Cubic Tetra

MMT meshing always starts from this type of tetra. The tetra is known from the
BCC mesh and from crystal mesh theory. We name it cubic.

You can tile the entire 3-space with cubic tetras. But more:
You can divide the cubic tetra to 8 smaller ones of the same type.

For all tetras: from divideability follows tileability ,
but   from tileability does not follow divideability .

Which tetras can be devided to 8 smaller ones of the same type?

A simple answer could be to look at the list of tileable tetras.
But today a complete list is not known.

Goldberg (1972) extended Sommervilles' (1923) list of tileable tetras and added 3 new parameter dependant classes.
 quality Sommerville 1 (cubic) sqrt 0.5 Sommerville 2 0.35355 Sommerville 3 0.25 Sommerville 4 0.17678 Sommerville 5 0.35355 Goldberg 1/2/3 ...

Tetra quality is defined as Volume / max edge length **3 and is normalized such that the quality of the regular tetra is 1.0 .

Two views of Sommerville 1-5 Tetras

Sommerville 3 is a half of a pyramid. It can be divided to 8 smaller ones of the same type by "3 fold: split the longest edge". If you divide this tetra as you divide the cubic one, you get 8 tetras of 2 kinds of an average quality of 0.364 which is better than 0.25 .

2D: Every triangle can be devided to 4 smaller ones of the same type.
3D: There are at least 2 tetra, which can be devided to 8 smaller ones of the same type.
4D: Conjecture: There is no 4D simplex, which can be devided to 16 smaller ones of the same type.

Is there any 4D simplex, which is bounded by 5 cubic tetras? Let

P1 = (0,0,0,0)
P2 = (1,0,0,0)
P3 = (0.5,0.5 sqrt(2),0,0)
P4 = (0.5,0,0.5 sqrt(2),0)

define a 3D cubic tetra with 4th coordinate 0, let this tetra be the first of these bounds.

Where is the 5th vertex? From symmetry reasons we try P5 = (0.5,0,0,0.5 sqrt(2))

The bounding tetras are

T5 = P1-P2-P3-P4
T4 = P1-P2-P3-P5
T3 = P1-P2-P4-P5
T2 = P1-P3-P4-P5
T1 = P2-P3-P4-P5

edge sizes:

P1 P2 = 1.0
P1 P3 = 0.5 sqrt(3)
P1 P4 = 0.5 sqrt(3)
P1 P5 = 0.5 sqrt(3)
P2 P3 = 0.5 sqrt(3)
P2 P4 = 0.5 sqrt(3)
P2 P5 = 0.5 sqrt(3)
P3 P4 = 1.0
P3 P5 = 1.0
P4 P5 = 1.0

T5, T4, T3 are cubic: 2 edges are 1.0 and 4 edges are 0.5 sqrt(3)

T1, T2 cannot be cubic since 3 edges are 1.0 and 3 edges are 0.5 sqrt(3)

=> This P5 does not work.