A regular triangular prism is given with a base edge length of $3.2 \mathrm{~cm}$ and height $5 \mathrm{~cm}$. We are wrapping its lateral surface with a chessboard film, which consists of opaque and transparent square fields with side lengths of $1 \mathrm{~cm}$. The beginning of the film aligns with the $\mathrm{s}$ edge of the prism (see figure) and the length of the film is just enough for wrapping the entire lateral surface twice.

What percentage of the prism's lateral surface will be visible through the film after wrapping? Neglect the thickness of the film.

(K. Pazourek)

![](https://cdn.mathpix.com/cropped/2024_04_17_9fc9244c7ab9f9f8aa42g-2.jpg?height=583&width=782&top_left_y=1730&top_left_x=640)

Hint. How many of which fields are on the lateral surface after the first wrapping of the film?
The mantle of the prism consists of three identical rectangles with dimensions $3.2 \, \text{cm} \times 5 \, \text{cm}$, the unfolded mantle is a rectangle with dimensions $9.6 \, \text{cm} \times 5 \, \text{cm}$, and the area of the mantle is $48 \, \text{cm}^2$.

After the first wrapping, the unfolded mantle looks as follows (we use different shades of gray to distinguish the layers):

![](https://cdn.mathpix.com/cropped/2024_04_17_9fc9244c7ab9f9f8aa42g-3.jpg?height=485&width=937&top_left_y=343&top_left_x=568)

On the length of $9.6 \, \text{cm}$, the chessboard pattern appears as follows: 22 transparent fields are complete squares, 3 transparent fields are rectangles with a width of $0.6 \, \text{cm}$ (and a height of $1 \, \text{cm}$). The missing $0.4 \, \text{cm}$ is due to the film being shifted during the second wrapping:

![](https://cdn.mathpix.com/cropped/2024_04_17_9fc9244c7ab9f9f8aa42g-3.jpg?height=488&width=939&top_left_y=1064&top_left_x=564)

All transparent fields are thus reduced by $0.4 \, \text{cm}$ in their width. After the double wrapping, the transparent part consists of 22 rectangles with dimensions $0.6 \, \text{cm} \times 1 \, \text{cm}$ and 3 rectangles with dimensions $0.2 \, \text{cm} \times 1 \, \text{cm}$. Through the film, $13.8 \, \text{cm}^2$ of the prism's mantle is visible $(22 \cdot 0.6 + 3 \cdot 0.2 = 13.8)$, which represents $28.75\%$ of the total area of $48 \, \text{cm}^2$ $(13.8 : 48 = 0.2875)$.