| | |
Summary: The Number of Halving Circles
Federico Ardila
1. INTRODUCTION. We say that a set S of 2n + 1 points in the plane is in general
position if no three of the points are collinear and no four are concyclic. We call a
circle halving with respect to S if it has three points of S on its circumference, n - 1
points in its interior, and n - 1 in its exterior. The goal of this paper is to prove the
following surprising fact: any set of 2n + 1 points in general position in the plane has
exactly n2
halving circles.
Our starting point is the following problem, which appeared in the 1962 Chinese
Mathematical Olympiad [7].
Problem 1. Prove that any set of 2n + 1 points in general position in the plane has a
halving circle.
For the rest of sections 1 and 2, n is a fixed positive integer and S signifies an arbitrary
set of 2n + 1 points in general position in the plane.
There are several solutions to Problem 1. One possible approach is the following.
Let A and B be two consecutive vertices of the convex hull of S. We claim that some
circle going through A and B is halving. All circles through A and B have their centers
on the perpendicular bisector of the segment AB. Pick a point O on that lies on
the same side of AB as S and is sufficiently far away from AB that the circle with
|
