Difference between revisions of "2022 AMC 10A Problems/Problem 18"
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− | + | ==Problem== | |
+ | |||
+ | Let <math>T_k</math> be the transformation of the coordinate plane that first rotates the plane <math>k</math> degrees counter-clockwise around the origin and then reflects the plane across the <math>y</math>-axis. What is the least positive | ||
+ | integer <math>n</math> such that performing the sequence of transformations <math>T_1, T_2, T_3, \cdots, T_n</math> returns the point <math>(1,0)</math> back to itself? | ||
+ | |||
+ | <math>\textbf{(A) } 359 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 719 \qquad \textbf{(D) } 720 \qquad \textbf{(E) } 721 </math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | Note that since we're reflecting across the <math>y</math>-axis, if the point ever makes it to <math>(-1,0)</math> then it will flip back to the original point. Note that after <math>T_1</math> the point will be <math>1</math> degree clockwise from the negative <math>x</math>-axis. Applying <math>T_2</math> will rotate it to be <math>1</math> degree counterclockwise from the negative <math>x</math>-axis, and then flip it so that it is <math>1</math> degree clockwise from the positive <math>x</math>-axis. Therefore, after every <math>2</math> transformations, the point rotates <math>1</math> degree clockwise. To rotate it so that it will rotate <math>179</math> degrees clockwise will require <math>179 \cdot 2 = 358</math> transformations. Then finally on the last transformation, it will rotate on to <math>(-1,0)</math> and then flip back to it's original position. Therefore, the answer is <math>358+1 = 359 = \boxed{A}</math> | ||
+ | |||
+ | ~KingRavi | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>A_{n}</math> be the point <math>(\cos n^{\circ}, \sin n^{\circ})</math>. | ||
+ | |||
+ | Starting with <math>n=0</math>, the sequence goes <cmath>A_{0}\rightarrow A_{179}\rightarrow A_{359}\rightarrow A_{178}\rightarrow A_{358}\rightarrow A_{177}\rightarrow A_{357}\rightarrow\cdots</cmath> | ||
+ | |||
+ | 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== | ||
+ | |||
+ | https://youtu.be/QQrsKTErJn8 | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC10 box|year=2022|ab=A|num-b=17|num-a=19}} | ||
+ | {{AMC12 box|year=2022|ab=A|num-b=17|num-a=19}} | ||
+ | {{MAA Notice}} |
Revision as of 05:17, 19 November 2022
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 1
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 2
Let be the point
.
Starting with , the sequence goes
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 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.