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>. | 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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~inventivedant | ~inventivedant | ||
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+ | ==Solution 3== | ||
+ | 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== | ||
+ | https://youtu.be/edpZKDxMsAc | ||
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+ | ==Video Solution by Challenge 25== | ||
+ | https://www.youtube.com/watch?v=OQ8QcJB2hEA | ||
==See Also== | ==See Also== |
Latest revision as of 11:57, 10 September 2023
Contents
Problem
Let be real numbers such that and . Among all such -tuples of numbers, the greatest value that can achieve is , where and are relatively prime positive integers. Find .
Solution 1
To find the greatest value of , must be as large as possible, and must be as small as possible. If is as large as possible, . If is as small as possible, . The other numbers between and equal to . Let , . Substituting and into and we get: ,
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Solution 2
Define to be the sum of all the negatives, and to be the sum of all the positives.
Since the sum of the absolute values of all the numbers is , .
Since the sum of all the numbers is , .
Therefore, , so and since is negative and is positive.
To maximize , we need to make as small of a negative as possible, and as large of a positive as possible.
Note that is greater than or equal to because the numbers are in increasing order.
Similarly, is less than or equal to .
So we now know that is the best we can do for , and is the least we can do for .
Finally, the maximum value of , so the answer is .
(Indeed, we can easily show that , , and works.)
~inventivedant
Solution 3
Because the absolute value sum of all the numbers is , and the normal sum of all the numbers is , the positive numbers must add to and negative ones must add to . To maximize , we must make as big as possible and as small as possible. We can do this by making , where (because that makes the smallest possible value), and , where similarly (because it makes its biggest possible value.) That means , and . and , and subtracting them . .
~heheman
Video Solution by The Power of Logic
Video Solution by Challenge 25
https://www.youtube.com/watch?v=OQ8QcJB2hEA
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.