Kind of makes you wonder how many you'd need. Now ignoring some cheats, like just putting 6 vertices on the equator and calling both hemisphere a hexagon, a sh

Kind of makes you wonder how many you'd need. Now ignoring some cheats, like just putting 6 vertices on the equator and calling both hemisphere a hexagon, a shape consisting solely of 'h' hexagons and 'p' pentagons will have (h+p) faces, (6h+5p)/2 edges and at most (6h+5p)/3 vertices (we're forbidding the construction where you just cut an edge into two and claim you've created a vertex, so each vertex is part of at least 3 faces). This gives an Euler characeristic of at most: (6h+5p)/3 - (6h+5p)/2 + (h+p) = (6h+5p)/6 - (h+p) = (1/6) p. Since a sphere has a characteristic of 2 you need at least twelve pentagons. And apparently you can achieve this lower bound by subdividing the triangular faces of an icosahedron, leaving you with 12 pentagons at the corners of the icosahedron.