12.394 Two equal circles are inscribed in a segment of a circle of radius $R$, touching each other, the arc of the segment, and its chord. Find the radii of these circles if the central angle subtended by the arc of the segment is $\alpha(\alpha<\pi)$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-284.jpg?height=267&width=377&top_left_y=325&top_left_x=71)

Let $P_{1}$ and $P_{2}$ be the centers of the circles of radius $x$ inscribed in the segment $A B$ of the given circle with center $O$, and $L_{1}$ and $L_{2}$ be the points of tangency of the circles $P_{1}$ and $P_{2}$ with the circle $O$. Then

$$
O P_{1}=O P_{2}=O L_{1}-P_{1} L_{1}=O L_{2}-P_{2} L_{2}=R-x, P_{1} P_{2}=2 x .
$$

Since $P_{1}$ and $P_{2}$ are equidistant from $A B$, then $P_{1} P_{2} \| A B$.

Let $O K$ be the height of the isosceles $\triangle O R P_{2}$, intersecting the chord $A B$ at point $C$. Then $A B \perp O K, C K=x, O C=O A \cdot \cos \angle A O C=R \cdot \cos \frac{\alpha}{2} \Rightarrow$

$$
\begin{gathered}
\Rightarrow O P_{1}^{2}=(R-x)^{2}=P_{1} K^{2}+O K^{2}=x^{2}+\left(R \cdot \cos \frac{\alpha}{2}+x\right)^{2} \Rightarrow \\
\Rightarrow x=\left(-2 \cos ^{2} \frac{\alpha}{4}+2 \cos \frac{\alpha}{4}\right) \cdot R=4 R \cdot \cos \frac{\alpha}{4} \cdot \sin ^{2} \frac{\alpha}{8}
\end{gathered}
$$

Answer: $4 R \cdot \cos \frac{\alpha}{4} \cdot \sin ^{2} \frac{\alpha}{8}$.