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Difference between revisions of "2016 AIME II Problems/Problem 15"

(Solution)
 
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For <math>1 \leq i \leq 215</math> let <math>a_i = \dfrac{1}{2^{i}}</math> and <math>a_216 = \dfrac{1}{2^{215}}</math>. Let <math>x_1, x_2, ..., x_215</math> be positive real numbers such that <math>\sum_{i=1}^{215} x_i=1</math> and <math>\sum_{i \leq i < j \leq 216 x_ix_j = \dfrac{107}{215} + \sum_{i=1}^{216} \dfrac{a_i x_i^{2}}{2(1-a_i)}. The maximum possible value of </math>x_2=\dfrac{m}{n}<math>, where </math>m<math> and </math>n<math> are relatively prime positive integers. Find </math>m+n<math>.
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For <math>1 \leq i \leq 215</math> let <math>a_i = \dfrac{1}{2^{i}}</math> and <math>a_216 = \dfrac{1}{2^{215}}</math>. Let <math>x_1, x_2, ..., x_215</math> be positive real numbers such that <math>\sum_{i=1}^{215} x_i=1</math> and <math>\sum_{i \leq i < j \leq 216} x_ix_j = \dfrac{107}{215} + \sum_{i=1}^{216} \dfrac{a_i x_i^{2}}{2(1-a_i)}</math>. The maximum possible value of <math>x_2=\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 
==Solution==
 
==Solution==
−
Replace </math>\sum x_ix_j<math> with </math>\frac12\left(\left(\sum x_i\right)^2-\sum x_i^2\right)<math> and the second equation becomes </math>\sum\frac{x_i^2}{1-a_i}=\frac{1}{215}<math>. Conveniently, </math>\sum 1-a_i=215<math> so we get </math>\left(\sum 1-a_i\right)\left(\sum\frac{x_i^2}{1-a_i}\right)=1=\left(\sum x_i\right)^2<math>. This is the equality case of Cauchy so </math>x_i=c(1-a_i)<math> for some constant </math>c<math>. Using </math>\sum x_i=1<math>, we find </math>c=\frac{1}{215}<math> and thus </math>x_2=\frac{3}{860}$.
+
Replace <math>\sum x_ix_j</math> with <math>\frac12\left(\left(\sum x_i\right)^2-\sum x_i^2\right)</math> and the second equation becomes <math>\sum\frac{x_i^2}{1-a_i}=\frac{1}{215}</math>. Conveniently, <math>\sum 1-a_i=215</math> so we get <math>\left(\sum 1-a_i\right)\left(\sum\frac{x_i^2}{1-a_i}\right)=1=\left(\sum x_i\right)^2</math>. This is the equality case of Cauchy so <math>x_i=c(1-a_i)</math> for some constant <math>c</math>. Using <math>\sum x_i=1</math>, we find <math>c=\frac{1}{215}</math> and thus <math>x_2=\frac{3}{860}</math>.

Revision as of 21:04, 17 March 2016

For $1 \leq i \leq 215$ let $a_i = \dfrac{1}{2^{i}}$ and $a_216 = \dfrac{1}{2^{215}}$. Let $x_1, x_2, ..., x_215$ be positive real numbers such that $\sum_{i=1}^{215} x_i=1$ and $\sum_{i \leq i < j \leq 216} x_ix_j = \dfrac{107}{215} + \sum_{i=1}^{216} \dfrac{a_i x_i^{2}}{2(1-a_i)}$. The maximum possible value of $x_2=\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Replace $\sum x_ix_j$ with $\frac12\left(\left(\sum x_i\right)^2-\sum x_i^2\right)$ and the second equation becomes $\sum\frac{x_i^2}{1-a_i}=\frac{1}{215}$. Conveniently, $\sum 1-a_i=215$ so we get $\left(\sum 1-a_i\right)\left(\sum\frac{x_i^2}{1-a_i}\right)=1=\left(\sum x_i\right)^2$. This is the equality case of Cauchy so $x_i=c(1-a_i)$ for some constant $c$. Using $\sum x_i=1$, we find $c=\frac{1}{215}$ and thus $x_2=\frac{3}{860}$.

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