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In a $21$ sides regular polygon, how many points inside it are intersection of its diagonal?

I found that a polygon with $n$ sides has $\dfrac{n(n - 3)}{2}$ diagonals, but I feel this is not so useful to the problem solution. I've been trying for $3$ hours without success.

What's the correct solution?

This is part of a contest that is already finished (the solutions have not been released yet).

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    @Victor: It's a$n$ italian contest, part of "Mathematics Olympics". In particular it's from "Festa della Matematica 2012", held in Turin on Friday 9 march. You'll find more information here: http://www.festadellamatematica.bussola.it/. Unfortunately it's in Italian. There are two contests: "Gara tra Istituti", and "Gara per il pubblico". This problem is from the former (a more difficult competition).2012-03-10

2 Answers 2

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The general case is solved here: http://www-math.mit.edu/~poonen/papers/ngon.pdf Although the general case is solved in a rather complicated way, with use of 'heavy casework', the result for an $n$-gon, when $n$ is prime, or is a product of two prime numbers, as it is in your case can be obtained easily studying a bit the article.

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    Thank you! The formula and its proof are quite complicated, I'm not sure I can fully understand it. I'll let you know.2012-03-10
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You may find the following idea useful. Take a convex $n$-gon. Suppose that there is no point inside the $n$-gon at which three diagonals meet. Then there are $\binom{n}{4}$ intersection points of diagonals inside the $n$-gon.

There are various ways to get at this result, but only one simplest one. Choose $4$ vertices. Exactly one of the pairs of lines determined by these $4$ points meets in the interior of the $n$-gon, and therefore the total number of intersection points in the interior of the $n$-gon is $\binom{n}{4}$.

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    Yes, you're right. Sorry, I should had tried that before commenting.2012-03-11