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Forum URL: http://www.cut-the-knot.org/cgi-bin/dcforum/forumctk.cgi
Forum Name: College math
Topic ID: 546
Message ID: 2
#2, RE: difficult sequence, areas partitioned in circle ,
Posted by mr_homm on Dec-07-05 at 04:47 PM
In response to message #0
I don't think that this question has a unique answer, at least as it was stated. For instance, if I take 4 points forming a perfect square concentric with the circle, there will be 9 regions (like a tic-tac-toe board), but if I tilt the two vertical sides until their point of intersection of comes inside the circle, I will get 10 regions, and if I do the same for the horizontal sides as will, 11 regions.

Therefore, the number of regions depends on the exact arrangement of the points. If you are looking for the maximum number of partitions, then moving the points very near the center (or equivalently, making the circle very large) so that all the intersection points are within the circle will give the largest number of regions. In that case, it is best to simply push the circle out to infinity and ask how many regions the whole plane is cut into by lines joining pairs of points.

Even if no three points are collinear, it is still possible to have coincidences where two lines are parallel, or more than two lines are concurrent at a common point. These cases will reduce the number of regions, and so if you want the maximum, these must also be excluded. In this case, you are assuming that the lines are in general position, so you don't need to say anything about the original set of n points, except that n points will determine n(n-1)/2 lines.

Therefore, this is a special case of the more general problem: how many regions is the plane cut into by m lines in general position? Setting m = n(n-1)/2 would then give the solution to your problem.

--Stuart Anderson