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Problem 1758

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.1 Prove it Team Selection Test for IMO · Turkey · 2007

The acute triangle $ABC$ and the triangle $A_1B_1C_1$ , whose vertices $B_1$ , $C_1$ and $A_1$ lie on the rays $AC$ , $BA$ and $CB$ , respectively, are similar. Prove that the orthocenter of the triangle $A_1B_1C_1$ and the circumcenter of the triangle $ABC$ coincide.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $H$ be the orthocenter of the triangle $A_1B_1C_1$ . By the similarity of the triangles $ABC$ and $A_1B_1C_1$ , we have $\angle A_1B_1C_1 = \angle ABC$ . From this, $\angle A_1HC_1 = 180^\circ - \angle A_1B_1C_1 = 180^\circ - \angle ABC = \angle A_1BC_1$ is obtained. Therefore the points $A_1$ , $B$ , $H$ , $C_1$ are concyclic. It follows that $\angle HBA = \angle HA_1C_1$ . Similarly, $\angle HAC = \angle HC_1B_1$ .

Also since $A_1$ , $B$ , $H$ , $C_1$ are concyclic, the point $H$ does not lie in the triangle $A_1C_1B$ . Similarly, $H$ does not lie in the triangles $B_1A_1C$ and $C_1B_1A$ . So $H$ lies inside the triangle $ABC$ .

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Then, $\angle HAB = \angle CAB - \angle HAC = \angle C_1A_1B_1 - \angle HC_1B_1 = \angle C_1A_1B_1 - \angle HA_1B_1 = \angle HA_1C_1 = \angle HBA$ , and consequently, $HA = HB$ . Similarly, $HA = HC$ . Therefore the point $H$ is the circumcenter of the triangle $ABC$ .

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