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 $2\sum_{1\leq i<j\leq n} x_ix_j=\left(\sum_{i=1}^{n}x_i\right)^2-\left(\sum_{j=1}^{n} x_i^2\right)=-1$ , so the sum of the $x_i$ taken two at a time is $-1/2$ . Now consider the following sum:

$\sum_{A\subseteq N}=S_A^2=2^{n-1}\left(\sum_{j=1}^{n} x_i^2\right)+2^{n-1}\left(\sum_{1\leq i<j\leq n} x_ix_j\right)=2^{n-2}.$

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 $\lambda=1/\sqrt{2}$ .

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2012 USAMO Problems/Problem 6

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