Task 3. (15 points) Point $A$ lies on side $L M$ of triangle $K L M$ with an angle of $60^{\circ}$ 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 $A O=7, A F=4$.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_6d5fca905c76f828a3d8g-07.jpg?height=594&width=782&top_left_y=183&top_left_x=637)

Fig. 1

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

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

By the Pythagorean theorem, we find $F O: F O=\sqrt{A F^{2}+A O^{2}}=\sqrt{16+49}=\sqrt{65}$.

The 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 we have

$\angle F K O=\angle F K A+\angle O K A=\frac{1}{2} \angle L K A+\frac{1}{2} \angle M K A=\frac{1}{2} \angle L K M=\frac{1}{2} \cdot 60^{\circ}=30^{\circ}$.

By the sine rule 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 data: $\frac{\sqrt{65}}{\sin 30^{\circ}}=2 R \Rightarrow R=\frac{\sqrt{65}}{2 \sin 30^{0}}=\frac{\sqrt{65}}{2 \cdot \frac{1}{2}}=\sqrt{65}$.

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

Answer. $R=\sqrt{65}$.