A CYCLOLOGIC THEOREM RELATED TO EXCENTRAL TRIANGLE.

[APH]

Excentral version

Let ABC be a triangle, IaIbIc the excentral triangle and P a point..

Denote

Pa, Pb, Pc = same to P points of IaBC, IbCA, IcAB, resp.

ABC, PaPbPc are circumcyclologic

Cyclologic center (ABC, PaPbPc) = Q = ? (on the circumcircle of ABC)
Cyclologic center (PaPbPc, ABC) = Q* = ? (on the circumcircle of PaPbPc)

[Ercole Suppa]

1. P on the Euler Line:

Q = X(100)
Locus of Q* as P moves on the Euler line: K086 2. P on the Brocard axis:

Q = X(101)

[Bernard Gibert]

If P is on a line through X(3) and a strong point M then the locus seems to be a circular cK(#X1,R) with singular focus F.

When M = X2, you get K086.

When M = X6, you get K040.

[APH]

For the case of the cyclologic center (ABC, PaPbPc) M can be any point strong or not.
That is:
Let ABC be a triangle, IaIbIc the excentral triangle, M a fixed point and P a point on the line OM.

Denote:

Pa, Pb,Pc = same to P points of IaBC,IbCA,IcAB, resp.

The triangles ABC, PaPbPc are circumcyclologic.

As P moves on the line OM:
The cyclologic center (ABC, PaPbPc) is a fixed point Q on the circumcircle.
The locus of the cyclologic center (PaPbPc, ABC) is a cubic.

Let's see the Q's. The class of the cubics is a subject of Bernard Gibert.
1. P on the Euler line
Q = X(100) = Reflection point of IO = X(1)X(3) line = Reflection point of Euler line of INTOUCH triangle (pedal triangle of I).

2. P on the Brocard axis
Q = X(101) = Reflection point of X(1)X(7) line = Reflection point of Brocard axis of INTOUCH triangle (pedal triangle of I).

Generalization

Let M be a fixed Point and P be a point on the line OM = L
Denote: P' = the same to P point of the INTOUCH triangle.
L' = the same to L line of the INTOUCH triangle.

Then Q is the reflection point of the L' line = IP' line of ABC

Note:
Reflection point of a line L:
The reflections La, Lb, Lc of L in the siedelines BC, CA, AB, resp. bound a triangle A*B*C*.
ABC, A*B*C* are perspective.The perspector, lying on the circumcircle, is called "Reflection pont of the line L"
It is the incenter (or an excenter) of the triangle A*B*C*.

Case of OM with M = I= X(1)

1. P = X(1) = O of INTOUCH triangle.
Pa, Pb, Pc = X(1) of IaBC, IbCA, IcAB, resp.
Cyclologic center (ABC, PaPbPc) = Q

2. P = X(354) = X(2) of the INTOUCH triangle
Pa, Pb, Pc = X(2) of the intouch triangles of IaBC, IbCA, IcAB, resp.
Cyclologic center (ABC, PaPbPc) = Q1

3. P = X(942) = X(5) of the INTOUCH triangle
Pa, Pb, Pc = X(5) of the intouch triangles of IaBC, IbCA, IcAB, resp.
Cyclologic center (ABC, PaPbPc) = Q2

Q, Q1, Q2 coincide.

Q is the reflection point of the IO line of the INTOUCH triangle.
It is the line passing thrpugh the incenters of ABC and the intouch triangle.