Coordinate Geometry - Goat in a Circular Field
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 a starting value for x at about 0.7 radians 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 You'll find more answers to interesting 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 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/
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