A person is waiting at the $A$ tram stop. Tired of waiting, they start walking towards the next $B$ tram stop. When they have covered $1 / 3$ of the distance between $A$ and $B$, they spot a train approaching the $A$ stop at $30 \mathrm{~km} /$ hour. If they start running at full speed towards either the $A$ or the $B$ stop, they will just catch the train. What is the maximum speed at which they can run?
I. Solution. Let the sought speed be $v \mathrm{~km} /$ hour, the length of the $AB$ segment be $x \mathrm{~km}$, and let $C$ be the point from which the person spots the train. Then $CA = x / 3$, $CB = 2x / 3$, and the times required to travel these distances are $x / 3v$ hours and $2x / 3v$ hours, respectively. In the same amount of time, the train also reaches $A$ from the point where it was spotted, and $B$ from the point where it was spotted, so the time required for the train to travel the $AB$ segment is the difference between these two times: $x / 3v$ hours. This time, expressed in terms of the train's speed, is $x / 30$, so from the equality of these two expressions, $v = 10$. Our person can run at a maximum speed of $10 \mathrm{~km} /$ hour.

![](https://cdn.mathpix.com/cropped/2024_05_02_09724024ecbd505f33e2g-1.jpg?height=198&width=641&top_left_y=386&top_left_x=731)

II. Solution. The problem can also be solved without equations. By the time the person reaches $A$ from $C$, the train also reaches $A$. Thus, if the person runs towards $B$, in the time it takes for the train to reach $A$, he covers the same distance as the distance from $C$ to $A$, which is one-third of the $AB$ segment. Therefore, he has already covered two-thirds of the way to $B$, and only one-third remains. While covering this third, the train travels from $A$ to $B$, covering the entire $AB$ distance. Thus, the person's speed is one-third of the train's speed, which is $10 \mathrm{~km} /$ hour.

Remarks. 1. The solution does not provide the length of the considered segments. The only given relationship is sufficient to determine the unknown sought.

2. However, it can be deduced from the problem's conditions - and this can also be used in the solution - that at the moment of spotting (point $D$ on the diagram), the train is as far from $A$ as $A$ is from $B$, since the traveler covers twice the distance running to $B$ as he does running to $A$, so the distances the train travels to $B$ and to $A$ are also in this ratio.