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Geometry Difficulty 6.1 National olympiad Prove it Belarus

Let $\Gamma$ be the incircle of a none-isosceles triangle $ABC$ , $I$ be its center. Let $A_1$ , $B_1$ , $C_1$ be the tangency points of $\Gamma$ with the sides $BC$ , $AC$ , $AB$ , respectively. Let $A_2 = \Gamma \cap AA_1$ , $M = C_1B_1 \cap AI$ , $P$ and $Q$ be the other (different from $A_1$ , $A_2$ ) intersection points of $A_1M$ , $A_2M$ and $\Gamma$ , respectively.
Prove that $A$ , $P$ , $Q$ are collinear.

Solution

We first show that the quadrilateral $A_1IMA_2$ is cyclic. Indeed, $C_1B_1 \perp AI$ and $IC_1 \perp AB$ . So, $C_1M$ is an altitude in the right-angled triangle $AIC_1$ , whence $AM \cdot AI = AC_1^2$ . Further, by the power of a point theorem, $AC_1^2 = A_2A \cdot AA_1$ . So,
$AM \cdot AI = AC_1^2 = A_2A \cdot AA_1,$
hence $A_1IMA_2$ is cyclic. Therefore, $\angle IMA_1 = \angle IA_2A_1$ , $\angle A_2MI + \angle IA_1A_2 = 180^\circ$ . Since $IA_2 = IA_1$ we have $\angle IA_2A_1 = \angle IMA_1$ . Hence,
$\angle QMI = 180^\circ - \angle A_2MI = \angle IA_1A_2 = \angle IA_2A_1 = \angle IMA_1. \quad (1)$
From (1) it follows that the symmetrical image of the line $A_2Q$ with respect to the line $AI$ is $PA_1$ . Hence $P$ is the image of $A_2$ and $Q$ is the image of $A_1$ . Therefore the line $PQ$ is the image of $A_2A_1$ with respect to the line $AI$ , thus the statement follows.

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