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In two dimensions, it is not hard to see that the $n$-vertex polygon of maximum area that fits into a unit circle is the regular $n$-gon whose vertices lie on the circle: For any other vertex configuration, it is always possible to shift a point in a way that increases the area.

In three dimensions, things are much less clear. What is the polyhedron with $n$ vertices of maximum volume that fits into a unit sphere? All vertices of such a polyhedron must lie on the surface of the sphere (if one of them does not, translate it outwards along the vector connecting it to the sphere's midpoint to get a polyhedron of larger volume), but now what? Not even that the polyhedron must be convex for every $n$ is immediately obvious to me.

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    If the vertices are on the surface of the sphere the polyhedron will necessarily be convex - it will be the convex hull of the vertices. Because the sphere itself is convex the convex hull will lie entirely within it. – Mark Bennet Oct 18 '14 at 16:41
  • @MarkBennet: Good point, that settles this part at least. –  Oct 18 '14 at 16:46
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    I believe this is an open problem for $n > 8$. – achille hui Oct 21 '14 at 06:57
  • @achille hui: do you know solutions for n = 7, 8? One can check directly that cube is not even a local maximum, having in fact surprisingly poor performance. – Incnis Mrsi Oct 23 '14 at 19:50
  • A stickler point about your proof for polygons: given that the space of such polygons is compact... – Max Oct 23 '14 at 20:42
  • Contrary to achille’s perception, I think that for certain large n, namely n = 2 + 10 | z |², where z is an Eisenstein integer, a solution is not especially hard. At least I know how to find a local maximum in such case and can offer some arguments that global maximum for any n of mentioned form will be among maxima found. But I have no insights how to solve the problem for most moderately large odd n, such as 9, 11, or 13. – Incnis Mrsi Oct 23 '14 at 22:17
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    @IncnisMrsi $n = 7$ is easy, it is a pentagonal bipyramid. For $n = 8$, I don't know how to describe it, I've added an answer which contains the coordinates. – achille hui Oct 24 '14 at 06:26
  • In addition to the papers cited by Achille, there is a more recent paper that may shed more light on the subject: Mutoh, N. (2003). The Polyhedra of Maximal Volume Inscribed in the Unit Sphere and of Minimal Volume Circumscribed about the Unit Sphere. In: Akiyama, J., Kano, M. (eds) Discrete and Computational Geometry. JCDCG 2002. Lecture Notes in Computer Science, vol 2866. Springer, Berlin, Heidelberg. https://doi.org/10.1007/978-3-540-44400-8_22 (There is a preview available here: https://link.springer.com/chapter/10.1007/978-3-540-44400-8_22) It covers n up to 30. – LarsH Aug 19 '24 at 15:25
  • This question, and Don Grace's paper on the search for n=8, is featured in this new Stand-up Maths video: https://www.youtube.com/watch?v=XZy3rXr2yeM – LarsH Aug 19 '24 at 15:26

1 Answers1

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This is supposed to be a comment but I would like to post a picture.

For any $m \ge 3$, we can put $m+2$ vertices on the unit sphere

$$( 0, 0, \pm 1) \quad\text{ and }\quad \left( \cos\frac{2\pi k}{m}, \sin\frac{2\pi k}{m}, 0 \right) \quad\text{ for }\quad 0 \le k < m$$

Their convex hull will be a $m$-gonal bipyramid which appear below.

Up to my knowledge, the largest $n$-vertex polyhedron inside a sphere is known only up to $n = 8$.

  • $n = 4$, a tetrahedron.
  • $n = 5$, a triangular bipyramid.
  • $n = 6$, a octahedron = a square bipyramid
  • $n = 7$, a pentagonal bipyramid.
  • $n = 8$, it is neither the cube ( volume: $\frac{8}{3\sqrt{3}} \approx 1.53960$ ) nor the hexagonal bipyramid ( volume: $\sqrt{3} \approx 1.73205$ ). Instead, it has volume $\sqrt{\frac{475+29\sqrt{145}}{250}} \approx 1.815716104224$.
    Let $\phi = \cos^{-1}\sqrt{\frac{15+\sqrt{145}}{40}}$, one possible set of vertices are given below: $$ ( \pm \sin3\phi, 0, +\cos3\phi ),\;\; ( \pm\sin\phi, 0,+\cos\phi ),\\ (0, \pm\sin3\phi, -\cos3\phi),\;\; ( 0, \pm\sin\phi, -\cos\phi). $$ For this set of vertices, the polyhedron is the convex hull of two polylines. One in $xz$-plane and the other in $yz$-plane. Following is a figure of this polyhedron, the red/green/blue arrows are the $x/y/z$-axes respectively. The polyhedron has $D_{2}$ symmetry; it may be viewed as a square antiprism modified by buckling the bases along a pair of diagonals.

$\hspace0.75in$ Max volume polyhedron with 8 vertices (Berman Hanes version)

For $n \le 8$, above configurations are known to be optimal. A proof can be found in the paper

Joel D. Berman, Kit Hanes, Volumes of polyhedra inscribed in the unit sphere in $E^3$
Mathematische Annalen 1970, Volume 188, Issue 1, pp 78-84

An online copy of the paper is viewable at here (you need to scroll to image 84/page 78 at first visit).

For $n \le 130$, a good source of close to optimal configurations can be found under N.J.A. Sloane's web page on Maximal Volume Spherical Codes. It contains the best known configuration at least up to year 1994. For example, you can find an alternate set of coordinates for the $n = 8$ case from the maxvol3.8 files under the link to library of 3-d arrangements there.

Oscar Lanzi
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achille hui
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    That's plenty of information, and I'm willing to give you the bounty since you have answered my question ("Open problem for $n\ge 9$") but I'd like some more information for the $n=8$ case if possible. The cube not being optimal is already surprising since one might expect the 2D argument to somehow transfer to regular polyhedra, but where do the coordinates come from? You say that $n=8$ is known exactly but the coords just look like the result of a numerical optimization run. I'd expect there to be exact polar or cartesian coordinates or at least some formal description of the polyhedron. –  Oct 25 '14 at 07:44
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    That the $n=7$ solution is a pentagonal bipyramid also mildly surprised me. In my mental picture of it, the concentration of vertices in the plane spanned by the pyramids' shared base seems a little too high for the configuration to be optimal. But 3D is hard to imagine accurately, of course... –  Oct 25 '14 at 07:48
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    @pew look at Berman and Hanes paper (linked in updated answer) for a proof. – achille hui Oct 25 '14 at 09:16
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    I asked essentially the same question on MathOverflow (not being aware of this MSE one). If you are still active, perhaps you would like to copy your answer there, or simply link to it, so I can accept it? It seems to be as good as I could hope for. – Gro-Tsen Sep 01 '22 at 21:51
  • @gro-tsen I went a little further, finding some higher-dimensional results. See my update. – Oscar Lanzi Sep 02 '22 at 22:44
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    Worth mentioning that, as Berman and Hanes note, the $n=8$ polyhedron was found by Donald Grace, Search for largest polyhedra, in the early 1960s, although Grace did not prove optimality. (I learned this fact from Stand-up Maths.) – Timothy Chow Aug 17 '24 at 13:32
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    Another surprise for me is that the regular dodecahedron is not an optimal solution for = 20. It feels like the optimal solutions should be the most symmetrical. But I guess that shouldn't be surprising, since the cube (another Platonic solid) isn't optimal. Also, https://mathoverflow.net/questions/429497/known-configurations-maximizing-the-volume-of-the-convex-hull-of-n-points-on-the says it's been proven that all solutions for n >= 4 are polyhedra with only triangular faces. So that eliminates the cube, the regular dodecahedron, and all of the Archimedean solids. – LarsH Aug 19 '24 at 17:22
  • does this solid have a name? – Yukulélé Sep 14 '24 at 22:16