Difference between revisions of "2024 AMC 12B Problems/Problem 12"
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Using <math>A=\frac{a*b*sinC}{2}</math>, we get that <math>[OZ_1Z_2]=\frac{2*4*sin(\theta)}{2}=4sin(\theta)</math>, so <math>4sin(\theta)=3</math>, giving <math>sin(\theta)=\frac{3}{4}</math>. | Using <math>A=\frac{a*b*sinC}{2}</math>, we get that <math>[OZ_1Z_2]=\frac{2*4*sin(\theta)}{2}=4sin(\theta)</math>, so <math>4sin(\theta)=3</math>, giving <math>sin(\theta)=\frac{3}{4}</math>. | ||
− | Thus, <math>Im(z)=|z|sin(\theta)=2(\frac{3}{4})=\boxed{\textbf{( | + | Thus, <math>Im(z)=|z|sin(\theta)=2(\frac{3}{4})=\boxed{\textbf{(D) }\frac{3}{2}}</math>. |
~nm1728 | ~nm1728 |
Revision as of 01:41, 14 November 2024
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 (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