Problem 3. (15 points) Point $A$ lies on side $L M$ of triangle $K L M$ with an angle of $120^{\circ}$ at vertex $K$. Circles are inscribed in triangles $A K L$ and $A K M$ with centers $F$ and $O$ respectively. Find the radius of the circumcircle of triangle $F K O$, if $A F=3, A O=6$.
# Solution.

$\mathrm{L}$

![](https://cdn.mathpix.com/cropped/2024_05_06_6d5fca905c76f828a3d8g-12.jpg?height=403&width=1237&top_left_y=544&top_left_x=361)

Fig. 1

The center of the circle inscribed in an angle lies on the bisector of this angle, so rays $A F$ and $A O$ are the bisectors of $\angle L A K$ and $\angle M A K$ respectively (Fig. 1).

The angle between the bisectors of adjacent angles is a right angle, by a corollary of the theorem on adjacent angles: $\angle F A O=90^{\circ}$, i.e., $\triangle F A O$ is a right triangle.

By the Pythagorean theorem, we find $F O: F O=\sqrt{A F^{2}+A O^{2}}=\sqrt{6+36}=\sqrt{45}=3 \sqrt{5}$.

Rays $K F$ and $K O$ are the bisectors of $\angle L K A$ and $\angle M K A$ respectively, by the theorem on the property of two tangents to a circle, then

$\angle F K O=\angle F K A+\angle O K A=\frac{1}{2} \angle L K A+\frac{1}{2} \angle M K A=\frac{1}{2} \angle L K M=\frac{1}{2} \cdot 120^{\circ}=60^{\circ}$.

By the Law of Sines for $\triangle F K O: \frac{F O}{\sin \angle F K O}=\frac{F K}{\sin \angle F O K}=\frac{K O}{\sin \angle K F O}=2 R$.

Substitute the known values: $\frac{3 \sqrt{5}}{\sin 60^{\circ}}=2 R, R=\frac{3 \sqrt{5}}{2 \sin 60^{\circ}}=\frac{3 \sqrt{5}}{2 \cdot \frac{\sqrt{3}}{2}}=\sqrt{\frac{45}{3}}=\sqrt{15}$.

Remark. Instead of the Law of Sines, the theorem on inscribed and central angles could have been used.

Answer. $R=\sqrt{15}$.