Problem 8. In triangle $A B C$, side $B C=28$. The bisector $B L$ is divided by the point of intersection of the bisectors of the triangle in the ratio $4: 3$, counting from the vertex. Find the radius of the circumscribed circle around triangle $A B C$, if the radius of the inscribed circle in it is 12.
Answer: 50.

Solution. Let $I$ be the center of the inscribed circle in triangle $ABC$ (i.e., the point of intersection of the angle bisectors). Noting that $CI$ is the angle bisector in triangle $BLC$, by the angle bisector theorem, we have: $BC: CL = BI: IL = 4: 3$, from which $CL = 3BC / 4 = 21$.

Next, $BC \cdot CL \cdot \sin \angle C = 2 S_{\triangle BCL} = 2 S_{\triangle BIC} + 2 S_{\triangle CIL} = BC \cdot r + CL \cdot r = (BC + CL) \cdot r$, where $r$ is the radius of the inscribed circle in triangle $ABC$. Thus, $28 \cdot 21 \cdot \sin \angle C = (28 + 21) \cdot 12$, i.e., $\sin \angle C = 1, \angle C = 90^\circ$.

By the angle bisector theorem for $AI$ in triangle $ALB$, we have $BA: AL = BI: IL = 4: 3$. Letting $AL = 3x$, we have $AB = 4x$. By the Pythagorean theorem: $AC^2 + BC^2 = AB^2$, i.e., $28^2 + (21 + 3x)^2 = (4x)^2$, from which $x = 25$, and $R = AB / 2 = 4 \cdot 25 / 2 = 50$.