2020 AMC 8 Problems/Problem 9

Problem

Akash'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 to the front face.

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 $4x4$ rubik's cube, there are exactly $24$ 'edge' pieces, $8$ 'corners', and $24$ 'center' pieces. Edge pieces have two 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

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

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

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