Difference between revisions of "2010 AIME II Problems/Problem 7"
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== Problem 7 == | == Problem 7 == | ||
<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let <math>P(z)=z^3+az^2+bz+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> | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let <math>P(z)=z^3+az^2+bz+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> | ||
Revision as of 14:46, 9 August 2018
Contents
Problem
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 
.
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. 
