2013 IMO Problems/Problem 3
Problem
Let the excircle of triangle opposite the vertex be tangent to the side at the point . Define the points on and on analogously, using the excircles opposite and , respectively. Suppose that the circumcentre of triangle lies on the circumcircle of triangle . Prove that triangle is right-angled.
Proposed by Alexander A. Polyansky, Russia
Solution
Let the excenters opposite be . Let the midpoint of be , which lies on , the nine-point circle of ; analogously define .
and , so (SAS), thus is equidistant from , with analogous results for . It follows that the circumcentre of is one of ; WLOG, suppose it is .
By isogonal conjugacy, concur at the Bevan point of . is the common perpendicular bisector of and , so . is the circle on diameter , so by Reim's theorem, .
Hence , as required.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
2013 IMO (Problems) • Resources) | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |