We have a regular pentagonal prism and a square plate $ABCD$, all of whose edges are equal and of unit length. The prism is placed on the (horizontally placed) plate, first with one of its side faces, such that the edge $AB$ coincides with a side edge of the prism. Then, it is placed again with another side face, such that the edge $AD$ coincides with a side edge of the prism.

What is the volume of the spatial region covered by the two positions of the prism?
Our usual base plates of the parallelepiped, according to the settings, are vertical, let's call these walls, and continuing this analogy, the top two plates are roof plates. By taking twice the volume $V_{h}$ of one of our parallelepipeds, we get as much more than the sought volume $V$ as the volume $V_{m}$ of the solid that is inside the parallelepiped in both positions, hence (Figures 1-2)

$$
V=2 V_{h}-V_{m}
$$

The solid is between all four walls and below all four roof plates, it can be decomposed into a (right) parallelepiped with base $A B C D$ and a regular quadrilateral pyramid, whose apex is the common midpoint of the two roof ridge edges (above the center of the base).

It is straightforward that

$$
V_{h}=\frac{5}{4} \tan 54^{\circ}=1.7205 \text { volume units. }
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_66a193b36253534ec963g-1.jpg?height=509&width=507&top_left_y=629&top_left_x=733)

Figure 1
![](https://cdn.mathpix.com/cropped/2024_05_02_66a193b36253534ec963g-1.jpg?height=882&width=502&top_left_y=1187&top_left_x=778)
Figure 2

The side face of the mentioned parallelepiped is a rectangle inscribed in the wall (Figure 3), its height, based on the sine theorem, is

$$
m=\frac{\sin 108^{\circ}}{\sin 54^{\circ}}=2 \cos 54^{\circ}=\frac{1}{2} \sqrt{10-2 \sqrt{5}}=1.1756
$$

the height of the pyramid is $M-m$, where $M$ is the symmetry axis of the wall,

$$
M=\frac{1}{2} \tan 72^{\circ}=\frac{1}{2} \sqrt{5+2 \sqrt{5}}=1.5388
$$

and from these

$$
V_{m}=m+\frac{1}{3}(M-m)=\frac{2 m+M}{3}=1.2967
$$

Finally, from all of this

$$
V=2.144 \text { volume units. }
$$

Remark. We used that $\sin 18^{\circ}=(\sqrt{5}-1) / 4$, and from this, the numerical expression for each trigonometric function used can be obtained independently of the tables.