Difference between revisions of "2002 AMC 12B Problems/Problem 14"
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==Solution 2== 6 | ==Solution 2== 6 | ||
| − | ==Solution 3== | + | ==Solution 3== |
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| + | Pick a circle any circle- <math>4</math> ways. Then, pick any other circle- <math>3</math> ways. For each of these circles, there will be <math>2</math> intersections for a total of <math>4*3*2</math> = <math>24</math> intersections. However, we have counted each intersection twice, so we divide for overcounting. Therefore, we reach a total of <math>\frac{24}{2}=\boxed{12}</math>, which corresponds to <math>\text{(D)}</math>. | ||
== See also == | == See also == | ||
Revision as of 13:03, 12 July 2021
- The following problem is from both the 2002 AMC 12B #14 and 2002 AMC 10B #18, so both problems redirect to this page.
Problem
== Solution 1== 6
==Solution 2== 6
Solution 3
Pick a circle any circle- 
ways. Then, pick any other circle- 
ways. For each of these circles, there will be 
intersections for a total of 
= 
intersections. However, we have counted each intersection twice, so we divide for overcounting. Therefore, we reach a total of 
, which corresponds to 
.
See also
| 2002 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 AMC 10 Problems and Solutions | ||
| 2002 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 13 |
Followed by Problem 15 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
