4. Can a regular tetrahedron with edge length 1 be divided into regular tetrahedrons and octahedrons, each with edge lengths less than $1 / 100$?
4. By marking the midpoints of the edges of a regular tetrahedron, we easily obtain a partition of the regular tetrahedron into a regular octahedron and 4 regular tetrahedra. In other words, the small tetrahedra are obtained from the larger one by homotheties with a coefficient of $1 / 2$ and centers at the vertices of the larger tetrahedron (Fig. $88, a$ ).

![](https://cdn.mathpix.com/cropped/2024_05_21_6455f48598646f41a890g-212.jpg?height=480&width=422&top_left_y=644&top_left_x=193)
a)

![](https://cdn.mathpix.com/cropped/2024_05_21_6455f48598646f41a890g-212.jpg?height=478&width=523&top_left_y=645&top_left_x=691)

b)

Fig. 88

It is a bit more complicated to understand how to divide a regular octahedron. As in the case of the tetrahedron, subject the octahedron to homotheties with a coefficient of $1 / 2$ and a center at a vertex of the octahedron. Considering all 6 vertices of the octahedron, we get 6 small octahedra, and by cutting them out of the larger octahedron, we see that 8 regular tetrahedra remain, adjacent to the faces of the larger octahedron. One of the vertices of each of these tetrahedra is the center of the original octahedron, and the other vertices are the midpoints of its edges (Fig. 88, b).

After the first step (where only the tetrahedron is divided), we get an octahedron and tetrahedra with edge lengths equal to $1 / 2$. After the second step (dividing 4 tetrahedra and the octahedron), we get octahedra and tetrahedra with edge lengths of $1 / 4$, and so on. After the 7th step, the edges of the tetrahedra and octahedra will be equal to $1 / 128 < 1 / 100$.