The medians of triangle $ABC$, drawn from vertices $B$ and $C$, are equal to 6 and 9 and intersect at point $M$. It is known that $\angle BMC=120^{\circ}$. Find the sides of the triangle.
Let $B D$ and $C E$ be the medians of triangle $A B C$. Then $B M=\frac{2}{3} B D, C M=\frac{2}{3} C E$.

## Solution

Let $B D$ and $C E$ be the medians of triangle $A B C$. Then

$$
B M=\frac{2}{3} B D=4, C M=\frac{2}{3} C E=6
$$

Using the cosine theorem in triangles $M B C, B M E$, and $C M D$, we find that

$$
\begin{gathered}
B C^{2}=B M^{2}+C M^{2}-2 B M \cdot C M \cos \angle 120^{\circ}=16+36+24=76 \\
B E^{2}=B M^{2}+E M^{2}-2 B M \cdot E M \cos \angle 60^{\circ}=16+9-12=13 \\
D C^{2}=D M^{2}+C M^{2}-2 D M \cdot C M \cos \angle 60^{\circ}=4+36-12=28
\end{gathered}
$$

Therefore,

$$
B C=\sqrt{76}=2 \sqrt{19}, A B=2 B E=2 \sqrt{13}, A C=2 C D=2 \sqrt{28}=4 \sqrt{7}
$$

## Answer

$4 \sqrt{7} ; 2 \sqrt{13} ; 2 \sqrt{19}$