From: Jonathan Kirwan
Subject: Re: Complex number quiz
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NNTP-Posting-Date: Wed, 11 Sep 2002 06:21:48 GMT
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Date: Wed, 11 Sep 2002 06:21:48 GMT
On Tue, 10 Sep 2002 11:36:55 +0200, Nicolas Matringe
>> 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?
I'm sorry, but I've been out sick today and didn't try to tackle this
one until tonight, when I felt somewhat better.
I started out with the obvious, finding the intersection of the two
circles, calling those points A=(Xa,Ya) and B=(Xb,Yb), and taking a
rectilinear integration from Xa to Xb. It got messy, fast. It's
barely workable, but I went far enough to see that there had to be an
easier way. Hand-algebra (I don't trust the darned programs, yet)
would take perhaps 8 pages or more to work and then clean up. I
didn't like that idea.
So I stopped and looked again. Turns out that it actually IS an arc,
plus two equal chord segments.
Imagine the circular field has radius, R1, and that the tether has
length, R2. I placed the circular field in quadrant 1, with its
C1= ( R1, R1 )
The tether is pinned at:
C2= ( R1, 0 )
and sweeps it's own possible circle of R2 radius from that point. The
intersections of the two circles, the critical points I mentioned
above called A and B, will then be at:
assume dy= R2^2 / (2*R1)
dx= sqrt( (R2+dy) * (R2-dy) )
A= ( R1 - dx, dy )
B= ( R1 + dx, dy )
(Those are derived quite simply by setting the two circle equations
equal to each other and solving for the intersections.)
Given this picture, there is an angle formed from the tether point,
the angle formed from the interior of A..C2..B. That arc represents a
very simple to compute area, but it's only a part of the total. There
are two other areas, which are circle segments and they are equal in
area, so we only need to compute one of them. Interestingly, the
length of the chord for these segments is exactly and always R2. That
forms the hypotenuse for a triangle we can use to figure the arc's
angle. The triangle is formed by points C2, A, and (R1-dx,0). A
similar triangle is formed by points C2, B, and (R1+dx,0). The angle
of A..C2..(R1-dx,0) is:
If you subtract two of these from PI, you get the arc's angle and that
can be taken in ratio with 2*PI to compute the area it sweeps.
We still need the two segments added in, though. Those are the
segments found within those two triangles just mentioned. That is
handled easily knowing the radius of the circle (which is R1 in this
case) and the sagitta. TO find that, we get back to those same
triangles, noting that (1/2) of its hypotenuse forms one leg of
another triangle with a vertical R1 as the hypotenuse. From that, we
can compute the length of the other leg and that allows the
computation of the segment cut off by the chord of length R2.
Two of those plus the arc does the rest. And we are done.
That turns out to be a lot easier to compute. Actually, there are
many other viewpoints, as well, including some of that compute
overlapping areas and then remove out the redundancies. Interesting,
The short of it is, the area sculpted out is:
Set R1 to 50m, then solve R2 for equality to PI*R1^2/2.
The value for R2 then is 28.968211825453m.
Call it 29m.