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On a regular hendecagon (2 dimensions), how many non-symmetric ways can you draw a maximal set of non-intersecting diagonals?

This puzzle has been driving me crazy! I've been doing it by hand.

-The maximal set is defined as the largest number of diagonals that do not intersect, except at the endpoints

-Please also note reflects and rotations are not valid. (i.e. only counts as one)

Any ideas of how to go about problem solving would be helpful!

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    Can the diagonals intersect at the vertices? And by "maximal", do you mean that the set has the maximal number of diagonals, or just that it can't be extended by adding another non-intersecting diagonal to it?2012-08-20
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    Evidently means 11 sides. Who knew?2012-08-20
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    Dear Nat, on math.SE, the proper way to notify users is to add the symbol `@` before their usernames, e.g. @Nat2012-08-20
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    @joriki It can intersect only at vertices (end points), but not inside the hendecagon. Maximal for a hendecagon is 8 without intersecting, I believe2012-08-20
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    @WillJagy I know, right? 11-gon.2012-08-20
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    @JenniferDylan Thank-you so much!2012-08-20
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    I'm positive I saw exactly the same question asked on this site not more than a week ago (with the word *hendecagon* too, no less). In the comments people brought up Bell numbers, but apparently dealing with rotations and reflections makes it problematic. I can't find the old question any more; perhaps it has been deleted?2012-08-20
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    @RahulNarain - I've seen this problem around as well - along with catalan numbers. I find the rotations/reflections also problematic because I have found there's no definite way to separate unless it's by hand or visual2012-08-20
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    It could also mean a figure bounded by ten hens.2012-08-20
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    It is true that the maximum number of diagonals is $8$. For an $n$-gon it is $n-3$, as can be proved by induction. True for $n=3$. When you draw a diagonal, you partition the $n$-gon into an $m$-gon and $p$-gon with $m+p=n+2$ (as the new diagonal counts as two sides), so you can draw $1+(m-3)+(p-3)=n-3$ total diagonals.2012-08-20
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    @Ross: Ah, so my question didn't make sense, since you can prove in the same way that there's no maximal set of non-intersecting diagonals with less than the maximal number of non-intersecting diagonals. Then I wonder why Bell and Catalan numbers are being brought up -- is there something wrong with the counting in my answer?2012-08-20
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    I think the Catalan numbers are what you want except that rotations and reflections are considered different. I found it in http://oeis.org/A033282, where the right side of the triangle is the Catalan numbers.2012-08-20

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