4. Let $A B C D E F$ be a regular hexagon and $M \in(A C), N \in(C E)$ such that $\frac{A M}{A C}=\frac{C N}{C E}=r$. For what values of $r$ are the points $B, M, N$ collinear?

## MATHEMATICAL OLYMPIAD

- LOCAL STAGE 28.02.2015 -


## GRADE 9 SOLUTIONS AND ORIENTATIVE SCORING GUIDELINES

Note: Each problem is scored from 0 to 7 points.

Any other solution is assimilated according to the scoring guidelines.
## Subject 4.

| Solution Details | Associated Scoring |
| :---: | :---: |
| Let $\overrightarrow{A B}=\overrightarrow{v_{1}}$ and $\overrightarrow{B C}=\overrightarrow{v_{2}}$. Then from $\triangle A B C \Rightarrow \overrightarrow{B M}=(r-1) \overrightarrow{v_{1}}+r \cdot \overrightarrow{v_{2}} \ldots \ldots \ldots \ldots \ldots \ldots$ | $1 \text{p}$ |
| ![](https://cdn.mathpix.com/cropped/2024_06_07_9ec4707e635a4963fccag-3.jpg?height=164&width=1549&top_left_y=1152&top_left_x=185) | 2p |
| From $\triangle B C E \Rightarrow \overrightarrow{B N}=(1-r) \overrightarrow{B C}+r \cdot \overrightarrow{B E}=-2 r \cdot \overrightarrow{v_{1}}+(1+r) \cdot \overrightarrow{v_{2}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ | $2 \text{p}$ |
| $\overrightarrow{B M}, \overrightarrow{B N}$ are collinear vectors $\Rightarrow \overrightarrow{B M}=\alpha \cdot \overrightarrow{B N} \Rightarrow \frac{r-1}{-2 r}=\frac{r}{r+1}$ | $1 \text{p}$ |
| Obtain $r=\frac{1}{\sqrt{3}}$ | $1 \text{p}$ |
| Alternative solution: Points $\mathrm{B}, \mathrm{M}, \mathrm{N}$ are collinear $\Leftrightarrow A[B M C]+A[C M N]=A[B C N]$ |  |