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 $x_2=\dfrac{m}{n}$ , 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 $x_2=\frac{3}{860}$ . Thus, the desired answer is $860+3=\boxed{863}$ .

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2016 AIME II Problems/Problem 15

Solution