Difference between revisions of "2022 AMC 12B Problems/Problem 5"
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<math>\textbf{(A)}\ (-3, -4) \qquad \textbf{(B)}\ (0,5) \qquad \textbf{(C)}\ (2,-1) \qquad \textbf{(D)}\ (4,3) \qquad \textbf{(E)}\ (6,-3)</math> | <math>\textbf{(A)}\ (-3, -4) \qquad \textbf{(B)}\ (0,5) \qquad \textbf{(C)}\ (2,-1) \qquad \textbf{(D)}\ (4,3) \qquad \textbf{(E)}\ (6,-3)</math> | ||
| − | == Solution 1 == | + | == Solution 1 (Cartesian Plane) == |
| − | <math>(-1,-2)</math> is 4 units west and 3 units south of <math>(3,1)</math>. Performing a counterclockwise rotation of <math>270^{\circ}</math>, which is equivalent to a clockwise rotation of <math>90^{\circ}</math>, the answer is 3 units west and 4 units north of <math>(3,1)</math>, or <math>\boxed{\textbf{(B)}\ (0,5)}</math>. | + | <math>(-1,-2)</math> is <math>4</math> units west and <math>3</math> units south of <math>(3,1)</math>. Performing a counterclockwise rotation of <math>270^{\circ}</math>, which is equivalent to a clockwise rotation of <math>90^{\circ}</math>, the answer is <math>3</math> units west and <math>4</math> units north of <math>(3,1)</math>, or <math>\boxed{\textbf{(B)}\ (0,5)}</math>. |
~[[User: Bxiao31415 | Bxiao31415]] | ~[[User: Bxiao31415 | Bxiao31415]] | ||
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| + | == Solution 2 (Complex Numbers) == | ||
| + | We write <math>(-1, -2)</math> as <math>-1-2i.</math> We'd like to rotate about <math>(3, 1),</math> which is <math>3+i</math> in the complex plane, by an angle of <math>270^{\circ} = \frac{3\pi}{2} \text{ rad}</math> counterclockwise. | ||
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| + | The formula for rotating the complex number <math>z</math> about the complex number <math>w</math> by an angle of <math>\theta</math> counterclockwise is given as <math>(z-w)e^{\theta i} + w.</math> Plugging in our values <math>z = -1-2i, w = 3+i, \theta = \frac{3\pi}{2}</math>, we evaluate the expression as <math>((-1-2i) - (3+i))e^{\frac{3\pi i}{2}} + (3+i) = (-4-3i)(-i) + (3+i) = 4i-3i^2+3+i = 4i-3+3+i = 5i,</math> which corresponds to <math>\boxed{\textbf{(B)}\ (0,5)}</math> on the Cartesian plane. | ||
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| + | ~sirswagger21 | ||
==Video Solution 1== | ==Video Solution 1== | ||
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~Education, the Study of Everything | ~Education, the Study of Everything | ||
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| + | ==Video Solution(1-16)== | ||
| + | https://youtu.be/SCwQ9jUfr0g | ||
| + | |||
| + | ~~Hayabusa1 | ||
== See also == | == See also == | ||
{{AMC12 box|year=2022|ab=B|num-b=4|num-a=6}} | {{AMC12 box|year=2022|ab=B|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 12:32, 5 April 2023
Contents
Problem
The point 
is rotated 
counterclockwise about the point 
. What are the coordinates of its new position?
Solution 1 (Cartesian Plane)
is 
units west and 
units south of 
. Performing a counterclockwise rotation of , which is equivalent to a clockwise rotation of 
, the answer is units west and units north of , or 
.
Solution 2 (Complex Numbers)
We write as 
We'd like to rotate about 
which is 
in the complex plane, by an angle of 
counterclockwise.
The formula for rotating the complex number 
about the complex number 
by an angle of 
counterclockwise is given as 
Plugging in our values 
, we evaluate the expression as 
which corresponds to on the Cartesian plane.
~sirswagger21
Video Solution 1
~Education, the Study of Everything
Video Solution(1-16)
~~Hayabusa1
See also
| 2022 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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. 
