3. Let trapezium have bases $A B$ and $C D$, measuring $5 \mathrm{~cm}$ and $1 \mathrm{~cm}$, respectively. Let $M$ and $N$ be points on $A D$ and $B C$ such that the segment $M N$ is parallel to the base $A B$, and the area of quadrilateral $A B N M$ is twice the area of quadrilateral $C D M N$. How many centimeters does the segment $M N$ measure?
II/3. Let the length of $M N$ be $x$, the height of trapezoid $A B C D$ be $v$, and the height of trapezoid $A B N M$ be $h$. Then, $S_{A B N M}=\frac{(x+5)}{2} \cdot h$ and $S_{C D M N}=\frac{x+1}{2} \cdot(v-h)$. Since $S_{A B N M}=2 S_{C D M N}$ and $S_{A B N M}+S_{C D M N}=S_{A B C D}$, we get the equations

$$
\begin{aligned}
\frac{(x+5)}{2} \cdot h & =2 \frac{x+1}{2} \cdot(v-h) \\
\frac{(x+5)}{2} \cdot h+\frac{x+1}{2} \cdot(v-h) & =\frac{1+5}{2} v
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_e3e00bf556075403c4d7g-08.jpg?height=403&width=588&top_left_y=444&top_left_x=1271)

From the first equation, we express $h=\frac{x+1}{3 x+7} \cdot 2 v$ and substitute it into the second. The resulting equation can be divided by $v$, and a brief calculation shows that $x^{2}=9$ or $x=3$. Therefore, the segment $M N$ measures $3 \mathrm{~cm}$.

If the contestant expresses $S_{A B N M}$ and $S_{C D M N}$ in terms of $x, v$, and $h$ (or other dependent quantities): $\mathbf{1 + 1}$ point. If the contestant expresses the conditions $S_{A B N M}=2 S_{C D M N}$ and $S_{A B N M}+S_{C D M N}=S_{A B C D}$ in terms of $x, v$, and $h: 1+1$ point. Solution $x=3: 3$ points.