2018 AIME II Problems/Problem 5

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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

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