How much wood is needed to frame a rectangular window with a perimeter of $k \mathrm{~cm}$, if the cross-section of the wooden strip is an isosceles right triangle with legs of $1 \mathrm{~cm}$?
In the vertices of the rectangle, there are 8 pyramids with a base of 1 unit square and a height of $1 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a215a2762a5a58794c26g-1.jpg?height=423&width=401&top_left_y=241&top_left_x=840)

Thus, their total volume is $8 \cdot \frac{1 \cdot 1}{3}=\frac{8}{3} \mathrm{~cm}^{3}$. By stacking the remaining triangular-based prisms one after another, we get half of a square column with a unit base and a height of $k-8 \mathrm{~cm}$, which represents $\frac{(k-8)}{2} \mathrm{~cm}^{2}$ of gitt.

$$
\frac{k-8}{2}+\frac{8}{3}=\frac{3 k-24}{6}+\frac{16}{6}=\frac{3 k-8}{6}
$$

Therefore, the required amount of gitt is

$$
\left(\frac{k}{2}-\frac{4}{3}\right) \mathrm{cm}^{3}
$$

József Kövecs (Esztergom, I. István g. III. o. t.)