Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/17/84; site dspo.UUCP Path: utzoo!watmath!clyde!burl!ulysses!allegra!bellcore!decvax!genrad!panda!talcott!harvard!seismo!cmcl2!lanl!dspo!tallman From: tallman@dspo.UUCP Newsgroups: net.puzzle Subject: Re: YAIQ (cow puzzle - SPOILER) Message-ID: <207@dspo.UUCP> Date: Tue, 2-Apr-85 11:07:28 EST Article-I.D.: dspo.207 Posted: Tue Apr 2 11:07:28 1985 Date-Received: Fri, 5-Apr-85 02:44:23 EST References: <1347@amdahl.UUCP> Distribution: net Organization: Los Alamos National Laboratory Lines: 64 > A farmer has a fenced 10-meter by 20-meter field the occupant > of which is a "point cow" {see footnote}. When tethering the > cow to a corner-post of the field, how long does the rope need > to be so that the cow can eat half of the grass in the field? > Kris Stephens (408-746-6047) {whatever}!amdahl!krs I can see why this is a good interview question - it is tempting to try to integrate the "slice out of a circle" shape, which gets very messy. An easier way is to divide the shape into a circular wedge and a triangle. ----------------- - - The o's mark the circular arc - - of the rope's maximum extent. - - The *'s mark a line from the -ooo - tether corner to the point of 2x - oo - intersection of the arc with - o - the fence. - o - Let r be the rope length, x - r * - the length of the short side (10m), - * - and phi the angle marked. -* phi - 0 < phi < pi/4. ----------------- x phi = arccos(x/r) r = x/cos(r) area of triangle = x*x*tan(phi)/2 area of circular wedge = pi*r^2*(1/4 - phi/(2*pi)) [fraction of a circle] desired area = x*(2*x)/2 = x^2 x^2 = x^2*tan(phi)/2 + pi*(x^2/cos^2(phi))*(1/4 - phi/(2*pi)) 1 = tan(phi)/2 +(pi/cos^2(phi))*(1/4 - phi/(2*pi)) cos^2(phi) = sin(phi)*cos(phi)/2 + pi*(1/4 - phi/(2*pi)) phi = sin(phi)*cos(phi) - 2*cos^2(phi) + pi/2 phi = sin(2*phi)/2 - (cos(2*phi) + 1) + pi/2 let theta = 2*phi, 0 < theta < pi/2 theta/2 = sin(theta)/2 - cos(theta) - 1 + pi/2 theta = sin(theta) - 2*cos(theta) - 2 + pi > This is a deceptive problem. The only answers I've seen were from > Engineers, who offered to tell me the answer "to as many decimal places > as I want" by successive approximation, which, as we all know, is _not_ > the answer. I have no problem with an honest numerical approximation if the solution can't be put in closed form (and very few real-world problems can be). By Newton's method I found: theta = 1.07927646 radians r = 10/cos(theta/2) = 11.646443 meters Perhaps the problem is a trick question because it does not say the cow must only reach exactly half the field. In this case, any length greater than this solution will do. Say 25 meters, for example. C. David Tallman - dspo!tallman@LANL or ucbvax!unmvax!lanl!dspo!tallman Los Alamos National Laboratory - E-10/Data Systems Los Alamos, New Mexico - (505) 667-8495 -- C. David Tallman - dspo!tallman@LANL or ucbvax!unmvax!lanl!dspo!tallman Los Alamos National Laboratory - E-10/Data Systems Los Alamos, New Mexico - (505) 667-8495