6. Point $A$ lies on side $L M$ of triangle $K L M$ with an angle of $120^{\circ}$ at vertex $K$. Incircles 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 $A O=2, A F=7$.
Answer: $\sqrt{\frac{53}{3}}$.

Solution. 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 angles $L A K$ and $M A K$. Since the angle between the bisectors of adjacent angles is a right angle, $\angle F A O=90^{\circ}$, and then by the Pythagorean theorem, we find that $F O=\sqrt{F A^{2}+O A^{2}}=\sqrt{53}$.

Since $K F$ and $K O$ are the bisectors of angles $L K A$ and $M K A$, then

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

By the generalized sine theorem for triangle $F K O$, we find that the desired radius is $\frac{F O}{2 \sin \angle F K O}=\sqrt{\frac{53}{3}}$.