Harvard-MIT Mathematics Tournament

Solution:

Let the center of $\omega_{i}$ be $O_{i}$ for $i=1,2,3$ and let $O$ denote the center of $\Gamma$. Then $O$, $D$, and $O_{1}$ are collinear, as are $O$, $E$, and $O_{2}$. Denote by $F$ the point of tangency between $\Gamma$ and $\omega_{3}$; then $F$, $O$, and $O_{3}$ are collinear. Writing $r$ for the radius of $\Gamma$ we have $OO_{1} = r + 2$, $OO_{2} = r + 1$, $OO_{3} = 3 - r$. Now since $O_{1}O_{3} = 1$ and $O_{3}O_{2} = 2$, we apply Stewart's theorem:
$$\begin{aligned} OO_{1}^{2} \cdot O_{2}O_{3} + OO_{2}^{2} \cdot O_{1}O_{3} & = OO_{3}^{2} \cdot O_{1}O_{2} + O_{1}O_{3} \cdot O_{3}O_{2} \cdot O_{1}O_{2} \\ 2(r+2)^{2} + (r+1)^{2} & = 3(3-r)^{2} + 1 \cdot 2 \cdot 3 \end{aligned}$$
We find $r = \frac{6}{7}$. Now the key observation is that the circumcircle of triangle $CDE$ is the incircle of triangle $OO_{1}O_{2}$. We easily compute the sides of $OO_{1}O_{2}$ to be $\frac{13}{7}$, $\frac{20}{7}$, and $3$. By Heron's formula, the area of $OO_{1}O_{2}$ is $\frac{18}{7}$, but the semiperimeter is $\frac{27}{7}$, so the desired radius is $\frac{2}{3}$.