Can we count number of Regular tetrahedrons formed out of Unit cube?
If vertices of Unit cube are taken as $(0,0,0,)$, $(0,0,1)$, $(0,1,1)$, $(0,1,0)$, $(1,0,0)$, $(1,0,1)$, $(1,1,1)$ and $(1,1,0)$
I have seen a post where any four vertices satisfying $x \le y \le z$ forms a Regular Tetrahedron.
But if i choose vertices as $(0,0,0)$, $(1,1,1)$,$(0,0,1)$,$(0, 1,1)$
will it form Regular tet?
Can i know any logical way to choose vertices of cube to form Tet?
On
Whenever a polyhedron has faces with even counts of sides only, such as the cube, then you can apply a vertex alternation. Say you mark one initial vertex in red, all edge connected neighbours of red vertices in green, and conversely all edge connected neighbours of green vertices in red. This then describes two enantiomorph alternated polyhedra, defined as the convex hull of the red vertices only, and by the green ones only respectively. - For the cube this results in two mutually inverted copies of regular tetrahedra.
Any other choice of 4 vertices from the vertex set of a cube clearly combinatorically could be structured to become a simplex as well. But those simplices would not be regular tetrahedra (as those would use at least 2 different edge length out of $\{1, \sqrt{2}, \sqrt{3}\}$, i.e. edge of cube, diagonal of square face, body diagonal of cube. In some cases such simplices even would become degenerate (when choosing all 4 vertices from a single square face).
--- rk
No, because, for instance, the distance from $(0,0,0)$ to $(1,1,1)$ is different from the distance from $(0,0,0)$ to $(0,0,1)$.
You'll get a tetrahedron if you take those vertices with an odd number of $0$'s and another one if you take those vertices with an even number of $0$'s.