| {"year": "2003", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "APMO", "problem": "Let $a, b, c, d, e, f$ be real numbers such that the polynomial\n\n$$\np(x)=x^{8}-4 x^{7}+7 x^{6}+a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x+f\n$$\n\nfactorises into eight linear factors $x-x_{i}$, with $x_{i}>0$ for $i=1,2, \\ldots, 8$. Determine all possible values of $f$.", "solution": "From\n\n$$\nx^{8}-4 x^{7}+7 x^{6}+a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x+f=\\left(x-x_{1}\\right)\\left(x-x_{2}\\right) \\ldots\\left(x-x_{8}\\right)\n$$\n\nwe have\n\n$$\n\\sum_{i=1}^{8} x_{i}=4 \\quad \\text { and } \\quad \\sum x_{i} x_{j}=7\n$$\n\nwhere the second sum is over all pairs $(i, j)$ of integers where $1 \\leq i<j \\leq 8$. Since this sum can also be written\n\n$$\n\\frac{1}{2}\\left[\\left(\\sum_{i=1}^{8} x_{i}\\right)^{2}-\\sum_{i=1}^{8} x_{i}^{2}\\right]\n$$\n\nwe get\n\n$$\n14=\\left(\\sum_{i=1}^{8} x_{i}\\right)^{2}-\\sum_{i=1}^{8} x_{i}^{2}=16-\\sum_{i=1}^{8} x_{i}^{2}\n$$\n\nso\n\n$$\n\\sum_{i=1}^{8} x_{i}^{2}=2 \\quad \\text { while } \\quad \\sum_{i=1}^{8} x_{i}=4 . \\quad[3 \\text { marks }]\n$$\n\nNow\n\n$$\n\\sum_{i=1}^{8}\\left(2 x_{i}-1\\right)^{2}=4 \\sum_{i=1}^{8} x_{i}^{2}-4 \\sum_{i=1}^{8} x_{i}+8=4(2)-4(4)+8=0\n$$\n\nwhich forces $x_{i}=1 / 2$ for all $i$. [3 marks] Therefore\n\n$$\nf=\\prod_{i=1}^{8} x_{i}=\\left(\\frac{1}{2}\\right)^{8}=\\frac{1}{256} . \\quad[1 m x]\n$$\n\nAlternate solution: After obtaining (1) [3 marks], use Cauchy's inequality to get\n\n$$\n16=\\left(x_{1} \\cdot 1+x_{2} \\cdot 1+\\cdots+x_{8} \\cdot 1\\right)^{2} \\leq\\left(x_{1}^{2-}+x_{2}^{2}+\\cdots+x_{8}^{2}\\right)\\left(1^{2}+1^{2}+\\cdots+1^{2}\\right)=8 \\cdot 2=16\n$$\n\nor the power mean inequality to get\n\n$$\n\\frac{1}{2}=\\frac{1}{8} \\sum_{i=1}^{8} x_{i} \\leq\\left(\\frac{1}{8} \\sum_{i=1}^{8} x_{i}^{2}\\right)^{1 / 2}=\\frac{1}{2} . \\quad[2 \\text { marks }]\n$$\n\nEither way, equality must hold, which can only happen if all the terms $x_{i}$ are equal, that is, if $x_{i}=1 / 2$ for all $i$. [1 mark] Thus $f=1 / 256$ as above. [ 1 mark]", "metadata": {"resource_path": "APMO/segmented/en-apmo2003_sol.jsonl", "problem_match": "\n1. ", "solution_match": "# Solution."}} |
| {"year": "2003", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "APMO", "problem": "Suppose $A B C D$ is a square piece of cardboard with side length $a$. On a plane are two parallel lines $\\ell_{1}$ and $\\ell_{2}$, which are also $a$ units apart. The square $A B C D$ is placed on the plane so that sides $A B$ and $A D$ intersect $\\ell_{1}$ at $E$ and $F$ respectively. Also, sides $C B$ and $C D$ intersect $\\ell_{2}$ at $G$ and $H$ respectively. Let the perimeters of $\\triangle A E F$ and $\\triangle C G H$ be $m_{1}$ and $m_{2}$ respectively. Prove that no matter how the square was placed, $m_{1}+m_{2}$ remains constant.", "solution": "Let $E H$ intersect $F G$ at $O$. The distance from $G$ to line $F D$ and line $E F$ are both $a$. So $F G$ bisects $\\angle E F D$. Similarly, $E H$ bisects $\\angle B E F$. So $O$ is an excentre of $\\triangle A E F$. Similarly, $O$ is an excentre of $\\triangle C G H$. [2 marks] Construct these excircles with centre $O$. Let $M, N, P, Q$ be on sides $A B, B C, C D, D A$ respectively, where these excircles touch the square. Then $O M \\perp A B, O N \\perp B C, O P \\perp C D$, and $O Q \\perp D A$. Since $A B \\| C D$ and $A D \\| B C, M, O, P$ are collinear and $N, O, Q$ are collinear. Now $M P=N Q=a$. [2 marks] Using the fact that the two tangents from a point to a circle have the same length, we get $E F=E M+F Q$ and $G H=G N+H P$. [1 mark] Then\n\n$$\nm_{1}=A E+A F+E F=A E+A F+(E M+F Q)=A M+A Q=O Q+O M\n$$\n\nand\n\n$$\nm_{2}=C G+C H+G H=C G+C H+(G N+H P)=C N+C P=O P+O N . \\quad[1 \\text { mark }]\n$$\n\nTherefore\n\n$$\nm_{1}+m_{2}=(O Q+O M)+(O P+O N)=M P+N Q=2 a . \\quad[1 \\mathrm{mark}]\n$$", "metadata": {"resource_path": "APMO/segmented/en-apmo2003_sol.jsonl", "problem_match": "\n2. ", "solution_match": "\nSolution 1."}} |
| {"year": "2003", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "APMO", "problem": "Suppose $A B C D$ is a square piece of cardboard with side length $a$. On a plane are two parallel lines $\\ell_{1}$ and $\\ell_{2}$, which are also $a$ units apart. The square $A B C D$ is placed on the plane so that sides $A B$ and $A D$ intersect $\\ell_{1}$ at $E$ and $F$ respectively. Also, sides $C B$ and $C D$ intersect $\\ell_{2}$ at $G$ and $H$ respectively. Let the perimeters of $\\triangle A E F$ and $\\triangle C G H$ be $m_{1}$ and $m_{2}$ respectively. Prove that no matter how the square was placed, $m_{1}+m_{2}$ remains constant.", "solution": "Extend $A B$ to $I$ and $D C$ to $J$ so that $A E=B I=C J$. Let $\\ell_{2}$ intersect $I J$ at $M$, and let $K$ lie on $I J$ so that $G K \\perp I J$. Then, since $A E=G K, \\triangle A E F$ and $\\triangle K G M$ are congruent. [1 mark] Thus, since $G K=C J$ and $G C=K J$,\n\n$$\nm_{1}+m_{2}=\\operatorname{perimeter}(K G M)+\\operatorname{perimeter}(C G H)=\\operatorname{perimeter}(H M J) . \\quad[\\mathbf{2} \\text { marks }]\n$$\n\nLet $L$ lie on $C D$ so that $E L \\perp C D$. Then a circle with centre $E$ and radius $a$ will touch $D C$ at $L, I J$ at $I$, and the interior of $H M$ at some point $N$, so\n\n$$\n\\operatorname{perimeter}(H M J)=J H+(H N+N M)+J M=(J H+H L)+(M I+J M)=J L+I J=a+a=2 a\n$$\n\n[4 marks] Thus $m_{1}+m_{2}=2 a$.", "metadata": {"resource_path": "APMO/segmented/en-apmo2003_sol.jsonl", "problem_match": "\n2. ", "solution_match": "# Solution 2."}} |
| {"year": "2003", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "APMO", "problem": "Suppose $A B C D$ is a square piece of cardboard with side length $a$. On a plane are two parallel lines $\\ell_{1}$ and $\\ell_{2}$, which are also $a$ units apart. The square $A B C D$ is placed on the plane so that sides $A B$ and $A D$ intersect $\\ell_{1}$ at $E$ and $F$ respectively. Also, sides $C B$ and $C D$ intersect $\\ell_{2}$ at $G$ and $H$ respectively. Let the perimeters of $\\triangle A E F$ and $\\triangle C G H$ be $m_{1}$ and $m_{2}$ respectively. Prove that no matter how the square was placed, $m_{1}+m_{2}$ remains constant.", "solution": "Without loss of generality, assume the square has side $a=1$. Let $\\theta$ be the acute angle between $\\ell_{1}$ (or $\\ell_{2}$ ) and the sides $A B$ and $C D$ of the square. Then, letting $E F=x$ and $G H=y$, we have\n\n$$\nE A=x \\cos \\theta, \\quad A F=x \\sin \\theta, \\quad C H=y \\cos \\theta, \\quad C G=y \\sin \\theta\n$$\n\nThus\n\n$$\nm_{1}+m_{2}=(x+y)(\\sin \\theta+\\cos \\theta+1) . \\quad[2 \\text { marks }]\n$$\n\nDraw lines parallel to $\\ell_{1}, \\ell_{2}$ through $A$ and $C$ respectively. The distance between these lines is $\\sin \\theta+\\cos \\theta$ [1 mark], as can be seen by drawing a mutual perpendicular to these lines through $B$, say. Also, the altitudes from $A$ to $E F$ and from $C$ to $G H$ have lengths $x \\sin \\theta \\cos \\theta$ and $y \\sin \\theta \\cos \\theta$ respectively [ 1 mark]. Therefore the distance between $\\ell_{1}$ and $\\ell_{2}$ must be\n\n$$\n(\\sin \\theta+\\cos \\theta)-x \\sin \\theta \\cos \\theta-y \\sin \\theta \\cos \\theta\n$$\n\nBut we are given that this distance is $a=1$, so\n\n$$\n(x+y) \\sin \\theta \\cos \\theta+1=\\sin \\theta+\\cos \\theta\n$$\n\nor\n\n$$\nx+y=\\frac{\\sin \\theta+\\cos \\theta-1}{\\sin \\theta \\cos \\theta} \\cdot \\quad[1 \\text { mark }]\n$$\n\nTherefore, by (1),\n\n$$\n\\begin{aligned}\nm_{1}+m_{2} & =\\frac{(\\sin \\theta+\\cos \\theta-1)(\\sin \\theta+\\cos \\theta+1)}{\\sin \\theta \\cos \\theta} \\\\\n& =\\frac{\\left(\\sin ^{2} \\theta+\\cos ^{2} \\theta+2 \\sin \\theta \\cos \\theta\\right)-1}{\\sin \\theta \\cos \\theta} \\\\\n& =\\frac{1+2 \\sin \\theta \\cos \\theta-1}{\\sin \\theta \\cos \\theta}=2 . \\quad[2 \\text { marks }]\n\\end{aligned}\n$$", "metadata": {"resource_path": "APMO/segmented/en-apmo2003_sol.jsonl", "problem_match": "\n2. ", "solution_match": "\nSolution 3."}} |
| {"year": "2003", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "APMO", "problem": "Let $k \\geq 14$ be an integer, and let $p_{k}$ be the largest prime number which is strictly less than $k$. You may assume that $p_{k} \\geq 3 k / 4$. Let $n$ be a composite integer. Prove:\n(a) if $n=2 p_{k}$, then $n$ does not divide $(n-k)$ !;\n(b) if $n>2 p_{k}$, then $n$ divides $(n-k)$ !.", "solution": "(a) Note that $n-k=2 p_{k}-k<2 p_{k}-p_{k}=p_{k}$, so $p_{k} \\nmid(n-k)$ !, so $2 p_{k} \\nless(n-k)$ !. [1 mark]\n(b) Note that $n>2 p_{k} \\geq 3 k / 2$ implies $k<2 n / 3$, so $n-k>n / 3$. So if we can find integers $a, b \\geq 3$ such that $n=a b$ and $a \\neq b$, then both $a$ and $b$ will appear separately in the product $(n-k)!=1 \\times 2 \\times \\cdots \\times(n-k)$, which means $n \\mid(n-k)!$. Observe that $k \\geq 14$ implies $p_{k} \\geq 13$, so that $n>2 p_{k} \\geq 26$.\n\nIf $n=2^{\\alpha}$ for some integer $\\alpha \\geq 5$, then take $a=2^{2}, b=2^{\\alpha-2}$. [ 1 mark] Otherwise, since $n \\geq 26>16$, we can take $a$ to be an odd prime factor of $n$ and $b=n / a$ [1 mark], unless $b<3$ or $b=a$.\n\nCase (i): $b<3$. Since $n$ is composite, this means $b=2$, so that $2 a=n>2 p_{k}$. As $a$ is a prime number and $p_{k}$ is the largest prime number which is strictly less than $k$, it follows that $a \\geq k$. From $n-k=2 a-k \\geq$ $2 a-a=a>2$ we see that $n=2 a$ divides into $(n-k)$. [ 2 marks]\n\nCase (ii): $b=a$. Then $n=a^{2}$ and $a>6$ since $n \\geq 26$. Thus $n-k>n / 3=a^{2} / 3>2 a$, so that both $a$ and $2 a$ appear among $\\{1,2, \\ldots, n-k\\}$. Hence $n=a^{2}$ divides into $(n-k)!$. [2 marks]", "metadata": {"resource_path": "APMO/segmented/en-apmo2003_sol.jsonl", "problem_match": "\n3. ", "solution_match": "\nSolution."}} |
| {"year": "2003", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "APMO", "problem": "Let $a, b, c$ be the sides of a triangle, with $a+b+c=1$, and let $n \\geq 2$ be an integer. Show that\n\n$$\n\\sqrt[n]{a^{n}+b^{n}}+\\sqrt[n]{b^{n}+c^{n}}+\\sqrt[n]{c^{n}+a^{n}}<1+\\frac{\\sqrt[n]{2}}{2}\n$$", "solution": "Without loss of generality, assume $a \\leq b \\leq c$. As $a+b>c$, we have\n\n$$\n\\frac{\\sqrt[n]{2}}{2}=\\frac{\\sqrt[n]{2}}{2}(a+b+c)>\\frac{\\sqrt[n]{2}}{2}(c+c)=\\sqrt[n]{2 c^{n}} \\geq \\sqrt[n]{b^{n}+c^{n}} \\quad \\quad[2 \\text { marks }]\n$$\n\nAs $a \\leq c$ and $n \\geq 2$, we have\n\n$$\n\\begin{aligned}\n\\left(c^{n}+a^{n}\\right)-\\left(c+\\frac{a}{2}\\right)^{n} & =a^{n}-\\sum_{k=1}^{n}\\binom{n}{k} c^{n-k}\\left(\\frac{a}{2}\\right)^{k} \\\\\n& \\leq\\left[1-\\sum_{k=1}^{n}\\binom{n}{k}\\left(\\frac{1}{2}\\right)^{k}\\right] a^{n} \\quad\\left(\\text { since } c^{n-k} \\geq a^{n-k}\\right) \\\\\n& =\\left[\\left(1-\\frac{n}{2}\\right)-\\sum_{k=2}^{n}\\binom{n}{k}\\left(\\frac{1}{2}\\right)^{k}\\right] a^{n}<0\n\\end{aligned}\n$$\n\nThus\n\n$$\n\\sqrt[n]{c^{n}+a^{n}}<c+\\frac{a}{2} . \\quad[3 \\text { marks }]\n$$\n\nLikewise\n\n$$\n\\sqrt[n]{b^{n}+a^{n}}<b+\\frac{a}{2} \\quad \\quad[1 \\text { mark }]\n$$\n\nAdding (1), (2) and (3), we get\n\n$$\n\\sqrt[n]{a^{n}+b^{n}}+\\sqrt[n]{b^{n}+c^{n}}+\\sqrt[n]{c^{n}+a^{n}}<\\frac{\\sqrt[n]{2}}{2}+c+\\frac{a}{2}+b+\\frac{a}{2}=1+\\frac{\\sqrt[n]{2}}{2} . \\quad[1 \\text { mark }]\n$$", "metadata": {"resource_path": "APMO/segmented/en-apmo2003_sol.jsonl", "problem_match": "\n4. ", "solution_match": "# Solution."}} |
| {"year": "2003", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "APMO", "problem": "Given two positive integers $m$ and $n$, find the smallest positive integer $k$ such that among any $k$ people, either there are $2 m$ of them who form $m$ pairs of mutually acquainted people or there are $2 n$ of them forming $n$ pairs of mutually unacquainted people.", "solution": "Let the smallest positive integer $k$ satisfying the condition of the problem be denoted $r(m, n)$. We shall show that\n\n$$\nr(m, n)=2(m+n)-\\min \\{m, n\\}-1\n$$\n\nObserve that, by symmetry, $r(m, n)=r(n, m)$. Therefore it suffices to consider the case where $m \\geq n$, and to prove that\n\n$$\nr(m, n)=2 m+n-1 . \\quad[1 \\text { mark }]\n$$\n\nFirst we prove that\n\n$$\nr(m, n) \\geq 2 m+n-1\n$$\n\nby an example. Call a group of $k$ people, every two of whom are mutually acquainted, a $k$-clique. Consider a set of $2 m+n-2$ people consisting of a $(2 m-1)$-clique together with an additional $n-1$ people none of whom know anyone else. (Call such people isolated.) Then there are not $2 m$ people forming $m$ mutually acquainted pairs, and there also are not $2 n$ people forming $n$ mutually unacquainted pairs. Thus $r(m, n) \\geq$ $(2 m-1)+(n-1)+1=2 m+n-1$ by the definition of $r(m, n)$. [1 mark]\n\nTo establish (1), we need to prove that $r(m, n) \\leq 2 m+n-1$. To do this, we now show that\n\n$$\nr(m, n) \\leq r(m-1, n-1)+3 \\quad \\text { for all } m \\geq n \\geq 2\n$$\n\nLet $G$ be a group of $t=r(m-1, n-1)+3$ people. Notice that\n\n$$\nt \\geq 2(m-1)+(n-1)-1+3=2 m+n-1 \\geq 2 m \\geq 2 n\n$$\n\nIf $G$ is a $t$-clique, then $G$ contains $2 m$ people forming $m$ mutually acquainted pairs, and if $G$ has only isolated people, then $G$ contains $2 n$ people forming $n$ mutually unacquainted pairs. Otherwise, there are three people in $G$, say $a, b$ and $c$, such that $a, b$ are acquainted but $a, c$ are not. Now consider the group $A$ obtained byremoving $a, b$ and $c$ from $G$. A has $t-3=r(m-1, n-1)$ people, so by the definition of $r(m-1, n-1)$, A either contains $2(m-1)$ people forming $m-1$ mutually acquainted pairs, or else contains $2(n-1)$ people forming $n-1$ mutually unacquainted pairs. In the former case, we add the acquainted pair $a, b$ to $A$ to form $m$ mutually acquainted pairs in $G$. In the latter case, we add the unacquainted pair $a, c$ to $A$ to form $n$ mutually unacquainted pairs in $G$. This proves (2). [3 marks]\n\nTrivially, $r(s, 1)=2 s$ for all $s[\\mathbf{1}$ mark], so $r(m, n) \\leq 2 m+n-1$ holds whenever $n=1$. Proceeding by induction on $n$, by (2) we obtain\n\n$$\nr(m, n) \\leq r(m-1, n-1)+3 \\leq 2(m-1)+(n-1)-1+3=2 m+n-1\n$$\n\nwhich completes the proof. [1 mark]\nNote. Give an additional 1 mark to any student who gets at most 5 marks by the above marking scheme, but in addition gives a valid argument that $r(2,2)=5$.", "metadata": {"resource_path": "APMO/segmented/en-apmo2003_sol.jsonl", "problem_match": "\n5. ", "solution_match": "# Solution."}} |
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