Difference between revisions of "2024 AMC 12B Problems/Problem 12"
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==Problem== | ==Problem== | ||
− | + | 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\textbf{(B) }1\qquad\textbf{(C) }\frac{4}{3}\qquad\textbf{(D) }\frac{3}{2}\qquad\textbf{(E) }\frac{5}{3}</math> | <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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[[File:2024_12B_Q12.png|600px]] | [[File:2024_12B_Q12.png|600px]] | ||
− | ==Solution 1 | + | ==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>. | ||
− | + | 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 <math>15</math>, we have <math>20\sin\theta=15</math>, so <math>\sin\theta=\frac{3}{4}</math>. | |
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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>. | ||
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~nm1728 | ~nm1728 | ||
− | ==Solution 2 ( | + | ==Solution 2 (Shoelace Theorem)== |
We have the vertices: | 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> | |
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The Shoelace formula for the area is: | The Shoelace formula for the area is: | ||
<cmath> | <cmath> | ||
− | + | = \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> | ||
<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> | ||
<cmath> | <cmath> | ||
− | + | = \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> | ||
<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> | |
+ | <cmath> | ||
+ | = \frac{1}{2} \left|8\sin(2\theta - \theta) + 32\sin(2\theta - \theta) \right| | ||
+ | </cmath> | ||
<cmath> | <cmath> | ||
− | + | = \frac{1}{2} \left| 8\sin \theta + 32\sin \theta \right| | |
− | + | </cmath> | |
− | + | <cmath>= \frac{1}{2} \left| 40\sin \theta \right| | |
</cmath> | </cmath> | ||
Given that the area is 15: | Given that the area is 15: | ||
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20|\sin \theta| = 15 \implies |\sin \theta| = \frac{3}{4}. | 20|\sin \theta| = 15 \implies |\sin \theta| = \frac{3}{4}. | ||
</cmath> | </cmath> | ||
− | Since<math> \theta </math> corresponds to a complex number<math> z </math> with a positive imaginary part, we have: | + | Since <math> \theta </math> corresponds to a complex number <math> z </math> with a positive imaginary part, we have: |
<cmath> | <cmath> | ||
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<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
Contents
[hide]Problem
Suppose is a complex number with positive imaginary part, with real part greater than , and with . In the complex plane, the four points , , , and are the vertices of a quadrilateral with area . What is the imaginary part of ?
Diagram
Solution 1
By making a rough estimate of where , , and are on the complex plane, we can draw a pretty accurate diagram (like above.)
Here, points , , and lie at the coordinates of , , and respectively, and is the origin.
We're given , so and . This gives us , , and .
Additionally, we know that (since every power of rotates around the origin by the same angle.) We set these angles equal to .
We have that
Since this is equal to , we have , so .
Thus, .
~nm1728
Solution 2 (Shoelace Theorem)
We have the vertices:
at , at , at , at
The Shoelace formula for the area is: Given that the area is 15: Since corresponds to a complex number with a positive imaginary part, we have:
Solution 3 (No Trig)
Let , so and . Therefore, converting from complex coordinates to Cartesian coordinates gives us the following.
The Shoelace Theorem tells us that the area is
We know that , so . Substituting this gives us this:
In other words,
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 |
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