Given any 5 points, no three collinear and no four concyclic, show that just 4 of the 10 circles through 3 points contain just one of the other two points.
Solution
Use inversion.
Let the 4 points be O, A, B, C, D. Invert wrt O so that O goes to infinity and A, B, C, D go to P, Q, R, S. Then a circle through O and two of A - D which contains just one of the other two, inverts into a line through two of P - S which has the other two on opposite sides. A circle through three of A - D which contains just one of the fourth and O goes into a circle through three of P - S which contains the fourth (because it certainly does not contain O).
If the convex hull of P, Q, R, S is a triangle, then without loss of generality S lies inside the triangle PQR. In that case it is easy to see that we have just four configurations described above: the lines SP, SQ, SR and the circle PQR. If the convex hull is a quadrilateral, then we also have four configurations: the two diagonals of the quadrilateral, one of the circles PQR and PQS (PQR iff ∠PRQ < ∠PSQ), and one of the circles RSP and RSQ.
© John Scholes
jscholes@kalva.demon.co.uk
10 Oct 2002