Difference between revisions of "2024 AMC 12B Problems/Problem 12"
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+ | ==Solution 3 (No Trig)== | ||
+ | |||
+ | Let <math>z = a + bi</math>, so <math>z^2 = a^2 + 2abi - b^2</math> and <math>z^3 = a^3 + 3a^2 bi - 3ab^2 - b^3 i</math>. Therefore, converting <math>0, z, z^2, z^3</math> from complex coordinates to Cartesian coordinates gives us the following. | ||
+ | |||
+ | <cmath>(0, 0)</cmath> | ||
+ | |||
+ | <cmath>(a, b)</cmath> | ||
+ | |||
+ | <cmath>(a^2 - b^2, 2ab)</cmath> | ||
+ | |||
+ | <cmath>(a^3 - 3ab^2, 3a^2 b - b^3)</cmath> | ||
+ | |||
+ | The Shoelace Theorem tells us that the area is | ||
+ | |||
+ | <cmath>\frac{1}{2} \Bigg| \Big[ (0)(b) + (a)(2ab) + (a^2 - b^2)(3a^2 b - b^3) + (a^3 - 3ab^2)(0) \Big] - \Big[ (0)(a) + (b)(a^2 - b^2) + (2ab)(a^3 - 3ab^2) + (3a^2 b - b^3)(0) \Big] \Bigg|</cmath> | ||
+ | |||
+ | <cmath>= \frac{1}{2} \Bigg| \Big[ (0) + (2a^2 b) + (3a^4 b - a^2 b^3 - 3a^2 b^3 + b^5) + (0) \Big] - \Big[ (0) + (a^2 b - b^3) + (2a^4 b - 6a^2 b^3) + (0) \Big] \Bigg|</cmath> | ||
+ | |||
+ | <cmath>= \frac{1}{2} \Big| [3a^4 b - 4a^2 b^3 + b^5 + 2a^2 b] - [2a^4 b - 6a^2 b^3 + a^2 b - b^3] \Big|</cmath> | ||
+ | |||
+ | <cmath>= \frac{1}{2} | a^4 b + 2a^2 b^3 + b^5 + a^2 b + b^3 |.</cmath> | ||
+ | |||
+ | We know that <math>|z| = |a + bi| = \sqrt{a^2 + b^2} = 2</math>, so <math>a^2 = 4 - b^2</math>. Substituting this gives us this: | ||
+ | |||
+ | <cmath>\frac{1}{2} \Big| (4 - b^2)^2 b + 2(4 - b^2)b^3 + b^5 + (4 - b^2)b + b^3 \Big|</cmath> | ||
+ | |||
+ | <cmath>= \frac{1}{2} \Big| (16b - 8b^3 + b^5) + (8b^3 - 2b^5) + b^5 + (4b - b^3) + b^3 \Big|</cmath> | ||
+ | |||
+ | <cmath>= \frac{1}{2} | 0b^5 + 0b^3 + 20b|</cmath> | ||
+ | |||
+ | <cmath>= 15.</cmath> | ||
+ | |||
+ | In other words, | ||
+ | |||
+ | <cmath>|10b| = 15</cmath> | ||
+ | |||
+ | <cmath>b = \textbf{(D) } \frac{3}{2}.</cmath> | ||
==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 18:31, 14 November 2024
Contents
[hide]Problem
Let be a complex number with real part greater than and . 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:
1. at
2. at
3. at
4. 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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