Hubert Shutrick

It will be shown that the proof that Dao Thanh Oai's condition is valid can be reduced to a proof of a property of circles.

Consider points A,B,C,D,E',F and let E be the intersection of the line DE' with the conic through A,B,C,D,F . It is assumed that the conic is an ellipse and that the points are arranged as in the posting referred to above. Let G = AB\cap DE, H = BC\cap EF and I = CD \cap FA as in the diagram in the posting. Further, let P = AC\cap DFand Q = AD\cap CF . Since the line PQ does not intersect the conic or the segments involved, the projectivity that makes it the line at infinity preserves the value of the invariant and so does the one that converts the ellipse into a circle. The quadrilateral ADFC becomes a rectangle as the result of the two projectivities and I becomes the centre of the circle.

 Attach:DTOsinvariant.png Δ

The condition that the invariant is 1 can be written

AG \cdot BH \cdot CI \cdot DG \cdot EH \cdot FI = GB\cdot HC \cdot ID \cdot GE \cdot HF \cdot IA

Since ADFC is a rectangle, the terms including I cancel and, since the points are on a circle, we also have AG \cdot GB = DG \cdot EG and BH \cdot HC = EH \cdot HF , which can be used to eliminate AG and EH and the result simplifies to

DG \cdot BH = BG \cdot FH

Let F' be the point on the line BF such that \angle HFF' = \angle HF'F . The triangles GBD and HBF' are similar because \angle GBD = \angle HBF' since they subtend chords of the same length and \angle GDB = \angle HFF' = \angle HF'F since DBFE is cyclic. Hence,

DG \cdot BH = BG \cdot F'H = BG \cdot FH
as required.

This proves that the condition is necessary. Note however, if the condition holds for the points A,B,C,D,E,F , it can't hold for A,B,C,D,E',F because \angle BFH would change but not the others. Hence, Dao Thanh Oai's condition is necessary and sufficient.

Although this method of proof would work if A,B,C,D,F lie on a parabola or on the same branch of a hyperbola since it can be transformed into an ellipse by choosing a line that doesn't intersect it as new line at infinity, it will not work if they lie on both branches of a hyperbola. However, if coordinates are introduced, the proof above implies that the condition is algebraically equivalent to the condition that the coordinates of the six points satisfy a second order equation even if they lie on a pair of lines or on a hyperbola. The invariant will even work for Pappus' theorem.