A new pyramid-like shape always lands the same side up

That 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.

Ah, I see. I saw that but disregarded it because if it's meant be an actual proof and not just a back of the envelope argument, it seems to be missing a few steps. On the face of it, the blanket assertion that at least two faces must be stable is clearly contradicted by these current results. To be valid, Goldberg would needed at least to have established that his argument was applicable to all tetrahedra of uniform density, and ideally to have also conceded that it may not be applicable to tetrahedra not of uniform density, don't you think?

This piqued my curiosity, which Google so tantalizingly drew out by indicating a paper (dissertation?) entitled "Phenomenal Three-Dimensional Objects" by Brennan Wade which flatly claims that Goldberg's proof was wrong. Unfortunately I don't have access to this paper so I can't investigate for myself. [Non working link: https://etd.auburn.edu/xmlui/handle/10415/2492 ] But Gemini summarizes that: "Goldberg's proof on the stability of tetrahedra was found to be incorrect because it didn't fully account for the position of the tetrahedron's center of gravity relative to all its faces. Specifically, a counterexample exists: A tetrahedron can be constructed that is stable on two of its faces, but not on the faces that Goldberg's criterion would predict. This means that simply identifying the faces nearest to the center of gravity is not sufficient to determine all the stable resting positions of a tetrahedron." Without seeing the actual paper, this could be a LLM hallucination so I wouldn't stand by it, but does perhaps raise some issues.

That's very interesting! I agree Goldberg's proof is not very persuasive. I hope Auburn university will fix their electronic dissertation library.

There's a 1985 paper by Robert Dawson, _Monostatic simplexes_ (The American Mathematical Monthly, Vol. 92, No. 8 (Oct., 1985), pp. 541-546) which opens with a more convincing proof, which it attributes to John H. Conway:

> Obviously, a simplex cannot tip about an edge unless the dihedral angle at that edge is obtuse. As the altitude, and hence the height of the barycenter, is inversely proportional to the area of the base for any given tetrahedron, a tetrahedron can only tip from a smaller face to a larger one.

Suppose some tetrahedron to be monostatic, and let A and B be the largest and second-largest faces respectively. Either the tetrahedron rolls from another face, C, onto B and thence onto A, or else it rolls from B to A and also from C to A. In either case, one of the two largest faces has two obtuse dihedral angles, and one of them is on an edge shared with the other of the two largest faces.

The projection of the remaining face, D, onto the face with two obtuse dihedral angles must be as large as the sum of the projections of the other three faces. But this makes the area of D larger than that of the face we are projecting onto, contradicting our assumption that A and B are the two largest faces