2022 AIME II Problems/Problem 6


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: \[25a - 16b = 1\] \[25a + 16b = 0\] $a = \frac{1}{50}$, $b = -\frac{1}{32}$

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


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.)


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 $\dfrac{1}{50} - \left( -\dfrac{1}{32}\right) = \dfrac{41}{800}$. $41 + 800 = 841$.


Video Solution by The Power of Logic


Video Solution by Challenge 25


See Also

2022 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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