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Difference between revisions of "2010 AIME II Problems/Problem 7"

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== Problem 7 ==
 
== Problem 7 ==
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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>.  
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<!-- 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 <math>a</math>, <math>b</math>, and <math>c</math> 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 ==
 
−
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.
+
== Solution (vieta's) ==
 +
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>.
  
−
Start with a, since it's the easiest one to do: <math>y+3+y+9+2y=0, y=-3</math>
+
Since <math>a,b,c\in{R}</math>, the imaginary part of <math>a,b,c</math> must be <math>0</math>.
  
−
and therefore: <math>x_1 = x</math>, <math>x_2 = x+6i</math>, <math>x_3 = 2x-4-6i</math>
+
Start with a, since it's the easiest one to do: <math>y+3+y+9+2y=0, y=-3</math>,
  
−
now, do the part where the imaginery part of c is 0, since it's the second easiest one to do:  
+
and therefore: <math>x_1 = x</math>, <math>x_2 = x+6i</math>, <math>x_3 = 2x-4-6i</math>.
−
<math>x(x+6i)(2x-4-6i)</math>, the imaginery part is: <math>6x^2-24x</math>, which is 0, and therefore x=4, since x=0 don't work
 
  
−
so now, <math>x_1 = 4, x_2 = 4+6i, x_3 = 4-6i</math>
+
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 <math>x=4</math>, since <math>x=0</math> doesn't work.
  
−
and therefore: <math>a=-12, b=84, c=-208</math>, and finally, we have <math>|a+b+c|=|-12+84-208|=136</math>.
+
So now, <math>x_1 = 4, x_2 = 4+6i, x_3 = 4-6i</math>,
 +
 
 +
and therefore: <math>a=-12, b=84, c=-208</math>. Finally, we have <math>|a+b+c|=|-12+84-208|=\boxed{136}</math>.
 +
 
 +
=== Solution 1b ===
 +
 
 +
Same as solution 1 except that when you get to <math>x_1 = x</math>, <math>x_2 = x+6i</math>, <math>x_3 = 2x-4-6i</math>, you don't need to find the imaginary part of <math>c</math>. We know that <math>x_1</math> is a real number, which means that <math>x_2</math> and <math>x_3</math> are complex conjugates. Therefore, <math>x=2x-4</math>.
 +
 
 +
== Solution 2 (casework) ==
 +
 
 +
Note that at least one of <math>w+3i</math>, <math>w+9i</math>, or <math>2w-4</math> is real by complex conjugate roots. We now separate into casework based on which one.
 +
 
 +
Let <math>w=x+yi</math>, where <math>x</math> and <math>y</math> are reals.
 +
 
 +
Case 1: <math>w+3i</math> is real. This implies that <math>x+yi+3i</math> is real, so by setting the imaginary part equal to zero we get <math>y=-3</math>, so <math>w=x-3i</math>. Now note that since <math>w+3i</math> is real, <math>w+9i</math> and <math>2w-4</math> are complex conjugates. Thus <math>\overline{w+9i}=2w-4</math>, so <math>\overline{x+6i}=2(x-3i)-4</math>, implying that <math>x=4</math>, so <math>w=4-3i</math>.
 +
 
 +
Case 2: <math>w+9i</math> is real. This means that <math>x+yi+9i</math> is real, so again setting imaginary part to zero we get <math>y=-9</math>, so <math>w=x-9i</math>. Now by the same logic as above <math>w+3i</math> and <math>2w-4</math> are complex conjugates. Thus <math>\overline{w+3i}=2w-4</math>, so <math>\overline{x-6i}=2(x-9i)-4</math>, so <math>x+6i=2x-4-18i</math>, which has no solution as <math>x</math> is real.
 +
 
 +
Case 3: <math>2w-4</math> is real. Going through the same steps, we get <math>y=0</math>, so <math>w=x</math>. Now <math>w+3i</math> and <math>w+6i</math> are complex conjugates, but <math>w=x</math>, which means that <math>\overline{x+3i}=x+6i</math>, so <math>x-3i=x+6i</math>, which has no solutions.
 +
 
 +
Thus case 1 is the only one that works, so <math>w=4-3i</math> and our polynomial is <math>(z-(4))(z-(4+6i))(z-(4-6i))</math>. Note that instead of expanding this, we can save time by realizing that the answer format is <math>|a+b+c|</math>, so we can plug in <math>z=1</math> to our polynomial to get the sum of coefficients, which will give us <math>a+b+c+1</math>. Plugging in <math>z=1</math> into our polynomial, we get <math>(-3)(-3-6i)(-3+6i)</math> which evaluates to <math>-135</math>. Since this is <math>a+b+c+1</math>, we subtract 1 from this to get <math>a+b+c=-136</math>, so <math>|a+b+c|=\boxed{136}</math>.
 +
 
 +
~chrisdiamond10
 +
 
 +
== Solution 3 (skibid) ==
 +
 
 +
By vieta's we know the sum of the roots must be <math>-a</math>, a real number. That means <math>4w+12i-4</math> is a real number, meaning <math>w</math> has an imaaginary component of <math>-3i</math>.
 +
 
 +
 
 +
Now we write <math>w = x-3i</math>. Then, <math>w+3i</math> is the real root, meaning the other two are complex conjugates. We have <math>\overline{x+6i} = 2x-4-6i</math>, and solving, we get <math>x=4</math>. Then, <math>f(x) = (x-4)(x-4-6i)(x-4+6i) = (x-4)(x^2-8x+52)</math>.
 +
 
 +
 
 +
We get <math>|a+b+c| = |-12+84-208| = \boxed{136}</math>.
 +
 
 +
 
 +
-skibbysiggy
 +
 
 +
== See also ==
 +
{{AIME box|year=2010|num-b=6|num-a=8|n=II}}
 +
{{MAA Notice}}

Latest revision as of 15:40, 19 September 2024

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 (vieta's)

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 $|a+b+c|=|-12+84-208|=\boxed{136}$.

Solution 1b

Same as solution 1 except that when you get to $x_1 = x$, $x_2 = x+6i$, $x_3 = 2x-4-6i$, you don't need to find the imaginary part of $c$. We know that $x_1$ is a real number, which means that $x_2$ and $x_3$ are complex conjugates. Therefore, $x=2x-4$.

Solution 2 (casework)

Note that at least one of $w+3i$, $w+9i$, or $2w-4$ is real by complex conjugate roots. We now separate into casework based on which one.

Let $w=x+yi$, where $x$ and $y$ are reals.

Case 1: $w+3i$ is real. This implies that $x+yi+3i$ is real, so by setting the imaginary part equal to zero we get $y=-3$, so $w=x-3i$. Now note that since $w+3i$ is real, $w+9i$ and $2w-4$ are complex conjugates. Thus $\overline{w+9i}=2w-4$, so $\overline{x+6i}=2(x-3i)-4$, implying that $x=4$, so $w=4-3i$.

Case 2: $w+9i$ is real. This means that $x+yi+9i$ is real, so again setting imaginary part to zero we get $y=-9$, so $w=x-9i$. Now by the same logic as above $w+3i$ and $2w-4$ are complex conjugates. Thus $\overline{w+3i}=2w-4$, so $\overline{x-6i}=2(x-9i)-4$, so $x+6i=2x-4-18i$, which has no solution as $x$ is real.

Case 3: $2w-4$ is real. Going through the same steps, we get $y=0$, so $w=x$. Now $w+3i$ and $w+6i$ are complex conjugates, but $w=x$, which means that $\overline{x+3i}=x+6i$, so $x-3i=x+6i$, which has no solutions.

Thus case 1 is the only one that works, so $w=4-3i$ and our polynomial is $(z-(4))(z-(4+6i))(z-(4-6i))$. Note that instead of expanding this, we can save time by realizing that the answer format is $|a+b+c|$, so we can plug in $z=1$ to our polynomial to get the sum of coefficients, which will give us $a+b+c+1$. Plugging in $z=1$ into our polynomial, we get $(-3)(-3-6i)(-3+6i)$ which evaluates to $-135$. Since this is $a+b+c+1$, we subtract 1 from this to get $a+b+c=-136$, so $|a+b+c|=\boxed{136}$.

~chrisdiamond10

Solution 3 (skibid)

By vieta's we know the sum of the roots must be $-a$, a real number. That means $4w+12i-4$ is a real number, meaning $w$ has an imaaginary component of $-3i$.


Now we write $w = x-3i$. Then, $w+3i$ is the real root, meaning the other two are complex conjugates. We have $\overline{x+6i} = 2x-4-6i$, and solving, we get $x=4$. Then, $f(x) = (x-4)(x-4-6i)(x-4+6i) = (x-4)(x^2-8x+52)$.


We get $|a+b+c| = |-12+84-208| = \boxed{136}$.


-skibbysiggy

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

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