# Difference between revisions of "2020 AMC 10A Problems/Problem 19"

## 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? $[asy] import graph; unitsize(5cm); pair A = (0.082, 0.378); pair B = (0.091, 0.649); pair C = (0.249, 0.899); pair D = (0.479, 0.939); pair E = (0.758, 0.893); pair F = (0.862, 0.658); pair G = (0.924, 0.403); pair H = (0.747, 0.194); pair I = (0.526, 0.075); pair J = (0.251, 0.170); pair K = (0.568, 0.234); pair L = (0.262, 0.449); pair M = (0.373, 0.813); pair N = (0.731, 0.813); pair O = (0.851, 0.461); path[] f; f[0] = A--B--C--M--L--cycle; f[1] = C--D--E--N--M--cycle; f[2] = E--F--G--O--N--cycle; f[3] = G--H--I--K--O--cycle; f[4] = I--J--A--L--K--cycle; f[5] = K--L--M--N--O--cycle; draw(f[0]); axialshade(f[1], white, M, gray(0.5), (C+2*D)/3); draw(f[1]); filldraw(f[2], gray); filldraw(f[3], gray); axialshade(f[4], white, L, gray(0.7), J); draw(f[4]); draw(f[5]); [/asy]$ $\textbf{(A) } 125 \qquad \textbf{(B) } 250 \qquad \textbf{(C) } 405 \qquad \textbf{(D) } 640 \qquad \textbf{(E) } 810$

## 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 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 $5 \cdot 9 \cdot 2 \cdot 9 = \boxed{\textbf{(E) } 810}$.

## Solution 2

Swap the faces as vertices and the vertices as faces. Then, this problem is the same as 2016 AIME I #3 which had an answer of $\boxed{\textbf{(E) } 810}$. $\textbf{\textbf{- Emathmaster}}$

## Solution 3 (Similar to Solution 1)

From the top, we can go down in five different ways to the five faces underneath the first face. From here we can go down or go to the adjacent faces. From the face you went down from the top face, you can either go clockwise or counterclockwise $1$, $2$, $3$,or $4$ times, or you can go straight down. Then from there, you go down into the lower row, which you have two choices, left or right down. From here we have $5 \cdot 9 \cdot 2$ ways multiplied by the ways you can move from the bottom ring to the bottom face, but we don't need to know that since from here we can see that $\boxed{(E)}$ is the only answer choice divisible by $90$, so our answer is $\boxed{(E)}$. ~Terribleteeth

## Video Solutions

### Video Solution 1

https://youtu.be/rH0MexSWafo ~DSA_Catachu

### Video Solution 2

Education, The Study of Everything

## See Also

 2020 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 20 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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