Is a facet always a maximal area section of a simplex?

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 $$T$$?

• 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 $(1,0,0.01),(-1,0,0.01),(0,1,-0.01),(0,-1,-0.01)$) Jan 31 at 17:40
• @Saul In other words, the width of a simplex is not its shortest height. Yes, it is not, even for a regular simplex. Jan 31 at 18:41

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.