30 points in the plane 12 red segments and 17 blue segments from each point

Question

Given 30 points in the plane no 3 of which are collinear, we color the segments formed by the pairs of these points red and blue such that: from every point there emerge 12 red segments and 17 blue segments. How many triangles that have all three sides of the same colour are there?

N.B. i think the phrasing of the problem is bad - the number of monochromatic triangles surely cannot be constant right? However it’s rather hard for me to construct two examples with a different number of triangles. Maybe the problem wanted the minimum number of such triangles

Answer 1

Lemma 1 in the Lorden paper shows that if node $i$ has $r_i$ red incident edges and $n-1-r_i$ blue incident edges, the number of monochromatic triangles is $$\binom{n}{3}-\frac{1}{2}\sum_{i=1}^n r_i (n-1-r_i).$$ For $n=30$ and $r_i=12$ for all $i$, this formula yields $$\binom{30}{3}-\frac{30 \cdot 12 \cdot 17}{2} = 4060 - 3060 = 1000$$ monochromatic triangles