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Revision as of 11:07, 4 July 2013

A solid cube of side length $1$ is removed from each corner of a solid cube of side length $3$. How many edges does the remaining solid have?


$\textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad$

Solution

We can use Euler's polyhedron formula that says that $F+V=E+2$. We know that there are originally $6$ faces on the cube, and each corner cube creates $3$ more. $6+8(3) = 30$. In addition, each cube creates $7$ new vertices while taking away the original $8$, yielding $8(7) = 56$ vertices. Thus $E+2=56+30$, so $E=\boxed{\textbf{(D) }84}$

2013 AMC 10A (Problems • Answer Key • Resources)
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