What is the maximum number of parts the plane can be divided into by $n$ rectangles with parallel sides?
A rectangle divides the plane into two parts. Our goal is to place the second (and any subsequent) rectangle in such a way that the number of resulting plane sections is maximized. According to the first and second figures, the number of plane sections increased by 2 and 3, respectively. It is easy to see that the number of new plane sections will be the largest if the perimeter of the second rectangle intersects the perimeter of the first rectangle at four points. In this case, a new plane section is formed above each side of the rectangle defined by the four intersection points, as indicated by the dashed lines in the third figure.
![](https://cdn.mathpix.com/cropped/2024_05_02_0d0642e9dfb1f442fa61g-1.jpg?height=852&width=426&top_left_y=388&top_left_x=836)

Thus, two rectangles can divide the plane into a maximum of $2+4$ sections. Applying the previous reasoning, the third rectangle can create at most $4 \cdot 2$ new plane sections by intersecting each of the already placed rectangles at 4 points, so three rectangles can divide the plane into a maximum of $2+4 \cdot 1+4 \cdot 2$ sections. We conjecture that $n$ rectangles can divide the plane into a maximum of $S_{n}=2+4 \cdot 1+4 \cdot 2+\ldots+4(n-1)$ sections. $S_{n}$ can be written as follows: $S_{n}=2+4(1+2+\ldots+n-1)=2+4 \cdot \frac{n(n-1)}{2}$, and thus $S_{n}=2 n^{2}-2 n+2$, which can be proven by mathematical induction.

For $n=1$, $S_{1}=2$, so the statement is true. Suppose the formula is true for $n$. The $(n+1)$-th rectangle, intersecting each of the previous ones at 4 points, creates $4 \cdot n$ intersection points, and thus at most $4 \cdot n$ new plane sections. Therefore, $S_{n+1}=$ $2 n^{2}-2 n+2+4 n=2(n+1)^{2}-2(n+1)+2$, so the statement is also true for $n+1$.

![](https://cdn.mathpix.com/cropped/2024_05_02_0d0642e9dfb1f442fa61g-1.jpg?height=532&width=618&top_left_y=1710&top_left_x=735)

Figure 4

Our solution also shows how to divide the plane into the maximum number of sections using $n$ rectangles with parallel sides. In Figure 4, we have drawn the case for $n=5$.