Solution to Problem of the Month, June 2006.We consider a (convex) polyhedron satisfying the conditions:(i) Faces are either pentagons or hexagons (ii) Two hexagons and one pentagon meet at each vertex. Let p and h be the number of pentagons and hexagons
and F, V, and E the number of faces, vertices and edges respectively.
p + h + (5p + 6h)/3 = (5p + 6h)/2 +2 p(1 + 5/3 -5/2) + h(1 + 2 - 3) = 2 p/6 = 2, p = 12. Note that we have not yet used any information from condition (ii) above. Equations (2) and (4) are completely general conditions and the number 3 in (3) is necessary, since you can only glue 3 pentagons and/or hexagons together to get a polyhedron vertex. Note also that since h cancels out, these equations do not give any information on h.
Condition (ii) states that each pentagon corresponds to 5 football vertices and each hexagon corresponds to 6/2=3 football vertices . Hence
Actually only one of equations (5) and (6) is needed to get this result together with (1) - (4). Remarks. In order to get a proof (not demanded by the stated problem) that the given solution can be realized as a regular (Archimedean) polyhedron, we observe that the football can be constructed by truncation of the regular icosahedron.
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