Number of extreme points of a convex polyhedron In $\mathbb{R}_+^n$ $(n\geq 1)$, let $S$ be the $(n-1)$-simplex:
$$S=\left\{x\in\mathbb{R}_+^n,\,\sum_{i=1}^n x_i=1\right\}$$
whose relative interior is denoted $\mathring S$, and $E$ be a linear subspace of dimension $n-2$ in $\mathbb{R}^n$, such that $E\cap\mathring S\neq\varnothing$. Hence $E\cap\mathring S$ is a convex polyhedron with relative interior interior of dimension $n-3$ and has hence a minimum of $n-2$ extremal points.
Question: what is the maximal number of extreme points for $E\cap S$?
 A: This is not a full solution but more of an educated guess. Hopefully it can be turned into one with enough work, or at the very least give some useful ideas.
Reduction to counting triangles:
Start by projecting the simplex into $E^\perp$, a 2-dimensional space. Now all of $E$ has collapsed into a point $O$ (which will be our "origin" on $E^\perp$).
If we assume that $E$ was in a somewhat generic position (not parallel to a facet, hopefully you can wiggle extremal solutions around). Then the number of extremal points of $E\cap \mathring S$ is the number of intersections between $E$ and $2-$dimensional facets of $S$. At the projection level, this becomes a combinatorial question:

You are given $n$ non-zero distinct points in $\mathbb{R}^2$, which generate $n\choose 3$ triangles. How many of them can contain the origin inside?

By scaling, you can even assume the $n$ points are on the unit circle! At this point it almost feels like a discrete version of a famous Putnam question: https://www.youtube.com/watch?v=OkmNXy7er84. If we believe the analogy, we expect to get an asymptotic of $\frac 1 4 \frac {n^3} {6}  $.
(not) Solving the counting problem I don't know the answer to this problem, but we can guess it, by hoping that the extremizer comes from a regular polygon.
Start by assuming $n=2k+1$ is odd. Fix one vertex. We will have to multiply by $n$ (choice of vertex) and divide by 3 (triangle has 3 vertices). The other 2 vertices will be one on the "left" and one on the "right". Let $j$ be the number of "hops" to the right you have to do to get to the right vertex (so $j=1 \dots k$). Then you have $k+j-1$ options for the left vertex. Then
$$
\text{Tri}(2k+1) = \frac{2k+1}{3}\sum_{j=1}^k k-j+1 = \frac{2k+1}{3}\left(k^2 -\frac{k^2+k}{2}-k\right) = \frac{k(k+1)(2k+1)}{6}
$$
A similar computation, where you count points that are in the boundary with a factor of $1/2$ (because, what else?) gives
$$
\text{Tri}(2k) = \frac 13 k(k^2-1)
$$
Is our guess any good? Through more rudimentary methods and still assuming somewhat generic positions (which amount to checking all possible configurations) you can do the computation for small $n$, and I got:
$$
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
n & 3 &4 & 5 & 6 & 7 &8 &9&10&11\\ \hline
 \text{Extrema}(n)& 1 &2 & 5&8&14&20 &?&?&? \\ \hline
\text{Tri}(n) & 1 & 2& 5 &8& 14 &20& 30 &40& 55\\ \hline
\end{array}
$$
