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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.

Octagon

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.

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    $\begingroup$ Wow... what a nice question, with a beautiful picture no less. Kudos! $\endgroup$ – wltrup Aug 10 '15 at 20:19
  • $\begingroup$ I think you mean "polygon" where it says "polyhedron"? $\endgroup$ – joriki Aug 10 '15 at 22:08
  • $\begingroup$ @wltrup Thanks! I blame Mathematica for that $\endgroup$ – arax Aug 10 '15 at 22:32
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    $\begingroup$ possible duplicate of to find the intersection points of diagonals of a regular polygon $\endgroup$ – Venus Aug 11 '15 at 5:24
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    $\begingroup$ @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 $\endgroup$ – arax Aug 11 '15 at 20:50
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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.

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