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Algebra Difficulty 5.1 AIME, harder Find the answer

Let a,b,c,x,ya, b, c, x, y, and zz be complex numbers such that a=b+cx2,b=c+ay2,c=a+bz2a=\frac{b+c}{x-2}, \quad b=\frac{c+a}{y-2}, \quad c=\frac{a+b}{z-2}. If xy+yz+zx=67x y+y z+z x=67 and x+y+z=2010x+y+z=2010, find the value of xyzx y z.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Manipulate the equations to get a common denominator: a=b+cx2x2=a=\frac{b+c}{x-2} \Longrightarrow x-2= b+cax1=a+b+ca1x1=aa+b+c\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 $\frac{1}{x-1}+\frac{1}{y-1}+\frac{1}{z-1} =1 \Rightarrow (y-1)(z-1)+(x-1)(z-1)+(x-1)(y-1) =(x-1)(y-1)(z-1) \Rightarrow x y+y z+z x-2(x+y+z)+3 =x y z-(x y+y z+z x)+(x+y+z)-1 \Rightarrow x y z-2(x y+y z+z x)+3(x+y+z)-4 =0 \Rightarrow x y z-2(67)+3(2010)-4 =0 \Rightarrow x y z =-5892

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