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Old 07-26-2007   #6
Dr Spock
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Someone who uses the CRUDE approximation 22/7 for Ï€ when setting his question (and truncates it at that) is hardly in a position to insist on MATHEMATICAL PROOF in the responses! (The area of the field is in fact Ï€*10^4 / 4 = 7853.9816...sq ft. and NOT 7857.143 sq ft.) 'ironduke8159' has the correct answer, though it's a pity he rounded his result too severely at the end. The length of the tether is 57.936423...ft ~ 57ft 11.24" or ~ 57ft 11 1/4".I worked it out by considering an original circle ("circle 1") of radius 1, centred at the origin O, and another circle ("circle R") of radius R centred on C on the perimeter. The two circles intersect at A and B, and AB intersects OC at point D, where OD = d. In my method, 'd' was the primary variable to be solved for. Note that d = cos θ. Almost everything needed (except for just that ONE angle, θ = cos^(-1) d, could be expressed as ALGEBRAIC functions of 'd.')The tethered area consists of two unequal "segments,""circle slices" or "lenticular areas" (for want of better words) ABC (subtending an angle 2θ at O) and ABO (subtending an angle 2Ï* at C). Similar formal formulae apply to each circle slice, because we're adding two segments, albeit asymmetrical.LEMMA. For a circle of radius r, and a segment subtending an angle 2α at the centre, the area of the full 2α sector minus the relevant triangle gives a segment area of (α - sinα cosα) r^2.(This is easy to derive, and readily checked at special values of α, such as α = 0, Ï€/2, and Ï€.) APPLICATIONS. From the lemma, the area of circle 1's segment is simply θ - sinθ cos θ;while circle R's segment is (Ï* - sinÏ* cosÏ*) R^2.So the full area accessible to our ravenous kid (a "reduced goat") in the reduced problem isθ - sinθ cos θ + (Ï* - sinÏ* cosÏ*) R^2,and this must equal Ï€/2.Several geometrical relationships enable this to be expressed largely as an algebraic function of d, with just one transendental function cos^(-1)d in addition:The cosine formula applied to triangle OAC (or OBC) shows that R^2 = 2 - 2 cosθ = 2(1 - d). Use of the sine formula applied to that same triangle, or alternatively the fact that the sum of its angles shows that:2Ï* + θ = Ï€ lead to sinÏ* cosÏ* = (1 - d^2)^(1/2) / 2.Putting this all together and requiring that the sum of the two segments be Ï€/2 gives a formula involving just cos^(-1) d and algebraic functions of d (I'll still write θ for cos^(-1) d):θ - d(1 - d^2)^(1/2) + [Ï€ - θ - (1 - d^2)^(1/2)](1 - d) =Ï€/2.With θ = cos^(-1) d, we can iterate this equation to obtain the solution d (= cosθ) = 0.328674167.Since R^2 = 2(1 - d), that gives r^2 = 1.342651665... , orR = 1.15872847... .Applied to the problem posed, where the orginal circle has radius 50 ft, that gives a tether length of 50 x 1.15872847... ft = 57.936423... ft.This was a challenging problem! The solution by 'ironduke8159' is more elegant than mine, and I congratulate him.Live long and prosper.