1. **Problem statement:** We have a parallelogram $OABC$ with vectors $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. Point $M$ lies on $BC$ such that $BM : MC = 3 : 2$. The line $AB$ is extended to $N$ so that points $O$, $M$, and $N$ are collinear. We need to find the ratio $AB : BN$ in the form $k : 1$. 2. **Key vectors and points:** - Since $OABC$ is a parallelogram, $\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$. - Vector $\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \mathbf{c} - (\mathbf{a} + \mathbf{c}) = -\mathbf{a}$. 3. **Coordinates of $M$ on $BC$:** Since $BM : MC = 3 : 2$, $M$ divides $BC$ in ratio $3:2$ starting from $B$ to $C$. Using section formula: $$\overrightarrow{OM} = \overrightarrow{OB} + \frac{3}{3+2} \overrightarrow{BC} = (\mathbf{a} + \mathbf{c}) + \frac{3}{5}(-\mathbf{a}) = \mathbf{a} + \mathbf{c} - \frac{3}{5} \mathbf{a} = \frac{2}{5} \mathbf{a} + \mathbf{c}$$ 4. **Parametric form of line $AB$ extended to $N$:** - Vector $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (\mathbf{a} + \mathbf{c}) - \mathbf{a} = \mathbf{c}$. - Let $N$ be on line $AB$ extended beyond $B$, so: $$\overrightarrow{ON} = \overrightarrow{OA} + t \overrightarrow{AB} = \mathbf{a} + t \mathbf{c}$$ 5. **Collinearity of $O$, $M$, and $N$:** Points $O$, $M$, and $N$ are collinear, so $\overrightarrow{OM}$ and $\overrightarrow{ON}$ are linearly dependent: $$\overrightarrow{OM} = s \overrightarrow{ON}$$ for some scalar $s$. Substitute: $$\frac{2}{5} \mathbf{a} + \mathbf{c} = s (\mathbf{a} + t \mathbf{c}) = s \mathbf{a} + s t \mathbf{c}$$ Equate coefficients: - For $\mathbf{a}$: $\frac{2}{5} = s$ - For $\mathbf{c}$: $1 = s t$ From the first, $s = \frac{2}{5}$. From the second, $1 = \frac{2}{5} t \Rightarrow t = \frac{5}{2} = 2.5$. 6. **Find ratio $AB : BN$:** - Since $N$ lies on line $AB$ extended beyond $B$, and $t$ is the parameter from $A$ to $N$ along $\mathbf{c}$. - $t=1$ corresponds to point $B$. - The segment $BN$ corresponds to $t - 1 = 2.5 - 1 = 1.5$ units. - The segment $AB$ corresponds to $1$ unit. Therefore, $$AB : BN = 1 : 1.5 = \frac{2}{3} : 1$$ 7. **Final answer:** $$\boxed{\frac{2}{3} : 1}$$ This means the ratio $AB : BN$ is $2 : 3$ when expressed with integer terms, but since the problem asks for $k : 1$, the answer is $\frac{2}{3} : 1$.