9-point Circle as a Locus of Concurrency: What Is This About?
A Mathematical Droodle

The Simson lines of two diametrically opposite points on a circumcircle of a triangle intersect on the 9-point circle of that triangle. A generalization of this fact is known as Griffiths' theorem: Let the line joining two points P and Q pass through the circumcenter of ΔA₁A₂A₂. Then the pedal circles of points P and Q meet on the 9-point circle of ΔA₁A₂A₂, i.e. the three circles are concurrent. This is known as Griffiths' theorem. In fact the point of concurrency only depends on the line the two points lie on. In other words, the pedal circles of the points on a line through the circumcenter of a triangle all concur at the same point on its 9-point circle. The point is known as Griffiths' point.

(It is worthwhile to note that the 9-point circle is the pedal circle of the circumcenter as well as that of the orthocenter.)

Griffiths' theorem admits a proof based on complex numbers.

Copyright © 1996-2008 Alexander Bogomolny

Since C₁ is collinear A₂ and A₃

From |a₂| = 1, a'₂ = 1/a₂ and a'₃ = 1/a₃. So we have

from which

or

If the indices i, j, k are distinct and come from the set {1, 2, 3} then, more generally,

On the other hand, PC_k is perpendicular to A_iA_j, so that

which, using |a_i| = |a_j| = 1, we modify as follows:

or, with p = p',

Finally, we obtain

A comparison of (2) and (1) gives

For the concyclicity of C₁, C₂, C₃, W we have to check that

which we shall prove in another form

First of all, from (3) and (4),

From here, using |a_i| = 1,

where a = a₁a₂a₃. Also

and

Similar expressions are obtained for the differences in (5) that include c₃. The left hand side in (5) becomes

or, after reduction,

which is symmetric in a₂ and a₃ showing that the right hand side in (5) equals the same expression, thus proving (5) and, with it, Griffiths' theorem.

(Line l passing through the circumcenter intersects the circumcircle at two points. For each of these, the three projections C₁, C₂, C₃ are collinear and lie on the simsons of ΔA₁A₂A₃. It follows that the two simsons meet at the Griffiths point associated with l.)

Assume now there are four concyclic points A₁, A₂, A₃, and A₄. 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 the given line l there correspond Griffiths point W_m. Using the same technique as above we can show (after J. Tabov) that the four Griffiths points W₁, W₂, W₃, W₄ are collinear. It is sufficient to show that

This and an additional feature of the configuration is considered on a separate page.

References

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

Copyright © 1996-2008 Alexander Bogomolny