I'll try only the case when x,y and z are all +ve, while noting that if (x,y,z) is a solution, so is (-x,-y,-z).
In that case the equations are consistent with the following situation:
Consider a triangle ABC, with AB = a2, BC = b2 and CA = c2.
Consider the point P in the interior of the triangle such that Angles APB, BPC and and CPA are all 120o. Let CP=x, AP =y and BP = z.
From the cosine formula, we can see that x, y and z satisfy
z2 + zy + y2 = a2
z2 + zx + x2 = b2 and
x2 + xy + y2 = c2
The area of triangle ABC can be got in two ways:
1. By Heron's formula for a triangle with sides a2, b2 and c2
S = sqrt(s(s-a2)(s-b2)(s-c2) where s = a2+b2+c2/2
2. By sine formula as 0.5*(xysin1200 + yzsin1200 + zxsin1200)
=
3/4
xy
Call
x
2 as M and
xy as N
Adding the three equations given above we get 2M+N = a2+b2+c2
From the area equation, N = 2S/
3.
Hence (x+y+z)
2 = M+2N = 1/2(2M+N) + 3N/2 = (a
2+b
2+c
2)/2 +S
3
Hence (x+y+z) =
sqrt(M+2N) =
sqrt[1/2(a
2+b
2+c
2)+S
3)] =
sqrt(s/2 + S
3)
[Minus sign for the soln (-x,-y,-z) ]
Moderation Team
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