An express train overtakes a freight train. The speed of the express is as many times greater than that of the freight train as the time it takes for them to pass each other side by side is greater than the time it would take if they were passing each other head-on. What is this ratio?
Let the speeds of the freight and express trains be arbitrary but in the same unit $c$, and $x c$ where $x$ is the sought ratio - and let the sum of the lengths of the two trains be $h$ (in the length unit used for the speeds). Whether the two trains are moving in the same direction or opposite to each other, the passing each other lasts until the end of the train that was together at the meeting point is $h$ distance away from the other. The time $t_{1}$ for this (in the time unit used for the speeds) in the case of opposite direction is from the equation $c t_{1}+x c t_{1}=h$ we get $t_{1}=h /(x c+c)$, because the locomotives are $ct$ and $xct$ away from the meeting point in opposite directions. In the case of moving in the same direction, the locomotive of the express train (and every point of it) travels $h$ more distance from the overtaking point during the passing time $t_{2}$ (in the same unit as $t_{1}$) than the end point of the freight train, so from $x c t_{2}-c t_{2}=h$ we get $t_{2}=h /(x c-x)$. According to the problem,

$$
x=\frac{t_{2}}{t_{1}}=\frac{h}{(x-1) c}: \frac{h}{(x+1) c}=\frac{x+1}{x-1}
$$

and from this - since $x-1 \neq 0-$

$$
\begin{gathered}
x^{2}-2 x-1=0 \\
x_{1}=1+\sqrt{2}, \quad x_{2}=1-\sqrt{2}
\end{gathered}
$$

The solution $x_{2}$ does not fit the content of the problem, so the sought ratio is $1+\sqrt{2} \approx 2.414$.

György Baráti (Budapest, I. István Gym. I. o. t.)

Remarks. 1. Our thought can also be expressed by relying on the kinematic concept of relative motion: examining the phenomenon in the coordinate system of the freight train, the relative speed of the express train in the case of opposite and same direction motion is $x c+c$ and $x c-c$, therefore the times required to travel the distance $h$ are the above $t_{1}$ and $t_{2}$.

2. The opposite or the same direction of the trains' motion can also be expressed by the sign assigned to the two speeds instead of words. Thus, the negative root also makes sense, but does not bring anything essentially new. It only swaps the cases of moving in the same direction and opposite direction, and the express train becomes slower than the freight train.