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For $n\geq 4$, let $V_n$ be the maximum volume of the convex hull of $n$ points on the unit sphere (in $\mathbb{R}^3$, although information on higher dimensions is welcome as well). I'm sure the problem of computing $V_n$ has been extensively studied and has a standard name: what is this name?

For which values of $n$ is the exact value of $V_n$, and/or a configuration realizing it, known (at least combinatorially)? And, for small $n$, what are the best known configurations even if they are not proved to be optimal? Essentially, I'm looking for a list analogous to what Wikipedia lists for the Thomson problem but for the volume of the convex hull instead of electrostatic potential energy.

PS: the dual problem of finding the least possible volume cut by $n$ (hyper)planes all tangent to the unit sphere (i.e., the smallest convex polytope exscribed around a sphere, rather than the largest inscribed in one as described above) also seems interesting, so if it has a standard name I'm also curious to know.

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    $\begingroup$ It would be nice if the answers included the regular polytopes for the appropriate dimensions & number of points/vertices, but I see that the Thomson problem for 3-space and $n=8$ is the square antiprism, not the cube, and for $n=20$ it's not the dodecahedron. $\endgroup$
    – Brian Hopkins
    Commented Aug 31, 2022 at 12:47
  • $\begingroup$ The cube does not work here for $n=8$ either. Any prism is beaten by the antiprism obtained by twisting either basal face in-plane. $\endgroup$
    – Oscar Lanzi
    Commented Aug 31, 2022 at 12:50

1 Answer 1

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The problem is elementary for $n=5$.

We may regard that case as a combination of two triangular pyramids sharing a base $\triangle ABC$. Then the volume is always bounded by one-third the area of the common base times the distance between the two remaining points $D,E$ (the latter is always greater than or equal to the sum of the pyramid altitudes). Then this bound is saturated and both factors maximized by placing $D$ and $E$ at opposite points ("poles") and distributing $A,B,C$ uniformly along the "equator" (a maximally symmetric triangular bipyramid).

Let's look more broadly here, as other broader answers have yet to appear.

We begin with this 2014 answer from Math Stack Exchange, which concentrates on four to eight points in three dimensions. Summarizing those results gives:

$n=4$ -- regular tetrahedron

$n=5$ -- triangular bipyramid, as above

$n=6$ -- regular octahedron

$n=7$ -- pentagonal bipyramid (this would seem to overcrowd the "equator" with five points, but plausible alternatives such as the capped octahedron would be worse)

$n=8$ -- not a cube, which isn't even close; rather a $D_2$-symmetry object with triangular faces and vertices of order $4$ and $5$.

The linked answer cites Ref. [1] for proof.

Horváth and Lángi[2] give a lemma that all solutions with four or more points in three dimensions are polyhedra with triangular faces; and the faces are also simplices in higher dimensions. The above claims conform with this, as do well-known (but AFAIK not rigorously proved) arrangements for larger $n$ such as the laterally tricapped trigonal prism for $n=9$ and the regular icosahedron for $n=12$. The triangular face requirement excludes the regular dodecahedron, which in any event has long since been beaten for $20$ points.

Reference [2] also considers higher dimensionalities. For $n=d+1$ points on the surface of a $d$-dinensional ball the optimum is the regular simplex. With $n=d+2$ the optimum is described as combining regular-simplex cross-sections in two complementary subsets of the dimensions. Going back to the triangular bipyramid with $n=5$ and $d=3$, the relevant cross-sections are an equilateral triangle (the base of the bipyramid) in two of the dimensions, and a line segment (the axis) in the remaining dimension. With higher dimensionality the two orthogonal cross-sections should have as nearly equal dimensionalities as possible. Thus for $(n,d)=(6,4)$ we would select an equilateral triangle in two dimensions and another equilateral triangle in the remaining two dimensions, not a three-dimensional tetrahedron and a one-dimensional line segment; so the $(6,4)$ optimum is not a higher-dimensional bipyramid. For $n=d+3$ with $d$ odd the optimum is guaranteed by using three orthogonal sim-plectic cross-sections, such as the line segments which are diagonals of the regular octahedron for $n=6, d=3$.

References

  1. Joel D. Berman, Kit Hanes (1970), "Volumes of polyhedra inscribed in the unit sphere in E³", Mathematische Annalen, 188, 1, 78-84.

  2. Horváth, Ákos G.; Lángi, Zsolt (2016), "Maximum volume polytopes inscribed in the unit sphere", Monatsh. Math. 181, No. 2, 341-354. ZBL1354.52016.

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  • $\begingroup$ What about configurations with three points in the southern hemisphere and two points in the northern hemisphere? $\endgroup$
    – user44143
    Commented Aug 31, 2022 at 13:30
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    $\begingroup$ You reorient the equatorial plane. If D and E are two points in the northern hemisphere and the the others are in the south with A having its longitude closest to D, you interchange A with D. $\endgroup$
    – Oscar Lanzi
    Commented Aug 31, 2022 at 13:34
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    $\begingroup$ Related: math.stackexchange.com/questions/979660/…. This identifies bipyramidal solution for all of $n=5,6,7$ including the regular one for $n=6$. $\endgroup$
    – Oscar Lanzi
    Commented Aug 31, 2022 at 13:36
  • $\begingroup$ 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, who asks if this is the first shape ever discovered with the help of an electronic computer.) $\endgroup$
    – Timothy Chow
    Commented Aug 17 at 13:34

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