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

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Note that <math>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</math>, so the sum of the <math>x_i</math> taken two at a time is <math>-1/2</math>. Now consider the following sum:
 
Note that <math>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</math>, so the sum of the <math>x_i</math> taken two at a time is <math>-1/2</math>. Now consider the following sum:
  
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<cmath>\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}.</cmath>
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<cmath>\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}.</cmath>
  
 
Since <math>S_A^2\geq 0</math>, it follows that at most <math>2^{n-2}/\lambda^2</math> sets <math>A\subseteq N</math> have <math>|S_A|\geq \lambda</math>.
 
Since <math>S_A^2\geq 0</math>, it follows that at most <math>2^{n-2}/\lambda^2</math> sets <math>A\subseteq N</math> have <math>|S_A|\geq \lambda</math>.
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Note that if equality holds, every subset <math>A</math> of <math>N</math> has <math>S_A\in\{-\lambda,0,\lambda\}</math>. It immediately follows that <math>(x_1,x_2,\ldots , x_n)</math> is a permutation of <math>(\lambda,-\lambda,0,0,\ldots , 0)</math>. Since we know that <math>\sum_{i=1}^{n} x_i^2=1</math>, we have that <math>\lambda=1/\sqrt{2}</math>.
 
Note that if equality holds, every subset <math>A</math> of <math>N</math> has <math>S_A\in\{-\lambda,0,\lambda\}</math>. It immediately follows that <math>(x_1,x_2,\ldots , x_n)</math> is a permutation of <math>(\lambda,-\lambda,0,0,\ldots , 0)</math>. Since we know that <math>\sum_{i=1}^{n} x_i^2=1</math>, we have that <math>\lambda=1/\sqrt{2}</math>.
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==See Also==
 
==See Also==

Revision as of 19:23, 10 December 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

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

See Also

2012 USAMO (Problems • Resources)
Preceded by
Problem 5 		Last Problem
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
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