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From: larsen@brahms (Michael Larsen)
Newsgroups: sci.math.stat
Subject: Re: Triangles in Space
Message-ID: <183@cartan.Berkeley.EDU>
Date: Wed, 5-Nov-86 13:48:34 EST
Article-I.D.: cartan.183
Posted: Wed Nov  5 13:48:34 1986
Date-Received: Wed, 5-Nov-86 22:36:46 EST
References: <1550001@hpcnof.UUCP> <200@clan.UUCP>
Sender: daemon@cartan.Berkeley.EDU
Reply-To: larsen@brahms (Michael Larsen)
Distribution: net
Organization: Math Dept. UC Berkeley
Lines: 26
Keywords: triangles, probabilistic geometry, silly Euclidean preferences

In article <2664@curly.ucla-cs.ARPA> cc@LOCUS.UCLA.EDU (Mitch) writes:
>
>   If we must think of it in terms of pure geometry, why this silly 
>preference for a space with zero curvature?  If we assume positive
>curvature, as for Riemannian spherical geometry, we lend an entirely 
>new twist to the problem.
>   Seriously, consider three points chosen in such a space - I think the
>problem is a little more interesting.
>
>   Mitch Gunzler
>
	Unlike the plane, the sphere admits a uniform probability distribution.
Choosing 3 points at random, each point of the resulting spherical triangle
has an expected angle of pi/2.  Indeed, let the first point be the north pole
(without loss of generality).  Let the second point fall (again without loss
of generality) on the Greenwich longitude line.  The longitude of the third
point is uniformly distributed between -pi and pi.  The expected absolute
value of this angle is thus pi/2.

	It follows that the expected total angle of random spherical triangles
is 3pi/2.  This is fine in a positively curved space.  In fact, by a classical
spherical geometry theorem, this is equivalent to the statement that the
average spherical triangle on a sphere has area 1/8 of the total area of the
sphere.  Anyone see a method for computing the constant 1/8 directly?

larsen @ berkeley.edu.brahms