# Classifying two-faces of four-polytopes

Motivation: This question is related to my study of hyperbolic Coxeter polytopes. In general, if one put some restrictions on the type of their dihedral angles (say, all dihedral angles are equal to $\pi/2$), an the type of the polytope (all its vertices are proper, or all are ideal, or of mixed type), then one obtains some restrictions on the combinatorics. At the moment, I try to see if there are compact right-angled four-dimensional polytopes with volume bounded above by some (reasonably small) constant.

I have come to the following (probably, quite technical) question (or, better say, the following type of questions).

Question: Is there a convex simple polytope (in $\mathbb{R}^4$) which has $n=2,3,...,12$ hexagonal two-faces, while all other two-faces are pentagons.

Or, more generally, let $n_k \geq 0$ be the number of $k$-gonal faces ($k\geq 6$). Does there exist a convex simple polytope (in $\mathbb{R}^4$) which has given amounts $n_6$, $n_7$, ..., of, respectively, $6$-, $7$-, ...-gonal two-faces, while all other two-faces are pentagons?

I'm curious to know if there are general methods to study these (or similar) problems. Also, computational methods can be of interest (e.g. one can impose an upper bound on the number of three-facets of such a polytope and try to get a computer seek possible polytopes).

Thank you in advance for any help/references/etc!

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