Griffiths Points in Cyclic Quadrilateral

Given a cyclic quadrilateral ABCD and a straight line l through its circumcenter, Griffiths' theorem supplies four Griffiths points for line l and four triangles ABC, BCD, CDA, DAB. As J. Tabov observed, the four points are collinear. In addition, the applet shows that the four pedal circles of any point on l with respect to the four triangles are concurrent. This is true for any complete quadrangle ABCD. For the cyclic quadrilateral, the point of concurrency lies on the line determined by the Griffiths points.

(The applet clear shows that the 9-point circles of triangles ABC, BCD, CDA, DAB also concur. This result is treated elsewhere.)

Proof

References

1. J. Tabov, Four Collinear Griffiths Points, Mathematics Magazine, v. 68, n 1, February 1995, pp. 61-64

Copyright © 1996-2008 Alexander Bogomolny

I shall follow the convention used in establishing the existence of Griffiths points.

Assume there are four concyclic points A₁, A₂, A₃, and A₄, with the circumcenter at the origin and the circumradius of 1. Taken by three, they form four triangles T_m = ΔA_iA_jA_k, where I, j, k, m are distinct elements from {1, 2, 3, 4}. To each T_m and a given line l there correspond Griffiths point W_m, m = 1, 2, 3, 4, which we found to be represented by complex numbers as

To establish their collinearity we have to check that

Define a = a₁a₂a₃a₄ for convenience. Easily

Taking the conjugates gives

which implies

Obviously, the result will be the same if we replace w₂ with w₃ or w₄.

... to be continued ...

Copyright © 1996-2008 Alexander Bogomolny