\begin{aligned} & \text { Tangent Circles } \\ & \text { [Pythagorean Theorem (direct and inverse).] }\end{aligned}

On a segment of length $2 R$ as a diameter, a semicircle is constructed. A circle of radius $R / 2$ is inscribed in the resulting figure. Find the radius of the circle that is tangent to the constructed circle, the semicircle, and the given segment.
Let $O$ be the center of the semicircle constructed on the segment $AB$ as its diameter, $O_1$ be the center of the circle of radius $\frac{R}{2}$ inscribed in the semicircle, $O_2$ be the center of the circle of unknown radius $x$ inscribed in the resulting figure, $M$ be the point of tangency of this circle with the segment $AB$, and $K$ be the point of tangency with the semicircle.

The circle of radius $\frac{R}{2}$ is tangent to the semicircle and its diameter, so the point of tangency with the diameter coincides with the center $O$ of the semicircle. Therefore, $OM$ is a segment of the common tangent to the touching circles of radii $\frac{R}{2}$ and $x$. Thus,

$$
OM = 2 \sqrt{\frac{R}{2} \cdot x} = \sqrt{2Rx}
$$

The line of centers of two touching circles passes through their point of tangency, so

$$
OO_2 = OK - O_2K = R - x
$$

In the right triangle $OO_2M$, it is known that

$$
OO_2 = R - x, O_2M = x, OM = \sqrt{2Rx}
$$

By the Pythagorean theorem,

$$
OO_2^2 = OM^2 + O_2M^2, (R - x)^2 = 2Rx + x^2
$$

from which $x = \frac{R}{4}$.

## Answer

$\frac{R}{4}$