Is it possible that a clock's three hands divide the clock face into 3 equal parts? And is there a proof/disproof out of intuition? Here we ignore the structure of the clock and suppose the hands move with constant speed.
 A: Let $\theta$ be the angle of the hour hand past 12 o'clock. Then the angle of the minute hand is $12\theta$ and the angle of the second hand is $720\theta$.
If the face of the clock is to be divided into three equal parts, we need the angle between two hands to be $\frac{2\pi}{3}$ for any two hands. Hence,
$$720\theta - 12\theta \equiv 12\theta - \theta \equiv \theta-720\theta \equiv \pm\frac{2\pi}{3} \pmod {2\pi}$$
or, calculating a little,
$$708\theta \equiv 11\theta \equiv -719\theta \equiv \pm\frac{2\pi}{3}\pmod {2\pi}.$$
It immediately follows that $708\theta-11\theta =697\theta\equiv 0 \pmod {2\pi}$ as well as $11\theta-(-719)\theta = 730\theta \equiv 0\pmod {2\pi}$.
Thus $\gcd(730\theta,697\theta)=\theta\equiv 0\pmod {2\pi}$, which is not a proper solution.
Thus, there is no such configuration of the hands of a clock as required by your problem.
A: In a 12 hour period, which I will take to be the unit of time, the hands start out and end aligned. The minutes hand overtakes the hours hand $11$ times during this period, and the seconds hand does so $12\times60-1=719$ times; these give their relative speeds with respect to the hours hand in revolutions per unit time. Note that $11$ and $719$ are relatively prime, which means there is no intermediate moment in time $0<t<1$ at which the hands are all three aligned: this would meen $a=11t$ and $b=719t$ are integers, so $719a=11b$ is a common multiple of $11$ and $719$, therefore a multiple of $\operatorname{lcm}(11,719)=11\times719$, which would force $t$ to be integer (but it isn't).
Now if $t$ is a time at which the relative positions of all hands are all multiples of $\frac13$ of a revolution, then by uniformity of their motion they will all be aligned again at time $3t$. By what we've just seen this means that $t$ is a multiple of $\frac13$ unit of time. But with the unit of time being 12 hours, $\frac13$ unit of time is precisely $4$ hours, at which points in time the minutes and seconds hands are aligned. So it never happens that the hands are precisely in relative positions $0,\frac13,\frac23$ of a revolution.
