A line parallel to side $AB$ of triangle $ABC$ intersects side $BC$ at point $M$ and side $AC$ at point $N$. The area of triangle $MCN$ is twice the area of trapezoid $ABMN$. Find $CM: MB$.
The areas of similar triangles are in the ratio of the squares of the corresponding sides.

## Solution

Let $S_{A B M N}=S$. Then $S_{C M N}=2 S$ and $S_{A B C}=3 S$.

Triangle $M C N$ is similar to triangle $B C A$ with a similarity ratio of $\frac{C M}{C B}=\sqrt{\frac{S_{M C N}}{S_{B C A}}}=\frac{\sqrt{2}}{\sqrt{3}}$.

Therefore, $\frac{C M}{M B}=\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}=2+\sqrt{6}$.

## Answer

$2+\sqrt{6}$