[ Inscribed and described circles ]

In triangle $A B C$ with sides $A B=\sqrt{3}, B C=4, A C=\sqrt{7}$, the median $B D$ is drawn. The circles inscribed in triangles $A B D$ and $B D C$ touch $B D$ at points $M$ and $N$ respectively. Find $M N$.

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The distance from the vertex of a triangle to the nearest point of tangency with the inscribed circle is equal to the difference between the semiperimeter and the opposite side.

## Solution

Since $M$ and $N$ are the points of tangency of the given circles with the common side $BD$ of triangles $ABD$ and $ACD$, then

\[
\begin{gathered}
MN=|DM-DN|= \\
=\left|\left(\frac{AB+BD+AD}{2}-AB\right)-\left(\frac{BC+BD+CD}{2}-BC\right)\right|= \\
\left|\left(\frac{\sqrt{3}+\frac{\sqrt{7}}{2}+BD}{2}-\sqrt{3}\right)-\left(\frac{4+\frac{\sqrt{7}}{2}+BD}{2}-4\right)\right|= \\
=2-\frac{\sqrt{3}}{2} .
\end{gathered}
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-17.jpg?height=301&width=669&top_left_y=1959&top_left_x=700)

Answer

$2-\frac{\sqrt{3}}{2}$.

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