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

Revision as of 23:03, 11 December 2022

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 1

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

Solution 2

Let $x_i=x_1,x_2,...,x_n$ It is evident that $x_i = \frac{(-1)^i}{\sqrt{n}}$ for evens because of the second equation and $x_i=\frac{(-1)^i}{\sqrt{n-1}}$ for odds(one term will be 0 to maintain the first condition). . We may then try and get an expression for the maximum number of sets that satisfy this which occur when $\lambda = \frac{1}{\sqrt{n}}$ : since it will be $x_1 + x_2 + ... + x_n$ for any choice of A we pick, it will have to be greater than $\frac{1}{\sqrt{n}}$ which means we can either pick 0 negative $x_m$ or $\frac{j}{2}-1$ negatives for j positive terms which gives us: $\sum_{k=0}^{\frac{n}{2}-1} \sum_{i=0}^{\frac{n}{2}-k-1} \binom{\frac{n}{2} - k-1}{i}$ and $\sum_{k=0}^{\frac{n}{2} - 1} 2^{\frac{n}{2} - k - 1}$ Thus, we have $2^{\frac{n}{2}} - 1$ choices for our set A for even values: $2^{\frac{n}{2}}-1 \le n2^{n-3}$ For odd values, let it be the same as the last even valued sequence where n is even(i.e. the same as the sequence before it but with an extra 0 in one of the spots). Then, the following is apparent: $2^{\lfloor\frac{n}{2}\rfloor}-1 \le n2^{n-3}$ Thus, we may say that this holds to be true for all $n \ge 2$ . Note that equality holds when $S_A \in \{\lambda,0,-\lambda\}$ for all i which occurs when $x_i= \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0,....$ $\blacksquare$

@@ Line 9: / Line 9: @@
 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 1==
 For convenience, let <math>N=\{1,2,\dots,n\}</math>.
@@ Line 22: / Line 22: @@
 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>.
+==Solution 2==
+Let <math>x_i=x_1,x_2,...,x_n</math> It is evident that <math>x_i = \frac{(-1)^i}{\sqrt{n}}</math> for evens because of the second equation and <math>x_i=\frac{(-1)^i}{\sqrt{n-1}}</math> for odds(one term will be 0 to maintain the first condition). .
+We may then try and get an expression for the maximum number of sets that satisfy this which occur when <math>\lambda = \frac{1}{\sqrt{n}}</math>:
+since it will be
+<cmath>x_1 + x_2 + ... + x_n </cmath>
+for any choice of A we pick, it will have to be greater than <math>\frac{1}{\sqrt{n}}</math> which means we can either pick 0 negative <math>x_m</math> or <math>\frac{j}{2}-1</math> negatives for j positive terms  which gives us:
+<cmath>\sum_{k=0}^{\frac{n}{2}-1} \sum_{i=0}^{\frac{n}{2}-k-1} \binom{\frac{n}{2} - k-1}{i}</cmath>
+and
+<cmath>\sum_{k=0}^{\frac{n}{2} - 1} 2^{\frac{n}{2} - k - 1}</cmath>
+Thus, we have <math>2^{\frac{n}{2}} - 1</math> choices for our set A for even values:
+<cmath>2^{\frac{n}{2}}-1 \le n2^{n-3}</cmath>
+For odd values, let it be the same as the last even valued sequence where n is even(i.e. the same as the sequence before it but with an extra 0 in one of the spots). Then, the following is apparent:
+<cmath>2^{\lfloor\frac{n}{2}\rfloor}-1 \le n2^{n-3}</cmath>
+Thus, we may say that this holds to be true for all <math>n \ge 2</math>. Note that equality holds when <math>S_A \in \{\lambda,0,-\lambda\}</math> for all i which occurs when <math>x_i= \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0,....</math>
+<math>\blacksquare</math>
 ==See Also==
 *[[USAMO Problems and Solutions]]

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

Revision as of 23:03, 11 December 2022

Contents

Problem

Solution 1

Solution 2

See Also