We have counted the number of faces (F), the number of vertices (V) and the number of edges (E) on five regular solids. These solids are regular because the number of edges bounding each face (m) is the same for each face. And the number of edges meeting at each vertex (n) is the same for each vertex. This table is a summary of the situation.
(The astute reader will have noticed apparent links between the cube and the octahedron and between the dodecahedron and the icosahedron. Each pair has the same number of edges and for each pair values of V and F and of m and n have been interchanged. Only the tetrahedron stands alone. But for the tetrahedron V = F and m = n. There’s more to be said about this pattern, but I’m not going to follow it up here.)
We also established the facts that in regular solids, the number of edges can be expressed in two ways.
We left Part One with the challenge of finding more regular solids.
I certainly hope you found none, because I am going to devote this and following blogs to convincing you that these five are the only ones. That seems an enormous job if you think about it: how do I know that there is not some massive regular solid out there with thousands of faces, edges and vertices that still remains to be discovered? After all, I cannot check every possible solid.
The beauty of the argument I am going to give you is that it avoids extensive checking. But before we begin, take a look at that table and compute F + V – E for each solid. 2, right? Is there a rule here? Does F + V – E equal 2 for just regular solids? Let’s look at the solid in this diagram.
I’ve taken a slightly wilted tetrahedron and lopped off the top. The result is a solid that is certainly not regular. We have two triangular faces and three quadrilateral faces. There are nine edges and six vertices. So
F + V – E = 5 + 6 – 9 = 2.
Is it a general rule that for any three-dimensional polytope, F + V – E = 2 ?
Well, as it turns out, yes. The proof is not hard, but it does involve a lot of diagrams.
Theorem: In any three-dimensional polytope, F + V – E = 2.
To get to a proof of this, we first establish that in the two-dimensional equivalent of a polytope, F + V – E =1.
The equivalent of a polytope in two-dimensions is what I call a planar graph. Here’s an example. I have labelled the vertices so you can keep track.
Notice that there is a vertex whenever two edges meet. No edges cross without a vertex at their intersection.
Let’s not count faces, vertices and edges for the time being. Just let V + F – E = X, whatever X turns out to be. I am going to modify this diagram quite a bit. V or F or E may change, but, at each stage, X will stay the same.
First I am going to break the polygons into triangles. Starting with Face 4, I draw in a new edge GC.
New X = V + new F – new E
= V + (F + 1) – (E + 1)
= V + F + 1 – E – 1
= V + F – E = X.
X, whatever it is, has not changed, even though the original F and E have each increased by 1.
This means I can draw extra edges across all the faces I want to without changing X. So let’s put in AC and CH.
The important point here is that I can do this with any planar graph, no matter how complex.
I now start on the second phase of modification — deleting triangles.
If I delete vertex I and the adjacent edges HI and IF and the face they border, I get the graph below.
(V-1) + (F-1) – (E-2)
= V + F – 2 – E + 2
= V + F – E .
X is unchanged.
There is another way of simplifying the figure and that is to delete one edge and a face. Let’s chop HF and face HFC.
We can continue this process of deleting edges, faces and vertices without changing the value of X until we are left with a single triangle.
Now, throughout the modification process, X has not changed; and X for a triangle is easy.
X = V + F – E = 3 + 1 – 3 = 1.
So X was 1 all along.
But we could have applied the process to any planar graph, no matter how complex. So for any such figure V + F – E is 1.
In our original figure (Remember that? Here it is again),
Which we could have calculated a long time ago.
The point is, V + F – E = 1 for any planar graph.
But the original plan — remember — was to show that in a three-dimensional polytope, V + F – E = 2. How does what we have so far help us? To see this, go back to that irregular truncated tetrahedron we saw earlier. Here it is again. I have labelled the vertices to clarify the next step.
Now cut away the face bounded by vertices D, E and F. Leave the vertices and the edges in place; just remove the face.
Now without breaking anything, stretch the sides outwards and down so that they lie on the table beside the base ABC.
Sure, the lengths of the edges and the shapes of the faces have been distorted, but the relationships between the edges, faces and vertices are unaltered. And we know that for this figure, V + F – E = 1 . Now lift up the distorted parts and restore the original solid by replacing the face we removed. That increases F by 1, so that in the solid, V + F – E = 2.
Since we can go through this whole tedious process for any solid with flat faces and straight edges, we can say that for that solid
V + F – E = 2.
Note that the solid does not have to be regular. The faces can be of different shapes and we can have different numbers of edges meeting at different vertices. We’ll get to regular solids next blog.
The value of V + F – E is called the Euler characteristic or Euler number after the Swiss mathematician Leonhard Euler (1707-1783). “Euler” is pronounced “Oiler”.