Angles Formed By The Hands Of A Clock Given a clock with hour, minute, and second hands that each move continuously (i.e., no “ticking” occurs), show whether there exists a time at which the lesser angle formed by each pair of the hands is $\frac{2 \pi}{3}$.
I ask this because, though I expect the solution to be fairly simple in retrospect, I have never before resolved a question of this form and do not know how to phrase any proof I had.
 A: First consider just the hour and minute hands.  They are together at noon.  How long does it take for the minute hand to gain $120^\circ$ on the hour hand?
The minute hand travels $6^\circ$ per minute, and the hour hand travels $\frac12^\circ$ per minute, so we need the solution to $$6m={m\over 2}+120$$ which gives $$m={240\over11} = 21{9\over11}\text{ minutes}$$  In $m$ minutes the minute hand travels $130{10\over11}^\circ$ and the hour hand travels $10{10\over11}^\circ.$  After $m$ minutes the minute hand is $120^\circ$ ahead.  After $2m$ minutes, it is $240^\circ$ ahead and the angle is $120^\circ$ again.  After $3m$ minutes, the hands are together again.  In any event, the only possibilities are $nm$ minutes after noon, for some integer $n$.  
How far does the second hand travel in $m$ minutes?  Since it makes a whole number of revolutions in $21$ minutes, we only need be concerned with how far it travels in ${9\over11}$ minutes, which is $294{6\over11}^\circ.$
Looking at the fractions, we see that $n$ must be a multiple of $11$ if there is to be a whole number of degree between the minute hand and the second hand.  However, $$11m=240\text{ minutes } = 4\text{ hours,}$$ so the minute and second hands will coincide.
Thus, it is impossible.
Try doing the problem on a $24$-hour clock.
A: There are 11 times in a 12 hour period when the hour hand an the minute hand form an angle of $\frac {2\pi}{3}$
Exactly 4:00 is one such time.
$H =(\frac {12}{11}k + \frac 23)$
Where $H$ is expressed in fractions of a revolution.  In these units the position of the minute hand is:
$M = \{12 H\} = \frac {1}{11} k$ where $\{x\} = x  - \lfloor x \rfloor $ or the fractional part of the number.
$S = \{60 M\} = {\frac {5}{11} k}$
$S-M$ will always be expressed as $\frac {1}{11}$'s and will never equal $\frac {1}{3}$
