Difference between revisions of "2015 AMC 8 Problems/Problem 12"

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<math>\text{(A) }6 \quad\text{(B) }12 \quad\text{(C) } 18 \quad\text{(D) } 24 \quad \text{(E) } 36</math>
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<math>\textbf{(A) }6\qquad\textbf{(B) }12\qquad\textbf{(C) }18\qquad\textbf{(D) }24\qquad \textbf{(E) }36</math>
 
 
  
 
==Solutions==
 
==Solutions==

Latest revision as of 01:32, 2 March 2024

Problem

How many pairs of parallel edges, such as $\overline{AB}$ and $\overline{GH}$ or $\overline{EH}$ and $\overline{FG}$, does a cube have?

[asy] import three; currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1));  draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); label("$D$",(0,0,0),S); label("$A$",(0,0,1),N); label("$H$",(0,1,0),S); label("$E$",(0,1,1),N); label("$C$",(1,0,0),S); label("$B$",(1,0,1),N); label("$G$",(1,1,0),S); label("$F$",(1,1,1),N); [/asy]

$\textbf{(A) }6\qquad\textbf{(B) }12\qquad\textbf{(C) }18\qquad\textbf{(D) }24\qquad \textbf{(E) }36$

Solutions

Solution 1

We first count the number of pairs of parallel lines that are in the same direction as $\overline{AB}$. The pairs of parallel lines are $\overline{AB}\text{ and }\overline{EF}$, $\overline{CD}\text{ and }\overline{GH}$, $\overline{AB}\text{ and }\overline{CD}$, $\overline{EF}\text{ and }\overline{GH}$, $\overline{AB}\text{ and }\overline{GH}$, and $\overline{CD}\text{ and }\overline{EF}$. These are $6$ pairs total. We can do the same for the lines in the same direction as $\overline{AE}$ and $\overline{AD}$. This means there are $6\cdot 3=\boxed{\textbf{(C) } 18}$ total pairs of parallel lines.

Solution 2

Look at any edge, let's say $\overline{AB}$. There are three ways we can pair $\overline{AB}$ with another edge. $\overline{AB}\text{ and }\overline{EF}$, $\overline{AB}\text{ and }\overline{HG}$, and $\overline{AB}\text{ and }\overline{DC}$. There are 12 edges on a cube. 3 times 12 is 36. We have to divide by 2 because every pair is counted twice, so $\frac{36}{2}$ is $\boxed{\textbf{(C) } 18}$ total pairs of parallel lines.

Solution 3

We can use the feature of 3-Dimension in a cube to solve the problem systematically. For example, in the 3-D of the cube, $\overline{AB}$, $\overline{BC}$, and $\overline{BF}$ have $4$ different parallel edges respectively. So it gives us the total pairs of parallel lines are $\binom{4}{2}\cdot3 =\boxed{\textbf{(C) } 18}$.

--LarryFlora

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/iJC0Wqd1ZcU

~Education, the Study of Everything

Video Solution 2

https://youtu.be/7bgsUa62d4g

~savannahsolver

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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 AJHSME/AMC 8 Problems and Solutions

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