# Smallest containing simplex

Let $V_n$ be the least real number such that for every convex subset of $\mathbb{R}^n$ with hypervolume $1$ there is a containing simplex with hypervolume $V_n$. What is known about $V_n$? Is there a known general formula? If not, then what are the known best bounds for $V_n$?

• Do you have an example with $V_2 > 2$? – Michael Biro Dec 1 '11 at 3:10
• No, I don't have one. – Vladimir Reshetnikov Dec 1 '11 at 5:02

The problem seems to be still open even for $n=3$:
Weisstein, Eric W. "Tetrahedron Circumscribing."

The paper Parallelotopes of Maximum Volume in a Simplex by Lassak gives the maximum possible volume of a parallelotope in a simplex as $n!/n^n$ times the volume of the simplex. This gives us a bound of $V_n \geq n^n/n!$, which I suspect is tight.

In the paper Minimum area of circumscribed polygons, in Elemente der Mathematik Vol. 28 (1973), Chakerian proved the following:

Any convex body K in Euclidean n-space is contained in a simplex T of volume not more than n^{n-1} times that of K.

Nothing is said there about the extremal cases, so it is possible that the bound is not tight.

http://www.um.es/beca/NoLiFA/charlas/talk_Merzbacher_NoLiFA.pdf

Please check out these nice slides with lots of references; they get an asymptotic of root n.

• To be clear, the asymptotic of $n^{1/2}$ is for $V_n^{1/n}$. – Yoav Kallus Jul 3 '18 at 3:18