| {"year": "1996", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "Prove that the average of the numbers $n \\sin n^{\\circ}$ for $n=2,4,6, \\ldots, 180$ is $\\cot 1^{\\circ}$.", "solution": " Because $$ n \\sin n^{\\circ}+(180-n) \\sin \\left(180^{\\circ}-n^{\\circ}\\right)=180 \\sin n^{\\circ} $$ So enough to show that $$ \\sum_{n=0}^{89} \\sin (2 n)^{\\circ}=\\cot 1^{\\circ} $$ Let $\\zeta=\\cos 2^{\\circ}+i \\sin 2^{\\circ}$ be a primitive root. Then $$ \\begin{aligned} \\sum_{n=0}^{89} \\frac{\\zeta^{n}-\\zeta^{-n}}{2 i} & =\\frac{1}{2 i}\\left[\\frac{\\zeta^{90}-1}{\\zeta-1}-\\frac{\\zeta^{-90}-1}{\\zeta^{-1}-1}\\right] \\\\ & =\\frac{1}{2 i}\\left[\\frac{-2}{\\zeta-1}-\\frac{-2}{\\zeta^{-1}-1}\\right] \\\\ & =\\frac{1}{-i} \\frac{\\zeta^{-1}-\\zeta}{(\\zeta-1)\\left(\\zeta^{-1}-1\\right)}=i \\cdot \\frac{\\zeta+1}{\\zeta-1} \\end{aligned} $$ Also, $$ \\begin{aligned} \\cot 1^{\\circ} & =\\frac{\\cos 1^{\\circ}}{\\sin 1^{\\circ}}=\\frac{\\left(\\cos 1^{\\circ}\\right)^{2}}{\\cos 1^{\\circ} \\sin 1^{\\circ}} \\\\ & =\\frac{\\frac{\\cos 2^{\\circ}+1}{2}}{\\frac{\\sin 2^{\\circ}}{2}}=\\frac{\\frac{1}{2}\\left(\\zeta+\\zeta^{-1}\\right)+1}{\\frac{1}{2 i}\\left(\\zeta-\\zeta^{-1}\\right)} \\\\ & =i \\cdot \\frac{(\\zeta+1)^{2}}{\\zeta^{2}-1}=i \\cdot \\frac{\\zeta+1}{\\zeta-1} \\end{aligned} $$ So we're done.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1996-notes.jsonl"}} |
| {"year": "1996", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "For any nonempty set $S$ of real numbers, let $\\sigma(S)$ denote the sum of the elements of $S$. Given a set $A$ of $n$ positive integers, consider the collection of all distinct sums $\\sigma(S)$ as $S$ ranges over the nonempty subsets of $A$. Prove that this collection of sums can be partitioned into $n$ classes so that in each class, the ratio of the largest sum to the smallest sum does not exceed 2 .", "solution": " By induction on $n$ with $n=1$ being easy. For the inductive step, assume $$ A=\\left\\{a_{1}>a_{2}>\\cdots>a_{n}\\right\\} $$ Fix any index $k$ with the property that $$ a_{k}>\\frac{\\sigma(A)}{2^{k}} $$ (which must exist since $\\frac{1}{2}+\\frac{1}{4}+\\cdots+\\frac{1}{2^{k}}<1$ ). Then - We make $k$ classes for the sums between $\\frac{\\sigma(A)}{2^{k}}$ and $\\sigma(A)$; this handles every set which has any element in $\\left\\{a_{1}, \\ldots, a_{k}\\right\\}$. - We make $n-k$ classes via induction hypothesis on $\\left\\{a_{k+1}, \\ldots, a_{n}\\right\\}$. This solves the problem.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1996-notes.jsonl"}} |
| {"year": "1996", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "Let $A B C$ be a triangle. Prove that there is a line $\\ell$ (in the plane of triangle $A B C$ ) such that the intersection of the interior of triangle $A B C$ and the interior of its reflection $A^{\\prime} B^{\\prime} C^{\\prime}$ in $\\ell$ has area more than $\\frac{2}{3}$ the area of triangle $A B C$.", "solution": " All that's needed is: Claim - If $A B C$ is a triangle where $\\frac{1}{2}<\\frac{A B}{A C}<1$, then the $\\angle A$ bisector works. $$ \\frac{[A B D]}{[A B C]}=\\frac{B D}{B C}=\\frac{A B}{A B+A C} $$ by angle bisector theorem. In general, suppose $x<y<z$ are sides of a triangle. Then $\\frac{1}{2}<\\frac{y}{z}<1$ by triangle inequality as needed.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1996-notes.jsonl"}} |
| {"year": "1996", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "An $n$-term sequence $\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)$ in which each term is either 0 or 1 is called a binary sequence of length $n$. Let $a_{n}$ be the number of binary sequences of length $n$ containing no three consecutive terms equal to $0,1,0$ in that order. Let $b_{n}$ be the number of binary sequences of length $n$ that contain no four consecutive terms equal to $0,0,1,1$ or $1,1,0,0$ in that order. Prove that $b_{n+1}=2 a_{n}$ for all positive integers $n$.", "solution": " Consider the map from sequences of the latter form to sequences of the first form by $$ \\left(y_{1}, \\ldots, y_{n+1}\\right) \\mapsto\\left(y_{1}+y_{2}, y_{2}+y_{3}, \\ldots, y_{n}+y_{n+1}\\right) . $$ It is 2-to-1. The end.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1996-notes.jsonl"}} |
| {"year": "1996", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "Let $A B C$ be a triangle, and $M$ an interior point such that $\\angle M A B=10^{\\circ}, \\angle M B A=$ $20^{\\circ}, \\angle M A C=40^{\\circ}$ and $\\angle M C A=30^{\\circ}$. Prove that the triangle is isosceles.", "solution": " Let $\\theta=\\angle M B C<80^{\\circ}$. By trig Ceva, we get $$ \\frac{\\sin 10^{\\circ}}{\\sin 40^{\\circ}} \\cdot \\frac{\\sin \\theta}{\\sin 20^{\\circ}} \\cdot \\frac{\\sin 30^{\\circ}}{\\sin \\left(80^{\\circ}-\\theta\\right)}=1 $$ This simplifies to $$ \\sin \\theta=4 \\sin \\left(80^{\\circ}-\\theta\\right) \\sin 40^{\\circ} \\cos 10^{\\circ} $$ Claim - We have $\\theta=60^{\\circ}$. $$ \\begin{aligned} 4 \\sin 20^{\\circ} \\sin 40^{\\circ} \\cos 10^{\\circ} & =2\\left(\\cos 20^{\\circ}-\\cos 60^{\\circ}\\right) \\cos 10^{\\circ} \\\\ & =2 \\cos 20^{\\circ} \\cos 10^{\\circ}+\\sin 80^{\\circ} \\\\ & =\\left(\\cos 30^{\\circ}-\\cos 10^{\\circ}\\right)+\\sin 80^{\\circ}=\\cos 30^{\\circ} \\end{aligned} $$ as desired.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1996-notes.jsonl"}} |
| {"year": "1996", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "USAMO", "problem": "Determine with proof whether there is a subset $X \\subseteq \\mathbb{Z}$ with the following property: for any $n \\in \\mathbb{Z}$, there is exactly one solution to $a+2 b=n$, with $a, b \\in X$.", "solution": "The idea is generating functions, but extra care is required since exponents will be in $\\mathbb{Z}$ rather than in $\\mathbb{Z}_{\\geq 0}$. However, consider formally the limit $$ f(x)=\\prod_{k \\geq 0}\\left(1+x^{(-4)^{k}}\\right) $$ For size reasons, this indeed converges formally to a power series, in the sense that the coefficient of any $x^{k}$ is eventually zero or one for all partial sums. We claim $X=\\left\\{n:\\left[x^{n}\\right] f(x)=1\\right\\}$ works. For a given $n$, we can truncate the sum at some large $N$ again for size reasons. For convenience assume $N$ is even. Now set $$ f_{N}(x)=\\prod_{k=0}^{N}\\left(1+x^{(-4)^{k}}\\right) $$ Next, we compute $$ f_{N}(x) f_{N}\\left(x^{2}\\right)=\\frac{(1+x)\\left(1+x^{2}\\right) \\ldots\\left(1+x^{2^{2 N+1}}\\right)}{x^{2+8+\\cdots+2^{2 N-1}}}=\\cdots+x^{-2}+x^{-1}+1+x+x^{2}+\\ldots $$ as desired.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1996-notes.jsonl"}} |
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