Annual Harvard-MIT Mathematics Tournament

Solution:
Manipulate the equations to get a common denominator: $a = \frac{b+c}{x-2} \Longrightarrow x-2 = \frac{b+c}{a} \Longrightarrow x-1 = \frac{a+b+c}{a} \Longrightarrow \frac{1}{x-1} = \frac{a}{a+b+c}$; similarly, $\frac{1}{y-1} = \frac{b}{a+b+c}$ and $\frac{1}{z-1} = \frac{c}{a+b+c}$. Thus
$$\begin{aligned} \frac{1}{x-1} + \frac{1}{y-1} + \frac{1}{z-1} & = 1 \\ (y-1)(z-1) + (x-1)(z-1) + (x-1)(y-1) & = (x-1)(y-1)(z-1) \\ xy + yz + zx - 2(x + y + z) + 3 & = xyz - (xy + yz + zx) + (x + y + z) - 1 \\ xyz - 2(xy + yz + zx) + 3(x + y + z) - 4 & = 0 \\ xyz - 2(67) + 3(2010) - 4 & = 0 \\ xyz & = -5892 \end{aligned}$$