Quadrilaterals related to the Incentral Triangle, ABC, Excentral Triangle

Quadrilaterals related to ABC

In the three-ness of triangles there exists the four-ness of projective geometry. Each point P(l:m:n) in ABC creates a quadrilateral formed from the 4 lines ±x/l ± y/m ± z/n = 0, three of which are the edges of the Cevian tiangle of P in ABC. But these four lines also form the edges of the ex-Cevian triangles as well, so that these four triangles are bounded by the four sides of a quadrilateral, each triangle resulting from these lines taken 3 at a time.

There are two other ways to form a complete quadrilateral. The first is to choose an arbitrary 4th line to join the edges of ABC. One then considers this line as the tripolar of a point related to ABC.

Next, an arbitrary line along with ABC.

The cevian and excevian triangles of ABC are bounded by four lines. This means that they form a complete quadrilateral for which ABC is the diagonal triangle.

1. For the incenter in ABC the sides of the incentral triangle and the antiorthic axis (the tripolar of Io) form this quadrilateral.

2. But for ABC in the excentral triangle the quadrilateral is formed from the edges of ABC and the orthic axis of the excentral triangle, which is the same as the antiorthic axis of ABC.

3. For symmetry we can add a complete quadrilateral whose sides are those of the extangent triangle with its orthic axis (this line shared by all three quadrilaterals). The exorthic triangle of the excentral tirangle seems to be the diagonal triangle for this.

All three of these quadrilaterals share one line and three points, which are the intersection of the incentral triangle edges, and ABC edges, and the excentral triangle edges with the common line.

These quadrilaterals all have what I call Steiner sturucture. The edges, taken 3 at a time, form triangles which have orthocenters, circumcenters, etc. See here for a description of Steiner structure.

Each quadrangle has a full complement of Steiner stuff, meaning that there is an orthic line containing all four orthocenters, a Miquel circle containing the 4 circumcenters, and the Miquel point, which is the point of concurrence of the 4 circumcircles. The is also a coaxal system of circles generated by the incenters. All a strikingly beautiful system.

Paul Yiu on his website has a paper on generalizations of the Apollonian circles which covers this and Jean-Pierre published a paper in Forum Geometricorum describing this (and giving the formulas for case 3).

This leads to lots of relationships and a very full picture.

In case 1 we get the 4 extraversions of the orthocenter of the incentral triangle to be colinear and the extraversions of the circumcircle to be concyclic, a very interesting result. The extraversions of its incenter form a coaxal system of circles, an even more interesting result. This moves the incentral triangle from being intractable, to having way more structrure than most.

This picture shows the orthocenter line (red) the circumcircles of the 4 triangles concuring at the Miquel point M, and the circle through the circumcenters and the Miquel point.

 

Case 2

In case 2, the same structures exist, with the orthocentral line containing the orthocenter of ABC and the Miquel circle the circumcircle of ABC.

This picture shows the orthocenter line (red) the circumcircles of the 4 triangles concuring at the Miquel point M, and the circle through the circumcenters and the Miquel point.

 

I am just beginning with case 3.

This picture shows the orthocenter line (red) the circumcircles of the 4 triangles concuring at the Miquel point M, and the circle through the circumcenters and the Miquel point.

 

And here are the coaxal incenters.

 

An interesting (and suspicious result) is that the three Miquel points are colinear and always on a single bisector of ABC. Being on a bisector is suspicious because an unsymmetric structure is breading symmetry, but it is hard the get the concurrance of 5 circles if you are doing something wrong.