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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.

Nandakumar R
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    A pentagonal pyramid can have five congruent faces, and one different one. – Gerry Myerson Jul 14 '22 at 07:28
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    ${n-1,1}$, $n \ge 4$ is achievable by a pyramid (generalization of @GerryMyerson's example). ${n-2,2}$, $n \ge 5$ is achievable by a cylinder. – Joseph O'Rourke Jul 14 '22 at 12:09
  • Thanks for pointing out these further examples and the error in the question statement. Corrected. – Nandakumar R Jul 14 '22 at 12:42
  • ${2 k, k } = {\frac{2}{3} n, \frac{1}{3} n }$ ($k \ge 3$, $n \ge 9$ a multiple of $3$) achievable by a bipyramid with an equitorial band of, say, squares. E.g., the Johnson solid elongated square bipyramid – Joseph O'Rourke Jul 14 '22 at 13:16
  • And the bipyramid example generalizes to ${ 2k,k,k,\ldots,k }$, $k \ge 3$, just by using differently shaped rectangles on multiple equitorial bands. And instead of bi-pyramid, could be a pyramid, so it realizes ${ k,k,\ldots,k, 1 }$. – Joseph O'Rourke Jul 14 '22 at 13:28
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    Along the same lines: {k, k} can be achieved with a bipyramid with different heights on top and bottom. – Yoav Kallus Jul 14 '22 at 16:45
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    Zonohedra provide many other examples. Given this plethora, perhaps the inverse question should be considered: Which tuples ${ k_1,k_2,\ldots}$ cannot be realized? – Joseph O'Rourke Jul 15 '22 at 00:26
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    Yes. I would guess that even characterizing pairs {k1, k2} which cannot be realized would be interesting. – Nandakumar R Jul 15 '22 at 13:59
  • E.g., realize ${3,5}$. – Joseph O'Rourke Jul 15 '22 at 21:26
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    @JosephO'Rourke Take a triangular bipyramid and construct a pyramid on one of its faces. – Ilya Bogdanov Jul 16 '22 at 12:11
  • @IlyaBogdanov: Nice! – Joseph O'Rourke Jul 16 '22 at 22:39
  • Even if the required polyhedron is constrained to be convex, {4,5} can be easily realized by patching nearly flat square pyramid to a cube. Likewise, a convex {3,7} results when a flat triangular pyramid is attached to an octahedron. However, {3,5} is much harder and may even be impossible. – Nandakumar R Aug 06 '22 at 03:56
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    No, ${3,5}$ is handled by Ilya's answer. Start with two reg tetrahedra base-to-base. That's $6$ congruent faces. Replace one with a shallow tetrahedron. So now ${5,3}$. – Joseph O'Rourke Aug 07 '22 at 00:48
  • Thank you for correcting that the oversight! It might be of interest to explicitly show that some specific {a,b} cannot be realized as a convex polyhedron. – Nandakumar R Aug 08 '22 at 08:34

1 Answers1

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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...

Ilya Bogdanov
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  • Nice... one guesses that interesting classes of impossible pairs of integers {n1, n2} would emerge when we restrict to convex polyhedrons. – Nandakumar R Jul 17 '22 at 21:45