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This page is at the same time a summary of the two articles mentioned below and a compilation of many complementary properties. How pivotal cubics intersect the circumcircle Asymptotic Directions of Pivotal Isocubics Let's start by recalling some definitions and defining the notations that will be used in the following. As usual, gX, tX, cX and aX are the isogonal, isotomic, complement and anticomplement of a point or a curve X. X x Y and X ÷ Y are barycentric product and quotient. X/Y is the XCeva conjugate of Y. psK cubics are defined and studied here. Let K = pK(Ω, P) be the pivotal cubic with pole Ω = p:q:r, pivot P = u:v:w and isopivot Q = p/u:q/v:r/w. If K' = pK(Ω', P') is another cubic with isopivot Q', we will denote by : • KLΩ, KLP, KLQ the loci of Ω', P', Q' such that K and K' meet the line at infinity (L∞) at the same points called Lpoints, • KOΩ, KOP, KOQ the loci of Ω', P', Q' such that K and K' meet the circumcircle (O) at the same points called Opoints, • KOQ2 the complement of KOQ and Z = cevapoint(Ω, X6 x P) we will meet in the sequel. Note that all cubics mentioned above are circumcubics. *** This first table gives the characterizations of these cubics and points in the first column that lie on them.




Notes : MaMbMc is the medial triangle. GaGbGc and KaKbKc are the anticevian triangles of X(2), X(6). Z1 = X(6) x aCeva(G, gP), Z2 = X(6) x a(X6 x gΩ). When Ω = P^2, K decomposes into the cevian lines of P. See circumcevians pKs in CL072. 



KLP and KOQ that are psKs In the table, five of the seven cubics are always psKs. • KLP becomes a psK if and only if either : – P lies on the line GΩ in which case KLP is in fact a pK having pole G. In this case, KLΩ = KLQ. – or P lies on the circumconic with center Ω and perspector G/Ω, equivalently Ω lies on the bicevian conic C(G, P). • KOQ becomes a psK if and only if either : – P lies on the line X6tgΩ in which case KOQ is in fact a pK having pole X(32). In this case, KOQ2 = gKOP. – or P lies on the circumconic with perspector (Ω/X32)÷X6, equivalently Ω lies on the circumconic with perspector (P/X6)xX6. • KLP and KOQ are then simultaneously psKs (but not necessarily pKs) when P is the trilinear pole of the line passing through G/Ω and (Ω/X32)÷X6 which is a rather complicated point. *** Cubics that are pKs Recall that a psK that passes through its pseudopivot is actually a pK. • It is clear that KLΩ and KLQ (and then also KLP) are pKs if and only if G, Ω, P are collinear. In particular, this will be the case when K is a pK(G,P) or a pK(Ω,G) or a pK(Ω,Ω). • KOΩ, KOP, KOQ2 (and then also KOQ) are pKs if and only if X(6), tgΩ, P are collinear, or equivalently, X(32), gtP = P x X(6), Ω are collinear. In particular, this will be the case when K is a pK(X32,P) or a pK(Ω,X6). • Combining the two conditions above, we see that all seven cubics will be pKs simultaneously if and only if P = GΩ /\ X(6)tgΩ = F(Ω), or equivalently, Ω = GP /\ X(32)gtP = F^{1}(P). With X = x:y:z, we have F(X) = b^4 c^2 x^2b^2 c^4 x^2a^4 c^2 x y+b^2 c^4 x y+a^4 b^2 x zb^4 c^2 x za^4 b^2 y z+a^4 c^2 y z : : . F^{1}(X) = a^2 (b^4 x^2c^4 x^2a^2 b^2 x y+c^4 x yb^4 x z+a^2 c^2 x z+a^2 b^2 y za^2 c^2 y z) : : . These are the quadratic transformations denoted BGF and invBGF in ETC, introduced in the preamble just before X(41231). See also the related cubics K1182 and K1183. *** Properties of F • singular points : X(2), X(6), X(32). Every point on lines {2,6}, {2,32}, {6,32} is mapped onto X(6), X(83), X(2) respectively. • fixed points : those on the circumconic having perspector X(669), which passes through X(6) and X32). • every other line through X(2) is mapped onto itself. In particular, the Euler line is mapped onto the Euler line. • every point on the circumconic having perspector X(512) – which passes through X(2) and X6) – is mapped onto a point on the Kiepert hyperbola. • every point on the circumconic having perspector X(688) – which passes through X(2) and X32) – is mapped onto a point on the circumconic having perspector X(512) above. • Note that the three circumconics above are the unique proper circumconics passing through two of the three singular points hence they are the unique circumconics which are transformed into another circumconic. 



Seven special groups of cubics The properties of F above give rise to the examination of four situations in which the seven cubics are all pKs. • Ω on the line X(2)X(6) hence P = F(Ω) = X(6). K = pK(Ω, P) is a cubic of the pencil K102, K368, K659, K739, K790, K1016, having basepoints A, B, C, X(2), X(6), X(194) and the cevians of X(6). • Ω on the line X(2)X(32) hence P = F(Ω) = X(83). K = pK(Ω, P) is a cubic of the pencil containing K644 and pK(X1799, X83), having basepoints A, B, C, X(2), X(6), X(83) and the cevians of X(83). • Ω on the line X(6)X(32) hence P = F(Ω) = X(2). K = pK(Ω, P) is a cubic of the pencil K002, K043 (see Table 35), having basepoints A, B, C, X(2), X(3), X(6) and the cevians of X(2) i.e. midpoints of ABC. • Ω on the circumconic having perspector X(669) hence P = Ω. K = pK(Ω, P) is a cubic of the pencil K102, K787, K1014, K1179, having basepoints A, B, C, X(2), X(6), X(3224) with tangents at A, B, C concurring at X(2). *** In a similar way, the properties of F^{1}give rise to the examination of three more situations in which the seven cubics are all pKs. Indeed, the singular points of F^{1} are X(2), X(6), X(83) and every point on lines {2,6}, {2,83}, {6,83} is mapped onto X(6), X(32), X(2) respectively. • P on the line X(2)X(6) hence Ω = F^{1}(P) = X(6). K = pK(Ω, P) is a cubic of the pencil K002, K102, (see Table 13) having basepoints A, B, C, X(1), X(2), X(6) and the excenters. • P on the line X(2)X(83) hence Ω = F^{1}(P) = X(32). K = pK(Ω, P) is a cubic of the pencil K177, pK(X32, X315) (see Table 34) having basepoints A, B, C, X(2), X(6), X(32) and the vertices of the tangential triangle. • P on the line X(6)X(83) hence Ω = F^{1}(P) = X(2). K = pK(Ω, P) is a cubic of the pencil K141, pK(X2, X880) having basepoints A, B, C, X(2), X(6), X(76) and the vertices of the antimedial triangle. Remark : in each of the seven situations above, the cubics KLΩ and KLQ coincide. 



A few remarkable examples The following table shows a selection of these groups of seven cubics associated with a given K = pK(Ω, P). 



Remarks • The cubics K in the light blue cells are associated with the same group of four cubics KO. This is true for every cubic K whose pole Ω lies on K177 and pivot P on K141 such that P = (G/Ω) ÷ X(6). Other examples are pK(X206, X6) and pK(X3162, X4). • The cubics K in the light green cells are the only cubics for which K, KLΩ, KLQ, KOQ2 coincide. • When the cubics K are the isogonal pKs with pivot on the line GK (see Table 13), KLΩ, KLQ, KOQ2 coincide as in the pink cells.




pKs with given points on (L∞) and (O) The nonisogonal cubics pK(Ω, P) meeting (L∞) and (O) at the same points as two isogonal pKs with respective pivots P1, P2 are obtained when both KLΩ and KOΩ pass through X(6). This occurs when X(4), P, Ω÷X(3) – equivalently X(6), Ω, P x X(3) – are collinear. In this case : – KOQ2 also passes through X(6), – KLP and KOP pass through X(4), – KLQ and KOQ pass through X(3). For example, if K is an isotomic pK (hence Ω = G), then P must lie on the line (L) = X(4)X(69). The two corresponding isogonal pKs have their pivots P1, P2 on the lines (L1) = X(2)X(6) and (L2) = X66)X(69) respectively. When P traverses (L), the line (𝞴1) = PP1 envelopes an ellipse (E), the line (𝞴2) = PP2 envelopes a hyperbola (H) and the line (𝞴3) = P1P2 envelopes a parabola (P).


(E) passes through X(2), X(315), X(671) and is tangent to the lines (L), (L1), {X6, X76}, {X316, X524}. (H) passes through X(76), X(2979) and is tangent to the lines (L), (L2). (P) passes through X(6), X(2854) and is tangent to the lines (L1), (L2). Its focus is F on the lines {X67, X524}, {X110, X1499}. Its directrix (D) is the line {X110, X1296}. Note that (E) and (H) are both tangent at X(315) and X(76) respectively to (L). Hence, for any P on (L) different from these two latter points, (𝞴1) and (𝞴2) are the second tangents drawn through P to (E) and (H) respectively. These meet the lines (L1), (L2) at the requested pivots P1, P2. When P = X(76), K = K141 and P1 = X(6), P2 = X(69) so that the isogonal pKs are K102 and K169. 

Back to the general case with the same notations as above, • (L1) is the line passing through taΩ and a(Ω ÷ X4) with equation : ∑ p ( p + q + r) (c^2 q  b^2 r) x = 0. • (L2) is the line passing through ta(X32 ÷ Ω) and a(X184 ÷ Ω) with equation : ∑a^2 (a^4 q r + b^4 r p + c^4 p q) (c^2 q  b^2 r) x = 0. • (𝞴1) = PP1 and (𝞴2) = PP2 envelope two conics (C1), (C2) respectively and the line (𝞴3) = P1P2 envelopes a parabola (P). *** Let Q be the midpoint of P1, P2. The cubics spK(P1, Q) and pK(Ω, P) share the same nine points on (L∞) and (O). See CL055 for spK cubics. These cubics coincide if and only if they have a tenth common point. This occurs in two situations, one of them being obtained when P = X4, see below. The other is obtained when spK(P1, Q) is a K0, or equivalently, X(6), P1, Q are collinear. *** Two special cases When two of the three points X(4), P, Ω÷X(3) – equivalently X(6), Ω, P x X(3) – coincide, one can find infinitely many K = pK(Ω, P) meeting (L∞) and (O) at the same points as two isogonal pKs with respective pivots P1, P2. Obviously, with X(4), Ω÷X(3), pK(Ω, P) is already isogonal and this leaves two other cases. • when P = X(4), K = pK(Ω, H) is a member of CL019. P1 = a(Ω ÷ H) and P2 = at(cevapoint(Ω ÷ X6, H)). example : Ω = X(25), K = pK(X25, X4) = K233 hence the two cubics are pK(X6, X69) = K169 and pK(X6, X193). • when Ω = P x X(3), K = pK(Ω, P) is a member of CL074. Its isopivot is X(3) hence its tangents at A, B, C pass through X(3). P1 = a(ctP ÷ H) and P2 = agP. example : P = X(2), K = pK(X3, X2) = K168 hence the two cubics are pK(X6, X193) and pK(X6, X69) = K169. The second case For a given Ω ≠ X(6), there is one and only one P ≠ X(4) on the line X(4), Ω÷X(3) such that spK(P1, Q) is a K0, hence a pK. Its pole lies on the bicevian conic C(X4, X112) and its pivot lies on an unicursal circumquintic passing through X(2), X(4), X(2592), X(2593), X(8743). In other words, for every Ω on the bicevian conic C(X4, X112), one can find two pKs with same pole Ω, pivots X(4) and a point S on the quintic. Each one is a spK. The pK with pivot X(4) is spK(P', Q') where P' lies on the Jerabek hyperbola, the reflection of P' in Q' lies on the bicevian conic C(X4, X648). The pK with pivot S is more complicated. Here is a selection of these pairs of cubics. Example 1 with Ω = X(25) : pK(X25,X4) = spK(X69,X6) = K233 and pK(X25,X8743) = spK(X159,X206). Example 2 with Ω = X(571) : pK(X571,X4) = spK(X68,X5) and pK(X571,X2) = spK(X45794,X343). Example 3 : pKs with pivot X(4) pK(X53,X4) = spK(X3,X389) = K049, pK(X427,X4) = spK(X6,X9969) = K517, pK(X1841,X4) = spK(X72,X44547) = K1184. *** The Grebe butterfly Q184 Let us recall that pK(Ω, P on the line X4,Ω÷X3) and spK(P1, Q) share the same nine points on (L∞) and (O). There are two points P for which spK(P1, Q) is also a pK. These points coincide if and only if Ω lies on Q184, in which case only one pivot is found. Example 1 : Ω = X(2), pK(X2, X76) = spK(X6, X141) = K141. Example 2 : Ω = X(32), pK(X32, X2) = spK(X69, X141) = K177, isogonal transform of K141. 
