IMO Team Selection Test 1

We consider the configuration in which $Y$ lies between $A$ and $X$; the other configurations are treated analogously. Let $C$ be the intersection of ${\ell }_2$ and ${\Gamma }_1$. Then $C$ is the reflection of $A$ in $O_1{O}_2$. We get $$\begin{array}{rlrl}\angle O_{1}YX& =180^{\circ }-\angle O_{1}YA& & \text{(straight angle)}\\ & =180^{\circ }-\angle YA{O}_1& & \text{(O1YA is isosceles)}\\ & =180^{\circ }-\angle XA{O}_1& & \text{(A is the reflection of C in O1X)}\\ & =180^{\circ }-\angle XC{O}_1& & \text{(A is the reflection of C in O1X)}\end{array}$$ which yields that $O_{1}CXY$ is cyclic.

Now let $S$ be the intersection of the line through $Y$ tangent to ${\Gamma }_1$, and the line ${\ell }_2$. Then both $SC$ and $SY$ are tangent to ${\Gamma }_1$, hence we have $\angle SC{O}_1=90^{\circ }=\angle SY{O}_1$, and $O_{1}CSY$ is cyclic.

We see that both $X$ and $S$ lie on the circle through $O_1$, $C$, and $Y$. Therefore, we have $\angle SX{O}_1=\angle SY{O}_1=90^{\circ }$. We conclude that $SX$ is perpendicular to $O_1{O}_2$. Analogously, for the intersection $S^{\prime }$ of the line through $Z$ tangent to ${\Gamma }_2$, and the line ${\ell }_2$, we can deduce that $S^{\prime }X$ is perpendicular to $O_1{O}_2$. Because $S$ and $S^{\prime }$ both lie on ${\ell }_2$, we have $S=S^{\prime }$. Hence the two tangents intersect each other on ${\ell }_2$. $\square$