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Problem A. 540. (September 2011)

A. 540. Let A1A2A3 be a non-equilateral triangle, let point M be its orthocenter, let F be its Feuerbach point, and let k be the circumcircle of the triangle. For i=1,2,3 denote by ki the circle that is internally tangent to k and tangent to the sides AiAi+1 and AiAi+2. (The indices are considered modulo 3, i.e. A4=A1 and A5=A2.) Let Ti be the point of tangency between k and ki. Prove that the lines A1T1, A2T2, A3T3 and MF are concurrent.

Proposed by: Gábor Damásdi and Márton Mester, Budapest

(5 pont)

Deadline expired on 10 October 2011.

Solution. Denote by b the incircle and by f the 9-point circle and let H be the external homothety center between f and k.

Apply Monge's circle theorem to the circles b, k and ki. The three external homothety centers between b and ki, k and ki, b and k are Ai, Ti and H, respectively; by the theorem these points are collinear. Therefore the line AiTi passes through H (i=1,2,3).

Now apply Monge's theorem to the circles b, f and k. The three homotety centers between them are M, F and H. Therefore the line FM also passes through H.


6 students sent a solution.
5 points:Ágoston Tamás, Gyarmati Máté, Mester Márton, Omer Cerrahoglu, Szabó 928 Attila.
0 point:1 student.

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