Can a cube be decomposed into tetrahedrons for any configuration of diagonals? Suppose we have a cube and configuration of diagonals of the faces of this cube. Is there a way to separate this cube into tetrahedrons, such that for each edge of the each tetrahedron, if the edge lies on the surface of the cube, then it's either an edge of the original cube, or one of the diagonals? Is it possible for any configuration of diagonals?
 A: Yes, this is possible for any configuration of diagonals, and it's pretty easy to accomplish.
The chosen diagonals, together with the sides of the cube, determine a triangulation of the surface of the cube having $12$ triangles, $18$ edges, and $8$ vertices.
Now let $\mathcal O$ denote the center of the cube. Extend the triangulation of the surface to a tetrahedral decomposition of the cube by coning from $\mathcal O$. In detail:

*

*There are $9$ vertices, namely the original $8$ on the surface of the cube plus $\mathcal O$ at the center of the cube.

*There are $26$ edges, namely the original $18$ edges on the surface of the cube plus one edge connecting $\mathcal O$ to each of the $8$ vertices on the surface of the cube.

*There are $30$ triangles, namely the original $12$ triangles on the surface of the cube, plus one triangle having $\mathcal O$ as a vertex and having, as its opposite edge, one of the $18$ edges on the surface of the cube.

*There are $12$ tetrahedra, namely one having $\mathcal O$ as a vertex and having, as its opposite face, one of the $12$ triangles on the surface of the cube.

