Best Answer bonanova, 13 May 2013 - 10:07 AM

OK I have it.

Spoiler for Euler and the pentagon

If we draw blue lines to connect dimples with their nearest neighbors and red lines as perpendicular bisectors of the blue lines, the red lines form the edges of a polyhedron. They make hexagons around dimples with six nearest neighbors and pentagons around dimples with five. Each edge is common to two faces, and each vertex is common to three faces.

In forming the polyhedron, hexagons and pentagons each contribute 1 face, but they contribute differently to the edge and vertex count: each hexagon associates with (6/2=3) edges and (6/3=2) vertices, while each pentagon pairs with (5/2) edges and (5/3) vertices. Euler says these numbers grow and decrease in lock-step. They are not independent. For any polyhedron,

Let's add things up.

6

6

6

Note that Euler places no restrictions on the number of hexagons: they can tile the infinite plane.

But to close that tiled surface into a polyhedron requires us to add somewhere to the mix exactly 12 pentagons.

Go to the full post
If we draw blue lines to connect dimples with their nearest neighbors and red lines as perpendicular bisectors of the blue lines, the red lines form the edges of a polyhedron. They make hexagons around dimples with six nearest neighbors and pentagons around dimples with five. Each edge is common to two faces, and each vertex is common to three faces.

In forming the polyhedron, hexagons and pentagons each contribute 1 face, but they contribute differently to the edge and vertex count: each hexagon associates with (6/2=3) edges and (6/3=2) vertices, while each pentagon pairs with (5/2) edges and (5/3) vertices. Euler says these numbers grow and decrease in lock-step. They are not independent. For any polyhedron,

__faces + vertices - edges = 2.__

Let's add things up.

*f*=

*+*

**h***(total faces = #hexagons + #pentagons)*

**p***v*= 2

**h***+ (5/3)*

*(total vertices)*

**p***e*= 3

*+ (5/2)*

**h***(total edges)*

**p**6

*f*= 6

*+ 6*

**h**

**p**6

*v*= 12

*+ 10*

**h**

**p**6

*e*= 18

*+ 15*

**h**

**p**__6(__= (6+12-18)

*) = 6(2)**f*+*v*-*e**+ (6+10-15)*

**h**

**p***= 12.*

**p**Note that Euler places no restrictions on the number of hexagons: they can tile the infinite plane.

But to close that tiled surface into a polyhedron requires us to add somewhere to the mix exactly 12 pentagons.

So on that golf ball, populated mainly by dimples having 6 nearest neighbors,

somewhere there are 12 dimples that have 5 nearest neighbors.