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]]}}
 
{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #14]] and [[2002 AMC 10B Problems|2002 AMC 10B #18]]}}
 
== Problem ==
 
== Problem ==
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Four distinct [[circle]]s are drawn in a [[plane]]. What is the maximum number of points where at least two of the circles intersect?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
== Solution 1== 6
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<math>\mathrm{(A)}\ 8
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\qquad\mathrm{(B)}\ 9
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\qquad\mathrm{(C)}\ 10
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\qquad\mathrm{(D)}\ 12
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\qquad\mathrm{(E)}\ 16</math>
  
==Solution 2== 6
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== Solution 1==
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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>.
  
==Solution 3== 6
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[[Image:2002_12B_AMC-14.png]]
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==Solution 2==
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Because a pair or circles can intersect at most <math>2</math> times, the first circle can intersect the second at <math>2</math> points, the third can intersect the first two at <math>4</math> points, and the fourth can intersect the first three at <math>6</math> points. This means that our answer is <math>2+4+6=\boxed{\mathrm{(D)}\ 12}.</math>
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==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>.
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== See also ==
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{{AMC10 box|year=2002|ab=B|num-b=17|num-a=19}}
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{{AMC12 box|year=2002|ab=B|num-b=13|num-a=15}}
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[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 14:04, 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

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 1

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

Solution 2

Because a pair or circles can intersect at most $2$ times, the first circle can intersect the second at $2$ points, the third can intersect the first two at $4$ points, and the fourth can intersect the first three at $6$ points. This means that our answer is $2+4+6=\boxed{\mathrm{(D)}\ 12}.$

Solution 3

Pick a circle any circle- $4$ ways. Then, pick any other circle- $3$ ways. For each of these circles, there will be $2$ intersections for a total of $4*3*2$ = $24$ intersections. However, we have counted each intersection twice, so we divide for overcounting. Therefore, we reach a total of $\frac{24}{2}=\boxed{12}$, which corresponds to $\text{(D)}$.

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

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