2015 AIME I Problems/Problem 11
Problem
Triangle has positive integer side lengths with . Let be the intersection of the bisectors of and . Suppose . Find the smallest possible perimeter of .
Solution
Let be the midpoint of . Then by SAS Congruence, , so .
Now let , , and .
Then
and .
Cross-multiplying yields .
Since , must be positive, so .
Additionally, since has hypotenuse of length , .
Therefore, given that is an integer, the only possible values for are , , , and .
However, only one of these values, yields an integral value for , so we conclude that and .
Thus the perimeter of must be .
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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All AIME Problems and Solutions |
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