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

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==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{\textbf{(A) } 359}</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 its original position. Therefore, the answer is <math>358+1 = 359 = \boxed{\textbf{(A) } 359}</math>.
  
 
~KingRavi
 
~KingRavi
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We see that it takes <math>2</math> steps to downgrade the point by <math>1^{\circ}</math>. Since the <math>1</math>st point in the sequence is <math>{179}</math>, the answer is <math>1+2(179)=\boxed{\textbf{(A) } 359}.</math>
 
We see that it takes <math>2</math> steps to downgrade the point by <math>1^{\circ}</math>. Since the <math>1</math>st point in the sequence is <math>{179}</math>, the answer is <math>1+2(179)=\boxed{\textbf{(A) } 359}.</math>
 +
 +
==Solution 4 (Simple)==
 +
We can consider the rotations and reflections separately. For the rotations, each rotation turns it by the next natural number. Thus the total number of degrees turned would be a triangle number. We test the smallest number, <math>359</math> first, and we get that it turns <math>\frac{(1+359)359}{2} = 180n</math>, where <math>n</math> is an integer. Thus, the point would be rotated to <math>(-1,0)</math>. We may be tempted to dismiss this option but we haven't considered the reflections. Each reflection acts as a <math>180^{\circ}</math> rotation, so every two reflections cancel. However, <math>359</math> is odd so we have to reflect <math>(-1,0)</math>, taking us to <math>(1,0)</math>, which is what we want. Thus we get <math>\boxed{\textbf{(A) } 359}</math>.
 +
 +
==Solution 5 (Complex)==
 +
Rotations and reflections in 2D are very nice to describe using complex numbers in polar form. Reflections and rotations also don't affect the length (modulus or absolute value) of a complex number. This motivates us to set <math>z = \exp (i\theta)</math>, starting with <math>z=1+0i=\exp(i0)</math>.
 +
Then, we can describe rotations and reflections about the <math>y</math>-axis using these formulas:
 +
<cmath>
 +
\text{Rotation}(\exp(i\theta), k): \exp(i\theta) \mapsto \exp(i(\theta + k))
 +
</cmath>
 +
<cmath>
 +
\text{Reflection}(\exp(i\theta)): \exp(i\theta) \mapsto \exp i (180^\circ - \theta) = -\exp(i(-\theta))
 +
</cmath>
 +
 +
If we apply two successive iterations, we see a simplification:
 +
 +
<cmath>
 +
T_k(\exp(i\theta)) = -\exp(-i(\theta+k))
 +
</cmath>
 +
 +
\begin{aligned}
 +
T_{k+1}(T_k(\exp(i\theta))) &= T_{k+1}(-\exp(i(-\theta-k))
 +
\\
 +
&= --\exp(-i(-\theta-k+(k+1))
 +
\\
 +
&= \exp i(\theta-1)
 +
\end{aligned}
 +
 +
 +
We also can calculate <math>T_1(\exp i0) = \exp 179^\circ</math>. Thus, the point <math>(1,0)</math> gets sent back to when all double iterations after <math>1</math> cancel <math>179^\circ</math>. <math>179-\left(\frac{n-1}{2}\right)=0</math> so <math>n = 1 + 2 \cdot 179 = \boxed{\textbf{(A) }359}</math>.
  
 
==Video Solution==
 
==Video Solution==
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~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==
 
==See also==

Latest revision as of 10:19, 9 November 2024

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 $1+2(179)=\boxed{\textbf{(A) } 359}.$

Solution 4 (Simple)

We can consider the rotations and reflections separately. For the rotations, each rotation turns it by the next natural number. Thus the total number of degrees turned would be a triangle number. We test the smallest number, $359$ first, and we get that it turns $\frac{(1+359)359}{2} = 180n$, where $n$ is an integer. Thus, the point would be rotated to $(-1,0)$. We may be tempted to dismiss this option but we haven't considered the reflections. Each reflection acts as a $180^{\circ}$ rotation, so every two reflections cancel. However, $359$ is odd so we have to reflect $(-1,0)$, taking us to $(1,0)$, which is what we want. Thus we get $\boxed{\textbf{(A) } 359}$.

Solution 5 (Complex)

Rotations and reflections in 2D are very nice to describe using complex numbers in polar form. Reflections and rotations also don't affect the length (modulus or absolute value) of a complex number. This motivates us to set $z = \exp (i\theta)$, starting with $z=1+0i=\exp(i0)$. Then, we can describe rotations and reflections about the $y$-axis using these formulas: \[\text{Rotation}(\exp(i\theta), k): \exp(i\theta) \mapsto \exp(i(\theta + k))\] \[\text{Reflection}(\exp(i\theta)): \exp(i\theta) \mapsto \exp i (180^\circ - \theta) = -\exp(i(-\theta))\]

If we apply two successive iterations, we see a simplification:

\[T_k(\exp(i\theta)) = -\exp(-i(\theta+k))\]

\begin{aligned} T_{k+1}(T_k(\exp(i\theta))) &= T_{k+1}(-\exp(i(-\theta-k)) \\ &= --\exp(-i(-\theta-k+(k+1)) \\ &= \exp i(\theta-1) \end{aligned}


We also can calculate $T_1(\exp i0) = \exp 179^\circ$. Thus, the point $(1,0)$ gets sent back to when all double iterations after $1$ cancel $179^\circ$. $179-\left(\frac{n-1}{2}\right)=0$ so $n = 1 + 2 \cdot 179 = \boxed{\textbf{(A) }359}$.

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 (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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