8,9 The area of a triangle is $6 \sqrt{6}$, its perimeter is 18, and the distance from the center of the inscribed circle to one of the vertices is $\frac{2 \sqrt{42}}{3}$. Find the smallest 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 $A M=p-B C$. Use this equality, and then apply the Law of Cosines.

## Solution

Let $O$ be the center of the circle inscribed in the given triangle $A B C$, $r$ its radius, $S=6 \sqrt{6}$ the area, $2 p=$ 18 the perimeter, and $M$ the point of tangency with side $A C$.

Since $S=p \cdot r$, then

$$
r=\frac{S}{p}=\frac{12 \sqrt{6}}{9}=\frac{2 \sqrt{6}}{3}
$$

From the right triangle $AOM$, we find that

$$
A M=\sqrt{A O^{2}-O M^{2}}=\sqrt{\left(\frac{2 \sqrt{42}}{3}\right)^{2}-\left(\frac{2 \sqrt{6}}{3}\right)^{2}}=\frac{2}{3} \cdot \sqrt{42-6}=4
$$

From the equality $A M=p-B C$, we find that $B C=9-4=5$. Let $C M=x$. From the equality $C M=p-A B$, we find that $A B=p-C M=9-x$.

By Heron's formula

$$
S=\sqrt{p(p-B C)(p-A C)(p-A B)} \text {, or } 6 \sqrt{6}=\sqrt{9 \cdot 4 \cdot(5-x) \cdot x} \text {. }
$$

From this equation, we find that $x=2$ or $x=3$. In the first case, $A B=7, A C=6$. In the second case, $A B=6, A C=7$. Therefore, in each of these cases, the side $B C=5$ is the smallest.

![](https://cdn.mathpix.com/cropped/2024_05_06_49e49f22191b45b40a57g-31.jpg?height=686&width=606&top_left_y=709&top_left_x=726)

Answer

5. Send a comment

Problem