If 2n points are evenly placed on the circumference of a circle, and then a (non-convex) 2n-gon is drawn joining up these points, prove that it must have two parallel sides.
If 2n points are evenly placed on the circumference of a circle, and then a (non-convex) 2n-gon is drawn joining up these points, prove that it must have two parallel sides.
I put this problem in the Algebra section as my work so far has led me to understand the sets of parallel edges as equivalence classes modulo 2n.
If we number the vertices 1..2n around the circle. then the classes are the sets of pairs of vertices who's numbers have the same sum modulo 2n.
For instance the set containing {1,2n} also has {2,2n-1},{3,2n-3}... the sum of each of these pairs =1 (modulo 2n)
These classes can be broken down into 2 types also, the sets with sums equal to an odd number (mod 2n) I would describe as even, this is because each of the edges within has an even number of points around the circle between the two vertices (or 0 for instance in {1,2n} where the points are adjacent). There are n even sets, each containing n pairs.
The sets who's sum of each pair = an even number (mod 2n) I describe as odd, as there are an odd number of vertices between the two vertices, for instance {1,2n-1} has just the vertex numbered 2n between the two vertices. There are n odd sets, each containing n-1 elements.
Now we have these 2n congruence classes, the problem now is to establish that taking one unorded pair from each class as defined above, we can not have a complete tour of the 2n vertices visiting each once and only once.
I understand this is a little wordy, please ask me if anything needs clarifying