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 $V_n$?
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asked Dec 1, 2011 at 0:02
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$\begingroup$ Do you have an example with $V_2 > 2$? $\endgroup$– Michael BiroCommented Dec 1, 2011 at 3:10
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$\begingroup$ No, I don't have one. $\endgroup$– Vladimir ReshetnikovCommented Dec 1, 2011 at 5:02
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4 Answers
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The problem seems to be still open even for $n=3$:
Weisstein, Eric W. "Tetrahedron Circumscribing."
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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.
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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.
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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.
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$\begingroup$ To be clear, the asymptotic of $n^{1/2}$ is for $V_n^{1/n}$. $\endgroup$ Commented Jul 3, 2018 at 3:18