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Difference between revisions of "2022 AIME II Problems/Problem 6"

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

Revision as of 05:06, 19 February 2022

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, $x_{76} = x_{77} = x_{78} = \dots = x_{100} > 0$. If $x_{16}$ is as small as possible, $x_{16} = x_{15} = x_{14} = \dots = x_{1} < 0$. The other numbers between $x_{16}$ and $x_{76}$ equal to $0$. Let $a = x_{76}$, $b = x_{16}$. 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

See Also

2022 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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