G47 (23-5, Netherlands) Given that $A C, C E$ are two diagonals of a regular hexagon $A B C D E F$, and points $M, N$ internally divide $A C, C E$ respectively, such that $\frac{A M}{A C}=\frac{C N}{C E}=r$. If $B, M, N$ are collinear, find the value of $r$.

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The problem is to find the value of $r$ given that points $M$ and $N$ divide the diagonals $A C$ and $C E$ of a regular hexagon $A B C D E F$ in the ratio $r$, and that points $B, M, N$ are collinear.
Solution 1 As shown in Figure 4-46, connect $D N$. It is easy to see that $A M=C N, \triangle A M B \cong \triangle C N D$.
Thus $\angle B N D=\angle 3+\angle 4$
$$
\begin{array}{l}
=\angle 1+\angle 4 \\
=\angle 2+\angle 4 \\
=180^{\circ}-\angle A C E \\
=120^{\circ} .
\end{array}
$$

Since $\angle B C D=120^{\circ}, C B=C D$, draw a circle with $C$ as the center and $B C$ as the radius. The inscribed angle on the same side of the chord $B D$ is $60^{\circ}$, and the inscribed angle on the opposite side of the chord $B D$ is $120^{\circ}$. $N$ satisfies this condition, so $N$ lies on the circle, and $C N=C D$.
Thus
$$
r=\frac{C N}{C E}=\frac{C D}{C E}=\frac{\sqrt{3}}{3} .
$$

Solution 2 Let the side length of the regular hexagon $A B C D E F$ be 1, as shown in Figure 4-47, connect $B E$ intersecting $A C$ at $G$, it is easy to see that $B E$ $=2, G B=\frac{1}{2}$.
$$
\begin{array}{c}
C E=\sqrt{3}, \\
C N=\sqrt{3} r . \\
C M=E N=\sqrt{3}(1-r), \\
G M=\frac{1}{2} A C-C M \\
=\frac{\sqrt{3}}{2}-\sqrt{3}(1-r) \\
=\frac{\sqrt{3}}{2}(2 r-1) .
\end{array}
$$

By the Menelaus theorem, we have:
$$
\frac{E B}{B G} \cdot \frac{G M}{M C} \cdot \frac{C N}{N E}=1 .
$$

Substituting, we get the equation
$$
\frac{2 r-1}{2(1-r)} \cdot \frac{r}{1-r}=\frac{1}{4} \text {. }
$$

Solving, we get
$$
r=\frac{1}{\sqrt{3}} .
$$