Let $A B C D E F$ be a regular hexagon, $M$ a point on the diagonal $A C$ and $N$ a point on $C E$ such that

$$
\frac{A M}{A C}=\frac{C N}{C E}=r
$$

What can be said about $r$ if $B, M$ and $N$ are collinear?
Let's start by drawing the figure, adding G as the intersection point of $A C$ and $B E$ (by the symmetries of the hexagon, $G$ is the midpoint of $A C$):

![](https://cdn.mathpix.com/cropped/2024_05_10_59677aa68548da326d08g-110.jpg?height=440&width=380&top_left_y=211&top_left_x=840)

Consider the triangle CEG: the points $M, N$, and $B$ are on $G C, C E$, and $E G$ respectively, and are collinear, so we can use Menelaus' theorem:

$$
\frac{\mathrm{GM}}{\mathrm{MC}} \cdot \frac{\mathrm{CN}}{\mathrm{NE}} \cdot \frac{\mathrm{EB}}{\mathrm{BG}}=-1
$$

Let's calculate the three ratios:

$$
\begin{gathered}
\left.\mathrm{GM}=\mathrm{AM}-\mathrm{AG}=\mathrm{r} A C-\frac{1}{2} A C \text { so } \frac{\mathrm{GM}}{\mathrm{MC}}=\frac{(\mathrm{r}-1 / 2) A C}{(1-r) A C}\right)=\frac{2 \mathrm{r}-1}{2-2 r} \\
\frac{\mathrm{CN}}{\mathrm{NE}}=\frac{\mathrm{rCE}}{\mathrm{CE}-\mathrm{rCE}}=\frac{\mathrm{r}}{1-\mathrm{r}}
\end{gathered}
$$

For the last ratio, we will calculate $E B$ and $B G$ using the properties of the regular hexagon. Let $c$ be the side length of the hexagon, $EB$ is a major diagonal and measures $2 c$ (we divide the hexagon into 6 equilateral triangles and it is obvious), and $B G=c / 2$ (since $A G$ is the axis of symmetry of one of the small equilateral triangles), $\frac{E B}{B G}=-4$. Therefore, $r$ satisfies the following equation:

$$
\begin{aligned}
& -4 \cdot \frac{2 r-1}{2-2 r} \cdot \frac{r}{1-r}=-1 \\
& \text { if and only if } \quad 2(2 r-1) r=(1-r)^{2} \\
& \text { if and only if } \quad 4 r^{2}-2 r=r^{2}-2 r+1 \\
& \text { if and only if } \\
& r= \pm \frac{\sqrt{3}}{3} .
\end{aligned}
$$

The real number $r$ is either $\sqrt{3} / 3$ or $-\sqrt{3} / 3$.