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Let a polygon have $n$ vertices, $n \in\Bbb N$, any with no three diagonals concurrent.

  1. How many intersections of diagonals are there?

  2. How many triangles are there whose vertices are vertices of the polygon or intersections of diagonals, while their sides are either sides or diagonals of the polygon.

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1 Answer 1

I think by "tops" you mean "sides" of the polygon (or maybe vertices). To have an intersection of diagonals, you have to select four points from the set of vertices. There will be one way of selecting which pairs to join to form a crossing. So there are ${n \choose 4}$ intersections of diagonals.

I am not sure I understand 2. Are you looking for the number of triangles formed when all the diagonals are drawn? In that case, you get one triangle for each selection of vertices, ${n \choose 3}$

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you can tell me the result of $C_x^n$ I don't understand $()$ –  LevanDokite Oct 20 '12 at 16:13
@LevanDokite: $\binom{n}{k}$ is the number of ways to choose $k$ objects from $n$ distinct objects. Now you know. –  André Nicolas Oct 20 '12 at 16:18
ok i understand the first answer I accept but the second I think somethings is wrong here –  LevanDokite Oct 21 '12 at 6:49
@LevanDokite: the second gives the number of triangles formed by the original vertices. If you want the number formed by all the diagonals it is much higher. For a regluar n-gon (where many diagonals can intersect in a point) it is If there are no common intersections it is –  Ross Millikan Nov 5 '12 at 14:18

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