Edge-attaching numerous hexagons outcomes in a plane. Edge-attaching pentagons returns a dodecahedron.

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Is there some insight into why the alternation of 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 with each other to create continual sphere-like surfaces?



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

Suppose we usage $f_3$ triangles, $f_4$ squares, $f_5$ pentagons, etc. Every edge meets exactly two faces, and also an edge of form $f_n$ meets $n$ faces, therefore let\"s double-count the variety of pairs of one edge and also a face 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 right into Euler\"s formula provides $V - \\fracf_3 + 2f_4 + 3f_5 + ...2 = 2$. If in addition the polyhedron is convex and also the polygons are regular, there are constraints ~ above the deals with that can accomplish at each vertex comes from the reality that the angles should sum increase to less than $360^\\circ$. (This is one method to prove the group of Platonic solids.) because that example, at many $5$ faces can meet at each vertex if we enable arbitrary faces; this way $3f_3 + 4f_4 + ... \\le 5V$. (If friend really desire to, you can permit six triangles to touch in ~ one point, however I would simply count this as a hexagon.) If us don\"t allow triangles, exactly $3$ encounters meet at each vertex; this means $4f_4 + 5f_5 + ... = 3V$.

Here is an application in chemistry: a fullerene is a certain form of molecule made from carbon atoms. (One of these, the buckyball, looks as with a soccer ball.) It provides a convex polyhedron in which each confront is either a continuous pentagon or hexagon. This provides $V - \\frac3f_5 + 4f_62 = 2$ on the one hand, and also $3V = 5f_5 + 6f_6$ ~ above the other. With each other these equations provide $f_5 = 12$ and $V - 2f_6 = 20$; in other words, any fullerene must have exactly twelve pentagons (Twelve Pentagon theorem because that fullerene).

(Hexagons space special. One means to translate this an outcome is that an infinite airplane can it is in tiled v hexagons, therefore hexagons correspond to zero curvature, whereas because pentagons have a smaller sized angle at every vertex they exchange mail to positive curvature. What the over statement says, roughly, is that the total amount the curvature is a constant. This is a simple form of the Gauss-Bonnet theorem, i beg your pardon is very closely related come Euler\"s formula.)

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

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

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This is already most of the method to the group of Platonic solids. If you\"re interested in learning an ext about Euler\"s formula, I very recommend David Richeson\"s Euler\"s Gem. Extremely well-written and also informative. You might also enjoy David Eppstein\"s Nineteen methods to prove Euler\"s formula.