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 "2015 AMC 8 Problems/Problem 12"

(→‎Solution)
m
Line 18: Line 18:
 
label("$F$",(1,1,1),N);
 
label("$F$",(1,1,1),N);
 
</asy>
 
</asy>
−
==Solution==
+
==Solution 1==
 
We first count the number of pairs of parallel lines that are in the same direction as <math>\overline{AB}</math>.  The pairs of parallel lines are <math>\overline{AB}\text{ and }\overline{EF}</math>, <math>\overline{CD}\text{ and }\overline{GH}</math>, <math>\overline{AB}\text{ and }\overline{CD}</math>, <math>\overline{EF}\text{ and }\overline{GH}</math>, <math>\overline{AB}\text{ and }\overline{GH}</math>, and <math>\overline{CD}\text{ and }\overline{EF}</math>.  These are <math>6</math> pairs total.  We can do the same for the lines in the same direction as <math>\overline{AE}</math> and <math>\overline{AD}</math>.  This means there are <math>6\cdot 3=\boxed{\textbf{(C) } 18}</math> total pairs of parallel lines.
 
We first count the number of pairs of parallel lines that are in the same direction as <math>\overline{AB}</math>.  The pairs of parallel lines are <math>\overline{AB}\text{ and }\overline{EF}</math>, <math>\overline{CD}\text{ and }\overline{GH}</math>, <math>\overline{AB}\text{ and }\overline{CD}</math>, <math>\overline{EF}\text{ and }\overline{GH}</math>, <math>\overline{AB}\text{ and }\overline{GH}</math>, and <math>\overline{CD}\text{ and }\overline{EF}</math>.  These are <math>6</math> pairs total.  We can do the same for the lines in the same direction as <math>\overline{AE}</math> and <math>\overline{AD}</math>.  This means there are <math>6\cdot 3=\boxed{\textbf{(C) } 18}</math> total pairs of parallel lines.
  
−
==Solution==
+
==Solution 2==
 
Look at any edge, let's say <math>\overline{AB}</math>. There are three ways we can pair <math>\overline{AB}</math> with another edge. <math>\overline{AB}\text{ and }\overline{EF}</math>, <math>\overline{AB}\text{ and }\overline{HG}</math>, and <math>\overline{AB}\text{ and }\overline{DC}</math>. 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 <math>\frac{36}{2}</math> is (C) 18 total pairs of parallel lines.
 
Look at any edge, let's say <math>\overline{AB}</math>. There are three ways we can pair <math>\overline{AB}</math> with another edge. <math>\overline{AB}\text{ and }\overline{EF}</math>, <math>\overline{AB}\text{ and }\overline{HG}</math>, and <math>\overline{AB}\text{ and }\overline{DC}</math>. 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 <math>\frac{36}{2}</math> is (C) 18 total pairs of parallel lines.
  

Revision as of 20:43, 12 November 2019

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


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

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 (C) 18 total pairs of parallel lines.

-NoisedHens

Please add the latex for the box. Delete this line when finished.

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
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

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=2015_AMC_8_Problems/Problem_12&oldid=111356"