What is the smallest number of tetrahedrons into which a cube can be partitioned?

Answer 1

The smalles number is 5.

A cube has 12 edges, which should also belong to edges of polyhedra. Cube edges are aligned in 3 directions. If we take 4 or more edges of a cube, then at least two of them will be parallel. A polyhedron does not have parallel edges. Hence, one polyhedron can "cover" not more than 3 edges of a cube, so that the total number of polyhedra can not be less than 12/3=4.

Consider a unit cube with two opposite faces A,B,C,D and A1, B1, C1, D1. Take a polyhedron with the edges are AB, AD, and A A1, so that it covers three edges of the cube. The area of the triangle ABD is 1/2. The height of the polyhedron is |A A1|=1. The volume of the polyhedron is 1/6. We see that four polyhedra, which cover edges of a unit cube have the total volume of 2/3. Since this is not sufficient to partition the cube, we need at least 5 polyhedra.

Partitioning of a cube in 5 polyhedra can be done explicitely. Take 4 polyhedra covering the cube edges:
(B,A,C,B1), (C1, C, B1, D1), (A1,A,B1,D1), (D,A,B1,D1).
The remaining, inner part of the cube is the 5th polyhedron (A, B1,A1,D1).

Answer 2

I'm going with 4.

A TH has 4 corners and 6 edges, while a cube has 8 corners and 12 edges.
So there is no way 2 TH's can work, otherwise they must be completely disconnected which results in unfilled spaces in the cube.

A TH has 4 triangular faces compared to 6 square faces for the cube. At best 1 TH can touch three 1/2 faces of the cube. In order for the TH to touch four faces of the cube, two of its sides must be parallel, which is never the case in a TH.
So if one TH touches three 1/2 faces of the cube, theoretically 4 THs can completely fill up the cube.

My answer: 4

Answer 3

Well, any prism(a cube included) can be
partioned into 3 congruent pyramids.
So I would guess the answer is 3, but
I have no proof that 2 would do.