Ambiguous Times

Theorem
Let $T$ be a time of day in $12$-hour clock form.

Then $T$ is an ambiguous time :


 * $T = 12:00 + n \times 5 \tfrac 5 {143} \mathrm {min}$

where:
 * $n \in \set {1, 2, \ldots, 142}$
 * the hour hand and minute hand are pointing in different directions.

Proof
Let $T$ be an ambiguous time.

Let $T$ be specified in hours $h$ and minutes $m$, where:
 * $1 \le h \le 12$ is an integer
 * $0 \le m < 60$ is a real number

whether a.m. or p.m. is immaterial.

At this time $T$:


 * let $\theta \degrees$ be the angle made by the minute hand with respect to twelve o'clock


 * let $\phi \degrees$ be the angle made by the hour hand with respect to twelve o'clock.

$\theta$ and $\phi$ give a valid time indication by definition.

From Condition for Valid Time Indication:


 * $12 \phi \mod 360 = \theta$

By definition of ambiguous time, we have that if the hour hand and minute hand were reversed, the time of day indicated is also a valid time indication.

That is:
 * $12 \theta \mod 360 = \phi$

Hence:

It remains to calculate $\dfrac {360 k} {143}$ and convert the number each time into a valid time indication:

and so forth:

These are the ambiguous times corresponding to the times of day between $12:00$ and $01:00$.

It remains to calculate:
 * $(1) \quad$ the angles made by the minute hand at these times


 * $(2) \quad$ work out the times of day corresponding to the hour hand at these angles.

Let us take the ambiguous time $12 : 05 \tfrac 5 {143}$.

The hour hand is at $\dfrac {360} {143} \degrees$ while the minute hand is at $\dfrac {12 \times 360} {143} \degrees = 30 \tfrac {30} {143} \degrees$.

Exchanging the hour hand and minute hand gives us a time soon after $1:00$ where the minute hand is at $\dfrac {360} {143} \degrees$.

From Speed of Minute Hand, the minute hand moves at $6 \degrees$ per minute.

This leads us to a time of $\dfrac {360} {6 \times 143}$ minutes after $1:00$, or $\dfrac {60} {143}$ minutes after $1:00$.

To check our arithmetic, we investigate the hour hand position at $01 : 00 \tfrac {60} {143}$

A time of $\dfrac {60} {143}$ minutes past the hour corresponds to an angle of $\dfrac 1 2 \times \dfrac {60} {143} \degrees$, or $\dfrac {30} {143} \degrees$ past $30 \degrees$.

This agrees with our calculation of the position of the minute hand.

The remaining corresponding times are evaluated as multiples of this time.