On a square sheet of size $N$, we construct 5 regular quadrilateral pyramids, each with its lateral faces inclined at the same angle to the base plane. The bases of the four pyramids are the four smaller squares into which the two medians of $N$ divide it, while the base square of the fifth pyramid has its vertices at the midpoints of the sides of $N$. Calculate the volume of the spatial region filled by the 5 pyramids.
The sought volume is less than the sum of the volumes of the four small pyramids and the large pyramid by the volume of the common part of the large pyramid and the small pyramids. It is obvious that the figure is symmetric with respect to the perpendicular bisector planes of the $N$ sides and diagonals, so the volumes of the four small pyramids are equal, as are the volumes of the four common parts.

![](https://cdn.mathpix.com/cropped/2024_05_02_002ac6149610613e39d9g-1.jpg?height=793&width=1266&top_left_y=311&top_left_x=427)

Let the side of $N$ be $a$, then the base edge of the large pyramid $B C=a / \sqrt{2}$, and the base edge of the small pyramid $A B=a / 2$, which is $\sqrt{2}$ times smaller. The same ratio applies to the heights of the two pyramids, with the height of the large pyramid denoted by $O F=m$, and the height of the small pyramid $K f=m / \sqrt{2}$. Indeed, by cutting both with the perpendicular bisector plane of one of their base edges, the cross-sections are similar isosceles triangles, because the angle at the base is the slant angle, and the base and height of the cross-section triangle are equal to the base edge and height of the respective pyramid.

The common part of the large and small pyramids is a triangular pyramid, the area of the base triangle $O B C$ is half the area of the base of the small pyramid; the upper vertex $f_{1}$ is cut off from the side edge $f O$ of the small pyramid by the symmetry axis $F K$ of the side face $B C F$ of the large pyramid.

The height $f_{1} L$ of the vertex $f_{1}$ is calculated from the similar right triangles $f_{1} O L$ and $f O K$, as well as $f_{1} K L$ and $F K O$, and the equality $O L+L K=O K$:

$$
\begin{gathered}
O L=O K \cdot \frac{f_{1} L}{f K}, \quad L K=O K \cdot \frac{f_{1} L}{F O}, \quad f_{1} L\left(\frac{1}{f K}+\frac{1}{F O}\right)=1, \\
f_{1} L=\frac{f K \cdot F O}{f K+F O}=\frac{m}{\sqrt{2}+1}=m(\sqrt{2}-1) .
\end{gathered}
$$

Accordingly, the volume $V_{k}$ of the common part $O B C f_{1}$, and the volume $V$ of the entire figure:

$$
\begin{gathered}
V_{k}=\frac{1}{3} \cdot \frac{a^{2}}{8} m(\sqrt{2}-1) \\
V=\frac{1}{3} \cdot \frac{a^{2}}{2} m+\frac{4}{3} \cdot \frac{a^{2}}{4} \cdot \frac{m}{\sqrt{2}}-4 V_{k}=\frac{a^{2} m}{3}\left(\frac{1}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{2}-1}{2}\right)=\frac{a^{2} m}{3}
\end{gathered}
$$

is exactly twice the volume of the large pyramid, in other words, it is as much as the volume of a pyramid with base $N$ and height $F O$.