TITLE: Why are there 12 pentagons and 20 hexagons on a soccer ball?
QUESTION [32 upvotes]: Edge-attaching many hexagons results in a plane. Edge-attaching pentagons yields a dodecahedron.
Is there some insight into why the alternation of pentagons and hexagons yields an approximated sphere? Is this special, or are there an arbitrary number of assorted n-gons sets that may be joined together to create regular sphere-like surfaces?

REPLY [2 votes]: An icosahedron is a platonic solid having 20 congruent equilateral triangular faces ($F=20$), 30 edges ($E=30$) each five meets at each vertex & 12 identical vertices ($V=12$) lying on the spherical surface (with certain radius). 
When an icosahedron is (partially) truncated at all its 12 vertices then each of 20 triangular faces become a hexagon & each of 12 vertices produces a new petagonal face. Thus a truncated icosahedron has 12 regular pentagons & 20 regular hexagons. 
While the process of truncation produces $12\times 5=60$ new vertices & $12\times 5=60$ new edges in addition to its 30 original edges. Thus it has total $60+30=90$ edges. For a truncated icosahedron, $F=12+20=32$, $E=90$ & $V=60$ which duly satisfies Euler's formula ($F+V=E+2$). It is also called Archimedean solid.    
A soccer ball is analogous to a truncated icosahedron. Hence it has 12 (regular) petagons & 20 (regular) hexagons.