Difference between revisions of "2022 AIME II Problems/Problem 6"
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==Problem== | ==Problem== | ||
+ | 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>. | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | 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 = 0</cmath> | ||
+ | <math>a = \frac{1}{50}</math>, <math>b = -\frac{1}{32}</math> | ||
+ | |||
+ | <math>x_{76} - x_{16} = a - b = \frac{1}{50} + \frac{1}{32} = \frac{41}{800}</math>. <math>m+n = \boxed{\textbf{841}}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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. | ||
+ | |||
+ | Since the sum of the absolute values of all the numbers is <math>1</math>, <math>|s_N|+|s_P|=1</math>. | ||
+ | |||
+ | Since the sum of all the numbers is <math>0</math>, <math>s_N=-s_P\implies |s_N|=|s_P|</math>. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | Similarly, <math>x_{1}+x_{2}+\cdots+x_{16}=-\frac 12</math> is less than or equal to <math>16x_{16}</math>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | (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.) | ||
+ | |||
+ | ~inventivedant | ||
− | + | ==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>. | ||
+ | |||
+ | ~heheman | ||
+ | |||
+ | ==Video Solution by The Power of Logic== | ||
+ | https://youtu.be/edpZKDxMsAc | ||
− | ==Solution== | + | ==Video Solution by Challenge 25== |
+ | 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}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:57, 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:
,
.
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.