Team Selection Test for EGMO 2023

Let the circumcircle of the triangle $O_1{O}_2{O}_3$ be $\Gamma$. Let the tangency point of ${\omega }_1$ and ${\omega }_2$ be $D$, ${\omega }_1$ and ${\omega }_3$ be $E$, ${\omega }_2$ and ${\omega }_3$ be $F$. Let the incenter of $O_1{O}_2{O}_3$ be $I$, then it's easy to see that $ID,IE,IF$ are perpendicular to the respective sides of $O_1{O}_2{O}_3$ since we have $|O_{1}D|=|O_{1}E|$, etc. The radical axis of circles: ${\omega }_1,{\omega }_2,\Gamma$ are concurrent, let's say at point $X$ where $Y,Z$ are defined similarly. Then $X$ lies on the perpendicular $IF$ since it is the radical axis of the tangent circles ${\omega }_1,{\omega }_2$. Let $O$ be the center of $\Gamma$ and $O{O}_1\cap A_1{B}_1=P$, then $O{O}_1\perp A_1{B}_1$ and $X,D,P,O_1$ is cyclic. Therefore, $\angle DX{O}_1=\angle D{O}_{1}O$ and similarly $\angle DX{O}_2=\angle D{O}_{2}O$ and $XI$ is an angle bisector of the triangle $XYZ$, results for $Y$ and $Z$ follow similarly.