1. Two circles of radius 12 have their centers on each other. As shown in the figure, $A$ is the center of the left circle, and $A B$ is a diameter of the right circle. A smaller circle is constructed tangent to $A B$ and the two given circles, internally to the right circle and externally to the left circle, as shown. Find the radius of the smaller circle.
Let $R$ be the common radius of the larger circles, and $r$ that of the small circle. Let $C$ and $D$ be the centers of the right large circle and the small circle, respectively. Let $E, F$ and $G$ be the points of tangency of the small circle with $A B$, the left large circle, and the right large circle, respectively. Since the centers of tangent circles are collinear with the point of tangency, then $A-F-D$ and $C-D-G$ are collinear.
From $\triangle A E D, A E^{2}=(R+r)^{2}-r^{2}=R^{2}+2 R r$. Therefore, $C E=A E-R=$ $\sqrt{R^{2}+2 R r}-R$.
From $\triangle C E D, C E^{2}=(R-r)^{2}-r^{2}=R^{2}-2 R r$.
Therefore, $\sqrt{R^{2}+2 R r}-R=\sqrt{R^{2}-2 R r}$. Solving this for $r$ yields $r=\frac{\sqrt{3}}{4} R$. With $R=12$, we get $r=3 \sqrt{3}$.