We need the following basic fact from spherical trigonometry:
if we normalize the surface area of a sphere to 4 pi,
and look at any triangle defined by great circle arcs on the sphere,
the sum of the three interior angles is pi+a where a (the *excess*
of the triangle, is equal to the surface area of the triangle.
(E.g. see Wells, p. 238).

To translate our question on polyhedra to one of spherical geometry, first triangulate the polyhedron; each new edge increases E and F by one each, so V-E+F is left unchanged. Now perform a similar light-shining experiment to the one described on the index page: place a light source at an interior point of the polyhedron, and place a spherical screen outside the polyhedron having the light source as its centerpoint. The shadows cast on the screen by the polyhedron edges will form a spherical triangulation. Since every edge is on two triangles and every triangle has three edges, 2E=3F.Sommerville attributes this proof to Legendre. Because of its connections with geometric topology, this is the proof used by Weeks, who also gives an elegant proof of the spherical angle-area relationship based on inclusion-exclusion of great-circular double wedges.We now add up the angles of all the triangles; by the spherical trigonometry described above, the sum is (4+F) pi. Adding the same angles another way, in terms of the vertices, gives a total of 2 V pi. Since these two sums measure the same set of angles, F=2V-4 and combining this with the other equation 2E=3F yields the result.

The relation A*k = 2 pi (V-E+F) on a surface of constant curvature k such as the sphere is a form of the Gauss-Bonnet formula from differential geometry.

Proofs of Euler's Formula.

From the Geometry Junkyard,
computational
and recreational geometry pointers.

David Eppstein,
Theory Group,
ICS,
UC Irvine.

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