How many pairs of diagonals of a $2n+1$ sided regular polygon intersect within the interior of the polygon?

(By interior I mean it shouldn't intersect on the vertex)

For a triangle there is no diagonal.

For a Pentagon five pair of diagonals intersect inside the interior of the Pentagon.

For a heptagon there are so many diagonals that I got confused.

Can someone help me out.

  • $\begingroup$ This might be helpful --> wikihow.com/Find-How-Many-Diagonals-Are-in-a-Polygon $\endgroup$ – bigfocalchord Jul 3 '16 at 3:33
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    $\begingroup$ All of them intersect at least one other diagonal. What are you really asking? $\endgroup$ – Ross Millikan Jul 3 '16 at 3:49
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    $\begingroup$ Are you asking how many (unordered pairs) of diagonals intersect in the interior of the polygon? That is an interesting question with a suprisingly short answer. $\endgroup$ – André Nicolas Jul 3 '16 at 4:18
  • $\begingroup$ @AndréNicolas Yes, I am exactly asking the same. I wrote in the question it intersect inside the polygon. $\endgroup$ – Babai Jul 3 '16 at 6:44
  • $\begingroup$ @RossMillikan Yes, they surely do intersect with sides, which is quite obvious. My question was how many pairs of diagonals intersect within the interior of the polygon. $\endgroup$ – Babai Jul 3 '16 at 6:48

There are a number of different ways to count. This is the quickest I know.

Choose $4$ vertices from the $2n+1$ available. Exactly one pair of diagonals determined by dividing these vertices into two pairs meet in the interior of the polygon.

Thus the required number of pairs of diagonals is $\binom{2n+1}{4}$.


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