[net.math] UCSB competition problem

cochon@sun.UUCP (Philippe Lacroute) (11/15/83)

I recently participated in a prize competition sponsored by the College
of Creative Studies, UC Santa Barbara.  I can't figure out one of the
problems:

	Solve the system of equations

		 2               2
		x   = 1 + (y - z)

		 2               2
		y   = 2 + (x - z)

		 2               2
		z   = 3 + (x - y)

Note: no higher math (above high school trig) should be necessary.
	There is probably some clever trick to solve it which I can't
	figure out.

				Philippe Lacroute
				...decvax!sun!cochon

rjnoe@ihlts.UUCP (11/15/83)

Actually, only algebra is needed.  Rearrange to get:
		x^2 - (y - z)^2  =  1
		y^2 - (x - z)^2  =  2
		z^2 - (x - y)^2  =  3
and let
		A =  x + y - z
		B =  x - y + z
		C = -x + y + z
Then the system of equations is
		AB = 1
		CA = 2
		BC = 3
Which can be quickly solved to get A = sqrt(2/3), B = sqrt(3/2), and
C = sqrt(6).  Of course, the negatives of these work just as well.  Now
		2x = A + B = 5/sqrt(6)
		2y = A + C = 4*sqrt(2/3)
		2z = B + C = 3*sqrt(3/2)
therefore
		        5
		x = ---------
		    2*sqrt(6)

		y = 2*sqrt(2/3)


		z = (3/2) ^ (3/2)

As I've said, there is one other solution, taking the additive inverses of
all of these.
-- 
		Roger Noe		...ihnp4!ihlts!rjnoe

rwhaas@ihuxf.UUCP (Roy W. Haas) (11/16/83)

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

sonnens@mprvaxa (11/18/83)

Here's a solution to the following recently posed problem:

		 2              2
		x  = 1 + (y - z)

		 2              2
		y  = 2 + (x - z)

		 2              2
		z  = 3 + (x - y)


If b = sqrt(24), then (x,y,z) = (5/b,8/b,9/b) or -(5/b,8/b,9/b).

The trick is to write each equation as a difference of squares, which
can then be factored.  There are a lot of identical factors, which can
be cancelled out to reduce this to a system of three linear equations
(e.g., 2(x + y - z)(x - y + z) = (y + x - z)(y - x + z)).

Dan Sonnenschein
Microtel Pacific Research
...microsoft!uw-beaver!ubc-visi!mprvaxa!sonnens