angleproof-writing
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.
If the minute hand has made $x$ revolutions around the clock ($60x$ minutes), the hour hand has made $\frac1{12}x$ revolutions; the difference between these two values represents an angle. Both hands start at 0 revolutions.
For (1), the relevant angle is $\frac14$ of the circle: $$x=\frac14+\frac1{12}x$$ Solving, we get $x=\frac3{11}$, which corresponds to a time of $\frac{180}{11}$ minutes after noon.
For (2) we replace $\frac14$ with 1, since the minute hand has lapped the hour hand. Then $$x=1+\frac1{12}x$$ Solving, we get $x=\frac{12}{11}$, i.e. $\frac{720}{11}$ minutes after noon.
Therefore, both your solutions are correct.
The equation for the angle between the hands:
Best Answer
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.