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 a,b,c must be 0. | + | Since <math>a,b,c\in{R}</math>, the imaginary part of <math>a,b,c</math> must be 0. |
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>, | ||
Revision as of 22:06, 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 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. 
