2022 AMC 10A Problems/Problem 18

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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 $T_k$ 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, where $\theta$ is in degrees.

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 $(r,\theta)\rightarrow(r,180^{\circ}-\theta).$ Note that \begin{align*} T_k(P)&=(r,180^{\circ}-\theta-k^{\circ}), \\ T_{k+1}(T_k(P)) &= (r,\theta -1^{\circ}). \end{align*} We start with $(1,0^{\circ})$ in polar coordinates. For the sequence of transformations $T_1, T_2, T_3, \cdots, T_k,$ it follows that

  • After $T_1,$ we have $(1,179^{\circ}).$
  • After $T_2,$ we have $(1,-1^{\circ}).$
  • After $T_3,$ we have $(1,178^{\circ}).$
  • After $T_4,$ we have $(1,-2^{\circ}).$
  • After $T_5,$ we have $(1,177^{\circ}).$
  • After $T_6,$ we have $(1,-3^{\circ}).$
  • ...
  • After $T_{2k-1},$ we have $(1,180^{\circ}-k^{\circ}).$
  • After $T_{2k},$ we have $(1,-k^{\circ}).$

The least such positive integer $k$ is $180.$ Therefore, the least such positive integer $n$ is $2k-1=\boxed{\textbf{(A) } 359}.$

~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 its original position. Therefore, the answer is $358+1 = 359 = \boxed{\textbf{(A) } 359}$.

~KingRavi

Solution 3

In degrees:

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

We see that it takes $2$ steps to downgrade the point by $1^{\circ}$. Since the $1$st point in the sequence is ${179}$, the answer is $1+2(179)=\boxed{\textbf{(A) } 359}.$

Video Solution

https://youtu.be/QQrsKTErJn8

~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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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