Dear Antreas,

In Hyacinthos message #7858, you wrote (partly):

>> Let ABC be a triangle and AHa, BHb, CHc its altitudes.

>>

>> Special case #1:

>>

>> Let Ab, Ac be the reflections of A in Hc, Hb, resp.

>> Let Bc, Ba be the reflections of B in Ha, Hc, resp.

>> Let Ca, Cb be the reflections of C in Hb, Ha, resp.

>>

>> The Euler Lines of the Triangles AHbHc, BHcHa, CHaHb

^^^^^^^^^^^^^^^^^^^

I guess you mean AAbAc, BBcBa, CCaCb.

The Euler lines of these triangles concur since the

orthocenter H of triangle ABC is their common

circumcenter.

>> concur (at H of ABC).

>> Problem: Are their Nine Point Circles concurrent?

Yes. In fact, the nine-point circle of a triangle

ABC is the reflection of the circumcircle of triangle

AB'C' in the line B'C', where A', B', C' are the

midpoints of the sides BC, CA, AB. This yields that

the nine-point circle of triangle AAbAc is the

reflection of the circumcircle of triangle AHbHc in

the line HbHc. Similarly for the other two nine-point

circles. Since the circles AHbHc, BHcHa, CHaHb concur

at H, the three nine-point circles concur at the

antigonal conjugate of H with respect to triangle

HaHbHc.

NOTE. Antigonal conjugates? If P is a point in the

plane of a triangle ABC, then the reflections of the

circles PBC, PCA, PAB in the sidelines BC, CA, AB

have a common point. This point is called antigonal

conjugate of P with respect to triangle ABC. But I

see (Hyacinthos message #7826) that you know this,

maybe with another name.

The point where your nine-point circles concur is

not in the ETC.

>> Generalization:

>>

>> Let Pac, Pab be points on AB, AC resp. such that:

>>

>> APac / AHc = APab / AHb = t

>>

>> Similarly we define the points Pba, Pbc; Pcb, Pca

>>

>> The Euler Lines of the Triangles APabPac, BPbcPba,

>> CPcaPcb are concurrent

Yes, they are: the proof is simple. Moreover, if

J = X(125) is the intersection of the Euler lines of

triangles AHbHc, BHcHa, CHaHb, then the Euler lines

of triangles APabPac, BPbcPba, CPcaPcb meet at the

point W lying on the line HJ and dividing the segment

HJ in the ratio

HW 2-t

-- = ---.

WJ t-1

Remark that the line HJ passes through X(74).

Sincerely,

Darij Grinberg