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Problem 1194

AIME late

Geometry Difficulty 5.1 Prove it Saudi Arabian Mathematical Competitions · Saudi Arabia · 2015

Let $ABC$ be a triangle, $H_{a}, H_{b}$ and $H_{c}$ the feet of its altitudes from $A, B$ and $C$ , respectively, $T_{a}, T_{b}, T_{c}$ its intouch points on the sides $BC, CA$ and $AB$ , respectively. The circumcircles of triangles $A H_{b} H_{c}$ and $A T_{b} T_{c}$ intersect again at $A'$ . The circumcircles of triangles $B H_{c} H_{a}$ and $B T_{c} T_{a}$ intersect again at $B'$ . The circumcircles of triangles $C H_{a} H_{b}$ and $C T_{a} T_{b}$ intersect again at $C'$ . Prove that the points $A', B', C'$ are collinear.

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 $H$ and $I$ be the orthocenter and the incenter of triangle $ABC$ , respectively. Because $\angle A H_{b} H = \angle H H_{c} A = 90^{\circ}$ , $AH$ is a diameter of the circumcircle of $A H_{c} H H_{b}$ , and therefore $\angle A A' H = 90^{\circ}$ . Because $\angle A T_{b} I = \angle I T_{c} A = 90^{\circ}$ , $AI$ is a diameter of the circumcircle of $A T_{c} I T_{b}$ , and therefore $\angle A A' I = 90^{\circ}$ . We deduce that point $A'$ is on the line $HI$ .

Similarly, we prove that the points $B'$ and $C'$ are on the line $HI$ , which prove that the points $A', B', C'$ are collinear.

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