Edge-attaching many hexagons results in a plane. Edge-attaching pentagons returns a dodecahedron.

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Is over there some understanding into why the alternation the pentagons and also hexagons returns an approximated sphere? Is this special, or space there one arbitrary variety of assorted n-gons to adjust that might be joined together to create consistent sphere-like surfaces?

The possible ways to put polygons together to type a sphere-like object are constrained by Euler"s formula \$V - E + F = 2\$ (where \$V\$ is the number of vertices, \$E\$ is the variety of edges, and \$F\$ is the variety of faces). Equivalently you can think of this as a statement around planar graphs.

Suppose we use \$f_3\$ triangles, \$f_4\$ squares, \$f_5\$ pentagons, etc. Every sheet meets exactly two faces, and also an edge of form \$f_n\$ meets \$n\$ faces, so let"s double-count the number of pairs of an edge and a challenge next to it: ~ above the one hand, this is \$2E\$, and on the various other hand, this is

\$\$3f_3 + 4f_4 + 5f_5 + ...\$\$

Plugging this into Euler"s formula provides \$V - fracf_3 + 2f_4 + 3f_5 + ...2 = 2\$. If in enhancement the polyhedron is convex and also the polygons are regular, there are constraints on the faces that can accomplish at every vertex comes from the fact that the angles must sum up to much less than \$360^circ\$. (This is one method to prove the group of Platonic solids.) for example, at most \$5\$ faces can fulfill at each vertex if we permit arbitrary faces; this way \$3f_3 + 4f_4 + ... le 5V\$. (If girlfriend really want to, girlfriend can enable six triangle to touch at one point, however I would simply count this as a hexagon.) If we don"t enable triangles, exactly \$3\$ deals with meet at each vertex; this method \$4f_4 + 5f_5 + ... = 3V\$.

Here is an application in chemistry: a fullerene is a certain kind of molecule do from carbon atoms. (One of these, the buckyball, looks similar to a football ball.) It gives a convex polyhedron in which each challenge is one of two people a continuous pentagon or hexagon. This provides \$V - frac3f_5 + 4f_62 = 2\$ ~ above the one hand, and also \$3V = 5f_5 + 6f_6\$ top top the other. Together these equations provide \$f_5 = 12\$ and \$V - 2f_6 = 20\$; in various other words, any type of fullerene must have actually exactly twelve pentagons (Twelve Pentagon theorem because that fullerene).

(Hexagons room special. One method to translate this an outcome is that an infinite plane can it is in tiled through hexagons, for this reason hexagons correspond to zero curvature, whereas due to the fact that pentagons have actually a smaller angle at each vertex they correspond to positive curvature. What the over statement says, roughly, is that the full amount that curvature is a constant. This is a simple form of the Gauss-Bonnet theorem, i m sorry is carefully related come Euler"s formula.)

Here are some other things you deserve to prove, again under the assumptions of convexity and also regularity:

If you only use triangle (as protest to triangles and hexagons), then \$f_3 le 20\$ and \$4 | f_3\$.If you just use squares, then \$f_4 = 6\$.

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This is currently most the the way to the category of Platonic solids. If you"re interested in learning much more about Euler"s formula, I very recommend David Richeson"s Euler"s Gem. Extremely well-written and informative. You might also enjoy David Eppstein"s Nineteen means to prove Euler"s formula.