Let $A$ be a point in the plane and $3$ lines which pass through this point divide the plane in $6$ regions. In each region there are $5$ points. We know that no three of the $30$ points existing in these regions are collinear. Prove that there exist at least $1000$ triangles whose vertices are points of those regions such that $A$ lies either in the interior or on the side of the triangle.
Problem
Source: 2017 Greece National Olympiad Problem 2
Tags: combinatorics
Borbas
02.05.2017 19:09
Name $R_1,R_2,...,R_6$ the regions cyclic. We seperate the triangles in $2$ categories: 1) Those whose points are in the regions: $(R_1,R_3,R_5)$ or $(R_2,R_4,R_6)$ We can easily notice that all these triangles are acceptable. The number of ways to pick those points are: $2\binom{5}{1}^3=250$ 2) Those whose $2$ of their points are in regions $(R_i,R_j)$ such that $|i-j|=3$ Fixing those points let them be $B,C$. The third point should be in the same half-plane with $A$ over the line $BC$. There are at least $10$ ways to choose those points from the $2$ regions that satisfy the condition. The number of ways to choose them are: $3\times 10\binom{5}{1}^2=750$ So there are at least $1000$ triangles which can be formed by those points.
khbghvb
20.12.2017 14:50
How do you prove that the first triangles are acceptable?
khbghvb
20.12.2017 15:21
Is that because of the angels the points make with A that is bigger?