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From: ech@spuxll.UUCP (Ned Horvath)
Newsgroups: net.math
Subject: Re: another clock problem
Message-ID: <627@spuxll.UUCP>
Date: Sun, 17-Feb-85 00:33:40 EST
Article-I.D.: spuxll.627
Posted: Sun Feb 17 00:33:40 1985
Date-Received: Mon, 18-Feb-85 05:39:17 EST
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Kirk Pruhs (pruhs@uwvax.UUCP) writes:


>     Here is another rather interesting clock problem that was on the Putnam
>last year. If you have a standard clock with hand lengths 3 and 4, how far
>apart are the tips of the hands when the distance between them is increasing 
>most rapidly? A freshmen calculas student has the tools to solve this but
>a little care needs to be taken or you can fill  a wall with the equations.

One need only remember the high school formula for the third side of a
triangle given two sides and the angle between them:
	c^2 = a^2 + b^2 - 2ab(cos C)

Where C(t) is just 2pi*(11/12)t, t in hours.  I mention that just to get
rid of t -- C is just a linear multiple of t.  But then the change in
distance is given by 
	d(c^2)/dC = 2ab(sinC)
which is, of course, maximized at C=pi/2.  When the hands are at right angles
I can calculate c in my head...

=Ned=