1995 USAMO Problems/Problem 3
Problem
Given a nonisosceles, nonright triangle let denote its circumcenter, and let and be the midpoints of sides , and respectively. Point is located on the ray so that is similar to . Points and on rays and respectively, are defined similarly. Prove that lines and are concurrent.
Solution
LEMMA 1: In with circumcenter , .
PROOF of Lemma 1: The arc equals which equals . Since is isosceles we have that . QED
Define s.t. . Since , . Let and . Since we have , we have that . Also, we have that . Furthermore, , by lemma 1. Therefore, . Since is the midpoint of , is the median. However tells us that is just reflected across the internal angle bisector of . By definition, is the -symmedian. Likewise, is the -symmedian and is the -symmedian. Since the symmedians concur at the symmedian point, we are done.
QED
See Also
1995 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
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