Optimal 8-vertex isoperimetric polyhedron?

Question

I know from Marcel Berger's Geometry Revealed: A Jacob's Ladder to Modern Higher Geometry (p.531) that it is not yet established which polyhedron in $\mathbb{R}^3$ on 8 vertices achieves the optimal isoperimetric ratio $A^3/V^2$, where $A$ is the surface area and $V$ the volume. Berger says "We also know that the cube ... [is] not the best for $v=8$" (where $v$ is the number of vertices).

Many other aspects of isoperimetry for polyhedra are unresolved, but this one especially interests me. It is not even clear to me that it is known that there is an optimal polyhedron for each $v$. I've been trying to imagine what would be a strong candidate for an optimal 8-vertex polyhedron. I've been unsuccessful in finding information on this, although it seems likely to have been explored computationally. Does anyone have a candidate, or know of one proposed/calculated? A pointer or reference would be greatly appreciated. Thanks!

Addendum. From the reference Igor provided (Nobuaki Mutoh, "The Polyhedra of Maximal Volume Inscribed in the Unit Sphere and of Minimal Volume Circumscribed about the Unit Sphere," 2009), here is a piece of Mutoh's Fig.1, which computationally verifies the earlier derivation of the max volume inscribed 8-vertex polyhedron by Berman and Haynes ("Volumes of polyhedra inscribed in the unit sphere in $\mathbb{R}^3$," Math. Ann., 188(1): 78-84, 1970, doi: 10.1007/BF01435416, eudml), as mentioned in the comments:
          
This is surely a candidate for achieving the min of $A^3/V^2$! I thank Jean-Marc, Igor, and Anton for the rapid convergence to what I sought.

...And then a bit later to Henry for showing that this candidate does not in fact achieve the best ratio! Here is Henry's polyhedron, if I have interpreted him correctly:

Answer 1

An $8$-vertex polyhedron can achieve an isoperimetric ratio of $A^3/V^2 = 159.3243297053\dots$, and based on some quick experiments I'm pretty confident this is optimal (although I wouldn't be shocked if it could be beaten).

To construct it, let $V_\alpha$ denote the squashed tetrahedron with vertices $(\pm \sqrt{1-\alpha^2},0,\alpha)$ and $(0,\pm \sqrt{1-\alpha^2},-\alpha)$. Then the optimal $8$-vertex polyhedron seems to be the union of $V_\alpha$ and $-\beta V_\gamma$, with $\alpha = 0.2272117725\dots$, $\beta = 0.87345300464\dots$, and $\gamma = 0.83792301859\dots$. The optimal values of $\alpha$, $\beta$, and $\gamma$ are algebraic, but they're pretty complicated and I haven't computed their minimal polynomials.

For comparison, the maximum volume polyhedron inscribed in a sphere has a worse isoperimetric ratio, namely $162.248792\dots$. For the cube, it's $216$.

In general there's no reason to expect the optimal polyhedron to be inscribed in a sphere. The $5$-vertex case is a particularly nice example: it consists of an equilateral triangle on the equator of the unit sphere together with $1/\sqrt{2}$ times the north and south poles. This achieves an isoperimetric ratio of $243$, and I'd be very surprised if that's not optimal. Five vertices is few enough that a rigorous proof may be possible, but I can't think of a non-painful way to do it.

Answer 2

Henry investigated objects with $D_{2d}$ symmetry. Here the $D_{3d}$ alternative is explored.

$D_{3d}$ symmetry includes an improper sixfold rotation axis, so the eight-vertex polyhedron must have a set of six vertices equivalent under this symmetry plus two polar vertices lying along the principal axis. This allows either of two graphical structures indicated below:

In both structures, the outward lines with arrowheads go to infinity where the second polar vertex is projected (the first one being at the center of each graph).

Structure (A) is topologically equivalrnt to the $D_{6h}$ hexagonal bipyramid, and this effectively forces the optimum to that figure. We can adjust the ratio of polar to equatorial diameter to minimize the isopetimetric ratio; the end result is that the optimum diameter ratio is $\sqrt6/2$ and this gives an isoperimetric ratio of $162$.

We move on to structure (B), in which the maximal symmetry is redued back to $D_{3d}$. Polydedra conforming with this graphical strcture and symmetry will have two sets of six congruent isosceles triangular faces: one ringing the "equator" withapex angles $\theta_1$, the others meeting at the poles with apex angles $\theta_2$. When the isopetimetric ratio is rendered as a function of these two angular parameters we note the following results:

$\theta_1=90°,\theta_2=90°\implies 216\tag{cube}$

$\theta_1=60°,\theta_2=120°\implies 187.06\tag{octahedron}$

$\theta_1=66°,\theta_2=114°\implies 171.05\tag{minimum}$

We may regard the minimal case (for integer-degree vertex angles) with graphical structure (B) as a modified version of the octahedron that meets the requirement of having twelve tangency points on the inscribed sphere; the octahedron itself actually has only eight such points.

Nonetheless, the overall $D_{3d}$ optimum, and likely the optimum that has a center of symmetry at all, is the hexagonal bipyramid from graphical structure (A), which has two vertices of order six rather than four of order five. This suggests that the fully optimized eight-vertex polyhedron may also have one or more vertices of order six, a possibility not allowed under the $D_{2d}$ symmetry considered in Henry's bitetrahedral trials.

Answer 3

For algebraic exact values see this Math.SE Q&A, and for history on first discovery see this Youtube video with further links to papers.