## Task 4 - 330834

On a route $A B$, a cyclist $X$ travels from $A$ to $B$, and a second cyclist $Y$ travels from $B$ to $A$. Both start at the same time. Upon reaching $B$ or $A$, they immediately turn around and travel the same route back to $A$ or $B$, respectively, and then end their journey. It is assumed that each cyclist maintains a constant speed and that the time used for turning around can be neglected.

On the outbound trip, they met 30 minutes after the start at a point that is $7.5 \mathrm{~km}$ from $A$. Another 30 minutes later, the cyclists were together again at a point between $A$ and $B$.

Determine all possibilities for how, according to this description,

a) the length of the route $A B$,

b) the speeds of the cyclists $X$ and $Y$ can be!
Given that $X$ traveled a distance of $11 \mathrm{~km}$ in the time $\frac{1}{2} \mathrm{~h}$ until the first meeting, $X$ drove at a speed of $15 \frac{\mathrm{km}}{\mathrm{h}}$. If $X$ and $Y$ had met for the second time while driving towards each other, $X$ would have had to reach $B$ and change direction between the first and second meetings, meaning the distance from the first meeting point to $B$ would be shorter than $12 \mathrm{~km}$.

Therefore, $Y$ would have a lower speed than $X$ and would not have even reached $A$ in the second half hour, let alone turned around and met $X$.

Therefore, there are only two possibilities:

I. In the time $\frac{1}{2} \mathrm{~h}$ from the first to the second meeting, $X$ traveled another $\frac{15}{2} \mathrm{~km}$ in the same direction and was caught up by $Y$ (who had previously reached $A$ and changed direction).

This means: $Y$ had to drive a distance three times as long in this time and thus had a speed of $45 \frac{\mathrm{km}}{\mathrm{h}}$.

Furthermore, it follows that: from the start to the first meeting, $Y$ had traveled a distance three times as long as $X$. Therefore, the total length of $A B$ is $4 \cdot \frac{15}{2} \mathrm{~km} = 30 \mathrm{~km}$.

II. Between the first and second meetings, $X$ arrived at $B$, changed direction, and then caught up with $Y$ at the second meeting.

The sequence of meetings is the same as in Case I, but with the roles of $X$ and $Y$ reversed. Thus, $X$ now has a speed three times that of $Y$, i.e., $Y$ has a speed of $5 \frac{\mathrm{km}}{\mathrm{h}}$. Accordingly, in Case II, the length of $A B$ is one-third of the value in Case I; i.e., in Case II, $A B$ has a length of $10 \mathrm{~km}$.