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Difference between revisions of "2020 AMC 8 Problems/Problem 9"
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its very easy to get trolled here if you dont read "no icing on the bottom"
 
its very easy to get trolled here if you dont read "no icing on the bottom"
 
==Solution 1==
 
==Solution 1==
−
Notice that all the faces with the exception of the bottom faces have the two center edge pieces with 2 faces with icing on them. This is <math>8\cdot 2 = 16</math>. Additionally, on the bottom face, the corners have 2 faces with icing, as the bottom face does not have icing. This is <math>4</math> cubes. The total is <math>16+4 = 20, \textbf{(D) }20</math>
+
Notice that all the faces with the exception of the bottom faces have the two center edge pieces with 2 faces with icing on them. This is <math>8\cdot 2 = 16</math>. Additionally, on the bottom face, the corners have 2 faces with icing, as the bottom face does not have icing. This is <math>4</math> cubes. The total is <math>16+4 = 20, \textbf{(D) }20</math>. Note we just need to get The total is <math>16+4 = 20, \textbf{(D) \textbf{(D) }20</math> }20<math>
  
−
~Windigo
+
~Windgo and minor edits made by oceanxia
  
 
==Solution 2==
 
==Solution 2==
Line 60: Line 60:
 
https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi82Lzk1NjIyMDEyYzEwMmU2MGRhM2U3OGMzYzA0MDNmOGFmZjdkMDk3LnBuZw==&rn=YW1jIDggbm8gOS5wbmc
 
https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi82Lzk1NjIyMDEyYzEwMmU2MGRhM2U3OGMzYzA0MDNmOGFmZjdkMDk3LnBuZw==&rn=YW1jIDggbm8gOS5wbmc
  
−
The face on the opposite side of the front face (hidden) is an exact copy of the front face. So the answer is <math>8+4+8=\textbf{(D)}20</math>.
+
The face on the opposite side of the front face (hidden) is an exact copy of the front face. So the answer is </math>8+4+8=\textbf{(D)}20$.
  
 
-franzliszt
 
-franzliszt

Revision as of 23:10, 19 November 2020

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] $\textbf{(A) }12 \qquad \textbf{(B) }16 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24$

its very easy to get trolled here if you dont read "no icing on the bottom"

Solution 1

Notice that all the faces with the exception of the bottom faces have the two center edge pieces with 2 faces with icing on them. This is $8\cdot 2 = 16$. Additionally, on the bottom face, the corners have 2 faces with icing, as the bottom face does not have icing. This is $4$ cubes. The total is $16+4 = 20, \textbf{(D) }20$. Note we just need to get The total is $16+4 = 20, \textbf{(D) \textbf{(D) }20$ (Error compiling LaTeX. Unknown error_msg) }20$~Windgo and minor edits made by oceanxia

==Solution 2== This is just careful casework. Consider the following diagram: https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi82Lzk1NjIyMDEyYzEwMmU2MGRhM2U3OGMzYzA0MDNmOGFmZjdkMDk3LnBuZw==&rn=YW1jIDggbm8gOS5wbmc

The face on the opposite side of the front face (hidden) is an exact copy of the front face. So the answer is$ (Error compiling LaTeX. Unknown error_msg)8+4+8=\textbf{(D)}20$.

-franzliszt

Franz Liszt is in 12th grade

Video Solution

https://youtu.be/WyvmQUfxTfo

~savannahsolver

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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