2002 AMC 10B Problems/Problem 18: Difference between revisions
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== Solution == | == Solution == | ||
We know that <math>2</math> distinct circles can intersect at no more than <math>2</math> points. Thus <math>4</math> circles can intersect at <math>2 \times 4= \boxed{\textbf{(D)}\ 8}</math> points total. | |||
Revision as of 02:24, 29 January 2011
Problem
Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?
Solution
We know that 
distinct circles can intersect at no more than points. Thus 
circles can intersect at 
points total.
