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

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Let <math>x_1, x_2, ... , x_6</math> be nonnegaative 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 \le \frac{1}{540}</math>. Let p and q be positive relatively prime integers such that \frac{p}{q} is the maximum possible value of
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Let <math>x_1, x_2, ... , x_6</math> be nonnegaative 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 \le \frac{1}{540}</math>. Let p and q 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 p+q.
 
<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 p+q.

Revision as of 17:14, 31 March 2011

Let $x_1, x_2, ... , x_6$ be nonnegaative real numbers such that $x_1 +x_2 +x_3 +x_4 +x_5 +x_6 =1$, and $x_1 x_3 x_5 +x_2 x_4 x_6 \le \frac{1}{540}$. 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.