     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 semi-circle 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.  