Given a regular triangular pyramid $S A B C$. Point $S$ is the vertex of the pyramid, $A B=1, A S=2, B M$ is the median of triangle $A B C$, $A D$ is the bisector of triangle $S A B$. Find the length of the segment $D M$.

#
The bisector of a triangle divides its side into segments proportional to the other two sides, so $\frac{B D}{D S}$ $=\frac{1}{2} \cdot$ Therefore,

$$
B D=\frac{1}{3} B S=\frac{1}{3} A S=\frac{2}{3} .
$$

Let $S H$ be the height of the pyramid. Denote $\angle S B H=\angle S A H=\alpha$. From the right triangle $S B H$, we find that

$$
\cos \alpha=\frac{B H}{S B}=\frac{3^{2 B M}}{B S}=\frac{\frac{\sqrt{3}}{3}}{2}=\frac{\sqrt{3}}{6}
$$

By the cosine theorem,

$$
\begin{gathered}
D M=\sqrt{B M^{2}+B D^{2}-2 B M \cdot B D \cos \alpha}= \\
=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{2}{3}\right)^{2}-2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{2}{3} \cdot \frac{\sqrt{3}}{6}}=\sqrt{\frac{3}{4}+\frac{4}{9}-\frac{1}{3}}=\frac{\sqrt{31}}{6}
\end{gathered}
$$

## Answer

$\frac{\sqrt{31}}{6}$