Two trucks start simultaneously from $A$ and $B$ heading towards each other, and they meet at $8^{\mathrm{h}} 40^{\mathrm{m}}$, $62 / 5 \mathrm{~km}$ away from $B$. If the first truck had started $1 / 12$ of an hour later, and the second truck had started $1 / 8$ of an hour earlier, they would still have met at $8^{\mathrm{h}} 40^{\mathrm{m}}$, but this time $62 / 5 \mathrm{~km}$ away from $A$. What were their speeds and what is the length of the $A B$ route?
I. solution. Let the length of the $AB$ road be $s(\mathrm{~km})$, the time until they meet be $t$ (hours), the speeds of the two cars be $v_{1}$ and $v_{2}$ (km/h), the meeting point be $T$, and the point in the imagined scenario be $M$. From the actual and imagined movements, we can derive two equations for $s$ and $t$:

![](https://cdn.mathpix.com/cropped/2024_05_02_28629a4105d4c128400ag-1.jpg?height=146&width=428&top_left_y=312&top_left_x=846)
$v_{1}=\frac{AT}{t}=\frac{AM}{t-\frac{1}{12}}$,
thus $\quad \frac{s-6.4}{t}=\frac{6.4}{t-\frac{1}{12}}$,
$v_{2}=\frac{BT}{t}=\frac{MB}{t+\frac{1}{8}}$
$\frac{6.4}{t}=\frac{s-6.4}{t+\frac{1}{8}}$

Indeed, the speeds did not change, and due to the simultaneity of the two meetings, the travel times to $M$ in the imagined scenario are $t-1/12$ and $t+1/8$ hours, respectively.

By swapping the sides of (2) and dividing the new equation by (1), each side by the corresponding side:

$$
\frac{t+\frac{1}{8}}{t}=\frac{t}{t-\frac{1}{12}}, \quad \text { from which } \quad t=\frac{1}{4} \text { hours. }
$$

Thus, from (1) $s=16 \mathrm{~km}$, and finally $v_{1}=38.4 \mathrm{~km}/$ hour, $v_{2}=25.6 \mathrm{~km}/$ hour.

Judit Bendl (Sopron, Széchenyi I. g. I. o. t.)

II. solution. We can achieve the goal without writing equations. The distance $MT$ between the actual and imagined meeting points is covered by the car starting from $B$ in $1/8$ hours, since the increase in travel time by $1/8$ hours results in exactly the additional distance $MT$. The car starting from $A$ takes $1/12$ hours to cover the same segment (since it travels in the actual manner instead of the imagined one). The speeds are inversely proportional to the times required to cover the same distance, so the speed ratio of the car starting from $A$ to the car starting from $B$ is $(1/8):(1/12)=3:2$. Accordingly, while the car starting from $B$ (in the actual movement) covers $6.4 \mathrm{~km}$ to $T$, the car starting from $A$ covers $3/2$ times that, $9.6 \mathrm{~km}$, so $AB=9.6+6.4=16 \mathrm{~km}$. Furthermore, $MT=3.2 \mathrm{~km}$, and the two cars cover 12 times and 8 times this distance per hour, respectively: $38.4 \mathrm{~km}-\mathrm{t}$, and $25.6 \mathrm{~km}-\mathrm{t}$.

László Monostory (Budapest, XIX. district, Ady E. úti Ált. Isk., 8. o. t.)