# USAMO 1996 Solution Notes Evan Chen《陳誼廷》 15 April 2024 This is a compilation of solutions for the 1996 USAMO. The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me. These notes will tend to be a bit more advanced and terse than the "official" solutions from the organizers. In particular, if a theorem or technique is not known to beginners but is still considered "standard", then I often prefer to use this theory anyways, rather than try to work around or conceal it. For example, in geometry problems I typically use directed angles without further comment, rather than awkwardly work around configuration issues. Similarly, sentences like "let $\mathbb{R}$ denote the set of real numbers" are typically omitted entirely. Corrections and comments are welcome! ## Contents 0 Problems ..... 2 1 Solutions to Day 1 ..... 3 1.1 USAMO 1996/1 ..... 3 1.2 USAMO 1996/2 ..... 4 1.3 USAMO 1996/3 2 Solutions to Day 2 ..... 6 2.1 USAMO 1996/4 ..... 6 2.2 USAMO 1996/5 ..... 7 2.3 USAMO 1996/6 ..... 8 ## §0 Problems 1. Prove that the average of the numbers $n \sin n^{\circ}$ for $n=2,4,6, \ldots, 180$ is $\cot 1^{\circ}$. 2. 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 . 3. 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 $A B C$. 4. 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 $n$. 5. 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. 6. 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 $a, b \in X$. ## §1 Solutions to Day 1 ## §1.1 USAMO 1996/1 Available online at https://aops.com/community/p353049. ## Problem statement Prove that the average of the numbers $n \sin n^{\circ}$ for $n=2,4,6, \ldots, 180$ is $\cot 1^{\circ}$. 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. ## §1.2 USAMO 1996/2 Available online at https://aops.com/community/p353051. ## Problem statement 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. 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. ## §1.3 USAMO 1996/3 Available online at https://aops.com/community/p353052. ## Problem statement 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 $A B C$. 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. Proof. Let the $\angle A$-bisector meet $B C$ at $D$. The overlapped area is $2[A B D]$ and $$ \frac{[A B D]}{[A B C]}=\frac{B D}{B C}=\frac{A B}{A B+A C} $$ by angle bisector theorem. In general, suppose $x