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

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\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
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==Solution 2 (Not legit)==
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There's a symmetry between <math>a_1, a_3, a_5</math> and <math>a_2,a_4,a_6</math>. Therefore, a good guess is that <math>x = a_1 = a_3 = a_5</math> and <math>y = a_2 = a_4 = a_6</math>, at which point we know that <math>a+b = 1/3</math>, <math>a^3+b^3 \geq 1/540</math>, and we are trying to maximize <math>3a^2b+3ab^2</math>. Then,
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<cmath>3a^3b+3ab^2 = (a+b)^3-a^3-b^3 \leq 1/27 - 1/540 = 19/540</cmath>, which is the answer.
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This solution is extremely lucky; if you attempt to solve for <math>x</math> and <math>y</math> you receive complex answers (which contradict the problem statement), but the final answer is correct.
  
 
==See also==
 
==See also==

Revision as of 13:44, 16 February 2020

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*}

Solution 2 (Not legit)

There's a symmetry between $a_1, a_3, a_5$ and $a_2,a_4,a_6$. Therefore, a good guess is that $x = a_1 = a_3 = a_5$ and $y = a_2 = a_4 = a_6$, at which point we know that $a+b = 1/3$, $a^3+b^3 \geq 1/540$, and we are trying to maximize $3a^2b+3ab^2$. Then,

\[3a^3b+3ab^2 = (a+b)^3-a^3-b^3 \leq 1/27 - 1/540 = 19/540\], which is the answer.

This solution is extremely lucky; if you attempt to solve for $x$ and $y$ you receive complex answers (which contradict the problem statement), but the final answer is correct.

See also

2011 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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