Difference between revisions of "2013 AMC 10A Problems/Problem 14"
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== Solution == | == 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 | + | 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=\boxed{\textbf{(D) }84}</math> |
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| − | 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=\boxed{\textbf{(D) }84}</math> | ||
{{AMC10 box|year=2013|ab=A|num-b=13|num-a=15}} | {{AMC10 box|year=2013|ab=A|num-b=13|num-a=15}} | ||
Revision as of 15:27, 8 February 2013
A solid cube of side length 
is removed from each corner of a solid cube of side length 
. How many edges does the remaining solid have?
Solution
We can use Euler's polyhedron formula that says that 
. We know that there are originally 
faces on the cube, and each corner cube creates more. 
. In addition, each cube creates 
new vertices while taking away the original 
, yielding 
vertices. Thus 
, so 
| 2013 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
