Difference between revisions of "2015 AMC 8 Problems/Problem 12"
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==Solution== | ==Solution== | ||
| − | 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>. | + | 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 |
==See Also== | ==See Also== | ||
Revision as of 20:00, 12 November 2019
How many pairs of parallel edges, such as 
and 
or 
and 
, 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]](http://latex.artofproblemsolving.com/f/1/b/f1b4f8e43db0ad2a38957a7e452bb8c8688d5906.png)
Solution
We first count the number of pairs of parallel lines that are in the same direction as . The pairs of parallel lines are 
, 
, 
, 
, 
, and 
. These are 
pairs total. We can do the same for the lines in the same direction as 
and 
. This means there are 
total pairs of parallel lines.
Solution
Look at any edge, let's say . There are three ways we can pair with another edge. , 
, and 
. 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 
is
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. 
