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.

  • 4
    $\begingroup$ 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. $\endgroup$ Oct 18 '14 at 16:41
  • $\begingroup$ @MarkBennet: Good point, that settles this part at least. $\endgroup$
    – user139000
    Oct 18 '14 at 16:46
  • $\begingroup$ I believe this is an open problem for $n > 8$. $\endgroup$ Oct 21 '14 at 6:57
  • $\begingroup$ @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. $\endgroup$ Oct 23 '14 at 19:50
  • $\begingroup$ A stickler point about your proof for polygons: given that the space of such polygons is compact... $\endgroup$
    – Max
    Oct 23 '14 at 20:42

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.

$\hspace0.75in$ 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.

  • $\begingroup$ 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. $\endgroup$
    – user139000
    Oct 25 '14 at 7:44
  • $\begingroup$ 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... $\endgroup$
    – user139000
    Oct 25 '14 at 7:48
  • $\begingroup$ @pew look at Berman and Hanes paper (linked in updated answer) for a proof. $\endgroup$ Oct 25 '14 at 9:16

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