Re: [math-fun] quasi-icosahedral frustration
Wouter writes: << when building a 3D set of points using a simple rule, I accidentally found a subset that seems to approach icosahedral symmetry. Why 'approach' ? because all points in my set are rationals, and those can never be icosahedral since they lack the necessary Sqrt[5] irrationality. The procedure is simple enough: start from the tetrahedron (zeroth generation) {0, 0, 0}, {1, 1, 0}, {1, 0, 1}, {0, 1, 1} and on each triangular face, add the point to form the fourth vertex of a new tetrahedron: . . . . . . . . . with the n-th generation having denominators 3^n, and the number of points tripling each generation. The quasi-icosahedron becomes apparent when selecting all points at Sqrt[2] from the origin. . . . . . . I didn't succeed in understanding where the underlying connection to their pseudo-icosahedral grouping comes from. You got to plot them to believe it
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 _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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
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Dan Asimov -
wouter meeussen