Difference between revisions of "2024 AMC 12B Problems/Problem 12"
(→Solution 2 (shoelace theorem)) |
(simplify solution 1) |
||
Line 8: | Line 8: | ||
[[File:2024_12B_Q12.png|600px]] | [[File:2024_12B_Q12.png|600px]] | ||
− | ==Solution 1 | + | ==Solution 1== |
By making a rough estimate of where <math>z</math>, <math>z^2</math>, and <math>z^3</math> are on the complex plane, we can draw a pretty accurate diagram (like above.) | By making a rough estimate of where <math>z</math>, <math>z^2</math>, and <math>z^3</math> are on the complex plane, we can draw a pretty accurate diagram (like above.) | ||
Line 18: | Line 18: | ||
Additionally, we know that <math>\angle{Z_1OZ_2}\cong\angle{Z_2OZ_3}</math> (since every power of <math>z</math> rotates around the origin by the same angle.) We set these angles equal to <math>\theta</math>. | Additionally, we know that <math>\angle{Z_1OZ_2}\cong\angle{Z_2OZ_3}</math> (since every power of <math>z</math> rotates around the origin by the same angle.) We set these angles equal to <math>\theta</math>. | ||
− | + | We have that | |
+ | \begin{align*} | ||
+ | [OZ_1Z_2Z_3]&=[OZ_1Z_2]+[OZ_2Z_3] \ | ||
+ | &=\frac{1}{2}\cdot2\cdot4 \sin\theta+\frac{1}{2}\cdot4\cdot8 \sin\theta \ | ||
+ | &=4\sin\theta+16\sin\theta \ | ||
+ | &=20 \sin\theta | ||
+ | \end{align*} | ||
− | + | Since this is equal to <math>15</math>, we have <math>20\sin\theta=15</math>, so <math>\sin\theta=\frac{3}{4}</math>. | |
− | |||
− | |||
− | |||
− | |||
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>. | ||
Line 110: | Line 112: | ||
<cmath>b = \textbf{(D) } \frac{3}{2}.</cmath> | <cmath>b = \textbf{(D) } \frac{3}{2}.</cmath> | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | |||
==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 22:55, 14 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
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.