Problem 6. Yura drew a rectangle (along the grid lines) on graph paper and painted a picture on it. After that, he drew a frame one cell wide around the picture (see figure). It turned out that the area of the picture is equal to the area of the frame.

![](https://cdn.mathpix.com/cropped/2024_05_06_569060baad9798200f98g-05.jpg?height=281&width=245&top_left_y=1416&top_left_x=1032)
What dimensions could Yura's picture have? (List all possible options and prove that there are no others.)

[8 points] (T.I. Golenishcheva-Kutuzova)
Answer: $3 \times 10$ or $4 \times 6$ cells.

Solution: It is obvious that the width of the picture is more than one cell. Let's draw another frame inside the picture with a width of one cell (see the figure). Then, in the small frame, as in the large one, there will be four corner cells (they are shaded), and each side will be two cells shorter. Therefore, the small frame will have 8 cells less than the large one (these cells are hatched). Thus, the 8 cells will form the rectangle inside the small frame. It is clear that a rectangle with an area of 8 cells can have dimensions of $2 \times 4$ or $1 \times 8$ cells. Hence, we get the answer.

![](https://cdn.mathpix.com/cropped/2024_05_06_569060baad9798200f98g-06.jpg?height=436&width=356&top_left_y=402&top_left_x=918)

Comment: Those who are already familiar with algebra can obtain the answer in another way. If the picture is a rectangle of $a \times b$ cells, then the picture in the frame is already a rectangle of $(a+2) \times (b+2)$ cells. The area of the picture in the frame is twice the area of the picture without the frame, so $(a+2)(b+2) = 2ab$. Expanding the brackets, we get $2a + 2b + 4 = ab$. We can now transform this equation to $(a-2)(b-2) = 8$ and get that either one of the numbers $a-2$ and $b-2$ is 1 and the other is 8, or one of the numbers $a-2$ and $b-2$ is 2 and the other is 4. These options lead to the correct answers.

## 7th grade