Difference between revisions of "2002 AMC 10B Problems/Problem 18"

m (Problem)
(Solution)
Line 6: Line 6:
  
 
== 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 03:24, 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?

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

Solution

We know that $2$ distinct circles can intersect at no more than $2$ points. Thus $4$ circles can intersect at $2 \times 4= \boxed{\textbf{(D)}\ 8}$ points total.