Cyclic $d$-polytope -- does every facet border every other facet?

Question

A cyclic $d$-polytope is "neighborly," where every set of vertices sized $d/2$ together make up a face. What I'm interested in: in a cyclic $d$-polytope with $N$ facet ($d-1$ dimensional faces), does every facet "border" every other facet (i.e., share a $d-2$ dimensional face)?

Why I think it might: we can use the upper bound theorem in dual form (https://www.cs.mcgill.ca/~fukuda/soft/polyfaq/node12.html) to find the total number of $d-2$ dimensional faces for a $d$-polytope with $N$ facets. If you use the formula in the provided link, we find that there are $N(N-1)/2$ such $d-2$ dimensional faces. Since every such face borders $2$ facets, this total number of $d-2$ dimensional faces suggests that the property holds.

Answer 1

The cyclic polytope C(4,6) has nine tetrahedron facets. Each of these facets can only border four other facets, so not every facet borders all other facets.

Answer 2

This unfortunately fails in even the simplest case: when $d = 2$.

Consider the cyclic polytope $C_{2}(T) \subset \mathbb{R}^{2}$ where $T = \{ t_{1}, t_{2}, t_{3}, t_{4} \}_{<} \subset \mathbb{Z}$. It has facets $F_{i, i + 1} = \mathrm{conv}\{(t_{i}, t_{i}^{2}), (t_{i + 1}, t_{i + 1}^{2}) \}$, and it's not hard to see that facets $F_{i, j}, F_{k, \ell}$ border each other if and only if $\{i, j \} \cap \{ k, \ell \}$ is a singleton set.

Consider the facets $F_{1, 4}, F_{2, 3}$. They do not meet at a vertex of $C_{2}(T)$ since their indexing sets are disjoint.