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