# How many sewings are there on a soccer ball?

A soccer ball is obtained by sewing $20$ hexagonal pieces of leather and $12$ pieces of leather of pentagonal shape.

A sewing joins together the sides of two adjacent pieces. How many sewings are there ?

My effort

I was able to solve this problem by realizing that if I count the number of sewings adjacent to the hexagons and the ones adjacent to the pentagons I will be counting each sewing twice.

So, the number of sewings is $$\cfrac{120 + 60}{2}=90.$$

If we count the sewings adjacent to the pentagons we have $12 \cdot 5 =60$ sewings ,now to count the rest of the sewings I just observe that any other sewing starts at the edge of some pentagon,so I have $60$ other sewings,for a total of $120$ sewings .

However this doesn't quite work, but if I look at the picture I have posted above it seems to be correct as I don't have any pentagon sharing a sewing with another pentagon.

What am I missing?

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Is your picture correct? I'm looking at a google image of a soccer ball, and the adjacencies are NOT the same. – Michael Burr Feb 14 at 16:49
@MichaelBurr The image is actually from the problem,so yes it's correct(if you want I can give you the link). – Mr. Y Feb 14 at 16:49
The picture is slightly off, traditionally for a truncated icosahedron, one has every pentagon surrounded on all sides by a hexagon, and each hexagon surrounded on three sides by other hexagons, and the other three sides by pentagons in an alternating fashion. In the picture, we see a hexagon with four neighboring hexagons in the center. It makes me curious to check if a ball with the above pattern is guaranteed to have 20 hexagons and 12 pentagons like it is guaranteed in the usual ball. Regardless, all the information needed is in the number of pentagons and hexagons to solve the problem. – JMoravitz Feb 14 at 16:52
@JMoravitz I might add that this soccer ball has been stitched by hand. – Mr. Y Feb 14 at 16:57
@Mr.Y Irrelevant since it must still follow the Euler Characteristic of a convex polyhedron. $2=\#V-\#E+\#F$, where $\#V,\#E,\#F$ represent the number of vertices, edges, and faces of the polyhedron respectively. – JMoravitz Feb 14 at 16:59

What's wrong is that your diagram does not have the usual pattern of pentagons and hexagons that a soccer ball usually has. The seams of a soccer ball are given by projecting centrally the edges of the truncated icosahedron onto a sphere:

In particular, no vertex on a truncated icosahedron is shared by three hexagons, but that is not the case for the polyhedron in the diagram, which (per pjs36's answer) is called a chamfered dodecahedron. (A truncated icosahedron is also the shape of the molecular structure of buckminsterfullerene a.k.a. buckyballs.)

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The issue is that the picture depicts not the conventional soccer ball (a truncated icosahedron) but rather something a little different, the chamfered dodecahedron, also known as a truncated rhombic triacontahedron. This actually does have 120 edges.

So, in a sense, you were right both times, but just thinking about different polyhedra!

It's interesting to note that these are both examples of Goldberg Polyhedra, polyhedra made from only pentagons and hexagons -- although the faces are not necessarily regular (and in the chamfered dodecahedron, they are not).

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This is quite interesting: One can choose the hexagons to be equilateral, and the angular defect from regularity is small, so it's not apparent at a glance: "The hexagon faces can be equilateral but not regular with D2 symmetry. The angles at the two vertices with vertex configuration 6.6.6 are $\arccos\left(-\frac{1}{\sqrt 5}\right) \approx 116.565^{\circ}$..." – Travis Feb 14 at 17:15
@Travis It is interesting! I've had "play with Goldberg polyhedra" on my to-do list for a while, but haven't gotten around to it (I was only able to find the name of the depicted polyhedron by remembering the name of Goldberg polyhedra and looking at examples). – pjs36 Feb 14 at 17:19

The problem is in the picture. In the picture, there are five hexagons adjacent to each pentagon. Moreover, each hexagon is adjacent to exactly two pentagons. This gives that $$\frac{12\cdot5}{2}=30$$ hexagons are needed, not $20$. Therefore, the given figure is does not match the given data.

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It's not that the given figure itself is impossible---one can construct it by appropriately shearing a dodecahedron $42$ times---it's that it's not the one described in the text. – Travis Feb 14 at 17:05
@Travis Ok, you are correct, the figure is possible, but it doesn't match the given data. – Michael Burr Feb 14 at 17:06
@Travis Just out of curiosity:how do you know that you can make that figure by shearing a dodecahedron $42$ times ? – Mr. Y Feb 14 at 17:19
@Mr.Y That's the number of faces of the solid in the diagram. In fact I meant to say $30$, which is the number of hexagonal faces, as one can cut the original solid in such a way that the pentagon faces are subsets of the faces of the original dodecahedron. – Travis Feb 15 at 9:23