
Solution. We will show that $t$ is equal to $4 / 11$ hours, independent of the ratio $k$ of the lengths of the hands.

![](https://cdn.mathpix.com/cropped/2024_04_17_dea7690a2ad723913678g-24.jpg?height=734&width=820&top_left_y=181&top_left_x=618)

Fig. 2

Denote by $c$ the circumference of the watch, by $S$ its center, and by $M$ the tip of the hour hand (Fig. 2). Let us first explain why for each fixed $M$ there exist precisely two equilateral triangles $M X Y$ with vertices $X, Y$ lying on the circle $c$. Since the line $S M$ must be the axis of the chord $X Y$, hence, also the bisector of the angle $X M Y$, both lines $M X, M Y$ meet the line $S M$ at the same angle of $30^{\circ}$. Thus the triangle $M X Y$ coincides with one of the equilateral triangles $M V_{1} V_{2}, M V_{3} V_{4}$ shown on Fig. 2 .

The points $V_{i}$ divide the circle $c$ into four arcs. At points on the arc $V_{2} V_{3}$, the segment $V_{2} V_{3}$ subtends an angle of the same size as $\angle V_{2} V_{4} V_{3}$, i.e. $60^{\circ}$; similarly for the $\operatorname{arc} V_{4} V_{1}$. Thus

$$
\angle V_{2} S V_{3}=\angle V_{4} S V_{1}=120^{\circ} \text {. }
$$

Consequently,

$$
\angle V_{1} S V_{2}+\angle V_{3} S V_{4}=120^{\circ}
$$

It follows that both angles $V_{1} S V_{2}$ and $V_{3} S V_{4}$ are smaller than $120^{\circ}$.

We may visualize the situation by thinking of the hour hand as being motionless and the minute hand rotating around the circle $S$ at the angular speed of $(360-30)^{\circ}=$ $330^{\circ}$ per hour. We have seen that the phenomenon we are interested in occurs precisely when the tip $V$ of the minute hand coincides with one of the four points $V_{i}$. Between two consecutive occurrences of the phenomenon, the minute hand thus advances by an angle of either $120^{\circ}$, or $\angle V_{1} S V_{2}$ or $\angle V_{3} S V_{4}$ which are both less than $120^{\circ}$ (and depend on the ratio $k$ ). The greatest possible period $t$ is thus independent of the ratio $k$ and equals $120 / 330$ hours.


5. At some moment, Charles noticed that the tip of the minute hand, the tip of the hour hand, and a suitable point of the circumference of his watch formed three vertices of an equilateral triangle. A period $t$ of time elapsed before the next occurrence of this phenomenon. Find the greatest possible $t$ for a given watch, given the ratio $k$ of the lengths of the two hands $(k>1)$, provided that the radius of the circumference of the watch is equal to the length of the minute hand.
