[  Angles between angle bisectors  Law of Cosines

[ Area of a triangle (using semiperimeter and radius of inscribed or excircle

In triangle $A B C$, the angle at vertex $B$ is $\frac{\pi}{3}$, and the segments connecting the incenter to vertices $A$ and $C$ are 4 and 6, respectively. Find the radius of the circle inscribed in triangle $A B C$.

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Let $O$ be the center of the circle inscribed in triangle $ABC$. Then $\angle AOC = 120^{\circ}$. Apply the Law of Cosines and express the area of triangle $AOC$ in two ways.

## Solution

Let $O$ be the center of the circle inscribed in triangle $ABC$. Since $AO$ and $BO$ are the angle bisectors of $\angle BAC$ and $\angle ACB$, respectively, we have

$$
\angle AOC = 180^{\circ} - \frac{180^{\circ} - \angle ABC}{2} = 90^{\circ} + \frac{1}{2} \angle ABC = 90^{\circ} + 30^{\circ} = 120^{\circ}
$$

By the Law of Cosines in triangle $AOC$, we find that

$$
AC = \sqrt{OA^2 + OC^2 - 2 \cdot OA \cdot OC \cdot \cos 120^{\circ}} = \sqrt{16 + 36 + 2 \cdot 4 \cdot 6 \cdot \frac{1}{2}} = 2 \sqrt{19}
$$

$$
S_{\triangle AOC} = \frac{1}{2} AC \cdot r = r \sqrt{19}, \text{ and } S_{\triangle AOC} = \frac{1}{2} OA \cdot OC \cdot \sin 120^{\circ} = \frac{1}{2} \cdot 4 \cdot 6 \cdot \frac{\sqrt{3}}{2} = 6 \sqrt{3}
$$

From the equation $r \sqrt{19} = 6 \sqrt{3}$, we find that $r = \frac{6 \sqrt{3}}{\sqrt{19}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6ed8f1205731b80a0039g-04.jpg?height=486&width=580&top_left_y=266&top_left_x=745)

## Answer

$\frac{6 \sqrt{3}}{\sqrt{19}}$.