AMC 10 · 2020 · #20

Grade 8 geometry-2d

Archetype Casework by Position / Vertex Type └ Sub-archetype Vertex or Element Role

transformations-compositionrotation-isometryreflection-symmetrymodular-arithmeticcombinations-basic caseworksystematic-enumerationphysical-representation ↑ Prerequisites: rotation-isometryreflection-symmetry

📏 Long solution 💡 3 insights

Problem

There are five rigid motions: three rotations about the origin and two reflections across the axes. Pick an ordered triple from them, repetition allowed. Count how many of those triples, applied one after another, return a given triangle to its starting position.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Identify Subproblems

Tool #7 (Subproblems): the triangle is asymmetric, so orientation parity splits the problem cleanly — even number of reflections required, giving Case 1 (0 reflections, 3 rotations) and Case 2 (2 reflections, 1 rotation). Tool #10 (Physical): cut out a paper triangle, label one side, and manipulate to verify each composition. Tool #2 (Systematic List): list all rotation-only triples summing to a multiple of 360°. Tool #5 (Pattern): two distinct axis-reflections compose to R₁₈₀ — this collapses Case 2 to a single unordered set {S_x, S_y, R₁₈₀}.

1STEP 1

Reflections come in pairs

Orientation forces an even number of reflections.

#reflections ∈ {0, 2}

2STEP 2

The rotations-only case

The angles must add to a whole turn.

n₁ + n₂ + n₃ ≡ 0 (mod 4), n_i ∈ {1, 2, 3}

3STEP 3

Count the rotations-only case

Count the triples that fit.

|Case 1| = 3 + 3 = 6

4STEP 4

Two reflections and one rotation

Two reflections compose to a half turn.

S_y ° S_x = R₁₈₀; R ° R₁₈₀ = I → R = R₁₈₀

5STEP 5

Count that case

That fixes the remaining rotation.

|Case 2| = 3! = 6

6STEP 6

Add the two cases

Adding gives 12.

Total = 6 + 6 = 12 → (A)

Answer

Sanity. 12 out of 125 is about 10%, plausible for a tight identity constraint. Spot-check (R₉₀, R₉₀, R₁₈₀) from Case 1: total CCW rotation = 90 + 90 + 180 = 360°, identity ✓. Spot-check (S_x, R₁₈₀, S_y) from Case 2: (x, y) → (x, -y) → (-x, y) → (x, y), identity ✓. Spot-check a non-winner (R₉₀, S_x, S_x): (x, y) → (-y, x) → (-y, -x) → (-y, x) — not identity ✓ (matches our "same reflection twice + rotation fails" reasoning). The trap choice (E) 25 = 5² would be "any rotation choice once two reflections are placed" without filtering — clearly an over-count.

💡Key takeaway

This AMC 12 problem only needs Grade 8 properties of rotations and reflections you already know — even number of reflections (so 0 or 2); three rotations summing to 360° or 720° give 6 ways; two different axis-reflections plus an R₁₈₀ give another 6 ways; 6 + 6 = 12.