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Topic: Clock Watcher (Read 968 times) |
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ThudnBlunder
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Clock Watcher
« on: Oct 7th, 2008, 7:27pm » |
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One afternoon a clock watcher made two lines on a clock face to mark the angle then formed by the two hands. Some time later he noticed that the two hands exactly trisected the angle he had marked.
1) In how short a time could this have happened?
2) How soon after 3 o'clock could he have made his lines in order to to observe the trisection in this short a time?
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Sir Col
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Re: Clock Watcher
« Reply #1 on: Oct 8th, 2008, 12:14am » |
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Not sure if I'm understandnig this properly....
Just after one o'clock the minute hand could be very close to the hour hand when the marks are added to the clockface such that the trisection will occur an arbitrarily short time later.
If I've done my calculations correctly then by marking the clockface k minutes before the hands intersect (60/11 minutes past one) the angle will be trisected at (60-4k)/11 minutes past one, or 7k/11 minutes later.
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rmsgrey
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Re: Clock Watcher
« Reply #2 on: Oct 8th, 2008, 12:04pm » |
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on Oct 8th, 2008, 12:14am, Sir Col wrote:| Not sure if I'm understandnig this properly.... |
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My reading is that, having marked the current positions, if you then drew in the two lines that trisect the angle, you would then wait until the hands were one on each of those two lines, rather than simply waiting for either of the hands to cross a line...
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ThudnBlunder
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Re: Clock Watcher
« Reply #3 on: Oct 8th, 2008, 9:06pm » |
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on Oct 8th, 2008, 12:04pm, rmsgrey wrote:
That is correct.
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Immanuel_Bonfils
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Re: Clock Watcher
« Reply #4 on: Nov 18th, 2008, 4:09pm » |
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a) (660/715) h ?
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ThudnBlunder
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Re: Clock Watcher
« Reply #5 on: Nov 26th, 2008, 7:51am » |
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on Nov 18th, 2008, 4:09pm, Immanuel_Bonfils wrote:
Yes, or 12/13 hours when simplified.
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ThudnBlunder
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Re: Clock Watcher
« Reply #6 on: Dec 15th, 2008, 10:00pm » |
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1) Let A be the angle in minutes between the hands, with the minute hand ahead of the hour hand.
If the hour hand moves through 2/3 of angle A, the minute hand must move through (60 - A) + (A/3) = 60 - (2A/3)
Since the minute hand moves 12 times as fast as the hour hand, we have 12*(2A/3) = 60 - (2A/3), giving A = 90/13 minutes.
Hence elapsed time = 12*(2/3)*(90/13) = 720/13 minutes = 12/13 hours.
Interestingly, we get the same result if we let the hour hand move only 1/3 of the angle and the minute hand ends up ahead of it.
Then hour hand moves through B/3 minutes and minute hand moves through (60 - B) + (2B/3) = 60 - (B/3)
And we have 12*(B/3) = 60 - (B/3); so B = 180/13 minutes.
Hence elapsed time = 12*(1/3)*(180/13) = 720/13 minutes = 12/13 hours, as before.
2) 
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| « Last Edit: Dec 17th, 2008, 3:02am by ThudnBlunder » |
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