Difference between revisions of "2022 AMC 12A Problems/Problem 18"
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We see that it takes <math>2</math> turns to downgrade the point by <math>1^{\circ}</math>. Since the fifth point in the sequence is <math>A_{177}</math>, the answer is <math>5+2(177)=\boxed{\textbf{(A)}~359}</math> | We see that it takes <math>2</math> turns to downgrade the point by <math>1^{\circ}</math>. Since the fifth point in the sequence is <math>A_{177}</math>, the answer is <math>5+2(177)=\boxed{\textbf{(A)}~359}</math> | ||
| − | ==Video Solution | + | ==Video Solution== |
https://youtu.be/QQrsKTErJn8 | https://youtu.be/QQrsKTErJn8 | ||
| + | |||
| + | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==See also== | ==See also== | ||
Revision as of 20:32, 11 November 2022
Contents
Problem
Let 
be the transformation of the coordinate plane that first rotates the plane 
degrees counter-clockwise around the origin and then reflects the plane across the 
-axis. What is the least positive
integer 
such that performing the sequence of transformations 
returns the point 
back to itself?
Solution
Let 
be the point 
.
Starting with 
, the sequence goes ![\[A_{0}\rightarrow A_{179}\rightarrow A_{359}\rightarrow A_{178}\rightarrow A_{358}\rightarrow A_{177}\rightarrow A_{357}\rightarrow\cdots\]](http://latex.artofproblemsolving.com/4/c/d/4cd41a8dfb6dc8ece2e6c4a6296da01367aff9ee.png)
We see that it takes 
turns to downgrade the point by 
. Since the fifth point in the sequence is 
, the answer is 
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
| 2022 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 17 |
Followed by Problem 19 |
| 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 | |
| 2022 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
