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From: Fred Bloggs
X-Mailer: Mozilla 4.7 [en]C-CCK-MCD EBM-Compaq1 (Win95; U)
Subject: Re: Complex number quiz
References: <3D7B8D10.5070603@BOGUS.earthlink.net> <3D7CFD55.firstname.lastname@example.org> <6p3f9.201882$8aG1.email@example.com> <3D7D6CF1.firstname.lastname@example.org> <3D7DBD37.27C9BDEC@ipricot.com>
Date: Tue, 10 Sep 2002 16:54:50 GMT
NNTP-Posting-Date: Tue, 10 Sep 2002 09:54:50 PDT
Organization: EarthLink Inc. -- http://www.EarthLink.net
Nicolas Matringe wrote:
> markp wrote:
> > If you want to while away a few more hours, how about this one:
> > A square field 50m x 50m has a cow in it. The cow is tethered to a
> > rope that is fixed exactly half way down one side of the field.
> > The rope is just long enough for the cow to eat exactly half the
> > grass in the field. How long's the rope?
> Another one now (I have never found the answer, even on the internet ;o)
> It's exwactly the same problem except that the field is circular and 50m in
> diameter. How long is the rope?
About 35.25 meters or so with a hand calculator zero-solving. There is
nothing profound here. You add up some arc areas and one triangle area
and you end up with an equation of the form:
8( x^2+0.25-x*sin(theta))+ psi/2-4*x*sin(theta)=pi
where cos(theta)=x and cos(psi)=1-2*x^2
and x=R/D where R=length of rope and D= diameter of circle.
Solve for x.
I may have made mistake but this is essentially it.
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