IMO 1999 SL G2

Origin: JAP | Category: Geometry

Problem

A circle is called a separator for a set of five points in a plane if it passes through three of these points, it contains a fourth point in its interior, and the fifth point is outside the circle. Prove that every set of five points such that no three are collinear and no four are concyclic has exactly four separators.

Solution

Let A, B, C, and D be inverses of four of the five points, with the fifth point being the pole of the inversion. A separator through the pole trans- forms into a line containing two of the remaining four points such that the remaining two points are on opposite sides of the line. A separator not containing the pole transforms into a circle through three of the points with the fourth point in its interior. Let K be the convex hull of A, B, C, and D. We observe two cases: (i) K is a quadrilateral, for example ABCD. In that case the four sep- arators are the two diagonals and two circles ABC and ADC if \angleA + \angleC < 180◦, or BAD and BCD otherwise. The remaining six viable circles and lines are clearly not separators. (ii) K is a triangle, for example ABC with D in its interior. In that case the separators are lines DA, DB, DC and the circle ABC. No other lines and circles qualify. We have thus shown that any set of five points satisfying the stated con- ditions will have exactly four separators.