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### Coordinate Geometry - Goat in a Circular Field

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Date: 8/22/96 at 21:39:48
From: Anonymous
Subject: A Fun Coordinate Geometry Problem

I recently stumbled on your page and was immediately hopeful in
finding a worked solution to a problem that has taunted me for over 30
years. I first came across it in my first job as an engineering
apprentice in England. The works instructor posed the problem but
refused to tell us the answer (I don't think he actually knew the
answer, I think he just wanted to show off). Anyway many years later I
came across a reference to it in a Calculus text book, it gave the
answer but did not give a worked proof. I hope you can put my mind at
rest after all these years.

The text book in question is "Elementary Calculus and Coordinate
Geometry" by C.G.Nobbs. On page 364 he poses this question:

"The following is a very well known problem; possibly you have seen it
in the puzzle corner of a magazine.

A goat is tethered by a rope of length L to a point on the
circumference of a circular field of radius R. Find L in terms of R,
if the goat can graze exactly half the area of the field."

He then gives the answer as 1.16R, but no proof.

Can you help?

Many thanks,
Brian Booth
```

```
Date: 8/23/96 at 8:32:33
From: Doctor Anthony
Subject: A Fun Coordinate Geometry Problem

Since one cannot draw a decent diagram with ascii, I suggest you
draw the following diagram on a piece of paper and refer to it while
I go through the work. Draw a circle with suitable radius r. Now take
a point C on the circumference and with a slightly larger radius R
draw an arc of a circle to cut the first circle in points A and B.
Join AC and BC. Let O be the centre of the first circle of radius r.
Let angle OCA = x (radians). This will also be equal to angle OCB. The
area we require is made up of a sector of a circle radius R with angle
2x at the centre, C, of this circle, plus two small segments of the
first circle of radius r cut off by the chords AC and BC.

Area of sector of circle R is (1/2)R^2*2x = R^2*x

area of two segments = 2[(1/2)r^2(pi-2x) - (1/2)r^2sin(pi-2x)]
= r^2[pi - 2x - sin(2x)]

We also have R = 2rcos(x), so R^2*x = 4r^2*x*cos^2(x)

We add the two elements of area and equate to (1/2)pi*r^2

4r^2*x*cos^2(x) + r^2[pi-2x-sin(2x)] = (1/2)pi*r^2   divide out r^2

4x*cos^2(x) + pi - 2x - sin(2x)  = (1/2)pi

4x*cos^2(x) + (1/2)pi - 2x - sin(2x) = 0

We must solve this for x and we can then find R/r from R/r = 2cos(x)

Newton-Raphson is a suitable method for solving this equation, using

The solution I get is x = 0.95284786466 and from this
cos(x) = 0.579364236509

and so finally  R/r = 2cos(x) = 1.15872847

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

```
Date: 05/25/97
From: Doctor Sarah

grazing, some with illustrative diagrams, by searching the Dr. Math
archives for the words "goat" and "cow" (just the word, not the
quotes).

Here are two more answers by Dr. Anthony:

"Goat on a Rope":

http://mathforum.org/dr.math/problems/nicholls.8.6.96.html

"The Goat in the Field Problem"

http://mathforum.org/dr.math/problems/miller5.24.97.html

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

```
Associated Topics:
College Euclidean Geometry

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