AMC 10 · 2020 · #23

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
From the 5 rigid motions {R₉0, R₁80, R₂70, S_x, S_y} (CCW rotations about the origin and reflections across the axes), pick an ordered triple (with repetition) of size 3 — there are 5³ = 125 such ordered triples. How many of them, when applied one after another, return triangle T with vertices (0,0), (4,0), (0,3) to its starting position?

Pick an answer.

(A)
12
(B)
15
(C)
17
(D)
20
(E)
25

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

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

T's sides 3, 4, 5 are distinct, so the composition must be the identity; identity preserves orientation, so reflections number 0 or 2.

#reflections ∈ {0, 2}
2STEP 2

Case 1 (three rotations): as quarter-turns 1, 2, 3, the sum must be a multiple of 4, so only sums 4 or 8 work.

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

Sum 4 comes only from {1,1,2} and sum 8 only from {2,3,3}; each has 3!2!\frac{3!}{2!} = 3 orderings, so Case 1 gives 6 sequences.

|Case 1| = 3 + 3 = 6
4STEP 4

Case 2 (two reflections, one rotation): same-axis pairs cancel, so use S_x and S_y; they compose to R₁80, so the rotation is R₁80.

S_y ° S_x = R₁80; R ° R₁80 = I → R = R₁80
5STEP 5

The set {S_x, S_y, R₁80} has three distinct, pairwise-commuting elements, so all 3! orderings work — Case 2 gives 6.

|Case 2| = 3! = 6
6STEP 6

The two cases are disjoint (0 vs 2 reflections), so add: 6 + 6 = 12.

Total = 6 + 6 = 12 → (A)
Answer
12
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 10 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.

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