The area of the triangle is $4 \sqrt{21}$, the perimeter is 24, and the segment of the bisector from one of the vertices to the center of the inscribed circle is $\frac{\sqrt{30}}{3}$. Find the largest side of the triangle.
If the inscribed circle touches the side $A C$ of triangle $A B C$ at point $M$, and $p$ is the semiperimeter of the triangle, then $A M=p-B C$.

## Solution

Let $O$ be the center of the circle inscribed in the given triangle $A B C, r$ be its radius, $S=4 \sqrt{21}$ be the area, $2 p$

$=24$ be the perimeter, and $M$ be the point of tangency with side $A C$.

Since $S=p \cdot r$, then

$$
r=\frac{2 S}{P}=\frac{8 \sqrt{21}}{24}=\frac{\sqrt{21}}{3}
$$

From the right triangle $A O M$, we find that

$$
A M=\sqrt{A O^{2}-O M^{2}}=\sqrt{\left(\frac{\sqrt{30}}{3}\right)^{2}-\left(\frac{\sqrt{21}}{3}\right)^{2}}=1
$$

Let $\angle B A C=\alpha$. Then

$$
\operatorname{tg} \frac{\alpha}{2}=\frac{O M}{A M}=\frac{\sqrt{21}}{3}
$$

Therefore,

$$
\cos \alpha=\frac{1-\operatorname{tg}^{2} \frac{\alpha}{2}}{1+\operatorname{tg}^{2} \frac{\alpha}{2}}=\frac{1-\frac{21}{9}}{1+\frac{21}{9}}=-\frac{2}{5}
$$

From the equality $A M=p-B C$, we find that $B C=12-1=11$.

Since $\cos \angle B A C=\cos \alpha=-\frac{2}{5}<0$, the angle $B A C$ is obtuse, and therefore the side $B C$ $=11$ is the largest side of triangle $A B C$.

## Answer

11.