Progetto Olimpiadi della Matematica - GARA di FEBBRAIO

Solution:

The answer is $\mathbf{( E )}$. Let $O, O_{1}, O_{2}$ be the centers of $\Gamma, \Gamma_{1}, \Gamma_{2}$ respectively. Since $\Gamma_{1}, \Gamma_{2}$ are externally tangent, the distance $O_{1} O_{2}$ equals the sum of the radii of $\Gamma_{1}, \Gamma_{2}$, that is $O_{1} O_{2}=5$. Similarly, the distance $O O_{1}$ equals the difference of the radii of $\Gamma$ and $\Gamma_{1}$ (and is thus equal to $6-3=3$), and the distance $O O_{2}$ equals $6-2=4$. Since $O O_{2}^{2}+O O_{1}^{2}=3^{2}+4^{2}=5^{2}=O_{1} O_{2}^{2}$, the triangle $O O_{1} O_{2}$ is right-angled at $O$, and it follows that the triangle $A_{1} O A_{2}$ is an isosceles right triangle.

Now let $O^{\prime}$ be the center of the circle circumscribed about $A_{1} A A_{2}$. The triangle $A_{1} O^{\prime} A_{2}$ is certainly isosceles on base $A_{1} A_{2}$; we want to show that it is also right-angled at $O^{\prime}$, that is, that $O^{\prime}$ is the reflection of $O$ with respect to $A_{1} A_{2}$, hence that $O^{\prime} A_{1}=O^{\prime} A_{2}=6$.

We have $\widehat{A_{1} O^{\prime} A_{2}}=2\left(180^{\circ}-\widehat{A_{1} A A_{2}}\right)$ (central angle subtending the same arc as $\widehat{A_{1} A A_{2}}$, but on the opposite side); on the other hand, a quick angle computation gives $180^{\circ}-\widehat{A_{1} A A_{2}}=\widehat{A_{1} A O_{1}}+\widehat{A_{2} A O_{2}}=\widehat{A A_{1} O_{1}}+\widehat{A A_{2} O_{2}}=\widehat{A O_{1} O}/2+\widehat{A O_{2} O}/2=45^{\circ}$, which precisely implies the desired orthogonality.