OME 2013 Problem 3

Let $k,n$ be positive integers with $n \geq k \geq 3$ . We consider $n+1$ points on the real plane with none three of them on the same line. We colour any segment between the points with one of $k$ possibilities. We say that an angle is a "bicolour angle" iff its vertex is one of the $n+1$ points and the two segments that define it are of different colours. Show that there is always a way to colour the segments that makes more than $n \Big\lfloor{\frac{n}{k}}\Big\rfloor^2 \frac{k(k-1)}{2}$ bicolour angles.