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

Latest revision as of 19:44, 20 July 2025

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 $z_{1}^4=625(-\frac{3}{5}+\frac{4i}{5})=-375+500i$

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

Solution 3

Clearly, we want $\arg((1+2i)z^3) = \pi + \arg(z^5)$ . This is equivalent to $\arg(1+2i) = \pi + 2 \arg(z)$ by the additive property of the argument in complex multiplication. Because we want to find $z^4$ , we want an expression for $\arg(z^4)$ . We now have $2\arg(1+2i) = 2\pi + 4 \arg(z) \rightarrow \arg(-3+4i) = 4\arg(z)$ . Thus, $z^4$ is in the direction of $-3 + 4i$ . To achieve a magnitude of 5^4, we need to multiple $-3+4i$ by 125, so $z^4 = 125(-3+4i)$ and the answer is thus $\boxed{125}$ .

@@ Line 16: / Line 16: @@
 To maximize this value, we must have that <math>z^2</math> is in the opposite direction of <math>1+2i</math>. The unit vector in the complex plane in the desired direction is <math>\frac{-1}{\sqrt{5}} + \frac{-2}{\sqrt{5}} i</math>. Furthermore, we know that the magnitude of <math>z^2</math> is <math>25</math>, because the magnitude of <math>z</math> is <math>5</math>. From this information, we can find that <math>z^2 = \sqrt{5} (-5 - 10i)</math>
-Squaring, we get <math>z^4 = 5 (25 - 100 + 100i) = -375 + 500i</math>. Finally, <math>c+d = -375 + 500 = 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>
+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{4i}{5})=-375+500i</math>
+<math>-375+500=\boxed{125}</math>
+== Solution 3 ==
+Clearly, we want <math>\arg((1+2i)z^3) = \pi + \arg(z^5)</math>. This is equivalent to <math>\arg(1+2i) = \pi + 2 \arg(z)</math> by the additive property of the argument in complex multiplication. Because we want to find <math>z^4</math>, we want an expression for <math>\arg(z^4)</math>. We now have <math>2\arg(1+2i) = 2\pi + 4 \arg(z) \rightarrow \arg(-3+4i) = 4\arg(z)</math>. Thus, <math>z^4</math> is in the direction of <math>-3 + 4i</math>. To achieve a magnitude of 5^4, we need to multiple <math>-3+4i</math> by 125, so <math>z^4 = 125(-3+4i)</math> and the answer is thus <math>\boxed{125}</math>.
 == See Also ==
 {{AIME box|year=2012|n=II|num-b=5|num-a=7}}
 {{MAA Notice}}

2012 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search