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X(1), X(99), X(512), X(2142) = E(700), X(2143) = E(698), E(697), E(699)

excenters

Brocard points Ω1 and Ω2

details and other points below

K021 is the isogonal circular pK with pivot X(512) and singular focus X(98).

Locus properties :

  1. Locus of point M such that the cevian triangles of M and its isogonal conjugate M* have the same (algebraic) area. See Cubics associated with triangles of equal area, Clark Kimberling, Forum Geometricorum, vol.1, 2001. This cubic is also called EAC1 = equal areas (first) cevian cubic. When these areas are opposite, we obtain an isogonal nK with root X(39). See below.
  2. Denote by Oa, Ob, Oc the circumcenters of triangles PBC, PCA, PAB respectively. The circumcenter of OaObOc lies on the perpendicular at O to the Brocard line if and only if P lies on K021, together with C(O,R). Compare with K001, property 3.

 

K021a

K021 contains the intersections of the sidelines of the antimedial triangle GaGbGc and the tangential triangle KaKbKc.

These points are labelled Ab, Ac, Ba, Bc, Ca, Cb and lie on the cevian lines of the two Brocard points.

K021b

E(697) is the third point of the line X(1)X(99) on K021. It is the cevian quotient of the pivot X(512) and X(1).

E(699) is the isogonal conjugate of E(697).

These two points are weak points hence K021 contains their six extraversions.

The figure shows the extraversions A1, B1, C1 of E(697).

Note that ABC and A1B1C1 are perspective at E(699).

 

K021sister

The locus of of point M such that the cevian triangles of M and its isogonal conjugate M* have opposite (algebraic) areas is the non-pivotal isogonal cubic with root X(39) that passes through the six already mentioned points Ab, Ac, Ba, Bc, Ca, Cb.

This cubic and K021 generate a pencil that contains the two decomposed cubics which are the union of the sidelines of the anticevian triangles of G and K.

These two cubics are mentioned in a paper by C. E. Noble namely :

An Anallagmatic Cubic, AMM vol. 55, Jan. 1948, pp. 7-14.

The trilinear equations given there are

(by+cz)(cz+ax)(ax+by) ± (bx+ay)(cx+az)(cy+bz) = 0,

and the cubics are called "minus-sign cubic" for K021 and "plus-sign cubic" for the other.