HYACINTHOS 24768

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and IaIbIc the antipedal triangle of I (excentral triangle).

Denote:

A', B', C' = the reflections of Ia, Ib, Ic in BC, CA, AB, resp.

Ma, Mb, M = the midpoints of AA', BB', CC', resp.

1. ABC, A'B'C' are cyclologic.

ie
the circumcircles of AB'C', BC'A, CA'B' are concurrent at X.

the circumcircles of A'BC, B'CA', C'AB are concurrent at Y.

 

2. ABC, MaMbMc are cyclologic.

ie
the circumcircles of AMbMc, BMcMa, CMaMb are concurrent at Z.

the circumcircles of MaBC, MbCA, McAB are concurrent at T.

 

Y = Z ?

[Peter Moses]:

Hi Antreas,

 

1).

X = a^2 (a^4-2 a^2 b^2+b^4-a^3 c+a^2 b c+a b^2 c-b^3 c-a^2 c^2-b^2 c^2+a c^3+b c^3) (a^4-a^3 b-a^2 b^2+a b^3+a^2 b c+b^3 c-2 a^2 c^2+a b c^2-b^2 c^2-b c^3+c^4) (a^6-3 a^4 b^2+3 a^2 b^4-b^6+2 a^4 b c+2 a^3 b^2 c-2 a^2 b^3 c-2 a b^4 c-3 a^4 c^2+2 a^3 b c^2-2 a^2 b^2 c^2+2 a b^3 c^2+b^4 c^2-2 a^2 b c^3+2 a b^2 c^3+3 a^2 c^4-2 a b c^4+b^2 c^4-c^6):: {{1,915},{84,224},{155,3435},{ 913,5190},...}.

On K269.

 

Y = X(80).

 

2).

Z = X(80).

T = a (b+c) (a^4-2 a^2 b^2+b^4+a^2 b c+a b^2 c-a b c^2-c^4) (a^4-b^4+a^2 b c-a b^2 c-2 a^2 c^2+a b c^2+c^4):: {{3,191},{4,2778},{54,5494},{ 65,125},{67,3827},{69,2836},{ 72,7068},{73,2632},{74,2766},{ 110,960},{265,517},{518,895},. ..}.

midpoint of X(3448) & X(3869).

reflection of X(i) in X(j) for these {i,j}: {{65, 125}, {110, 960}}.

on Jerabek.

on K685, K720.

X(3724)-crossconjugate of X(37).

X(i)-isoconjugate of X(j) for these {i,j}: {{1,1325},{58,5080},{ 162,2850}}.

X(3)-crosssum of X(2948).

trilinear pole of line X(647) X(2092).

 

Best regards,

Peter Moses.


[APH]:

 

Thanks, Peter !

 X(80) is the antigonal conjugate of the incenter I.

It is true for any point P:

Let ABC be a triangle and PaPbPc the antipedal triangle of P.

Denote:

A', B', C' = the reflections of Pa,Pb, Pc in BC, CA, AB, resp.

ABC, A'B'C' are cyclologic.

The cyclologic center (A'B'C', ABC) is the antigonal conjugate of P.

APH