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 1 (No Trig)
Let and the foot of the altitude from to be point and . Since ABC is isosceles, is on . By Pythagorean Theorem, . Let and . By angle bisector theorem, . Also, . Solving for , we get . Then, using Pythagorean Theorem on we have . Simplifying, we have . Factoring out the , we have . Adding 1 to the fraction and simplifying, we have . Crossing out the , and solving for yields . Then, we continue as Solution 2 does.
Solution 2 (Trig)
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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