The figure below represents 4 circular bicycle circuits. The four cyclists start at noon and travel different circles, one at a speed of $6 \mathrm{~km}$ per hour, another at a speed of $9 \mathrm{~km}$ per hour, another at a speed of $12 \mathrm{~km}$ per hour, and finally, the fourth at a speed of $15 \mathrm{~km}$ per hour. They agree to cycle until they all meet simultaneously at the center of the figure, which is where they initially started. The perimeter of each circumference is exactly one third of a kilometer. When will they meet again for the fourth time?

![](https://cdn.mathpix.com/cropped/2024_05_01_20684006194a352111ccg-11.jpg?height=417&width=420&top_left_y=905&top_left_x=915)

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Solution

If cyclists $A, B, C$ and $D$ have speeds of $6 \mathrm{~km} / \mathrm{h}, 9 \mathrm{~km} / \mathrm{h}, 12 \mathrm{~km} / \mathrm{h}$ and $15 \mathrm{~km} / \mathrm{h}$, respectively, we have the following table with the times they will spend to complete one full lap:

Times in hours to complete one lap

| Cyclist A | $\frac{1}{3} \div 6=\frac{1}{18} h$ |
| :--- | :--- |
| Cyclist B | $\frac{1}{3} \div 9=\frac{1}{27} h$ |
| Cyclist C | $\frac{1}{3} \div 12=\frac{1}{36} h$ |
| Cyclist D | $\frac{1}{3} \div 15=\frac{1}{45} h$ |

Since the speed of cyclist $C$ is double the speed of cyclist $A$, for every two laps that cyclist $C$ completes, cyclist $A$ completes only one. Thus, they only meet after an even number of laps by cyclist $C$. After $2 \cdot \frac{1}{36}=\frac{1}{18} h$, cyclists $A$ and $C$ meet at the center, but cyclists $B$ and $C$ are not there. The next possible meeting between $A$ and $C$ is after $4 \cdot \frac{1}{36}=\frac{1}{9} h$. For this value of elapsed time, cyclist $B$ will have completed exactly $\frac{1}{27} \div \frac{1}{9}=3$ laps and cyclist $D$ will have completed exactly $\frac{1}{45} \div \frac{1}{9}=5$ laps, meaning all will meet for the first time. Note that the next meeting will be after exactly $\frac{1}{9} h$, as from the first meeting point, everything practically happens as if all were starting the race again. Repeating the reasoning for the third and fourth meetings, we find that the fourth meeting will be after $4 \cdot \frac{1}{9} h=\frac{80}{3} m=26 m 40$ s.