Question

设 $x_1 \le x_2 \le \cdots \le x_{100}$ 为实数，满足 $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ 且 $x_1 + x_2 + \cdots + x_{100} = 0$ 。在所有这样的 $100$ 元数组中， $x_{76} - x_{16}$ 能达到的最大值为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质的正整数。求 $m + n$ 。

Let $x_1 \le x_2 \le \cdots \le x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0.$ Among all such $100$ -tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：841

知识点：不等式绝对值最优化

难度评级：2600

解答：

因为各项之和为 $0$ ，而绝对值之和为 $1$ ，正项之和为 $\frac{1}{2}$ ，负项之和为 $-\frac{1}{2}$ 。若 $x_{16} \lt -\frac{1}{32}$ ，则 $x_1, \ldots, x_{16}$ 都小于 $-\frac{1}{32}$ ，它们的和会小于 $-\frac{1}{2}$ ，矛盾；因此 $x_{16} \ge -\frac{1}{32}$ 。类似地，若 $x_{76} \gt \frac{1}{50}$ ，则 $x_{76}, \ldots, x_{100}$ 这 $25$ 项都大于 $\frac{1}{50}$ ，总和超过 $\frac{1}{2}$ ；因此 $x_{76} \le \frac{1}{50}$ 。

所以 $x_{76} - x_{16} \le \frac{1}{50} + \frac{1}{32}$ $= \frac{16 + 25}{800}$ $= \frac{41}{800}$ ，这个值可以通过取 $x_1 = \cdots = x_{16} = -\frac{1}{32}$ ， $x_{17} = \cdots = x_{75} = 0$ ，以及 $x_{76} = \cdots = x_{100} = \frac{1}{50}$ 。实现。

因为 $\gcd(41, 800) = 1$ ，答案为 $41 + 800 = 841$ 。

Since the terms sum to $0$ while their absolute values sum to $1,$ the positive terms sum to $\frac{1}{2}$ and the negative terms sum to $-\frac{1}{2}.$ If $x_{16} \lt -\frac{1}{32},$ then $x_1, \ldots, x_{16}$ are all less than $-\frac{1}{32}$ and would sum below $-\frac{1}{2},$ a contradiction; hence $x_{16} \ge -\frac{1}{32}.$ Similarly, if $x_{76} \gt \frac{1}{50}$ then $x_{76}, \ldots, x_{100}$ are $25$ terms each exceeding $\frac{1}{50},$ summing above $\frac{1}{2};$ hence $x_{76} \le \frac{1}{50}.$

Therefore $x_{76} - x_{16} \le \frac{1}{50} + \frac{1}{32}$ $= \frac{16 + 25}{800}$ $= \frac{41}{800},$ and this is achieved by taking $x_1 = \cdots = x_{16} = -\frac{1}{32},$ $x_{17} = \cdots = x_{75} = 0,$ and $x_{76} = \cdots = x_{100} = \frac{1}{50}.$

Since $\gcd(41, 800) = 1,$ the answer is $41 + 800 = 841.$

2022 AIME II 第 6 题

答案：841

难度评级：2600

解答：

其他年份的第 6 题