MathDB

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.

Problem Statement