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

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<math>\textbf{(A) } 125 \qquad \textbf{(B) } 250 \qquad \textbf{(C) } 405 \qquad \textbf{(D) } 640 \qquad \textbf{(E) } 810</math>
 
<math>\textbf{(A) } 125 \qquad \textbf{(B) } 250 \qquad \textbf{(C) } 405 \qquad \textbf{(D) } 640 \qquad \textbf{(E) } 810</math>
  
−
== Solution ==
+
== Diagram ==
 +
<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>
  
 +
== 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 <math>5</math> choices for which face we visit first on the top ring. From there, we have <math>9</math> choices for how far around the top ring we go before moving down: <math>1,2,3,</math> or <math>4</math> faces around clockwise, <math>1,2,3,</math> or <math>4</math> faces around counterclockwise, or immediately going down to the lower ring without visiting any other faces in the upper ring.
 +
 +
We then have <math>2</math> choices for which lower ring face to visit first (since every upper-ring face is adjacent to exactly <math>2</math> lower-ring faces) and then once again <math>9</math> 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 <math>5 \cdot 9 \cdot 2 \cdot 9 = \boxed{\textbf{(E) } 810}</math>.
 +
 +
== Solution 2 ==
 +
Swap the faces as vertices and the vertices as faces. Then, this problem is the same as
 +
[https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3 2016 AIME I #3]
 +
which had an answer of <math>\boxed{\textbf{(E) } 810}</math>.
 +
<math>\textbf{\textbf{- Emathmaster}}</math>
 +
 +
==Video Solution 1==
 +
https://youtu.be/rH0MexSWafo ~DSA_Catachu
 +
 +
==Video Solution 1==
 +
 +
Education, The Study of Everything
 +
 +
https://youtu.be/av1hZOm5ELU
 +
 +
==Video Solution 2==
 +
https://youtu.be/RKlG6oZq9so
 +
 +
~IceMatrix
 +
 +
==Video Solution 3==
 +
 +
https://www.youtube.com/watch?v=Y0gezpr8Mrk&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=1 ~ MathEx
  
 
==See Also==
 
==See Also==
−
 
+
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3
 
{{AMC10 box|year=2020|ab=A|num-b=18|num-a=20}}
 
{{AMC10 box|year=2020|ab=A|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:54, 7 November 2020

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?

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

Diagram

[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]

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 $\boxed{\textbf{(E) } 810}$. $\textbf{\textbf{- Emathmaster}}$

Video Solution 1

https://youtu.be/rH0MexSWafo ~DSA_Catachu

Video Solution 1

Education, The Study of Everything

https://youtu.be/av1hZOm5ELU

Video Solution 2

https://youtu.be/RKlG6oZq9so

~IceMatrix

Video Solution 3

https://www.youtube.com/watch?v=Y0gezpr8Mrk&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=1 ~ MathEx

See Also

https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3

2020 AMC 10A (Problems • Answer Key • Resources)
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Problem 18 		Followed by
Problem 20
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