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2020 AMC 10A Problem 19

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2020 AMC 10A Problem 19AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Counting · difficulty 4 of 5 · about 6 min · position P16-20

As shown in the figure below, a regular dodecahedron (the polyhedron consisting of 1212 congruent regular pentagonal faces) floats in empty 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? figure

  1. A)

    125125

  2. B)

    250250

  3. C)

    405405

  4. D)

    640640

  5. E)

    810810

Answer

E

Key insight

Routes factor into stages: entry face on the top ring (5), walk along it (9), step down (2), walk the bottom ring (9): 810.

Solution

Each face of the top ring touches the top face, its two ring neighbors, and two faces of the bottom ring (its remaining two edges). Likewise each bottom-ring face touches the bottom face, two ring neighbors, and two top-ring faces.

Since a route cannot return to the top face and cannot climb from the bottom ring back to the top ring, every route has the shape: top face, some faces of the top ring, some faces of the bottom ring, bottom face.

Stage 1: choose the first top-ring face: 55 ways.

Stage 2: wander along the top ring without repeating. The ring is a 55-cycle, so from the entry face you can stop immediately (11 way) or move 1,2,3,1, 2, 3, or 44 steps clockwise or counterclockwise (2⋅4=82 \cdot 4 = 8 ways): 99 ways in all, each ending at a definite face.

Stage 3: from the final top-ring face, step down to one of its 22 adjacent bottom-ring faces: 22 ways.

Stage 4: wander along the bottom ring: again 99 ways.

Stage 5: step onto the bottom face: 11 way.

Total: 5⋅9⋅2⋅9⋅1=8105 \cdot 9 \cdot 2 \cdot 9 \cdot 1 = 810.

The answer is (E) 810\boxed{\textbf{(E)}\ 810}.

Why this works

The "no upward moves" rule makes the route pass through the layers in order, so the count factors into independent stages. Within a cycle of nn faces, the number of simple walks from a fixed start is 1+2(n−1)1 + 2(n-1), because any other face can be reached going either direction. Model the solid as a layered graph and the geometry disappears.

The trap

Counting only 5 ways to traverse a ring (one per exit face) instead of 9, since every non-entry face can be reached going either way around, which gives 250.

Common mistakes

  • Counting only 5 ways to traverse a ring (one per exit face) instead of 9, since every non-entry face can be reached going either way around, which gives 250.
  • Assuming each top-ring face touches only one bottom-ring face, or forgetting that a route may pass through several faces of the same ring.

Techniques

Set up the equation/formula and compute; no special trick needed · Exploit symmetry to reduce work or pair up objects