18. If $A C$ and $C E$ are two diagonals of a regular hexagon $A B C D E F$, and points $M$ and $N$ internally divide $A C$ and $C E$ such that $A M: A C = C N: C E = r$, if $B$, $M$, and $N$ are collinear, find $r$.
18. Let $A C, B E$ intersect at $K$. By the collinearity of $B, M, N$ and Menelaus' theorem, we have $\frac{C N}{N E} \cdot \frac{E B}{B K} \cdot \frac{K M}{M C}=1$.

Assume the side length of the regular hexagon is 1, then $A C=C E=\sqrt{3}$. And $\frac{C N}{N E}=\frac{C N}{C E-C N}=\frac{r}{1-r}$,
also $\frac{E B}{B K}=\frac{2}{-\frac{1}{2}}=4$, and $\frac{K M}{M C}=\frac{A M-A K}{A C-A M}=\frac{r-\frac{1}{2}}{1-r}$.
Substituting the above three expressions into (10), simplifying yields $3 r^{2}=1, r=\frac{1}{\sqrt{3}}$.