2024 AMC 12B Problems/Problem 12
Contents
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 \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 , 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 |
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All AMC 12 Problems and Solutions |
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