[ [ Pythagorean Theorem (direct and inverse). ]

Given triangle $ABC$ with base $AB$, equal to $\frac{\sqrt{3}}{2}$, and height $CH$, dropped to this base and equal to $\frac{\sqrt{6}}{3}$. It is known that point $H$ lies on $AB$ and $AH: HB=2: 1$. A circle is inscribed in angle $ABC$ of triangle $ABC$, with its center lying on the height $CH$. Find the radius of this circle.
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,

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

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-30.jpg?height=386&width=366&top_left_y=930&top_left_x=846)

## Answer

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