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2010 AIME II Problems/Problem 7

Revision as of 10:51, 3 April 2010 by Sinkokuyou (talk | contribs)

set $w=x+yi$, so $x_1 = x+(y+3)i$, $x_2 = x+(y+9)i$, $x_3 = 2x-4+2yi$. Since $a,b,c\in{R}$, the imaginary part of a,b,c must be 0. Start with a, since it's the easiest one to do: $y+3+y+9+2y=0, y=-3$ and therefore: $x_1 = x+i$, $x_2 = x+6i$, $x_3 = 2x-4-6i$

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