The goat problem (or bull-tethering problem) considers a fenced circular field of radius 
with a goat (or bull, or other animal) tied to a point on the interior or exterior
of the fence by means of a tether of length 
, and asks for the solution to various problems concerning
how much of the field can be grazed.
Tieing a goat to a point on the interior of the fence with radius 
1 using a chain of length 
, consider the length of chain that must be used in order
to allow the goat to graze exactly one half the area of the field. The answer is
obtained by using the equation for a circle-circle
intersection
 |  |  |
(1)
|
Taking 
gives
 |
(2)
|
plotted above. Setting 
(i.e., half of 
)
leads to the equation
 |
(3)
|
which cannot be solved exactly, but which has approximate solution
 |
(4)
|
(OEIS A133731).
Now instead consider tieing the goat to the exterior of the fence (or equivalently, to the exterior of a silo whose horizontal cross section
is a circle) with radius 
. Assume that 
, so that the goat is not able to reach further around
than the point on the fence opposite his starting point (Hoffman 1998, where we have
replaced Hoffman's bull with a more prosaic goat). The goat may obviously graze inside
the interior of a semicircle of radius 
whose diameter is tangent to the fence. In addition, the goat
may graze two area on either side of the semicircle that have the fence as the inner
boundary and a circle evolute as the outer boundary.
To find the area of this region, assume the fence is oriented so that the farthest
point around the circumference that the goat can reach is at position 
. Now, note that the equation for a circle
involute is given by
 |  |  |
(5)
|
 |  |  |
(6)
|
From geometry, the goat will transition between being radially bound and being bound by pulling tangent to the circle at the point where 
, so
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(7)
|
 |  |  |
(8)
|
 |  |  |
(9)
|
Equating (8) and (9) and solving for 
then shows that this occurs at parameter
. The area of the involute portion
that the goat can graze is then given by
 |  |  |
(10)
|
 |  |  |
(11)
|
 |  |  |
(12)
|
Adding twice this area to the area of a semicircle of radius 
then gives the total area which the
goat can graze as
 |
(13)
|
The grazable areas are illustrated above for a number of ratios of 
. Note that the case 
forms a curve that resembles, but is not equivalent to,
a cardioid.
