TITLE: Given $2015$ points, show that it is possible to separate them such that $1007$ of them lie inside the circle
QUESTION [14 upvotes]: Suppose there are $2015$ points on a plane, no three collinear. Show that you can draw a circle such that exactly $1007$ lie inside the circle.

Here is my solution:

First draw a line that divides the plane into two parts, one containing $1007$ points and the other containing $1008$ points. Now draw a circle tangent to this line, such that the centre of the circle is on the side containing $1007$ points. Note that $1007$ points cover only a finite amount of area. But we can make this circle as big as we like, and thus make sure that the points are inside the circle.

I have two questions:
1) Where are we using the "... no three collinear" part of the statement? I think that since a point has no size, we can always divide the plane into two parts, one having $1007$ points and the other having $1008$ points.
2)How rigorous is my solution?

REPLY [7 votes]: (I originally left this as a comment, since it doesn't really answer the questions 1) and 2) that rah4927 asked. But due to popular demand I'm posting it as an answer.)
The assumption of "no three collinear" isn't needed. For each pair of points draw their perpendicular bisector. This produces a finite set of lines. Pick some new point not on any of these lines (nor any of the original points), and consider a growing circle centred on this point. The points will pass into it one at a time, so at some time it will contain 1007 of them.
(Note that we do use the unstated assumption that the original 2015 points are all distinct. Otherwise they could all just be the same point and we wouldn't be able to separate any of them.)