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.


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.


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.


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.


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.


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.