Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AMC 10A Problems/Problem 14"

(→‎Solution)
m
 
(5 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 +
==Problem==
 
A solid cube of side length <math>1</math> is removed from each corner of a solid cube of side length <math>3</math>.  How many edges does the remaining solid have?
 
A solid cube of side length <math>1</math> is removed from each corner of a solid cube of side length <math>3</math>.  How many edges does the remaining solid have?
  
  
 
<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>
 +
[[Category: Introductory Geometry Problems]]
 +
== Solution 1 ==
  
−
== 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=\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 <math>12</math> edges and there are eight vertices, the number of edges will be <math>12 + 9 \times 8 = \boxed{\textbf{(D) } 84}</math>.
  
 +
==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}}

Latest revision as of 20:46, 28 December 2019

Problem

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 1

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

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 $12 + 9 \times 8 = \boxed{\textbf{(D) } 84}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_14&oldid=113627"