Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.3 AIME, harder Find the answer

[ 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 $M N$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

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

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

Answer

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

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