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Difference between revisions of "2011 AIME II Problems/Problem 9"
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==Solution==
 
==Solution==
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Note that neither the constraint nor the expression we need to maximize involves products <math>x_i x_j</math> with <math>i \equiv j \pmod 3</math>. Factoring out say <math>x_1</math> and <math>x_4</math> we see that 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 this upper bound can in fact be achieved by ensuring that the constraint expression is equal to <math>\scriptstyle\frac1{540}</math> with <math>x_1 + x_4 = x_2 + x_5 = x_3 + x_6 = \scriptstyle\frac13</math>&mdash;for example, by choosing <math>x_1</math> and <math>x_2</math> small enough&mdash;so our answer is <math>540 + 19 = \fbox{559}.</math>
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Note that neither the constraint nor the expression we need to maximize involves products <math>x_i x_j</math> with <math>i \equiv j \pmod 3</math>. Factoring out say <math>x_1</math> and <math>x_4</math> we see that 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 this upper bound can in fact be achieved by ensuring that the constraint expression is equal to <math>\scriptstyle\frac1{540}</math> with <math>x_1 + x_4 = x_2 + x_5 = x_3 + x_6 = \scriptstyle\frac13</math>&mdash;for example, by choosing <math>x_1</math> and <math>x_2</math> small enough&mdash;so our answer is <math>540 + 19 = \fbox{559}.</math>
  
 
An example is:
 
An example is:

Revision as of 18:26, 28 December 2013

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 neither the constraint nor the expression we need to maximize involves products $x_i x_j$ with $i \equiv j \pmod 3$. Factoring out say $x_1$ and $x_4$ we see that 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 this upper bound can in fact be achieved by ensuring that the constraint expression is equal to $\scriptstyle\frac1{540}$ with $x_1 + x_4 = x_2 + x_5 = x_3 + x_6 = \scriptstyle\frac13$—for example, by choosing $x_1$ and $x_2$ small enough—so our answer is $540 + 19 = \fbox{559}.$

An example is: \begin{align*} x_3 &= x_6 = \frac16 \\ x_1 &= x_2 = \frac{5 - \sqrt{20}}{30} \\ x_5 &= x_4 = \frac{5 + \sqrt{20}}{30} \end{align*}

Another example is \begin{align*} x_1 = x_3 = \frac{1}{3} \\ x_2 = \frac{19}{60}, \ x_5 = \frac{1}{60} \\ x_4 &= x_6 = 0 \end{align*}

See also

2011 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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All AIME Problems and Solutions

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