75th NMO Selection Tests

Let $\{P_3\}={\Gamma }_1\cap {\Gamma }_2$ be the second point of intersection of the circles ${\Gamma }_1$ and ${\Gamma }_2$, and let $O_1,O_2,O_3$ be the centers of the circles ${\Gamma }_1,{\Gamma }_2$, and ${\Gamma }_3$, respectively. Denote by $O_4$ the intersection point of the common tangents to the circles ${\Gamma }_1$ and ${\Gamma }_2$.

$$\begin{array}{rl}\widehat{P_{1}Z{P}_2}& =\frac{1}{2}(\widehat{AB}+\widehat{P_1{P}_2})=\frac{1}{2}(180^{\circ }+\widehat{P_1{O}_3{P}_2})\\ & =\frac{1}{2}(180^{\circ }+180^{\circ }-\widehat{P_1{O}_2{P}_3}-\widehat{P_2{O}_1{P}_3})=\\ & =\frac{1}{2}(180^{\circ }-\widehat{P_1{O}_2{P}_3})+\frac{1}{2}(180^{\circ }-\widehat{P_2{O}_1{P}_3})\\ & =\widehat{O_1{P}_3{P}_2}+\widehat{O_2{P}_3{P}_1}=180^{\circ }-\widehat{P_1{P}_3{P}_2},\end{array}$$ so the quadrilateral $Z{P}_1{P}_3{P}_2$ is cyclic.

We will prove that $X,P_3$, and $Z$ are collinear. It suffices to show that $\widehat{X{P}_3{O}_1}=\widehat{Z{P}_3{O}_2}$. Since triangle $O_{1}X{P}_3$ is isosceles and $B{P}_2\perp AX$ (because $AB$ is a diameter), we have: $$\begin{array}{rl}\widehat{X{P}_3{O}_1}& =\frac{1}{2}(180^{\circ }-\widehat{X{O}_1{P}_3})=90^{\circ }-\frac{1}{2}\widehat{X{P}_3}=\widehat{X{P}_{2}B}-\widehat{X{P}_2{P}_3}=\widehat{B{P}_2{P}_3}\\ & =\widehat{Z{P}_2{P}_3}=\frac{1}{2}\widehat{Z{P}_1{P}_3}.\end{array}$$ Because $O_1{O}_2$ is tangent to the circumcircle of triangle $P_1{P}_2{P}_3$, it follows that $\widehat{P_1{P}_3{O}_2}=\frac{1}{2}\widehat{P_1{P}_3}$. Therefore, $\widehat{Z{P}_3{O}_2}=\widehat{Z{P}_3{P}_1}+\widehat{P_1{P}_3{O}_2}=\frac{1}{2}\widehat{Z{P}_1}+\frac{1}{2}\widehat{P_1{P}_3}=\frac{1}{2}\widehat{Z{P}_1{P}_3}=\widehat{X{P}_3{O}_1}$. Hence, points $X,Z$, and $P_3$ are collinear. Similarly, one can show that $Y,Z$, and $P_3$ are collinear, and thus $X,Y$, and $Z$ are collinear as well.