The question is in the title:

Question: Which inscribed $n$-dimensional polytope (inscribed in the unit sphere) with $2^n$ vertices has the largest possible volume?

Is it the $n$-dimensional cube? If not, how much larger can its volume be?

  • 3
    $\begingroup$ This seems to be false in dimension 3 : the convex hull of the north pole, the south pole, and a regular hexagon inscribed in the equator has volume $\sqrt{3}$, larger than the volume of the inscribed cube. $\endgroup$ Commented Feb 15, 2022 at 13:29
  • 2
    $\begingroup$ I guess a natural follow-up question would be. does the cube maximise volume among inscribed polytopes combinatorically equivelant to the cube? $\endgroup$
    – Nick L
    Commented Feb 16, 2022 at 13:08
  • $\begingroup$ @NickL more generally, we can ask if the cube maximizes volume among inscribed $n$-polytopes with $2n$ facets. This looks plausible indeed. We can also ask whether the regular cross-polytope maximizes volume among inscribed polytopes with $2n$ vertices. The answers to both questions are positive if we restrict a priori to centrally symmetric polytopes, but it is not obvious why maximizers should be symmetric. $\endgroup$ Commented Feb 16, 2022 at 13:39
  • 1
    $\begingroup$ @Nick L Did you ever find an answer to your question about polytopes combinatorially equivalent to the hypercube? I am also interested in this question, not just for volume, but for surface area as well. $\endgroup$
    – user3816
    Commented Jul 20, 2022 at 22:50

1 Answer 1


For $n=3$, the maximal volume polytope with 8 vertices is described in that paper.

Berman, J. D.; Hanes, K., Volumes of polyhedra inscribed in the unit sphere in (E^3), Math. Ann. 188, 78-84 (1970). ZBL0187.19604.The link is paywalled, but a summary with the optimal shape drawn is available in this answer.

It is combinatorially very different from a cube: it is simplicial, and each vertex has degree 4 or 5. The correct optimal polyhedron also has only $D_2$ symmetry.

Edit. An inscribed polytope which maximizes the volume given the number of vertices is always simplicial (see Lemma 1 in Horváth, Ákos G.; Lángi, Zsolt, Maximum volume polytopes inscribed in the unit sphere, Monatsh. Math. 181, No. 2, 341-354 (2016). ZBL1354.52016.), so the answer is "no" as well in any dimension $>2$.

The gap is even quantitative. Let $\mu_n$ be the largest volume of an $n$-dimensional polytope with $2^n$ vertices inscribed in the ball of radius $\sqrt{n}$ (this normalization is better since it counterbalances the fact that the ball of unit radius has a tiny volume for large $n$). The cube gives the lower bound $\mu_n \geq 2^n$. By taking direct products, we have $\mu_{m+n} \geq \mu_m\mu_n$, and therefore $\mu_{3n} \geq \mu_3^n \geq (2+\epsilon)^{3n}$, exponentially better that the cube. We cannot do much better since the upper bound $\mu_n \leq C^n$ for some constant $C$ holds trivially by comparing with the ball itself.

  • $\begingroup$ Greate answer, thank you! The question that remains for me: can we put an upper bound on $2+\epsilon$ better than the obvious upper bound $2\pi e$ coming from the volume of the ball? $\endgroup$
    – M. Rumpy
    Commented Feb 18, 2022 at 16:06
  • $\begingroup$ Is it rather $\sqrt{2 \pi e}$ ? This is a very good question. At the moment I cannot find an argument giving an estimate $\mu_n \leq (\sqrt{2 \pi e} - \delta)^n$ for some $\delta > 0$. $\endgroup$ Commented Feb 21, 2022 at 9:15
  • $\begingroup$ What might actually work is the dual proposition using cross polytopes (similar to the octahedron in three dimensions): $2n$ vertices at the surface of a ball of $n$ dimensions. The regular cross polytope clearly has maximal volume in both dimensions $2$ and $3$, and to the degree that I can visualize things in higher dimensions the generalization "looks right". $\endgroup$ Commented Sep 1, 2022 at 13:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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