I have the following problem:

I'm given convex polygon with 18 vertices. It is known that no 3 diagonals of the polygon intersect in a single point.

How many points of intersection of the diagonals has a polygon?

I've found this article, that is very near to my problem, but it is about regular polygon, where diagonals can intersect more than 2 times in one point. It seems to me that my problem has more complex solution.

Can you give me any hint how to solve this problem for convex polygon?

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  • $\begingroup$ The convex polygon with $18$ sides has many diagonals, where by diagonal we mean a line segment joining two non-adjacent vertices. There is a simple solution where we take the above interpretation of diagonal. $\endgroup$ – André Nicolas Apr 10 '13 at 12:32
  • $\begingroup$ I think instead of "none of the 3 diagonals of polygon intersects at one point" you meant "no 3 diagonals of the polygon intersect in a single point". $\endgroup$ – hmakholm left over Monica Apr 10 '13 at 12:39

We interpret diagonal as meaning a line segment that joins non-adjacent vertices.

There are $\dbinom{18}{4}$ ways to choose $4$ vertices. If the polygon is strictly convex, then exactly one pair of the lines joining the vertices meets in the interior of the polygon. So the required number of intersection points of diagonals is $\dbinom{18}{4}$.

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  • $\begingroup$ Clever! $ (+1) $ $\endgroup$ – hmakholm left over Monica Apr 10 '13 at 12:40
  • $\begingroup$ Please, explain why we choose 4 vertices? This is because we are examining separately two diagonals in each moment and these diagonals have only one point of intersection and contain 4 vertices? $\endgroup$ – Ann Orlova Apr 10 '13 at 12:53
  • $\begingroup$ Yes, I've understood! Thank you for your clever solution! $\endgroup$ – Ann Orlova Apr 10 '13 at 12:54
  • $\begingroup$ @HenningMakholm: Certainly it is not cleverness on my part, at least not today. It's a golden oldie. $\endgroup$ – André Nicolas Apr 10 '13 at 13:01

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