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Difference between revisions of "2022 AIME II Problems/Problem 6"
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==Problem==
 
==Problem==
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Let <math>x_1\leq x_2\leq \cdots\leq x_{100}</math> be real numbers such that <math>|x_1| + |x_2| + \cdots + |x_{100}| = 1</math> and <math>x_1 + x_2 + \cdots + x_{100} = 0</math>. Among all such <math>100</math>-tuples of numbers, the greatest value that <math>x_{76} - x_{16}</math> can achieve is <math>\tfrac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
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==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_{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:
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<cmath>25a - 16b = 1</cmath>
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<cmath>25a + 16b = 0</cmath>
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<math>a = \frac{1}{50}</math>, <math>b = -\frac{1}{32}</math>
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<math>x_{76} - x_{16} = a - b = \frac{1}{50} + \frac{1}{32} = \frac{41}{800}</math>. <math>m+n = \boxed{\textbf{841}}</math>
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~[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
  
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Let <math>x_1\leq x_2\leq \cdots\leq x_{100}</math> be real numbers such that <math>|x_1| + |x_2| + \cdots + |x_{100}| = 1</math> and <math>x_1 + x_2 + \cdots + x_{100} = 0</math>. Among all such <math>100</math>-tuples of numbers, the greatest value that <math>x_{76} - x_{16}</math> can achieve is <math>\tfrac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
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==Solution 3==
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Because the absolute value sum of all the numbers is <math>1</math>, and the normal sum of all the numbers is <math>0</math>, the positive numbers must add to <math>\frac12</math> and negative ones must add to <math>-\dfrac12</math>. To maximize <math>x_{76} - x_{16}</math>, we must make <math>x_{76}</math> as big as possible and <math>x_{16}</math> as small as possible. We can do this by making <math>x_1 + x_2 + x_3 \dots x_{16} = -\dfrac{1}{2}</math>, where <math>x_1 = x_2 = x_3 = \dots = x_{16}</math> (because that makes <math>x_{16}</math> the smallest possible value), and <math>x_{76} + x_{77} + x_{78} + \dots + x_{100} = \dfrac{1}{2}</math>, where similarly <math>x_{76} = x_{77} = \dots = x_{100}</math> (because it makes <math>x_{76}</math> its biggest possible value.) That means <math>16(x_{16}) = -\dfrac{1}{2}</math>, and <math>25(x_{76}) = \dfrac{1}{2}</math>. <math>x_{16} = -\dfrac{1}{32}</math> and <math>x_{76} = \dfrac{1}{50}</math>, and subtracting them <math>\dfrac{1}{50} - \left( -\dfrac{1}{32}\right) = \dfrac{41}{800}</math>. <math>41 + 800 = 841</math>.
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~heheman
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==Video Solution by The Power of Logic==
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https://youtu.be/edpZKDxMsAc
  
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==Solution==
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==Video Solution by Challenge 25==
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https://www.youtube.com/watch?v=OQ8QcJB2hEA
  
 
==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}}

Latest revision as of 11:57, 10 September 2023

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

Solution 3

Because the absolute value sum of all the numbers is $1$, and the normal sum of all the numbers is $0$, the positive numbers must add to $\frac12$ and negative ones must add to $-\dfrac12$. To maximize $x_{76} - x_{16}$, we must make $x_{76}$ as big as possible and $x_{16}$ as small as possible. We can do this by making $x_1 + x_2 + x_3 \dots x_{16} = -\dfrac{1}{2}$, where $x_1 = x_2 = x_3 = \dots = x_{16}$ (because that makes $x_{16}$ the smallest possible value), and $x_{76} + x_{77} + x_{78} + \dots + x_{100} = \dfrac{1}{2}$, where similarly $x_{76} = x_{77} = \dots = x_{100}$ (because it makes $x_{76}$ its biggest possible value.) That means $16(x_{16}) = -\dfrac{1}{2}$, and $25(x_{76}) = \dfrac{1}{2}$. $x_{16} = -\dfrac{1}{32}$ and $x_{76} = \dfrac{1}{50}$, and subtracting them $\dfrac{1}{50} - \left( -\dfrac{1}{32}\right) = \dfrac{41}{800}$. $41 + 800 = 841$.

~heheman

Video Solution by The Power of Logic

https://youtu.be/edpZKDxMsAc

Video Solution by Challenge 25

https://www.youtube.com/watch?v=OQ8QcJB2hEA

See Also

2022 AIME II (Problems • Answer Key • Resources)
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