USAMO 2005 Problem #5
by KingSmasher3, Mar 29, 2013, 10:09 PM
Let
be an integer greater than 1. Suppose
points are given in the plane, no three of which are collinear. Suppose
of the given
points are colored blue and the other
colored red. A line in the plane is called a balancing line if it passes through one blue and one red point and, for each side of the line, the number of blue points on that side is equal to the number of red points on the same side.
Prove that there exist at least two balancing lines.
_______
Consider the convex hull of the
points. If it consists of both red and blue points, then there exists two edges such that each edge connects a red and blue vertex. It is easy to see that the lines containing these edges are balancing lines since one side of the line has no points. Thus, we just have to consider the remaining case where the convex hull consists of only one color, say blue.
Let the convex hull contain
points and let one of them be
Draw the rays from
to the
other blue points. This creates
potential regions for the red points. Consider the right-most region. If there is more than
red point in this region, then we have a balancing line from
to of the red points. Similarly, if there are more than
red points in the two right-most regions, we again have a balancing line from
. We can extend this result to the
right-most regions for all
When
we have
red points being placed in these regions, so there must exist a balancing line through
Since
we have at least
choices for
so there are at least
balancing lines for this case.





Prove that there exist at least two balancing lines.
_______
Consider the convex hull of the

Let the convex hull contain


















This post has been edited 2 times. Last edited by KingSmasher3, Mar 29, 2013, 11:15 PM