Doesn't it suffice to prove that nC3 <= \frac{2}{3}(n^2-n) Well, my latex doesn't seem to be working!!!

consider a graph wich its vertices are the n points and its edges are the segments between these points . every edge can give us at max 4 triangles which the area is 1 ( because no three are collinear ) and every triangle with area 1 gives us exactly 3 edges ,if m be the number af triangles with area 1 , then with double counting we have that m is smaller or equal to 2n(n-1)/3 .

The poster didn't post the correct statement of problem. Please edit it so that other readers don't get confused.

Actually AMPARVARDI didn'y post the real problem. The real problem is this : There are $ n $ points in the page such that no three of them are collinear. Prove that number of triangles that vertices of them are chosen from these points and their area is $ 1 $ , is not greater than $ \frac{2}3(n^{2}-n) $ .

the way i see it, four triangles with area 1 can share at most an edge, so if we multiply the number of edges by 4 we will identify a 1-triangle as a unique triplet of edges, not sharing any of its edges with others. So we switch the initial graph with a multigraph having 4 edges where there was one before. The number of edges cannot exceed 2N(N-1) now. At this point we count triangles. For each triangle we count we discard three edges. Thus the number of triangles cannot exceed 2N(N-1)/3.

Consider any pair of points A and B. The loci of points C such that the area of ABC is 1 are two lines of a certain distance from AB. There are at most two points on each line, hence there are at most 4 such triangle for each pair of points. In total, there are at most 4C(n,2) triangles, but each triangle is counted three times, hence there are at most 4C(n,2)/3 triangles.

Johan Gunardi wrote: Consider any pair of points A and B. The loci of points C such that the area of ABC is 1 are two lines of a certain distance from AB. There are at most two points on each line, hence there are at most 4 such triangle for each pair of points. In total, there are at most 4C(n,2) triangles, but each triangle is counted three times, hence there are at most 4C(n,2)/3 triangles. Agreed! =D Finally I can see your post =)

Just wondering, does equality hold here and how? Thanks!

Goutham wrote: Doesn't it suffice to prove that nC3 <= \frac{2}{3}(n^2-n) Well, my latex doesn't seem to be working!!! You need to put american dollar sign before and after. ${ n \choose 3} \leq \frac{2}{3}(n^2-n) $

When does equality hold? Doesn't this look like a weak inequality?

Let $A,B$ be $2$ point's with distance $d$. Assume $2$ lines with distance $\frac{2}{d}$ from $AB$. Note that all triangles which one of their points lies on this $2$ lines and other points are $A,B$ have area $1$. Because of condition "no three of them are collinear" we have at most 4 points possible so we can at most have 4 triangles for $A,B$ but Note that for each triangle we count it 3 times, so the answer is : $\frac{4}{3}{ n \choose 2} = \frac{2}{3}n^2 - n$.