Four vehicles start from $A$ at the same time, and arrive at $B$ in succession at equal time intervals. The speeds of the fastest and the slowest vehicle are $v_{1}$ and $v_{4}$, respectively. What are the speeds of the other two vehicles?
If the fastest vehicle covers the distance $AB = s$ in time $t_{1}$, then $t_{1} = \frac{s}{v_{1}}$. The slowest vehicle covers this distance in time $t_{4}$, $t_{4} = \frac{s}{v_{4}}$.

Since the vehicles arrive at $B$ at equal intervals, the time interval $t_{4} - t_{1}$ must be divided into 3 equal parts to find the time between the arrivals of the individual vehicles.

$$
\frac{t_{4} - t_{1}}{3} = \frac{\frac{s}{v_{4}} - \frac{s}{v_{1}}}{3} = \Delta t
$$

The travel time of the 2nd vehicle is therefore

$$
t_{2} = t_{1} + \Delta t = \frac{s}{v_{1}} + \frac{\frac{s}{v_{4}} - \frac{s}{v_{1}}}{3} = \frac{1}{3} \left( \frac{2s}{v_{1}} + \frac{s}{v_{4}} \right)
$$

and its speed is

$$
v_{2} = \frac{s}{t_{2}} = \frac{s}{\frac{1}{3} \left( \frac{2s}{v_{1}} + \frac{s}{v_{4}} \right)} = \frac{3 v_{1} v_{4}}{2 v_{4} + v_{1}}
$$

The travel time of the 3rd vehicle is similarly

$$
t_{3} = t_{4} - \Delta t = \frac{s}{3} \left( \frac{2}{v_{4}} + \frac{1}{v_{1}} \right)
$$

and its speed is

$$
v_{3} = \frac{s}{t_{3}} = \frac{3 v_{1} v_{4}}{v_{4} + 2 v_{1}}
$$

Remark. The problem can be generalized to $n$ vehicles. In this case, the size of the intervals is $\Delta t = \frac{t_{n} - t_{1}}{n-1} = \frac{\frac{s}{v_{n}} - \frac{s}{v_{1}}}{n-1}$, and the speed of the $k$-th vehicle is

$$
v_{k} = \frac{s}{t_{k}} = \frac{(n-1) v_{1} v_{n}}{(n-1) v_{n} + (k-1) v_{1} - (k-1) v_{k}} = \frac{(n-1) v_{1} v_{n}}{(n-k) v_{n} + (k-1) v_{1}}
$$

Béla Gajdos (Beregszász, Bethlen G. Gymnasium, 12th grade)