1995 USAMO Problems/Problem 3
Problem
Given a nonisosceles, nonright triangle let denote the center of its circumscribed circle, 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, i.e. these three lines intersect at a point.
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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