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Extending the idea of W. Jagy, this is a Mathematica code visualizing that 33 spheres with radius 1/2 centered at the origin and the midpoints of the faces of a soccerball with circumradius 3/4 cover the unitsphere.

coord = PolyhedronData["TruncatedIcosahedron", "VertexCoordinates"]; faces = PolyhedronData["TruncatedIcosahedron", "FaceIndices"]; f6 = Select[faces, Length[#] == 6 &]; f5 = Select[faces, Length[#] == 5 &]; len = Norm[coord[[1]]Norm[coord[[1]]] // Simplify; Graphics3D[{{Opacity[0.3], Sphere[{0, 0, 0}, len*4/3], Sphere[ Mean[coord[[#]]], len*2/3] & /@ f6, Sphere[ Mean[coord[[#]]], len*2/3] & /@ f5, Opacity[1]Opacity[1], Sphere[{0, 0, 0}, len*2/3]}, {Opacity[0.4], PolyhedronData["TruncatedIcosahedron" , "Faces"]}}, Boxed -> False]
[Graphic from the above code added by J.O'Rourke:]

Plotting only three outer spheres as in

Graphics3D[{{Opacity[0.3], Sphere[{0, 0, 0}, len*4/3], Sphere[ Mean[coord[[#]]], len*2/3] & /@ Take[f6, 2], Sphere[ Mean[coord[[#]]], len*2/3] & /@ Take[f5, 1], Opacity[1]Opacity[1], Sphere[{0, 0, 0}, len*2/3]}, {Opacity[0.4], PolyhedronData["TruncatedIcosahedron" , "Faces"]}}, Boxed -> False]
[Graphic from the above code added by J.O'Rourke:]

shows that the two intersection points of three neighboring outer spheres lie either outside the sphere with radius 1 or inside the sphere of radius 1/2. This can easily be made rigorous by calculation:

mcl = Chop[ Join[Mean[coord[[#]]] & /@ Take[f6, 2], Mean[coord[[#]]] & /@ Take[f5, 1]]]; erg = Solve[(Norm[{x, y, z} - #] == len*2/3) & /@ mcl, {x, y, z}] // N; ((Norm[{x, y, z}]/len*3/4) /. #) & /@ erg

that yields the distances

{0.216794, 1.06042}

1

Extending the idea of W. Jagy, this is a Mathematica code visualizing that 33 spheres with radius 1/2 centered at the origin and the midpoints of the faces of a soccerball with circumradius 3/4 cover the unitsphere.

coord = PolyhedronData["TruncatedIcosahedron", "VertexCoordinates"]; faces = PolyhedronData["TruncatedIcosahedron", "FaceIndices"]; f6 = Select[faces, Length[#] == 6 &]; f5 = Select[faces, Length[#] == 5 &]; len = Norm[coord[[1]]] // Simplify; Graphics3D[{{Opacity[0.3], Sphere[{0, 0, 0}, len*4/3], Sphere[ Mean[coord[[#]]], len*2/3] & /@ f6, Sphere[ Mean[coord[[#]]], len*2/3] & /@ f5, Opacity[1], Sphere[{0, 0, 0}, len*2/3]}, {Opacity[0.4], PolyhedronData["TruncatedIcosahedron" , "Faces"]}}, Boxed -> False]

Plotting only three outer spheres as in

Graphics3D[{{Opacity[0.3], Sphere[{0, 0, 0}, len*4/3], Sphere[ Mean[coord[[#]]], len*2/3] & /@ Take[f6, 2], Sphere[ Mean[coord[[#]]], len*2/3] & /@ Take[f5, 1], Opacity[1], Sphere[{0, 0, 0}, len*2/3]}, {Opacity[0.4], PolyhedronData["TruncatedIcosahedron" , "Faces"]}}, Boxed -> False]

shows that the two intersection points of three neighboring outer spheres
lie either outside the sphere with radius 1 or inside the sphere of radius 1/2. This can easily be made rigorous by calculation:

mcl = Chop[ Join[Mean[coord[[#]]] & /@ Take[f6, 2], Mean[coord[[#]]] & /@ Take[f5, 1]]]; erg = Solve[(Norm[{x, y, z} - #] == len*2/3) & /@ mcl, {x, y, z}] // N; ((Norm[{x, y, z}]/len*3/4) /. #) & /@ erg

that yields the distances

{0.216794, 1.06042}