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The Euler pencil of cubics is formed by the isogonal pKs with pivot P = X(i) on the Euler line. Each cubic contains A, B, C, I, Ia, Ib, Ic, O, H. The table at the bottom of this page gives a selection of such cubics. See also the related CL037 and a generalization in Table 54. 



Points on these cubics Each cubic with given pivot P defined by OP = k OH (vectors, k real number or infinity) contains the already mentioned points and : (1) the isogonal conjugate P* of P (P* is the 6th point of the cubic on the Jerabek hyperbola) (2) the cevian quotient P/M of any point M on the cubic and its isogonal conjugate (P/M)*. Note that P*, M and P/M are three collinear points on the cubic. In particular :
(3) the 6 vertices of the (Kiepert) isosceles triangles erected on the sides of ABC having a base angle π such that cos(2π) = (1 β k) / (2k) <=> k = 1 / (4cos^{2} π β 1) = (cot^{2} π + 1) / (3 cot^{2} π  1) <=> tan^{2} π = 3 β 4 / (k + 1). These triangles can be drawn externally (triangle AeBeCe) or internally (triangle AiBiCi) and are real if and only if k is not in ]1;1/3[ i.e. P is not in ]X(20);X(2)[. These six points obviously lie on the perpendicular bisectors of ABC. See Table 32 for other curves related to these triangles. See also rotated cevian lines below. (4) the perspectors Pi (resp. Pe) of the triangles ABC and AeBeCe (resp. AiBiCi) and their isogonal conjugates Pe*, Pi*. Note that Pe, Pi lie on the Kiepert hyperbola and on a line passing through K and P/O. Hence Pe*, Pi* are the 2nd and 3rd intersections of the cubic with the Brocard line OK. The tangents to the cubic at Ae, Be, Ce, Pe* (resp. Ai, Bi, Ci, Pi*) concur on the curve. (5) the vertices Pa, Pb, Pc of the cevian triangle of P. (6) the points Ha = AH /\ Pa P/O, Hb, Hc (on the altitudes) and their isogonal conjugates Oa = AO /\ Pa P/H, Ob, Oc (on the cevian lines of O). P/H is the perspector of OaObOc and PaPbPc and P/O is the perspector of HaHbHc and PaPbPc. 

These points Ha, Hb, Hc are collinear if and only if : β P = X(20) in which case they are the infinite points of the altitudes, β P = X(14709) or P = X(14710). These are the common points of the Euler line and the circumconic through X(74) and X(252). The corresponding points Ha, Hb, Hc lie on the axes of the ellipse K (the inconic with center K, perspector H). The two cubics pK(X6, X14709) and pK(X6, X14710) have a flex at O and the axes of the ellipse K are the harmonic polars of O in the cubics. The inflexional tangents at O are parallel to the asymptotes of the Jerabek hyperbola. This has been studied by Fred Lang (Hyacinthos #4382) who called these cubics the Darboux sisters. See figure opposite. 

(7) the perspector IO of IaIbIc and OaObOc (the 3rd point of the cubic on the line IH) and its extraversions IOa, IOb, IOc. (8) the perspector IH of IaIbIc and HaHbHc (the 3rd point of the cubic on the line IO) and its extraversions IHa, IHb, IHc. Thus, IH* is the 6th point of the cubic on the Feuerbach hyperbola and IHa*, IHb*, IHc* are the 6th points of the cubic on the Boutin hyperbolas. (9) the common points β apart A, B, C β of the cubic and the circumcircle (O) of ABC are detailed in a separate section below. 



Construction of these triangles AeBeCe and AiBiCi with a given pivot P 

First, we construct the polar conic C(O) of O in the cubic. C(O) is a rectangular hyperbola passing through O and K, having its asymptotes parallel to those of the Jerabek hyperbola. It also contains the harmonic conjugate of O with respect to H and P. See below for further properties of C(O). The perpendicular bisector of BC meets C(O) at O and another point A'. The circle with center the midpoint Ma of BC which is orthogonal to the circle with diameter OA' meets the perpendicular bisector of BC at the requested points Ae and Ai. Similarly we can construct Be and Bi, Ce and Ci. This easily gives all the other mentioned points on the cubic. 



A remarkable property 

Let M be a point and t a real number. The homothety h(M, t) transforms A, B, C into At, Bt, Ct. The perpendiculars at At, Bt, Ct to the lines AM, BM, CM meet the sidelines BC, CA, AB at A', B', C' respectively. These points A', B', C' are collinear (on a line L) if and only if M lies on a cubic of the Euler pencil whose pivot Pt is the barycenter of {X3, 2t}, {X4, 1}. In this case, the six remaining common points of the perpendiculars and the sidelines of ABC are on a same conic. When M describes the cubic, the envelope of L is generally a complicated sextic. The figure opposite is obtained with t = 1/2 and M = X(1138). The cubic is the Neuberg cubic K001. 

The table below gives a selection of these cubics with the corresponding value of t and associated pivot Pt. 



P1, P2 (on the Euler line) lie on the lines passing through X(6) and X(2545), X(2544) respectively. When t = Β± cosA cosB cosC, we find Pt = X(25) and X(1593). 



Rotated cevian lines 

With the same notations as in (3) above, let us consider a variable point M on the cubic with pivot P on the Euler line such that OP = 1 / (1 + 2 cos 2 π) OH (vectors). Recall that P lies outside ]X(2), X(20)[. The cevian line AM is rotated about A with angles +/β π producing two lines meeting BC at A1, A2 respectively. The corresponding points B1, B2 on AC and C1, C2 on AB are defined similarly. One remarkable property is that, for any M on the cubic, these six points always lie on a same conic. 



Orthic line and poles of the orthic line 

The orthic line of any cubic of the pencil (except K003 since it is a stelloid) is the Euler line. In other words, the polar conic of any point on the Euler line is a rectangular hyperbola. For a given cubic different of K003, these hyperbolas form a pencil and meet at four points called the poles of the Euler line in the cubic. The locus of the poles when the pivot traverses the Euler line is a quartic passing through X(3), X(4), X(6), X(1670), X(1671), X(2574), X(2575), X(3413), X(3414). It has four real asymptotes parallel to those of the Kiepert and Jerabek hyperbolas. It is bitangent at O and H to the Euler line. The figure represents the Orthocubic K004 and the four corresponding poles, one of them being H. The polar conic of O is the Jerabek hyperbola and that of H is the diagonal hyperbola through H and the in/excenters. Note that the locus of the poles of the line PP* is the cubic K511. 



Circular polar conics 

For any cubic with pivot P of the Euler pencil, there is one and only one point Q whose polar conic is a circle. This point is in a way the Lemoine point of the cubic and we call it the Stuyvaert point of the cubic. See K609 for further details and the table below. When P traverses the Euler line, the locus of Q is a rectangular hyperbola (H) passing through X(2), X(3), X(6), X(110), X(154), X(354), X(392), X(1201), X(2574), X(2575), X(3167) and the three vertices of the Thomson triangle. These points are the common points (apart A, B, C) of the Thomson cubic K002 and the circumcircle of ABC. (H) is actually the Jerabek hyperbola for this Thomson triangle. Its center X(5642), labelled X on the figure, is the midpoint of GX(110). Q coincides with the points X(2), X(3), X(6), X(110) when P is X(3), X(20), X(4), X(30) corresponding to the cubics K003, K004, K006, K001 respectively. Note that two antipodes Q, Q' on (H) correspond to two points P, P' on the Euler line which are inverse in the circle with center O, radius 3R. 

The points Q related to K002 and K005 were added to ETC in June 2014. They are respectively : X(5544) = a^2*(a^4  4*a^2*b^2 + 3*b^4  4*a^2*c^2  26*b^2*c^2 + 3*c^4) : : , SEARCH = 2.1291814336, X(5643) = a^2*(a^4  3*a^2*b^2 + 2*b^4  3*a^2*c^2  11*b^2*c^2 + 2*c^4) : : , SEARCH = 1.8976292080. 



Other polar conics 

Polar conic of O Recall that the polar conic C(O) of the circumcenter O in every pK of the Euler pencil is a rectangular hyperbola passing through O, K, X(2574), X(2575) hence having its asymptotes parallel to those of the Jerabek hyperbola. It also contains the harmonic conjugate P' of O with respect to H and P. All these polar conics are centered on the line X(2), X(98), X(110), X(114), X(125), etc. For example, with P = X(3), X(4), X(631) we obtain the Stammler, Jerabek, JerabekThomson hyperbolas respectively. See Q002 for other related properties and also K920. It follows that the two parallels at O to these asymptotes meet the pK again at two pairs of (not always real nor distinct) points {P1, P2), {P3, P4} with same midpoint O. These four points also lie on the circumconics which are the isogonal transforms of the parallels to the asymptotes at Q, the PCeva conjugate of O. Q obviously lies on the pK. A pK meets its associated polar conic C(O) at 6 points (one of them is O counted twice) lying on the quintic Q150. 

There are three points on the Euler line such that C(O) is a degenerate rectangular hyperbola. One of them is X(20) corresponding to the Darboux cubic K004, a central cubic with center O. C(O) splits into the line at infinity and the inflexional tangent at O, namely the Brocard axis. The two remaining points P1, P2 lie on the circumconic passing through X(74), X(252). See "Points on these cubics, (6)" above. The corresponding cubics pK1, pK2 have an point of inflexion at O with inflexional tangent passing through X(2574), X(2575) respectively. Remark that the inflexional tangent of pK1 (resp. pK2) meets pK2 (resp. pK1) at two points on the circle with center O passing through H. C(O) splits into this tangent and its perpendicular at K. Note that the centers Q1, Q2 of these decomposed conics lie on the Brocard circle. 

The polar lines of any point M in all these hyperbolas C(O) concur at M* which is the isogonal conjugate of M with respect to the diagonal triangle of the system. This triangle is actually improper since one vertex is X(511) on the line at infinity. Nevertheless, an isogonal conjugation may be defined in this triangle. See Table 62. 

Polar conic of H The polar conic C(H) of the orthocenter H in every pK of the Euler pencil is a rectangular hyperbola passing through H, the harmonic conjugate of H with respect to O and P, and three other (blue) fixed points which lie : β’ on the Jerabek hyperbola (J) corresponding to K003, β’ on the diagonal hyperbola (D) passing through the in/excenters of ABC corresponding to K006. These three points also lie on the circle (C) with center X(9409), orthogonal to the circumcircle (O) and to the Brocard circle. (C) passes through X(15), X(16), X(74), X(112), etc. The image of (C) under the homothety h(H, 1/2) is the circle (C'), the locus of the centers of all the polar conics of H. (C') is orthogonal to the circumcircle (O) and to the nine point circle (N). It passes through X(98), X(107), X(125), X(132), X(5000), X(5001), etc, and its center is X(6130). 

A pK meets its associated polar conic C(H) at 6 points (one of them is H counted twice) lying on the quintic Q151. See also the related K1096. 



Intersection of a cubic of the Euler pencil and (O) 

The common points β apart A, B, C β of the cubic K(P) = pK(X6, P) and the circumcircle (O) of ABC are the vertices Q1, Q2, Q3 of a triangle denoted T(P). These points are not necessarily real nor all distinct. The more general case for any pK is studied in the paper How pivotal cubics intersect the circumcircle. 

Construction of T(P) It is not possible with ruler and compass only hence at least another conic β apart (O) β is needed. Since P is the orthocenter of T(P), the Euler line of ABC is also the Euler line of T(P) and its isogonal transform β with respect to T(P) β is the Jerabek hyperbola J(P) of T(P). J(P) passes through X(3), X(110), X(2574), X(2575), P and meets (O) again at the requested points Q1, Q2, Q3. Note that the center of J(P) is the midpoint of X(110)P and that J(P) is homothetic to the Jerabek hyperbola of ABC. K(P) meets the sidelines of T(P) again at R1, R2, R3 which lie on the bicevian conic C(P) passing through the midpoints A', B', C' of ABC and the cevians Pa, Pb, Pc of P. The lines passing through the isogonal conjugate P* of P and these points R1, R2, R3 are parallel to the asymptotes of K(P) and perpendicular at P* to the sidelines of T(P). 

Conics inscribed in T(P) β’ when P traverses the Euler line, the sidelines of T(P) envelope a parabola (P) which is the reflection in O of the Kiepert parabola (K) of ABC. (P) is the parabola with focus X(74) and directrix the Euler line. See lefthand figure below. β’ since the triangles ABC and T(P) are inscribed in (O), they must circumscribe a same conic (C1) which is the inconic of ABC with center P' = ccP, the homothetic of P under h(X2, 1/4), and perspector tcP. When P traverses the Euler line, (C1) remains tangent to the line (L1) which is the perpendicular bisector of X(4)X(74) and also the trilinear polar of X(525), passing through X(122), X(125), X(684), X(1650), X(2972), etc. See righthand figure below. 

β’ for any P, one can find a conic (C2) inscribed in T(P) and in the CircumTangential triangle CT. Its center P" is the homothetic of P under h(X3, 1/4). When P traverses the Euler line, (C2) remains tangent to the line (L2) which is the perpendicular bisector of X(3)X(74). 

Remarks : The parabola (P) is inscribed in some special triangles T(P) for some points P on the Euler line : β’ P = X(2), T(P) is the Thomson triangle, K(P) is K002, (C1) is the Steiner inellipse, β’ P = X(3), T(P) is the CircumNormal triangle, K(P) is K003, (C1) is the inconic with perspector X(69), (C2) is the circle with center X(3), radius R/2. β’ P = X(20), T(P) is the X(3)reflection triangle of ABC, K(P) is K004.
When P = X(4), (C1) is the McBeath inconic with foci X(3) and X(4), K(P) is K006. See the related K187. 

In/excenters and Lemoine point of T(P) When P traverses the Euler line, the in/excenters and Lemoine point KP of T(P) lie on SRH, the Stammler Reflection Hyperbola as mentioned in the preamble just before X(33573) in ETC. SRH is the reflection in O of the Stammler hyperbola. The list below gives (P,KP) = (X(i),X(j)) for these (i,j) : (2,5646), (3,3), (4,11472), (20,1350), (21,33538), (23,10117), (30,10620), (546,33539), (550,33544), (3090, 33540), (3091,33537), (3146,64), (3529,12163), (3627,33541), (6912,33536), (7464,12302), (12086,32345), (12103,33542), (14156,2574), (15157,2575), (15704,12307), (17538,33543), (33557,40). SRH passes through the vertices of these triangles: β’ A'B'C' : hexyl (see TCCT 6.36) β’ excentral triangle of Thomson triangle (see Q113) β’ R1R2R3 : tangential triangle of Thomson triangle Q1Q2Q3 (see K172) β’ antiHutson intouch triangle (see X(11363))


SRH = Feuerbach hyperbola of tangential triangle of Thomson triangle SRH = isogonal conjugate of the Euler line of ABC wrt the tangential triangle of Thomson triangle SRH passes through X(i) for these i : 3, 40, 64, 1350, 2574, 2575, 5373, 5646, 9914, 10117, 10620, 11472, 12163, 12301, 12302, 12307, 14926, 15622, 16010, 19376, 22549, 32345, 33535, 33536, 33537, 33538, 33539, 33540, 33541, 33542, 33543. The center of SRH is X(74), the Feuerbach point of R1R2R3. The circumcircle (C) of R1R2R3 passes through X(2930) and its center is X(33532), the reflection of X(31861) in X(3). The antipode of X(2930) lies on SRH; it is the reflection of the orthocenter X(11472) of R1R2R3 in X(74). 



Table of cubics of the Euler pencil 

Recall that P is the pivot on the Euler line and Q is the Stuyvaert point (the point with a circular polar conic) on the Jerabek hyperbola (JT) of the Thomson triangle. Remark 1 : two points P, P' on the Euler line defined by OP = k OH and OP' = k' OH such that 1/k + 1/k' = 2 correspond to two angles π, π' with inverse tangents hence π + π' = Ο/2 i.e. π' = Ο/2  π. These points are inverse in the circle with diameter OH (abbreviated OHinverses) or, equivalently, harmonic conjugates with respect to O and H. Some of these pairs of points are highlighted in two cells of same colour in the colum P of the table below. Note that the corresponding points Q, Q' are collinear with X(373). See X384 and X5999 for example where π = Ο, the Brocard angle. Remark 2 : two points P, P' on the Euler line, inverse in the circle C(O, 3R), correspond to two points Q, Q' antipodes on (JT). *** The cells highlighted in blue in the last column correspond to complex values of π since the point P is between X(20) and X(2). They are given for completeness but they have no geometrical signification. Table built with the cooperation of Peter Moses. 



Notes : P152 = (a^2b^2c^2) (a^82 a^4 b^4+b^8+20 a^4 b^2 c^24 b^6 c^22 a^4 c^4+6 b^4 c^44 b^2 c^6+c^8): : , on lines {2,3}, {64,141}, {69,185}, {216,7738}, {388,1040}, {497,1038}. P152 and X7386 are OHinverses. For this point, we have cos(2π) = cosA cosB cosC / 4. P152 is now X(10996) in ETC (20161120). *** Z1 = midpoint of X7924, X9855, on lines {2, 3}, {99, 736}, {187, 5152}, {316, 5149}, {325, 8290}, {385, 698}, {511, 4027}, {1691,1916}, {3094,3407}, {3095,10131}, {3314,4048}, {3329,5116}, {5017,7766}, {5104,8289}, {7761,10000}. Z2 on lines {2,3}, {99,8295}, {3095,3406}, {3398,3399}, {9983,10104}. Z3 on lines {2,3}, {6,2546}, {76,1671}, {83,1343}, {2547,5085}, {6248,8161}. Z4 on lines {2,3}, {6,2547}, {76,1670}, {83,1342}, {2546,5085}, {6248,8160}. {Z1, Z2} and {Z3, Z4} are pairs of OHinverses. They are related to Table 38. Z1, Z2, Z3, Z4 are now X(10997), X(10998), X(10999), X(11000) in ETC (20161120). *** X5067' is the OHinverse of X5067, on lines {2,3}, {40,4669}, {388,4324}, {485,6486}, {486,6487}, {497,4316}, {515,4677}, {516,7967}, etc. X5067' is now X(11001) in ETC (20161121). *** Additional angles : X3146 : +/ 1/2 arccos(1/3). X9855 : +/ 1/2 arccos[(4  5 cos 2Ο) / (5  4 cos 2Ο)] and X10486 : +/ 1/2 arccos[(4  5 cos 2Ο) / (5  4 cos 2Ο)], these two points are OHinverses.

