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:

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:

which is a full rotation.

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