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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

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