AMC 10 · 2020 · #23

Grade 8 geometry-2d

Archetype Branching State Enumeration └ Sub-archetype Constrained Sequence or Path Count

transformations-compositionrotation-isometryreflection-symmetryparity pattern-recognitioncaseworksymmetry-argument ↑ Prerequisites: transformations-compositionreflection-symmetry

📏 Long solution 💡 3 insights

Problem

A unit-square labeled ABCD undergoes a sequence of 20 moves, each chosen from {L, R, H, V} (90° rotations counterclockwise/clockwise and reflections across the x- and y-axes). How many of the 4²⁰ sequences return every labeled vertex to its starting position?

Pick an answer.

(A)

$2^{37}$

(B)

$3\cdot 2^{36}$

(C)

$2^{38}$

(D)

$3\cdot 2^{37}$

(E)

$2^{39}$

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

Tool #9 (Easier Problem): work out the count for n = 2 moves, then n = 4, and look for the rule. Tool #5 (Pattern): the small cases reveal that exactly $\frac{1}{4}$ of all sequences of even length return home — a constant ratio regardless of n (for n ≥ 2). Tool #2 (Systematic List): enumerate the 4 × 4 = 16 length-2 sequences and check which give identity. Tool #7 (Subproblems): split the 20 moves into 10 pairs of consecutive moves; each pair has 4 equiprobable effects on the square. Tool #6 (Guess and Check): the simple guess $\frac{4²⁰}{4}$ = 4¹⁹ = 2³⁸ is verified by both the pair-induction and the small-case computation. Tool #1 (Diagram) anchors the four moves as concrete permutations; Tool #3 (Eliminate) cross-checks the answer against the choice list.

1STEP 1

Write each move as a vertex permutation: L, R, H, V each send every labeled vertex to an adjacent position, never the diagonal corner.

L = (A B C D), R = (A D C B), H = (A D)(B C), V = (A B)(C D)

2STEP 2

Each move toggles A between diagonal classes {A,C} and {B,D}, so returning home needs an even length — and 20 qualifies.

A_after even ∈ {A, C}, A_after odd ∈ {B, D}

3STEP 3

List all sixteen length-two sequences: exactly LR, RL, HH, VV compose to the identity, so 4 of 16 return — a ratio of $\frac{1}{4}$ .

N(2) = 4,

\frac{N(2)}{4²}

\frac{1}{4}

4STEP 4

Induction: appending a pair to a length-n sequence lands the identity with chance one in four, so N(n+2) = 4·N(n).

N(n+2) = 4 · N(n), N(n) = 4ⁿ⁻¹ for n ≥ 2 even

5STEP 5

Directly: each pair gives one of four effects (a 4-element subgroup), one the identity, so the return probability is exactly $\frac{1}{4}$ .

P(identity) =

\frac{1}{4}

, N(20) =

\frac{4²⁰}{4}

= 4¹⁹ = 2³⁸

6STEP 6

Compute: 4¹⁹ = (2²)¹⁹ = 2³⁸, which is choice (C).

N(20) = 2³⁸ → (C)

7STEP 7

Cross-check: total 2⁴⁰ times ratio $\frac{1}{4}$ gives 2³⁸ (C); ratios $\frac{1}{8}$ or $\frac{1}{2}$ would give (A) or (E), ruled out by four effects per pair.

Ratio =

\frac{1}{4}

→ N =

\frac{2⁴⁰}{4}

= 2³⁸

Answer

2³⁸

Sanity: total 4²⁰ = 2⁴⁰ ≈ 10¹², identity count 2³⁸ ≈ 2.75 × 10¹¹, ratio exactly

\frac{1}{4}

. Small case verification: n = 2 gives 4 identity sequences (LR, RL, HH, VV);

\frac{4}{16}

\frac{1}{4}

✓. n = 4: enumeration (or pair-induction) gives 4³ = 64 identity sequences out of 256;

\frac{64}{256}

\frac{1}{4}

✓. Choice (C) confirmed.

💡Key takeaway

This AMC 10 problem only needs Grade 8 transformations: each pair of moves has exactly 4 possible effects on the labeled square (one of which is the identity), so the ratio of identity sequences is $\frac{1}{4}$ , giving $\frac{4²⁰}{4}$ = 4¹⁹ = 2³⁸.

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