Question

The vertices of a regular 20-sided polygon are divided into 10 pairs, and each pair of vertices is connected by a line segment. Show that there must be a pair of these line segments having equal length.

Answer 1

I think there are exactly 10 different lengths possible. This is because of symmetry. If we imagine shooting rays from a vertex to all other vertices, the rays that are on either side of the diameter will have same lengths.

Answer 2

Somehow we need to prepare the ground to apply pigeonhole principle. It doesn't look easy hmm

Answer 3

I agree with that, if you just start with a blank polygon and want to draw in one line, you can start it anywhere, so might as well start it from the top. Since it's an even number of vertices, we have a mirror plane from that to the bottom and all the possible lines we can draw end up being half of that like you say.

I tried doing some kind of recursion but realized that it's not gonna be possible in a nice way. This thing holds for square and hexagon but not octagon, here's a counter example:

Answer 4

In trying to do some kind of pidgeon hole thingy, I was imagining taking all the different lengths which I will call 1,2,..., 10 since each one's vertices on the polygon have that many edges you have to walk to get to them.

I first thought, OK the 10, longest length one, will cut the polygon exactly in half and act as a kind of symmetric thing to look at. Then I realize the 1 will have to have both of its vertices on one side or the other cause it's too small to bridge past, and the 9 will have to have one vertex on each side, like this:



Then I was thinking, OK the left side has 6 open spots and the right side has 8 open spots. But that's where my dead end is, cause I still have to place 2,..,8 somewhere. I guess maybe the right side now is forced to have at least 1 entire piece on it to balance out but past that idk.

Answer 5

This approach is probably not good haha.