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Difference between revisions of "2012 AIME II Problems/Problem 6"

m (→‎Solution)
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Squaring, we get <math>z^4 = 5 (25 - 100 + 100i) = -375 + 500i</math>. Finally, <math>c+d = -375 + 500 = \boxed{125}</math>
 
Squaring, we get <math>z^4 = 5 (25 - 100 + 100i) = -375 + 500i</math>. Finally, <math>c+d = -375 + 500 = \boxed{125}</math>
 +
 +
== Solution 2 ==
 +
 +
WLOG, let <math>z_{1}=5(\cos{\theta_{1}}+i\sin{\theta_{1}})</math> and
 +
 +
<math>z_{2}=1+2i=\sqrt{5}(\cos{\theta_{2}+i\sin{\theta_{2}}})</math>
 +
 +
This means that
 +
 +
<math>z_{1}^3=125(\cos{3\theta_{1}}+i\sin{3\theta_{1}})</math>
 +
 +
<math>z_{1}^4=625(\cos{4\theta_{1}}+i\sin{4\theta_{1}})</math>
 +
 +
Hence, this means that
 +
 +
<math>z_{2}z_{1}^3=125\sqrt{5}(\cos({\theta_{2}+3\theta_{1}})+i\sin({\theta_{2}+3\theta_{1}}))</math>
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 +
And
 +
 +
<math>z_{1}^5=3125(\cos{5\theta_{1}}+i\sin{5\theta_{1}})</math>
 +
 +
Now, common sense tells us that the distance between these two complex numbers is maxed when they both are points satisfying the equation of the line <math>yi=mx</math>, or when they are each a <math>180^{\circ}</math> rotation away from each other.
 +
 +
Hence, we must have that <math>5\theta_{1}=3\theta_{1}+\theta_{2}+180^{\circ}\implies\theta_{1}=\frac{\theta_{2}+180^{\circ}}{2}</math>
 +
 +
Now, plug this back into <math>z_{1}^4</math>(if you want to know why, reread what we want in the problem!)
 +
 +
So now, we have that
 +
<math>z_{1}^4=625(\cos{2\theta_{2}}+i\sin{2\theta_{2}})</math>
 +
 +
Notice that <math>\cos\theta_{2}=\frac{1}{\sqrt{5}}</math> and <math>\sin\theta_{2}=\frac{2}{\sqrt{5}}</math>
 +
 +
Then, we have that <math>\cos{2\theta_{2}}=\cos^2{\theta_{2}}-\sin^2{\theta_{2}}=-\frac{3}{5}</math> and <math>\sin{2\theta_{2}}=2\sin{\theta_{2}}\cos{\theta_{2}}=\frac{4}{5}</math>
 +
 +
Finally, plugging back in, we find that <math>z_{1}^4=625(-\frac{3}{5}+\frac{4}{5})=-375+500i</math>
 +
 +
<math>-375+500=\boxed{125}</math>
 +
 +
 +
  
 
== See Also ==
 
== See Also ==
 
{{AIME box|year=2012|n=II|num-b=5|num-a=7}}
 
{{AIME box|year=2012|n=II|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:46, 8 December 2017

Problem 6

Let $z=a+bi$ be the complex number with $\vert z \vert = 5$ and $b > 0$ such that the distance between $(1+2i)z^3$ and $z^5$ is maximized, and let $z^4 = c+di$. Find $c+d$.

Solution

Let's consider the maximization constraint first: we want to maximize the value of $|z^5 - (1+2i)z^3|$ Simplifying, we have

$|z^3| * |z^2 - (1+2i)|$

$=|z|^3 * |z^2 - (1+2i)|$

$=125|z^2 - (1+2i)|$

Thus we only need to maximize the value of $|z^2 - (1+2i)|$.

To maximize this value, we must have that $z^2$ is in the opposite direction of $1+2i$. The unit vector in the complex plane in the desired direction is $\frac{-1}{\sqrt{5}} + \frac{-2}{\sqrt{5}} i$. Furthermore, we know that the magnitude of $z^2$ is $25$, because the magnitude of $z$ is $5$. From this information, we can find that $z^2 = \sqrt{5} (-5 - 10i)$

Squaring, we get $z^4 = 5 (25 - 100 + 100i) = -375 + 500i$. Finally, $c+d = -375 + 500 = \boxed{125}$

Solution 2

WLOG, let $z_{1}=5(\cos{\theta_{1}}+i\sin{\theta_{1}})$ and

$z_{2}=1+2i=\sqrt{5}(\cos{\theta_{2}+i\sin{\theta_{2}}})$

This means that

$z_{1}^3=125(\cos{3\theta_{1}}+i\sin{3\theta_{1}})$

$z_{1}^4=625(\cos{4\theta_{1}}+i\sin{4\theta_{1}})$

Hence, this means that

$z_{2}z_{1}^3=125\sqrt{5}(\cos({\theta_{2}+3\theta_{1}})+i\sin({\theta_{2}+3\theta_{1}}))$

And

$z_{1}^5=3125(\cos{5\theta_{1}}+i\sin{5\theta_{1}})$

Now, common sense tells us that the distance between these two complex numbers is maxed when they both are points satisfying the equation of the line $yi=mx$, or when they are each a $180^{\circ}$ rotation away from each other.

Hence, we must have that $5\theta_{1}=3\theta_{1}+\theta_{2}+180^{\circ}\implies\theta_{1}=\frac{\theta_{2}+180^{\circ}}{2}$

Now, plug this back into $z_{1}^4$(if you want to know why, reread what we want in the problem!)

So now, we have that $z_{1}^4=625(\cos{2\theta_{2}}+i\sin{2\theta_{2}})$

Notice that $\cos\theta_{2}=\frac{1}{\sqrt{5}}$ and $\sin\theta_{2}=\frac{2}{\sqrt{5}}$

Then, we have that $\cos{2\theta_{2}}=\cos^2{\theta_{2}}-\sin^2{\theta_{2}}=-\frac{3}{5}$ and $\sin{2\theta_{2}}=2\sin{\theta_{2}}\cos{\theta_{2}}=\frac{4}{5}$

Finally, plugging back in, we find that $z_{1}^4=625(-\frac{3}{5}+\frac{4}{5})=-375+500i$

$-375+500=\boxed{125}$



See Also

2012 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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All AIME Problems and Solutions

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