Ok here is how i tried.
Angle of hands given as a funtion of time:
\theta_x = \frac{2\pi}{60} t
\theta_y = \frac {2 \pi} {3600} t
\theta_z = \frac {2 \pi} {43200} t
where x,y,z denote second minute and hour hands respectively.
The relationship between the angles between the hands can be given as so:
\theta_x,y = \theta_x - \theta_y = 2 \pi t \frac{59} {3600}
and similarly for x,z and y,z pairs.
\theta_x,z = \theta_x - \theta_z = 2 \pi t \frac{719} {43200}
\theta_y,z = \theta_y - \theta_z = 2 \pi t \frac{11} {43200}
It is clear that these differences must be an integer multiple of \frac{2}{3} \pi if they are to satisfy the condition set out in the original post. So take the y,z pair equation and place that equal to \frac{2}{3} \pi as so.
2 \pi t \frac{11}{43200} = n \frac{2} {3} \pi
This gives the time as (if we assume t=0 is exactly 12:00),
t = n \frac{43200}{33} = n\times 1309.091s
Now every third integer will correspond to when the minute and hour hand coincide so we neglect them. What about the second hand? We can find its angle wrt the minute hand via the x,y angle equation above by entering the times we get when multiplying by the integer. The angle we obtain should be a multiple of \frac{2}{3} \pi.
m = 3 t \frac{59}{3600}
And after fiddling about in excell the only time m is an integer is when the second and minute hand coincide at four o'clock eight o'clock and twelve o'clock. Thats where my proof fizzles out
Perhaps someone could tie it up for just a lowly silly physicist.
