2024 AMC 12B Problems/Problem 12

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 $z$ ?

Diagram

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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2024 AMC 12B Problems/Problem 12

Problem

Diagram

Solution 1 (similar triangles)