If every black pentagon shares a face with 5 hexagons, and every white hexagon shares a face with 3 pentagons, can you determine the ratio of hexagons to pentagons?

(I haven't solved this problem, but I think it is doable.)

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gus_massa 3 days ago | parent [-]

Tip: Count the edges. Imagine each edge has two sides/ribons. One side/ribon is white and the other is black.

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C-x_C-f a day ago | parent [-]

You can even find the exact number of pentagons (or hexagons, or edges, or vertices) using the equation

V - E + F = 2

where V, E, F are the number of vertices, edges, and faces, respectively.

This holds for any polyhedron (and other shapes have similar equations possibly with a different right hand side) and the left hand side is called the Euler characteristic of the soccer ball (or any polyhedron).

Spoiler warning: the Wikipedia article for the Euler characteristic [0] has a worked out example specifically for the soccer ball.

[0] https://en.wikipedia.org/wiki/Euler_characteristic

	▲	gus_massa 13 hours ago \| parent [-]
		Side note: From Wikipedia: > The surfaces of nonconvex polyhedra can have various Euler characteristics: It's strange because the examples use weird faces, but in most (all?) of them it is possible to split the weird faces into a few poligonal faces and get V-E+F=2. For example https://en.wikipedia.org/wiki/Small_stellated_dodecahedron use faces that are stars that intersect other faces and the intersection is not an edge. Replacing each star with 5 triangles the Euler characteristic is 2.