2022 AMC 10A Problems/Problem 18
- The following problem is from both the 2022 AMC 10A #18 and 2022 AMC 12A #18, so both problems redirect to this page.
Contents
Problem
Let be the transformation of the coordinate plane that first rotates the plane degrees counterclockwise 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 1
Let be a point in polar coordinates, where is in degrees.
Rotating by counterclockwise around the origin gives the transformation Reflecting across the -axis gives the transformation Note that We start with in polar coordinates. For the sequence of transformations it follows that
- After we have
- After we have
- After we have
- After we have
- After we have
- After we have
- ...
- After we have
- After we have
The least such positive integer is Therefore, the least such positive integer is
~MRENTHUSIASM
Solution 2
Note that since we're reflecting across the -axis, if the point ever makes it to then it will flip back to the original point. Note that after the point will be degree clockwise from the negative -axis. Applying will rotate it to be degree counterclockwise from the negative -axis, and then flip it so that it is degree clockwise from the positive -axis. Therefore, after every transformations, the point rotates degree clockwise. To rotate it so that it will rotate degrees clockwise will require transformations. Then finally on the last transformation, it will rotate on to and then flip back to its original position. Therefore, the answer is .
~KingRavi
Solution 3
In degrees:
Starting with , the sequence goes
We see that it takes steps to downgrade the point by . Since the st point in the sequence is , the answer is
Solution 4 (Simple)
We recognize that the reflection actually just makes it turn clockwise. So, each time it turns the step degree. Looking at the answer choices, the closest number to make a multiple of 360 is 359, which leads us to our answer D.
This is because (1+359)359/2 = something with 360, so we get the reflection back to the spot.
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution (Simple and Fun!!!)
https://youtu.be/7yAh4MtJ8a8?si=2UC_9X7DjkL8UW5C&t=4968
~Math-X
See also
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.