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07262007  #1 
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mathematics?
You MUST offer mathematical proof to sustain your answer to this question:Stake a goat out anywhere on the perimiter of a field that is 100 ft. in diameter. ( area of field = 7857.143 sq.ft. )What will be the length of his tether to allow him to eat exactly half of the grass in the area of the field. ( 3928.571 sq.ft. )It's not 50 ft.

07262007  #2 
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Call the rope length r and field radius R=50.We want the followinggoat area) / (whole area) = 1/2(Ï€rÂ²) / (Ï€RÂ²) = 1/2So we simply solverÂ²) / (RÂ²) = 1/2rÂ² = RÂ²/2âˆš(rÂ²) = âˆš(RÂ²/2)r = R/âˆš(2) = 50 / âˆš(2)approximately 35.3553391 ft

07262007  #3 
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You know that the area of the field is 7857.143sq ft. This means that the area you want the goat to be able to eat on is half of that, or 3928.571 sq ft. Know that A= Ï€RÂ²You can simply solve3928.571 = Ï€RÂ²âˆš(3928.571/Ï€) = RR = 35.3624 feet

07262007  #4 
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Area of asymmetrical lens. Can't remember how to compute 1/cos after 35 years  but this is the method. (See Reference for formula)r > 50 ft probably 5460 ftDoorrat and cheeser are wrong because they do not figure area from perimeter
mathworld.wolfram.com 
07262007  #5 
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What people seem to be missing is that the original field is circular, and that now the goat is being staked out at the edge. So he doesn't get to eat within a circular region, but within the intersection of his ropecircle with the original fieldcircle.I'm afraid I don't have the drawing tools to explain this very well, but I can do the calculation for you:a) Goat's rope is tied at the leftmost end of the circular field. If he restricts himself to the region between radius r and radius (r + dr), he can eat within a partial annulus with radial depth dr, and with angular extension +/ theta, where theta is the angle at which he runs out of field. Through some simple geometry, which I cannot illustrate for you:r/2 = Rsin(A/2)A = 2*Arcsin(r/(2R))2*theta + A = pitheta = (pi  A)/2 , sotheta = pi/2  Arcsin(r/(2R))(Note that when r = 0, theta = pi/2; and when r = 2R, theta = 0. That's right!)So the area the goat can eat in is, for rope length L:integral [r*(2 theta) dr] {from r = 0 to r = L}= integral [pi*r dr]  integral [Arcsin(r/(2R)*rdr]= pi*r^2/2  integral [r * Arcsin(r/(2R) dr]= pi*L^2/2  integral [r * Arcsin(r/(2R) dr]To do the integral:Let z = r/2R , so thenr = 2RzLet y = Arcsin(r/(2R)) = Arcsin(z) , so thensin(y) = z and thenr = 2R sin(y) anddr = 2R cos(y) dySo: integral [r * Arcsin(r/(2R) dr] = integral [2Rsin(y)*y*2Rcos(y)dy]= 4R^2 integral [y sin(y)cos(y)dy] {y = 0 to Arcsin(L/(2R))}= 2R^2 integral [y * 2 sin(y) cos(y) dy]= R^2 integral [y * sin(2y) d(2y)]= R^2 integral [u * sin(u) du]/2 {u = 0 to u = 2*Arcsin(L/(2R))}= (R^2/2) integral [u sin(u) du]But we can integrate this by parts:integral [u sin(u) du] = integral [ u*d(cos(u))] = u*cos(u) + integral [cos(u) du] =  u*cos(u) + sin(u)At u = 0, this is 0At u = 2*Arcsin(L/(2R))cos(u) = cos(2*Arcsin(L/(2R)) = (cos(Arcsin(L/(2R))^2  sin(Arcsin(L/(2R))^2 = 1  2*(sin(Arcsin(L/2(R))^2 = 1  2* (L/(2R))^2sin(u) = 1  (cos(u))^2 = 1  (1  2*(L/(2R))^2)^2) = 1  (1 4(L/(2R))^2 + 4(L/(2R))^4) = 4(L/(2R))^2  4 (L/(2R))^4 = 4(L/(2R))^2(1  (L/(2R))^2)So this integral is: 2*Arcsin(L/(2R) )(1  2((L/(2R))^2) + 4(L/(2R))^2 (1  (L/(2R))^2)This is pretty awful:Multiply by R^2/2: R^2(Arcsin(L/(2R)) (1  2 (L/(2R))^2) + 2R^2(L/(2R))^2(1(L/(2R))^2)Subtract from pi*L^2/2i*L^2/2 + R^2 (Arcsin(L/(2R))(1  2(L/(2R))^2)  2R^2 (L/(2R)^2(1  (L/(2R))^2)Sanity check: If L = 2R, this givesi(2R)^2/2 + R^2(Arcsin(1))(11)  2R^2(1)(11)= 2piR^2Nope, the answer should be pi*R^2, the original area of the field. I can't do long calculations using this keyboard, and I can't draw pictures. So I've made a mistake somewhere.I'll describe the calculation in words, and maybe you can do it on your own:a) On a very tight rope, the goat can get a little bit of the farwestern bit of the original circular lawn. The angle he can traverse is from south to north: +/ 90degrees.b) On a looser rope, he can get farther in towards the center. But his angle is restricted, because he runs out of lawn on north and south.c) When his rope has length sqrt(2) R, he can just reach the northmost and southmost bits of the lawn, so his angle of range is +/ 45 degrees.d) When his rope gets even longer, he can cover more and more of the lawn, but his angle of range starts to decrease, because he's running out of lawn.e) Finally, when his rope has length = 2R, he's eaten the entire lawn, and his range is just the easternmost tip of the lawn. So, the solution to the problem is when the total integrated range = pi*R^2/2.

07262007  #6 
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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 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.

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