Thursday, February 1, 2024

1969.Conics Intersecting Cubic Curves At a Fixed Point

 Let P be a triangle center on K001-Neuberg Cubic of ABC.

Any line through P intersect the  K001-Neuberg Cubic at P1 and P2.

*As P1, P2 varies on K001-Neuberg Cubic, the conic {A,B,C,P1,P2} intersect the K001-Neuberg Cubic at a fixed point.

**If P=X(1)-Incenter of ABC, fixed point is X(7164).
    If P=X(3)-Circumcenter of ABC, fixed point is X(1138).
    If P=X(4)-Orthocenter of ABC, fixed point is X(8431).
    If P=X(13)-1st Fermat-Toricelli point of ABC, fixed point is X(8445).
    If P=X(399) of ABC, fixed point is X(4).  

Thursday, January 25, 2024

1965. A Type Of Circlecevian and Concyclic Points

1964. A Construction For X(18569)

 Let A1B1C1 reflection triangle of X(3)-circumcenter on the side lines of ABC.

Tangents to circumcircle of A1B1C1 at A1, B1, C1 form a triangle A2B2C2.*

*Circumcenter of A2B2C2 lies on Euler line of ABC. It's X(18569).

1963. A Triangle Inscribed In Nine-Point Circle

 Let A1B1Cmedian triangle of ABC.

A2: Inverse of X(3)-Circumcenter of ABC in the circle with diameter B1C1
Define B2, C2 cyclically.

*A2B2C2 is inscribed in the Nine-Point Circle of ABC.

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

1978. Symmedian Points On Euler Line