AIME 2016 II · 第 15 题

Solution

Note that

$$\begin{aligned}\sum_{1 \leq i < j \leq 216} x_ix_j &= \frac12\left(\left(\sum_{i=1}^{216} x_i\right)^2-\sum_{i=1}^{216} x_i^2\right)\\&=\frac12\left(1-\sum x_i^2\right).\end{aligned}$$ Substituting this into the second equation and collecting $x_i^2$ terms, we find

$$\sum_{i=1}^{216}\frac{x_i^2}{1-a_i}=\frac{1}{215}.$$ Conveniently, $\sum_{i=1}^{216} 1-a_i=215$ so we find

$$\left(\sum_{i=1}^{216} 1-a_i\right)\left(\sum_{i=1}^{216}\frac{x_i^2}{1-a_i}\right)=1=\left(\sum_{i=1}^{216} x_i\right)^2.$$ This is the equality case of the Cauchy-Schwarz Inequality, so $x_i=c(1-a_i)$ for some constant $c$. Summing these equations and using the facts that $\sum_{i=1}^{216} x_i=1$ and $\sum_{i=1}^{216} 1-a_i=215$, we find $c=\frac{1}{215}$ and thus $x_2=c(1-a_2)=\frac{1}{215}\cdot \left(1-\frac{1}{4}\right)=\frac{3}{860}$. Hence the desired answer is $860+3=\boxed{863}$.

Video Solution

https://youtu.be/mjtM-Coav4k

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