# Thread: An accurate 12-hour analog clock

1. ## An accurate 12-hour analog clock

An accurate 12-hour analog clock has an hour hand, a minute hand, and a second hand that are aligned at 12:00 o'clock and make one revolution in 12 hours, 1 hour, and 1 minute, respectively. It is well known, and not difficult to prove, that there is no time when the three hands are equally spaced around the clock, with each separating angle $\frac{2\pi}{3}$ . Let $f(t), g(t), h(t)$ be the respective absolute deviations of the separating angles from $\frac{2\pi}{3}$ at $t$ hours after 12:00 o'clock. What is the minimum value of $max\{f(t), g(t), h(t)\}?$

I can't seem to understand this problem...it's from IMO Long list 1989, Can anybody help ?

2. ## Re: An accurate 12-hour analog clock

The separating angles are the angles between the various hands of the clock. Let $\theta_1(t), \theta_2(t)$ and $\theta_3(t)$ be the angle of the hour, minute, and second hand, resectively, as measured from the 12:00 o'clock position, modulo $2\pi$. Then the separating angles will be $\theta_1(t)-\theta_2(t), \theta_1(t)-\theta_3(t$, and $\theta_2(t)-\theta_3(t)$. Hence $f(t)=| \theta_1(t)-\theta_2(t) -2\pi/3 |,\cdots$ etc. Find expressions for the $\theta_i$'s using the angular speeds $\omega_i$ of the various hands. You should then be able to find $\text{max}\{f(t),g(t),h(t)\}$.