I am trying to develop an integral in $R^n$ and I am faced with the following problem:

Given a polyhedron $P$ in $R^n$ of diameter d, define the "compactness" of $P$ as the quotient of the volume of $P$ by $d^n$. It is clear that that the compactness is invariant under a change of scale.

I am certain that the following assertion is true, but it is difficult for me to prove it:

"Among all the 2-d polyhedra with 4 vertices, the square owns the largest compactness; among all the 3-d polyhedra with 8 vertices, the cube owns the largest compactness, and in general, the hypercube is the unique n-d polyhedron of $2^n$ vertices whose compactness is maximal."

Theoretically, it should be possible to solve this problem using differential calculus, but such a direct approach seems awful. A more clever way would be to seek the maximal polyhedron of $2^n$ vertices whose diameter is given, using, say, Lagrange multipliers, but this seems awful too.

My impression is that a proof should begin with: "Assume that the polyhedron is not a cube. Then by moving a bit some vertex(es), it is possible to obtain a polyhedron with the same diameter and with larger volume".

N.B: 1.the volume of the n-d polyhedron is the sum of the volumes of the n-d simplexes composing it. 2. the diameter of a polyhedron is necessarily the largest distance between two vertexes, among all the pairs of vertexes.

  • $\begingroup$ Hold on: doesn't an octagon have more surface area than a square with the same diagonal? $\endgroup$ – Riccardo Orlando May 2 '16 at 22:35

The conjecture already fails at $3$-dimension.

For $3$-dimension, consider the square anti-prism with vertices at

$$\left( \pm \sqrt{3}, \pm \sqrt{3}, \sqrt{2} \right),\quad ( \pm \sqrt{6}, 0, -\sqrt{2} )\quad\text{ and }\quad( 0, \pm \sqrt{6}, -\sqrt{2})$$

It has volume $V = 16(\sqrt{2}+1)$ and diameter $d = \sqrt{20 + 6\sqrt{2}}$. Its "compactness" $$\frac{V}{d^3} = \frac{16(\sqrt{2}+1)}{(20+6\sqrt{2})^{3/2}} \approx 0.25407673354018117$$ is bigger than that of a cube $\frac{1}{3^{3/2}} \approx 0.1924500897298753$.

  • $\begingroup$ Extremely interesting achille hui. I turns out that I've asked a too broad question. I need less for what I need. So, I will open another thread to ask a less elegant but more restricted question. $\endgroup$ – MikeTeX May 3 '16 at 5:43
  • $\begingroup$ Can I ask you where did you found these computations of volume and diameter ? $\endgroup$ – MikeTeX May 3 '16 at 5:46
  • $\begingroup$ @MikeTeX For the symbolic values of the volume and diameter, I just compute them by hand using the coordinates. However, it is very easy to make mistakes that way. The volumes I give in the first version of this answer is completely wrong. At the end I also compute them numerically (essentially using the algorithm you mention in question) to verify the answer. $\endgroup$ – achille hui May 3 '16 at 5:53

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