AIME 2010 II · 第 7 题

Solution 1 (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

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 imaginary 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