Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2018 AIME II Problems/Problem 5

Revision as of 18:05, 12 April 2018 by Coolgeo (talk | contribs) (→‎Solution 3)

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 1

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

Solution 2

Dividing the first equation by the second equation given, we find that $\frac{xy}{yz}=\frac{x}{z}=\frac{-80-320i}{60}=-\frac{4}{3}-\frac{16}{3}i \implies x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)$. Substituting this into the third equation, we get $z^2=\frac{-96+24i}{-\frac{4}{3}-\frac{16}{3}i}=3\cdot \frac{-24+6i}{-1-4i}=3\cdot \frac{(-24+6i)(-1+4i)}{1+16}=3\cdot \frac{-102i}{17}=-18i$. Taking the square root of this is equivalent to halving the argument and taking the square root of the magnitude. Furthermore, the second equation given tells us that the argument of $y$ is the negative of that of $z$, and their magnitudes multiply to $60$. Thus we have $z=\sqrt{-18i}=3-3i$ and $3\sqrt{2}\cdot |y|=60 \implies |y|=10\sqrt{2} \implies y=10+10i$. To find $x$, we can use the previous substitution we made to find that $x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)=-\frac{4}{3}\cdot (3-3i)(1+4i)=-4(1-i)(1+4i)=-4(5+3i)=-20-12i$ Therefore, $x+y+z=(-20+10+3)+(-12+10-3)i=-7-5i \implies a^2+b^2=(-7)^2+(-5)^2=49+25=\boxed{074}$ Solution by ktong

2018 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png


Solution 3

We are given that $xy=-80-320i$. Thus $y=\frac{-80-320i}{x}$. We are also given that $xz= -96+24i$. Thus $z=\frac{-96+24i}{x}$. We are also given that $yz$ = $60$. Substitute $y=\frac{-80-320i}{x}$ and $z=\frac{-96+24i}{x}$ into $yz$ = $60$. We have $\frac{(-80-320i)(-96+24i)}{x^2}=60$. Multiplying out $(-80-320i)(-96+24i)$ we get $(1920)(8+15i)$. Thus $\frac{1920(8+15i)}{x^2} =60$. Simplifying this fraction we get $\frac{32(8+15i)}{x^2}=1$. Cross-multiplying the fractions we get $x^2=32(8+15i)$ or $x^2= 256+480i$. Now we can rewrite this as $x^2-256=480i$. Let $x= (a+bi)$.Thus $x^2=(a+bi)^2$ or $a^2+2abi-b^2$. We can see that $a^2+2abi-b^2-256=480i$ and thus $2abi=480i$ or $ab=240$.We also can see that $a^2-b^2-256=0$ because there is no real term in $480i$. Thus $a^2-b^2=256$ or $(a+b)(a-b)=256$. Using the two equations $ab=240$ and $(a+b)(a-b)=256$ we solve by doing system of equations that $a=-20$ and $b=-12$. And $x=a+bi$ so $x=-20-12i$. Because $y=\frac{-80-320i}{x}$, then $y=\frac{-80-320i}{-20-12i}$. Simplifying this fraction we get $y=\frac{-80(1+4i)}{-4(5+3i)}$ or $y=\frac{20(1+4i)}{(5+3i)}$. Multiplying by the conjugate of the denominator ($5-3i$) in the numerator and the denominator and we get $y=\frac{20(17-17i)}{34}$. Simplifying this fraction we get $y=10-10i$. Given that $yz$ = $60$ we can substitute $(10-10i)(z)=60$ We can solve for z and get $z=3+3i$. Now we know what $x$, $y$, and $z$ are, so all we have to do is plug and chug. $x+y+z= (-20-12i)+(10+10i)+(3-3i)$ or $x+y+z= -7-5i$ Now $a^2 +b^2=(-7)^2+(-5)^2$ or $a^2 +b^2 = 74$. Thus $074$ is our final answer.(David Camacho)

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_5&oldid=93915"