Difference between revisions of "2022 AMC 12A Problems/Problem 18"

Revision as of 15:01, 12 November 2022

Problem

Let $T_k$ be the transformation of the coordinate plane that first rotates the plane $k$ degrees counter-clockwise around the origin and then reflects the plane across the $y$ -axis. What is the least positive integer $n$ such that performing the sequence of transformations $T_1, T_2, T_3, \cdots, T_n$ returns the point $(1,0)$ back to itself?

Solution 1

Note that since we're reflecting across the $x$ -axis, if the point ever makes it to $(-1,0)$ then it will flip back to the original point. Note that after $T_1$ the point will be $1$ degree clockwise from the negative $x$ -axis. Applying $T_2$ will rotate it to be $1$ degree counterclockwise from the negative $x$ -axis, and then flip it so that it is $1$ degree clockwise from the positive $x$ -axis. Therefore, after every $2$ transformations, the point rotates $1$ degree clockwise. To rotate it so that it will rotate $179$ degrees clockwise will require $179 \cdot 2 = 358$ transformations. Then finally on the last transformation, it will rotate on to $(-1,0)$ and then flip back to it's original position. Therefore, the answer is $358+1 = 359 = \boxed{B}$

~KingRavi

Solution 2

Let $A_{n}$ be the point $(\cos n^{\circ}, \sin n^{\circ})$ .

Starting with $n=0$ , the sequence goes $A_{0}\rightarrow A_{179}\rightarrow A_{359}\rightarrow A_{178}\rightarrow A_{358}\rightarrow A_{177}\rightarrow A_{357}\rightarrow\cdots$

We see that it takes $2$ turns to downgrade the point by $1^{\circ}$ . Since the fifth point in the sequence is $A_{177}$ , the answer is $5+2(177)=\boxed{\textbf{(A)}~359}$

Video Solution

https://youtu.be/QQrsKTErJn8

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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.

@@ Line 4: / Line 4: @@
 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?
-==Solution==
+==Solution 1==
+Note that since we're reflecting across the <math>x</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{B}</math>
+~KingRavi
+==Solution 2==
 Let <math>A_{n}</math> be the point <math>(\cos n^{\circ}, \sin n^{\circ})</math>.

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