In triangle $ABC$, it is known that $AB=20, AC=24$. It is also known that vertex $C$, the center of the inscribed circle of triangle $ABC$, and the intersection point of the angle bisector of angle $A$ with side $BC$ lie on a circle, the center of which lies on side $AC$. Find the radius of the circumcircle of triangle $ABC$.
Express the angles of the triangle in terms of $\alpha=\angle A C B / 2$ and use the Law of Sines.

## Solution

Let $O$ be the center of the inscribed circle of triangle $A B C$, $Q$ be the center of the circle passing through points $C$, $O$, and $M$ (the intersection of the angle bisector of $\angle A$ with side $BC$), and $CP$ be the diameter of this circle. Denote $\angle O C A = \angle O C B = \alpha$. Then

$$
\begin{gathered}
\angle O M P = \angle O C P = \alpha, \angle A M C = \angle A M P + \angle P M C = \alpha + 90^{\circ} \\
\angle M A C = 180^{\circ} - \angle A M C - \angle M C A = 180^{\circ} - (\alpha + 90^{\circ}) - 2 \alpha = 90^{\circ} - 3 \alpha \\
\angle B A C = 2 \cdot \angle M A C = 180^{\circ} - 6 \alpha \\
\angle A B C = 180^{\circ} - \angle A C B - \angle B A C = 180^{\circ} - 2 \alpha - (180^{\circ} - 6 \alpha) = 4 \alpha
\end{gathered}
$$

By the Law of Sines in triangle $A B C$, we have:

$$
A B / \sin 2 \alpha = A C / \sin 4 \alpha, \text{ or } 20 / \sin 2 \alpha = 24 / (2 \cdot \sin 2 \alpha \cdot \cos 2 \alpha)
$$

From this, we find that

$$
\cos 2 \alpha = 12 / 20 = 3 / 5, \sin 2 \alpha = 4 / 5
$$

If $R$ is the radius of the circumcircle of triangle $A B C$, then

$$
R = A B / (2 \cdot \sin 2 \alpha) = 25 / 2
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_8027e864227276d12944g-50.jpg?height=437&width=728&top_left_y=1828&top_left_x=665)

## Answer

$12.5$.