Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2 (shoelace theorem))
 
(5 intermediate revisions by 3 users not shown)
Line 8: Line 8:
 
[[File:2024_12B_Q12.png|600px]]
 
[[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.)  
Line 18: Line 18:
 
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 similarity (since <math>\frac{OZ_2}{OZ_1}=\frac{OZ_3}{OZ_2}=2</math>.)
+
We have that  
 +
\begin{align*}
 +
[OZ_1Z_2Z_3]&=[OZ_1Z_2]+[OZ_2Z_3] \\
 +
&=\frac{1}{2}\cdot2\cdot4 \sin\theta+\frac{1}{2}\cdot4\cdot8 \sin\theta \\
 +
&=4\sin\theta+16\sin\theta \\
 +
&=20 \sin\theta
 +
\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>.
 
−
 
 
−
Using <math>A=\frac{ab\sin(C)}{2}</math>, we get that <math>[OZ_1Z_2]=\frac{2\cdot4\cdot \sin(\theta)}{2}=4\sin(\theta)</math>, so <math>4\sin(\theta)=3</math>, giving <math>\sin(\theta)=\frac{3}{4}</math>.
 
  
 
Thus, <math>\text{Im}(z)=|z|\sin(\theta)=2(\frac{3}{4})=\boxed{\textbf{(D) }\frac{3}{2}}</math>.
 
Thus, <math>\text{Im}(z)=|z|\sin(\theta)=2(\frac{3}{4})=\boxed{\textbf{(D) }\frac{3}{2}}</math>.
Line 30: Line 32:
 
~nm1728
 
~nm1728
  
−
==Solution 2 (shoelace theorem)==
+
==Solution 2 (Shoelace Theorem)==
 
We have the vertices:
 
We have the vertices:
  
Line 46: Line 48:
 
</cmath>
 
</cmath>
 
<cmath>
 
<cmath>
−
= \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>
+
= \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>
 
<cmath>
 
= \frac{1}{2} \left|8\sin(2\theta - \theta)  + 32\sin(2\theta - \theta)  \right|
 
= \frac{1}{2} \left|8\sin(2\theta - \theta)  + 32\sin(2\theta - \theta)  \right|
Line 110: Line 112:
  
 
<cmath>b = \textbf{(D) } \frac{3}{2}.</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==
 
==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}}

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 \begin{align*} [OZ_1Z_2Z_3]&=[OZ_1Z_2]+[OZ_2Z_3] \\ &=\frac{1}{2}\cdot2\cdot4 \sin\theta+\frac{1}{2}\cdot4\cdot8 \sin\theta \\ &=4\sin\theta+16\sin\theta \\ &=20 \sin\theta \end{align*}

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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12B_Problems/Problem_12&oldid=234509"