Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2010 AIME II Problems/Problem 7: Difference between revisions

← Older editNewer edit →
Mingxu (talk | contribs)
27 edits
Mingxu (talk | contribs)
27 edits
Line 5: Line 5:
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>,
Line 11: Line 11:
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, since it's the second easiest one to do:  
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>6x^2-24x</math>, which is 0, and therefore x=4, since x=0 doesn't work.
<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 22:07, 13 August 2017

Problem 7

Let $P(z)=z^3+az^2+bz+c$, where a, b, and c are real. There exists a complex number $w$ such that the three roots of $P(z)$ are $w+3i$, $w+9i$, and $2w-4$, where $i^2=-1$. Find $|a+b+c|$.

Solution

Set $w=x+yi$, so $x_1 = x+(y+3)i$, $x_2 = x+(y+9)i$, $x_3 = 2x-4+2yi$.

Since $a,b,c\in{R}$, the imaginary part of $a,b,c$ must be $0$.

Start with a, since it's the easiest one to do: $y+3+y+9+2y=0, y=-3$,

and therefore: $x_1 = x$, $x_2 = x+6i$, $x_3 = 2x-4-6i$.

Now, do the part where the imaginary part of c is 0 since it's the second easiest one to do: $x(x+6i)(2x-4-6i)$. The imaginary part is $6x^2-24x$, which is 0, and therefore $x=4$, since $x=0$ doesn't work.

So now, $x_1 = 4, x_2 = 4+6i, x_3 = 4-6i$,

and therefore: $a=-12, b=84, c=-208$. Finally, we have $|a+b+c|=|-12+84-208|=\boxed{136}$.

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

These problems are copyrighted © by the Mathematical Association of America.

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_7&oldid=87023"