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

(Solution 3)
 
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===Solution 3===
 
===Solution 3===
We can use the feature of 3-Dimension in a cube to solve the problem systematically. In the 3-D of the cube, <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{BF}</math> have <math>4</math> different parallel edges respectively. So it gives us the total pairs of parallel lines are <math>\binom{4}{2}*3 =\boxed{\textbf{(C) } 18}</math>. -----LarryFlora
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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, <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{BF}</math> have <math>4</math> different parallel edges respectively. So it gives us the total pairs of parallel lines are <math>\binom{4}{2}*3 =\boxed{\textbf{(C) } 18}</math>. -----LarryFlora
+
 
 
===Video Solution===
 
===Video Solution===
 
https://youtu.be/Zhsb5lv6jCI?t=1306
 
https://youtu.be/Zhsb5lv6jCI?t=1306

Latest revision as of 13:54, 6 July 2021

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]

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


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. -NoisedHens

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 $\binom{4}{2}*3 =\boxed{\textbf{(C) } 18}$. -----LarryFlora

Video Solution

https://youtu.be/Zhsb5lv6jCI?t=1306

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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