hi Dan et al. your argument about initial tetrahedral symmetry holds if we measure distances from the center of the original tetrahedron. But I don't: I look at all points exactly Sqrt[2] from *one* vertex at {0,0,0}. So I'm measuring from a point at {-1/2,-1/2,-1/2} from the center of the original tetrahedron. I get 12 patches of points, each with radius less then 1. Building 8 generations for a total of 768=12*64 points, then the exact centres of the patches are at: {{57703/59049, 57703/59049, -5812/59049}, {-6464/6561, -6464/6561, 9478/190269}, {-6464/6561, 9478/190269, -6464/6561}, {57703/59049, -5812/59049, 57703/59049}, {-5812/59049, 57703/59049, 57703/59049}, {9478/190269, -6464/6561, -6464/6561}, {-268921/209952, 158621/419904, 158621/419904}, {64016/51759, -207967/465831, -207967/465831}, {-207967/465831, -207967/465831, 64016/51759}, {158621/419904, 158621/419904, -268921/209952}, {-207967/465831, 64016/51759, -207967/465831}, {158621/419904, -268921/209952, 158621/419904}} These centers (patch-averages) do not sit at Sqrt[2] from the origin, (though all the points in a patch do), but they form a pretty cute approximation to an icosahedron. The count of points in the patches is not exactly 64 each time: {63, 58, 58, 63, 63, 58, 64, 71, 71, 64, 71, 64} Wouter. ----- Original Message ----- From: "Dan Asimov" <dasimov@earthlink.net> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Sunday, March 23, 2008 7:06 PM Subject: Re: [math-fun] quasi-icosahedral frustration
Wouter writes:
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Interesting, but I don't see how this procedure, whose input is solely the vertices of a regular tetrahedron, can have anything but the original tetrahedral symmetry at each stage and therefore in the limit as well.
(E.g., given a regular tetrahedron, one can find a regular icosahedron inscribed in it -- each of 4 faces of the icosahedron is a portion of one face of the tetrahedron -- but the icosahedron is not determined
uniquely.)
--Dan