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Recently stumbled across the idea of monostatic polytopes, and I was reminded of an old book I'd read that gave a "proof" that every polyhedron has at least one stable face. Since such a polyhedron would continuously fall over, it could be used to create a perpetual motion machine, so it can't exist. It's certainly a clever argument, but it doesn't sound very rigorous to me. So I'm wondering what the actual proof of this fact is, and whether the book's argument could be made into a rigorous proof.

eyeballfrog
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    I suppose the following is rigorous formulation of this question: is it the case that every polyhedron has a face so that the orthogonal projection of the center of mass of the polyhedron onto the affine span of this face lies in (the interior of) the face? – M. Winter Mar 15 '21 at 14:45
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    The physics approach can probably be formalized as follows: choose the supporting plane of $P$ whose distance to the center of mass is minimal. Show that it supports $P$ at a face. Show that this face is stable by using that if the polyhedron could tip over, then this leads to supporting planes even closer to the origin (a contradiction). – M. Winter Mar 15 '21 at 14:50

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Taking M. Winter's proposed formulation, yes, we can rigorously prove this. In fact we show a more general fact:

For any nondegenerate polyhedron $P$ and any point $x$ in the interior of $P$, there exists at least one face $F$ of $P$ such that the orthogonal projection of $x$ onto the plane of $F$ lies in the interior of $F$.

Taking $x$ to be the center of mass recovers the initial claim.

Proof: Consider the set of $r>0$ such that a sphere of radius $r$ centered at $x$ is contained within $P$. Let $R$ be the maximum such $r$ (if we take $P$ to be closed, then by compactness such a maximum will exist). It follows that there exists some $y$ on the boundary of $P$ at distance $R$ from $x$. For any face $F$ on which $P$ lies (there might be more than one), all points at a distance of less than $R$ from $x$ lie on the same side of the plane containing $F$, so the sphere of radius $R$ must be tangent to the plane, and hence the vector from $x$ to $y$ is orthogonal to the plane. It follows that the projection from $x$ onto $F$ is exactly the point $y$.

The only remaining concern is that $y$ might be on the boundary, rather than the interior, of the face $F$. If so, then $y$ would be contained in two or more different faces of $P$. But then the vector from $x$ to $y$ would be normal to both of the planes defined by those faces, which would imply the faces were either parallel (in which case they meet and are the same face) or oppositely oriented (in which case they form a dihedral angle of $0^\circ$ and form a degenerate polyhedron).

As an aside, this more general fact can also be proved by physical intuition if we imagine the polyhedron to be weighted such that almost all the mass is concentrated at $x$.

RavenclawPrefect
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