# 1994 AHSME Problems/Problem 11

## 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 $[asy] import three; currentprojection = orthographic(5,-30,20); triple[] P = {(0,0,0),(1,0,0),(1,1,0),(0,1,0),(0,0,1),(1,0,1),(1,1,1),(0,1,1)},P2={(1,0.5,1),(0.5,0,1),(0.5,0.5,1),(1,0,1.5),(1,0.5,1.5),(0.5,0.5,1.5),(0.5,0,1.5)},P3={(0.25,0,1),(0.25,0.25,1),(0.5,0.25,1),(0.25,0,1.25),(0.25,0.25,1.25),(0.5,0.25,1.25),(0.5,0,1.25)}; void drawFrontFace(int w, int x, int y, int z){ draw(P[w]--P[x] -- P[y] -- P[z] -- cycle, linewidth(0.7)); /* fill(P[x] -- P[y] -- P[z] -- cycle, rgb(0.7,0.7,0.7)); */ } void drawBackFace(int w, int x, int y, int z){ draw(P[w]--P[x] -- P[y] -- P[z] -- cycle, linetype("2 6")); } drawFrontFace(4,5,6,7); drawFrontFace(0,1,5,4); drawFrontFace(1,2,6,5); drawBackFace(0,1,2,3); drawBackFace(3,3,7,7); draw(P2--P2--P2--P2--cycle,linewidth(0.7)); draw(P--P2--P2--P2--cycle,linewidth(0.7)); draw(P--P2--P2--P2--cycle,linewidth(0.7)); draw(P--P2--P2--P2--cycle,linetype("2 6")); draw(P2--P2--P2--P2--cycle,linetype("2 6")); draw(P3--P2--P3--P3--cycle,linetype("2 6")); draw(P3--P3--P3--P3--cycle,linetype("2 6")); draw(P3--P2--P3--P3--cycle,linewidth(0.7)); draw(P2--P3--P3--P3--cycle,linewidth(0.7)); draw(P3--P3--P3--P3--cycle,linewidth(0.7));[/asy]$

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 $45+4+3+16+4=\boxed{\textbf{(D) }72.}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 