Difference between revisions of "2010 AIME II Problems/Problem 7"
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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-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>. | ||
− | Since <math>a,b,c\in{R}</math>, the imaginary part of <math>a,b,c</math> must be 0. | + | Since <math>a,b,c\in{R}</math>, the imaginary part of <math>a,b,c</math> must be <math>0</math>. |
Start with a, since it's the easiest one to do: <math>y+3+y+9+2y=0, y=-3</math>, | Start with a, since it's the easiest one to do: <math>y+3+y+9+2y=0, y=-3</math>, | ||
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and therefore: <math>x_1 = x</math>, <math>x_2 = x+6i</math>, <math>x_3 = 2x-4-6i</math>. | and therefore: <math>x_1 = x</math>, <math>x_2 = x+6i</math>, <math>x_3 = 2x-4-6i</math>. | ||
− | Now, do the part where the imaginary part of c is 0 | + | Now, do the part where the imaginary part of c is 0 since it's the second easiest one to do: |
− | <math>x(x+6i)(2x-4-6i)</math>. The imaginary part is | + | <math>x(x+6i)(2x-4-6i)</math>. The imaginary part is <math>6x^2-24x</math>, which is 0, and therefore <math>x=4</math>, since <math>x=0</math> doesn't work. |
So now, <math>x_1 = 4, x_2 = 4+6i, x_3 = 4-6i</math>, | So now, <math>x_1 = 4, x_2 = 4+6i, x_3 = 4-6i</math>, |
Revision as of 23:07, 13 August 2017
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.