From a certain community $C$, a car departs at 7:58 AM at a speed of $60 \mathrm{~km} /$ hour heading east on a straight national road towards community $B$. At 8:00 AM, a cyclist departs from the same place heading north on a national road at a speed of 18 $\mathrm{km} /$ hour towards community $A$, which is 10 $\mathrm{~km}$ away. A pedestrian had already departed from $A$ at 6:44 AM on the straight national road $AB$, which forms a $60^{\circ}$ angle with the national road $AC$, heading towards $B$ at a speed of $6 \mathrm{~km} /$ hour. When will the cyclist be at an equal distance from the pedestrian and the car by air distance?
Let the sought time be $8+x$ hours and let the car be at point $A_{1}$, the cyclist at point $B_{1}$, and the pedestrian at point $C_{1}$ (see Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_e9b01bb9f0024f0d8033g-1.jpg?height=309&width=452&top_left_y=236&top_left_x=823)

According to the problem, in kilometers:

$$
\begin{gathered}
C B_{1}=18 x, \quad A B_{1}=10-18 x, \quad C A_{1}=2+60 x \\
A C_{1}=6 \cdot \frac{76}{60}+6 x=7.6+6 x
\end{gathered}
$$

By the Pythagorean theorem,

$$
B_{1} A_{1}^{2}=C B_{1}^{2}+C A_{1}^{2}=324 x^{2}+(2+60 x)^{2}
$$

By the cosine rule,

$$
\begin{gathered}
B_{1} C_{1}^{2}=A B_{1}^{2}+A C_{1}^{2}-2 A B_{1} \cdot A C_{1} \cos 60^{\circ}= \\
=(10-18 x)^{2}+(7.6+6 x)^{2}-(10-18 x)(7.6+6 x)
\end{gathered}
$$

According to the problem, the left sides of (1) and (2) are equal, so the right sides are also equal, that is,

$$
324 x^{2}+(2+60 x)^{2}=(10-18)^{2}+(7.6+6 x)^{2}-(10-18 x)(7.6+6 x)
$$

Expanding the parentheses, rearranging, and simplifying by 432,

$$
8 x^{2}+x-0.18=0
$$

from which (ignoring the negative root),

$$
x=\frac{1}{10} \text { hour, }
$$

which means the sought time is 8 hours and 6 minutes.

János Borbély (Pápa, Türr István g. III. o. t.)