On polyhedrons with specified numbers of congruent faces

Question

Basic question: Given 3 integers n, n1 and n2 such that n1+n2 = n, to form an n-face polyhedron such that n1 of its faces are mutually congruent and the remaining n2 faces are different but congruent among themselves.

Simple examples: It is easy to form tetrahedrons with (1) 3 faces mutually congruent and the remaining 1 face different or (2) with faces grouped 2+2 where each members of each pair are congruent but different from the faces in the other pair.

Hexahedrons with 4 faces mutually congruent and the other 2 different and mutually congruent are also easy to make.

A 'buckyball' has n=32 and n1, n2 = 20, 12. Some further examples are at https://en.wikipedia.org/wiki/Semiregular_polyhedron. Please note that the present question does not insist each face is a regular polygon.

General question: Given an integer n and a set of integers, m1, m2,... which add to n, to decide whether we can form an n-faced polyhedron with m1 faces congruent among themselves, another m2 faces congruent among themselves and so on.

An earlier discussion which could be of interest: Convex polyhedra with non-congruent faces

Remark added on July 20th 2022: Ilya Bogdanov has given nice constructions below to the case where the polyhedron being constructed is allowed to be non-convex. One feels however that restricting it to convex would lead to many non-realizable pairs of {n1,n2} and characterizing them would be of interest.

Answer 1

I assume that the polyhedra are not required to be convex.

Construction 1 ($n_1\equiv n_2\mod 2$).

Gluing together some regular tetrahedra, one can reach an arbitrary even number $2k$ of equal faces.

Attach a regular pyramid $P$ to one of its faces. Now, one can attach a bipyramid consisting of two pyramids congruent to $P$ to a lateral face of $P$ and proceed further in a similar way. In this fashion, we get all pairs of the form

$\bullet$ $(2k-1,4\ell-1)$ with $k\geq 2, \ell\geq 1$.

Attaching bipyramids to two/three/four faces, we get the pairs

$\bullet$ $(2k-2,4\ell-2)$ for $k,\ell\geq 2$;

$\bullet$ $(2k-3, 4\ell+1)$ for $k,\ell\geq 2$;

$\bullet$ $(2k-4,4\ell+4)$ for $k,\ell\geq 2$.

This covers all pairs of numbers of the same parity except for $(1,5)$ (a pyramid), $(4,8)$ (two pyramids on a cube), $(4,4), $(8,8)$ and $(5,5)$ (a bipyramid),

Construction 2 ($n_2$ even, $n_1$ odd). The cases $n_2=2$ or $n_1=1$ are covered in the comments.

It is not hard to construct a parallelepiped with 6 congruent rhombic faces which are not squares. Gluing together copies of such, we reach $4k+2$ congruent faces for $k\geq 1$. Now, one may attach a pyramid to its face, and start attaching isohedral tetrahedra to its lateral face, this way obtaining

$\bullet$ $(4k+1, 2\ell)$ for $k\geq 1, \ell\geq 2$.

Starting with 3 pyramids attached, we get

$\bullet$ $(4k-1, 2\ell)$ for $k\geq 1$, $\ell\geq6$.

Moreover, while doing that, we could take the lateral face of a pyramid such that two of the three pyramids share a lateral face (this should be done at the edge where two faces of parallelepipeds form a nonconvex dihedral angle). The same can be done with all three pyramids, achieving values $\ell=4$ and $\ell=5$.

Leftovers. So the cases left are $(4k-1,4)$ and $(4k-1,6)$. For $(3,4)$ one can take a triangular prism with a tetrahedron on top...