2018 AIME II Problems/Problem 5

Problem

Suppose that $x$ , $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ , $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$ .

Solution

First we evaluate the magnitudes. $|xy|=80\sqrt{17}$ , $|yz|=60$ , and $|zx|=24\sqrt{17}$ . Therefore, $|x^2y^2z^2|=17\cdot80\cdot60\cdot24$ , or $|xyz|=240\sqrt{34}$ . Divide to find that $|z|=3\sqrt{2}$ , $|x|=40\sqrt{34}$ , and $|y|=10\sqrt{2}$ . $[asy] draw((0,0)--(4,0)); dot((4,0),red); draw((0,0)--(-4,0)); draw((0,0)--(0,-4)); draw((0,0)--(-4,1)); dot((-4,1),red); draw((0,0)--(-1,-4)); dot((-1,-4),red); draw((0,0)--(4,4),red); draw((0,0)--(4,-4),red); [/asy]$ This allows us to see that the argument of $y$ is $\frac{\pi}{4}$ , and the argument of $z$ is $-\frac{\pi}{4}$ . We need to convert the polar form to a standard form. Simple trig identities show $y=10+10i$ and $z=3-3i$ . More division is needed to find what $x$ is. $x=-20-12i$ $x+y+z=-7-5i$ $(-7)^2+(-5)^2=\boxed{074}$ $QED\blacksquare$ Written by a1b2

2018 AIME II (Problems • Answer Key • Resources)
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2018 AIME II Problems/Problem 5

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