Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2011 AIME II Problems/Problem 9"
(→‎Solution)
Line 4: Line 4:
  
 
==Solution==
 
==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>
+
Note that neither the constraint nor the expression we need to maximize involves products <math>x_i x_j</math> with <math>i - j \equiv 3 \pmod 6</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 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:
 +
<cmath>
 +
\begin{align*}
 +
x_3 &= x_6 = \frac16 \\
 +
x_1 &= x_2 = \frac{15 - \sqrt{220}}{30} \\
 +
x_5 &= x_4 = \frac{15 + \sqrt{220}}{30}
 +
\end{align*}
 +
</cmath>

Revision as of 00:38, 2 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 neither the constraint nor the expression we need to maximize involves products $x_i x_j$ with $i - j \equiv 3 \pmod 6$. 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 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{15 - \sqrt{220}}{30} \\ x_5 &= x_4 = \frac{15 + \sqrt{220}}{30} \end{align*}

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_9&oldid=38007"