AMC 10 · 2020 · #19

Grade 7 counting

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

📏 Long solution 💡 3 insights 📊 Diagram

Problem

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?

$\textbf{(A) } 125 \qquad \textbf{(B) } 250 \qquad \textbf{(C) } 405 \qquad \textbf{(D) } 640 \qquad \textbf{(E) } 810$

Pick an answer.

(A)

125

(B)

250

(C)

405

(D)

640

(E)

810

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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

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

#1 Draw a Diagram 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 $B$.

$$T \to (U\text{'s}) \to (L\text{'s}) \to B$$

💡 Picture the dodecahedron as four layers stacked vertically.

#7 Identify Subproblems 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.

$$\text{Entry choices} = 5$$

💡 Five doors to enter the top ring.

#5 Look for a Pattern 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 $U$.

$$\text{Top-ring sub-paths} = 1 + 2 \cdot 4 = 9$$

💡 Either stop right away, or wander up to 4 faces in one of 2 directions.

#7 Identify Subproblems 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.

$$\text{Drop choices} = 2$$

💡 Each top-ring pentagon hangs over exactly 2 bottom-ring pentagons.

#5 Look for a Pattern 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.

$$\text{Bottom-ring sub-paths} = 1 + 2 \cdot 4 = 9$$

💡 Same shape as the top ring — flip the dodecahedron upside down.

#7 Identify Subproblems 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.

$$\text{Finish} = 1$$

💡 All bottom-ring faces share a border with the bottom face.

#7 Identify Subproblems 7.SP.C.8 Step 7

Multiply the independent choices: $5 \cdot 9 \cdot 2 \cdot 9 \cdot 1 = 810$.
That's choice (E).

$$5 \cdot 9 \cdot 2 \cdot 9 \cdot 1 = 810 \;\Rightarrow\; \textbf{(E)}$$

💡 Multiplication principle — independent stages, multiply the choices.

[1] #1 5.G.A.2 Draw the layout: top face $T$ touches 5 top-ring faces $U_1, U_2, U_3, U_4, U_5$

[2] #7 7.SP.C.8 Choose which top-ring face to enter first. $T$ is adjacent to all 5 of $U_1, \ld

[3] #5 7.SP.C.8 Walk through the top ring. After entering at some $U_i$, you can either drop str

[4] #7 7.SP.C.8 From the final top-ring face $U_j$ visited, drop into the bottom ring. Each $U_j

[5] #5 7.SP.C.8 Walk through the bottom ring. The geometry mirrors the top ring: starting at the

[6] #7 7.SP.C.8 Step to the bottom face. Every bottom-ring face is adjacent to $B$, so from the

[7] #7 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.2 Represent 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.8 Find 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 $\textbf{(E)} \; 810$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 7)

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