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 : P/I, 6th point on the conic through O, H, Ia, Ib, Ic and perspector of IaIbIc and PaPbPc, P/O, 6th point on the Stammler hyperbola (the diagonal conic passing through the in/excenters and O). Note that O, P* and P/O are collinear. P/H, 6th point on the Steiner or Don Wallace hyperbola (the diagonal conic passing through the in/excenters and H). Note that H, P* and P/H are collinear. P/P*. Note that P/O, P/H and (P/P*)* are collinear. (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 / (4cos2 π β 1) = (cot2 π + 1) / (3 cot2 π - 1) <=> tan2 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}. 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.
 t 1/2 -1 β 1 -1/2 0 2 -2 cos2Ο -cos2Ο cosΟ -cosΟ sinΟ -sinΟ Pt X(30) X(2) X(3) X(20) X(5) X(4) X(376) X(631) X(5999) X(384) X(11000) X(10999) P1 P2 pK K001 K002 K003 K004 K005 K006 K243 K832 K422 K020
 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 with 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, Jerabek-Thomson 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 circum-conics which are the isogonal transforms of the parallels to the asymptotes at Q, the P-Ceva conjugate of O. Q obviously lies on the pK. 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 circum-conic 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.
 Polar conic of H The polar conic 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).
 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 left-hand 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 right-hand 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.
 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 OH-inverses) 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.
 cubic P = X(i) Q= X(i) centers on the cubic : X(1), X(3), X(4) and X(i) for i = ... / remark π K002 2 5544 see the page 0 K003 3 2 see the page / the only equilateral cubic of the pencil K006 4 6 see the page +/- Ο/4 K005 5 5643 see the page +/- Ο/6 K004 20 3 see the page / the only central cubic of the pencil +/- Ο/2 21 21, 65 22 22, 66, 159, 8270 23 23, 67, 2930 24 24, 68, 9937 25 25, 69, 1716, 3186, 3504, 7093 26 26, 70 K109 27 19, 27, 63, 71, 226, 284, 579, 1751, 1780 28 28, 72, 1724 29 29, 73 K001 30 110 see the page / the only circular cubic of the pencil +/- Ο/3 140 140, 1173, 3337, 7161 +/- 1/2 arccos(3/2) 186 186, 265, 2931 237 237, 290, 3511 297 248, 297 K243 376 5646 376, 3426, 5119, 7284 +/- 1/2 arccos(-2) 381 5645 381, 3431 +/- arctan(β(3/5)) 382 9716 382 +/- arctan(β(5/3)) K020 384 see the page / contains the vertices of the 1st and 3rd Brocard triangle +/- Ο 401 401, 1987 404 8, 56, 404, 3216 407 407, 2647, 9398 408 408, 2662 412 412, 1715 427 427, 1176 429 429, 1798 442 442, 1175 450 450, 1092, 1093, 1942 452 452, 2213 468 468, 895 474 474, 3361, 4866 550 5888 550 +/- 1/2 arccos(-3/2) K832 631 631, 3338, 3527, 7162 +/- 1/2 arccos(2) 851 851, 1758, 2648, 2655, 2656 858 858, 1177 1005 1005, 1242 1006 1006, 1243, 5902 1009 1009, 1244 1010 10, 58, 386, 1010, 1245 1011 1011, 1246 1012 1012, 2093 1113 1113, 2574 1114 1114, 2575 1325 1325, 2948, 10693 1585 493, 1585, 3068, 6413 1586 494, 1586, 3069, 6414 1656 1656 +/- arctan(1/β7) 1657 7712 1657 +/- arctan(β7) 1816 1816, 1896 1817 610, 1172, 1214, 1817, 1903, 2184 2071 2071, 2935 2409 2409, 2435 2475 191, 267, 2475, 6597 2476 1454, 2476 2675 2671, 2672, 2673, 2674, 2675 +/- arctan(3 β β5) 3091 5644 3091, 8888 +/- arctan(1/β2) 3146 3167 3146, 3532, 5128, 7285 +/- arctan(β2) 3522 1697, 3522, 7091 3523 3333, 3523, 7160 3529 154 3529 3542 1079, 3542, 7042 3559 225, 283, 920, 921, 3559 3651 35, 79, 3651 4184 4184, 8049, 8053 4188 1420, 3680, 4188 4203 1403, 4203, 7155 4220 171, 256, 4220 5056 3311, 3312, 3316, 3317, 5056 +/- arctan(1/2) K156 5059 1131, 1132, 1151, 1152, 5059 +/- arctan(2) 5067 5067, 6419, 6420, 10194, 10195 +/- 1/2 arccos(4/5) 5125 34, 78, 1714, 5125 K422 5999 98, 147, 182, 262, 511, 5999 +/- (Ο/2 β Ο) 6353 6353, 6391, 8854, 8855 6820 393, 394, 6820 6840 2949, 5535, 6326, 6840 6905 36, 80, 6905 6909 104, 517, 6909 6940 5559, 5563, 6940 6986 942, 943, 6986 7411 7, 55, 7411 7437 885, 2283, 7437 7471 110, 523, 7471 7488 2917, 6145, 7488 7513 580, 581, 7513 7541 1807, 1870, 7541 7580 165, 2947, 3062, 7580 8613 216, 275, 577, 2052, 8612, 8613 K751 9855 187, 574, 598, 671, 8591 tan π = 3 tan Ο 10486 575, 576, 7607, 7608, 7609, 7616, 8914, 10486 cot π = 3 tan Ο K152 P152 P152 = X(10996), see below Z1 1691, 1916, 3094, 3407, 8782, Z1 +/- 2Ο Z2 3095, 3398, 3399, 3406 +/- (Ο/2 β 2Ο) Z3 1343, 1671, 1676, 5404 +/- Ο/2 Z4 1342, 1670, 1677, 5403 +/- (Ο/2 β Ο/2) 5067' 1327, 1328, 6200, 6396 +/- 1/2 arccos(β4/5)
 Notes : P152 = (a^2-b^2-c^2) (a^8-2 a^4 b^4+b^8+20 a^4 b^2 c^2-4 b^6 c^2-2 a^4 c^4+6 b^4 c^4-4 b^2 c^6+c^8): : , on lines {2,3}, {64,141}, {69,185}, {216,7738}, {388,1040}, {497,1038}. P152 and X7386 are OH-inverses. For this point, we have cos(2π) = cosA cosB cosC / 4. P152 is now X(10996) in ETC (2016-11-20). *** 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 OH-inverses. They are related with Table 38. Z1, Z2, Z3, Z4 are now X(10997), X(10998), X(10999), X(11000) in ETC (2016-11-20). *** X5067' is the OH-inverse 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 (2016-11-21). *** 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 OH-inverses.