Is there a term for a polyhedron with $n$ faces (or, similarly, $n$ vertices) that maximises the enclosed volume for a given surface area (equivalently, minimises the surface area for a given volume)?

The equivalent in polygons produces the regular polygons, and can be considered a property of regular polygons. However, as we all know, there are only five "regular polyhedra" (not counting the star polyhedra), the platonic solids. I expect that these five solids maximise for either their respective numbers of faces or vertices (probably not both, as that would result in the dodecahedron and icosahedron having identical volume for the same surface area, which seems counterintuitive), but I'm interested in what the maximising shape would be for, for instance, a pentahedron.

  • $\begingroup$ I've been reading a lot about conjectures lately, so I'll make one: the shape for the maximal-volume $n$-hedron places all of the vertices on a single sphere. This does not imply any regularity of the faces or edges; there also is likely not to be a unique solution for other than the regular polyhedra. It's also not immediately obvious to me that this proposition is true... $\endgroup$ Apr 20, 2013 at 2:22
  • $\begingroup$ "This proposition"? You mean your conjectures, or my suggestion regarding the platonic solids? As for the prediction that all of the vertices will lie on the surface of one sphere, that's my expectation, too... but it's not obvious how they'll be organised. For instance, for the pentahedron, there are two possible shapes that I can see - square based pyramid (which is self-dual), and triangular prism (whose dual is a hexahedron). $\endgroup$
    – Glen O
    Apr 20, 2013 at 2:39
  • $\begingroup$ Oh, my proposition... It may be that the number of solution figures is small for small $n$, but tends upward generally as $n$ increases (except for the Platonic solids). I haven't made a search yet, but I'd be surprised if no one had ever looked into this question before. $\endgroup$ Apr 20, 2013 at 4:58
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    $\begingroup$ I wrote this question and answer to solve the case of $5$ vertices. Note that it disproves your conjecture that the vertices of the optimal polyhedra lie on a sphere. $\endgroup$
    – joriki
    Apr 20, 2013 at 12:41
  • $\begingroup$ Your answer is certainly interesting regarding 5 vertices. It doesn't disprove RecklessReckoner's conjecture, though, because that was about the optimal solution for $n$ faces, not $n$ vertices. The 5-vertex optimum is a hexahedron, and is in fact dual to the triangular-based prism that I suspect might be the optimal pentahedron, which has an isoperimetric ratio of $2\cdot 3^{9/2}\approx 280.6$. The other obvious possibility is the optimal square-based pyramid, whose isoperimetric ratio is 288, and thus not optimal for the pentahedron. $\endgroup$
    – Glen O
    Apr 20, 2013 at 13:45

1 Answer 1


This will wind up similar to the Thomson problem of determining the minimum energy configuration of N electrons on the surface of a sphere.

Those point configurations, which are available from the Wikipedia link, can at least provide a starting point. Put hulls around the points, then calculate surface area and volume. Then perturb the points slightly and recalculate a million times, and see if a better solution pops up. Repeat until perturbing finds a new local minimum. My bet is that all the Thomson configurations will already be local minima. Thomson will be one of these 1. always optimal, 2. always nonoptimal, 3. optimal in certain cases.

  • $\begingroup$ It's probably hard to control the number of faces of the convex hull of a fixed number of points. But the dual polyhedron to the convex hull has exactly $N$ faces, and might be a candidate. $\endgroup$
    – M. Winter
    Apr 23, 2020 at 9:42

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