Name: Murray

Who is asking: Student Level: Secondary

If you conect all the vertices of a regular n-gon with lines you will have (n-3)(n/2) lines inside the polygon. I want to find out how many sections these lines divide the polygon into and how many intersections they have with each other. Could you please give me hint to set me on the right track and guess how complicated the solution might be. (But don't tell me it!) Thanks alot, I really appreciate all your help; you guys are great.

Hi Murray,

The lead is Euler's formula: If F is your total number of faces (including the outside face), E the total number of edges and V the total number of vertices, then

F + V = E + 2.

So, if there is no obvious way to count the faces, then there might be ways to count the vertices and edges, and come up with a formula.

There is a classical problem that looks like that: Instead of starting with a regular polygon, you start with a circle, put n points on it, and then join these by straight lines. The question is: What is the maximum number of faces in the circle that you can obtain? This question is fun because the sequence starts like this:

1 point: 1 face at most
2 points: 2 faces at most
3 points: 4 faces at most
4 points: 8 faces at most
5 points: 16 faces at most

at this point you get an idea of what might come next, but here comes the surprise:

6 points: 31 faces at most
7 points: 57 faces at most

So, what seemed to be an obvious pattern turns out to be the wrong lead, and you need to use Euler's formula to get the right pattern.

To get 31 faces with 6 points, you need to place the points irregularly, to get a small triangle in the middle. If you place the 6 points as in a regular hexagon, you will get 6 edges meeting at a point in the middle, and this will give you at most 30 faces. In fact, the same thing worries me about your problem: The fact that your n-gon is regular implies that you will often have more than 4 edges meeting at a point, so it may be hard to count the edges. I am not sure how to solve this difficulty. Euler's formula may turn out to be the wrong approach.

Go to Math Central