One farmer made cheese in the form of an irregular pentagonal prism, while another made it in the form of a regular quadrilateral pyramid, the height of which is half the length of the side of the base. At night, mice ate all the cheese particles from the vertices of these polyhedra that were within 1 cm of the corresponding vertex. The eaten pieces of cheese did not share any particles. Which farmer suffered greater damage and by how many times was the damage greater?
On the extension of the lateral edge $A A 1$ of the pentagonal prism $A B C D E A 1 B 1 C 1 D 1 E 1$, all edges of which are greater than 2, mark point $F$ beyond vertex $A$ (Fig.1). When the three-dimensional angle at vertex $A 1$ of the prism is translated by the vector $A 1 A$, it will transform into the three-dimensional angle $A F B E$ with vertex $A$. Consider a sphere of radius 1 centered at point $A$. Then, the sum of the volumes of the cheese pieces cut off from vertices $A$ and $A 1$ is equal to the volume of the intersection of this sphere with the dihedral angle $A A 1$ of the prism. If the linear angle of this dihedral angle is $\alpha 1$, then the volume of the intersection is $\frac{\alpha_{1}}{2 \pi} \cdot \frac{\bar{t}}{3} \pi=\frac{2}{3} \alpha 1$

If $\alpha 2, \alpha 3, \alpha 4$ and $\alpha 5$ are the linear angles of the dihedral angles $B B 1, C C 1, D D 1$ and $E E 1$ of the prism, then the sum of the volumes of all the cheese pieces cut off in this case is

$$
V 1=\frac{2}{3} \alpha 1+\frac{2}{3} \alpha 2+\frac{2}{3} \alpha 3+\frac{2}{3} \alpha 4+\frac{2}{3} \alpha 5=\frac{2}{3}(\alpha 1+\alpha 2+\alpha 3+\alpha 4+\alpha 5)=\frac{2}{3} \pi(5-2)=2 \pi
$$

since $\alpha 1+\alpha 2+\alpha 3+\alpha 4+\alpha 5$ is the sum of the interior angles of the pentagon, which is a perpendicular cross-section of the prism.

Consider the cube KLMNK1L1M1N1, whose edge is greater than 2 (Fig.2). Let $S$ be the center of this cube. Then the quadrilateral pyramid $S K L M N$ with vertex $S$ is a regular quadrilateral pyramid, as mentioned in the problem statement.

The volume of the cheese piece cut off from vertex $S$ is one-sixth of the volume of a unit sphere centered at $S$, i.e., $\frac{1}{6} \cdot \frac{\frac{1}{3}}{3} \pi=\frac{2 \pi}{9}$, since 6 equal regular quadrilateral pyramids meet at this vertex, with their bases being the 6 faces of the cube.

The volume of the cheese piece cut off from vertex $M$ is one-third of the volume of the intersection of a unit sphere centered at $M$ with the trihedral angle of the cube at vertex $M$, i.e., $\frac{1}{3} \cdot \frac{1}{8} \cdot \frac{\frac{1}{3}}{3} \pi=\frac{\pi}{18}$, since three equal trihedral angles of the regular quadrilateral pyramids $S K L M N, S L M M 1 L 1$ and $S M N N 1 M 1$ with a common vertex $S$ meet at this vertex. Similarly for vertices $K, L$ and $N$. Therefore, the total volume of the cheese pieces cut off in this case is

$$
V 2=\frac{2 \pi}{9}+4 \cdot \frac{\pi}{18}=\frac{4 \pi}{9}
$$

Thus, the first farmer suffered a greater loss, and

$$
\frac{V_{1}}{V_{2}}=\frac{2 \pi}{\frac{4 \pi}{9}}=\frac{9}{2}
$$

## Answer

The loss of the first farmer is 4.5 times greater than the loss of the second.

$60^{\circ}$.