Showing posts with label Perspective Triangles. Show all posts
Showing posts with label Perspective Triangles. Show all posts

Sunday, July 16, 2023

1955. A Construction For the Perspector Of 2nd Lemoine Circle

Let K=X(6)-Symmedian point of ABC. Antiparallel from K to BC intersect the AB, AC at Ac, Ab resp.
Define Ba, Bc, Ca, Cb cyclically.
KA: Antipode of K respect to circle (KBaCa). Define KB, Kcyclically.

* ABC and KAKB,KC  are perspective triangles. Perspector is X(3527) = ISOGONAL CONJUGATE OF X(631)=perspector of 2nd Lemoine circle

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?

Wednesday, April 27, 2022

1832. Inverted Triangles and Circumcenters On Euler Line

 Let ABC be a triangle and A1B1C1 is perspective with ABC at a point P and homothetic to ABC.

A2 is inversion of A1 respect to circumcircle of ABC. Define B2, C2 cyclically.

* Circumcenter of A2B2C2 lies on the line OP.

Application. Let P be a point on Euler line of ABC. A1,B1,C2 midpoints of AP, BP, CP resp.

A2 is inversion of A1 respect to circumcircle of ABC. Define B2, C2 cyclically.

** O',Circumcenter of A2B2C2 lies on the Euler Line of ABC.

For P=X(2), O'=X(33532)

For P=X(5), O'=X(?)

For P=X(20), O'=X(12084)

1996. A Collinearity