2013 AMC 10A Problems/Problem 14

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 $E=84$ , $\textbf{(D)}$

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2013 AMC 10A Problems/Problem 14

Solution