The altitudes of an acute-angled triangle $ABC$, drawn from vertices $B$ and $C$, are 7 and 9, respectively, and the median $AM$ is 8. Points $P$ and $Q$ are symmetric to point $M$ with respect to sides $AC$ and $AB$, respectively. Find the perimeter of quadrilateral $APMQ$.

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Segments $M Q$ and $M P$ are equal to the given heights.

## Solution

Let $B B_{1}$ and $C C_{1}$ be the altitudes of triangle $A B C$, with $B B_{1}=7$ and $C C_{1}=9$. Let $K$ and $N$ be the midpoints of segments $M P$ and $M Q$. Then $M K$ and $M N$ are the midlines of triangles $C B B_{1}$ and $C B C_{1}$, respectively, so $M P=2 M K=B B_{1}=7$, $M Q=C C_{1}=9$, and since $A Q=A M=8$ and $A P=A M=8$, the desired perimeter is $7+9+8+8=32$.

## Answer

32.