Difference between revisions of "2012 USAMO Problems/Problem 6"
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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? | 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>. | For convenience, let <math>N=\{1,2,\dots,n\}</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>. | ||
+ | ==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== | ==See Also== | ||
*[[USAMO Problems and Solutions]] | *[[USAMO Problems and Solutions]] |
Revision as of 22:03, 11 December 2022
Contents
[hide]Problem
For integer , let , , , be real numbers satisfying For each subset , define (If is the empty set, then .)
Prove that for any positive number , the number of sets satisfying is at most . For what choices of , , , , does equality hold?
Solution 1
For convenience, let .
Note that , so the sum of the taken two at a time is . Now consider the following sum:
Since , it follows that at most sets have .
Now note that . It follows that at most half of the such that are positive. This shows that at most sets satisfy .
Note that if equality holds, every subset of has . It immediately follows that is a permutation of . Since we know that , we have that .
Solution 2
Let It is evident that for evens because of the second equation and 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 : since it will be for any choice of A we pick, it will have to be greater than which means we can either pick 0 negative or negatives for j positive terms which gives us: and Thus, we have choices for our set A for even values: 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: Thus, we may say that this holds to be true for all . Note that equality holds when for all i which occurs when
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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