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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), as mentioned in the comments:
           MaxVol8
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:
Cohn

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There is a related and also interesting question: what is the polyhedron with 8 vertices, inscribed in a sphere, of max volume? There is a candiate in www.jstor.org/stable/2003644 but I don't know whether the question has been solved since then. –  Jean-Marc Schlenker Aug 28 '11 at 15:01
    
@Jean-Marc: It appears that question was settled by Berman and Haynes in "Volumes of polyhedra inscribed in the unit sphere in $\mathbb{R}^3$," Math. Ann. 1970, which I cannot access right now: springerlink.com/content/r7h7112424214257 –  Joseph O'Rourke Aug 28 '11 at 15:20
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@Jean-Marc and @Joseph: more is known, see www-users.cs.umn.edu/~shao/fulltext.pdf –  Igor Rivin Aug 28 '11 at 15:28
    
@Igor: Thanks! That reference includes a candidate, now included at the end of my question. –  Joseph O'Rourke Aug 28 '11 at 16:04
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"it is known that there is an optimal polyhedron for each v". Is it stated as an open question somewhere? $$ $$ It is not a problem to show existence for algebraic volume (i.e., with overlaps counted). Further this optimal polyhedron (for large $v$) has to be close to a sphere. It should follow that this optimal polyhedron has no overlaps... –  Anton Petrunin Aug 28 '11 at 17:02

1 Answer 1

up vote 13 down vote accepted

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.

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Wonderful, Henry!! I hope I interpreted your construction correctly in the figure I posted. Do you care to adumbrate the skeleton of your reasoning? Seems like it might be close to a proof... –  Joseph O'Rourke Aug 29 '11 at 23:57
    
Unfortunately, there's not much reasoning behind it: I wrote a program to optimize the isoperimetric ratio and ran it a bunch of times. I designed it for ease of programming rather than execution speed, so I couldn't gather a lot of evidence, but it found this configuration a number of times and never beat it (sometimes it got stuck in local optima). The program produced ugly coordinates but I played around with them and identified the structure described above. Given the three-parameter family, it's not hard to get 500 digits of accuracy, but that's not enough to identify the exact values. –  Henry Cohn Aug 30 '11 at 3:57
    
By the way, someone should carry out a serious numerical investigation of isoperimetric problems for polytopes with a fixed number of vertices (if it hasn't already been done). I don't currently have the time or energy to do this right, but if anyone would like to try, I'd be happy to give them my crude code and some suggestions for how to do it better. Just send me an e-mail. –  Henry Cohn Aug 30 '11 at 4:06
    
I proved your assertion about the 5-vertex case in this math.SE post: math.stackexchange.com/questions/367287. –  joriki Apr 20 '13 at 12:38
    
@joriki: That's great! –  Henry Cohn Apr 20 '13 at 16:57

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