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Math Help - An accurate 12-hour analog clock

  1. #1
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    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 ?
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  2. #2
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    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)\}.
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