AMC 10 · 2020 · #19
Grade 7 countingProblem
As shown in the figure below, a regular dodecahedron (the polyhedron consisting of 12 congruent regular pentagonal faces) floats in space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring?
Pick an answer.
AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: 12 faces total: 1 top, 5 top-ring, 5 bottom-ring, 1 bottom; Top face is adjacent to all 5 top-ring faces; Each top-ring face is adjacent to 2 neighbors in the top ring and 2 faces in the bottom ring; Bottom face is adjacent to all 5 bottom-ring faces (mirror image of the top half); Choices: (A) $125$, (B) $250$, (C) $405$, (D) $640$, (E) $810$
Unknowns: Number of valid walks from top face to bottom face
Understand
Restated: 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.
Givens: 12 faces total: 1 top, 5 top-ring, 5 bottom-ring, 1 bottom; Top face is adjacent to all 5 top-ring faces; Each top-ring face is adjacent to 2 neighbors in the top ring and 2 faces in the bottom ring; Bottom face is adjacent to all 5 bottom-ring faces (mirror image of the top half); Choices: (A) $125$, (B) $250$, (C) $405$, (D) $640$, (E) $810$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #5 Look for a Pattern
Tool #7 (Subproblems): every valid walk follows the same shape — Top $\to$ (some top-ring faces) $\to$ (some bottom-ring faces) $\to$ 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 $1 + 2 \cdot 4 = 9$ choices.
Execute — Answer: E
5.G.A.2 Step 1 - Draw the layout: top face $T$ touches 5 top-ring faces $U_1, U_2, U_3, U_4, U_5$ in a cycle.
- Each $U_i$ touches its two neighbors $U_{i-1}, U_{i+1}$ and exactly two bottom-ring faces.
- The bottom ring $L_1, \ldots, L_5$ is a mirror: each $L_j$ touches its two ring neighbors and the bottom face $B$.
- The rule says we start at $T$, walk through some $U$'s, jump down to the bottom ring, walk through some $L$'s, then finish at $B$.
💡 Picture the dodecahedron as four layers stacked vertically.
7.SP.C.8 Step 2 - Choose which top-ring face to enter first.
- $T$ is adjacent to all 5 of $U_1, \ldots, U_5$, so there are $5$ choices for the entry face.
💡 Five doors to enter the top ring.
7.SP.C.8 Step 3 - Walk through the top ring.
- After entering at some $U_i$, you can either drop straight down (1 option) or walk clockwise or counterclockwise (2 directions) for $1, 2, 3, $ or $4$ more steps (4 lengths — you can't take 5 steps because that would re-enter $U_i$, which is already visited).
- That's $1 + 2 \cdot 4 = 9$ possible top-ring sub-paths, each ending at a different $U$.
💡 Either stop right away, or wander up to 4 faces in one of 2 directions.
7.SP.C.8 Step 4 - From the final top-ring face $U_j$ visited, drop into the bottom ring.
- Each $U_j$ touches exactly 2 bottom-ring faces, so $2$ choices.
💡 Each top-ring pentagon hangs over exactly 2 bottom-ring pentagons.
7.SP.C.8 Step 5 - Walk through the bottom ring.
- The geometry mirrors the top ring: starting at the entry face $L_k$, you can step straight down to $B$ (1 option) or walk $1, 2, 3,$ or $4$ steps in either direction (8 options).
- Total: $1 + 2 \cdot 4 = 9$ bottom-ring sub-paths.
💡 Same shape as the top ring — flip the dodecahedron upside down.
7.SP.C.8 Step 6 - Step to the bottom face.
- Every bottom-ring face is adjacent to $B$, so from the final $L$ visited there is exactly $1$ way to land on the bottom.
💡 All bottom-ring faces share a border with the bottom face.
7.SP.C.8 Step 7 - Multiply the independent choices: $5 \cdot 9 \cdot 2 \cdot 9 \cdot 1 = 810$.
- That's choice (E).
💡 Multiplication principle — independent stages, multiply the choices.
5.G.A.2 Draw the layout: top face $T$ touches 5 top-ring faces $U_1, U_2, U_3, U_4, U_5$ 7.SP.C.8 Choose which top-ring face to enter first. $T$ is adjacent to all 5 of $U_1, \ld 7.SP.C.8 Walk through the top ring. After entering at some $U_i$, you can either drop str 7.SP.C.8 From the final top-ring face $U_j$ visited, drop into the bottom ring. Each $U_j 7.SP.C.8 Walk through the bottom ring. The geometry mirrors the top ring: starting at the 7.SP.C.8 Step to the bottom face. Every bottom-ring face is adjacent to $B$, so from the 7.SP.C.8 Multiply the independent choices: $5 \cdot 9 \cdot 2 \cdot 9 \cdot 1 = 810$. Tha Review
Reasonableness: 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 $5 \cdot 9 \cdot 2 \cdot 9 \cdot 1 = 810$ 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 $\times 2$ drop factor.
Alternative: Tool #2 (Systematic List): split by total path length. Length-3 walks ($T \to U \to L \to B$): $5 \cdot 2 = 10$ ways. Length-4 walks ($T \to U \to U \to L \to B$, or $T \to U \to L \to L \to B$): $5 \cdot 2 \cdot 2 + 5 \cdot 2 \cdot 2 = 40$. Continuing through length-11 and summing all lengths gives $810$. Multiplication-principle is faster, but listing confirms the count.
CCSS standards used (min grade 7)
5.G.A.2Represent real-world and mathematical problems by graphing points (Drawing the dodecahedron's four-layer face graph to visualize all valid walk shapes.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting walks by multiplying independent stage-choices: $5 \cdot 9 \cdot 2 \cdot 9 \cdot 1$.)
⭐ This AMC 10 problem only needs Grade 7 multiplication-principle counting you already know — break the walk into five independent stages ($5 \cdot 9 \cdot 2 \cdot 9 \cdot 1$) and multiply. The answer is $\textbf{(E)} \; 810$.
⭐ This AMC 10 problem only needs Grade 7 multiplication-principle counting you already know — break the walk into five independent stages ($5 \cdot 9 \cdot 2 \cdot 9 \cdot 1$) and multiply. The answer is $\textbf{(E)} \; 810$.
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