Problem 8. In triangle $A B C$, side $B C=35$. The center $I$ of the inscribed circle of the triangle divides the bisector $A L$ in the ratio $5: 2$, counting from the vertex. Find the length of side $A B$, if the radius of the inscribed circle of triangle $A B C$ is 10.
Answer: $A B=\frac{5}{12}(105 \pm 2 \sqrt{105})$.

Solution. Let $A H$ be the height of the triangle, $I S$ be the perpendicular from $I$ to the line $B C$. Since $I$ is the center of the inscribed circle, $I S=10$. Due to the similarity of triangles $A L H$ and $I L S, A H: I S=$ $A L: I L$, hence $A H=I S \cdot \frac{A I+I L}{I L}=10 \cdot\left(\frac{5}{2}+1\right)=35$. Then the area of triangle $A B C$ is $\frac{1}{2} A H \cdot B C=\frac{35^{2}}{2}$.

Furthermore, since $B I$ is the bisector of triangle $A B L$, then $A B: B L=A I: I L=5: 2$, similarly $A C: L C=5: 2$. Let $A B=5 x$, then $B L=2 x, C L=35-2 x, A C=175 / 2-5 x$. Then the perimeter of triangle $A B C$ is $35+5 x+(175 / 2-5 x)=245 / 2$

By Heron's formula:

$$
S^{2}=p(p-A B)(p-A C)(p-B C)
$$

where $p$ is the semi-perimeter of the triangle. Substituting:

$$
\frac{35^{4}}{4}=\frac{245}{4}\left(\frac{245}{4}-5 x\right)\left(\frac{245}{4}-\left(\frac{175}{2}-5 x\right)\right)\left(\frac{245}{4}-35\right)
$$

the roots of which are $\frac{105 \pm 2 \sqrt{105}}{12}, A B=\frac{5}{12}(105 \pm 2 \sqrt{105})$.

Comment. Such a complicated answer and the difference in the solution method in this version compared to the others is due to a typo. In the correct formulation, it should have been the bisector $B L$ instead of the bisector $A L$.