Hellenic Mathematical Olympiad

2. First we observe that for every selection of six points, one per each sector, an hexagon (convex or non-convex) is created containing A.
From these six points we can create totally $\binom{6}{3} = 20$ triangles. We will count how many of them contain A. For any two points lying in two opposite sectors the third vertex of the triangle can be selected in two ways.

Figure 3

Figure 4
For example, for the points B, C of figure 4 we have the possibility of taking the points form the two colored sectors.
There exist three pairs of opposite sectors. We have 5x5 selections for the base BC, whereas the third vertex can be selected by $2 \cdot 5 = 10$ ways. Therefore we have totally at least $3 \cdot 2 \cdot 5^3 = 6 \cdot 5^3$ such triangles containing A.

Figure 5
Considering points in non-succcessive and non-opposite sectors (see figure 5) in this case we have the triangles like CBD or EFG.
Like CBD there are $5 \cdot 5 \cdot 5 = 5^3$ triangles containing A and like EFG there are also $5 \cdot 5 \cdot 5 = 5^3$ triangles containing A. Totally in this case we have $2 \cdot 5^3$ triangle containing A.
Summing up all the above cases we count at least $6 \cdot 5^3 + 2 \cdot 5^3 = 8 \cdot 5^3 = 1000$

triangles containing A either in their interior or in their sides.