Difference between revisions of "2022 AMC 10A Problems/Problem 18"

Revision as of 05:02, 19 November 2022

The following problem is from both the 2022 AMC 10A #1 and 2022 AMC 12A #1, so both problems redirect to this page.

Problem

Let $T_k$ be the transformation of the coordinate plane that first rotates the plane $k$ degrees counterclockwise 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?

$\textbf{(A) } 359 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 719 \qquad \textbf{(D) } 720 \qquad \textbf{(E) } 721$

Solution 1

Let $P=(r,\theta)$ be a point in polar coordinates. Rotating $P$ by $k^{\circ}$ counterclockwise around the origin gives the transformation $(r,\theta)\rightarrow(r,\theta+k^{\circ}).$ Reflecting $P$ across the $y$ -axis gives the transformation $(r,\theta)\rightarrow(r,180^{\circ}-\theta).$

~MRENTHUSIASM

Solution 2

Note that since we're reflecting across the $y$ -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 $358+1 = 359 = \boxed{A}$

~KingRavi

Solution 3

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 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.

@@ Line 8: / Line 8: @@
 <math>\textbf{(A) } 359 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 719 \qquad \textbf{(D) } 720 \qquad \textbf{(E) } 721 </math>
+==Solution 1==
+Let <math>P=(r,\theta)</math> be a point in polar coordinates. Rotating <math>P</math> by <math>k^{\circ}</math> counterclockwise around the origin gives the transformation <math>(r,\theta)\rightarrow(r,\theta+k^{\circ}).</math> Reflecting <math>P</math> across the <math>y</math>-axis gives the transformation <math>(r,\theta)\rightarrow(r,180^{\circ}-\theta).</math>
+~MRENTHUSIASM
 ==Solution 2==

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