Difference between revisions of "2022 AMC 10A Problems/Problem 18"
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* After <math>T_6,</math> we have <math>(1,-3^{\circ}).</math> | * After <math>T_6,</math> we have <math>(1,-3^{\circ}).</math> | ||
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* After <math>T_{2k-1},</math> we have <math>(1,180^{\circ}-k^{\circ}).</math> | * After <math>T_{2k-1},</math> we have <math>(1,180^{\circ}-k^{\circ}).</math> |
Revision as of 14:52, 7 October 2023
- The following problem is from both the 2022 AMC 10A #18 and 2022 AMC 12A #18, so both problems redirect to this page.
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 it's 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
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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.