In triangle $A B C$, it is known that $A B=c, B C=a, \angle B=120^{\circ}$. Find the distance between the bases of the altitudes drawn from vertices $A$ and $C$.

#
Let $A M$ and $C N$ be the altitudes of triangle $A B C$. Since angle $A B C$ is obtuse, points $M$ and $N$ lie on the extensions of sides $B C$ and $A B$.

Triangle $M B N$ is similar to triangle $A B C$ with a similarity coefficient of $\cos 60^{\circ}=1 / 2$. Therefore, $M N^{2}=1 / 4 A C^{2}=1 / 4\left(A B^{2}+B C^{2}-A B \cdot B C \cos 120^{\circ}\right)=1 / 4\left(a^{2}+c^{2}+a c\right)$.

## Answer

$1 / 2 \sqrt{a^{2}+c^{2}+a c}$.