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
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. |