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

Latest revision as of 14:06, 16 January 2023

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 $x_1 x_3 x_5 +x_2 x_4 x_6 \ge {\frac{1}{540}}$ . 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

Solution 1

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 $x_1(x_3x_5) + x_4(x_2x_6) \ge {\frac1{540}}$ , 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 $\frac1{27}$ . Since we have added at least $\frac{1}{540}$ the desired maximum is at most $\frac1{27} - \frac1{540} =\frac{19}{540}$ . It is easy to see that this upper bound can in fact be achieved by ensuring that the constraint expression is equal to $\frac1{540}$ with $x_1 + x_4 = x_2 + x_5 = x_3 + x_6 =\frac13$ —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 $x_1, x_3, x_5$ and $x_2,x_4,x_6$ . Therefore, a good guess is that $a = x_1 = x_3 = x_5$ and $b = x_2 = x_4 = x_6$ , 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,

$3a^2b+3ab^2 = (a+b)^3-a^3-b^3 \leq \frac{1}{27} - \frac{1}{540} = \boxed{\frac{19}{540}}$ which is the answer.

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

@@ Line 1: / Line 1: @@
-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 \ge \frac{1}{540}</math>.
+==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 {\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>.
+==Solution==
+===Solution 1===
+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 {\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>\frac1{27}</math>. Since we have added at least <math>\frac{1}{540}</math> the desired maximum is at most <math>\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>\frac1{540}</math> with <math>x_1 + x_4 = x_2 + x_5 = x_3 + x_6 =\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{5 - \sqrt{20}}{30} \\
+x_5 &= x_4 = \frac{5 + \sqrt{20}}{30}
+\end{align*}
+</cmath>
+Another example is
+<cmath>
+\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*}
+</cmath>
+===Solution 2 (Not legit)===
+There's a symmetry between <math>x_1, x_3, x_5</math> and <math>x_2,x_4,x_6</math>. Therefore, a good guess is that <math>a = x_1 = x_3 = x_5</math> and <math>b = x_2 = x_4 = x_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,
+<cmath>3a^2b+3ab^2 = (a+b)^3-a^3-b^3 \leq \frac{1}{27} - \frac{1}{540} = \boxed{\frac{19}{540}}</cmath> which is the answer.
+This solution is extremely lucky; if you attempt to solve for <math>a</math> and <math>b</math> you receive complex answers (which contradict the problem statement), but the final answer is correct.
+==See also==
+{{AIME box|year=2011|n=II|num-b=8|num-a=10}}
+[[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

2011 AIME II (Problems • Answer Key • Resources)
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