Question

For $1 \le i \le 215$ let $a_i = \frac{1}{2^i}$ and $a_{216} = \frac{1}{2^{215}}.$ Let $x_1, x_2, \ldots, x_{216}$ be positive real numbers such that $\sum_{i=1}^{216} x_i = 1 \quad \text{and} \quad \sum_{1 \le i \lt j \le 216} x_i x_j = \frac{107}{215} + \sum_{i=1}^{216} \frac{a_i x_i^2}{2(1 - a_i)}.$ The maximum possible value of $x_2 = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

Solution:

Since $\sum x_i = 1,$ we have $2\sum_{i \lt j} x_i x_j = 1 - \sum x_i^2.$ Doubling the given equation and rearranging, $1 - \sum_{i=1}^{216} x_i^2 = \frac{214}{215} + \sum_{i=1}^{216} \frac{a_i x_i^2}{1 - a_i}, \qquad \text{so} \qquad \frac{1}{215} = \sum_{i=1}^{216}\left(1 + \frac{a_i}{1 - a_i}\right)x_i^2 = \sum_{i=1}^{216} \frac{x_i^2}{1 - a_i}.$

Now $\sum a_i = \left(\frac{1}{2} + \cdots + \frac{1}{2^{215}}\right) + \frac{1}{2^{215}} = 1,$ so $\sum (1 - a_i) = 216 - 1 = 215.$ By the Cauchy-Schwarz inequality, $1 = \left(\sum_{i=1}^{216} x_i\right)^2 \le \left(\sum_{i=1}^{216} \frac{x_i^2}{1 - a_i}\right)\left(\sum_{i=1}^{216} (1 - a_i)\right) = \frac{1}{215} \cdot 215 = 1.$

Equality holds, so $x_i$ is proportional to $1 - a_i,$ forcing $x_i = \frac{1 - a_i}{215}.$ The only, hence maximum, possible value of $x_2$ is $\frac{1 - \frac{1}{4}}{215} = \frac{3}{860},$ and $m + n = 3 + 860 = 863.$

2016 AIME II Problem 15

Solution:

Problem 15 in Other Years