The diagonals $AC$ and $CE$ of the regular hexagon ABCDEF are divided by the inner points M and N, respectively, so that $AM/AC =CN/CE = r$. Determine $r$ if $B, M,$ and $N$ are collinear.
The process to do this by coordinates is very tedious. Does anyone know how to do this with the use of complex number geometry specifically?
Consider the complex plane, let $\omega = e^{\pi/3}$ be an elementary sixth root of unity. We find that $\omega^2 = \omega - 1$.
Let $A = 1+\omega, B = \omega, C = 0, D = 1-\omega, E = 2-\omega, F = 2$. These six points form a regular hexagon.
Then $M = (1-r)A, N = rE$.
Let $X = M\bar N + N\bar B + B\bar M$, simplify to obtain $X = a(r) + b(r)\omega$ for some real values $a(r)$ and $b(r)$ depending on $r$.
From this question we find that $M, N, B$ are collinear iff $X$ is real, which is the case iff $b(r) = 0$. By solving that you should be done quickly.