Difference between revisions of "2024 AMC 12B Problems/Problem 12"

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==Problem==
 
==Problem==
Let <math>z</math> be a complex number with real part greater than <math>1</math> and <math>|z|=2</math>. In the complex plane, the four points <math>0</math>, <math>z</math>, <math>z^2</math>, and <math>z^3</math> are the vertices of a quadrilateral with area <math>15</math>. What is the imaginary part of <math>z</math>?
+
Suppose <math>z</math> is a complex number with positive imaginary part, with real part greater than <math>1</math>, and with <math>|z| = 2</math>. In the complex plane, the four points <math>0</math>, <math>z</math>, <math>z^{2}</math>, and <math>z^{3}</math> are the vertices of a quadrilateral with area <math>15</math>. What is the imaginary part of <math>z</math>?
  
<math>\textbf{(A) }\frac{3}{4}\qquad
+
<math>\textbf{(A) }\frac{3}{4}\qquad\textbf{(B) }1\qquad\textbf{(C) }\frac{4}{3}\qquad\textbf{(D) }\frac{3}{2}\qquad\textbf{(E) }\frac{5}{3}</math>
\textbf{(B) }1\qquad
 
\textbf{(C) }\frac{4}[3}\qquad
 
\textbf{(D) }\frac{3}{2}\qquad
 
\textbf{(E) }\frac{5}[3}</math>
 
  
 
==Diagram==
 
==Diagram==
  
[[File:2024_12B_Q12.png|400px]]
+
[[File:2024_12B_Q12.png|600px]]
  
==Solution 1 (similar triangles)==
+
==Solution 1==
  
 
By making a rough estimate of where <math>z</math>, <math>z^2</math>, and <math>z^3</math> are on the complex plane, we can draw a pretty accurate diagram (like above.)  
 
By making a rough estimate of where <math>z</math>, <math>z^2</math>, and <math>z^3</math> are on the complex plane, we can draw a pretty accurate diagram (like above.)  
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Additionally, we know that <math>\angle{Z_1OZ_2}\cong\angle{Z_2OZ_3}</math> (since every power of <math>z</math> rotates around the origin by the same angle.) We set these angles equal to <math>\theta</math>.
 
Additionally, we know that <math>\angle{Z_1OZ_2}\cong\angle{Z_2OZ_3}</math> (since every power of <math>z</math> rotates around the origin by the same angle.) We set these angles equal to <math>\theta</math>.
  
This gives us enough info to say that <math>\triangle{OZ_1Z_2}\sim\triangle{OZ_2Z_3}</math> by SAS (since <math>\frac{OZ_2}{OZ_1}=\frac{OZ_3}{OZ_2}=2</math>.)
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We have that  
 +
\begin{align*}
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[OZ_1Z_2Z_3]&=[OZ_1Z_2]+[OZ_2Z_3] \
 +
&=\frac{1}{2}\cdot2\cdot4 \sin\theta+\frac{1}{2}\cdot4\cdot8 \sin\theta \
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&=4\sin\theta+16\sin\theta \
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&=20 \sin\theta
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\end{align*}
  
It follows that <math>[OZ_1Z_2Z_3]=[OZ_1Z_2]+[OZ_2Z_3]=[OZ_1Z_2]+2^2[OZ_1Z_2]=5[OZ_1Z_2]</math> as the ratio of side lengths of the two triangles is 2 to 1.
+
Since this is equal to <math>15</math>, we have <math>20\sin\theta=15</math>, so <math>\sin\theta=\frac{3}{4}</math>.
  
This means <math>5[OZ_1Z_2]=15</math> or <math>[OZ_1Z_2]=3</math> as we were given <math>[OZ_1Z_2Z_3]=15</math>.
+
Thus, <math>\text{Im}(z)=|z|\sin(\theta)=2(\frac{3}{4})=\boxed{\textbf{(D) }\frac{3}{2}}</math>.
  
Using <math>A=\frac{a*b*sinC}{2}</math>, we get that <math>[OZ_1Z_2]=\frac{2*4*sin(\theta)}{2}=4sin(\theta)</math>, so <math>4sin(\theta)=3</math>, giving <math>sin(\theta)=\frac{3}{4}</math>.
+
~nm1728
 +
 
 +
==Solution 2 (Shoelace Theorem)==
 +
We have the vertices:
 +
 
 +
<math> 0 </math> at<math>(0, 0)</math> , <math> z </math> at<math>(2\cos \theta, 2\sin \theta)</math> , <math> z^2 </math> at<math>(4\cos 2\theta, 4\sin 2\theta)</math> , <math> z^3 </math> at<math>(8\cos 3\theta, 8\sin 3\theta)</math>
 +
 
 +
The Shoelace formula for the area is:
 +
<cmath>
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= \frac{1}{2} \left| x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1 - (y_1 x_2 + y_2 x_3 + y_3 x_4 + y_4 x_1) \right|.
 +
</cmath>
 +
<cmath>
 +
= \frac{1}{2} \left| 0 + 2\cos \theta \cdot 4\sin 2\theta + 4\cos 2\theta \cdot 8\sin 3\theta - (2\sin \theta \cdot 4\cos 2\theta + 4\sin 2\theta \cdot 8\cos 3\theta) \right|.
 +
</cmath>
 +
<cmath>
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= \frac{1}{2} \left| 8\cos \theta \sin 2\theta + 32\cos 2\theta \sin 3\theta - 8\sin \theta \cos 2\theta - 32\sin 2\theta \cos 3\theta \right|
 +
</cmath>
 +
<cmath>
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= \frac{1}{2} \left|(8\cos \theta \sin 2\theta - 8\sin \theta \cos 2\theta)  + (32\cos 2\theta \sin 3\theta - 32\sin 2\theta \cos 3\theta) \right|</cmath>
 +
<cmath>
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= \frac{1}{2} \left|8\sin(2\theta - \theta) + 32\sin(2\theta - \theta)  \right|
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</cmath>
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<cmath>
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= \frac{1}{2} \left| 8\sin \theta + 32\sin \theta \right|
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</cmath>
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<cmath>= \frac{1}{2} \left| 40\sin \theta \right|
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</cmath>
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Given that the area is 15:
 +
<cmath>
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\frac{1}{2} \left| 40\sin \theta \right| = 15.
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</cmath>
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<cmath>
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20|\sin \theta| = 15 \implies |\sin \theta| = \frac{3}{4}.
 +
</cmath>
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Since <math> \theta </math> corresponds to a complex number <math> z </math> with a positive imaginary part, we have:
 +
 
 +
<cmath>
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\sin \theta = \frac{3}{4}.
 +
</cmath>
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<cmath>
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\text{Imaginary part} = 2\sin \theta = 2 \times \frac{3}{4} = \boxed{\textbf{(D) }\frac{3}{2}}.
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</cmath>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 +
 
 +
==Solution 3 (No Trig)==
 +
 
 +
Let <math>z = a + bi</math>, so <math>z^2 = a^2 + 2abi - b^2</math> and <math>z^3 = a^3 + 3a^2 bi - 3ab^2 - b^3 i</math>. Therefore, converting <math>0, z, z^2, z^3</math> from complex coordinates to Cartesian coordinates gives us the following.
 +
 
 +
<cmath>(0, 0)</cmath>
 +
 
 +
<cmath>(a, b)</cmath>
  
Thus, <math>Im(z)=|z|sin(\theta)=2(\frac{3}{4})=\boxed{\textbf{(B) }\frac{3}{2}}</math>.
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<cmath>(a^2 - b^2, 2ab)</cmath>
  
~nm1728
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<cmath>(a^3 - 3ab^2, 3a^2 b - b^3)</cmath>
 +
 
 +
The Shoelace Theorem tells us that the area is
 +
 
 +
<cmath>\frac{1}{2} \Bigg| \Big[ (0)(b) + (a)(2ab) + (a^2 - b^2)(3a^2 b - b^3) + (a^3 - 3ab^2)(0) \Big] - \Big[ (0)(a) + (b)(a^2 - b^2) + (2ab)(a^3 - 3ab^2) + (3a^2 b - b^3)(0) \Big] \Bigg|</cmath>
 +
 
 +
<cmath>= \frac{1}{2} \Bigg| \Big[ (0) + (2a^2 b) + (3a^4 b - a^2 b^3 - 3a^2 b^3 + b^5) + (0) \Big] - \Big[ (0) + (a^2 b - b^3) + (2a^4 b - 6a^2 b^3) + (0) \Big] \Bigg|</cmath>
 +
 
 +
<cmath>= \frac{1}{2} \Big| [3a^4 b - 4a^2 b^3 + b^5 + 2a^2 b] - [2a^4 b - 6a^2 b^3 + a^2 b - b^3] \Big|</cmath>
 +
 
 +
<cmath>= \frac{1}{2} | a^4 b + 2a^2 b^3 + b^5 + a^2 b + b^3 |.</cmath>
 +
 
 +
We know that <math>|z| = |a + bi| = \sqrt{a^2 + b^2} = 2</math>, so <math>a^2 = 4 - b^2</math>. Substituting this gives us this:
 +
 
 +
<cmath>\frac{1}{2} \Big| (4 - b^2)^2 b + 2(4 - b^2)b^3 + b^5 + (4 - b^2)b + b^3 \Big|</cmath>
 +
 
 +
<cmath>= \frac{1}{2} \Big| (16b - 8b^3 + b^5) + (8b^3 - 2b^5) + b^5 + (4b - b^3) + b^3 \Big|</cmath>
 +
 
 +
<cmath>= \frac{1}{2} | 0b^5 + 0b^3 + 20b|</cmath>
 +
 
 +
<cmath>= 15.</cmath>
 +
 
 +
In other words,
 +
 
 +
<cmath>|10b| = 15</cmath>
 +
 
 +
<cmath>b = \textbf{(D) } \frac{3}{2}.</cmath>
 +
 
 +
==Solution 4==
 +
==Video Solution 1 by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=OMR5MYtu11s&t=0s
 +
 
 +
 
 +
==See also==
 +
{{AMC12 box|year=2024|ab=B|num-b=11|num-a=13}}
 +
{{MAA Notice}}

Latest revision as of 00:15, 16 November 2024

Problem

Suppose $z$ is a complex number with positive imaginary part, with real part greater than $1$, and with $|z| = 2$. In the complex plane, the four points $0$, $z$, $z^{2}$, and $z^{3}$ are the vertices of a quadrilateral with area $15$. What is the imaginary part of $z$?

$\textbf{(A) }\frac{3}{4}\qquad\textbf{(B) }1\qquad\textbf{(C) }\frac{4}{3}\qquad\textbf{(D) }\frac{3}{2}\qquad\textbf{(E) }\frac{5}{3}$

Diagram

2024 12B Q12.png

Solution 1

By making a rough estimate of where $z$, $z^2$, and $z^3$ are on the complex plane, we can draw a pretty accurate diagram (like above.)

Here, points $Z_1$, $Z_2$, and $Z_3$ lie at the coordinates of $z$, $z^2$, and $z^3$ respectively, and $O$ is the origin.

We're given $|z|=2$, so $|z^2|=|z|^2=4$ and $|z^3|=|z|^3 = 8$. This gives us $OZ_1=2$, $OZ_2=4$, and $OZ_3=8$.

Additionally, we know that $\angle{Z_1OZ_2}\cong\angle{Z_2OZ_3}$ (since every power of $z$ rotates around the origin by the same angle.) We set these angles equal to $\theta$.

We have that [OZ1Z2Z3]=[OZ1Z2]+[OZ2Z3]=1224sinθ+1248sinθ=4sinθ+16sinθ=20sinθ

Since this is equal to $15$, we have $20\sin\theta=15$, so $\sin\theta=\frac{3}{4}$.

Thus, $\text{Im}(z)=|z|\sin(\theta)=2(\frac{3}{4})=\boxed{\textbf{(D) }\frac{3}{2}}$.

~nm1728

Solution 2 (Shoelace Theorem)

We have the vertices:

$0$ at$(0, 0)$ , $z$ at$(2\cos \theta, 2\sin \theta)$ , $z^2$ at$(4\cos 2\theta, 4\sin 2\theta)$ , $z^3$ at$(8\cos 3\theta, 8\sin 3\theta)$

The Shoelace formula for the area is: \[= \frac{1}{2} \left| x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1 - (y_1 x_2 + y_2 x_3 + y_3 x_4 + y_4 x_1) \right|.\] \[= \frac{1}{2} \left| 0 + 2\cos \theta \cdot 4\sin 2\theta + 4\cos 2\theta \cdot 8\sin 3\theta - (2\sin \theta \cdot 4\cos 2\theta + 4\sin 2\theta \cdot 8\cos 3\theta) \right|.\] \[= \frac{1}{2} \left| 8\cos \theta \sin 2\theta + 32\cos 2\theta \sin 3\theta - 8\sin \theta \cos 2\theta - 32\sin 2\theta \cos 3\theta \right|\] \[= \frac{1}{2} \left|(8\cos \theta \sin 2\theta - 8\sin \theta \cos 2\theta)  + (32\cos 2\theta \sin 3\theta - 32\sin 2\theta \cos 3\theta)  \right|\] \[= \frac{1}{2} \left|8\sin(2\theta - \theta)  + 32\sin(2\theta - \theta)  \right|\] \[= \frac{1}{2} \left| 8\sin \theta + 32\sin \theta \right|\] \[= \frac{1}{2} \left| 40\sin \theta \right|\] Given that the area is 15: \[\frac{1}{2} \left| 40\sin \theta \right| = 15.\] \[20|\sin \theta| = 15 \implies |\sin \theta| = \frac{3}{4}.\] Since $\theta$ corresponds to a complex number $z$ with a positive imaginary part, we have:

\[\sin \theta = \frac{3}{4}.\] \[\text{Imaginary part} = 2\sin \theta = 2 \times \frac{3}{4} = \boxed{\textbf{(D) }\frac{3}{2}}.\]

~luckuso

Solution 3 (No Trig)

Let $z = a + bi$, so $z^2 = a^2 + 2abi - b^2$ and $z^3 = a^3 + 3a^2 bi - 3ab^2 - b^3 i$. Therefore, converting $0, z, z^2, z^3$ from complex coordinates to Cartesian coordinates gives us the following.

\[(0, 0)\]

\[(a, b)\]

\[(a^2 - b^2, 2ab)\]

\[(a^3 - 3ab^2, 3a^2 b - b^3)\]

The Shoelace Theorem tells us that the area is

\[\frac{1}{2} \Bigg| \Big[ (0)(b) + (a)(2ab) + (a^2 - b^2)(3a^2 b - b^3) + (a^3 - 3ab^2)(0) \Big] - \Big[ (0)(a) + (b)(a^2 - b^2) + (2ab)(a^3 - 3ab^2) + (3a^2 b - b^3)(0) \Big] \Bigg|\]

\[= \frac{1}{2} \Bigg| \Big[ (0) + (2a^2 b) + (3a^4 b - a^2 b^3 - 3a^2 b^3 + b^5) + (0) \Big] - \Big[ (0) + (a^2 b - b^3) + (2a^4 b - 6a^2 b^3) + (0) \Big] \Bigg|\]

\[= \frac{1}{2} \Big| [3a^4 b - 4a^2 b^3 + b^5 + 2a^2 b] - [2a^4 b - 6a^2 b^3 + a^2 b - b^3] \Big|\]

\[= \frac{1}{2} | a^4 b + 2a^2 b^3 + b^5 + a^2 b + b^3 |.\]

We know that $|z| = |a + bi| = \sqrt{a^2 + b^2} = 2$, so $a^2 = 4 - b^2$. Substituting this gives us this:

\[\frac{1}{2} \Big| (4 - b^2)^2 b + 2(4 - b^2)b^3 + b^5 + (4 - b^2)b + b^3 \Big|\]

\[= \frac{1}{2} \Big| (16b - 8b^3 + b^5) + (8b^3 - 2b^5) + b^5 + (4b - b^3) + b^3 \Big|\]

\[= \frac{1}{2} | 0b^5 + 0b^3 + 20b|\]

\[= 15.\]

In other words,

\[|10b| = 15\]

\[b = \textbf{(D) } \frac{3}{2}.\]

Solution 4

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=OMR5MYtu11s&t=0s


See also

2024 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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

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