# 2012 USAMO Problems/Problem 6

## Problem

For integer $n \ge 2$, let $x_1$, $x_2$, $\dots$, $x_n$ be real numbers satisfying $$x_1 + x_2 + \dots + x_n = 0, \quad \text{and} \quad x_1^2 + x_2^2 + \dots + x_n^2 = 1.$$ For each subset $A \subseteq \{1, 2, \dots, n\}$, define $$S_A = \sum_{i \in A} x_i.$$ (If $A$ is the empty set, then $S_A = 0$.)

Prove that for any positive number $\lambda$, the number of sets $A$ satisfying $S_A \ge \lambda$ is at most $2^{n - 3}/\lambda^2$. For what choices of $x_1$, $x_2$, $\dots$, $x_n$, $\lambda$ does equality hold?

## Solution

For convenience, let $N=\{1,2,\dots,n\}$.

Note that $2\sum_{1\leq i, so the sum of the $x_i$ taken two at a time is $-1/2$. Now consider the following sum: $\[\sum_{A\subseteq N}S_A^2=2^{n-1}\left(\sum_{j=1}^{n} x_i^2\right)+2^{n-1}\left(\sum_{1\leq i

Since $S_A^2\geq 0$, it follows that at most $2^{n-2}/\lambda^2$ sets $A\subseteq N$ have $|S_A|\geq \lambda$.

Now note that $S_A+S_{N/A}=0$. It follows that at most half of the $S_A$ such that $|S_A|\geq\lambda$ are positive. This shows that at most $2^{n-3}/\lambda^2$ sets $A\subseteq N$ satisfy $S_A\geq \lambda$.

Note that if equality holds, every subset $A$ of $N$ has $S_A\in\{-\lambda,0,\lambda\}$. It immediately follows that $(x_1,x_2,\ldots , x_n)$ is a permutation of $(\lambda,-\lambda,0,0,\ldots , 0)$. Since we know that $\sum_{i=1}^{n} x_i^2=1$, we have that $\lambda=1/\sqrt{2}$.