Exercise 6. Let $A B C D E F$ be a regular hexagon and $M \in[A, C], N \in[C, E]$. Suppose that $\frac{A M}{A C}$ and $\frac{C N}{C E}$ are equal to a number $r>0$, and that $B, M, N$ are collinear. Determine the value of $r$.

![](https://cdn.mathpix.com/cropped/2024_06_04_d2d43aab2d6adaa2fe91g-6.jpg?height=656&width=588&top_left_y=515&top_left_x=779)
## Solution.

We can assume that the hexagon is inscribed in a circle of radius 1. We place ourselves in a coordinate system such that the coordinates of $A, B, C$ are respectively $A=(0,1), B=\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ and $C=\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$.

Since $C E=\sqrt{3}$, we have $C N=r \sqrt{3}$, so $\overrightarrow{C N}=(-r \sqrt{3}, 0)$. We deduce that

$$
N=\left(\frac{\sqrt{3}}{2}-r \sqrt{3},-\frac{1}{2}\right)
$$

Since $\overrightarrow{A C}=\left(\frac{\sqrt{3}}{2},-\frac{3}{2}\right)$, we have $\overrightarrow{A M}=\left(\frac{\sqrt{3}}{2} r,-\frac{3 r}{2}\right)$, so

$$
M=\left(\frac{\sqrt{3}}{2} r, 1-\frac{3 r}{2}\right)
$$

We can easily calculate that

$$
\overrightarrow{B M}=\left(\frac{\sqrt{3}}{2}(r-1), \frac{1}{2}-\frac{3 r}{2}\right), \quad \overrightarrow{B N}=(-r \sqrt{3},-1)
$$

The fact that $B, M, N$ are collinear means that these two vectors are collinear, which is equivalent to

$$
\frac{\sqrt{3}}{2}(1-r)+r \sqrt{3}\left(\frac{1}{2}-\frac{3 r}{2}\right)=0
$$

which simplifies to $r=\frac{1}{\sqrt{3}}$.