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Algebra Difficulty 5.1 AIME, harder Prove it United States

Problem:
Let aa, bb, cc, xx, yy, and zz be complex numbers such that
a=b+cx2,b=c+ay2,c=a+bz2. a = \frac{b+c}{x-2}, \quad b = \frac{c+a}{y-2}, \quad c = \frac{a+b}{z-2}.
If xy+yz+zx=67xy + yz + zx = 67 and x+y+z=2010x + y + z = 2010, find the value of xyzxyz.

Solution

Solution:
Manipulate the equations to get a common denominator: a=b+cx2x2=b+cax1=a+b+ca1x1=aa+b+ca = \frac{b+c}{x-2} \Longrightarrow x-2 = \frac{b+c}{a} \Longrightarrow x-1 = \frac{a+b+c}{a} \Longrightarrow \frac{1}{x-1} = \frac{a}{a+b+c}; similarly, 1y1=ba+b+c\frac{1}{y-1} = \frac{b}{a+b+c} and 1z1=ca+b+c\frac{1}{z-1} = \frac{c}{a+b+c}. Thus
1x1+1y1+1z1=1(y1)(z1)+(x1)(z1)+(x1)(y1)=(x1)(y1)(z1)xy+yz+zx2(x+y+z)+3=xyz(xy+yz+zx)+(x+y+z)1xyz2(xy+yz+zx)+3(x+y+z)4=0xyz2(67)+3(2010)4=0xyz=5892 \begin{aligned} \frac{1}{x-1} + \frac{1}{y-1} + \frac{1}{z-1} & = 1 \\ (y-1)(z-1) + (x-1)(z-1) + (x-1)(y-1) & = (x-1)(y-1)(z-1) \\ xy + yz + zx - 2(x + y + z) + 3 & = xyz - (xy + yz + zx) + (x + y + z) - 1 \\ xyz - 2(xy + yz + zx) + 3(x + y + z) - 4 & = 0 \\ xyz - 2(67) + 3(2010) - 4 & = 0 \\ xyz & = -5892 \end{aligned}

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.