Given a *finite* set of *convex* $d$-dimensional polytopes $\mathcal P$, for some $d\ge 2$.



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> **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 $\mathcal P$?
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**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](https://en.wikipedia.org/wiki/Cauchy%27s_theorem_(geometry)), 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](https://en.wikipedia.org/wiki/Johnson_solid)).



Can I build infinitely many polytopes from only finitely many prescribed facets?
I think this is indeed true. This should follow from the rigidity of convex spherical polytopes and a few more lemmas. I'll explain how to deal with cases $d\ge 3$.


*Lemma 1, rigidity*. Two convex spherical cobminatorially equivalent polytopes with isomertric faces are isometric.


*Lemma 2, bounded volume.* The boundary of any convex subset of round $\mathbb S^n$ has volume bounded from above by the volume of round $\mathbb S^{n-1}$.


*Lemma 3, gap.* Suppose we have a spherical convex polytope in $S^n$ of diameter less than $\pi-\varepsilon$ then the volume of the dual polytope is bounded from below by $c(n,\varepsilon)>0$.


*Lemma 4, sum of dual solid angles.* Let $P$ be a convex $d+1$-polytope. For each of its vertices consider the dual cone. Then the sum of solid angles of such dual cones is equal to the volume of the round $\mathbb S^d$.


*Proof of finitness.* So, suppose we have a finite number of convex $d$-polytopes $P\_1,\ldots P\_k$. Denote by $S\_1,\ldots, S\_m$ the spherical $d-1$ polytopes corresponding to vertices of $P\_1,\ldots, P\_k$. 


Suppose we have a $d+1$-polytope $P$ with faces homothetic to $P\_1,\ldots P\_k$. Then for each vertex of $P$ we get a convex spherical $d$-polytope whose faces are composed from $S\_1,\ldots, S\_m$. Note the the number of such spherical polytopes is finite up to isometry, by Lemmas 1 and 2. Moreover, all of them have diameter less than $\pi-\varepsilon$ for some $\varepsilon$ . So by Lemma 3 the volumes of spherical polytopes dual to those (coming from vertices) are at least $c(d,\varepsilon)$. Hence by Lemma 4 the total number of vertices of $P$ is at most $vol(S^d)/c(d,\varepsilon)$. Hence we have only finite number of combinatorial types of $P$.


