TITLE: Can I build infinitely many polytopes from only finitely many prescribed facets?
QUESTION [16 upvotes]: Given a finite set of convex $d$-dimensional polytopes $\mathcal P$, for some $d\ge 2$.

Question: Is it true that there are only finitely many different convex $(d+1)$-dimensional polytopes whose facets are solely (scaled and rotated versions of) polytopes in $\mathcal P$?


Some clarifications
A face of a polytope is the intersection of the polytope with a touching hyperplane, so subdividing facets does not count here.
In general, two $(d+1)$-polytopes shall be considered as different if they differ not just in scale and orientation.
By the usual rigidity arguments, given the shape of facets and their connections, the metric of the polytope is uniquely determined. Hence, if we can built different polytopes, they will be combinatorially different as well.
Example
There are only finitely many polyhedra that can be built from any finite set of regular polygons, but as far as I know, this result is by enumeration  (see, e.g. Johnson solids).

REPLY [6 votes]: Not an answer, but an "almost counterexample" for $d=2$ that came to my mind when I saw that question. Starting with a regular dodecahedron, we can successively insert circular "belts" of hexagons, in a way that the whole polyhedron is convex at each stage. The illustrations should give an idea. The shapes of the hexagonal facets can come very close to each other (in the appropriate sense), though it is clear that there can only be so many of a precise shape.