Difference between revisions of "2002 AMC 12B Problems/Problem 14"

Revision as of 20:10, 3 June 2021

The following problem is from both the 2002 AMC 12B #14 and 2002 AMC 10B #18, so both problems redirect to this page.

== Solution 1== 6

==Solution 2== 6

==Solution 3== 6

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

@@ Line 6: / Line 6: @@
 ==Solution 2== 6
-==Solution 3==
+==Solution 3== 6
-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 ==