Antal and Béla set off from Hazul to Cegléd on their motorcycles. From the one-fifth point of the journey, Antal turns back for some reason, which causes him to "switch gears," and he manages to increase his speed by a quarter. He immediately sets off again from Hazul. Béla, traveling alone, reduces his speed by a quarter. They eventually meet again and travel the last part of the journey together at a speed of $48 \mathrm{~km} /$ hour, arriving 10 minutes later than planned. What can we calculate from this?
I. Solution: Due to the turnaround, by the time Béla caught up, Antal Béla had traveled 2/5 more of the entire distance. From the turnaround to the catch-up, the two travelers moved at 5/4 and 3/4 of their original speeds, respectively, or in other words, their speeds, measured in a suitable unit, were 5 and 3, respectively. Thus, from the turnaround to the catch-up, Antal traveled 2/5 more of the distance he himself had covered, compared to Béla. Since this difference is also 2/5 of the distance from home to Cegléd, Antal traveled the same distance from the separation to the meeting as the distance from home to Cegléd. One-fifth of this was the return journey, so the catch-up occurred at 4/5 of the entire distance; and at the time they were originally scheduled to arrive in Cegléd, because in this time Béla would have traveled 4/5 instead of 3/5 of the distance at their original speed, or the entire remaining distance at the moment of their separation.

Therefore, during the 10-minute, 1/6-hour delay, they traveled the last 1/5 of the distance, so its length is 48/6 = 8 km, and the total distance is 5 * 8 = 40 km, which is how far they live from Cegléd. Antal turned back after traveling the first 8 km.

A natural question would be what their original speed was, and what speed they exerted when traveling alone, as well as how long they spent on the journey, and how much later Antal set out from home the second time compared to the first. We cannot calculate this because the above reasoning is valid for any original speed up to the point of catch-up. At most, we can say about Antal's second departure that it happened when they should have already traveled 8 + 8 * 4/5 = 14.4 km.

Imre Soós (Budapest, Rákóczi F. g. II. o. t.)

![](https://cdn.mathpix.com/cropped/2024_05_02_92fa52488c391b7db95eg-1.jpg?height=881&width=485&top_left_y=814&top_left_x=798)

II. Solution: Let the distance from Cegléd to the travelers' home be $s \mathrm{~km}$, the planned speed $v \mathrm{~km} /$ hour, and the planned travel time $t = s / v$ hours. Furthermore, let Béla's distance traveled alone be $s_{1} \mathrm{~km}$, then Antal's distance traveled alone is $s_{1} + 2s / 5$. They traveled these distances at 3v/4 and 5v/4 speeds, respectively, in the same amount of time, so

$$
\frac{s_{1}}{\frac{3 v}{4}} = \frac{s_{1} + \frac{2 s}{5}}{\frac{5 v}{4}}
$$

from which

$$
s_{1} = 3s / 5
$$

and the time spent traveling alone is $s_{1} : 3v / 4 = 4t / 5$ hours. Therefore, the catch-up occurred at a distance of $s / 5 + s_{1} = 4s / 5 \mathrm{~km}$ from home, $t / 5 + 4t / 5 = t$ hours after (the first) departure - that is, exactly when they were supposed to arrive - and at this point, $s / 5 \mathrm{~km}$ of the journey still lay ahead of the travelers. They covered this in the remaining 1/6 hour from the actual travel time of $t + 1/6$ hours.

$$
\frac{s}{5} = 48 \frac{\mathrm{km}}{\text{hour}} \cdot \frac{1}{6} \text{ hour, }
$$

and from this, $s = 40 \mathrm{~km}$.

Zoltán Endreffy (Budapest, I. István Gymn. II. o. t.)

Remarks. 1. The speeds could not be determined because the only data given in absolute units (distance) is for the last segment of the journey. The motion graph cannot connect the segments before and after the catch-up.

2. The second solution is not substantially different from the first. Readers should compare the two to see how the need for setting up equations can be eliminated.