AMC 10 · 2022 · #18

Grade 8 arithmetic

rotation-isometryreflection-symmetrypattern-recognition easier-related-problempattern-recognitionconvert-to-algebra ↑ Prerequisites: rotation-isometry

📏 Long solution 💡 3 insights

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?

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

Pick an answer.

(A)

359

(B)

360

(C)

719

(D)

720

(E)

721

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Toolkit + CCSS Solution

Understand

Restated: Let $T_k$ rotate the plane $k°$ counterclockwise about the origin and then reflect across the $y$-axis. Find the least positive integer $n$ such that applying $T_1, T_2, T_3, \dots, T_n$ in order returns the point $(1, 0)$ to itself.

Givens: Each $T_k$ = (rotate by $k°$ CCW) then (reflect across $y$-axis); Starting point $(1, 0)$ has polar coordinates $(r=1, \theta=0°)$; Choices: (A) $359$, (B) $360$, (C) $719$, (D) $720$, (E) $721$

Unknowns: Least positive integer $n$ with $(1, 0) \xrightarrow{T_1 T_2 \cdots T_n} (1, 0)$

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #5 Look for a Pattern, #13 Convert to Algebra, #3 Eliminate Possibilities

Since only the angle matters (radius is preserved), track $\theta_n$ alone. Tool #9 (Easier Problem): compute $\theta_1, \theta_2, \theta_3, \theta_4$ explicitly using the rule that reflection across the $y$-axis sends angle $\phi$ to $180° - \phi$. Tool #5 (Look for a Pattern): the odd-index and even-index angles fall into two clean arithmetic sequences. Tool #13 (Convert to Algebra): set the closed-form expressions equal to a multiple of $360°$ and solve for the smallest valid $n$. Tool #3 (Eliminate): compare the two cases and pick the smaller.

Execute — Answer: A

#13 Convert to Algebra 8.G.A.3 Step 1

Translate one $T_k$ step into angle arithmetic.
A CCW rotation by $k°$ sends polar angle $\theta$ to $\theta + k°$; reflection across the $y$-axis sends polar angle $\phi$ to $180° - \phi$.
Composing them, $T_k$ sends $\theta$ to $180° - (\theta + k°) = 180° - \theta - k°$.

$$\theta_n = 180° - \theta_{n-1} - n°$$

💡 Reflection across $y$-axis is the angle map $\phi \mapsto 180° - \phi$ — that single fact closes the recurrence.

#9 Solve an Easier Related Problem 5.OA.B.3 Step 2

Compute the first few angles starting from $\theta_0 = 0°$.
This is the Easier-Problem move: see what happens before guessing a closed form.

$$\theta_1 = 179°,\ \theta_2 = -1°,\ \theta_3 = 178°,\ \theta_4 = -2°,\ \theta_5 = 177°,\ \theta_6 = -3°$$

💡 Just run the rule a few times and watch.

#5 Look for a Pattern 4.OA.C.5 Step 3

Spot the pattern by separating odd and even indices.

$$\theta_{2k-1} = 180° - k°,\qquad \theta_{2k} = -k°$$

💡 Odd steps land near $180°$ minus a counter; even steps land near $0°$ minus a counter.

#5 Look for a Pattern 5.OA.B.3 Step 4

Verify the pattern by induction-flavor check: assume $\theta_{2k} = -k$.
Then $\theta_{2k+1} = 180 - (-k) - (2k+1) = 180 + k - 2k - 1 = 179 - k = 180 - (k+1)$, matching the odd formula.
Next, $\theta_{2k+2} = 180 - (179 - k) - (2k+2) = 180 - 179 + k - 2k - 2 = -k - 1 = -(k+1)$, matching the even formula.
The closed forms are correct.

$\theta_{2k+1} = 180 - (k+1),\quad \theta_{2k+2} = -(k+1)$ — confirmed.

💡 One step of the recurrence carries the pattern from index $2k$ to $2k+1$ and then to $2k+2$.

#13 Convert to Algebra 7.NS.A.3 Step 5

The point $(1, 0)$ returns when $\theta_n$ is a multiple of $360°$.
Case A: $n$ is odd, $n = 2k - 1$.
Set $180° - k° \equiv 0 \pmod{360°}$, so $k \equiv 180 \pmod{360}$.
Smallest positive $k$ is $180$, giving $n = 2(180) - 1 = 359$.

$$n_{\text{odd}} = 359$$

💡 $180 - k = 0$ (the simplest multiple of $360$) gives $k = 180$ — the cheapest odd return.

#13 Convert to Algebra 7.NS.A.3 Step 6

Case B: $n$ is even, $n = 2k$.
Set $-k° \equiv 0 \pmod{360°}$, so $k \equiv 0 \pmod{360}$.
Smallest positive $k$ is $360$, giving $n = 720$.

$$n_{\text{even}} = 720$$

💡 Even-step angle is $-k$, so it can only hit $0 \pmod{360}$ when $k$ itself is a multiple of $360$.

#3 Eliminate Possibilities 4.NBT.A.2 Step 7

Compare the two candidates.
$359 < 720$, so the least positive integer is $359$, choice (A).

$$\min(359, 720) = 359 \;\Rightarrow\; \textbf{(A)}$$

💡 Pick the smaller of the two case minimums.

[1] #13 8.G.A.3 Translate one $T_k$ step into angle arithmetic. A CCW rotation by $k°$ sends pol

[2] #9 5.OA.B.3 Compute the first few angles starting from $\theta_0 = 0°$. This is the Easier-P

[3] #5 4.OA.C.5 Spot the pattern by separating odd and even indices.

[4] #5 5.OA.B.3 Verify the pattern by induction-flavor check: assume $\theta_{2k} = -k$. Then $\

[5] #13 7.NS.A.3 The point $(1, 0)$ returns when $\theta_n$ is a multiple of $360°$. Case A: $n$

[6] #13 7.NS.A.3 Case B: $n$ is even, $n = 2k$. Set $-k° \equiv 0 \pmod{360°}$, so $k \equiv 0 \p

[7] #3 4.NBT.A.2 Compare the two candidates. $359 < 720$, so the least positive integer is $359$,

Review

Reasonableness: Direct check at $n = 359$: this is odd with $k = 180$, so $\theta_{359} = 180° - 180° = 0°$ — point lands exactly back on $(1, 0)$. Direct check that no smaller $n$ works: for any odd $n = 2k - 1$ with $1 \le k \le 179$ we have $\theta_n = 180 - k \in \{1°, \dots, 179°\}$, none of which is a multiple of $360°$. For any even $n = 2k$ with $1 \le k \le 359$ we have $\theta_n = -k° \in \{-1°, \dots, -359°\}$, none of which is a multiple of $360°$. So $n = 359$ is indeed the minimum.

Alternative: Tool #10 (Physical Representation): on a sheet of paper mark $(1, 0)$, do $T_1$ (rotate $1°$, flip), do $T_2$ (rotate $2°$, flip), and watch the point flip back and forth between near-$0°$ and near-$180°$. After a handful of steps the symmetry is visible: every two applications shift the "near" target by $1°$, so the angle near $180°$ marches toward $0°$ at $1°$ per pair — reaching $0°$ after $180$ pairs, which is the odd index $n = 359$.

CCSS standards used (min grade 8)

8.G.A.3 Describe the effect of dilations, translations, rotations, and reflections on coordinates (Translating one $T_k$ application into the polar-angle update $\theta \mapsto 180° - \theta - k°$.)
5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships (Generating the angle sequence $\theta_0, \theta_1, \theta_2, \dots$ from the recurrence and confirming the odd/even pattern.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Spotting the two arithmetic sub-sequences $\theta_{2k-1} = 180 - k$ and $\theta_{2k} = -k$.)
7.NS.A.3 Solve real-world problems involving the four operations with rational numbers (Solving $180 - k \equiv 0 \pmod{360}$ and $-k \equiv 0 \pmod{360}$ for the smallest positive integer $k$ in each case.)
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Comparing $359 < 720$ to pick the least positive integer answer.)

⭐ Only the angle moves — the radius stays $1$. After running $T_k$ a few times you see two clean rules: odd steps land at $180° - k$, even steps land at $-k$. Solving "when is this a multiple of $360°$?" gives $n = 359$ (odd) and $n = 720$ (even), so the smallest is $\textbf{(A) }359$.