Difference between revisions of "2024 AMC 12B Problems/Problem 12"
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This means <math>5[OZ_1Z_2]=15</math> or <math>[OZ_1Z_2]=3</math> as we were given <math>[OZ_1Z_2Z_3]=15</math>. | This means <math>5[OZ_1Z_2]=15</math> or <math>[OZ_1Z_2]=3</math> as we were given <math>[OZ_1Z_2Z_3]=15</math>. | ||
− | Using <math>A=\frac{ | + | Using <math>A=\frac{ab\sin(C)}{2}</math>, we get that <math>[OZ_1Z_2]=\frac{2\cdot4\cdot \sin(\theta)}{2}=4\sin(\theta)</math>, so <math>4\sin(\theta)=3</math>, giving <math>\sin(\theta)=\frac{3}{4}</math>. |
− | Thus, <math>Im(z)=|z|sin(\theta)=2(\frac{3}{4})=\boxed{\textbf{(D) }\frac{3}{2}}</math>. | + | Thus, <math>\text{Im}(z)=|z|\sin(\theta)=2(\frac{3}{4})=\boxed{\textbf{(D) }\frac{3}{2}}</math>. |
~nm1728 | ~nm1728 |
Revision as of 18:18, 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:
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.