Question

Para $1 \le i \le 215$ sea $a_i = \frac{1}{2^i}$ y $a_{216} = \frac{1}{2^{215}}.$ Sean $x_1, x_2, \ldots, x_{216}$ números reales positivos tales que $\sum_{i=1}^{216} x_i = 1$ y $\begin{aligned} &\sum_{1 \le i \lt j \le 216} x_i x_j \\ &= \frac{107}{215} \\ &\quad {}+ \sum_{i=1}^{216} \frac{a_i x_i^2}{2(1 - a_i)}. \end{aligned}$ El máximo valor posible de $x_2 = \frac{m}{n},$ donde $m$ y $n$ son enteros positivos primos entre sí. Halla $m + n.$

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 $\sum_{i=1}^{216} x_i = 1$ and $\begin{aligned} &\sum_{1 \le i \lt j \le 216} x_i x_j \\ &= \frac{107}{215} \\ &\quad {}+ \sum_{i=1}^{216} \frac{a_i x_i^2}{2(1 - a_i)}. \end{aligned}$ The maximum possible value of $x_2 = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Respuesta

Solución:

Como $\sum x_i = 1,$ tenemos $2\sum_{i \lt j} x_i x_j = 1 - \sum x_i^2.$ Duplicando la ecuación dada y reordenando, $\begin{aligned} 1 - \sum_{i=1}^{216} x_i^2 &= \frac{214}{215} \\ &\quad {}+ \sum_{i=1}^{216} \frac{a_i x_i^2}{1 - a_i}, \end{aligned}$ así que $\begin{aligned} \frac{1}{215} &= \sum_{i=1}^{216}\left(1 + \frac{a_i}{1 - a_i}\right)x_i^2 \\ &= \sum_{i=1}^{216} \frac{x_i^2}{1 - a_i}. \end{aligned}$

Ahora $\sum a_i = \left(\frac{1}{2} + \cdots + \frac{1}{2^{215}}\right)$ $+ \frac{1}{2^{215}} = 1,$ así que $\sum (1 - a_i) = 216 - 1 = 215.$ Por la desigualdad de Cauchy-Schwarz, $\begin{aligned} 1 &= \left(\sum_{i=1}^{216} x_i\right)^2 \\ &\le \left(\sum_{i=1}^{216} \frac{x_i^2}{1 - a_i}\right) \\ &\quad {}\cdot \left(\sum_{i=1}^{216} (1 - a_i)\right) \\ &= \frac{1}{215} \cdot 215 = 1. \end{aligned}$

Se cumple la igualdad, así que $x_i$ es proporcional a $1 - a_i,$ forzando $x_i = \frac{1 - a_i}{215}.$ El único, y por tanto máximo, valor posible de $x_2$ es $\frac{1 - \frac{1}{4}}{215} = \frac{3}{860},$ y $m + n = 3 + 860 = 863.$

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, $\begin{aligned} 1 - \sum_{i=1}^{216} x_i^2 &= \frac{214}{215} \\ &\quad {}+ \sum_{i=1}^{216} \frac{a_i x_i^2}{1 - a_i}, \end{aligned}$ so $\begin{aligned} \frac{1}{215} &= \sum_{i=1}^{216}\left(1 + \frac{a_i}{1 - a_i}\right)x_i^2 \\ &= \sum_{i=1}^{216} \frac{x_i^2}{1 - a_i}. \end{aligned}$

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, $\begin{aligned} 1 &= \left(\sum_{i=1}^{216} x_i\right)^2 \\ &\le \left(\sum_{i=1}^{216} \frac{x_i^2}{1 - a_i}\right) \\ &\quad {}\cdot \left(\sum_{i=1}^{216} (1 - a_i)\right) \\ &= \frac{1}{215} \cdot 215 = 1. \end{aligned}$

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 Problema 15

Solución:

El Problema 15 en otros años