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>. | ||
+ | ==Solution== | ||
+ | |||
+ | 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>. | ||
+ | <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 <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> | ||
+ | <cmath>QED\blacksquare</cmath> | ||
+ | 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 , , and are complex numbers such that , , and , where . Then there are real numbers and such that . Find .
Solution
First we evaluate the magnitudes. , , and . Therefore, , or . Divide to find that , , and . This allows us to see that the argument of is , and the argument of is . We need to convert the polar form to a standard form. Simple trig identities show and . More division is needed to find what is. Written by a1b2
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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