The center of the specified circle divides the height $C H$ into segments proportional to the segments $B H$ and $C H$.

## Solution

Let $O$ be the center of the specified circle, and $r$ be its radius. Since the center of a circle inscribed in an angle lies on the angle bisector, $B O$ is the bisector of triangle $B H C$. Since $O H \perp A B$, $O H$ is the radius of the circle. In this triangle,

$$C H=\frac{\sqrt{6}}{3}, \quad B H=\frac{1}{3} A B=\frac{\sqrt{3}}{6}, \quad B C=\sqrt{C H^{2}+B H^{2}}=\sqrt{\frac{6}{9}+\frac{3}{36}}=\frac{\sqrt{3}}{2}.$$

By the property of the angle bisector of a triangle,

$$\frac{O H}{O C}=\frac{B H}{B C}=\frac{\frac{\sqrt{3}}{6}}{\frac{\sqrt{3}}{2}}=\frac{1}{3}$$

Therefore,

$$r=\frac{1}{4} \cdot C H=\frac{1}{4} \cdot \frac{\sqrt{6}}{3}=\frac{\sqrt{6}}{12}$$

!

## Answer

$\frac{\sqrt{6}}{12}$.