Angles Formed By The Hands Of A Clock

Question

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.

Answer 1

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.

Answer 2

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 $\frac {1}{3}$