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Given a regular hexagon $ABCDEF$. We draw diagonals $AC$ and $CE$. Then, we choose two random points inside the hexagon, 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.

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Where are you getting your questions? As @MvG points out, a regular hexagon's circumradius has nothing to do with $M$ and $N$. Your other recent geometry questions are also somewhat flawed. I had thought this might be due to a language issue ---and there's nothing wrong with that--- but this hexagon problem makes me wonder if there's more to it. –  Blue Mar 24 at 9:25
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By the way, since $AC=CE$, the proportionality condition says simply that $AM=CN$. Calling the common distance $r$, we can draw circles of radius $r$ about $A$ and $C$. For a given line $\ell$ through $B$, then, $M$ is one of the pts where $\ell$ meets $\bigcirc A$, and $N$ is one of the pts where $\ell$ meets $\bigcirc C$. This can happen in lots of ways. Based on previous questions, I wonder if the description intends (for large-enough $r$) that $M$ and $N$ are the two pts where $\bigcirc A$ and $\bigcirc C$ meet. (Note that the line joining these pts necessarily contains $B$ (and $E$).) –  Blue Mar 24 at 10:01
    
@Blue, I got it from geometry challenge book. But, the writer didn't insert the answer. So, I think this is good for us to become a problem. Thanks –  akusaja Mar 24 at 10:42
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Can you indicate the name of that book, and the page where this problem was stated? It would be interesting to know this. Even if the book isn't in English. –  MvG Mar 24 at 13:19
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@akusaja: Ah ... The fact that $M$ and $N$ are on the diagonals (and not simply "random points inside the hexagon") makes a BIG difference! (Be sure to give the original questions from now on. :) The figure is now uniquely determined. It's still odd to have a proportionality statement that reduces to $AM=CN$, but ok. If I can trust my figure, it appears that the $\bigcirc A$ and $\bigcirc C$ from my previous comment must pass through $B$ itself, which means that the side of the hexagon (and therefore, the radius of the circumcircle) is equal to the common distance $AM=CN$. Now to prove it! –  Blue Mar 24 at 20:57
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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:

Almost the desired configuration

$M$ 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?

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