## Task 4 - 160714

In a cycling race on a circular track of $1 \mathrm{~km}$ in length, at a certain point in time, cyclist $A$ had exactly $500 \mathrm{~m}$ lead over cyclist $B$. $B$ was riding at a speed of $50 \frac{\mathrm{km}}{\mathrm{h}}$, and $A$ at a speed of $45 \frac{\mathrm{km}}{\mathrm{h}}$.

a) After how many minutes from the given time would $B$ catch up to cyclist $A$ for the first time, assuming both rode at a constant speed?

b) After how many more minutes would $B$ catch up to cyclist $A$ for the second time (i.e., "overtake" him), if both cyclists continued to ride at a constant speed? How many laps would $A$ and $B$ have completed between the first and second overtaking?
a) $B$ must catch up a lead of $500 \mathrm{~m}$ within the required time. Within an hour, $B$ would have covered a distance $5 \mathrm{~km}=5000 \mathrm{~m}$ longer than $A$, i.e., a ten times larger lead than necessary. Therefore, he caught up the lead of $A$ in $\frac{1}{10}$ of an hour, i.e., in 6 minutes.

b) By the second time of overtaking, $B$ would have to catch up a lead of $1 \mathrm{~km}$. The time required for this would thus be twice as long as in a), i.e., 12 minutes. In this time, $A$ covered a distance of $\frac{1}{5} \cdot 45 \mathrm{~km}=9 \mathrm{~km}$, i.e., 9 laps, and $B$ therefore one more lap, i.e., 10 laps.

Solutions of the I. Round 1976 adopted from [5]

### 4.18.2 II. Round 1976, Class 7