2020 AMC 10A Problems/Problem 19

Revision as of 21:42, 31 January 2020 by Pcampbell (talk | contribs) (Solution)

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?

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

Solution

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 upper-ring 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 get $5 \cdot 9 \cdot 2 \cdot 9 = \boxed{(E) 810}$.

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 10 Problems and Solutions

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