The Geometry Junkyard

Twenty-one Proofs of Euler's Formula: \(V-E+F=2\)

Many theorems in mathematics are important enough that they have been proved repeatedly in surprisingly many different ways. Examples of this include the existence of infinitely many prime numbers, the evaluation of \(\zeta(2)\), the fundamental theorem of algebra (polynomials have roots), quadratic reciprocity (a formula for testing whether an arithmetic progression contains a square) and the Pythagorean theorem (which according to Wells has at least 367 proofs). This also sometimes happens for unimportant theorems, such as the fact that in any rectangle dissected into smaller rectangles, if each smaller rectangle has integer width or height, so does the large one.

This page lists proofs of the Euler formula: for any convex polyhedron, the number of vertices and faces together is exactly two more than the number of edges. Symbolically \(V-E+F=2\). For instance, a tetrahedron has four vertices, four faces, and six edges; \(4-6+4=2\).

Long before Euler, in 1537, Francesco Maurolico stated the same formula for the five Platonic solids (see Friedman). Another version of the formula dates over 100 years earlier than Euler, to Descartes in 1630. Descartes gives a discrete form of the Gauss-Bonnet theorem, stating that the sum of the face angles of a polyhedron is \(2\pi(V-2)\), from which he infers that the number of plane angles is \(2F+2V-4\). The number of plane angles is always twice the number of edges, so this is equivalent to Euler's formula, but later authors such as Lakatos, Malkevitch, and Polya disagree, feeling that the distinction between face angles and edges is too large for this to be viewed as the same formula. The formula \(V-E+F=2\) was (re)discovered by Euler; he wrote about it twice in 1750, and in 1752 published the result, with a faulty proof by induction for triangulated polyhedra based on removing a vertex and retriangulating the hole formed by its removal. The retriangulation step does not necessarily preserve the convexity or planarity of the resulting shape, so the induction does not go through. Another early attempt at a proof, by Meister in 1784, is essentially the triangle removal proof given here, but without justifying the existence of a triangle to remove. In 1794, Legendre provided a complete proof, using spherical angles. Cauchy got into the act in 1811, citing Legendre and adding incomplete proofs based on triangle removal, ear decomposition, and tetrahedron removal from a tetrahedralization of a partition of the polyhedron into smaller polyhedra. Hilton and Pederson provide more references as well as entertaining speculation on Euler's discovery of the formula. Confusingly, other equations such as \(e^{i\pi}=-1\) and \(a^{\varphi(n)}=1\bmod n\) also go by the name of "Euler's formula"; Euler was a busy man.

The polyhedron formula, of course, can be generalized in many important ways, some using methods described below. One important generalization is to planar graphs. To form a planar graph from a polyhedron, place a light source near one face of the polyhedron, and a plane on the other side.

shadow of a polyhedron

The shadows of the polyhedron edges form a planar graph, embedded in such a way that the edges are straight line segments. The faces of the polyhedron correspond to convex polygons that are faces of the embedding. The face nearest the light source corresponds to the outside face of the embedding, which is also convex. Conversely, any planar graph with certain connectivity properties comes from a polyhedron in this way.

Some of the proofs below use only the topology of the planar graph, some use the geometry of its embedding, and some use the three-dimensional geometry of the original polyhedron. Graphs in these proofs will not necessarily be simple: edges may connect a vertex to itself, and two vertices may be connected by multiple edges. Several of the proofs rely on the Jordan curve theorem, which itself has multiple proofs; however these are not generally based on Euler's formula so one can use Jordan curves without fear of circular reasoning.

Please send email if you know of a proof not listed here. I would especially appreciate proofs involving cohomology theory, toric varieties, or other higher mathematics. (Helena Verrill has shown that Euler's formula is equivalent to the fact that every toric variety over \(\mathbb{Z}_p\) has a number of points equal to \(1 \bmod p\) but is still missing a nice non-combinatorial proof of the latter fact.)

I imagine it would be possible to construct inductions based on the representation of convex polyhedra as intersections of halfspaces or convex hulls of points, but the need to handle inputs in non-general position would make the resulting proofs quite messy.

There also seems to be a potential connection to binomials: if one defines a polynomial \[p(t) = 1+Vt+Et^2+Ft3+t^4,\] the Euler formula can be interpreted as saying that \( p(t) \) is divisible by \(1+t\). But for simplices of any dimension, \(p(t)=(1+t)^{d+1}\) by the binomial formula. Perhaps there is a proof of Euler's formula that uses these polynomials directly rather than merely translating one of the inductions into polynomial form. Jim Propp asks similar questions for infinite-dimensional polytopes, interpreting \(p(t)\) as a power series (see also his expansion of these ideas).

From the Geometry Junkyard, computational and recreational geometry pointers.
David Eppstein, Theory Group, ICS, UC Irvine.