2022 AIME II 第 6 题

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

小提示：

两个条件迫使正项之和为

\frac{1}{2}

，负项之和为

-\frac{1}{2}

The two conditions force the positive terms to sum to

\frac{1}{2}

and the negative terms to sum to

-\frac{1}{2}

大提示：

若

x_{16}

小于

-\frac{1}{32}

，仅前

16

项之和就会小于

-\frac{1}{2}

；用同样方式由最后

25

项约束

x_{76}

If

x_{16}

were below

-\frac{1}{32},

the first

16

terms alone would sum below

-\frac{1}{2};

bound

x_{76}

the same way using the last

25

terms

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}

。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}

，即可取到这个值。

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

小提示：

大提示：

解答：

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