9. [7] Three unit circles $\omega_{1}, \omega_{2}$, and $\omega_{3}$ in the plane have the property that each circle passes through the centers of the other two. A square $S$ surrounds the three circles in such a way that each of its four sides is tangent to at least one of $\omega_{1}, \omega_{2}$ and $\omega_{3}$. Find the side length of the square $S$.
Answer: $\frac{\sqrt{6}+\sqrt{2}+8}{4}$
By the Pigeonhole Principle, two of the sides must be tangent to the same circle, say $\omega_{1}$. Since $S$ surrounds the circles, these two sides must be adjacent, so we can let $A$ denote the common vertex of the two sides tangent to $\omega_{1}$. Let $B, C$, and $D$ be the other vertices of $S$ in clockwise order, and let $P, Q$, and $R$ be the centers of $\omega_{1}, \omega_{2}$, and $\omega_{3}$ respectively, and suppose WLOG that they are also in clockwise order. Then $A C$ passes through the center of $\omega_{1}$, and by symmetry (since $A B=A D$ ) it
must also pass through the other intersection point of $\omega_{2}$ and $\omega_{3}$. That is, $A C$ is the radical axis of $\omega_{2}$ and $\omega_{3}$.
Now, let $M$ and $N$ be the feet of the perpendiculars from $P$ and $R$, respectively, to side $A D$. Let $E$ and $F$ be the feet of the perpendiculars from $P$ to $A B$ and from $R$ to $D C$, respectively. Then $P E A M$ and $N R F D$ are rectangles, and $P E$ and $R F$ are radii of $\omega_{1}$ and $\omega_{2}$ respectively. Thus $A M=E P=1$ and $N D=R F=1$. Finally, we have
$$
\begin{aligned}
M N & =P R \cdot \cos \left(180^{\circ}-\angle E P R\right) \\
& =\cos \left(180^{\circ}-E P Q-R P Q\right) \\
& =-\cos \left(\left(270^{\circ}-60^{\circ}\right) / 2+60^{\circ}\right) \\
& =-\cos \left(165^{\circ}\right) \\
& =\cos \left(15^{\circ}\right) \\
& =\frac{\sqrt{6}+\sqrt{2}}{4}
\end{aligned}
$$

Thus $A D=A M+M N+N D=1+\frac{\sqrt{6}+\sqrt{2}}{4}+1=\frac{\sqrt{6}+\sqrt{2}+8}{4}$ as claimed.