Difference between revisions of "2024 AMC 12B Problems/Problem 12"
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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: | ||
Line 65: | Line 62: | ||
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> |
Revision as of 21:10, 14 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 (similar triangles)
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 .
This gives us enough info to say that by SAS similarity (since .)
It follows that as the ratio of side lengths of the two triangles is 2 to 1.
This means or as we were given .
Using , we get that , so , giving .
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,
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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