2016 AIME II Problems/Problem 15

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 $\sum_{i \leq i < j \leq 216 x_ix_j = \dfrac{107}{215} + \sum_{i=1}^{216} \dfrac{a_i x_i^{2}}{2(1-a_i)}. The maximum possible value of$ (Error compiling LaTeX. Unknown error_msg)x_2=\dfrac{m}{n} $, where$ m $and$ n $are relatively prime positive integers. Find$ m+n$.

==Solution== Replace$ (Error compiling LaTeX. Unknown error_msg)\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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2016 AIME II Problems/Problem 15