Difference between revisions of "2013 AMC 10A Problems/Problem 14"
Scrabbler94 (talk | contribs) (→Solution: alternate solution) |
|||
| Line 8: | Line 8: | ||
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> | 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> | ||
| + | |||
| + | == Solution 2 == | ||
| + | The removal of each cube adds nine additional edges to the solid. Since a cube initially has 12 edges and there are eight vertices, the number of edges will be <math>12 + 9 \times 8 = \boxed{\textbf{(D) } 84}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2013|ab=A|num-b=13|num-a=15}} | {{AMC10 box|year=2013|ab=A|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:12, 17 July 2018
Contents
Problem
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 
Solution 2
The removal of each cube adds nine additional edges to the solid. Since a cube initially has 12 edges and there are eight vertices, the number of edges will be 
.
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
