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

Solution

Note that none of the expressions involve products $x_i x_j$ with $i - j \equiv 3 \pmod 6$. The constraint is $x_1(x_3x_5) + x_4(x_2x_6) \ge {\scriptstyle\frac1{540}}$, while the expression we want to maximize is $x_1(x_2x_3 + x_5x_6 + x_6x_2) + x_4(x_2x_3 + x_5x_6 + x_3x_5)$. Adding the left side of the constraint to the expression we get: $(x_1 + x_4)(x_2x_3 + x_5x_6 + x_6x_2 + x_3x_5) = (x_1 + x_4)(x_2 + x_5)(x_3 + x_6)$. 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 $\scriptstyle\frac1{27}$. Since we have added at least $\scriptstyle\frac1{540}$ the desired maximum is at most $\scriptstyle\frac1{27} - \frac1{540} = \frac{19}{540}$. It is easy to see that the maximum can in fact be achieved, so our answer is $540 + 19 = \fbox{559}.$

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