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Difference between revisions of "2013 AMC 10A Problems/Problem 14"
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<math> \textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad </math>
 
<math> \textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad </math>
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== Solution ==
  
 
We can use Euler's polyhedron formula that says that <math>F+V=E+2</math>.  We know that there are originally <math>6</math> faces on the cube, and each corner cube creates <math>3</math> more.  <math>6+8(3) = 30</math>.  In addition, each cube creates <math>7</math> new vertices while taking away the original <math>8</math>, yielding <math>8(7) = 56</math> vertices.  Thus <math>E+2=56+30</math>, so <math>E=84</math>, <math>\textbf{(D)}</math>
 
We can use Euler's polyhedron formula that says that <math>F+V=E+2</math>.  We know that there are originally <math>6</math> faces on the cube, and each corner cube creates <math>3</math> more.  <math>6+8(3) = 30</math>.  In addition, each cube creates <math>7</math> new vertices while taking away the original <math>8</math>, yielding <math>8(7) = 56</math> vertices.  Thus <math>E+2=56+30</math>, so <math>E=84</math>, <math>\textbf{(D)}</math>

Revision as of 17:27, 7 February 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 $E=84$, $\textbf{(D)}$

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