## 8. Diagonal Squares

Vlado covered the diagonal of a large square with a side length of $2020 \mathrm{~cm}$ using a row of squares with a side length of $4 \mathrm{~cm}$ cut from green collage paper. The diagonals of the green squares lie on the diagonal of the large square, and the intersection of any two consecutive green squares is a square with a side length of 1 cm. Calculate the perimeter of the shape formed by the green squares. Express the result in decimeters.

Result: $\quad 808$
Solution, first method.

Let's calculate how many squares of side length $4 \mathrm{~cm}$ there are in total.

Start by drawing the squares starting from the bottom left corner of the large square, which we will mark as $A_{0}$. The top right corner of the first drawn square will be marked as $A_{1}$, the top right corner of the second drawn square as $A_{2}$, and so on.

From the vertex $A_{0}$ to the vertex $A_{1}$, one can move $4 \mathrm{~cm}$ to the right and $4 \mathrm{~cm}$ up. To reach the vertex $A_{2}$, one needs to move another $3 \mathrm{~cm}$ to the right and another $3 \mathrm{~cm}$ up, and to reach the vertex $A_{3}$, another $3 \mathrm{~cm}$ to the right and another $3 \mathrm{~cm}$ up. Thus, after the first square, each subsequent square reaches a point that is $3 \mathrm{~cm}$ to the right and $3 \mathrm{~cm}$ up from the last point on the diagonal of the large square.

![](https://cdn.mathpix.com/cropped/2024_05_30_033072dbc63a78c728f4g-08.jpg?height=1022&width=1005&top_left_y=143&top_left_x=160)

Since adding squares moves equally upwards and to the right, it is sufficient to look at the rightward movements. From the starting point, the sum of all movements must equal the length of the side of the large square, which is $2020 \mathrm{~cm}$. For the first square, the movement is $4 \mathrm{~cm}$, and for each subsequent square, it is $3 \mathrm{~cm}$.

Let $n$ be the total number of squares of side length $4 \mathrm{~cm}$ in this sequence.

Then we have $4 \mathrm{~cm} + (n-1) \cdot 3 \mathrm{~cm} = 2020 \mathrm{~cm}$.

This simplifies to $3n + 1 = 2020$, from which we get $n = 673$.

The perimeter of the resulting shape can be obtained by subtracting the sum of the perimeters of the small squares of side length $1 \mathrm{~cm}$ that overlap from the sum of the perimeters of all squares of side length $4 \mathrm{~cm}$.

The desired perimeter is $673 \cdot 16 \mathrm{~cm} - 672 \cdot 4 \mathrm{~cm} = 8080 \mathrm{~cm} = 808 \mathrm{dm}$.

## Solution, second method

As in the first method, determine that the total number of squares is 673.

The first and last squares in the sequence contribute to the perimeter of the shape with $4+4+3+3=14 \mathrm{~cm}$, while all others contribute with $3+3+3+3=12 \mathrm{~cm}$. Therefore, the perimeter of the green shape is

$$
2 \cdot 14 \mathrm{~cm} + 671 \cdot 4 \mathrm{~cm} = 8080 \mathrm{~cm} = 808 \mathrm{dm}
$$

## Solution, third method.

Extend the sides of the green squares so that on each half of the large square, we get a square grid (as shown in the smaller example on the image).

Now we easily observe that the perimeter of the shape is equal to the perimeter of the large square. Indeed, the green shape is bounded by blue, pink, orange, and red segments. The total length of the blue segments is equal to the length of the blue side of the large square, and similarly for the pink, orange, and red segments.

The perimeter of the shape is equal to the perimeter of the square:

![](https://cdn.mathpix.com/cropped/2024_05_30_033072dbc63a78c728f4g-08.jpg?height=694&width=708&top_left_y=1863&top_left_x=1162)

$4 \cdot 2020 = 8080 \mathrm{~cm} = 808 \mathrm{dm}$.