Let $T\subset \mathbb{R}^n$ be a fixed simplex, $H\subset \mathbb{R}^n$ be a variable affine hyperplane. Is it true that the maximal area (i.e. the $(n1)$dimensional volume) of $T\cap H$ is attained when $H$ contains a facet of $T$?

$\begingroup$ In case someone is trying to prove this and at some point needs that for any vector $v$, the interval $\{\langle x,v\rangle;x\in T\}$ achieves minimal length when $v$ is perpendicular to some face: that is not true (consider the $3$simplex with vertices $(1,0,0.01),(1,0,0.01),(0,1,0.01),(0,1,0.01)$) $\endgroup$– Saúl RMJan 31 at 17:40

2$\begingroup$ @Saul In other words, the width of a simplex is not its shortest height. Yes, it is not, even for a regular simplex. $\endgroup$– Fedor PetrovJan 31 at 18:41
1 Answer
No, there is a $5$dimensional simplex with a hyperplane section which is larger than any of its facets, see Walkup, A simplex with a large crosssection, Am. Math. Monthly, January 1968. The idea is to squeeze a regular 5simplex along the common perpendicular of two opposite 2dimensional faces. The volume of the midsection between these faces does not change, but it turns out that all facets become smaller than this crosssection if the scaling coefficient is small enough.
For more information see also the PhD thesis of Dirksen.

3$\begingroup$ Thank you Vanya! The Dirksen's thesis also contains the information that for a regular simplex the claim is true. $\endgroup$ Feb 1 at 6:23