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Difference between revisions of "2020 AMC 8 Problems/Problem 9"

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Akash's birthday cake is in the form of a <math>4 \times 4 \times 4</math> inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into <math>64</math> smaller cubes, each measuring <math>1 \times 1 \times 1</math> inch, as shown below. How many of the small pieces will have icing on exactly two sides?
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==Problem==
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Max's birthday cake is in the form of a <math>4 \times 4 \times 4</math> inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into <math>64</math> smaller cubes, each measuring <math>1 \times 1 \times 1</math> inch, as shown below. How many of the small pieces will have icing on exactly two sides?
  
 
<asy>
 
<asy>
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</asy>
 
</asy>
 
<math>\textbf{(A) }12 \qquad \textbf{(B) }16 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24</math>
 
<math>\textbf{(A) }12 \qquad \textbf{(B) }16 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24</math>
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==Solution 1==
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Notice that, for a small cube which does not form part of the bottom face, it will have exactly <math>2</math> faces with icing on them only if it is one of the <math>2</math> center cubes of an edge of the larger cube. There are <math>12-4 = 8</math> such edges (as we exclude the <math>4</math> edges of the bottom face), so this case yields <math>2 \cdot 8 = 16</math> small cubes. As for the bottom face, we can see that only the <math>4</math> corner cubes have exactly <math>2</math> faces with icing, so the total is <math>16+4 = \boxed{\textbf{(D) }20}</math>.
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==Solution 2==
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The following diagram shows <math>12</math> of the small cubes having exactly <math>2</math> faces with icing on them; that is all of them except for those on the hidden face directly opposite the front face.
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[[File:Prob10-diagram.png|middle|center]]
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But the hidden face is an exact copy of the front face, so the answer is <math>12+8=\boxed{\textbf{(D) }20}</math>.
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==Solution 3==
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(For Rubik's Cubers)
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On a <math>4</math>x<math>4</math> rubik's cube, there are exactly <math>24</math>'edge' pieces, <math>8</math> 'corners', and <math>24</math> 'center' pieces. Edge pieces have <math>2</math> frosted faces (the ones on the bottom only have one, corners have <math>3</math> frosted faces, and centers have <math>1</math>. So since we have <math>24</math> edges pieces, we minus the <math>8</math> 'edge' pieces on the bottom (they only have one frosted face), and then we add the <math>4</math> bottom 'corner' pieces (they have also 2 frosted faces). we get <math>24-8+4=\boxed{\textbf{(D) }20}</math>.
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-Solution by MismatchedCubing and Andrew_Lu
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 +
==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=8hgK6rESdek&t=9s
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 +
~NiuniuMaths
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 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/UnVo6jZ3Wnk?si=9Se2Yd0UrYpxjhk3&t=1038
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 +
~Math-X
 +
 +
==Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/_diexvyeje4
 +
 +
~Education, the Study of Everything
 +
 +
==Video Solution by North America Math Contest Go-Go Go==
 +
 +
https://www.youtube.com/watch?v=6LbBcFUmBr0
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 +
~North America Math Contest Go Go Go
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/WyvmQUfxTfo
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 +
~savannahsolver
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 +
==Video Solution==
 +
https://youtu.be/61c1MR9tne8
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 +
==Video Solution by Interstigation==
 +
https://youtu.be/YnwkBZTv5Fw?t=355
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 +
~Interstigation
 +
 +
==Video Solution by STEMbreezy==
 +
https://youtu.be/U27z1hwMXKY?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=268
 +
 +
~STEMbreezy
 +
 +
==See also==
 +
{{AMC8 box|year=2020|num-b=8|num-a=10}}
 +
{{MAA Notice}}

Latest revision as of 17:46, 29 May 2024

Problem

Max's birthday cake is in the form of a $4 \times 4 \times 4$ inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into $64$ smaller cubes, each measuring $1 \times 1 \times 1$ inch, as shown below. How many of the small pieces will have icing on exactly two sides?

[asy] /* Created by SirCalcsALot and sonone Code modfied from https://artofproblemsolving.com/community/c3114h2152994_the_old__aops_logo_with_asymptote */ import three; currentprojection=orthographic(1.75,7,2); //++++ edit colors, names are self-explainatory ++++ //pen top=rgb(27/255, 135/255, 212/255); //pen right=rgb(254/255,245/255,182/255); //pen left=rgb(153/255,200/255,99/255); pen top = rgb(170/255, 170/255, 170/255); pen left = rgb(81/255, 81/255, 81/255); pen right = rgb(165/255, 165/255, 165/255); pen edges=black; int max_side = 4; //+++++++++++++++++++++++++++++++++++++++ path3 leftface=(1,0,0)--(1,1,0)--(1,1,1)--(1,0,1)--cycle; path3 rightface=(0,1,0)--(1,1,0)--(1,1,1)--(0,1,1)--cycle; path3 topface=(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle; for(int i=0; i<max_side; ++i){ for(int j=0; j<max_side; ++j){ draw(shift(i,j,-1)*surface(topface),top); draw(shift(i,j,-1)*topface,edges); draw(shift(i,-1,j)*surface(rightface),right); draw(shift(i,-1,j)*rightface,edges); draw(shift(-1,j,i)*surface(leftface),left); draw(shift(-1,j,i)*leftface,edges); } } picture CUBE; draw(CUBE,surface(leftface),left,nolight); draw(CUBE,surface(rightface),right,nolight); draw(CUBE,surface(topface),top,nolight); draw(CUBE,topface,edges); draw(CUBE,leftface,edges); draw(CUBE,rightface,edges); // begin made by SirCalcsALot int[][] heights = {{4,4,4,4},{4,4,4,4},{4,4,4,4},{4,4,4,4}}; for (int i = 0; i < max_side; ++i) { for (int j = 0; j < max_side; ++j) { for (int k = 0; k < min(heights[i][j], max_side); ++k) { add(shift(i,j,k)*CUBE); } } } [/asy] $\textbf{(A) }12 \qquad \textbf{(B) }16 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24$

Solution 1

Notice that, for a small cube which does not form part of the bottom face, it will have exactly $2$ faces with icing on them only if it is one of the $2$ center cubes of an edge of the larger cube. There are $12-4 = 8$ such edges (as we exclude the $4$ edges of the bottom face), so this case yields $2 \cdot 8 = 16$ small cubes. As for the bottom face, we can see that only the $4$ corner cubes have exactly $2$ faces with icing, so the total is $16+4 = \boxed{\textbf{(D) }20}$.

Solution 2

The following diagram shows $12$ of the small cubes having exactly $2$ faces with icing on them; that is all of them except for those on the hidden face directly opposite the front face.

Prob10-diagram.png

But the hidden face is an exact copy of the front face, so the answer is $12+8=\boxed{\textbf{(D) }20}$.

Solution 3

(For Rubik's Cubers) On a $4$x$4$ rubik's cube, there are exactly $24$'edge' pieces, $8$ 'corners', and $24$ 'center' pieces. Edge pieces have $2$ frosted faces (the ones on the bottom only have one, corners have $3$ frosted faces, and centers have $1$. So since we have $24$ edges pieces, we minus the $8$ 'edge' pieces on the bottom (they only have one frosted face), and then we add the $4$ bottom 'corner' pieces (they have also 2 frosted faces). we get $24-8+4=\boxed{\textbf{(D) }20}$.

-Solution by MismatchedCubing and Andrew_Lu

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=8hgK6rESdek&t=9s

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=9Se2Yd0UrYpxjhk3&t=1038

~Math-X

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/_diexvyeje4

~Education, the Study of Everything

Video Solution by North America Math Contest Go-Go Go

https://www.youtube.com/watch?v=6LbBcFUmBr0

~North America Math Contest Go Go Go

Video Solution by WhyMath

https://youtu.be/WyvmQUfxTfo

~savannahsolver

Video Solution

https://youtu.be/61c1MR9tne8

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=355

~Interstigation

Video Solution by STEMbreezy

https://youtu.be/U27z1hwMXKY?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=268

~STEMbreezy

See also

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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 AJHSME/AMC 8 Problems and Solutions

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

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