Difference between revisions of "2021 AMC 12B Problems/Problem 18"
Lopkiloinm (talk | contribs) (Created page with "==Problem 18== Let <math>z</math> be a complex number satisfying <math>12|z|^2=2|z+2|^2+|z^2+1|^2+31.</math> What is the value if <math>z+\frac 6z?</math> <math>\textbf{(A) }...") |
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− | ==Problem | + | ==Problem== |
− | Let <math>z</math> be a complex number satisfying <math>12|z|^2=2|z+2|^2+|z^2+1|^2+31.</math> What is the value | + | Let <math>z</math> be a complex number satisfying <math>12|z|^2=2|z+2|^2+|z^2+1|^2+31.</math> What is the value of <math>z+\frac 6z?</math> |
<math>\textbf{(A) }-2 \qquad \textbf{(B) }-1 \qquad \textbf{(C) }\frac12\qquad \textbf{(D) }1 \qquad \textbf{(E) }4</math> | <math>\textbf{(A) }-2 \qquad \textbf{(B) }-1 \qquad \textbf{(C) }\frac12\qquad \textbf{(D) }1 \qquad \textbf{(E) }4</math> | ||
− | + | ==Solution 1== | |
+ | Using the fact <math>z\bar{z}=|z|^2</math>, the equation rewrites itself as | ||
+ | <cmath>\begin{align*} | ||
+ | 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. | ||
+ | \end{align*}</cmath> | ||
+ | As the two quantities in the parentheses are real, both quantities must equal <math>0</math> so <cmath>z+\frac6z=z+\bar{z}=\boxed{\textbf{(A) }-2}.</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Let <math>z = a + bi</math>, <math>z^2 = a^2-b^2+2abi</math> | ||
+ | |||
+ | By the equation given in the problem | ||
+ | |||
+ | <cmath>12(a^2+b^2) = 2((a+2)^2 + b^2) + ((a^2-b^2+1)^2 + (2ab)^2) + 31</cmath> | ||
+ | |||
+ | <cmath>12a^2 + 12b^2 = 2a^2 + 8a + 8 + 2b^2 + a^4 + b^4 + 1 + 2a^2 - 2b^2 - 2a^2b^2 + 4a^2b^2 + 31</cmath> | ||
+ | |||
+ | <cmath>a^4 + b^4 - 8a^2 - 12b^2 + 2a^2b^2 + 8a + 40 = 0</cmath> | ||
+ | |||
+ | <cmath>(a^2+b^2)^2 - 12(a^2+b^2) + 4(a^2 + 2a + 1) + 36=0</cmath> | ||
+ | |||
+ | <cmath>(a^2 + b^2 - 6)^2 + 4(a+1)^2 = 0</cmath> | ||
+ | |||
+ | Therefore, <math>a^2 + b^2 - 6 = 0</math> and <math>a+1 = 0</math> | ||
+ | |||
+ | <math>a = -1</math>, <math>b^2 = 6-1 = 5</math>, <math>b = \sqrt{5}</math> | ||
+ | |||
+ | <cmath>z + \frac{6}{z} = \frac{ a^2 - b^2 + 6 + 2abi }{ a+bi } = \frac{ 1 - 5 + 6 + 2(-1)\sqrt{5} i }{ -1 + i \sqrt{5} } = \frac{ 2 - 2i \sqrt{5} }{-1 + i \sqrt{5}} = \boxed{\textbf{(A)} -2}</cmath> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==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 that <math>x</math> is real and <math>x^2<24</math>, so <math>z = \frac{x \pm i\sqrt{24-x^2}}{2}</math>. 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 = \left(\frac{x^2}{2}-5\right)^2 + \frac{x^2(24-x^2)}{4} = x^2 +25</cmath> | ||
+ | Plugging 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 = \boxed{\textbf{(A) }-2}</math>. | ||
+ | |||
+ | ~Sequoia | ||
+ | |||
+ | ==Solution 4 (Funny Observations)== | ||
+ | There are actually several ways to see that <math>|z|^2 = 6.</math> I present two troll ways of seeing it, and a legitimate way of checking. | ||
+ | |||
+ | Rewrite using <math>w \overline{w} = |w|^2</math> | ||
+ | |||
+ | <math>12z \overline{z} + 2(z+2)(\overline{z} + 2) + (z^2+1)(\overline{z}^2+1)+31</math> | ||
+ | <math>12 z \overline{z} = 2z \overline{z} + 4z + 4 \overline{z} + 8 + z^2 \overline{z}^2+z^2+\overline{z}^2 + 1 + 31.</math> | ||
+ | <math>12 z \overline{z} = 4(z + \overline{z}) + (z \overline{z})^2 + (z + \overline{z})^2 + 40.</math> | ||
+ | |||
+ | Symmetric in <math>z</math> and <math>\overline{z},</math> so if <math>w</math> is a sol, then so is <math>\overline{w}</math> | ||
+ | |||
+ | TROLL OBSERVATION #1: ALL THE ANSWERS ARE REAL. THUS, <math>z + \frac{6}{z} \in \mathbb{R},</math> which means they must be conjugates and so <math>|z|^2 = 6.</math> | ||
+ | |||
+ | TROLL OBSERVATION #2: Note that <math>z+\frac{6}{z} = \overline{z} + \frac{6}{\overline{z}}</math> because either solution must give the same answer! which means that <math>|z|^2 = 6.</math> | ||
+ | |||
+ | Alternatively, you can check: | ||
+ | Let <math>a = w + \overline{w} \in \mathbb{R},</math> and <math>r = |w|^2 \in \mathbb{R}.</math> Thus, we have <math>a^2+4a+40+r^2-12r=0,</math> and the discriminant of this must be nonnegative as <math>a</math> is real. Thus, <math>16-4(40+r^2-12r) \geq 0</math> or <math>(r-6)^2 \leq 0,</math> which forces <math>r = 6,</math> as claimed. | ||
+ | |||
+ | Thus, we plug in <math>z \overline{z} = 6,</math> and get: <math>72 = 4(z + \overline{z}) + 76 + (z + \overline{z})^2,</math> ie. <math>(z+\overline{z})^2 + 4(z + \overline{z}) + 4 = 0,</math> or <math>(z+\overline{z} + 2)^2 = 0,</math> which means <math>z + \overline{z} = \boxed{\textbf{(A) }-2}</math> and that's our answer since we know <math>\overline{z} = 6 / z</math> | ||
+ | |||
+ | - ccx09 | ||
+ | |||
+ | ==Solution 5== | ||
+ | Observe that all the answer choices are real. Therefore, <math>z</math> and <math>\frac{6}{z}</math> must be complex conjugates as this is the only way for both their sum (one of the answer choices) and their product (<math>6</math>) to be real. Thus <math>|z|=|\tfrac{6}{z}|=\sqrt{6}</math>. We will test all the answer choices, starting with <math>\textbf{(A)}</math>. Suppose the answer is <math>\textbf{(A)}</math>. If <math>z+\tfrac{6}{z}=-2</math> then <math>z^{2}+2z+6=0</math> and <math>z=\frac{-2\pm\sqrt{4-24}}{2}=-1\pm\sqrt{5}i</math>. Note that if <math>z=-1+\sqrt{5}i</math> works, then so does <math>-1-\sqrt{5}i</math>. It is relatively easy to see that if <math>z=-1+\sqrt{5}i</math>, then <math>12|z|^{2}=12\cdot 6=72, 2|z+2|^{2}=2|1+\sqrt{5}i|=2\cdot 6=12, |z^{2}+1|^{2}=|-3-2\sqrt{5}i|^{2}=29,</math> and <math>72=12+29+31</math>. Thus the condition <cmath>12|z|^{2}=2|z+2|^{2}+|z^{2}+1|^{2}+31</cmath> is satisfied for <math>z+\tfrac{6}{z}=-2</math>, and the answer is <math>\boxed{\textbf{(A) }-2}</math>. | ||
+ | |||
+ | ==Video Solution by OmegaLearn (Using Complex Number Identities)== | ||
+ | https://youtu.be/AEbMTTGEZV4 | ||
+ | ~pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/Yw-IJvfrT_U | ||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | (includes review of complex numbers) | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ==Video Solution by Punxsutawney Phil== | ||
+ | https://youtu.be/E0HkYqZzw3s | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC12 box|year=2021|ab=B|num-b=17|num-a=19}} | ||
+ | {{MAA Notice}} |
Latest revision as of 08:48, 21 October 2023
Contents
Problem
Let be a complex number satisfying
What is the value of
Solution 1
Using the fact , the equation rewrites itself as
As the two quantities in the parentheses are real, both quantities must equal
so
Solution 2
Let ,
By the equation given in the problem
Therefore, and
,
,
Solution 3
Let . Then
. From the answer choices, we know that
is real and
, so
. Then we have
Plugging the above back to the original equation, we have
So
.
~Sequoia
Solution 4 (Funny Observations)
There are actually several ways to see that I present two troll ways of seeing it, and a legitimate way of checking.
Rewrite using
Symmetric in and
so if
is a sol, then so is
TROLL OBSERVATION #1: ALL THE ANSWERS ARE REAL. THUS, which means they must be conjugates and so
TROLL OBSERVATION #2: Note that because either solution must give the same answer! which means that
Alternatively, you can check:
Let and
Thus, we have
and the discriminant of this must be nonnegative as
is real. Thus,
or
which forces
as claimed.
Thus, we plug in and get:
ie.
or
which means
and that's our answer since we know
- ccx09
Solution 5
Observe that all the answer choices are real. Therefore, and
must be complex conjugates as this is the only way for both their sum (one of the answer choices) and their product (
) to be real. Thus
. We will test all the answer choices, starting with
. Suppose the answer is
. If
then
and
. Note that if
works, then so does
. It is relatively easy to see that if
, then
and
. Thus the condition
is satisfied for
, and the answer is
.
Video Solution by OmegaLearn (Using Complex Number Identities)
https://youtu.be/AEbMTTGEZV4 ~pi_is_3.14
Video Solution
https://youtu.be/Yw-IJvfrT_U ~MathProblemSolvingSkills.com
(includes review of complex numbers)
Video Solution by Punxsutawney Phil
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.