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 Î¸. Almost everything needed (except for just that ONE angle, Î¸ = 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 2Î¸ 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 2Î± at the centre, the area of the full 2Î± sector minus the relevant triangle gives a segment area of (Î±  sinÎ± cosÎ±) r^2.(This is easy to derive, and readily checked at special values of Î±, such as Î± = 0, Ï€/2, and Ï€.) APPLICATIONS. From the lemma, the area of circle 1's segment is simply Î¸  sinÎ¸ cos Î¸;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 isÎ¸  sinÎ¸ cos Î¸ + (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 cosÎ¸ = 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* + Î¸ = 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 Î¸ for cos^(1) d):Î¸  d(1  d^2)^(1/2) + [Ï€  Î¸  (1  d^2)^(1/2)](1  d) =Ï€/2.With Î¸ = cos^(1) d, we can iterate this equation to obtain the solution d (= cosÎ¸) = 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.
