$A$ and $B$ run a race from $M$ to $N$ and back. $A$ reaches $N$ faster than $B$, and on the way back, $A$ meets $B$ $100 m$ before $N$. $A$ returns to $M$ 4 minutes earlier than $B$. If $A$ were to turn around from $M$ again, they would meet $B$ at $\frac{1}{5}$ of the $M N$ distance. What is the distance of $M N$, and how long does the race take?
I. solution. When they first met, $A$ traveled $x+100 \mathrm{~m}-\mathrm{t}, B$ traveled $x-100 \mathrm{~m}-\mathrm{t} . ~ A$ thus traveled $200 \mathrm{~m}-\mathrm{rel}$ more than $B$. By the time of their second meeting, $B$ would have a $400 \mathrm{~m}$ lead; thus, if the length of the $M N$ route is $x$, then

$$
\frac{2}{5} \cdot x=400, \text { from which } x=1000 \mathrm{~m}
$$

When $A$ returns to $M$ again, $B$ is still $363.64 \mathrm{~m}$ away from $M$; $B$ covers this distance in 4 minutes and thus covers the $2000 \mathrm{~m}$ distance in 22 minutes. $A$ covers the entire distance in 18 minutes.

(Hirsch Jenő, Szamosújvár.)

II. solution. Since the ratio of distances traveled in equal times is constant, therefore:

$$
(x+100):(x-100)=\left(2 x+\frac{x}{5}\right):\left(2 x-\frac{x}{5}\right)
$$

And since the sum of the first two terms is in the same ratio to their difference as the sum of the third and fourth terms is to their difference, therefore:

$$
2 x: 200=4 x: \frac{2 x}{5}, \text { from which } x=1000
$$

(Krámer Gusztáv, Budapest.)

III. solution. If $A$ takes $y$ minutes to cover the entire $2 x$ distance, then $A$'s speed is $\frac{2 x}{y}, B$'s speed is $\frac{2 x}{y+4}$. And since they cover the $x+100$ and $x-100$ distances in equal times, therefore $\left(t=\frac{s}{c}\right)$ :

$$
(x+100): \frac{2 x}{y}=(x-100): \frac{2 x}{y+4}, \quad \text { or } \quad(x+100) y=(x-100)(y+4)
$$

Similarly

$$
2 \frac{1}{5} x \cdot y=1 \frac{4}{5} x(y+4)
$$

from which equations again:

$$
x=1000 \mathrm{~m}, y=18 \mathrm{m}
$$

(Deutsch Imre, Györ.)

The problem was solved by: Haar A., Ligeti P, Popoviciu M., Raab R., Schwarz G., Söpkéz Gy., Szávay Z., Vámossy L.