2005 IMO Shortlist Problems/G3
(Ukraine) Let be a parallelogram. A variable line passing through the point intersects the rays and at points and , respectively. Let and be the centres of the excircles of triangles and , touching the sides and , respectively. Prove that the size of angle does not depend on the choice of .
This was also Problem 3 of the 2006 3rd German TST, and a problem at the 2006 India IMO Training Camp. It also appeared in modified form as Problem 3, Day 3 of the 2006 Iran TST.
Let be the interior angle bisectors of . Let be the exterior angle bisectors of . Then is the intersection of and is the intersection of .
Let us denote as the measures of , and denote . Then . Furthermore, since is the exterior angle bisector of , we know that the exterior angle at is , so . Similarly, . It follows that triangles are similar. Then since is a parallelogram,
Since we know , this implies that triangles are similar. This means that
which is independent of , since is always half the measure of . ∎
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.