Solution to Problem of the Month, June 2006.

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Let p and h be the number of pentagons and hexagons and F, V, and E the number of faces, vertices and edges respectively.
Then Euler's polyhedron formula yields:

(1)	F + V = E + 2	We also have:
(2)	p+h = F	(counting faces)
(3)	5p+6h = 3V	(counting vertices)
(4)	5p+6h = 2E	(counting edges )

Putting (2),(3),(4) into (1) gives:

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.
But we must have p =12, which rules out the 'hexagon ball' in the problem. Of course any tiling of hexagons must be plane.

Condition (ii) states that each pentagon corresponds to 5 football vertices and each hexagon corresponds to 6/2=3 football vertices .

Hence

(5)	5p = V	and
(6)	3h = V

so V=60     and h = 20.
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.

Note that each vertex, where 5 edges meet, gives rise to a pentagon, and that each triangular face becomes a hexagon. It is obvious that careful truncation can make these polygons regular. The figure shows an icosahedron with one truncated vertex.

Remake from Wikipedia, platonic solids. This link is also a good startingpoint for further polyhedron studies.

Correct solutions have been supplied by Lars Filipsson and Dan Petersen as well as by Henrik Melbéus and Magnus Andersson ( in connection with an earlier problem seminar ).

BG GJ