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Difference between revisions of "2010 AIME II Problems/Problem 7"

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

Revision as of 10:51, 3 April 2010

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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