2006 iTest Problems/Problem U5
The following problem is from the Ultimate Question of the 2006 iTest, where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted.
Problem
In triangle , points , , and are the feet of the angle bisectors of , , respectively. Let point be the intersection of segments and , and let denote the perimeter of . If , , and , then the value of can be expressed uniquely as where and are positive integers such that is not divisible by the square of any prime. Find .
Solution
According to the Steiner-Lehmus Theorem, . Thus, by SAS Congruency, so and .
Let and . Since and share an altitude and and share an altitude, we know that and . Also, since , . Similarly, and .
Since and share an altitude, we have
Now let and . By the Angle Bisector Theorem, , so . By using the Pythagorean Theorem on , we have , so . By using the Pythagorean Theorem on ,
Therefore, the . The perimeter of the triangle equals , and . Thus, , and .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
2006 iTest (Problems) | ||
Preceded by: Problem U4 |
Followed by: Problem U6 | |
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