Vendelín lives between two bus stops, at a distance of three eighths of their distance. Today, he set out from home and found that whether he ran to one or the other stop, he would arrive at the stop at the same time as the bus. The average speed of the bus is $60 \mathrm{~km} / \mathrm{h}$.

What is the average speed at which Vendelín is running today?

(L. Hozová)

Hint. Sketch the situation on a line and consider both cases separately.
Let the first bus stop on the route be $A$, the second $B$, and Vendelín's house $D$. If Vendelín runs to $A$, he will cover the distance $D A$ before the bus arrives. If Vendelín runs to $B$, by the time the bus reaches $A$, he will have covered the same distance $D A$ towards $B$. At this point, we realize that $D$ must be closer to $A$, and specifically, it is at three-eighths of the distance $A B$:
![](https://cdn.mathpix.com/cropped/2024_04_17_6aef7ef60e9f55f815aeg-4.jpg?height=454&width=1308&top_left_y=186&top_left_x=380)

Thus, when running to $B$, at the moment the bus arrives at $A$, Vendelín will still be $\frac{2}{8} = \frac{1}{4}$ of the distance $A B$ away from his destination. To arrive at $B$ at the same time as the bus, he must have a speed that is one-fourth of the bus's speed. Vendelín runs at an average speed of $60: 4 = 15(\mathrm{~km} / \mathrm{h})$.