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

(Added a GENERALIZED solution.)
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==Solution 1==
 
==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>
+
Let <math>P=(r,\theta)</math> be a point in polar coordinates, where <math>\theta</math> is in degrees.
 +
 
 +
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> Note that
 +
<cmath>\begin{align*}
 +
T_k(P)&=(r,180^{\circ}-\theta-k^{\circ}), \\
 +
T_{k+1}(T_k(P)) &= (r,\theta -1^{\circ}).
 +
\end{align*}</cmath>
 +
We start with <math>(1,0^{\circ})</math> in polar coordinates. For the sequence of transformations <math>T_1, T_2, T_3, \cdots, T_k,</math> it follows that
 +
 
 +
* After <math>T_1,</math> we have <math>(1,179^{\circ}).</math>
 +
 
 +
* After <math>T_2,</math> we have <math>(1,-1^{\circ}).</math>
 +
 
 +
* After <math>T_3,</math> we have <math>(1,178^{\circ}).</math>
 +
 
 +
* After <math>T_4,</math> we have <math>(1,-2^{\circ}).</math>
 +
 
 +
* After <math>T_5,</math> we have <math>(1,177^{\circ}).</math>
 +
 
 +
* After <math>T_6,</math> we have <math>(1,-3^{\circ}).</math>
 +
 
 +
* ...
 +
 
 +
* After <math>T_{2k-1},</math> we have <math>(1,180^{\circ}-k^{\circ}).</math>
 +
 
 +
* After <math>T_{2k},</math> we have <math>(1,-k^{\circ}).</math>
 +
 
 +
The least such positive integer <math>k</math> is <math>180.</math> Therefore, the least such positive integer <math>n</math> is <math>2k-1=\boxed{\textbf{(A) } 359}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
 
==Solution 2==
 
==Solution 2==
−
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>
+
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{\textbf{(A) } 359}</math>.
  
 
~KingRavi
 
~KingRavi
Line 23: Line 50:
 
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>
 
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>
+
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==
 
==Video Solution==

Revision as of 05:24, 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, 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 it's original position. Therefore, the answer is $358+1 = 359 = \boxed{\textbf{(A) } 359}$.

~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)

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

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