Grazing Goats

Consider a circular pasture with known radius. At the edge of this 
pasture is a pole with a rope attached to it. At the other end of 
the rope we attach a goat.  What length of rope is required if we 
want the goat to graze over exactly HALF the area of the pasture?
  
The exact solution to this commonly asked question cannot be expressed 
in what most people would call "closed form".  If x = L/R, where L is 
the length of the rope and R is the radius of the pasture, then x is 
a root of

  g(x) = x sqrt(1-x^2/4) + asin(1-x^2/2) + x^2 asin(x/2) - x^2 pi/2

Alternatively, x is a root of

  f(x) = - x sqrt(1-x^2/4) + acos(1-x^2/2) + x^2 acos(x/2) - pi/2

It's interesting that these two functions have relatively simple
derivatives.  Specifically, the derivative of g with respect to x is 

                    g'(x) = 2x asin(x/2) - x pi

and the derivative of f with respect to x is simply

                    f'(x) = 2x acos(x/2)

Incidentally, Jeff Adams emailed me to say that this problem had 
gained some notoreity in the last place he taught, which was the 
Univ. of Maryland's European Division, which conducted classes at
the undergraduate level on US military bases all over Europe.  This 
sheds light on something that had been puzzling me.  I was talking 
with a young fellow recently who had never before (and has never 
since) expressed any interest in mathematics, but when he heard about 
my math web site he immediately took me aside and said "Please tell
me this:  Suppose you have a goat penned up in a circlular pasture..." 
He was intent on learning the answer to THAT question, which I thought 
was rather odd.  Now I recall he mentioned that he had been in the 
military and had been stationed in Europe for a while, so I'm guessing
we have the Univ. of Maryland to thank for his interest in the subject.

It's worth noting that the standard goat grazing problem is really 
just one (and far from the most interesting) of a general class of 
area problems involving shapes whose boundaries are segments of 
circles.  Such shapes are called "lunes".  As an example, suppose 
instead of leaving 1/2 of the pasture untouched, we want to leave 
1/pi untouched.  In this case with a bit of thought we can show that
the solution is L/R = sqrt(2).  Of course, this is just the same as 
saying that x is the root of x^2 - 2 = 0, and whether this is 
considered a "closed form" solution depends on your point of view, 
since we can compute the actual value of sqrt(2) only by iterative 
procedures.  On the other hand, it can be constructed by Euclidean 
methods, i.e., by straightedge and compass, so this type of solution 
can claim to possess a very elementary type of "constructibility".

For more on lunes, see The Five Squarable Lunes.