Difference between revisions of "2010 AIME II Problems/Problem 7"
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and therefore: <math>a=-12, b=84, c=-208</math>. Finally, we have <math>|a+b+c|=|-12+84-208|=\boxed{136}</math>. | and therefore: <math>a=-12, b=84, c=-208</math>. Finally, we have <math>|a+b+c|=|-12+84-208|=\boxed{136}</math>. | ||
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+ | == Solution 1b == | ||
+ | |||
+ | Same as solution 1 except that when you get to <math>x_1 = x</math>, <math>x_2 = x+6i</math>, <math>x_3 = 2x-4-6i</math>, you don't need to find the imaginary part of <math>c</math>. We know that <math>x_1</math> is a real number, which means that <math>x_2</math> and <math>x_3</math> are complex conjugates. Therefore, <math>x=2x-4</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2010|num-b=6|num-a=8|n=II}} | {{AIME box|year=2010|num-b=6|num-a=8|n=II}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:26, 30 July 2020
Contents
Problem 7
Let , where a, b, and c are real. There exists a complex number such that the three roots of are , , and , where . Find .
Solution
Set , so , , .
Since , the imaginary part of must be .
Start with a, since it's the easiest one to do: ,
and therefore: , , .
Now, do the part where the imaginary part of c is 0 since it's the second easiest one to do: . The imaginary part is , which is 0, and therefore , since doesn't work.
So now, ,
and therefore: . Finally, we have .
Solution 1b
Same as solution 1 except that when you get to , , , you don't need to find the imaginary part of . We know that is a real number, which means that and are complex conjugates. Therefore, .
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.