Proving Euler's Polyhedral Formula by Deleting Edges
Proving Euler's Polyhedral Formula by Deleting Edges
1,3,1 is longer than depth of object.
1,3,2,1 is longer than depth of object.
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For a polyhedron , let be the number of vertices, the number of edges, and the number of faces. Then Euler's polyhedral formula of 1752 is .
P
V
E
F
V-E+F=2
There are more than a dozen ways to prove this. Here is one such proof.
First, cut apart along enough edges to form a planar net. (The unbounded polygonal area outside the net is a face.) Cutting an edge in this way adds 1 to and 1 to , so does not change.
P
E
V
V-E+F
Next, triangulate the bounded faces. For each edge added, a face is also added, so again does not change.
V-E+F
Now, whenever possible, find an outside triangle (one with two outside edges). Delete the edges, their common vertex, and the triangular face. Yet again does not change.
V-E+F
If it happens that there is no outside triangle, delete any outside edge and its triangular face (make sure the third vertex is not on the outside); once more does not change. After each such step, make sure to see if there are any outside triangles!
V-E+F
At each step at least one edge is taken away, so eventually the process stops, when there is only a triangle or only one edge. In either case (remember to count the unbounded face!), , which must have been true for the original net and for .
V-E+F=2
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