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| ==Problem== | | ==Problem== |
| Three cubes of volume <math>1, 8</math> and <math>27</math> are glued together at their faces. The smallest possible surface area of the resulting configuration is | | Three cubes of volume <math>1</math>, <math>8</math> and <math>27</math> are glued together at their faces. The smallest possible surface area of the resulting configuration is |
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| <math> \textbf{(A)}\ 36 \qquad\textbf{(B)}\ 56 \qquad\textbf{(C)}\ 70 \qquad\textbf{(D)}\ 72 \qquad\textbf{(E)}\ 74 </math> | | <math> \textbf{(A)}\ 36 \qquad\textbf{(B)}\ 56 \qquad\textbf{(C)}\ 70 \qquad\textbf{(D)}\ 72 \qquad\textbf{(E)}\ 74 </math> |
| ==Solution==
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| <asy>
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| import three; currentprojection = orthographic(5,-30,20);
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| triple[] P = {(0,0,0),(1,0,0),(1,1,0),(0,1,0),(0,0,1),(1,0,1),(1,1,1),(0,1,1)},P2={(1,0.5,1),(0.5,0,1),(0.5,0.5,1),(1,0,1.5),(1,0.5,1.5),(0.5,0.5,1.5),(0.5,0,1.5)},P3={(0.25,0,1),(0.25,0.25,1),(0.5,0.25,1),(0.25,0,1.25),(0.25,0.25,1.25),(0.5,0.25,1.25),(0.5,0,1.25)};
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| void drawFrontFace(int w, int x, int y, int z){
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| draw(P[w]--P[x] -- P[y] -- P[z] -- cycle, linewidth(0.7));
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| /* fill(P[x] -- P[y] -- P[z] -- cycle, rgb(0.7,0.7,0.7)); */ }
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| void drawBackFace(int w, int x, int y, int z){
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| draw(P[w]--P[x] -- P[y] -- P[z] -- cycle, linetype("2 6")); }
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| drawFrontFace(4,5,6,7);
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| drawFrontFace(0,1,5,4);
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| drawFrontFace(1,2,6,5);
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| drawBackFace(0,1,2,3);
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| drawBackFace(3,3,7,7);
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| draw(P2[3]--P2[4]--P2[5]--P2[6]--cycle,linewidth(0.7));
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| draw(P[5]--P2[3]--P2[6]--P2[1]--cycle,linewidth(0.7));
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| draw(P[5]--P2[0]--P2[4]--P2[3]--cycle,linewidth(0.7));
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| draw(P[5]--P2[0]--P2[2]--P2[1]--cycle,linetype("2 6"));
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| draw(P2[2]--P2[0]--P2[4]--P2[5]--cycle,linetype("2 6"));
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| draw(P3[0]--P2[1]--P3[2]--P3[1]--cycle,linetype("2 6"));
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| draw(P3[2]--P3[1]--P3[4]--P3[5]--cycle,linetype("2 6"));
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| draw(P3[0]--P2[1]--P3[6]--P3[3]--cycle,linewidth(0.7));
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| draw(P2[1]--P3[2]--P3[5]--P3[6]--cycle,linewidth(0.7));
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| draw(P3[3]--P3[6]--P3[5]--P3[4]--cycle,linewidth(0.7));</asy>
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| We reach the minimum surface area with the configuration above. The cubes from smallest to largest have edge lengths <math>1,2,3</math> respectively. For the cube with edge <math>3</math>, the five faces that are not in contact with another cube have total surface area of <math>9\cdot 5=45</math>. The top face of that cube has surface area <math>9-4-1=4</math>. The left face of the cube with edge length <math>2</math> has surface area <math>4-1=3</math>. The four remaining faces of that cube have total surface area <math>4\cdot 4=16</math>. The four faces of the smallest cube that are not in contact with another cube have total surface area <math>1\cdot 4=4</math>. Therefore, our total surface area is <math>45+4+3+16+4=\boxed{\textbf{(D) }72.}</math>
| | ==See Also== |
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| --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] | | {{AHSME box|year=1994|num-b=10|num-a=12}} |
| | {{MAA Notice}} |
Problem
Three cubes of volume 
, 
and 
are glued together at their faces. The smallest possible surface area of the resulting configuration is
See Also
These problems are copyrighted © by the Mathematical Association of America. 
, 
and 
are glued together at their faces. The smallest possible surface area of the resulting configuration is
