At 12 o'clock, the three hands of the clock meet. The question is, how many minutes later will the second hand bisect the angle formed by the hour and minute hands?
The speeds of the second, minute, and hour hands are in the ratio of $60: 1: \frac{1}{12}$; if, therefore, the second hand halves the desired angle after $x$ minutes, then the distances traveled by the hands are: $60x, x, \frac{x}{12}$; the angle between the minute and hour hands is thus $x - \frac{x}{12} = \frac{11}{12} x$. Since the second hand only reaches the desired position after one complete revolution, therefore:

$$
60 x = 60 + \frac{x}{2} + \frac{11 x}{24}
$$

from which

$$
x = 1 \frac{13}{1427} \text{ minutes } 0.54 \text{ seconds. }
$$

(János Boros, Kolozsvár.)

The problem was also solved by: Breuer M., Freibauer E., Glass M., Kárf J., Kohn B., Kornis Ö., Krausz B., Krisztián Gy., Pollák N., Sasvári G., Sasvári J., Steiner M., Steiner S., Weigl G., Weisz J.