# The distance between the origin and all intersections by the diagonals of a regular polygon

The geometric center of an n-sided regular polygon is point $O$. Connect all diagonals of the polygon. How many different distances between diagonal-diagonal intersections ($O$ itself is counted) and $O$ are there (i.e. how many concentric circles are in the graph below)?

For n = 6, ..., 16, the answer should be 4, 5, 7, 11, 14, 21, 29, 36, 37, 54, 57, if I'm not mistaken. But I don't know the answer for a general n. Multiple junctions are not easy to deal with.

The other graphs are being uploaded to imgur so there will be more graphic examples on the way. Thanks! Edit: Imgur down. Will try later.

• Wow... what a nice question, with a beautiful picture no less. Kudos! – wltrup Aug 10 '15 at 20:19
• I think you mean "polygon" where it says "polyhedron"? – joriki Aug 10 '15 at 22:08
• @wltrup Thanks! I blame Mathematica for that – arax Aug 10 '15 at 22:32
• possible duplicate of to find the intersection points of diagonals of a regular polygon – Venus Aug 11 '15 at 5:24
• @Lucian It doesn't answer my question directly if I didn't miss anything, because there might be $n$ or $2n$ intersections on the same circle. But it does seems that my question is a bit too complicated for casual discussion – arax Aug 11 '15 at 20:50

If $n$ is odd then there are no intersections of $3$ or more diagonals and the problem can be analysed quite simply. Let's define a diagonal to be of rank $k$ ($k=0, \ldots, (n-1)/2$) if between its two endpoints $k$ vertices are comprised. Of course rank $0$ diagonals are the polygon sides, which form the outer circle. A diagonal of rank $1$ intersects all the $n-3$ diagonals issued from the single point it comprises, so we have $n-3$ intersection points. But these points are disposed symmetrically on the diagonal, so they have the same distance from the center in pairs and belong to only $(n-3)/2$ different circles.
Let's now consider a rank $2$ diagonal. We want to count its intersections with the other diagonals having rank $2$ or more, because its intersections with rank $1$ diagonals have already been counted before. That diagonal intersects all the $2(n-5)$ diagonals of rank $\ge2$ issued from the two point it comprises, but by symmetry we have $2(n-5)/2$ different new circles.
This analysis can be repeated, so for the total number $N$ of circles we have: $$N=1+1{n-3\over2}+2{n-5\over2}+3{n-7\over2}+\ldots,$$ where the initial $1$ is due to the outer circle. That is: $$N=1+\sum_{k=1}^{(n-1)/2}k{n-2k-1\over2}=1+{(n+1)(n-1)(n-3)\over48}.$$ However, if your results are correct, this formula fails for $n=15$ but I haven't explicitly checked that case.
If $n$ is even the analysis becomes much more complicated due to the presence of multiple diagonal intersections, see the paper cited by Lucian above.