[math-fun] existence of the icosahedron
About 10 yeras ago I had a long conversation with Daniel Allcock about proving the existence of the icosahedron and/or dodecahedron "by continuity" -- in particular, without any calculation. Yesterday I was talking to Igor Pak and mentioned the proof of the existence of the icosahedron that we'd come up with, and Igor was hoping to find a reference to it in print, so as to give proper credit. (I'm sure the idea long predates my and Daniel's conversation.) Does anyone know the pedigree of the following line of reasoning? I haven't defined "by continuity," but I think the intent will be clear from the proof. 1) In 2d, all the regular n-gons certainly exist (by continuity). Take a circle and a chord thereof, and repeat the chord around the circle, end-to-end, n times. By adjusting the length of the chord, you can make the nth copy's endpoint hit the starting point, and you get a regular n-gon. [In particular, by this construction, all edges are identical and all vertices lie on a sphere.] 2) Make a pentagonal antiprism. That is, make a polyhedron whose top and bottom faces are pentagons lying in parallel planes such that the line joining their centers is perpendicular to those planes. If the pentagons are rotated relative to one another, we need ten triangles to close this polyhedron around the equator. Rotating one pentagon, we can make the triangles isosceles (by continuity); then by adjusting the distance between the two planes, we can make them equilateral (by continuity). 3) On each of the pentagonal faces of the antiprism, erect a five-sided pyramid whose apex is over the center of the pentagon. Adjusting the altitude, you can make those five triangles equilateral as well (by continuity). There you have it: the polyhedron so constructed clearly has twenty equilateral triangular faces meeting five at each vertex. (Mind you, this proof is still not entirely satisfying: it only endows the figure with dihedral, not icosahedral, symmetry. And indeed if you start with a 3-gon or 4-gon instead of a 5-gon, everything still works, but you don't magically get extra symmetry at the end.) Can anyone point to an origin for this? --Michael Kleber kleber@brandeis.edu
Oof. Forget I asked that... this is the proof that Coxeter gives on p.5 of _Regular Polytopes_. --Michael Kleber kleber@brandeis.edu On Tuesday, March 2, 2004, at 11:12 AM, Michael Kleber wrote:
Does anyone know the pedigree of the following line of reasoning? I haven't defined "by continuity," but I think the intent will be clear from the proof. [...] There you have it: the polyhedron so constructed clearly has twenty equilateral triangular faces meeting five at each vertex.
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Michael Kleber