# Difference between revisions of "2011 AIME II Problems/Problem 9"

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+ | ==Problem 9== | ||

Let <math>x_1, x_2, ... , x_6</math> be non-negative real numbers such that <math>x_1 +x_2 +x_3 +x_4 +x_5 +x_6 =1</math>, and <math>x_1 x_3 x_5 +x_2 x_4 x_6 \ge {\scriptstyle\frac{1}{540}}</math>. Let <math>p</math> and <math>q</math> be positive relatively prime integers such that <math>\frac{p}{q}</math> is the maximum possible value of | Let <math>x_1, x_2, ... , x_6</math> be non-negative real numbers such that <math>x_1 +x_2 +x_3 +x_4 +x_5 +x_6 =1</math>, and <math>x_1 x_3 x_5 +x_2 x_4 x_6 \ge {\scriptstyle\frac{1}{540}}</math>. Let <math>p</math> and <math>q</math> be positive relatively prime integers such that <math>\frac{p}{q}</math> is the maximum possible value of | ||

<math>x_1 x_2 x_3 + x_2 x_3 x_4 +x_3 x_4 x_5 +x_4 x_5 x_6 +x_5 x_6 x_1 +x_6 x_1 x_2</math>. Find <math>p+q</math>. | <math>x_1 x_2 x_3 + x_2 x_3 x_4 +x_3 x_4 x_5 +x_4 x_5 x_6 +x_5 x_6 x_1 +x_6 x_1 x_2</math>. Find <math>p+q</math>. | ||

+ | |||

+ | ==Solution== | ||

+ | Note that none of the expressions involve products <math>x_i x_j</math> with <math>i - j \equiv 3 \pmod 6</math>. The constraint is <math>x_1(x_3x_5) + x_4(x_2x_6) \ge {\scriptstyle\frac1{540}}</math>, while the expression we want to maximize is <math>x_1(x_2x_3 + x_5x_6 + x_6x_2) + x_4(x_2x_3 + x_5x_6 + x_3x_5)</math>. Adding the left side of the constraint to the expression we get: <math>(x_1 + x_4)(x_2x_3 + x_5x_6 + x_6x_2 + x_3x_5) = (x_1 + x_4)(x_2 + x_5)(x_3 + x_6)</math>. This new expression is the product of three non-negative terms whose sum is equal to 1. By AM-GM this product is at most <math>\scriptstyle\frac1{27}</math>. Since we have added at least <math>\scriptstyle\frac1{540}</math> the desired maximum is at most <math>\scriptstyle\frac1{27} - \frac1{540} = \frac{19}{540}</math>. It is easy to see that the maximum can in fact be achieved, so our answer is <math>540 + 19 = \fbox{559}.</math> |

## Revision as of 22:42, 1 April 2011

## Problem 9

Let be non-negative real numbers such that , and . Let and be positive relatively prime integers such that is the maximum possible value of . Find .

## Solution

Note that none of the expressions involve products with . The constraint is , while the expression we want to maximize is . Adding the left side of the constraint to the expression we get: . This new expression is the product of three non-negative terms whose sum is equal to 1. By AM-GM this product is at most . Since we have added at least the desired maximum is at most . It is easy to see that the maximum can in fact be achieved, so our answer is