A pedestrian and a cyclist set off at 8:00 AM for a city 12 km away. The pedestrian's speed is 6 km/h, and the cyclist's speed is 18 km/h. The cyclist waits for 20 minutes in the city and then heads back. Where and when will they meet? (A graphical solution - if the construction is accurate and the units are large enough - will be accepted as a full solution.)





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I. solution. The cyclist arrives in the city at 8:40 and leaves at 9:00. By then, the pedestrian has covered $6 \mathrm{~km}$, so the distance from the city is also $6 \mathrm{~km}$. This segment of the journey is covered by both until they meet. Their speed ratio is $1:3$, so the pedestrian covers the ${ }^{1} / 4$ part of the segment, and the cyclist covers the ${ }^{3} / 4$ part. The latter part is $4.5 \mathrm{~km}$, and they meet at this distance from the city. It takes the cyclist $1 / 4$ hour to cover this distance, so the meeting occurs at 9:15.

Baranyai Emóke (Miskolc, Herman O. lg. I. o. t.)

Note. We can obtain our results in reverse order by the following reasoning. Starting from 9:00, the travelers approach each other at a speed of $24 \mathrm{~km} /$ hour, so it takes ${ }^{1} / 4$ hour for the initial $6 \mathrm{~km}$ distance to reduce to 0. In this time, the cyclist reaches $4.5 \mathrm{~km}$ from the city.

Botár Borbála (Budapest, Fazekas M. gyak. g. II. o. t.)

II. solution. The two travelers together cover twice the distance from their homes to the city, which is $24 \mathrm{~km}$. Denoting the number of hours elapsed by $t$, the lengths of their individual journeys are $6 t$ and $18\left(t-{ }^{1} / 3\right)$, because the cyclist spent ${ }^{1} / 3$ hour stationary. From the equation

$$
6 t+18\left(t-\frac{1}{3}\right)=24
$$

we get $t={ }^{5} / 4$, and the meeting place is $6 t=7.5 \mathrm{~km}$ from their common starting point.

Pákh Gábor (Esztergom, I. István g. I. o. t.)

Note. Our equation $6 t+18 t=24+6$ can be interpreted as follows: if the cyclist had moved instead of spending 20 minutes stationary in the city, they would have covered an additional $6 \mathrm{~km}$, so together they would have covered $30 \mathrm{~km}$ until the meeting.

![](https://cdn.mathpix.com/cropped/2024_05_02_c0a6ff4cddc8fe9297cfg-1.jpg?height=450&width=577&top_left_y=1084&top_left_x=774)

III. solution. Plotting the graphs of the two travelers' movements on the same coordinate system, the meeting is represented by the intersection point $M$ of the two graphs, which is the point to which the same time and the same location correspond. The coordinates of $M$ are 9:15 and $7.5 \mathrm{~km}$ from the common starting point, as shown in the figure.

Bakos Ferenc (Miskolc, Villamosip. t. II. o. t.)