## Task $4 / 66$

Which regular polyhedron has the following properties?

1. If the centers of the faces are connected to each other accordingly, another regular polyhedron is formed.
2. If the midpoints of the edges of this second polyhedron are connected to each other accordingly, a third regular polyhedron is formed.
3. If the centers of the faces of this third polyhedron are connected to each other accordingly, a fourth regular polyhedron is formed. Which polyhedron is this?
There are five regular polyhedra. To solve this, we set up a table of these polyhedra with the number of vertices, faces, and edges:

| Polyhedron | Vertices | Faces | Edges |
| :---: | :---: | :---: | :---: |
| Tetrahedron | 4 | 4 | 6 |
| Hexahedron | 8 | 6 | 12 |
| Octahedron | 6 | 8 | 12 |
| Dodecahedron | 20 | 12 | 30 |
| Icosahedron | 12 | 20 | 30 |

The property required under 1 is exhibited by each of the polyhedra. However, when connecting the midpoints of the edges to form a polyhedron, the Dodecahedron and Icosahedron are excluded because there is no regular polyhedron with 12 or 30 vertices (the midpoints of the edges would have to become vertices of the third polyhedron).

But the Hexahedron and Octahedron are also excluded because 4 of the 12 midpoints of the edges of the Hexahedron or Octahedron lie in the same plane. Only the Tetrahedron, with its 6 edges, would result in a polyhedron with 6 vertices, which is an Octahedron. The midpoints of the faces of this Octahedron, when connected to each other, form a Hexahedron.

Thus, the only regular polyhedron that meets the required properties is the Tetrahedron, and the fourth "nested" polyhedron is a Hexahedron (cube).