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

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[[File:Prob10-diagram.png|middle|center]]
 
[[File:Prob10-diagram.png|middle|center]]
 
But the hidden face is an exact copy of the front face, so the answer is <math>12+8=\boxed{\textbf{(D) }20}</math>.
 
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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It is clearly observable that only the middle-edged pieces of each face will have the icing exactly on two sides. There are 4 such pieces on each face. Considering 5 faces of the cube (since bottom is not iced), we can state that the number of such pieces of dimensions 1x1x1 is going to be 5x4=20. Therefore our answer is 20
  
 
==Solution 3==
 
==Solution 3==

Latest revision as of 13:43, 14 August 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

It is clearly observable that only the middle-edged pieces of each face will have the icing exactly on two sides. There are 4 such pieces on each face. Considering 5 faces of the cube (since bottom is not iced), we can state that the number of such pieces of dimensions 1x1x1 is going to be 5x4=20. Therefore our answer is 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

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