5.1. From point $A$ to point $B$, which are 10 km apart, a car departed at 7:00. After traveling $2 / 3$ of the distance, the car passed point $C$, from which a cyclist immediately set off for point $A$. As soon as the car arrived at $B$, a bus immediately departed from $B$ in the opposite direction and arrived at $A$ at 9:00. At what distance from $B$ did the bus catch up with the cyclist, if the cyclist arrived at point $A$ at 10:00 and the speed of each participant in the movement was constant?
Answer: 6.

Solution. Let's plot the graphs of the car's movement (segment $K L$), the bus's movement (segment $L M$), and the cyclist's movement (segment $N P$) on the axes $(t ; s)$, where $t$ is time (in hours) and $s$ is the distance (in kilometers) from point $A$. Let $Q$ be the intersection point of $L M$ and $N P$. According to the problem, $M K=2$ and $P M=1$. Draw $M G \| N Q, G \in K L$, then by Thales' theorem, $\frac{N G}{G K}=\frac{P M}{M K}=\frac{1}{2}$. Therefore, if $N G=2 x$, then $G K=4 x$, and $L N=3 x$. Hence (again by Thales' theorem) $\frac{L Q}{Q M}=\frac{L N}{N G}=\frac{3 x}{2 x}=\frac{3}{2}$. Therefore, the required distance is $\frac{3}{5} \cdot 10=6$ km.

![](https://cdn.mathpix.com/cropped/2024_05_06_a656db1b08002c401269g-4.jpg?height=685&width=1405&top_left_y=814&top_left_x=334)

Participants familiar with Menelaus' theorem can solve the problem as follows. Points $N, Q$, and $P$ lie on the same line, so $\frac{K N}{N L} \cdot \frac{L Q}{Q M} \cdot \frac{M P}{P K}=1$, or $\frac{2}{1} \cdot \frac{L Q}{Q M} \cdot \frac{1}{3}=1$. Therefore, $\frac{L Q}{Q M}=\frac{3}{2}$.