[ Area of a triangle (using two sides and the angle between them).]

The area of triangle $ABC$ is $15 \sqrt{3}$. Angle $BAC$ is $120^{\circ}$. Angle $ABC$ is greater than angle $ACB$. The distance from vertex $A$ to the center of the circle inscribed in triangle $ABC$ is 2. Find the median of triangle $ABC$ drawn from vertex $B$.
Apply the Law of Cosines.

## Solution

Let $O$ be the center of the circle; $K, M, N$ be the points of tangency with sides $AC, BC, AB$ respectively. Then

$$
O K=r=A O \sin 60^{\circ}=\sqrt{3}, A K=A O \cos 60^{\circ}=1
$$

If $p$ is the semiperimeter of triangle $ABC$, then $S_{\Delta \mathrm{ABC}}=p r$. From this, we find that $p=15$.

Let $B M=B N=x, C M=C K=y$. Then

$$
\left\{\begin{array}{l}
x+y+1=15 \\
(x+y)^{2}=(x+1)^{2}+(y+1)^{2}+(x+1)(y+1)
\end{array}\right.
$$

From this system, we find that $x=9, y=5$ or $x=5, y=9$. Since $AC > AB$, the only solution that satisfies the problem's condition is the second solution of this system: $x=5, y=9$. Therefore, $AB=6, AC=10$.

If $P$ is the midpoint of $AC$, then by the Law of Cosines

$$
B P^{2}=A B^{2}+A P^{2}-2 A B \cdot A P \cos 120^{\circ}=36+25+2 \cdot 6 \cdot 5 \cdot \frac{1}{2}=91
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-08.jpg?height=349&width=834&top_left_y=2058&top_left_x=612)

Answer

$\sqrt{91}$.