In the regular hexagon $A B C D E F$, the diagonals $A C$ and $C E$ are divided by the internal points $M$ and $N$ respectively, such that $A M: A C=C N: C E=r$. Determine $r$ if we know that $B, M$ and $N$ lie on the same straight line.
Let's rotate the hexagon $120^{\circ}$ around its center $O$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b87f6edd5515cd52cf1fg-1.jpg?height=845&width=718&top_left_y=236&top_left_x=690)

In this case, vertex $A$ moves to $C$, vertex $B$ moves to $D$, and vertex $C$ moves to $E$. Therefore, the diagonal $AC$ rotates to the diagonal $CE$. Since point $M$ divides the diagonal $AC$ in the same ratio as point $N$ divides the diagonal $CE$, during the rotation, $M$ moves to $N$. Consequently, the line $BM$ rotated by $120^{\circ}$ becomes the line $DN$, and the angle between these two lines, $\angle BND$, is exactly $120^{\circ}$. This means that $N$ lies on the $120^{\circ}$-arc of the segment $BD$.

The circle centered at $C$ with radius $CB = CD$ passes through point $O$. Since $\angle BOD = 120^{\circ}$, the arc of this circle that lies within the hexagon is the $120^{\circ}$-arc of the segment $BD$. Point $N$ lies on this arc, and thus $CN = CD = CO$. The triangle $ACE$ is equilateral, and its side is $\sqrt{3}$ times the radius of the circumscribed circle: $CE = \sqrt{3} CO$. From this, we can express the desired ratio:

$$
r = \frac{CN}{CE} = \frac{CO}{\sqrt{3} CO} = \frac{1}{\sqrt{3}}
$$

Remark. Some people conjectured that $AM = CN = BC$, and they verified that if we place points $M$ and $N$ in this way, then $B$, $M$, and $N$ lie on the same line. In this case, to fully solve the problem, it must also be proven that the points $M$ and $N$ cannot be located elsewhere, and their positions are unique.