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both cubics : X(2), X(4) K070a (outer cubic) = pK(X4, X1586) : X(2), X(4), X(486), X(492), X(1586), X(3069), X(13390) K070b (inner cubic) = pK(X4, X1585) : X(2), X(4), X(485), X(491), X(1585), X(1659), X(3068) 

Let P be a point and PaPbPc be its cevian triangle. Erect outwardly or inwardly an arbelos (shoemaker's knife) on each side of triangle ABC and inscribe a circle centered at Oa having Xa as contact with the semicircle with diameter BC. See figure 1 and also Peter Woo, Simple Constructions of the Incircle of an Arbelos, Forum Geometricorum, vol.1 (2001), pp. 133 136. Define similarly Ob and Oc, Xb and Xc. The locus of point P such that ABC and XaXbXc are in perspective is a shoemaker's cubic. These two cubics are pivotal isocubics with pole H = X(4) and pivots the points X(1586) (outer cubic), X(1585) (inner cubic). They belong to a pencil that also contains K181, K663, K677.
K070a and K070b are the Kiepert Cevian Mates of the Orthocubic K006. See Table 32. They are anharmonically equivalent to K006. 

Remark : the corresponding loci of point P such that ABC and OaObOc are in perspective are two other cubics namely pK(X1131, X1131 ÷ X1132) and pK(X1132, X1132 ÷ X1131) where ÷ denotes a barycentric quotient.

