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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.   

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

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

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


- Doctor Rob, The Math Forum   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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