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The Goat In the Field Problem

Date: 05/24/97 at 11:02:38
From: Anonymous
Subject: The Goat In the Field Problem

This problem was first posed by a colleague during my Engineering 
Degree some 15 years ago. The problem looked simple enough but it was 
stipulated that calculus must NOT be used. I wouldn't like to guess 
how many hours I've spent on it. I decided it was probably a standard, 
recognised problem but could not find it in any recreational 
mathematics book. I eventually decided there was probably some 
'Euclidean' geometric rule I'd forgotten since school days but basic 
research has got me nowhere. Here's how the problem was presented:

A farmer owns a circular field of grass (radius=r). He tethers a goat 
via a length of rope (R) to the circumference of the field. What ratio 
r/R must the farmer choose so that the goat can only eat half the area 
of grass?

Many thanks.

Date: 05/25/97 at 16:59:13
From: Doctor Anthony
Subject: Re: The Goat In the Field Problem

Let O be the center of the circle of the goat's grazing range and let 
C be the point on the circumference where the goat is tethered.  Let 
CA and CB be the chords of length R giving the extreme positions on 
the circumference where the goat can reach.  Angle AOC = angle BOC = 
phi (radians).  We shall first calculate the area of the segment of 
circular field cut off by the chords CA and CB.

The area cut off is found by subtracting the area of triangle OAC from 
the sector OAC:
                = (1/2)r^2.phi - (1/2)r^2.sin(phi)

                = (1/2)r^2[phi - sin(phi)]

The area cut off by both AC and BC is double this: r^2[phi -sin(phi)].

This area must be added to the area of the sector of the circle of 
radius R between the radii CA and CB.  By simple geometry the angle 
ACB = (1/2)(2.pi-2phi)  =  pi-phi

Area of sector CAB = (1/2)R^2(pi-phi)

Total area available for the goat 
                      = r^2[phi-sin(phi)] + (1/2)R^2(pi-phi)

This must equal half the area of the field (1/2)pi.r^2

  r^2[phi-sin(phi)] + (1/2)R^2(pi-phi) = (1/2)pi.r^2      

Dividing through by (1/2)r^2:

(1)   2[phi-sin(phi)] + (R/r)^2(pi-phi) = pi   

Now we can get another relationship between r and R by drawing a 
perpendicular from O to AC to bisect AC.  This shows that: 

     r.sin(phi/2) = R/2  and  R/r = 2sin(phi/2)

Equation(1) can be written:

   2[phi-sin(phi)] + 4sin^2(phi/2)[pi-phi] - pi = 0

This can be solved for phi by Newton Raphson or by a handy TI-92 
calculator to give:   
                phi = 1.23589  radians

Then   R/r = 2sin(.617945) = 1.158723

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 05/25/97 at 06:19:58
From: Doctor Sarah
Subject: Re: The Goat In the Field Problem

For more on this question, see this "Classic Problem from the 
Dr. Math FAQ:

Grazing Animals


You'll also find more answers to problems about tethering and 
grazing, some with illustrative diagrams, by searching the Dr. Math 
archives for the words "goat" and "cow" (just the word, not the 

-Doctor Sarah,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 02/13/2000 at 22:36:41
From: David Gillies
Subject: Re: The Goat in the Field Problem

I was browsing the Dr. Math archives today and came across this old 
chestnut. You may be interested to know that two friends of mine, 
Simon Shepherd and Peter van Eetvelt, came up with a completely 
closed-form, analytic solution. It turns out this problem is very 
germane to communications physics, for instance determining the 
optimum siting of mobile telephone base stations (you can imagine the 
circles as the coverage areas of the radio signals). It also is useful 
in determining the optimum placement for a jammer transmitter so as to 
cause maximal disruption.

Basically the result is this:

Consider two circles of radii a and b separated by a distance c 
between their centres, which are located at (-c/2,0) and (+c/2,0) 
respectively. The area of intersection is given by

   I(a,b,c) = a^2 arccos ((c^2 - b^2 + a^2)/2ac) - (sqrt(2(a^2 b^2
                + b^2 c^2 + c^2 a^2) - a^4 - b^4 - c^4)/2)
                + b^2 arccos ((c^2 + b^2 - a^2)/2bc)

Note that I(a,b,c) = I(b,a,c), as expected, and the *real part* of this 
integral gives the correct answer in all three of the cases: 
  1) the circles do not overlap and I = 0 
  2) the smaller circle is entirely contained in the larger and 
     I = Pi * b^2 (or Pi * a^2) 
  3) the circles partially overlap. 

As far as I know, Shepherd and Eetvelt's proof is the only closed-form 
solution that yields the correct result in all three cases. They used it 
to calculate very accurately that the length L of the goat's rope such 
that he can graze exactly half the field of radius R is 
L = 1.15872847301812171... R.

The title of their paper is as follows:

   S. J. Shepherd and P. W. J. van Eetvelt, "On Goats and Jammers,"
   Bulletin of the IMA, 31, (5-6), May 1995, pp. 87-89.
Associated Topics:
College Euclidean Geometry
High School Euclidean/Plane Geometry

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