Number of triangles with property Look at the convex polygon with n vertices.No four of them lie on the same circle. How many triangles there are with the vertices in polygon's vertices such that remaining n-3 vertices lie outside of the circumcircle of that triangle.I have come so far:Every edge of the polygon belongs to exactly one such triangle(look at the angles which look at that edge from other n-2 vertices, only the biggest angle of n-2 such composes the desired triangle), now the question is how many triangles with the desired property consist of 2 polygonial edges and how many from one and from zero.For n=2014 result is 2012 triangles.
 A: All nodes and edges of the triangles that have this property form a graph that is :
1) Connected: because every edge of the polygon belongs to a triangle as you pointed out.
2) Planar: No 2 edges can cross over each other because then there would be 2 triangles that overlap. And that's impossible (lemma 1)
Therefore Eulers theorem says that $V-E+F = 2 $ Where $V$ is the number of vertices, $E$ the number of edges and $F$ the number of faces, including the exterior face.
We have also:
3) Every edge that's not a part of the convex polygon has 2 neighbouring triangles.(lemma2)
Therefore all faces but the exterior face $(F-1)$ have 3 edges. The exterior face has $V$ edges. And every edge has 2 neigbouring faces so we have:
$2E = 3(F-1) + V$ 
Combined with Euler's Theorem this gives: $F = V-1$ but we have to exclude the exterior face, because it is not a valid triangle, so there are $V-2$ triangles that satisfy the special property.
http://en.wikipedia.org/wiki/Euler_characteristic#Planar_graphs
lemma 1: No 2 triangles that satify the given property can overlap i.e. the intesection of their interiors is empty.
Suppose there are two such triangles, consider the 2 circles $A$ and $B$ that go through the points of the triangles. These circles have to overlap because there are points in their intersection. Therefore the circles intersect in two points that define a line $L$ But triangle 1 is on the left of this line, because all of its 3 points are on circle A but outside circle B. Similarly, the second triangle is on the right of the line. Therefore the 2 triangles cannot overlap
lemma 2: I think you can prove this yourself, since it similar to proving that every edge of the convex poligon is in exactly one triangle.
