31st Turkish Mathematical Olympiad

From the radical axis theorem on the circles $(ABC)$, $(BCP)$ and ${\omega }_A$, the tangent lines to ${\omega }_A$ passing through $P$ and $A_1$ intersect on the line $BC$, let us say at point $D$. From the radical axis theorem on the circles $(ABC)$, $(BCP)$ and ${\Gamma }_A$, we see that the line passing through $A_2$ and tangent to the circle $(ABC)$ passes through $D$ as well. Therefore, $D$ is the pole of the line $A_1{A}_2$. Define the points $E,F$ analogously. Looking at the similarity $DBP\sim DPC$ we have

$$\frac{BP}{PC}=\frac{DB}{DP}=\frac{DP}{DC}\text{and hence}{\left(\frac{BP}{PC}\right)}^2=\frac{DB}{DC}$$

$${\left(\frac{CP}{PA}\right)}^2=\frac{EC}{EA}\text{and}{\left(\frac{AP}{PB}\right)}^2=\frac{FA}{FB}$$ Therefore, by Menelaus' Theorem we get that the points $D,E,F$ are collinear. Thus, from La Hire Theorem we can deduce that the lines $A_1{A}_2$, $B_1{B}_2$ and $C_1{C}_2$ are concurrent. Let the common point of these lines be $M$. Since $M$ is the pole of the line $\overline{DEF}$ we get $OM\perp DE$. Therefore, in order to complete the solution we should prove that $O,M,P$ are collinear. Hence it suffices to show that $OP$ is perpendicular to $DE$. Let $r$ be the circumradius of the triangle $ABC$. From the power of the point $D$ with respect to the circle $(ABC)$ we have $D{P}^2=DB\cdot DC=D{O}^2-r^2$. Similarly, from the power of the point $E$ with respect to the circle $(ABC)$ we get $E{P}^2=E{O}^2-r^2$. Hence $D{P}^2+E{O}^2=D{O}^2+E{P}^2$ and we get $OP\perp DE$. We are done.