46*. Inscribed a cube in a dodecahedron so that all eight vertices of the cube are vertices of the dodecahedron. How many solutions does the problem have?
46. Consider a dodecahedron with center $S$ (Fig. 101). Due to the congruence of the faces, their diagonals are also congruent: $[A B] \cong|B C| \cong \cong[C D] \cong[D A]$. In the regular triangular pyramid $P A B Q$, the edge $[P Q]$ is perpendicular to the base edge $[A B]$ (see problem 8 from § 5). But $(P Q)\|(A D)\|(B C)$, and thus, $A B C D$ is a square. Similarly,

![](https://cdn.mathpix.com/cropped/2024_05_21_5cdadc99c18e78f1502ag-173.jpg?height=543&width=616&top_left_y=254&top_left_x=343)

Fig. 101

we will prove that $A B B^{\prime} A^{\prime}, B C C^{\prime} B^{\prime}$, $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, and $A D D^{\prime} A^{\prime}$ are also squares. Thus, the polyhedron $A B C D C^{\prime} D^{\prime} A^{\prime} B^{\prime}$ is a cube, whose edges are the diagonals of the dodecahedron.

If in the pentagon $A M N B P$ we take another diagonal $[A N]$, we get the square $A N E Q$, which is a face of the cube $A N E Q C^{\prime} N^{\prime} E^{\prime} Q^{\prime}$. From the other three diagonals $[B M \mid, [M P]$, and $\mid N P]$, we obtain three more cubes: $B M F^{\prime} Q D^{\prime} M^{\prime} F Q^{\prime}$, $M P D E^{\prime} M^{\prime} F^{\prime} B^{\prime} E$, and $N P C F N^{\prime} P^{\prime} A^{\prime} F^{\prime}$.

It is not difficult to verify that each diagonal of each face of the dodecahedron is a side of one and only one of these cubes. Therefore, there are exactly five such cubes (60 face diagonals are distributed 12 to each of the 5 cubes).