A goat grazes on a circular meadow, its tail is tacked to a stake of the fence - What's the length of the...

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[ Editor: two later threads, 1996. ]

John Fleming wrote:

> A goat is tethered to the perimeter of a circular field with a rope
> just long enough that he can eat grass from exactly half the area of
> the field.  How long is the rope?

Let the fence be centered (0,0), radius r.
Let the goat be tethered at (0,r), chain length R.

In the first quadrant, the circles intersect at (m,n)

(1) m^2 + n^2 = r^2
(2) m^2 + (r-n)^2 = R^2

from (1),

(3) m^2 = r^2 - n^2

from (3) and (2),

(4) n = r - R^2/(2r)

from (4) and (3),

(5) m = R * sqrt(1 - R^2/(4r^2))

Now, looking at the first quadrant, the area under the circle [center
(0,0) radius r] from X = 0 to X = m is

(6) A1 = m/2 * sqrt(r^2 - m^2) + r^2/2 * arcsin(m/r)

Any good calc textbook will have this result.

From (6), (5), and (4), the area between the circle and the line Y = n
is

(7) A2 = r^2/2 * arcsin(m/r) - m/2 * sqrt(r^2 - m^2)

We can similarly look at the area between the line Y=n and circle
[center (0,r) radius R] from X = 0 to X = m is

(8) A3 = R^2/2 * arcsin(m/R) - m/2 * sqrt(R^2 - m^2)

The area that the goat can eat is equal to A2 + A3.  For this to be
one-half of the farmer's field,

A2 + A3 = 1/2 * pi * r^2

Let r = 1 and simplify:

arcsin(R*foo) + R^2 * arcsin(foo) - R*foo(sqrt(1 - 3*R^2/4) + R^2/2)
  = pi/2

where foo = sqrt(1 - R^2/4)

This is just too ugly for me to deal with.  Solving this numerically,
we get

R = 1.111707 r

Can anyone find an analytic solution?

How about this alternative: Another farmer has a circular field.  This
farmer tethers his goat _outside_ the fence so that it can eat an area
equal to 1/2 the field.  How long is the chain?

Note: as the goat moves within the tangent, the chain begins to wrap
around the outside of the fence.

-- Gaku Sato <gaku.sato@leland.stanford.edu>  (March 1996)

[ Editor: this differs from the 1.15872847302 found above.  At
R = 1.111707 Schumacher's area formula gives 1.4676, less than
pi/2 = 1.5708, so the slip is in the simplified equation (which does
hold at R = 1.111707). ]

Atta Chui (aykc1@amtp.cam.ac.uk) wrote:

: I think the problem becomes even more interesting if the goat is
: placed outside the meadow, rather than inside.  Hence the following
: variation...
:
: A farmer builds a circular pen to contain a goat.  The farmer ties
: the goat with a rope attached to one of the fenceposts on the
: perimeter of the circle.  However, he left the goat OUTSIDE the pen.
: How long must the rope be to allow the goat to graze on an area
: exactly equal to that enclosed by the pen?
:
: Note: you see, as the goat 'turns around', the pen blocks its way
: and the rope has to follow the fence of the pen.  The boundary of the
: area which the goat can reach is therefore not a circular arc!

Let the fence be the circle with radius R centred at (R,0), the goat's
fence-post be the point (0,0), and the rope's length be r.  The point
(x,y) with x < 0 is within reach if its distance from (0,0) is less
than r.  For x > 0, on the other hand, a more complex analysis is
needed.  Thanks to symmetry only the case of y > 0 needs to be
concerned.

Let (x1,y1) be the point where the rope stops following the fence.
Let a be the angle between the radii from (0,0) to (R,0) and from
(x1,y1) to (R,0).  The point (x1,y1) is thus given by

x1 = R ( 1 - cos(a) )

y1 = R sin(a)

Let (x2,y2) be the end point of the rope.  It is given by

x2 = x1 + ( r - R a ) sin(a) = R ( 1 - cos(a) ) + ( r - R a ) sin(a)

y2 = y1 + ( r - R a ) cos(a) = R sin(a) + ( r - R a ) cos(a)

The area underneath the curve from 0 to X = R ( 1 - sin(r/R) ) (the
point where the entire rope follows the fence) is given by

    X          r/R
S = I y(x) dx = I y(a) x'(a) da
    0           0

Substituting the above expressions for x and y yields

    r/R
S =  I [ R sin(a) + ( r - R a ) cos(a) ] ( r - R a ) cos(a) da =
     0
       r/R                                     2
    1/2 I [ R ( r - R a ) sin(2a) + ( r - R a )  ( 1 + cos(2a) ) ] da =
        0
                     2                                        2
    1/2 [ ( r - R a )  sin(2a) / 2 - R ( r - R a ) cos(2a) - R  sin(2a) / 2 -

               3          r/R               3              2
    ( r - R a )  / ( 3 R ) ]  = R r / 2  + r  / ( 6 R ) - R  sin(2r/R) / 4
                           0

This also includes part of the circle with radius R bound by the
x-axis and the line x = x2.  If t is the angle between the x-axis and
the radius from (R,0) to (x2,y2), its area is given by

     2                                            2
s = R  t / 2 - R cos(t) R sin(t) / 2 = R r / 2 - R  sin(2r/R) / 4

                                                          3
The area underneath the curve outside the circle is thus r  / ( 6 R ),
and the total grazing area is twice that plus the area of the
half-circle with radius r for x < 0.  It is therefore given by

        2        3
A = pi r  / 2 + r  / ( 3 R )

If it is equal to the area of the circle with radius R, we have

    2        3                 2
pi r  / 2 + r  / ( 3 R ) = pi R

which can be rewritten as

   3         2
2 x  + 3 pi x  - 6 pi = 0

where x = r / R.

Solving numerically we get x = 1.2565689, i.e. r = 1.2565689 R.

-- Ilan Mayer <ilan@isgtec.com>  (December 1996)