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

Revision as of 22:39, 11 September 2025

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

So, you don't know polar coordinates (solution 2) nor came up with the solution 3 logic, then, here we go!

The easy solution is to notice that, without reflections and taking $ k=1 $, there are 360 turns that need to be made.

Now take $ k=2 $. We see that there are 180 rotations, so the number of rotations is just $ 360/k $.

Now include the reflections over $ y=x $.

Say that each time we reflect a point over $ y=x $, we draw a line.

Each time we reflect over $ y=x $, we see more and more lines that generate. For $ k=1 $, there are quite obviously 360 of these lines.

However, one can see that this works for every single case, because when we reflect it over $ y=x $ and make a $ k^\circ $ rotation, we fill up all the missed gaps until we reach the end of a sequence - I prove this on the remark section.

But wait! We overcounted by one $ y=x $ line! Notice how I said there were 360 lines, but the line $ y=x $ is already given to you! So our final answer is $ 360-1 = $ $\boxed{\textbf{(A) } 359}.$

~Pinotation

Proof 1.1

A way you can prove such an assumption is by taking ideas like addition. We know that eventually there exists a number $ N $ such that $ l \times N = m $, where $ m $ has a units digit of $ l $. Utilizing the same approach, we can find that this actually works.

~Pinotation

Solution 2

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 3

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 4

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 5 (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 6 (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 $n = 1 + 2 \cdot 179 = \boxed{\textbf{(A) }359}$ .

Solution 7 (Matrices)

Similar to the above solution, one can use rotation matrices to compute it.

(i.e. $\begin{bmatrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \end{bmatrix}$ )

For the reflection, use the reflection matrix (i.e. $\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$ )

Then proceed like Solution 6.

~QuantumMathTitan

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

@@ Line 24: / Line 24: @@
 However, one can see that this works for every single case, because when we reflect it over \( y=x \) and make a \( k^\circ \) rotation, we fill up all the missed gaps until we reach the end of a sequence - I prove this on the remark section.
-But wait! We overcounted by one \( y=x \) line! Notice how I said there were 360 lines, but the line \( y=x \) is already given to you! So our final answer is \( 360-1 = \)
+But wait! We overcounted by one \( y=x \) line! Notice how I said there were 360 lines, but the line \( y=x \) is already given to you! So our final answer is \( 360-1 = \) <math>\boxed{\textbf{(A) } 359}.</math>
 ~Pinotation

2022 AMC 10A (Problems • Answer Key • Resources)
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2022 AMC 12A (Problems • Answer Key • Resources)
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