## Task 3 - 090823

a) How many times in total do the hour and minute hands of a clock stand perpendicular to each other over the course of 24 hours (from 0:00 to 24:00)?

b) In particular, calculate all such times between 4:00 and 5:00!
(I) In the period from one overlap of the two hands until the next, the angle between the hands (measured clockwise from the hour hand to the minute hand) takes on all values from $0^{\circ}$ to $360^{\circ}$, each exactly once. Among these hand positions, there are exactly two of the sought ones, namely those with the angles $90^{\circ}$ and $270^{\circ}$.

(II) In the time from 0 o'clock to 12 o'clock, the minute hand completes exactly 12 full revolutions, and the hour hand exactly one in the same direction.

Therefore, this time period is divided by the times of the overlap of the two hands into 11 parts (and because of the uniformity of the movements, into equal lengths).

(III) From (I) and (II) it follows initially: The time from 0 o'clock to 24 o'clock is divided by the times of the overlap into 22 parts, and in each of these time intervals, there are exactly 2 of the sought times.

The total number of these is thus 44.

(IV) From (I), (II), and the uniformity of the movements, it further follows: The time from 0 o'clock to 12 o'clock is divided, by the times corresponding to the angles $0^{\circ}, 90^{\circ}, 180^{\circ}, 270^{\circ}$, into 44 equal parts.

These times thus result as the first 43 positive integer multiples of $\frac{12}{44} \mathrm{~h}=\frac{3}{11} \mathrm{~h}$. The times with the angles $90^{\circ}$ and $270^{\circ}$ are the odd-numbered among these multiples. Of these, those $n \cdot \frac{3}{11} \mathrm{~h}$ ( $n$ odd) that lie between 4:00 and 5:00 are exactly those for which $n$ satisfies the inequality $4<n \cdot \frac{3}{11}<5$ or equivalently $\frac{44}{3}<n<\frac{55}{3}$, which are the times

$$
15 \cdot \frac{11}{3} \mathrm{~h}=4 \mathrm{~h} 5 \frac{5}{11} \min \quad ; \quad 17 \cdot \frac{11}{3} \mathrm{~h}=4 \mathrm{~h} 38 \frac{2}{11} \min
$$