Thanks for this straightforward solution! Apologies if these two follow-up questions are naive: 1) How do we know that there exists a projective transformation sending $\phi$ to infinity that retains the property that $F$ is projectively removable? 2) Your solution finds a ridge such that some d of its vertices are final, but the original question asks about a single vertex. I presume this is a trivial difference -- ie, if we choose a point $\phi$ lying in $H$ instead of a (d-2)-flat and place it next to a desired vertex $v$ prior to applying the rotation, is it true that $v$ is final?

Am I correct that the argument can be stated as follows, or did I misunderstand? For any vertex of a polytope, there exists a projective transformation mapping that vertex to infinity. Thus, there exists a projective transformation T mapping the endpoint a of e within Q to infinity. Since a is the unique vertex being mapped to infinity, T(e) has infinite length. Thus, any infinitesimal perturbation p of T will place p(T(a)) in a finite position such that the edge p(T(e)) will be larger than any other edge f when they are both projected onto any vector n (since p(T(e)) is arbitrarily large).

Thanks, Kristal! I understand the argument broadly, but would appreciate a slight clarification. Apologies if these questions are naive. Is G intended to be the n-dimensional space with coordinates {(x,0): x ∈ B} and a' the point (a,0), ie, does it matter whether the coordinates are identical? And what do you mean, formally, by projection through the point a - simply that a lies between Q and G?

Thanks! This is indeed an easy way to get a loose upper bound. While it certainly isn't tight, your use of the upper bound theorem suggests a method for determining a tighter bound in individual dimensions; given that I have a system of linear equations for computing the size of the removed vertex set given the f-vector of the facet, it should be possible to use the upper-bound theorem (along with Euler's formula) to derive a valid f-vector maximizing its size. Certainly, in individual dimensions this should be easy to find as the solution to an optimization problem.

Under the conjecture that the number of vertices formed must always be greater than number of vertices removed (seems vaguely plausible), I'd be inclined to believe that the given upper bound of k minus a constant is tight across dimensions. Could you give an example of a construction with (k-d) cut off in higher dimensions, and we can try and generalize it?