The hour hand, the minute hand, and the second hand rotate around a common axis. Let's determine the times when the second hand bisects the angle formed by the hour hand and the minute hand.
I. solution. It is appropriate to choose the minute markings on the clock face as the unit of angle measurement, that is, the 60th part of a full revolution.

Two hands divide the clock face into two sectors, which are bisected by two half-lines originating from the same straight line. In the case of the hands overlapping, one sector covers the entire clock face, while the other passes through a radius of the circle; in this case, the straight line can be considered the angle bisector.

Starting from a situation $H_{1}$ that meets the conditions, let's determine the time (measured in seconds) it takes for the hands to reach the next suitable $H_{2}$ situation. The second hand moves $t$ markings in $t$ seconds; in the $H_{2}$ situation, the bisector is covered by the other half-line, as it was in the $H_{1}$ situation. We must also take into account that the position of the bisector has changed in the meantime. Examining the arc that extends from the hour hand to the minute hand in the direction of the hands' movement, the hour hand advances $\frac{5}{60 \cdot 60} \cdot t=\frac{t}{720}$ markings in $t$ seconds, and the minute hand advances $\frac{t}{60}$ markings.

![](https://cdn.mathpix.com/cropped/2024_05_02_97c4a084282e0b2337dcg-1.jpg?height=691&width=789&top_left_y=679&top_left_x=646)

If in the $H_{1}$ situation the hour hand is at marking $b_{0}$ and the minute hand is at $b_{p}$, then the position of the bisector was $\left(b_{0}+b_{p}\right) / 2$, and after $t$ seconds it is $\left(b_{0}+t / 720+b_{p}+t / 60\right) / 2=\left(b_{0}+b_{p}\right) / 2+13 t / 1440$, so the bisector has moved 13t/1440 markings. The fact that after $i$ time the second hand is covered by the half-line that was opposite to the one covered by the bisector in the $H_{1}$ situation is expressed by the equation

$$
i=\frac{13 i}{1440}+30
$$

From this, $i=\frac{1440}{1427} \cdot 30=30+\frac{390}{1427} \text{ seconds}$. This does not include the data characterizing the $H_{1}$ situation, so we have obtained that between any two consecutive situations that meet the conditions, the time $i$ elapses.

The position of the three hands at 0 hours 0 minutes 0 seconds satisfies the requirement, so all suitable time points are multiples of $i$:

$$
\begin{aligned}
& t=k \cdot\left(30+\frac{390}{1427}\right) \text{ seconds}= \\
& =\frac{43200 k}{1427} \text{ seconds}=\frac{720 k}{1427} \text{ minutes }
\end{aligned}
$$

where $k$ is an integer. If $k=1427$, then $t=720$ minutes $=12$ hours. And from this point on, the hand positions repeat; for $k=0,1,2, \ldots, 1426$ values, however, there are all different hand positions, since the hour hand makes one full revolution, so it is always in a different position.

II. solution. The time that elapses between two consecutive suitable situations can also be calculated as follows: We are looking for the time points of those situations in which the bisector line and the second hand line overlap. This happens as many times in 12 hours as the latter returns to its initial position more often than the former, considering that at the end of 12 hours, both lines return to their common initial position.

The second hand line returns to its original position every half minute, so 12 hours is $12 \cdot 60 \cdot 2=1440$ times. The bisector line, as we have seen, advances $13 / 1440$ markings per second, so in 12 hours it covers 12 . $60 \cdot 60 \cdot 13 / 1440=30 \cdot 13$ markings. During this time, it overlaps with its original position 13 times, which happens after 30 markings of rotation. The second hand line, therefore, meets the bisector line 1440-13=1427 times in 12 hours, and thus between two consecutive meetings, $12 \cdot 60 \cdot 60 / 1427=30+390 / 1427$ seconds elapse.[^0]


[^0]:    ${ }^{1}$ The 4th figure shows the rotations magnified.