Euler's characteristic can be an equation

The formula V-E+F=2 may be seen as a Diophantine equation to be solved for V and F not less than 4. Not every solution (i.e. triple (V, E, F)) corresponds to some convex polyhedron with exactly this number of vertices, edges and faces. For instance, the triple (4, 7, 5) describes no polyhedron. Nevertheless, for every fixed value of V or F there is a solution, as well as for E=6 or E not less than 8. We will give here no general hints on recognising those triples that do correspond to polyhedrons; instead, you will find below some results on more regular solids.

To obtain a Platonic polyhedron, i.e. one with the same number (say, p) of equal regular faces (say, with q sides) meeting at each vertex, it is sufficient to observe that pV=2E=qF and both p and q are not greater than 5. After eliminating V and F from the formula we get

which has five solutions, each of them corresponding to a desired polyhedron.

To obtain an Archimedean solid, i.e. one with the same cycle of regular faces at each vertex, more effort is needed, although the procedure is in fact even simpler. If si faces with li sides each meet at every vertex (for i=1,...,n) - we neglect their cyclic ordering! - then


After eliminating E and F we get

With utter light-mindedness we only care about the consistency of signs, which leads to the requirement that the expression within parentheses be positive, which means

Now we don't need much effort to see that n must be less than 4 and consequently we solve two Diophantine inequalities:


Those who remember geometry know that in each case si is 3, 4 or 5, so we have 6 inequalities to solve. If we look for solutions corresponding to polyhedrons, we must only remember that when l1 is odd, then either s1>2, or s2>1, or s3>1, which we leave for the reader to justify. There are two infinite sequences of solutions and 13 more. And, indeed, there are thirteen Archimedean solids (one in two forms), as well as Archimedean prisms (two regular n-gons joined by a strip of squares) and antiprisms (two regular n-gons joined by a strip of equilateral triangles).

There is a case, however, when the procedure is desperately inefficient. Namely, when the required objects are convex solids with equilateral triangular faces. If wi stands for the number of vertices at which i faces meet, we have

3w3+4w4+5w5=2E=3F and V=w3+w4+w5,

which after elimination of V, E and F in Euler's formula, yields


This equation, however, has 19 solutions, whereas there are only 8 polyhedrons of the required kind. It is not easy to prove that this is so: the fact is known under the name of Freudenthal-van der Waerden's Theorem.

Marek Kordos