Difference between revisions of "2024 AMC 12B Problems/Problem 12"
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Thus, <math>\text{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, ShortPeopleFartalot |
==Solution 2 (Shoelace Theorem)== | ==Solution 2 (Shoelace Theorem)== |
Latest revision as of 22:02, 25 December 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, ShortPeopleFartalot
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 |
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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