2022 AIME II Problems/Problem 6

Problem

Let $x_1\leq x_2\leq \cdots\leq x_{100}$ 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 $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution 1

To find the greatest value of $x_{76} - x_{16}$ , $x_{76}$ must be as large as possible, and $x_{16}$ must be as small as possible. If $x_{76}$ is as large as possible, $a = x_{76} = x_{77} = x_{78} = \dots = x_{100} > 0$ . If $x_{16}$ is as small as possible, $b = x_{16} = x_{15} = x_{14} = \dots = x_{1} < 0$ . The other numbers between $x_{16}$ and $x_{76}$ equal to $0$ . Substituting $a$ and $b$ into $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ we get: $25a - 16b = 1$ $25a + 16b = 0$ $a = \frac{1}{50}$ , $b = -\frac{1}{32}$

$x_{76} - x_{16} = a - b = \frac{1}{50} + \frac{1}{32} = \frac{41}{800}$ . $m+n = \boxed{\textbf{841}}$

~isabelchen

2022 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2022 AIME II Problems/Problem 6

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Solution 1

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