6
$\begingroup$

By "roundest" I mean having the lowest surface area for the highest volume, given a fixed number of faces $n$. There've been a few questions about them on here (including from me), but I'm not sure if any are particularly close to this one.

Roundest polyhedra are known to exist for every $n \ge 4$, and to have an insphere which is tangent to every face at its centre of mass. The known and conjectured examples are all "simple" (three faces meeting at a vertex), and this is expected to hold in general. The simple Platonic solids - tetrahedron, cube, dodecahedron - are known to be roundest. The only other proven examples (to my knowledge) are the triangular and pentagonal prisms.


Since the faces of a roundest polyhedron should be quite evenly dispersed, it's reasonable to believe that their number of edges shouldn't vary much. Michael Goldberg conjectured (in 1935) that it should vary as little as possible, e.g. for $n \ge 14$ that they should all be hexagons except for the topologically necessary 12 pentagons. He didn't have the advantage of computers, and Alan Schoen (in 1986) found a candidate for $n = 33$ with a heptagon, which has yet to be overturned.

Wayne Deeter has a list of best-known candidates for $n \le 492$ which can be found here (last updated about a year ago); the ones without any heptagons seem to thin out quickly around $n = 400$. Many of these are probably not optimal, but I'd find it quite surprising if the true roundest polyhedra just happened to be heptagon-free. Adding heptagons (and pentagons) seems to compensate for distortion in some sense.


An obvious question, then, is how to bound the edge count of the faces. For example, I wouldn't be too surprised if octagons were possible, but would be extremely surprised if 20-gons were. Intuitively, the higher the number of edges on a face, the more "locally" it should be possible to perturb the faces to find a better alternative with lower maximum edge count.

Perhaps because of the limited overall literature on this subject, this question seems to have barely been addressed. Dmytro Taranovsky, in Asymptotic Optimal Sphericity (2021), notes "Conjecturally, only $n = 4$ or $5$ has any triangles; only $n = 5$$11$, $13$ has any quadrilaterals; no faces have 8 or more edges" but it's not clear if these conjectures have been mentioned anywhere else or if they're just obvious guesses (and the one about octagons feels ambitous to me).

I wonder, how low a bound could be proven without undue effort? I'm really just a curious amateur, with no knowledge of the sorts of tricks one might use - hence the "intuitively..." handwaving above.

$\endgroup$
4
  • $\begingroup$ "Roundest polyhedra are known to exist for every $n\geq 4$": I'm trying to understand what this means. Are you just saying, it is known that there is a unique (presumably up to... isometry?) polytope on $n$ vertices which maximizes volume given a fixed surface area? $\endgroup$ Commented May 2, 2022 at 0:11
  • $\begingroup$ $n$ is the number of faces, rather than vertices, and I don't think uniqueness has been proven, but there should be at most a finite number of optima for any $n$. $\endgroup$ Commented May 2, 2022 at 0:14
  • $\begingroup$ So, sorry, again what is the content of the assertion that "roundest polyhedra are known to exist"? That the supremum volume is achieved? $\endgroup$ Commented May 2, 2022 at 0:18
  • 1
    $\begingroup$ Yes. It is known that there are polytopes with $n$ faces which maximize volume given a fixed surface area. And it's known that among polyhedra with a fixed combinatorial structure (e.g. all tetrahedra), if the structure is simple, then the centre-of-mass condition is not only necessary but also sufficient. So the roundest polyhedra may not be unique, but there should be finitely many - at least, if they are indeed simple. $\endgroup$ Commented May 2, 2022 at 0:24

0

You must log in to answer this question.