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Question: Are there convex polyhedra wherein all faces are convex polygons with same area and perimeter and no two faces are mutually congruent?

Remarks: If the answer to above is "no", then, one can ask if there are other weaker conditions which can say, ensure that at least a pair of faces are congruent and also about cases with quantities other than area and perimeter shared among the faces. And if the answer is "yes", one can ask if equalizing more but finitely many quantities among faces will guarantee at least one pair of congruent faces or all faces congruent.

And one could ask about polyhedrons with convex faces with every pair of faces non-congruent and faces constrained only to have same perimeter OR same area(*). Another direction in which the question can be 'relaxed': let the polyhedron be non-convex with its faces remaining convex and non-congruent.

(*) Guess: 'perturbed' pyramids with quadrilateral base (all four sides and all angles of base slightly different from each other) and 4 triangular faces can be built with faces pairwise non-congruent and all 5 faces having equal area or equal perimeter - but not both equal in any obvious way.

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Most likely. Start with something like this polytope and denote by $u_1,\ldots,u_{12}$ the unit normals to faces. As given this polytope has all faces equilateral triangles with the same area $a$ and the same perimeter. Perturb all $u_i$ but keep all areas equal to $a$. This is always possible by the Minkowski theorem. The equal perimeter condition gives $12$ equations. The space of perturbations of normals is $24$-dimensional, giving you $12$-dimensional freedom of choice (probably). To ensure that all faces are non-congruent you will need to kill the relatively small symmetry group of order $8$, which should be possible. An actual computation is needed in this case to give a convincing answer.

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  • $\begingroup$ Thank you. In that case, can one infer something like: "if a small and finite number of quantities (say, area, perimeter, diameter, longest side length,...) are equal among all faces of a convex polyhedron, at least two of the faces will be congruent and if a few more additional quantities are kept equal, it will automatically get all faces congruent"? Indeed, there appear to be strong constraints on the number of edges on faces because the average degree of planar graphs is <6. $\endgroup$
    – Nandakumar R
    Commented Jun 16, 2021 at 15:09

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