Euler's Gem Applied to Geodesic Domes
I once read a book called Euler's Gem and it referred to Euler's Polyhedral Identity as Euler's Gem. While I haven't seem this repeated anywhere else, I much prefer it as a name. The idea is that for all solid shapes then Faces - Edges + Vertices = 2 and it is very far reaching. There are plenty of extensions and a lot of topology is based upon it.
However I'm going to use it without proof in an effort to show one application. Imagine we have a 3D shape made up of just pentagons and hexagrams like this classic image of a football:
Now there are various numbers of pentagons and hexagons that you can make these semi-regular polyhedrons out of, but there is a surprising fact: the number of pentagons will always be exactly 12. We're going to show it using Euler's Gem F - E + V = 2
Let the number of pentagons = P and the number of hexagons = H
Now we know that each pentagon has 5 edges and each hexagon has 6 edges which would suggest a total number of edges of 5P + 6H, except that we have double counted. On the whole 3D shape each edge is shared by two 2D shapes, so the total number of Edges E = (5P + 6H)/2.
That gives us something to substitute into the Gem for E, how about F and V?
Well F is just the number of faces which is simply the number of Pentagons added to the number of Hexagons. So F = P + H
For vertices we need to know how many shapes meet at a vertex. Looking at the picture of the football above seems to suggest the answer 3, but is that always true?
Well a Hexagon has an internal angle of 120 degrees so two of them wouldn't provide the 360 degrees needed to surround a point. How about 4 or more? Well the smaller internal angle on the Pentagons is 108 degrees. 4 of those puts you over the 360 so that won't work, so it must always be 3 shapes meeting at a point.
The total number of Vertices from all the individual 2D shapes is 5P + 6H, but each vertex is being triple counted, so V = (5P + 6H)/3.
We can now put all three of our formulae into the Gem to get 2 = P + H - (5P + 6H)/2.+ (5P + 6H)/3.
Let's multiply by 6 ->
12 = 6P + 6H - 15P - 18 H + 10P + 12H
Which after gathering up the right hand side gives: 12 = P, so this remains invariant (unchanging) as we change the number of Hexagons.
For instance here is one where the number of Hexagons is zero:
But this is still true even when the number of hexagons is massive like on a golf ball. Try to find a face with only 5 faces surrounding it:
And here are the ones I could see after a cursory search (although there are 12, some will be behind):
One of the great things about doing stuff with polyhedra is that you can exchange the faces and the vertices of a shape and get another valid shape (after all F and V play the same role in Euler's Gem). For example if you start with a cube and put a vertex in place of each of its 6 faces and the face in place of each of its 8 vertices you get an octahedron. If you do this to an octahedron you get back a cube. We call this shapes duals of one another: here's a picture showing the classics:
With this idea we can reframe our Theorem that: "Any Polyhedron made entirely out of Pentagons and Hexagons, where every vertex is the meeting point of 3 faces, has exactly 12 pentagons" into its dual: "Any polyhedron made of triangles, where the number of faces meeting at each vertex is 5 or 6, has exactly 12 vertices of degree 5."
If you like you can come to this by plugging through with the algebra into Euler's Gem following the same method we used above, but I assure you that the duality argument is solid.
So if you are looking up at one of those fantastic Geodesic domes which were popularised by Buckminster Fuller (who I wrote about here) then you should be able to find a small number of points with only 5 edges rather than 6 pointing into them. Try to find a few.
Of course, if it is just a dome them you may only have half of the 12 special 5-points, but I can think of at least one glorious place on Earth where you would be able to find all 12: