1998 USAMO Problems/Problem 2
Contents
Problem
Let and be concentric circles, with in the interior of . From a point on one draws the tangent to (). Let be the second point of intersection of and , and let be the midpoint of . A line passing through intersects at and in such a way that the perpendicular bisectors of and intersect at a point on . Find, with proof, the ratio .
Solution
First, . Because , and all lie on a circle, . Therefore, , so . Thus, quadrilateral is cyclic, and must be the center of the circumcircle of , which implies that . Putting it all together,
Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol982.html
Solution 2
We will use signed lengths. WLOG let , , and . Then and . Therefore, and . By Power of a Point, , so , and . Since is on the perpendicular bisectors of and , we have and .
By Stewart's Theorem on and cevian ,
So either or . If , then and , so the perpendicular bisectors of and are the same line, and they do not intersect at a point. Therefore, and , so .
See Also
1998 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.