
Re: polygons thrugh a given set of points
Posted:
Jun 12, 2014 1:51 AM


quasi <quasi@null.set> wrote: > Let n be an integer, n >= 3. For a set of n points in the plane, > no 3 collinear, is there always at least one ngon having those > points as vertices?
Yes. For n = 3 this certainly holds.
For n > 3, choose two points p, q in the set such that the line between them divides the remaining n  2 points into two nonempty sets. For both such sets, build a smaller polygon out of {that set's elements} U {p, q} by induction, then join the two such polygons by equating the mutual pq edge. Since no three points are collinear, no two remaining edges are collinear, so it is easily seen that there are, in fact, n edges.
This seems too straightforward. Have I missed a catch?

