Time when Hour Hand and Minute Hand at Right Angle

Theorem

Let the time of day be such that the hour hand and minute hand are at a right angle to each other.

Then the time happens $22$ times in every $12$ hour period:

when the minute hand is $15$ minutes ahead of the hour hand

when the minute hand is $15$ minutes behind of the hour hand.

In the first case, this happens at $09:00$ and every $1$ hour, $5$ minutes and $27 . \dot 2 \dot 7$ seconds after

In the second case, this happens at $03:00$ and every $1$ hour, $5$ minutes and $27 . \dot 2 \dot 7$ seconds after.

Thus the times are, to the nearest second:

$\begin {array} 09:00:00 & 03:00:00 \\ 10:05:27 & 04:05:27 \\ 11:10:55 & 05:10:55 \\ 12:16:22 & 06:16:22 \\ 13:21:49 & 07:21:49 \\ 14:27:16 & 08:27:16 \\ 15:32:44 & 09:32:44 \\ 16:38:11 & 10:38:11 \\ 17:43:38 & 11:43:38 \\ 18:49:05 & 12:49:05 \\ 19:54:33 & 13:54:33 \\ \end{array}$

and times $12$ hours different.

Proof

Obviously the hands are at right angles at $3$ and $9$ o'clock.

Thus we only need to show that the angle between the hands will be the same after every $1$ hour, $5$ minutes and $27 . \dot 2 \dot 7$ seconds.

Note that:

\(\ds 1 h \ 5 m \ 27. \dot 2 \dot 7 s\)

\(=\)

\(\ds 65 m \ 27 \tfrac {27}{99} s\)

\(\ds \)

\(=\)

\(\ds 65 m \ 27 \tfrac 3 {11} s\)

\(\ds \)

\(=\)

\(\ds 65 m \ \frac {300} {11} s\)

\(\ds \)

\(=\)

\(\ds 65 m + \frac 5 {11} m\)

\(\ds \)

\(=\)

\(\ds \frac {720} {11} m\)

\(\ds \)

\(=\)

\(\ds \frac {12} {11} h\)

In $\dfrac {12} {11}$ hours:

The minute hand has rotated $\dfrac {12} {11} \times 360^\circ$

The hour hand has rotated $\dfrac {12} {11} \times 30^\circ$

Thus the angle between the hands has changed by:

\(\ds \frac {12} {11} \times 360^\circ - \frac {12} {11} \times 30^\circ\)

\(=\)

\(\ds \frac {12} {11} \times 330^\circ\)

\(\ds \)

\(=\)

\(\ds 360^\circ\)

which is a full rotation.

Hence after $\dfrac {12} {11}$ hours the angle between the hands would remain unchanged.

$\blacksquare$