13.286. The schooner is sailing from $A$ to $B$ across the lake, and from $B$ to $C$ upstream, and then it returns. The speed of the schooner relative to still water is maintained at $c$ km/h throughout. The schooner takes $\alpha$ hours to travel from $A$ to $C$, and the return trip takes $\beta$ hours, with the journey from $C$ to $B$ taking three times less time than the journey from $B$ to $A$. Find the distances $A B$ and $B C$.
Solution.

Let $x, y$ be the distances $AB$ and $BC$ respectively, and $V$ be the speed of the river current. According to the problem, we have

$$
\frac{x}{c}+\frac{y}{c-V}=\alpha, \text { (1) } \frac{y}{c+V}+\frac{x}{c}=\beta, \text { (2) } \frac{y}{c+V}=\frac{1}{3} \cdot \frac{x}{c} \text { (3). }
$$

From equations (2) and (3), we find $x=\frac{3 \beta c}{4}$, and from equations (1) and (3), knowing the expression for $x$, we find $y=\frac{\beta c(4 \alpha-3 \beta)}{4(2 \alpha-\beta)}$.

Answer: $AB=\frac{3 \beta c}{4}$ km; $BC=\frac{\beta c(4 \alpha-3 \beta)}{4(2 \alpha-\beta)}$ km $(\alpha>\beta)$.