7. Suppose $A B C D$ is an isosceles trapezoid in which $\overline{A B} \| \overline{C D}$. Two mutually externally tangent circles $\omega_{1}$ and $\omega_{2}$ are inscribed in $A B C D$ such that $\omega_{1}$ is tangent to $\overline{A B}, \overline{B C}$, and $\overline{C D}$ while $\omega_{2}$ is tangent to $\overline{A B}, \overline{D A}$, and $\overline{C D}$. Given that $A B=1, C D=6$, compute the radius of either circle.
Answer: $\frac{3}{7}$
Solution: Let the radius of both circles be $r$, and let $\omega_{1}$ be centered at $O_{1}$. Let $\omega_{1}$ be tangent to $\overline{A B}, \overline{B C}$, and $\overline{C D}$ at $P, Q$, and $R$ respectively. Then, by symmetry, $P B=$ $\frac{1}{2}-r$ and $R C=3-r$. By equal tangents from $B$ and $C, B Q=\frac{1}{2}-r$ and $Q C=3-r$. Now, $\angle B O_{1} C$ is right because $m \angle O_{1} B C+m \angle B C O_{1}=\frac{1}{2}(m \angle P B C+m \angle B C R)=90^{\circ}$. Since $\overline{O_{1} Q} \perp \overline{B C}, r^{2}=O_{1} Q^{2}=B Q \cdot Q C=\left(\frac{1}{2}-r\right)(3-r)=r^{2}-\frac{7}{2} r+\frac{3}{2}$. Solving, we find $r=\frac{3}{7}$.