alt Hacker NewsThat article doesn't prove what you say that it does. It just proves because a perpetuum mobile is impossible, it is trivial that a polyhedron must always eventually come to rest on one face. It doesn't assert that the face-down face is always the same face (unistable/monostable). It goes on to query whether or not a uniformly dense object can be constructed so as to be unistable, although if I understand correctly Guy himself had already constructed a 19-faced one in 1968 and knew the answer to be true.
It sounds as though you're talking about the solution to part (b) as given in that reference. Have a look at the solution to part (a) by Michael Goldberg, which I think does prove that a homogeneous tetrahedron must rest stably on at least two of its faces. The proof is short enough to post here in its entirety:
> A tetrahedron is always stable when resting on the face nearest to the center of gravity (C.G.) since it can have no lower potential. The orthogonal projection of the C.G. onto this base will always lie within this base. Project the apex V to V’ onto this base as well as the edges. Then, the projection of the C.G. will lie within one of the projected triangles or on one of the projected edges. If it lies within a projected triangle, then a perpendicular from the C.G. to the corresponding face will meet within the face making it another stable face. If it lies on a projected edge, then both corresponding faces are stable faces.