Problem 8. In triangle $A B C$, side $A B=40$. The center $I$ of the inscribed circle divides the bisector $A L$ in the ratio $5: 3$, counting from the vertex. Find the radius of the circumscribed circle around triangle $A B C$, if the radius of the inscribed circle is 15.
Answer: $85 / 2=42.5$.

Solution. Noticing that $BI$ is the bisector in triangle $ALB$, by the property of the bisector of a triangle, we have: $AB: BL=AI: IL=5: 3$, from which $BL=3AB / 5=24$.

Next, $AB \cdot BL \cdot \sin \angle B=2 S_{\triangle ABL}=2 S_{\triangle AIB}+2 S_{\triangle BIL}=AB \cdot r+BL \cdot r=(AB+BL) \cdot r$, where $r$ is the radius of the inscribed circle in triangle $ABC$. Thus, $40 \cdot 24 \cdot \sin \angle B=(40+24) \cdot 15$, i.e., $\sin \angle B=1, \angle B=90^{\circ}$.

By the property of the bisector $CI$ in triangle $ALC$, we have $AC: CL=AI: IL=5: 3$. Assuming $CL=3x$, we have $AC=5x$. By the Pythagorean theorem: $AB^{2}+BC^{2}=AC^{2}$, i.e., $40^{2}+(24+3x)^{2}=(5x)^{2}$, from which $x=17$, and $R=AC / 2=5x / 2=85 / 2$.