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Subject: Re: Complex number quiz
Date: Tue, 10 Sep 2002 11:42:10 +0100
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"Jonathan Kirwan" wrote in message
> On Tue, 10 Sep 2002 00:57:40 +0100, "markp"
> >A square field 50m x 50m has a cow in it. The cow is tethered to a rope
> >is fixed exactly half way down one side of the field. The rope is just
> >enough for the cow to eat exactly half the grass in the field. How long's
> >the rope?
> >Before you answer, I have absolutely no idea what the solution is (used
> >but long since forgotten).
> No one took this on? Why, it's just your basic semicircle less twice
> your half-a-chord-segment problem. The answer is obviously:
> >Oh, you must show all your working too...!
> hehe. That, too?
> It's obviously just a semicircle minus a chord segment, but...
> Start out by imagining that the tether has radius R (which includes
> the length of the cow's neck, when eating grass.) Since it is staked
> on one side, at the half-way point, the only eating that can take
> place is within a semicircle on the same side as the square field.
> The rest doesn't matter for now.
> The area of the semicircle is PI*R^2/2. Let's call the size of a side
> of the field is S. If R were exactly 1/2 of S, then the area would be
> (PI/8)*S^2. But since (PI/8) is less than .5, it's obvious that R
> must be greater. This forces things from a new angle.
> Now you have to imagine R's semicircle superimposed by the square
> field, with R bigger than S/2. That leaves two pieces, wings if you
> will (and not car fenders, to you silly Britishers), where the cow can
> graze but where this isn't a part of the square field. Each of these
> is exactly 1/2 of a segment formed by a chord. Since there are two,
> you might as well treat them as a single chord's segment area. So the
> problem becomes: "The area on the square available to the cow is the
> area of the semicircle of radius R minus the area of a chord of a
> circle of radius R, where the chord closest approach to the center is
> S/2." This difference must be set to S^2/2 and solved.
> This amounts to:
> (1/2)*S^2 = [(1/2)*PI*R^2] -
> [R^2*atn(sqrt((2*R/S)^2-1)) - (S/2)*sqrt*(R^2-S^2/4)]
> Numerically solving for R yields the obvious 29.141108m. The area of
> the semicircle is then ~1333.93 m^2 and of the chord ~83.93 m^2. The
> difference being the requisite 1250 m^2.
> I'll leave the integral needed to figure out the area of the chord as
> an exercise. It's tedious to write out, but easy to see. And the
> numerical solution is pure trivia. But if someone wants to work on
> the closed solution for R, I'd love to see it.
Well done! Perfect answer (I think..). It's also a rectangle plus a chord
problem too, the rectangle formed by the bottom side of the tether and the
points where the rope radius would cross the adjacent sides. And maybe a
segment plus two triangles problem. Either way I think it leads to a 'theta
equals a trig function of theta' type equation.
> Numerically solving for R yields the obvious 29.141108m
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