Theorem: Ambiguous Times

Let $T$ be a time of day in $12$-hour clock form.
Then $T$ is an ambiguous time {{iff}}:
:$T = 12:00 + n \times 5 \tfrac 5 {143} \mathrm {min}$
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:
{{begin-eqn}}
{{eqn | l = \phi
      | r = 12 \theta \mod 360
      | c = 
}}
{{eqn | l = \phi
      | r = 12 \theta + 360 {h_1}
      | c = for $h_1 \in \set {1, 2, \ldots, 12}$
}}
{{eqn | ll= \leadsto
      | l = \dfrac \phi {12} + 30 {h_1}
      | r = \theta
}}
{{eqn | ll= \leadsto
      | l = 12 \phi + 30 h_2
      | r = \dfrac \phi {12} + 30 {h_1}
      | c = for $h_2 \in \set {1, 2, \ldots, 12}$
}}
{{eqn | ll= \leadsto
      | l = 143 \phi
      | r = 360 \paren {h_1 - h_2}
      | c = for $h_1, h_2 \in \set {1, 2, \ldots, 12}$
}}
{{eqn | ll= \leadsto
      | l = \phi
      | r = \dfrac {360 \paren {h_1 - h_2} } {143}
      | c = 
}}
{{end-eqn}}
It remains to calculate $\dfrac {360 k} {143}$ and convert the number each time into a valid time indication:
{{begin-eqn}}
{{eqn | q = h_2 - h_1 = 0
      | l = 143 \phi
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \phi
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = T
      | r = 12 : 00
      | c = but of course $12:00$ the hands coincide, so this does not count
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | q = h_2 - h_1 = 1
      | l = \phi
      | r = \dfrac {360} {143}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = T
      | r = 12 : 05 \tfrac 5 {143}
      | c = as $\phi \degrees$ equates to $2 \phi$ minutes of time
}}
{{end-eqn}}
and so forth:
{{begin-eqn}}
{{eqn | q = h_2 - h_1 = 2
      | l = T
      | r = 12 : 10 \tfrac {10} {143}
      | c = 
}}
{{eqn | q = h_2 - h_1 = 3
      | l = T
      | r = 12 : 15 \tfrac {15} {143}
      | c = 
}}
{{eqn | q = h_2 - h_1 = 4
      | l = T
      | r = 12 : 20 \tfrac {20} {143}
      | c = 
}}
{{eqn | q = h_2 - h_1 = 5
      | l = T
      | r = 12 : 25 \tfrac {25} {143}
      | c = 
}}
{{eqn | q = h_2 - h_1 = 6
      | l = T
      | r = 12 : 30 \tfrac {30} {143}
      | c = 
}}
{{eqn | q = h_2 - h_1 = 7
      | l = T
      | r = 12 : 35 \tfrac {35} {143}
      | c = 
}}
{{eqn | q = h_2 - h_1 = 8
      | l = T
      | r = 12 : 40 \tfrac {40} {143}
      | c = 
}}
{{eqn | q = h_2 - h_1 = 9
      | l = T
      | r = 12 : 45 \tfrac {45} {143}
      | c = 
}}
{{eqn | q = h_2 - h_1 = 10
      | l = T
      | r = 12 : 50 \tfrac {50} {143}
      | c = 
}}
{{eqn | q = h_2 - h_1 = 11
      | l = T
      | r = 12 : 55 \tfrac {55} {143}
      | c = 
}}
{{end-eqn}}
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.
{{qed}}