The Goat In the Field ProblemDate: 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 http://mathforum.org/dr.math/faq/faq.grazing.html 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 quotes). -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. |
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