From: Jonathan Kirwan
Subject: Re: Complex number quiz
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NNTP-Posting-Date: Tue, 10 Sep 2002 03:39:06 GMT
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Date: Tue, 10 Sep 2002 03:39:06 GMT
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 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
>Before you answer, I have absolutely no idea what the solution is (used to,
>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.