Goat In a Circular Field

greatwhiteshark

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A goat is tethered by a rope of length L to a point on the circumference of a circular field of radius R. Find L in terms of R if the goat can graze exactly half the area of the field.
 
You posted this under geometry. I worked on it, off and on, for years, trying up thru calculus. In fact, it was how I got started here. I posted it, looking for a solution but it seems to have disapeared.
The best solution I ever came up is a hairy equation using trig that required the Newton-Raphson method to solve. I hope that if you find a (non trig) geometric solution you will share it with me.
I think I can reconstruct it if you would be interested but I was never happy with it and it is not geometry.
-------------------
Gene

PS Before I try, I found
http://mathworld.wolfram.com/GoatProblem.html
which has a different equation.
G.
 
It's not all that hard a problem. What's the area of a circular sector? If you've two circles, each with different radius, it shouldn't bee that hard. Draw a picture and see what you get.
 
Don't just say how easy it is. Show us :evil:
-------------------
Gene
 
I didn't say it would be EASY.

Put circles on a coordinate axis.

Center the FIELD circle at the ORIGIN

x^2 + y^2 = R^2

Center the GOAT circle on the positive y-axis

x^2 + (y-R)^2 = L^2

It is important to note that R < L < R*sqrt(2), or we'll never scope out 1/2 the field. If L = R, clearly we get much less than 1/2. [You tell me where the upper bound came from.]

Our two circles intersect at x = (1/2)*(L/R)*sqrt(4*R^2 - L^2) and x = -(1/2)*(L/R)*sqrt(4*R^2 - L^2). The y-coordinate is the same for each x-value, y = (1/2)*(1/R)*(2*R^2 - L^2)

The Calculus solution to the GOAT area is now simple, being (using symmetry) twice the integral from

0 to (1/2)*(L/R)*sqrt(4*R^2 - L^2)

of the difference of the equations of the two half-circles,

sqrt(R^2 - x^2) - [R - sqrt(L^2 - x^2)]

with respect to x, of course.

However, maybe there is an easier way. The GOAT circle intersects the x-axis at x = +/- sqrt(L^2 - R^2). The calculus SOLUTION, now, and NOT just the GOAT area, is found by setting to zero the integral (again using symmetry) from 0 to sqrt(L^2 - R^2) of R - sqrt(L^2 - x^2) (with respect to x) and solving for L.

Maybe it's a little harder than I thought. I'm still thinking it is conceptually not so difficult. It doesn't seem obvious at the moment how to calculate the area of those little not-quite-triangular sections on the outside - in a geometry forum. Someone will beat me to it.

gtg
 
It's not all that hard a problem.
The sum of the areas of the two little triangular pieces = the area of segment of the goat circle below the x axis, but that never helped me.
I knew the answer was around the average of those limits or R*1.2 but my best answers had terms something like
tan(arcsin(R/L)^2) or
R^2(theta+sin(theta))/2
Not friendly.
(Those are from a leaky memory and are not to be relied on. They are just for illustration.)
The web sites answer is
R*1.1587...
I would accept a calcus solution, but look at the mathworld website before you try your integration.
 
A few more definitions:

h = piece of radius on the y-axis and below the x-axis.
h = L - R

c = length of secant segment on the x-axis
c = 2*sqrt(L<sup>2</sup> - R<sup>2</sup>)

Q = angle measure of GOAT circle sector subtended by 'c'

A = Area of GOAT circle below the x-axis.
A = (1/2)*L<sup>2</sup>*(Q-sin(Q)) = L<sup>2</sup>*cos<sup>-1</sup>(R/L) - L*sqrt(2*L*h - h<sup>2</sup>)

Now it seems we can do that with the little not-quite-triangular pieces, subtracting the triangle and adding the circular sector back.

It remains a horrible mess, no question. OK, I get to back-peddle a bit and agree that this is a difficult problem, but certainly not from the conception. There are at least a couple of ways to solve it. Paying attention and failing to make errors poses the difficulty.
 
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