USAMO 2009 #5 July 19, 2009Posted by Damien Jiang in Problem-solving, Uncategorized.
Tags: geometry, olympiad math
I like Olympiad geometry. Therefore, I will give my solution to this year’s USAMO #5; I was rather happy with my solution.
5. Trapezoid , with , is inscribed in circle and point lies inside triangle . Rays and meet again at points and , respectively. Let the line through parallel to intersects and at points and , respectively. Prove that quadrilateral is cyclic if and only if bisects .
First, note that since the homothety centered at that takes to takes their circumcircles to each other (and therefore their centers), their tangents at are the same. I.e., the circumcircles of and are internally tangent.
Let . Then we know . By cyclic quadrilaterals, where is the intersection of the common tangent at and . Hence , or are cyclic.
This implies , i.e. . But then since lie on , we obtain, by Power of a Point, , or is cyclic as desired.
Only if direction:
Let be the radical center of the circles of . (There is no degenerate case here, which was a problem in #1 of this year, because we have an isosceles trapezoid!) Since are collinear, , so is cyclic. Hence, . But and , so as desired.