Janka drew a rectangle with a perimeter of $22 \mathrm{~cm}$ and side lengths expressed in centimeters as whole numbers. Then she divided the rectangle without any remainder into three rectangles, one of which had dimensions $2 \mathrm{~cm} \times 6 \mathrm{~cm}$. The sum of the perimeters of all three rectangles was $18 \mathrm{~cm}$ greater than the perimeter of the original rectangle. What could be the dimensions of the original rectangle? Find all solutions.

(M. Dillingerová)

Idea. Determine how Janka could have divided the rectangle; for each individual possibility, then express the given difference in perimeters using the lengths of the dividing lines.
All quantities $\mathrm{v}$ in the text are expressed $\mathrm{v}$ in centimeters, units will not be specified further. The lengths of the sides of Jančina's rectangle will be denoted by $x$ and $y$, which are natural numbers according to the problem statement.

First, we will determine how Jančina could have divided her rectangle. We have only the following two typical options (note that the images are schematic, i.e., we do not assume that $x>y$):
![](https://cdn.mathpix.com/cropped/2024_04_17_39f221c1ba6e0b54eafbg-6.jpg?height=228&width=876&top_left_y=1231&top_left_x=592)

The perimeter of the original rectangle is $2(x+y)=22$, so $x+y=11$. The sum of the perimeters of the three new rectangles is always greater than the perimeter of the original, specifically by twice the sum of the lengths of the dividing segments, which are marked with dashed lines in the image. This difference must be equal to 18.

I. Both dividing segments have the same length, namely $y$. Therefore, $4 y=18$, hence $y=4.5$. This is not possible because 4.5 is not an integer. Thus, Jančina did not divide the rectangle in this way.

II. The two dividing segments that lie on the same line have a combined length of $y$. The length of the third dividing segment is denoted by $z$. Then it must be that $2 y+2 z=18$, so $y+z=9$. This, together with the condition $x+y=11$, means that the dimension $x$ is 2 units larger than the dimension $z$. We will note this observation in the image:

![](https://cdn.mathpix.com/cropped/2024_04_17_39f221c1ba6e0b54eafbg-6.jpg?height=271&width=414&top_left_y=2055&top_left_x=821)

Now we will check which of the new rectangles can have dimensions $2 \times 6$ and how it can be placed - in total, we have these three options:
![](https://cdn.mathpix.com/cropped/2024_04_17_39f221c1ba6e0b54eafbg-7.jpg?height=460&width=1266&top_left_y=244&top_left_x=398)

Using the previously derived relationships between $x, y$, and $z$, we will express the dimensions of the rectangles in each case:

a) If $y=6$, then $x=5$ and $z=3$.

b) If $z=6$, then $x=8$ and $y=3$.

c) If $z=2$, then $x=4$ and $y=7$.

Jančina's rectangle could have dimensions $5 \times 6, 8 \times 3$, or $4 \times 7$.

Note. Using the same notation as above, from the requirement that Jančina's rectangle must contain a $2 \times 6$ rectangle, it follows that $x, y \geq 2$. Since $x$ and $y$ are natural numbers and $x+y=11$, the dimensions of Jančina's rectangle could be $2 \times 9, 3 \times 8, 4 \times 7$, or $5 \times 6$.

Now we can sequentially examine these four cases, i.e., place the $2 \times 6$ rectangle, discuss possible additional divisions, and check the requirement for the perimeters. This way, we quickly find that the only ways Jančina could have divided her rectangle are precisely the above options a), b), c).
![](https://cdn.mathpix.com/cropped/2024_04_17_39f221c1ba6e0b54eafbg-7.jpg?height=930&width=1554&top_left_y=1464&top_left_x=251)