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

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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
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==Solution 2==
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Define <math>s_N</math> to be the sum of all the negatives, and <math>s_P</math> to be the sum of all the positives.
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Since the sum of the absolute values of all the numbers is <math>1</math>, <math>|s_N|+|s_P|=1</math>.
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Since the sum of all the numbers is <math>0</math>, <math>s_N=-s_P\implies |s_N|=|s_P|</math>.
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Therefore, <math>|s_N|=|s_P|=\frac 12</math>, so <math>s_N=-\frac 12</math> and <math>s_P=\frac 12</math> since <math>s_N</math> is negative and <math>s_P</math> is positive.
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To maximize <math>x_{76}-x_{16}</math>, we need to make <math>x_{16}</math> as small of a negative as possible, and <math>x_{76}</math> as large of a positive as possible.
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Note that <math>x_{76}+x_{77}+\cdots+x_{100}=\frac 12</math> is greater than or equal to <math>25x_{76}</math> because the numbers are in increasing order.
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Similarly, <math>x_{1}+x_{2}+\cdots+x_{16}=-\frac 12</math> is less than or equal to <math>16x_{16}</math>.
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So we now know that <math>\frac 1{50}</math> is the best we can do for <math>x_{76}</math>, and <math>-\frac 1{32}</math> is the least we can do for <math>x_{16}</math>.
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Finally, the maximum value of <math>x_{76}-x_{16}=\frac 1{50}+\frac 1{32}=\frac{41}{800}</math>, so the answer is <math>\boxed{841}</math>.
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(Indeed, we can easily show that <math>x_1=x_2=\cdots=x_{16}=-\frac 1{32}</math>, <math>x_{17}=x_{18}=\cdots=x_{75}=0</math>, and <math>x_{76}=x_{77}=\cdots=x_{100}=\frac 1{50}</math> works.)
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~inventivedant
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2022|n=II|num-b=5|num-a=7}}
 
{{AIME box|year=2022|n=II|num-b=5|num-a=7}}
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[[Category:Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:45, 25 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

Solution 2

Define $s_N$ to be the sum of all the negatives, and $s_P$ to be the sum of all the positives.

Since the sum of the absolute values of all the numbers is $1$, $|s_N|+|s_P|=1$.

Since the sum of all the numbers is $0$, $s_N=-s_P\implies |s_N|=|s_P|$.

Therefore, $|s_N|=|s_P|=\frac 12$, so $s_N=-\frac 12$ and $s_P=\frac 12$ since $s_N$ is negative and $s_P$ is positive.

To maximize $x_{76}-x_{16}$, we need to make $x_{16}$ as small of a negative as possible, and $x_{76}$ as large of a positive as possible.

Note that $x_{76}+x_{77}+\cdots+x_{100}=\frac 12$ is greater than or equal to $25x_{76}$ because the numbers are in increasing order.

Similarly, $x_{1}+x_{2}+\cdots+x_{16}=-\frac 12$ is less than or equal to $16x_{16}$.

So we now know that $\frac 1{50}$ is the best we can do for $x_{76}$, and $-\frac 1{32}$ is the least we can do for $x_{16}$.

Finally, the maximum value of $x_{76}-x_{16}=\frac 1{50}+\frac 1{32}=\frac{41}{800}$, so the answer is $\boxed{841}$.

(Indeed, we can easily show that $x_1=x_2=\cdots=x_{16}=-\frac 1{32}$, $x_{17}=x_{18}=\cdots=x_{75}=0$, and $x_{76}=x_{77}=\cdots=x_{100}=\frac 1{50}$ works.)

~inventivedant

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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