Difference between revisions of "2024 AMC 12B Problems/Problem 12"
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− | ==Solution 2 ( | + | ==Solution 2 (Shoelace Theorem)== |
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− | = \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| |
Revision as of 21:12, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.