2000 USAMO Problems/Problem 5
is the intersection of the perpendicular bisectors of , and each of the centers lie on the perpendicular bisector of the side of the triangle that determines . It follows from that .
Since , and the perpendicular bisector of are fixed, the angle determines the position of (since lies on the perpendicular bisector). Let ; then, and together imply that .
Now (due to collinearility). Hence, we have the recursion , and so . Thus, .
implies that , and circles and are the same circle since they have the same center and go through the same two points.
Using the collinearity of certain points and the fact that is isosceles, we quickly deduce that From ASA Congruence we deduce that and are congruent triangles, and so , that is .
|2000 USAMO (Problems • Resources)|
|1 • 2 • 3 • 4 • 5 • 6|
|All USAMO Problems and Solutions|