Difference between revisions of "2015 AMC 8 Problems/Problem 12"
m |
m |
||
Line 22: | Line 22: | ||
==Solution 2== | ==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 <math>\boxed{\textbf{(C) } 18}</math> total pairs of parallel lines. |
-NoisedHens | -NoisedHens |
Revision as of 20:44, 12 November 2019
How many pairs of parallel edges, such as and or and , does a cube have?
Solution 1
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 2
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 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.