From two points, $A$ and $B$, two cars start at the same time heading towards each other. Their speeds are constant, with a ratio of $5: 4$ (the car starting from $A$ is the faster one). The two cars circulate between $A$ and $B$. They meet for the second time at the 145th km marker, and for the third time at the 201st km marker. At which km markers are $A$ and $B$ located?
I. Solution: Let's briefly call the two cars $a$ and $b$. Let $A$ be at the $x$ km marker and $B$ at the $(x+y)$ km marker. Denote the first, second, and third meeting points as $T_{1}, T_{2}, T_{3}$, respectively.

By the time of the first meeting, the two cars together have traveled $y \mathrm{~km}$. According to the problem, $A T_{1}: T_{1} B=5: 4$, from which

![](https://cdn.mathpix.com/cropped/2024_05_02_2458887329323ed32a03g-1.jpg?height=138&width=710&top_left_y=321&top_left_x=686)

Figure 7

$$
A T_{1}=\frac{5 y}{9} \quad \text{(Figure 7)}
$$

By the time of the second meeting, the two cars together have traveled $3 y \mathrm{~km}$, so $a$ has traveled $3 \cdot \frac{5 y}{9}=\frac{15 y}{9}=y+\frac{6 y}{9}=y+\frac{2 y}{3}$ km, and thus $T_{1}$ is at the $x+y-\frac{2 y}{3}=x+\frac{y}{3}$ km marker.

By the time of the third meeting, the two cars together have traveled $5 y$ km, from which $a$ has traveled $5 \cdot \frac{5 y}{9}=\frac{25 y}{9}=2 y+\frac{7 y}{9}$, so $T_{2}$ is at the $x+\frac{7 y}{9}$ km marker. According to the problem,

$$
\begin{aligned}
& x+\frac{y}{3}=145 \\
& x+\frac{7 y}{9}=201
\end{aligned}
$$

from which

$$
y=126 \quad \text{ and } \quad x=103 .
$$

Thus, $A$ is at the 103rd km marker, and $B$ is at the $103+126=229$th km marker.

It is useful to initially ignore the unknown $x$ when solving for $y$ by taking the difference between (2) and (1), which results in a simple equation that can be solved directly, as shown in the following solution.

II. Solution: Keeping the notation $a$ and $b$, by the time of the first meeting, $a$ has traveled $\frac{5}{9}$ of the $A B$ segment, and $b$ has traveled $\frac{4}{9}$ of it. For every subsequent two meetings, the two cars together travel twice the $A B$ segment, so $a$ travels $\frac{10}{9}$ of the segment, and $b$ travels $\frac{8}{9}$ of it. Thus, the second meeting occurs at $\frac{1}{3}$ of the segment from $A$, and the third meeting at $\frac{7}{9}$ of the segment from $A$. The distance of $56 \mathrm{~km}$ between the two meeting points is therefore $\frac{4}{9}$ of the $A B$ segment. From the third meeting point, $B$ is $\frac{2}{9}$ of the segment away, which is the 229th km marker. $A$ is $\frac{1}{3}=\frac{3}{9}$ of the segment, or $42 \mathrm{~km}$, before the second meeting point, which is the 103rd km marker.