6. Point $P$ lies on side $B C$ of triangle $A B C$ with an angle of $60^{\circ}$ at vertex $A$. Incircles are inscribed in triangles $A P B$ and $A P C$ with centers $D$ and $T$ respectively. Find the radius of the circumcircle of triangle $A D T$, if $P D=7, P T=4$.
Answer: $\sqrt{65}$.

Solution. The center of the circle inscribed in an angle lies on the bisector of this angle, so rays $P T$ and $P D$ are the bisectors of angles $C P A$ and $B P A$. Since the angle between the bisectors of adjacent angles is a right angle, $\angle D P T=90^{\circ}$, and then by the Pythagorean theorem, we find that $D T=\sqrt{P D^{2}+P T^{2}}=\sqrt{65}$.

Since $A D$ and $A T$ are the bisectors of angles $B A P$ and $C A P$, then

$$
\angle D A T=\angle D A P+\angle T A P=\frac{1}{2} \angle B A P+\frac{1}{2} \angle C A P=\frac{1}{2} \angle B A C=30^{\circ}
$$

By the generalized sine theorem for triangle $A D T$, we find that the desired radius is $\frac{D T}{2 \sin \angle D A T}=\sqrt{65}$.