2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $\frac{1}{3^{3/2}} \approx 0.1924500897298753$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.