$[$ The Law of Sines $]$ [ Angles between angle bisectors $]$

In triangle $A B C$, it is known that $A C=b, \angle A B C=\alpha$. Find the radius of the circle passing through the center of the inscribed circle of triangle $A B C$ and vertices $A$ and $C$.
Prove that side $AC$ is seen from the center of the inscribed circle in triangle $ABC$ at an angle of $90^{\circ}+\frac{\alpha}{2}$.

## Solution

Let $O$ be the center of the inscribed circle in triangle $ABC$, and $R$ be the desired radius.

$$
\begin{gathered}
\angle AOC = 180^{\circ} - \frac{1}{2} \angle BAC - \frac{1}{2} \angle BCA = 180^{\circ} - \frac{1}{2} (\angle BAC + \angle BCA) = \\
= 180^{\circ} - \frac{1}{2} \left(180^{\circ} - \angle ABC\right) = 90^{\circ} + \frac{\alpha}{2}
\end{gathered}
$$

Then

$$
R = \frac{AC}{2 \sin \angle AOC} = \frac{b}{2 \sin \left(90^{\circ} + \frac{\alpha}{2}\right)} = \frac{b}{2 \cos \frac{\alpha}{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-23.jpg?height=443&width=735&top_left_y=1&top_left_x=667)

## Answer

$\frac{b}{2 \cos \frac{\alpha}{2}}$.