A $d$ meter long shooting range, at which point will the sound of the weapon's firing and the sound of the bullet hitting the target arrive simultaneously, if the bullet's speed is $c \mathrm{~m} / \mathrm{sec}$, and the speed of sound is $s \mathrm{~m} / \mathrm{sec}$? Consider the scenario.

Consider the scenario.
Solution: Let the point $P$ that meets the requirement be $x$ units away from the shooting position and $d-x$ units away from the target. In this case, the sound of the shot reaches $P$ in $\frac{x}{s}$ seconds. The bullet hits the target after $\frac{d}{c}$ seconds, and the sound of this impact arrives at $P$ after an additional $\frac{d-x}{s}$ seconds. Therefore,

$$
\frac{x}{s}=\frac{d}{c}+\frac{d-x}{s}
$$

from which

$$
x=\frac{(c+s) d}{2 c}=\frac{1}{2}\left(1+\frac{s}{c}\right) d
$$

Discussion: The problem only makes sense if $x < d$, i.e., if $\frac{s}{c} \leq 1$; otherwise, if $x > d$, we arrive at a contradiction with $\frac{x}{s}=\frac{d}{c}+\frac{x-d}{s}$. In this case, the sound of the gun always reaches our ears first. Therefore,

$$
0<\frac{s}{c} \leq 1, \quad \text{and thus} \quad \frac{d}{2}<x \leq d
$$

which means that the point $P$ is between the midpoint of the shooting range and the target, and it is closer to the midpoint the smaller the positive ratio $\frac{s}{c}$ (i.e., the larger $c$), and closer to the target the closer the ratio $\frac{s}{c}$ is to 1. When $s=c$, $P$ is at the target. (It is obvious that in this case, the two sounds are heard simultaneously at every point on the extension of the shooting range beyond the target.)

András Szilárd (Bp. II., Rákóczi g. II. o. t.)