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For which n can we divide the surface of a sphere into n equal parts, such that each part has k neighboring parts, and the total grid has k-fold rotational symmetry around an axis from any of the parts center to the center of the sphere? And no part can be distinguished from another other then fixing a coordinate system.

Is it the same as for the plane ?

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    If you relax the constraints a bit and only require a transitive symmetry group, then you get more than the Platonic solids. For example the 3D [finite reflections groups](http://en.wikipedia.org/wiki/Weyl_group) give rise to subdivisions of the sphere into 48 or 120 congruent parts (with the possibility of pairing up adjacent parts to halve their number).2011-10-30

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Your choices are either $n\ge 2$ identical wedges each stretching from north pole to south pole (with $k=2$), or one of the five Platonic solids projected radially on their circumscribed sphere:

  • $k=3, n=4$ (tetrahedron)
  • $k=3, n=8$ (octahedron)
  • $k=3, n=20$ (icosahedron)
  • $k=4, n=6$ (cube)
  • $k=5, n=12$ (dodecahedron)

There is no solution where the parts have smaller angular diameter than for the icosahedron.

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    On the other hand, if you require merely that the tiles are pairwise congruent, you can get 120 faces by subdividing each face of a icosahedron or dodecahedron. Better than that will be difficult unless you relax the standards considerably.2011-10-30