2020 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 19 Hard

As shown in the figure below, a regular dodecahedron (the polyhedron consisting of 12 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?

  • A.

    125

  • B.

    250

  • C.

    405

  • D.

    640

  • E.

    810

Answer:E

Solution 1: Since we start at the top face and end at the bottom face without moving from the lower ring to the upper ring or revisiting a face, our journey must consist of the top face, a series of faces in the upper ring, a series of faces in the lower ring, and the bottom face, in that order.

We have 5 choices for which face we visit first on the top ring. From there, we have 9 choices for how far around the top ring we go before moving down: 1,2,3, or 4 faces around clockwise, 1,2,3, or 4 faces around counterclockwise, or immediately going down to the lower ring without visiting any other faces in the upper ring. We then have 2 choices for which lower ring face to visit first (since every upperring face is adjacent to exactly 2 lower-ring faces) and then once again 9 choices for how to travel around the lower ring. We then proceed to the bottom face, completing the trip. Multiplying together all the numbers of choices we have, we \text { get } 5 \cdot 9 \cdot 2 \cdot 9=(\mathbf{E}) 810 \text {. }

Solution 2:

Swap the faces as vertices and the vertices as faces. Then, this problem is the same as 2016 AIME I #3which had an answer (E) 810 .