Mathematics - Polyhedra
Name: Denise Roberts
Who is asking: Teacher
I'm trying to design a unit (possibly involving a Webquest) on Polyhedra and I cannot find a formula I once used. The formula involved the number of vertices (V), edges (E), and faces (F) of the polyhedra. Can you help? Thank you for your time.Hi Denise,
The formula you are remembering is usually called Euler's Formula.
If the polyhedron is like a sphere (rather than like a donut, etc.) the forumula is V - E + F = 2.
Now all the standard polyhedra fit this formula. These are the convex' polyhedra - polyhedra in which each face is a possible 'seat' - a plane which has all of the other vertices, edges etc on one side. [If you check the ORIGIN of the word polyhedron, you will find that Hedron is 'seat' - a polyhedron has many seats! Playing with words, you might enjoy finding the connection to another English word with the same root - CatHEDRAL.]
If you want to PROVE, or at least indicate a proof of this formula, my favourite one (also in a book by Donald (HSM) Coxeter, probably the leading geometer of the 20th Century) runs like this.
Consider any connected PLANAR GRAPH drawn on the sphere with NO crossings.
FIRST find a spanning tree of edges linking up all the vertices, but with no polygons (another way of saying tree). This will have V vertices, E1 = (V-1) Edges and will leave all of the rest of the sruface as a single region or Face.
Now add in the missing edges, one at a time. Each time you add another edge, you cut an existing face into to faces. When you finish, you have added E2 = (F-1) missing edges.
Together you have
E = E1 + E2 = (V-1) + (F-1) or E = V+F -2.
Visually, you can work with the original graph and a dual graph - with another kind of point inside each of the ORIGINAL faces or regions. The edges left in the original graph, after the first spanning tree, now correspond to crossing edges among the new 'facial vertices' which form a second spanning tree on these F vertices. The two treees mesh like the fingers on two hands to match up with all of the original vertices.
There are multiple proofs and investigations of this formula on the web, as well as some free software, such as Poly, which gives pictures of many of the stadard polyhedra with have some types of symmetry.Walter