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

Difference between revisions of "2021 AMC 12B Problems/Problem 18"

(→‎Solution 1)
(→‎Solution 2)
Line 15: Line 15:
 
==Solution 2==
 
==Solution 2==
 
The answer being in the form <math>z+\frac 6z</math> means that there are two solutions, some complex number and its complex conjugate. <cmath>a+bi = \frac{6}{a-bi}</cmath> <cmath>a^2+b^2=6</cmath> We should then be able to test out some ordered pairs of <math>(a, b)</math>. After testing it out, we get the ordered pairs of <math>(-1, \sqrt{5})</math> and its conjugate <math>(-1, -\sqrt{5})</math>. Plugging this into answer format gives us <math>\boxed{\textbf{(A) }-2}</math> ~Lopkiloinm
 
The answer being in the form <math>z+\frac 6z</math> means that there are two solutions, some complex number and its complex conjugate. <cmath>a+bi = \frac{6}{a-bi}</cmath> <cmath>a^2+b^2=6</cmath> We should then be able to test out some ordered pairs of <math>(a, b)</math>. After testing it out, we get the ordered pairs of <math>(-1, \sqrt{5})</math> and its conjugate <math>(-1, -\sqrt{5})</math>. Plugging this into answer format gives us <math>\boxed{\textbf{(A) }-2}</math> ~Lopkiloinm
 +
 +
==Solution 3==
 +
Let <math>x = z + \frac{6}{z}</math>. Then <math>z = \frac{x \pm \sqrt{x^2-24}}{2}</math>. From the answer choices we know <math>x</math> is real and <math>|x|<24</math>, so <math>z = \frac{x \pm i\sqrt{24-x^2}}{2}</math>. We'll take the plus sign for now since we know the answer is unique. Then we have
 +
<cmath> |z|^2 = 6</cmath>
 +
<cmath>  |z+2|^2 = (\frac{x}{2} + 2)^2 + \frac{24-x^2}{4} = 2x+10</cmath>
 +
<cmath>  |z^2+1|^2 = |xz -6 +1|^2 = (\frac{x^2}{2}-5)^2 + \frac{x^2(24-x^2)}{4} = x^2 +25</cmath>
 +
Plug the above back to the original equation, we have
 +
<cmath> 12*6 = 2(2x+10) + x^2 + 25 + 31</cmath>
 +
<cmath> (x+2)^2 = 0</cmath>
 +
So <math>x = -2</math>  <math>\boxed{\textbf{(A) }-2}</math>.
 +
 +
~Sequoia
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021|ab=B|num-b=17|num-a=19}}
 
{{AMC12 box|year=2021|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:17, 12 February 2021

Problem

Let $z$ be a complex number satisfying $12|z|^2=2|z+2|^2+|z^2+1|^2+31.$ What is the value of $z+\frac 6z?$

$\textbf{(A) }-2 \qquad \textbf{(B) }-1 \qquad \textbf{(C) }\frac12\qquad \textbf{(D) }1 \qquad \textbf{(E) }4$

Solution 1

Using the fact $z\bar{z}=|z|^2$, the equation rewrites itself as

\[12z\bar{z}=2(z+2)(\bar{z}+2)+(z^2+1)(\bar{z}^2+1)+31\] \[-12z\bar{z}+2z\bar{z}+4(z+\bar{z})+8+z^2\bar{z}^2+(z^2+\bar{z}^2)+32=0\] \[\left((z^2+2z\bar{z}+\bar{z}^2)+4(z+\bar{z})+4\right)+\left(z^2\bar{z}^2-12z\bar{z}+36\right)=0\] \[(z+\bar{z}+2)^2+(z\bar{z}-6)^2=0.\] As the two quantities in the parentheses are real, both quantities must equal $0$ so \[z+\frac6z=z+\bar{z}=\boxed{\textbf{(A) }-2}.\]

Solution 2

The answer being in the form $z+\frac 6z$ means that there are two solutions, some complex number and its complex conjugate. \[a+bi = \frac{6}{a-bi}\] \[a^2+b^2=6\] We should then be able to test out some ordered pairs of $(a, b)$. After testing it out, we get the ordered pairs of $(-1, \sqrt{5})$ and its conjugate $(-1, -\sqrt{5})$. Plugging this into answer format gives us $\boxed{\textbf{(A) }-2}$ ~Lopkiloinm

Solution 3

Let $x = z + \frac{6}{z}$. Then $z = \frac{x \pm \sqrt{x^2-24}}{2}$. From the answer choices we know $x$ is real and $|x|<24$, so $z = \frac{x \pm i\sqrt{24-x^2}}{2}$. We'll take the plus sign for now since we know the answer is unique. Then we have \[|z|^2 = 6\] \[|z+2|^2 = (\frac{x}{2} + 2)^2 + \frac{24-x^2}{4} = 2x+10\] \[|z^2+1|^2 = |xz -6 +1|^2 = (\frac{x^2}{2}-5)^2 + \frac{x^2(24-x^2)}{4} = x^2 +25\] Plug the above back to the original equation, we have \[12*6 = 2(2x+10) + x^2 + 25 + 31\] \[(x+2)^2 = 0\] So $x = -2$ $\boxed{\textbf{(A) }-2}$.

~Sequoia

See Also

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_18&oldid=146358"