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 $(n-1)$-dimensional volume) of $T\cap H$ is attained when $H$ contains a facet of $T$?
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$\begingroup$ 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
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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
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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 cross-section, Am. Math. Monthly, January 1968. The idea is to squeeze a regular 5-simplex along the common perpendicular of two opposite 2-dimensional faces. The volume of the mid-section between these faces does not change, but it turns out that all facets become smaller than this cross-section if the scaling coefficient is small enough.
For more information see also the PhD thesis of Dirksen.
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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