Difference between revisions of "2018 AIME II Problems/Problem 5"

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Suppose that <math>x</math>, <math>y</math>, and <math>z</math> are complex numbers such that <math>xy = -80 - 320i</math>, <math>yz = 60</math>, and <math>zx = -96 + 24i</math>, where <math>i</math> <math>=</math> <math>\sqrt{-1}</math>. Then there are real numbers <math>a</math> and <math>b</math> such that <math>x + y + z = a + bi</math>. Find <math>a^2 + b^2</math>.
 
Suppose that <math>x</math>, <math>y</math>, and <math>z</math> are complex numbers such that <math>xy = -80 - 320i</math>, <math>yz = 60</math>, and <math>zx = -96 + 24i</math>, where <math>i</math> <math>=</math> <math>\sqrt{-1}</math>. Then there are real numbers <math>a</math> and <math>b</math> such that <math>x + y + z = a + bi</math>. Find <math>a^2 + b^2</math>.
  
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==Solution==
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First we evaluate the magnitudes. <math>|xy|=80\sqrt{17}</math>, <math>|yz|=60</math>, and <math>|zx|=24\sqrt{17}</math>. Therefore, <math>|x^2y^2z^2|=17\cdot80\cdot60\cdot24</math>, or <math>|xyz|=240\sqrt{34}</math>. Divide to find that <math>|z|=3\sqrt{2}</math>, <math>|x|=40\sqrt{34}</math>, and <math>|y|=10\sqrt{2}</math>.
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<asy>
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draw((0,0)--(4,0));
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dot((4,0),red);
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draw((0,0)--(-4,0));
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draw((0,0)--(0,-4));
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draw((0,0)--(-4,1));
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dot((-4,1),red);
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draw((0,0)--(-1,-4));
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dot((-1,-4),red);
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draw((0,0)--(4,4),red);
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draw((0,0)--(4,-4),red);
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</asy>
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This allows us to see that the argument of <math>y</math> is <math>\frac{\pi}{4}</math>, and the argument of <math>z</math> is <math>-\frac{\pi}{4}</math>. We need to convert the polar form to a standard form. Simple trig identities show <math>y=10+10i</math> and <math>z=3-3i</math>. More division is needed to find what <math>x</math> is. <cmath>x=-20-12i</cmath> <cmath>x+y+z=-7-5i</cmath> <cmath>(-7)^2+(-5)^2=\boxed{074}</cmath>
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<cmath>QED\blacksquare</cmath>
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Written by [[User:A1b2|a1b2]]
 
{{AIME box|year=2018|n=II|num-b=4|num-a=6}}
 
{{AIME box|year=2018|n=II|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:44, 23 March 2018

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 (ProblemsAnswer KeyResources)
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