Problem 2. Rectangle $ABCD$ with side lengths $8 \mathrm{~cm}$ and $5 \mathrm{~cm}$ is divided by segment $MN$ into two quadrilaterals with equal perimeters of $20 \mathrm{~cm}$ each. Determine the length of segment $MN$.
Solution. First method. The perimeter of rectangle $ABCD$ is equal to $2 \cdot (8+5) = 26 \, \text{cm}$. The sum of the perimeters of the two quadrilaterals $AMND$ and $MBCN$ is equal to $2 \cdot 20 = 40 \, \text{cm}$, and this is obtained when the perimeter of the rectangle is added to twice the length of segment $MN$, (see the diagram on the right). Therefore, $26 + 2 \overline{MN} = 40$, i.e., $\overline{MN} = (40 - 26) : 2 = 7 \, \text{cm}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_3a3538b5fdc92afac337g-03.jpg?height=202&width=360&top_left_y=1663&top_left_x=1108)

Second method. Let $a = 8 \, \text{cm}$ and $b = 5 \, \text{cm}$. With the labels as in the diagram on the right, the following equalities hold:

\[
\begin{aligned}
& a = x + y = m + n \\
& L_{AMND} = x + \overline{MN} + m + b \\
& L_{MBCN} = y + b + n + \overline{MN}
\end{aligned}
\]

Therefore,

![](https://cdn.mathpix.com/cropped/2024_06_05_3a3538b5fdc92afac337g-04.jpg?height=207&width=389&top_left_y=245&top_left_x=1091)

\[
\begin{aligned}
L_{AMND} + L_{MBCN} & = x + \overline{MN} + m + b + y + b + n + \overline{MN} \\
& = 2 \overline{MN} + (x + y) + (m + n) + 2b \\
& = 2 \overline{MN} + 2(a + b)
\end{aligned}
\]

From this, we get $20 + 20 = 2 \overline{MN} + 2 \cdot (8 + 5)$, i.e., $\overline{MN} = (40 - 26) : 2 = 7 \, \text{cm}$.