Maximum Change in Number of Vertices When One Constraint is Added to Convex Polytope For $A\in \mathbb{R}^{m \times n}$, if $P=\{x \in \mathbb{R}^n:Ax \leqslant b\}$ has a certain number of vertices, what is the largest possible change in the number of vertices that occurs when one constraint $a_{m+1}^Tx \leqslant b_{m+1}$ is added?
 A: Long comment/partial answer
For $n=1$, each new restriction can add a maximum of one new vertex.
For $n=2$, each new restriction can add a maximum of $2$ new vertices. This is because the new vertices lie on a line and the convex polygon formed by the original restrictions. A line can cut a convex polygon at a maximum of $2$ points.
For $n\geq 3$, the thing changes. 
For $n=3$, each new vertex is on a plane and form a convex polygon. We can associate a vertex to one of the previous restrictions (because the number of vertices and the number of sides of a convex polygon are the same). Therefore we can get a maximum of $m$ new vertices. This maximum can be attained when the original restrictions form a cone with $m$ sides and the new restriction truncates it.
For $n=4$, the new vertices lie on a $3$-space and form a convex polyhedron $G$. Each face of this polyhedron corresponds to one of the $m$ original restrictions. To maximize we can assume that each one of the $m$ produces a face, i.e. intersects the new restriction. So, we need to find the maximum number of vertices that a polyhedron of $m$ faces can have. We know that $V\leq 2F-4=2m-4$, from Euler's formula. Can we attain this bound?
I suspect we might need to use Dehn–Sommerville equations for higher dimensions. But if this is as hard as (dual to?) the Upper bound theorem this is too hard for me.
