14 - Problem J: Solution

Summary

First, we should pause and think about what we mean by a polygon as a mathematical object. A polygon Pn is just a sequence of points (or vectors), say v1, … , vn, called the vertices of Pn, in the plane together with the edges (or line segments) joining v1 to v2, v2 to v3, … and, finally, vn to v1. We shall denote the line segment joining a to b by [a, b], so that an n-gon (a polygon with n vertices) can be considered as a union of the form

At the moment there is no need to insist that these segments do not cross each other, or that a polygon has an ‘inside’ and an ‘outside’. We should not make assumptions until they are forced upon us.

There is also no need to insist that a polygon is planar (that is, it lies in a plane). Indeed, as we can define the segment [a, b] that joins a to b in any vector space, we can define a polygon in any vector space. Here, we shall assume that the polygons lie in some Euclideanm-dimensional space Rm, which is just the set of all real m–tuples (x1, … , xm).

As always, it makes sense to try special cases first. The case n = 1 is trivial because then the polygon is [v1, v1], which is just the set ﹛v1﹜. Then a1 = (v1 + v1)/2 = v1, so there is a unique solution when n = 1, namely v1 = a1.

The case n = 2 is more interesting, and quite different. Here we have two vertices v1 and v2, two coincident edges [v1, v2] and [v2, v1], and two midpoints a1 = (v1 + v2)/2 and a2 = (v2 + v1)/2. Clearly, there is no solution unless a1 = a2. If a1 = a2 there are infinitely many solutions, for we can choose any vertex v1 and then let v2 = 2a1 − v1.

Thus for n = 2, there may be no solutions, or there may be infinitely many solutions. In the latter case, for any point b, there is a solution with a vertex at b.