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X(3), X(15329)

K658 is a member of the class CL046. It is an acnodal conico-pivotal cubic with node O.

Consider a circum-conic (Hm) passing through O and meeting the circumcircle (O) again at m. Let (Tm) and (Nm) be the tangent and the normal at O to (Hm).

If M is the second intersection of (Nm) and (Hm) then the locus of M when m traverses (O) is K658.

The reflection of the line (mM) about O meets (Tm) at N which is the isoconjugate of M under the isoconjugation with fixed points O and the vertices of the anticevian triangle of O. It follows that the line (MN) envelopes the pivotal conic (C) of K658.



(C) is the ellipse with focus O, directrix the line (F) passing through the inflexion points of K658, eccentricity (OH/R)^2, tangent at X(110) to (O). The center of (C) is the intersection of the lines X(2)X(49), X(3)X(110), X(5)X(578), X(6)X(1493), etc.

(F) is the polar line of X(156) in (O) where X(156) is the nine point center of the tangential triangle.

The envelope of the line (mM) is another ellipse (E) with center O, bitangent at X(1113), X(1114) to (O). Its eccentricity is sin(2t) with tan(t) = OH/R. This ellipse is tritangent to K658.

The projection of O onto the line (MN) lies on the principal circle of (C) which contains the contacts of (C) and K658. These latter points lie on (E).

Let R1 = X(2)X(216) /\ X(6)X(110) = X(15355) and R2 = X(2)X(6) /\ X(4)X(32) = X(7735). K658 meets :

– the line at infinity at the same points as nK0(X6, R1),

– the circumcircle at the same points as nK0(X6, R2).


The inverse in the circumcircle of K658 is a trifolium (in green on the figure) with a triple point at O.

This trifolium is the pedal curve (with respect to O) of the Steiner deltoid of the tangential triangle KaKbKc.

Its center is X(156), the nine point center of KaKbKc and the nine point circle is the circle with center X(156) passing through X(110). This circle is obviously inscribed in the deltoid.

The three cuspidal tangents are parallel to the asymptotes of the McCay cubic K003.

K658 is also related with the Turner quartic Q102.