Who is asking: Student
Question: If you have a regular polygon with n sides and you draw all (n-3)n/2 diagonals how many intersection will they form with each other and how many sections will they devide the polygon into.
I have heard that this question has only been solved recently and if this is so I would like to know were I can obtain a copy of the solution.
Thanks alot, Murray.
We have two replies to your question.
With the keywords "regular polygon", "diagonal" and "regions", I came up with the following entry in mathscinet:
98k:52027 52B05 (51M04)
Summary: "We give a formula for the number of interior intersection points made by the diagonals of a regular n-gon. The answer is a polynomial on each residue class modulo 2520. We also compute the number of regions formed by the diagonals, by using Euler's formula V-E+F=2."
So I was right about Euler's formula being involved, but it seems that counting the number of vertices involves some number theory. I have no idea why precisely the number 2520 would come up in such a calculation.Claude.
The solution is not all that recent -- it came from Garrit Bol in Dutch in 1933 (although the correct answer has been rediscovered many times since then). When n is odd, there can be no multiple intersections of diagonals of regular n-gons. In this case the maximum number of regions is described in the response to a previous question.. (This is a number-theory result whose proof is described as "long and tedious, but straightforward".) When n is even, then the longest diagonals all go through the centre; moreover there must be other points where 3 or more diagonals intersect (up to a maximum of 7). The problem is different for each even n; it requires solving trigonometric equations. Even the simplest nontrivial case, n=12, would be very hard to work out by hand.
A reference that might be useful is