1. In the plane 2014 lines are arranged in three groups of parallel lines. What is the largest number of triangles formed by the lines (each side of the triangle lies on one of the lines).
Solution. Let $a \geq b \geq c$ be the numbers of lines in the three groups that yield the maximum number of triangles. Then $a+b+c=2014$, and the maximum possible number of triangles is $a b c$ (when no three lines intersect at the same point). We will prove that $a \leq c+1$.

Assume the contrary, i.e., $a>c+1$. Then

$$
a b c<b(a c+a-c-1)=b(a-1)(c+1)
$$

which contradicts the choice of $a, b$, and $c$. It cannot be that $a=c$, because in that case $a=b=c=\frac{2014}{3}$ is not an integer. For $a, b$, and $c$ to be integers, it must be that $a=672$ and $b=c=671$, and the number of triangles is $672 \cdot 671^{2}$.