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2024 AMC 12B Problems/Problem 12: Difference between revisions

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~nm1728
~nm1728
==Solution 2 (shoelace theorem)==
We have the vertices:
1.<math> 0 </math> at<math>(0, 0)</math>
2.<math> z </math> at<math>(2\cos \theta, 2\sin \theta)</math>
3.<math> z^2 </math> at<math>(4\cos 2\theta, 4\sin 2\theta)</math>
4.<math> z^3 </math> at<math>(8\cos 3\theta, 8\sin 3\theta)</math>
The Shoelace formula for the area is:
<cmath>
\text{Area} = \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>
\text{Area} = \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>
\text{Area} = \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>
\text{Area} = \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>
\text{Area} = \frac{1}{2} \left|8\sin(2\theta - \theta)  + (32\sin(2\theta - \theta)  \right|.</cmath>
<cmath>
\text{Area} = \frac{1}{2} \left| 8\sin \theta + 32\sin \theta \right| = \frac{1}{2} \left| 40\sin \theta \right|.
</cmath>
Given that the area is 15:
<cmath>
\frac{1}{2} \left| 40\sin \theta \right| = 15.
</cmath>
<cmath>
20|\sin \theta| = 15 \implies |\sin \theta| = \frac{3}{4}.
</cmath>
Since<math> \theta </math> corresponds to a complex number<math> z </math> with a positive imaginary part, we have:
<cmath>
\sin \theta = \frac{3}{4}.
</cmath>
<cmath>
\text{Imaginary part} = 2\sin \theta = 2 \times \frac{3}{4} = \boxed{\textbf{(D) }\frac{3}{2}}.
</cmath>
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]


==See also==
==See also==
{{AMC12 box|year=2024|ab=B|num-b=11|num-a=13}}
{{AMC12 box|year=2024|ab=B|num-b=11|num-a=13}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 02:07, 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

Error creating thumbnail: File missing

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

~nm1728

Solution 2 (shoelace theorem)

We have the vertices:

1.$0$ at$(0, 0)$

2.$z$ at$(2\cos \theta, 2\sin \theta)$

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

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

The Shoelace formula for the area is: \[\text{Area} = \frac{1}{2} \left| x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1 - (y_1 x_2 + y_2 x_3 + y_3 x_4 + y_4 x_1) \right|.\] \[\text{Area} = \frac{1}{2} \left| 0 + 2\cos \theta \cdot 4\sin 2\theta + 4\cos 2\theta \cdot 8\sin 3\theta - (2\sin \theta \cdot 4\cos 2\theta + 4\sin 2\theta \cdot 8\cos 3\theta) \right|.\] \[\text{Area} = \frac{1}{2} \left| 8\cos \theta \sin 2\theta + 32\cos 2\theta \sin 3\theta - 8\sin \theta \cos 2\theta - 32\sin 2\theta \cos 3\theta \right|.\] \[\text{Area} = \frac{1}{2} \left|8\cos \theta \sin 2\theta - 8\sin \theta \cos 2\theta) + (32\cos 2\theta \sin 3\theta - 32\sin 2\theta \cos 3\theta)) \right|.\] \[\text{Area} = \frac{1}{2} \left|8\sin(2\theta - \theta) + (32\sin(2\theta - \theta) \right|.\] 

\[\text{Area} = \frac{1}{2} \left| 8\sin \theta + 32\sin \theta \right| = \frac{1}{2} \left| 40\sin \theta \right|.\]

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

See also

2024 AMC 12B (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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