$ (O_1) \equiv (PQB_1),$ $ (O_2) \equiv (PQA_1)$ and let $ A_2' \equiv PB_2 \cap (O_2).$ Reflection $ B_1B_2$ of $ QB_1$ is symmetric WRT ray $ B_1O_1,$ i.e. $ \triangle QB_1B_2$ is isosceles with base $ QB_2.$ Let $ R$ be the second intersection of the ray $ B_1Q$ with $(O_2).$ Since $ \angle B_1QB_2 = \angle QRA_1,$ it follows that $ RA_1 \parallel QB_2.$ Let $ S \equiv B_1B_2 \cap RA_1,$ note that $ PA_1SB_2$ is cyclic because of $ \angle PA_1R = \angle PB_2B_1$ $\Longrightarrow$ $ \angle B_2PA_1 = \angle B_2SR.$ But $ QRSB_2$ is an isosceles trapezoid with bases $ QB_2 \parallel RS$ $\Longrightarrow$ $\angle A_2'PA_1 = \angle A_1A_2'Q$ $\Longrightarrow$ Chords $ A_1Q$ and $ A_1A_2'$ are equal. As a result, $ A_1A_2'$ is the reflection of $ QA_1$ about $ O_2A_1$ $\Longrightarrow$ $ A_1A_2$ and $ A_1A_2'$ are identical, then $ A_2B_2$ goes through $ P.$ Let $ A_2A_3$ be the reflection of $ A_2A_1$ about $ O_2A_2$ and $ B_2B_3$ the reflection of $ B_2B_1$ about $ O_1B_2.$ Since $ QB_3B_1B_2$ and $ QA_3A_1A_2$ are isosceles trapezoids, we have $ \angle B_1QB_3 = \angle QB_1B_2 = \angle B_1PB_3.$ Analogously, $ \angle A_3PA_1 = \angle QA_1A_2,$ but $ \angle QB_1B_2 = \angle QA_1A_2,$ hence $ \angle A_3PA_1 = \angle B_3PB_1$ $ \Longrightarrow$ $A_3,B_3,P$ are collinear.

The key point of this problem is the collinearity fact of A1,P and B1. In physics sense it means that two material or immaterial points rotate along the given circles in the same direction starting in point Q then they meet again at Q. By applying one of those facts that two reflections is equivalent to one rotation (or something like that) to these two points with equal angular velocities will make all Ai, P and Bi to be collinear. M.T.

Because of the collinearity of the points $ A_{1},\ P,\ B_{1},$ we conclude that the isosceles triangles $ \bigtriangleup A_{1}QA_{2}\ \ ( A_{1}Q = A_{1}A_{2}$ $ )$ and $ \bigtriangleup B_{1}QB_{2}\ \ ( B_{1}Q = B_{1}B_{2}$ $ )$ are similar, from $ \angle QB_{2}B_{1} = \angle QPA_{1} = \angle QA_{2}A_{1}$ $ ,(1)$ So, from $ (1)$ $ \Longrightarrow$ $ \angle A_{1}PA_{2} = \angle A_{1}QA_{2} = \angle QA_{2}A_{1} = \angle QB_{2}B_{1} = \angle B_{1}QB_{2} = \angle B_{1}PB_{2}$ $ ,(2)$ and hence, we conclude that the points $ A_{2},\ P,\ B_{2},$ are collinear. By the same way, from the collinearity of $ A_{2},\ P,\ B_{2},$ we conclude the collinearity of the points $ A_{3},\ P,\ B_{3}$ and so on. That is, from the collinearity of the points $ A_{i - 1},\ P,\ B_{i - 1},$ we conclude the collinearity of the points $ A_{i},\ P,\ B_{i}$ and the proof is completed. Kostas Vittas.

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