Saturday, May 21, 2022

1857. Isogonal Curves and Isogonal Perspectors

 1.) Let P be a point and A1B1C1 cevian triangle of P.

A2: inverse of A1 in the circle PBC. Define B2, C2 cyclically.
*ABC and A2B2C2 are perspective if P lies on a curve C passing through X(2)-centroid, X(4)-Orthocenter and X(13)-1st Fermat-Toricelli point.
For X(2)-centroid, X(4)-Orthocenter and X(13)-1st Fermat-Toricelli point perspectors of ABC and A2B2C2 are X(524), X(403) and X(36211).

2.) Let P be a point and A1B1C1 circumcevian triangle of P.
A2: inverse of A1 in the circle PBC. Define B2, C2 cyclically.
**ABC and A2B2C2 are perspective if P lies on a curve C*  passing through X(6)-symmedian point, X(3)-Circumcenter and X(15)-1st Isodynamic point.
For  X(6)-symmedian point, X(3)-Circumcenter and X(15)-1st Isodynamic point perspectors of ABC and A2B2C2 are X(111), X(5504) and X(36209).

*** Do curves C and C* defined in problem 1 and 2 are isogonal conjugates?
**** Interestingly perspectors in problem 1 and perspectors in problem 2 are isogonal conjugate points. 
Does this property valid for all perspectors?

Saturday, May 14, 2022

1853. A Construction For X(5497) = 5th HATZIPOLAKIS-MONTESDEOCA POINT

 Let I=X(1) incenter of ABC. Circle BIC ıntersects the AC and AB at Ab, Ac. Define Ba,Bc, Ca,Cb cyclically.

Inversion of the line BC in the circle (IBcCb) is a cirlcle GammaA. Define GammB, GammaC cyclically.

*  GammaA, GammB, GammaC are concurrent circles. Concurrency point is X(5497) =  5th HATZIPOLAKIS-MONTESDEOCA POINT.

1987. Circumcenter On Euler Line

 Let H=X(4)-Orthocenter of ABC. A 1 B 1 C 1 cevian triangle of H. B1 A , C1 A are reflections of B1, C1 on A. L A , line through B1 A , C1...