35. Cubes. As is known, the entire space can be filled with equal cubes. At each vertex, eight cubes will meet. Therefore, it is possible - by appropriately cutting off the vertices of the cubes and gluing the adjacent cut parts into one body - to fill the space with regular octahedra (eight-faced bodies) and the bodies remaining from the cubes. What will these bodies be? If we maximally increase the octahedra, what part of the space will they occupy?
35. Bodies formed by truncating the eight corners of a cube have 14 faces (the number of faces of the cube + the number of vertices of the cube). Eight of them are triangles, and six are octagons (Fig. 50).

![](https://cdn.mathpix.com/cropped/2024_05_21_5d77f4829f88503a49eag-087.jpg?height=1062&width=1171&top_left_y=383&top_left_x=431)

Fig. 50.

When the octahedra are as large as possible, the faces of the tetradecahedron will be triangles and squares (Fig. 51). Since the octahedron arises

![](https://cdn.mathpix.com/cropped/2024_05_21_5d77f4829f88503a49eag-087.jpg?height=554&width=651&top_left_y=1733&top_left_x=351)

Fig. 51. from the cube by truncating eight tetrahedra (triangular pyramids) with a base equal to \(1 / 8\) of the side of the cube and a height equal to half the side of the cube, the tetrahedra will occupy \(8 \cdot \frac{1}{3} \cdot \frac{1}{8} \cdot \frac{1}{2} = \frac{1}{6}\) of the space. At each vertex, six bodies meet: four tetradecahedra and two octahedra (octahedra).