From four identical squares, an ornament was created as shown in the picture. The sides of the squares are $4 \mathrm{~cm}$ long, and they are either parallel or perpendicular to each other, intersecting either at their quarters or halves. Libor wanted to color the ornament and found that the cost of the paint for $1 \mathrm{~cm}^{2}$ of each contiguous area would be as many crowns as the number of squares that area is shared by.

How many crowns will it cost to paint the ornament?

(K. Pazourek)

![](https://cdn.mathpix.com/cropped/2024_04_17_96587d6cc949a3c812fag-4.jpg?height=429&width=734&top_left_y=2007&top_left_x=664)
The price of the color in each part corresponds to the number of squares that part is common to. This is the same as if all the squares were individually colored with a color of the same price.

Each square has an area of $4 \cdot 4 = 16 \mathrm{~cm}^{2}$, there are four squares, and $1 \mathrm{~cm}^{2}$ of color costs 1 crown. Therefore, the color for painting the ornament will cost $16 \cdot 4 = 64$ crowns.

Another solution. Different shades of gray distinguish how many squares each part of the ornament is common to. The ornament is divided into squares with a side length of $1 \mathrm{~cm}$:

![](https://cdn.mathpix.com/cropped/2024_04_17_96587d6cc949a3c812fag-5.jpg?height=432&width=734&top_left_y=595&top_left_x=661)

One square is common to all four squares, 6 squares are common to three squares, 12 squares are common to two squares, and 18 squares are common to one square. Therefore, the color for painting the ornament will cost

$$
1 \cdot 4 + 6 \cdot 3 + 12 \cdot 2 + 18 \cdot 1 = 64 \text{ crowns. }
$$