Let S4(n),S5(n),S6(n)- number of different
4-poligons,5-poligons and 6-poligons in n poligon.
I have AC using formula F(n)=4*S4(n)+5*S5(n)+S6(n).
But this formula can not be right because
in different 6-poligons 3 hords may intersect in
the same point and this triangle will be counted
more the once.

Your worry is superfluous
This condition (put mentally n points on its periphery at equal distances) can guarantee that there will not be different 6-poligons 3 hords may intersect in
the same point and this triangle will be counted
more the once

P.S My English is so poor...

My suspictions based on considering
ideal 12-poligon in which exist two ideal
6- sub-poligons which different and distinvished
by rotation and having common centre in
which their hord intersected.


Edited by author 01.11.2007 22:26

I doubt anyone can mentally put 2000 points on a circle =)

Hah, I on the other hand believe everyone can mentally (and not only) put infinitely many points on a circle. After all - isn't that the definition of a circle? (Infinitely many points with equal distance from one other point - the center?) :D

F(n)=4*S4(n)+5*S5(n)+S6(n)
Why is it true and how to guess about it?

a picture is needed to understand ! please assist

Yes, I was thinking about it too and I understand it now. In fact, it's said in statement that interesting triangle is not a triangle but "any three different chords from this set that intersect pairwise" and "at least one of their intersection points lies inside the circle". Therefore, three segments intersecting in one point are interesting triangle( and not a triangle).

It's like saying that interesting triangle is a four-sided figure =)