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From: Francois Grieu on 21 Jun 2010 18:14 I'm after the 3D polyhedron with N>3 triangular faces, containing the sphere of radius 1, that has the minimum volume. A face with M sides counts as M-2 triangles. The problem has sort of a practical application when trying to make a ball by sanding a polyhedron obtained using 3D printing, where typically shapes are defined as polyhedron with triangular faces, and price is per volume. <http://www.shapeways.com> For N=4, N=8, and N=20, I assert (without proof) that solutions are the regular tetrahedron, octahedron, and icosahedron with the center of the N equilateral triangles tangent to the unit sphere; vertices of the polyhedrons are on a sphere concentric to the inscribed unit sphere. For N=20*(k^2), I conjecture some slight deformation of the polyhedron obtained by dividing each triangle of the icosahedron into k*k equilateral triangles; even for N=80 where symmetry leaves only one unknown, I see no reason why the vertices would be exactly on a sphere. All in all, the problem looks more complex than the classic "points on a sphere" problems and I see no exact reduction from one to the other. <http://www.ogre.nu/sphere.htm> TIA for your pointers and ideas. Francois Grieu |