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== | ==Solution 1== | ||
Line 43: | Line 43: | ||
==Video Solution by The Power of Logic== | ==Video Solution by The Power of Logic== | ||
https://youtu.be/edpZKDxMsAc | https://youtu.be/edpZKDxMsAc | ||
+ | |||
+ | ==Video Solution by Challenge 25== | ||
+ | https://www.youtube.com/watch?v=OQ8QcJB2hEA | ||
==See Also== | ==See Also== |
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.