60th Belarusian Mathematical Olympiad

Let $m,n,k$ be respectively the numbers of red, green, and blue points. Let $A,B,C$ be respectively the sums of the distances between blue and green points, red and blue points, red and green points. From the inequality of triangle it follows that the following inequalities are necessary for the existence of the points on a plane: $$mA-kC\le nB\le mA+kC,mA-nB\le kC\le mA+nB,nB-kC\le mA\le nB+kC(\ast )$$ In our case $$25m-5n\le 31k,(1)$$ $$31k\le 25m+5n.(2)$$ By condition $m+n+k=15$. Since the number of the red points is the largest, we have $m\ge 6,k\le 6,n\le 6$. Suppose that $m\ge 7$. Then from (1) it follows $26k+(5k+5n+5m)\ge 30m$, hence $26k+75\ge 30m$, which gives $k\ge 6$. Therefore, $k=6$, so $m+n=9$, and we have $$31k\le 25m+5n\Rightarrow 31\cdot 6\le 25m+5n=20m+5(m+n)=20m+45.$$

Therefore, $m=6,k+n=9$. Now equality (1) has the form $$150\le 31k+5n=26k+5(k+n)=26k+45\Rightarrow 26k\ge 105,$$