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A beautiful polyhedron with 20 hexagons and 60 pentagons can be seen here: http://robertlovespi.wordpress.com/2013/11/03/a-polyhedron-with-80-faces/ . Euler formula and the corresponding Diophantine equation give a smaller possible combination: 7 hexagons and 20 pentagons adjacent by two to hexagons' vertices and by five in the pentagonal vertices. Does such a polyhedron really exist? I doubt but my only argument is "I was not able to compose it". At the same time I do know that the non-existence of polyhedra permitted by Euler equation is not elementary (cf. the not-existing polyhedron with 12 pentagons and 1 hexagon only).

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Here's a kludgy solution. Start with a "hat box" with pentagons for top and bottom and $5$ square sides. Inside each square side draw a pentagon, and connect four of the vertices of that pentagon to the four corners of the square. In two of those connecting segments from opposite vertices of the square, draw an extra vertex. You'll see that each square now consists of $4$ pentagons and $1$ hexagon. This gives you a total so far of $2+5\cdot4=22$ pentagons and $5\cdot1=5$ hexagons. To get the numbers you want, take any two pentagons that share an edge and draw an extra vertex on that edge.

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There are 13 Archimedean solids, none of them with your recipe.

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    $\begingroup$ The Archimedean solids are regular; the OP never imposed such a condition, and neither does the example he linked at. $\endgroup$ – Lucian Feb 24 '14 at 19:50
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I'm not sure if this leads anywhere, but I would ignore the hexagons and just deal with the region bounded by the planes that the pentagons lie on. This should result in a polyhedron with 60 triangles. Try to show that if these 60 triangles were edge-connected, that they would all lie on the same plane (and thus they do not form a polyhedron at all).

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