Re: [math-fun] polyhedra with all faces congruent
thanks dan! that settles the question for convex polyhedra. the case of non-convex polyhedra and positive genus is still in doubt, but perhaps grunbaum's method sheds some light on it. mike
Here's the Math Reviews review of a paper by Branko Grunbaum that answers the "Odd?" question in the negative.
<< Branko Grunbaum On polyhedra in E^3 having all faces congruent. Bull. Res. Council Israel Sect. F 8F 1960 215–218 (1960)
The polyhedra having all faces congruent include the reciprocals (duals) of all the “uniform” polyhedra, namely, the five Platonic solids, the dipyramids, the reciprocals of the thirteen Archimedian solids, and the reciprocals of the antiprisms. These last have, for any n, 2n kite-shaped faces. (It is unfortunate that the author uses, instead of “kite”, the name “deltoid”, which belongs more properly to a curve, the three-cusped hypocycloid.) The author proves that, if all the faces of a convex polyhedron are congruent, their number is even. Since the faces cannot all have more than five sides, and the result is obvious when the faces are triangular or pentagonal, the problem is quickly reduced to the case of quadrangular faces, and then to kites or parallelograms. The kites are dealt with by an ingenious combination of metrical and topological arguments. Finally, if the faces are parallelograms, their number is equal to the product of two consecutive integers [Coxeter, Regular polytopes, Methuen, London, 1948, p. 27; MR0027148 (10,261e)].
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Michael Reid