 XP Math - Forums - View Single Post - mathematics? 07-26-2007 #6 Dr Spock Guest   Posts: n/a Someone who uses the CRUDE approximation 22/7 for ﾃ鞘ぎ 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 ﾃ鞘ぎ*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 ﾃ篠ｸ. Almost everything needed (except for just that ONE angle, ﾃ篠ｸ = 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 2ﾃ篠ｸ at O) and ABO (subtending an angle 2ﾃ鞘* 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 2ﾃ篠ｱ at the centre, the area of the full 2ﾃ篠ｱ sector minus the relevant triangle gives a segment area of (ﾃ篠ｱ - sinﾃ篠ｱ cosﾃ篠ｱ) r^2.(This is easy to derive, and readily checked at special values of ﾃ篠ｱ, such as ﾃ篠ｱ = 0, ﾃ鞘ぎ/2, and ﾃ鞘ぎ.) APPLICATIONS. From the lemma, the area of circle 1's segment is simply ﾃ篠ｸ - sinﾃ篠ｸ cos ﾃ篠ｸ;while circle R's segment is (ﾃ鞘* - sinﾃ鞘* cosﾃ鞘*) R^2.So the full area accessible to our ravenous kid (a "reduced goat") in the reduced problem isﾃ篠ｸ - sinﾃ篠ｸ cos ﾃ篠ｸ + (ﾃ鞘* - sinﾃ鞘* cosﾃ鞘*) R^2,and this must equal ﾃ鞘ぎ/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 cosﾃ篠ｸ = 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:2ﾃ鞘* + ﾃ篠ｸ = ﾃ鞘ぎ lead to sinﾃ鞘* cosﾃ鞘* = (1 - d^2)^(1/2) / 2.Putting this all together and requiring that the sum of the two segments be ﾃ鞘ぎ/2 gives a formula involving just cos^(-1) d and algebraic functions of d (I'll still write ﾃ篠ｸ for cos^(-1) d):ﾃ篠ｸ - d(1 - d^2)^(1/2) + [ﾃ鞘ぎ - ﾃ篠ｸ - (1 - d^2)^(1/2)](1 - d) =ﾃ鞘ぎ/2.With ﾃ篠ｸ = cos^(-1) d, we can iterate this equation to obtain the solution d (= cosﾃ篠ｸ) = 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.