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From: rwhaas@ihuxf.UUCP (Roy W. Haas)
Newsgroups: net.math
Subject: Re: UCSB competition problem
Message-ID: <1633@ihuxf.UUCP>
Date: Wed, 16-Nov-83 10:06:42 EST
Article-I.D.: ihuxf.1633
Posted: Wed Nov 16 10:06:42 1983
Date-Received: Thu, 17-Nov-83 06:14:51 EST
References: <345@sun.UUCP>
Organization: AT&T Bell Labs, Naperville, Il
Lines: 22

Recall:

		x^2 = 1 + (y-z)^2
		y^2 = 2 + (z-x)^2
		z^2 = 3 + (x-y)^2

Actually, no trig is necessary to solve this system. The "trick" if
there is one, is to realize that any pair of the equations yield a
pair of linear equations. For example, the first two may be used
to solve for x and y in terms of z, and then the value of z is
found from the third equation. There are two solutions, namely

		(x,y,z)= +-(5,8,9)/sqrt(24)

the second solution comes from the obvious fact that all the signs
may be changed. The identity a^2-b^2 = (a+b)(a-b) is handy
in solving a given pair.


				Roy Haas
				ihuxf!rwhaas