Difference between revisions of "2010 AIME II Problems/Problem 7"
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== Problem 7 == | == Problem 7 == | ||
− | Let <math>P(z)=x^3+ax^2+bx+c</math>, where a, b, and c are real. There exists a complex number <math>w</math> such that the three roots of <math>P(z)</math> are <math>w+3i</math>, <math>w+9i</math>, and <math>2w-4</math>, where <math>i^2=-1</math>. Find <math>|a+b+c|</math>. | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let <math>P(z)=x^3+ax^2+bx+c</math>, where a, b, and c are real. There exists a complex number <math>w</math> such that the three roots of <math>P(z)</math> are <math>w+3i</math>, <math>w+9i</math>, and <math>2w-4</math>, where <math>i^2=-1</math>. Find <math>|a+b+c|</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
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== Solution == | == Solution == | ||
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>. | 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>. |
Revision as of 17:52, 27 March 2015
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 a,b,c must be 0.
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 x=4, since x=0 doesn't work.
So now, ,
and therefore: . Finally, we have .
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.