Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83; site mcgill-vision.UUCP Path: utzoo!linus!philabs!micomvax!musocs!mcgill-vision!mouse From: mouse@mcgill-vision.UUCP (der Mouse) Newsgroups: net.puzzle,net.math Subject: Re: Polar Bear Problem Sequel Message-ID: <327@mcgill-vision.UUCP> Date: Sun, 3-Nov-85 01:21:08 EST Article-I.D.: mcgill-v.327 Posted: Sun Nov 3 01:21:08 1985 Date-Received: Tue, 5-Nov-85 07:55:23 EST References: <855@whuxlm.UUCP> <593@hou2c.UUCP> <373@faron.UUCP>, <374@faron.UUCP> Organization: McGill University, Montreal Lines: 23 Xref: linus net.puzzle:1058 net.math:2108 [ selected lines ] > such that the radius of a great-circle running E-W is the same as that > of another great-circle running E-W which is 1 mile south. The only place > this happens is the great-circle 1/2 mile north of the equator. Moving > 1 mile south places you on the great-circle 1/2 mile south of the equator > Thus, of the entire set of great circles (cardinality C) only 1 satisfies > At latitude 90-theta, an East-West great circle has radius 2 PI r sin(theta) a great circle. This does not bring one back to the point where one started Then walking 1 mile north places one back on the original great circle, only Correct me if I'm wrong, but isn't a great circle a circle with its center at the center of the earth (yes, I know the earth isn't a sphere, but this discussion is pretending it is)? Everyone here seems to be using it to mean a circle of constant latitude. -- der Mouse {ihnp4,decvax,akgua,etc}!utcsri!mcgill-vision!mouse philabs!micomvax!musocs!mcgill-vision!mouse Hacker: One responsible for destroying / Wizard: One responsible for recovering it afterward