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X(3), X(4), X(15328), X(15329)

vertices of the circum-tangential triangle

infinite points of K024

feet U, V, W of the trilinear polar of X(1993) : U lies on the perpendicular at O to AO.

Oa = OA /\ HU, Ob, Oc similarly

Ha = HA /\ OU, Hb, Hc similarly

K105 is an isogonal nK60 with root X(1993), the orthocorrespondent of O. See a generalization below and also Special Isocubics ยง7.2.2.

It is a member of the class CL004 of cubics. Its three real asymptotes are parallel to the sidelines of the Morley triangle and bound an equilateral triangle with center G, homothetic to the circum-tangential triangle under h(H,2/3). See a generalization in the page K687.

K105 is :

  1. the locus of point M such that the polar lines (in the circumcircle) of M and its isogonal conjugate M* are perpendicular. Compare with K003, property 1.
  2. the locus of M such that the circle with diameter MM* passes through O, and equivalently, such that the lines OM, OM* are perpendicular.
  3. the locus of the intersections of a line through O with the isogonal conjugate of its perpendicular through O.
  4. the locus of point M such that the radical axis of the nine point circle and the pedal circle of M passes through X(5).

 

Other geometric properties :

  1. the tangent at O is perpendicular to the Euler line.
  2. the common tangential S of O and H lies on the Jerabek hyperbola. S is now X(15328) in ETC (2017-11-22)
  3. its isogonal conjugate S* is the third point of K105 on the Euler line. It also lies on the tangent T at X(110) to the circumcircle. S* is now X(15329) in ETC (2017-11-22).
  4. the isogonal conjugate of T is a circum-parabola P through S and X(476), the Tixier point. P and T meet at R1 and R2, isogonal conjugates lying on the polar conic of S.
  5. a line through O meets K105 at M, N and the perpendicular at O meets K105 at M*, N*. The lines MN*, NM* pass through H and the lines MM*, NN* intersect at R on T.

 

Generalization :

Let P be a finite fixed point not lying on the sidelines of ABC. The lines PM and PM* are perpendicular if and only if M lies on K(P) = nK(X6, Q, P) where Q is the orthocorrespondent of P.

K105 is obviouly K(X3), the only K60 of this type, and K(X1) is K407, a nodal cubic.