Difference between revisions of "2012 USAMO Problems/Problem 6"

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(If <math>A</math> is the empty set, then <math>S_A = 0</math>.)
 
(If <math>A</math> is the empty set, then <math>S_A = 0</math>.)
  
Prove that for any positive number <math>\lambda</math>, the number of sets <math>A</math> satisfying <math>S_A \ge \lambda</math> is at most <math>2^{n - 3}/\lambda^2</math>.  For what choices of <math>x_1</math>, <math>x_2</math>, \dots, <math>x_n</math>, <math>\lambda</math> does equality hold?
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Prove that for any positive number <math>\lambda</math>, the number of sets <math>A</math> satisfying <math>S_A \ge \lambda</math> is at most <math>2^{n - 3}/\lambda^2</math>.  For what choices of <math>x_1</math>, <math>x_2</math>, <math>\dots</math>, <math>x_n</math>, <math>\lambda</math> does equality hold?
  
 
==Solution==
 
==Solution==

Revision as of 17:56, 25 April 2012

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

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