# Can you compute the ratio of shapes on a soccer ball?

The traditional shape of a soccer ball is a combination of hexagons and pentagons. It is an example of an Archimedean solid, which means that although there are different shapes, the vertices are indistinguishable. At each vertex, two hexagons and one pentagon come together. It may seem at first that the ratio of shapes is two to one, but that is not quite right. Can you find the ratio?

Extra credit – Using Euler’s equation and a little algebra, it is possible to determine the number hexagons and pentagons. For any shape that covers the surface of a sphere, F – E + V = 2. F is the number of faces, i.e., the number of hexagons plus pentagons. E is the number of edges and V is the number of vertices.

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I think the answer to your question is “no”. Just visualising a ball like that is making my visual cortex short-circuit. And then go on strike demanding better working conditions.

Interesting question though.

Fyrius (14416 )

To keep people from who want to try the problem on their own from accidentally seeing the answer accidentally I am using Rot13 to encode it.

Hint:
Svaq gur ahzore bs iregvprf va gjb jnlf, bapr hfvat gur ahzore bs cragntbaf (rnfl) naq ntnva hfvat gur ahzore bs urkntbaf (fyvtugyl zber gevpxl).

Solution:
Yrg c or gur ahzore bs cragntbaf. Gura gur ahzore bs iregvprf I vf fvzcyl I = svir c. Vs u vf gur ahzore bs urkntbaf gura gur rkcerffvba fvk u tvirf gjvpr 2I orpnhfr rnpu iregrk vf pbhagrq gjvpr, fvapr rnpu iregrk orybatf gb gjb urkntbaf.

Fb jr unir I = svir c = fvk u/ gjb fb u/c = svir/guerr

Extra Credit:

u=svir/guerr c
Jr unir:
I=5c

R = (fvk u + svir c)/2. Qvivqr ol gjb fvapr rnpu rqtr obeqref gjb snprf.

R = guerr u + svir/gjb c = svir c+ svir/gjb c

S = u + c = c + svir/guerr c

Fhofgvghgvat vagb S – R + I = 2

c + 5/3 c – 5c – 5/2 c + 5c=2

Zhygvcylvat obgu fvqrf ol 6,
6c + 10 c -30c – 15c +30c=12

c = gjryir
u = svir/guerr c =gjragl

That is fantastic! Is there really no other solution? One that uses smaller pentagons and hexagons with the numeric same ratio.
How the heck did the fist person figure out that pattern for the soccer ball?

LuckyGuy (25434 )

Good questions. I have enough trouble visualizing in two dimensions. I have no idea how people worked out all the various regular 3 dimensional shapes.

Each hexagon has three pentagons and three hexagons adjacent to it, while each pentagon has five hexagons adjacent to it. For a given repeatable area the ratio should be 2.5:1, if you take a pentagon and assign half its circumference to it, and half to the pentagon below left of it.

That is my best estimate. Now to look at the answer…..

Let:
v be the number of vertices,
h be the number of hexagons,
p be the number of pentagons,
f be the number of faces,
and e be the number of edges.

Each pentagon meets 5 vertices and no two pentagons meet the same vertex, so v = 5*p.
Each hexagon meets 6 vertices, but each vertex is met by 2 hexagons, so v = 6*h/2 = 3*h.
Therefore 3*h = 5*p, so the ratio of hexagons to pentagons is 5:3.

f = h+p = v/3 +v/5. Each pentagon has 5 edges and no two pentagons share an edge, so the number of edges adjacent to a pentagon is 5*p = v. Each hexagon has 6 edges, every other of which must be shared with a pentagon; otherwise three hexagons would meet at a vertex. Therefore, 3 edges remain and each is shared with another hexagon, so the number of edges not adjacent to a pentagon is 3*h/2 = v/2, so e = 3*v/2.

Then -e+v = -v/2 = -15*v/30 and f = 10*v/30 + 6*v/30 = 16*v/30. From the given formula, v/30 = 2, so v = 60, e = 90, f = 32, h = 20 and p = 12.

The obvious means of counting edges eluded me.

ratboy (15107 )

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