2010 AIME II Problems/Problem 7
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 (vieta's)
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, .
Solution 2 (casework)
Note that at least one of , , and is real by complex conjugate roots. We now separate into casework based on which one.
Let , where and are reals.
Case 1: is real. This implies that is real, so by setting the imaginary part equal to zero we get , so . Now note that since is real, and are complex conjugates. Thus , so , implying that , so .
Case 2: is real. This means that is real, so again setting imaginary part to zero we get , so . Now by the same logic as above and are complex conjugates. Thus , so , so , which has no solution as is real.
Case 3: is real. Going through the same steps, we get , so . Now and are complex conjugates, but , which means that , so , which has no solutions.
Thus case 1 is the only one that works, so and our polynomial is . Note that instead of expanding this, we can save time by realizing that the answer format is , so we can plug in to our polynomial to get the sum of coefficients, which will give us . Plugging in into our polynomial, we get which evaluates to . Since this is , we subtract 1 from this to get , so .
~chrisdiamond10
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 | ||
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