AMC 10 · 2020 · #19

Grade 7 counting

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

spatial-visualizationsystematic-enumerationface-adjacencysymmetry-argumenttree-enumeration identify-subproblemscaseworksymmetry-argument ↑ Prerequisites: face-adjacencysystematic-enumeration

📏 Long solution 💡 3 insights 📊 Diagram

Problem

A regular dodecahedron sits with a flat top face and a flat bottom face. Around the top sits a ring of 5 slanted pentagons (top ring); around the bottom sits another ring of 5 slanted pentagons (bottom ring). Walk from the top face to the bottom face, stepping only between adjacent faces, visiting each face at most once, and never moving from a bottom-ring face back up to a top-ring face. Count the number of distinct walks.

Pick an answer.

(A)

125

(B)

250

(C)

405

(D)

640

(E)

810

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

How to solve

Strategy Identify Subproblems

Tool #7 (Subproblems): every valid walk follows the same shape — Top → (some top-ring faces) → (some bottom-ring faces) → Bottom. Cut the walk at the moment it leaves the top ring, and count each leg independently, then multiply. Tool #1 (Diagram): draw the top ring as a 5-cycle of pentagons labeled 1 through 5, mark the 2 bottom-ring neighbors of each, and the path choices become visual. Tool #5 (Pattern): inside each ring, after entering at one face, you can leave directly or walk 1, 2, 3, 4 steps clockwise or counterclockwise — that's always 1 + 2 · 4 = 9 choices.

1STEP 1

Every walk has one shape: top face T, some top-ring faces, some bottom-ring faces, then bottom face B. Count each leg, then multiply.

T → (U's) → (L's) → B

2STEP 2

T touches all 5 top-ring faces, so there are 5 choices for the entry face.

Entry choices = 5

3STEP 3

Inside the ring: drop straight down (1 way) or walk 1–4 faces clockwise or counterclockwise (2·4) — 9 sub-paths.

Top-ring sub-paths = 1 + 2 · 4 = 9

4STEP 4

Each top-ring face hangs over exactly 2 bottom-ring faces, so dropping down gives 2 choices.

Drop choices = 2

5STEP 5

Mirror of the top ring: from the entry face, drop to B (1 way) or walk 1–4 faces either direction — 9 sub-paths again.

Bottom-ring sub-paths = 1 + 2 · 4 = 9

6STEP 6

Every bottom-ring face borders B, so the last step to the bottom is just 1 way.

Finish = 1

7STEP 7

Multiply the five stage counts: 5 · 9 · 2 · 9 · 1 = 810, choice (E).

5 · 9 · 2 · 9 · 1 = 810 → (E)

Answer

810

Sanity-check the in-ring count 9: at the entry face, the choices are stop (1 way), walk 1, 2, 3, 4 clockwise (4 ways), walk 1, 2, 3, 4 counterclockwise (4 ways) — 1 + 4 + 4 = 9. The product 5 · 9 · 2 · 9 · 1 = 810 matches the only choice divisible by 9 in the list — the others (125, 250, 405, 640) come from under-counting the in-ring options or skipping the × 2 drop factor.

💡Key takeaway

This AMC 10 problem only needs Grade 7 multiplication-principle counting you already know — break the walk into five independent stages (5 · 9 · 2 · 9 · 1) and multiply. The answer is (E) 810.

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