$A$ and $B$ are cyclists who start from a point on a circular track at the same time, heading in opposite directions at speeds of $15 \mathrm{~km} /$ hour and $13 \mathrm{~km} /$ hour, respectively. At the same time, from the same point, $C$ a pedestrian starts walking along the diameter of the track. After 50 seconds (before $A$ and $B$ would have met), $A$, $B$, and $C$ are all equidistant from each other. What is the speed of the pedestrian and what is the diameter of the track?
After $50 \mathrm{mp}$, $A$ reaches point $A_{1}$, and $B$ reaches point $B_{1}$. Meanwhile, $C$ will be on the perpendicular bisector of the chord $A_{1} B_{1}$, as it is equidistant from $A_{1}$ and $B_{1}$. However, the perpendicular bisector of the chord is a diameter of the circle, so $C$ is at the intersection of two diameters, which is the center of the circle. According to the condition, at the same time, $A_{1}$ is also a radius distance away from $B_{1}$. Therefore, in $50 \mathrm{mp}$, $A$ and $B$ together either cover $\frac{1}{6}$ of the circumference or $\frac{5}{6}$ of the circumference. The first case must be excluded because it would imply that the pedestrian is almost as fast as the two cyclists combined. In the second case, denoting the diameter of the circle by $d$ meters, according to the problem,

$$
\frac{5}{6}\left(\frac{13000}{60}+\frac{15000}{60}\right)=\frac{5}{6} d \pi
$$

which simplifies to

$$
\frac{1400}{3}=d \pi
$$

from which

$$
d=\frac{1400}{3 \pi} \approx \frac{1400}{9.42} \approx 148.62 \mathrm{~m}
$$

The pedestrian covers a distance of $\frac{d}{2}=r=74.31 \mathrm{~m}$ in $\frac{5}{6}$ of a minute, so their speed in $\mathrm{km} /$ hour is

$$
60 \cdot \frac{6}{5} \cdot 0.07431=72 \cdot 0.07431 \approx 5.350 \mathrm{~km} / \text { hour }
$$

Ádám Antal (Bp. VIII., Széchenyi g. II. o. t.)