The lengths of the sides of a rectangle, expressed in cm, are two different integers. Its perimeter is the same number of cm as the number of $\mathrm{cm}^{2}$ of its area. What are the lengths of the sides?
I. solution: Let the two sides of the rectangle be $x$ and $y$ integers, where $x \neq y$.

According to the problem,

$$
x y=2 x+2 y
$$

from which

$$
x y-2 x-2 y=0, \quad x y-2 x-2 y+4=4
$$

thus

$$
(x-2)(y-2)=4
$$

Therefore, $x-2$ and $y-2$ are two different divisors of 4. For positive integers $x$ and $y$, the two factors are only negative when $x=y=1$, but in this case, the product is 1. Thus, only the 4 different, positive divisors need to be found. These are 1 and 4. If $x-2=1$ and $y-2=4$, then $x=3, y=6$. Swapping the two factors swaps the values of $x$ and $y$.

Papp Éva (Bp. VIII., Ságvári lg. II. o. t.)

II. solution: If we draw lines parallel to the sides of any rectangle at a distance of $1 \mathrm{~cm}$, then the area of the frame-like figure (concave octagon) thus obtained is 4 less than the perimeter of the rectangle in $\mathrm{cm}^{2}$, because each cm of the perimeter of the rectangle corresponds to one $\mathrm{cm}^{2}$ of the frame figure, but the square centimeters at the 4 corners are counted twice (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_f58295709ea79f91e864g-1.jpg?height=382&width=572&top_left_y=904&top_left_x=728)

Thus, the rectangle that satisfies our problem is the one where the remaining inner rectangle has an area of $4 \mathrm{~cm}^{2}$, and the sides are different integers. The only such rectangle is the one with sides $1 \mathrm{~cm}$ and $4 \mathrm{~cm}$, so the sides of the original rectangle are $1+2=3 \mathrm{~cm}$, and $4+2=6 \mathrm{~cm}$.