} \\cdot 15 u=10 u$ and $G E_{1}=A_{1} G-A_{1} E_{1}=10 u-\\frac{2}{5} \\cdot 15 u=4 u$, and\n\n$$\n\\frac{G E_{1}}{G A_{1}}=\\frac{4}{10}=\\frac{2}{5}\n$$\n\nSimilar results hold for the other medians, therefore $E_{1} E_{2} E_{3}$ and $A_{1} A_{2} A_{3}$ are homothetic with center $G$ and ratio $\\frac{2}{5}$.\nThen $D_{1} D_{2} D_{3}$ and $E_{1} E_{2} E_{3}$ are homothetic with center $G$ and ratio $-\\frac{2}{7}: \\frac{2}{5}=-\\frac{5}{7}$, and the ratio of their area is $\\left(\\frac{5}{7}\\right)^{2}=\\frac{25}{49}$.", "metadata": {"resource_path": "APMO/segmented/en-apmo1989_sol.jsonl", "problem_match": "# Problem 3", "solution_match": "\nSolution\n"}}
{"year": "1989", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "APMO", "problem": "Let $S$ be a set consisting of $m$ pairs $(a, b)$ of positive integers with the property that $1 \\leq a<$ $b \\leq n$. Show that there are at least\n\n$$\n4 m \\frac{\\left(m-\\frac{n^{2}}{4}\\right)}{3 n}\n$$\n\ntriples $(a, b, c)$ such that $(a, b),(a, c)$, and $(b, c)$ belong to $S$.", "solution": "Call a triple $(a, b, c)$ good if and only if $(a, b),(a, c)$, and $(b, c)$ all belong to $S$. For $i$ in $\\{1,2, \\ldots, n\\}$, let $d_{i}$ be the number of pairs in $S$ that contain $i$, and let $D_{i}$ be the set of numbers paired with $i$ in $S$ (so $\\left|D_{i}\\right|=d_{i}$ ). Consider a pair $(i, j) \\in S$. Our goal is to estimate the number of integers $k$ such that any permutation of $\\{i, j, k\\}$ is good, that is, $\\left|D_{i} \\cap D_{j}\\right|$. Note that $i \\notin D_{i}$ and $j \\notin D_{j}$, so $i, j \\notin D_{i} \\cap D_{j}$; thus any $k \\in D_{i} \\cap D_{j}$ is different from both $i$ and $j$, and $\\{i, j, k\\}$ has three elements as required. Now, since $D_{i} \\cup D_{j} \\subseteq\\{1,2, \\ldots, n\\}$,\n\n$$\n\\left|D_{i} \\cap D_{j}\\right|=\\left|D_{i}\\right|+\\left|D_{j}\\right|-\\left|D_{i} \\cup D_{j}\\right| \\leq d_{i}+d_{j}-n\n$$\n\nSumming all the results, and having in mind that each good triple is counted three times (one for each two of the three numbers), the number of good triples $T$ is at least\n\n$$\nT \\geq \\frac{1}{3} \\sum_{(i, j) \\in S}\\left(d_{i}+d_{j}-n\\right)\n$$\n\nEach term $d_{i}$ appears each time $i$ is in a pair from $S$, that is, $d_{i}$ times; there are $m$ pairs in $S$, so $n$ is subtracted $m$ times. By the Cauchy-Schwartz inequality\n\n$$\nT \\geq \\frac{1}{3}\\left(\\sum_{i=1}^{n} d_{i}^{2}-m n\\right) \\geq \\frac{1}{3}\\left(\\frac{\\left(\\sum_{i=1}^{n} d_{i}\\right)^{2}}{n}-m n\\right) .\n$$\n\nFinally, the sum $\\sum_{i=1}^{n} d_{i}$ is $2 m$, since $d_{i}$ counts the number of pairs containing $i$, and each pair $(i, j)$ is counted twice: once in $d_{i}$ and once in $d_{j}$. Therefore\n\n$$\nT \\geq \\frac{1}{3}\\left(\\frac{(2 m)^{2}}{n}-m n\\right)=4 m \\frac{\\left(m-\\frac{n^{2}}{4}\\right)}{3 n} .\n$$\n\nComment: This is a celebrated graph theory fact named Goodman's bound, after A. M. Goodman's method published in 1959. The generalized version of the problem is still studied to this day.", "metadata": {"resource_path": "APMO/segmented/en-apmo1989_sol.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution\n\n"}}
{"year": "1989", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "APMO", "problem": "Determine all functions $f$ from the reals to the reals for which\n(1) $f(x)$ is strictly increasing,\n(2) $f(x)+g(x)=2 x$ for all real $x$, where $g(x)$ is the composition inverse function to $f(x)$.\n(Note: $f$ and $g$ are said to be composition inverses if $f(g(x))=x$ and $g(f(x))=x$ for all real x.)\n\nAnswer: $f(x)=x+c, c \\in \\mathbb{R}$ constant.", "solution": "Denote by $f_{n}$ the $n$th iterate of $f$, that is, $f_{n}(x)=\\underbrace{f(f(\\ldots f}_{n \\text { times }}(x)))$.\nPlug $x \\rightarrow f_{n+1}(x)$ in (2): since $g\\left(f_{n+1}(x)\\right)=g\\left(f\\left(f_{n}(x)\\right)\\right)=f_{n}(x)$,\n\n$$\nf_{n+2}(x)+f_{n}(x)=2 f_{n+1}(x)\n$$\n\nthat is,\n\n$$\nf_{n+2}(x)-f_{n+1}(x)=f_{n+1}(x)-f_{n}(x)\n$$\n\nTherefore $f_{n}(x)-f_{n-1}(x)$ does not depend on $n$, and is equal to $f(x)-x$. Summing the corresponding results for smaller values of $n$ we find\n\n$$\nf_{n}(x)-x=n(f(x)-x) .\n$$\n\nSince $g$ has the same properties as $f$,\n\n$$\ng_{n}(x)-x=n(g(x)-x)=-n(f(x)-x) .\n$$\n\nFinally, $g$ is also increasing, because since $f$ is increasing $g(x)>g(y) \\Longrightarrow f(g(x))>$ $f(g(y)) \\Longrightarrow x>y$. An induction proves that $f_{n}$ and $g_{n}$ are also increasing functions.\nLet $x>y$ be real numbers. Since $f_{n}$ and $g_{n}$ are increasing,\n\n$$\nx+n(f(x)-x)>y+n(f(y)-y) \\Longleftrightarrow n[(f(x)-x)-(f(y)-y)]>y-x\n$$\n\nand\n\n$$\nx-n(f(x)-x)>y-n(f(y)-y) \\Longleftrightarrow n[(f(x)-x)-(f(y)-y)]<x-y\n$$\n\nSumming it up,\n\n$$\n|n[(f(x)-x)-(f(y)-y)]|<x-y \\quad \\text { for all } n \\in \\mathbb{Z}_{>0}\n$$\n\nSuppose that $a=f(x)-x$ and $b=f(y)-y$ are distinct. Then, for all positive integers $n$,\n\n$$\n|n(a-b)|<x-y\n$$\n\nwhich is false for a sufficiently large $n$. Hence $a=b$, and $f(x)-x$ is a constant $c$ for all $x \\in \\mathbb{R}$, that is, $f(x)=x+c$.\nIt is immediate that $f(x)=x+c$ satisfies the problem, as $g(x)=x-c$.", "metadata": {"resource_path": "APMO/segmented/en-apmo1989_sol.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution\n\n"}}