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This question is closely related to Convex polyhedra with non-congruent faces

It is known that if all faces of a tetrahedron ought to have same area (or same perimeter), then, the faces are necessarily mutually congruent (https://en.wikipedia.org/wiki/Disphenoid).

Question: What is the polyhedron with least number of faces such that that all faces have same area but are pair-wise non-congruent? The faces need only be non-congruent; they can have different numbers of sides etc.

Is it a pentahedron? And do such polyhedrons exist for all values of the number of faces above a minimal number? Does allowing the polyhedron to be non-convex have any implications?

Further questions: One can replace "same area" above with "same perimeter", "same area and same perimeter" (asked in post linked at the top; the answer there shows such polyhedrons almost certainly exist but does not find the one with least number of faces) etc.

Note 1: It is easy to see that even a tetrahedron can be made with all faces having the same diameter but are mutually non-congruent (take an equilateral triangle and add the 4th vertex a little 'above' a random interior point; with two such tetrahedrons, we form a hexahedron with the desired property). But not sure if for all n>4, polyhedrons with mutually non-congruent, equal diameter faces can be made.

Note 2: Another earlier question from the same ballpark: On polyhedrons with specified numbers of congruent faces. Recently, results have been obtained on tiling the plane with mutually non-congruent polygons, for example: https://www.sciencedirect.com/science/article/abs/pii/S0097316521000601

Nandakumar R
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    Minkowski's theorem (https://en.wikipedia.org/wiki/Minkowski_problem_for_polytopes) states that given a set of vectors in $\mathbb{R}^3$ that span $\mathbb{R}^3$ and have sum 0, then there is a unique (up to congruence) convex polyhedron $P$ with these vectors as its face normals, such that the area of the face normal to $v$ is the length of $v$. Thus if the vectors all have the same length, then all faces of $P$ have the same area. It seems plausible that if we take the vectors to be sufficiently generic, then none of the faces will be congruent. – Richard Stanley Mar 09 '24 at 15:31
  • Thanks. For number of faces = 4, the sum of the 4 vectors being constrained to be 0 seems to force the congruence of the faces. But your observation seems to imply that for higher values of number of faces, there might always be ways to keep the sum of the vectors 0 and the faces pairwise non-congruent. – Nandakumar R Mar 09 '24 at 18:47
  • And the argument in terms of vectors of equal magnitude appears to restrict the resulting convex polyhedrons to those inscribable in a sphere. It is not clear to me whether also considering convex polyhedrons that are not so inscribable might increase the chance of finding polyhedrons with the properties required by the present question. – Nandakumar R Mar 10 '24 at 17:14
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    The polyhedra don't have to be inscribable in a sphere. When I say that a vector $v\in\mathbb{R}^3$ is a face normal, I mean that the direction of the vector is normal (and points outward) to the face. I don't mean, for instance, that the face contains $v$. – Richard Stanley Mar 11 '24 at 02:48
  • Thanks for clarifying. Hope to see more on polyhedrons with non-congruent faces having quantities other than area (perimeter, diameter,...) equal. – Nandakumar R Mar 13 '24 at 04:26

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