4. In triangle $A B C$, angle $A$ is equal to $60^{\circ}$, the distances from vertices $B$ and $C$ to the center of the inscribed circle of triangle $A B C$ are 3 and 4, respectively. Find the radius of the circle circumscribed around triangle $A B C$.
Answer. $\sqrt{\frac{37}{3}}$ Indication. Let $O$ be the center of the inscribed circle. Then $\angle O B C+\angle B C O$ $=\frac{1}{2}(\angle B+\angle C)=\frac{1}{2} \cdot 120^{\circ}=60^{\circ}$, and therefore $\angle B O C=180^{\circ}-60^{\circ}=120^{\circ}$. By the cosine theorem, we get $B C^{2}=B O^{2}+O C^{2}-2 B O \cdot O C \cdot \cos \angle B O C=3^{2}+4^{2}-2 \cdot 3 \cdot 4 \cdot(-0.5)=37$. Therefore, for the radius $R$ of the circumscribed circle of triangle $A B C$, we will have $2 R \sin 60^{\circ}=\sqrt{37}$. Hence, $R=\sqrt{37 / 3}$.