View source for 2008 iTest Problems/Problem 50

==Problem==

As the Kubiks head out of town for vacation, Jerry takes the first driving shift while Hannah and most of the kids settle down to read books they brought along. Tony does not feel like reading, so Alexis gives him one of her math notebooks and Tony gets to work solving some of the problems, and struggling over others. After a while, Tony comes to [[2002 AMC 12B Problems/Problem 14|a problem he likes from an old AMC 10 exam]]:

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

Tony realizes that he can draw the four circles such that each pair of circles intersects in two points. After careful doodling, Tony finds the correct answer, and is proud that he can solve a problem from late on an AMC 10 exam.

"Mom, why didn't we all get Tony's brain?" Wendy inquires before turning he head back into her favorite Harry Potter volume (the fifth year).

Joshua leans over to Tony's seat to see his brother's work. Joshua knows that Tony has not yet discovered all the underlying principles behind the problem, so Joshua challenges, "What if there are a dozen circles?"

Tony gets to work on Joshua's problem of finding the maximum number of points of intersections where at least two of the twelve circles in a plane intersect. What is the answer to this problem? 

==Solution==

Each circle intersects another circle at maximum <math>2</math> points.  That means the second circle creates <math>2</math> more points, the third circle creates <math>4</math> more points, and so on.  That means the twelfth circle creates <math>22</math> more points, so the maximum number of points of intersection where at least two of the circles intersect for twelve distinct circles is <math>2+4+6+ \cdots 22 = \tfrac{24 \cdot 11}{2} = \boxed{132}</math>.

==See Also==
{{2008 iTest box|num-b=49|num-a=51}}

[[Category:Introductory Combinatorics Problems]]