Team Selection Test for EGMO 2024

Let ${\ell }_1\cap {\ell }_2=P$. Since $\angle AXY=\angle YAC$ and $\angle CXY=\angle YCA$ we have $\angle AXC+\angle AYC=180^{\circ }$. Therefore, by symmetry, we have $\angle AZC=\angle AYC=180^{\circ }-\angle AXC$ hence $A,X,C,Z$ are concyclic.

Now we will prove that $PX$ is tangent to this circle. Let the second intersection point of the line $PX$ with the circle ${\omega }_1$ be $Q$. Since $P$ is the center of the homothety sending ${\omega }_1$ to ${\omega }_2$, it sends $A$ to $C$ and $Q$ to $P$. Therefore, we have $AQ\parallel XC$ and $\angle XCA=\angle QAP$ and by the tangency we have $\angle QAP=\angle QXA$ hence $\angle XCA=\angle PXA$ which implies the desired tangency.

Since $P$ lies on the symmetry axis of the two circles, we have $PX=PY=PZ=\sqrt{PA\cdot PC}$ and $PZ$ is also tangent to the circle $(AXCZ)$. Letting $XZ\cap AC=T$, we then have that $(A,C;P,T)=-1$. Then we have $(KA,KC;KP,KT)=-1$. Let the second intersection of $PK$ and ${\omega }_1$ be $S$. Then by carrying this harmonic bundle to the circle ${\omega }_1$ we see that $(A,B;S,KT\cap {\omega }_1)=-1$. On the other hand, we know that $(A,B;S,K)=-1$ hence $KT$ is the tangent to ${\omega }_1$ at $K$.

Similarly we can prove that $LT$ is the line passing through $L$ and tangent to ${\omega }_2$, hence the proof is completed.