AMC 10 · 2020 · #19

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 with labelled vertices undergoes a sequence of 20 moves, each chosen from four: a counterclockwise and a clockwise quarter turn, and reflections across the two axes. Count how many of the sequences return every labelled 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)

$2^{39}$

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 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 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 the moves as permutations

Write all four as permutations.

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

2STEP 2

Where a vertex sits after an even count

After an even count only two positions are possible.

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

3STEP 3

Get the proportion for two moves

For two moves the proportion is one quarter.

N(2) = 4, N(2)/4² = 1/4

4STEP 4

Grow it two moves at a time

Every two more moves multiplies the count by four.

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

5STEP 5

See the proportion persists

The total also quadruples, so the proportion holds.

P(identity) = 1/4, N(20) = 4²⁰/4 = 4¹⁹ = 2³⁸

6STEP 6

Apply it to twenty moves

A quarter of all sequences works.

N(20) = 2³⁸ → (C)

7STEP 7

Write it as a power

Tidying gives two to the thirty-eighth.

Ratio = 1/4 → N = 2⁴⁰ / 4 = 2³⁸

Answer

2³⁸

Sanity: total 4²⁰ = 2⁴⁰ ≈ 10¹², identity count 2³⁸ ≈ 2.75 × 10¹¹, ratio exactly 1/4. Small case verification: n = 2 gives 4 identity sequences (LR, RL, HH, VV); 4/16 = 1/4 ✓. n = 4: enumeration (or pair-induction) gives 4³ = 64 identity sequences out of 256; 64/256 = 1/4 ✓. Choice (C) confirmed.

💡Key takeaway

This AMC 12 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 1/4, giving 4²⁰/4 = 4¹⁹ = 2³⁸.