Problem 1901

First, we prove that for every line $\ell$, there exist lines $d_1$, $d_2$ such that the triangle $\ell d_1 d_2$ has exactly one black region inside it. On one of the two sides of $\ell$, there is a black region. Without loss of generality, assume this region is above $\ell$. To see this, take the closest point to $\ell$ which is the vertex with the longest distance in the black region. Suppose this point is the intersection of $d_1$, $d_2$. The triangle $\ell d_1 d_2$ clearly has this property.

Now, let $x$ be the number of lines that have black regions on both sides. According to the point above, the number of good triangles is at least $\frac{2x + (n - x)}{3}$. Therefore, it suffices to prove that $x > \frac{n - 1}{2}$.

Consider all the lines that have a black region on only one of their sides, and consider the intersection of the sides of these lines in which these black regions exist. This intersection will be an $n - x$-gon which contains all the black regions inside it. In this case, the remaining $x$ lines eventually divide this polygon into $\frac{x(x + 1)}{2} + 1$ regions. Therefore, $\frac{n^2}{x} \le \frac{x(x + 1)}{2}$, which implies that $x > \frac{n - 1}{2}$.