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

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{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #14]] and [[2002 AMC 10B Problems|2002 AMC 10B #18]]}}
 
== Problem ==
 
== Problem ==
 
Four distinct [[circle]]s are drawn in a [[plane]]. What is the maximum number of points where at least two of the circles intersect?
 
Four distinct [[circle]]s are drawn in a [[plane]]. What is the maximum number of points where at least two of the circles intersect?
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== Solution ==
 
== Solution ==
For any given pair of circles, they can intersect at most <math>2</math> times. Since there are <math>{4\choose 2} = 6</math> pairs of circles, the maximum number of possible intersections is <math>6 \cdot 2 = 12</math>. We can construct such a situation as below, so the answer is <math>\mathrm{(D)}</math>.
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For any given pair of circles, they can intersect at most <math>2</math> times. Since there are <math>{4\choose 2} = 6</math> pairs of circles, the maximum number of possible intersections is <math>6 \cdot 2 = 12</math>. We can construct such a situation as below, so the answer is <math>\boxed{\mathrm{(D)}\ 12}</math>.
  
 
[[Image:2002_12B_AMC-14.png]]
 
[[Image:2002_12B_AMC-14.png]]
  
 
== See also ==
 
== See also ==
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{{AMC10 box|year=2002|ab=B|num-b=17|num-a=19}}
 
{{AMC12 box|year=2002|ab=B|num-b=13|num-a=15}}
 
{{AMC12 box|year=2002|ab=B|num-b=13|num-a=15}}
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]

Revision as of 16:15, 29 July 2011

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

Problem

Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?

$\mathrm{(A)}\ 8 \qquad\mathrm{(B)}\ 9 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 16$

Solution

For any given pair of circles, they can intersect at most $2$ times. Since there are ${4\choose 2} = 6$ pairs of circles, the maximum number of possible intersections is $6 \cdot 2 = 12$. We can construct such a situation as below, so the answer is $\boxed{\mathrm{(D)}\ 12}$.

2002 12B AMC-14.png

See also

2002 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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