USAMO 2005 Problem #5

by KingSmasher3, Mar 29, 2013, 10:09 PM

Let $n$ be an integer greater than 1. Suppose $2n$ points are given in the plane, no three of which are collinear. Suppose $n$ of the given $2n$ points are colored blue and the other $n$ 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 $2n$ 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 $m\le n$ points and let one of them be $A.$ Draw the rays from $A$ to the $n-1$ other blue points. This creates $n-2$ potential regions for the red points. Consider the right-most region. If there is more than $1$ red point in this region, then we have a balancing line from $A$ to of the red points. Similarly, if there are more than $2$ red points in the two right-most regions, we again have a balancing line from $A$. We can extend this result to the $x$ right-most regions for all $x.$ When $x=n-2,$ we have $n>n-2$ red points being placed in these regions, so there must exist a balancing line through $A.$ Since $m\ge 3,$ we have at least $3$ choices for $A,$ so there are at least $3$ balancing lines for this case.
This post has been edited 2 times. Last edited by KingSmasher3, Mar 29, 2013, 11:15 PM

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