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

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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>?
 
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>?
  
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<math>\textbf{(A) }\frac{3}{4}\qquad
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<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>
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\textbf{(B) }1\qquad
 
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\textbf{(C) }\frac{4}[3}\qquad
 
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\textbf{(D) }\frac{3}{2}\qquad
 
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\textbf{(E) }\frac{5}[3}</math>
 
  
 
==Diagram==
 
==Diagram==

Revision as of 01:40, 14 November 2024

Problem

Let $z$ be a complex number with real part greater than $1$ and $|z|=2$. 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 (similar triangles)

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 $\theta$.

This gives us enough info to say that $\triangle{OZ_1Z_2}\sim\triangle{OZ_2Z_3}$ by SAS (since $\frac{OZ_2}{OZ_1}=\frac{OZ_3}{OZ_2}=2$.)

It follows that $[OZ_1Z_2Z_3]=[OZ_1Z_2]+[OZ_2Z_3]=[OZ_1Z_2]+2^2[OZ_1Z_2]=5[OZ_1Z_2]$ as the ratio of side lengths of the two triangles is 2 to 1.

This means $5[OZ_1Z_2]=15$ or $[OZ_1Z_2]=3$ as we were given $[OZ_1Z_2Z_3]=15$.

Using $A=\frac{a*b*sinC}{2}$, we get that $[OZ_1Z_2]=\frac{2*4*sin(\theta)}{2}=4sin(\theta)$, so $4sin(\theta)=3$, giving $sin(\theta)=\frac{3}{4}$.

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

~nm1728

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