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Difference between revisions of "1994 AHSME Problems/Problem 11"

(Created page with "==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 <...")
 
(→‎Solution)
Line 4: Line 4:
 
<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==
 
==Solution==
 +
<asy>
 +
import three; currentprojection = orthographic(5,-30,20);
 +
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)};
 +
void drawFrontFace(int w, int x, int y, int z){
 +
draw(P[w]--P[x] -- P[y] -- P[z] -- cycle, linewidth(0.7));
 +
/* fill(P[x] -- P[y] -- P[z] -- cycle, rgb(0.7,0.7,0.7)); */ }
 +
void drawBackFace(int w, int x, int y, int z){
 +
draw(P[w]--P[x] -- P[y] -- P[z] -- cycle, linetype("2 6")); }
 +
drawFrontFace(4,5,6,7);
 +
drawFrontFace(0,1,5,4);
 +
drawFrontFace(1,2,6,5);
 +
drawBackFace(0,1,2,3);
 +
drawBackFace(3,3,7,7);
 +
draw(P2[3]--P2[4]--P2[5]--P2[6]--cycle,linewidth(0.7));
 +
draw(P[5]--P2[3]--P2[6]--P2[1]--cycle,linewidth(0.7));
 +
draw(P[5]--P2[0]--P2[4]--P2[3]--cycle,linewidth(0.7));
 +
draw(P[5]--P2[0]--P2[2]--P2[1]--cycle,linetype("2 6"));
 +
draw(P2[2]--P2[0]--P2[4]--P2[5]--cycle,linetype("2 6"));
 +
draw(P3[0]--P2[1]--P3[2]--P3[1]--cycle,linetype("2 6"));
 +
draw(P3[2]--P3[1]--P3[4]--P3[5]--cycle,linetype("2 6"));
 +
draw(P3[0]--P2[1]--P3[6]--P3[3]--cycle,linewidth(0.7));
 +
draw(P2[1]--P3[2]--P3[5]--P3[6]--cycle,linewidth(0.7));
 +
draw(P3[3]--P3[6]--P3[5]--P3[4]--cycle,linewidth(0.7));</asy>
 +
 +
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>
 +
 +
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]

Revision as of 22:06, 28 June 2014

Problem

Three cubes of volume $1, 8$ and $27$ are glued together at their faces. The smallest possible surface area of the resulting configuration is

$\textbf{(A)}\ 36 \qquad\textbf{(B)}\ 56 \qquad\textbf{(C)}\ 70 \qquad\textbf{(D)}\ 72 \qquad\textbf{(E)}\ 74$

Solution

[asy] import three; currentprojection = orthographic(5,-30,20); 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)}; void drawFrontFace(int w, int x, int y, int z){ draw(P[w]--P[x] -- P[y] -- P[z] -- cycle, linewidth(0.7)); /* fill(P[x] -- P[y] -- P[z] -- cycle, rgb(0.7,0.7,0.7)); */ } void drawBackFace(int w, int x, int y, int z){ draw(P[w]--P[x] -- P[y] -- P[z] -- cycle, linetype("2 6")); } drawFrontFace(4,5,6,7); drawFrontFace(0,1,5,4); drawFrontFace(1,2,6,5); drawBackFace(0,1,2,3); drawBackFace(3,3,7,7); draw(P2[3]--P2[4]--P2[5]--P2[6]--cycle,linewidth(0.7)); draw(P[5]--P2[3]--P2[6]--P2[1]--cycle,linewidth(0.7)); draw(P[5]--P2[0]--P2[4]--P2[3]--cycle,linewidth(0.7)); draw(P[5]--P2[0]--P2[2]--P2[1]--cycle,linetype("2 6")); draw(P2[2]--P2[0]--P2[4]--P2[5]--cycle,linetype("2 6")); draw(P3[0]--P2[1]--P3[2]--P3[1]--cycle,linetype("2 6")); draw(P3[2]--P3[1]--P3[4]--P3[5]--cycle,linetype("2 6")); draw(P3[0]--P2[1]--P3[6]--P3[3]--cycle,linewidth(0.7)); draw(P2[1]--P3[2]--P3[5]--P3[6]--cycle,linewidth(0.7)); draw(P3[3]--P3[6]--P3[5]--P3[4]--cycle,linewidth(0.7));[/asy]

We reach the minimum surface area with the configuration above. The cubes from smallest to largest have edge lengths $1,2,3$ respectively. For the cube with edge $3$, the five faces that are not in contact with another cube have total surface area of $9\cdot 5=45$. The top face of that cube has surface area $9-4-1=4$. The left face of the cube with edge length $2$ has surface area $4-1=3$. The four remaining faces of that cube have total surface area $4\cdot 4=16$. The four faces of the smallest cube that are not in contact with another cube have total surface area $1\cdot 4=4$. Therefore, our total surface area is $45+4+3+16+4=\boxed{\textbf{(D) }72.}$

--Solution by TheMaskedMagician

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