Difference between revisions of "2024 AMC 12B Problems/Problem 12"
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− | + | The quadrilateral can be broken down into two triangles: one between <math>z</math> and <math>z^2</math> and the other between <math>z^2</math> and <math>z^3</math>. Since the angle between each complex number is the same by De Moivre's Theorem, the area of these triangles is <math>\frac{1}{2} 2 * 4 \sin{\theta}</math> and <math>\frac{1}{2} 4 * 8 \sin{\theta}</math>. Thus, <math>\frac{1}{2} 2 * 4 \sin{\theta} + \frac{1}{2} 4 * 8 \sin{\theta} = 15</math>, and <math>\sin{\theta} = \frac{15}{20} = \frac{3}{4}. Since the imaginary part of </math>z<math> is equal to </math>|z|\sin{\theta}<math>, the imaginary part </math>= 2 * \frac{3}{4} = \boxed{\frac{3}{2}}$. | |
==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=B|num-b=11|num-a=13}} | {{AMC12 box|year=2024|ab=B|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 08:05, 15 November 2024
Contents
[hide]Problem
Suppose is a complex number with positive imaginary part, with real part greater than , and with . In the complex plane, the four points , , , and are the vertices of a quadrilateral with area . What is the imaginary part of ?
Diagram
Solution 1
By making a rough estimate of where , , and are on the complex plane, we can draw a pretty accurate diagram (like above.)
Here, points , , and lie at the coordinates of , , and respectively, and is the origin.
We're given , so and . This gives us , , and .
Additionally, we know that (since every power of rotates around the origin by the same angle.) We set these angles equal to .
We have that
Since this is equal to , we have , so .
Thus, .
~nm1728
Solution 2 (Shoelace Theorem)
We have the vertices:
at , at , at , at
The Shoelace formula for the area is: Given that the area is 15: Since corresponds to a complex number with a positive imaginary part, we have:
Solution 3 (No Trig)
Let , so and . Therefore, converting from complex coordinates to Cartesian coordinates gives us the following.
The Shoelace Theorem tells us that the area is
We know that , so . Substituting this gives us this:
In other words,
Solution 4
The quadrilateral can be broken down into two triangles: one between and and the other between and . Since the angle between each complex number is the same by De Moivre's Theorem, the area of these triangles is and . Thus, , and z|z|\sin{\theta}= 2 * \frac{3}{4} = \boxed{\frac{3}{2}}$.
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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