Difference between revisions of "2022 AIME II Problems/Problem 6"
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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>. | ||
| − | ==Solution== | + | ==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>a = x_{76} = x_{77} = x_{78} = \dots = x_{100} > 0</math>. If <math>x_{16}</math> is as small as possible, <math>b = x_{16} = x_{15} = x_{14} = \dots = x_{1} < 0</math>. The other numbers equal to <math>0</math>. | ||
| + | |||
| + | To be continued...... | ||
| + | |||
| + | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
==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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 05:51, 19 February 2022
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 equal to 
.
To be continued......
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. 
