On the segment $A B=2 R$ as a diameter, a circle is constructed. A second circle of the same radius as the first one has its center at point $A$. A third circle touches the first circle internally, the second circle externally, and the segment $A B$. Find the radius of the third circle.
Let $O$ be the center of the first circle, $O_{1}$ be the center of the third circle, $M$ be its point of tangency with the line $A B$, and $x$ be its radius. Then $O O_{1}=R-x, M O_{1}=x$,

$A O_{1}=R+x, O M^{2}=O_{1} O^{2}-O_{1} M^{2}=R^{2}-2 R x, A M^{2}+O_{1} M^{2}=O_{1} A^{2}$, or $\left(R+\sqrt{R^{2}-2 R x}\right)^{2}+x^{2}=(R+x)^{2}$, from which $x=\frac{R \sqrt{3}}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-18.jpg?height=300&width=712&top_left_y=353&top_left_x=679)

## Answer

$\frac{R \sqrt{3}}{4}$.