## Griffiths Points in Cyclic Quadrilateral

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

### References

- J. Tabov,
__Four Collinear Griffiths Points__,*Mathematics Magazine*, v. 68, n 1, February 1995, pp. 61-64

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Copyright © 1996-2018 Alexander Bogomolny

### Proof

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

Assume there are four concyclic points A_{1}, A_{2}, A_{3}, and A_{4}, with the circumcenter at the origin and the circumradius of 1. Taken by three, they form four triangles T_{m} = ΔA_{i}A_{j}A_{k}, where i, j, k, m are distinct elements from _{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

w_{m} = (a_{i} + a_{j} + a_{k} - a_{i} a_{j} a_{k}) / 2.

To establish their collinearity we have to check that

(w_{1} - w_{2}) / (w'_{1} - w'_{2}) = (w_{1} - w_{3}) / (w'_{1} - w'_{3}) = (w_{1} - w_{4}) / (w'_{1} - w'_{4}).

What if applet does not run? |

Define a = a_{1}a_{2}a_{3}a_{4} for convenience. Easily

w_{1} - w_{2} | = (a_{2} + a_{3} + a_{4} - a_{2} a_{3} a_{4}) / 2 - (a_{1} + a_{3} + a_{4} - a_{1} a_{3} a_{4}) / 2 |

= (a_{1} - a_{2})(a_{3} a_{4} - 1) / 2. |

Taking the conjugates gives

w'_{1} - w'_{2} | = (a'_{1} - a'_{2})(a'_{3} a'_{4} - 1) / 2 |

= (a_{1} - a_{2})(a_{3} a_{4} - 1) / 2a | |

= (w_{1} - w_{2}) / a. |

which implies

(w_{1} - w_{2}) / (w'_{1} - w'_{2}) = a.

Obviously, the result will be the same if we replace w_{2} with w_{3} or w_{4}. Thus indeed we have the condition of collinearity:

(w_{1} - w_{2}) / (w'_{1} - w'_{2}) = (w_{1} - w_{3}) / (w'_{1} - w'_{3}) = (w_{1} - w_{4}) / (w'_{1} - w'_{4}).

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Copyright © 1996-2018 Alexander Bogomolny

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