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1994 AHSME Problems/Problem 11

Revision as of 16:34, 9 January 2021 by Angelalz (talk | contribs) (→‎Solution)

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

See Also

1994 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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