Grazing Half of a Square Field

Date: 08/30/2001 at 12:36:07
From: Philip Roche
Subject: Grazing Animal

My puzzle consists of a cow tied to the middle of one side of a SQUARE 
field. Apart from that all the details shown in the Dr. Math Grazing 
Animals FAQ are the same. 

   http://mathforum.org/dr.math/faq/faq.grazing.html   

Can you tell me the length the rope should be to enable the cow to eat 
half the grass?

I tried solving the problem using geometry and integration but could 
never work it out. I did get to a point where I had an equation 
containing x and sin x but had no method of solving such an equation. 
Now I have forgotten most of what I once knew but I would be very 
interested in seeing a solution using whatever methods are required.

Thanks for your time.

Date: 08/30/2001 at 14:10:54
From: Doctor Rob
Subject: Re: Grazing Animal

Thanks for writing to Ask Dr. Math, Philip.

Let s be the length of the side of the square field. Let the length of 
the rope be r. Set up a coordinate system with origin at the point of 
tethering, and with the x-axis along one side of the square, so that 
the corners of the square are at (0,s/2), (-0,-s/2), (s,s/2), and
(s,-s/2). We are interested in finding the ratio t = r/s, which is
independent of r or s. Half the area of the square is s^2/2, and the
area the cow can graze is the area within the square but below the
circle y = sqrt(r^2-x^2). Thus you get the equation

                   s/2
   s^2/2 = INTEGRAL   sqrt(r^2-x^2) dx,
                  -s/2

                                                s/2
   s^2/2 = [x*sqrt(r^2-x^2)/2+r^2*arcsin(x/r)/2]
                                                x=-s/2

If you put r = s*t, you end up with the following equation to solve:

   4*t^2*arcsin(1/[2*t]) + sqrt(4*t^2-1) - 2 = 0,
   
   1 = 2*t*sin([2-sqrt(4*t^2-1)]/[4*t^2]).

This can only be solved numerically, not algebraically.  The answer
is

   r/s = t = 0.5828221624459...,
   r = s*0.5828221624459...,

approximately.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/