\section*{Problem 3 - V01203}

A clock, equipped with a synchronous motor, has ideally uniformly moving hands.

Determine exactly the times at which the hands are positioned such that one hour later, the angle between the hands is of the same size!

(Hint: The relevant angles are less than \(180^{\circ}\).)
}

Let the angle in the clockwise direction from 12 be \(\sigma(t)\) for the hour hand and \(\mu(t)\) for the minute hand. At a given starting time \(t=0\), the positions of the hands are \(\sigma_{0}\) and \(\mu_{0}\). The angle equations for the hands are then:

\[
\sigma(t)=\sigma_{0}+30^{\circ} \cdot t \quad ; \quad \mu(t)=\mu_{0}+360^{\circ} \cdot t
\]

Here, \(t\) is given in hours. Generally, one can state:

\[
|\mu(t)-\sigma(t)| \quad \bmod 360^{\circ}=\left|\mu_{0}-\sigma_{0}\right| \quad \bmod 360^{\circ}
\]

According to the problem statement, exactly one hour later, the same angle between the hands should be present again. This means:

\[
\frac{12}{11}\left(z-\frac{\mu_{0}-\sigma_{0}}{180^{\circ}}\right)=1
\]

Solving for \(\mu_{0}-\sigma_{0}\):

\[
\begin{gathered}
\mu_{0}-\sigma_{0}=180^{\circ}\left(z-\frac{11}{12}\right) \\
\mu_{0}-\sigma_{0}=180^{\circ} \cdot z-165^{\circ}
\end{gathered}
\]

Since \(z=2\) essentially gives the same result as \(z=0\), it follows that either:

\[
\mu_{0}=\sigma_{0}-165^{\circ} \quad \text { or } \quad \mu_{0}=\sigma_{0}+15^{\circ}
\]

The minute hand must therefore originally be either \(165^{\circ}\) behind the hour hand, or \(15^{\circ}\) ahead. This occurs every \(\frac{12}{11}\). The former would be the case, for example, at 11:30:00, but not only, as it would also be the case at a starting time of 12:35:27, etc. 15 \(5^{\circ}\) is the minute hand ahead of the hour hand at 05:30:00.

The same applies, i.e., other starting times would be, for example, 06:35:27, 07:40:55, and so on.