Given a regular hexagon $ABCDEF$. We draw diagonals $AC$ and $CE$.
Then, we choose two points in the hexagon's diagonals(AC and CE), call that $M$ and $N$, such that: $\frac{AM}{AC} =\frac{CN}{CE}$.
If $B, M$ and $N$ are collinear, how to find the circumradius of this hexagon? Thanks.
At least for your original question I don't see what $M$ and $N$ have to do with anything. Here is the described situation, or as close as I could get with only a bit of manual tweaking, no script:
$M$ and $N$ can be pretty much anywhere, and the circumcircle radius of this hexagon is the same as for any other regular hexagon: it's equal to the edge length. Could the rest be a red herring?