HYACINTHOS 23138

Antreas Hatzipolakis

 

 

Let ABC be a triangle

Denote:

Ha = the reflection of H in BC

Hab, Hac = the reflections of Ha in HB,HC, resp.

La = the Euler line of HaHabHac

Na = the NPC center of HaHabHac.

Similarly...............

I seem to recall I have posted this:
The NPCs (Na),(Nb), (Nc) are concurrent at X(1986).

But I do not remember if I have posted the following too:

1. The reflections of La,Lb,Lc in BC, CA, AB, resp. concur
(on the Euler line of ABC)

 

2. ABC, NaNbNc are cyclologic triangles

(The circumcircles of ANbNc, BNcNa, CNaNb, ABC are concurrent

and the circumcircles of NaBC, NbCA, NcAB, NaNbNc are concurrent).

By the way, if we have two cyclologic triangles A1B1C1, A2B2C2

with the additional property that the cyclologic centers lie on the

circumcircles of the triangles (ie the circumcircles of A1B2C2, B1C2A2,
C1A2B2, A1B1C1 are concurret and the circumcircles of A2B1C1, B2C1A1, C2A1B1, A2B2C2 are concurrent), then a good idea is to name them with

Some special name. How about circumcyclologic?

APH

 

[César Lozada]:

 

> The NPCs (Na),(Nb), (Nc) are concurrent at X(1986).

Confirmed.

 

1. The reflections of La,Lb,Lc in BC, CA, AB, resp. concur (on the Euler line of ABC)

At X(186)

 

2. ABC, NaNbNc are cyclologic triangles

Cyclologic centers:

Za = /\ Circle(A,Nb,Nc)

   = (cos(2*B)+cos(2*C))/(2*cos(A)*(cos(2*(B-C))+1)+cos(B-C)+cos(3*A)) (trilinears)

   = On K464 and lines (49,52), (389,3481)

   = ( -0.311973017213013, -0.19454865762634, 3.919339560516315 )

 

Zn = /\ Circle(Na,B,C)

   = 1/(sin(3*(B-C))*cos(2*A)+sin(2*(B-C))*(-cos(A)+2*cos(B-C))-sin(B-C)*cos(4*A)) (trilinears)

   = On circumcircle and line (1141,6240)

   = ( 13.356759169085460, 12.06828856472384, -10.878962602479260 )

 

Regards

César Lozada