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∑ a^4 SA (c^2 y^2  b^2 z^2) y z = 0 ∑ b^2 c^2 x^3 (c^2 SC y  b^2 SB z) = 0 

A, B, C which are inflexion points X(1), X(3), X(6), X(15), X(16), X(358), X(1135), X(1137), X(1155), X(2574), X(2575) excenters, extraversions of X(1155) 6 feet of bisectors common points of the Thomson cubic and the circumcircle i.e. vertices of the Thomson triangle 26 mates of X(358) (these are the isogonal conjugates of the perspectors of ABC and the 27 Morley triangles) See details, figures and other points below 

The EulerMorley quartic Q002 is called Q2 in "Orthocorrespondence and Orthopivotal Cubics" (see Downloads page) where a more complete description can be found. Its isogonal conjugate is the EulerMorley quintic Q003. See also the analogous quartic Q043 and Q067 and a generalization in CL009. Locus properties : Q002 is the locus of P (with isogonal conjugate P*) such that :
Other properties :


The trilinear polar of O meets BC at Oa. A variable line (L) passing through Oa meets AB and AC at B' and C'. C(B) is the conic passing through B with tangent OB, B' and the feet of the bisectors at B on the sideline AC. C(C) is defined likewise. C(B) and C(C) meet at four points Q1, Q2, Q3, Q4 which lie on Q002. The pencil of conics generated by C(B) and C(C) contains an analogous conic C(A) passing through A. It also contains a rectangular hyperbola H(P) passing through O, K, X(2574), X(2575). See below for another point of view related with cubics of the Euler pencil. See also A Remarkable Rational Transformation Related with Pivotal Cubics. 

This construction can be generalized for a certain analogous quartic Q(P, Q) where O = X(3) is replace with P, the feet of the bissectors with the vertices of the cevian triangle of Q and the traces of the trilinear polar of Q. In this case, if M* denotes the isoconjugate of M under the isoconjugation with fixed point Q (therefore with pole Q^2), then Q(P, Q) is the locus of M such that P, M, M*/M or M, M*, P*/M* are collinear. With P = u:v:w and Q = p:q:r the equation of Q(P, Q) is : ∑ u p^2 (r^2 y^2  q^2 z^2) y z = 0. For instance Q002 = Q(X3, X1), Q033 = Q(X3, X2), Q043 = Q(X6, X1), Q045 = Q(X8, X2), Q083 = Q(X4, X1).


100 points on Q002 The following table sums up the points that lie on this remarkable quartic (n is the number of associated points). See an explanation below. 



Q002 and the isogonal conjugates of the Morley perspectors Q002 contains the isogonal conjugates of the 27 perspectors of ABC and the Morley triangles. See Table 9. These are the points with barycentric coordinates : a cos (A/3 + k1 2pi/3) : b cos (B/3 + k2 2pi/3) : c cos (C/3 + k3 2pi/3) where k1, k2, k3 are integers in {1;0;1}. These 27 points lie on three groups of 9 lines passing through A, B, C. In particular, Q002 contains : X(358) obtained with k1=k2=k3=0, X(1135) obtained with k1=k2=k3=1, X(1137) obtained with k1=k2=k3=1. Note that the points X(16), X(358), X(1135), X(1137) are collinear. 

Q002 and the polar curves of the circumcenter O The polar line of O is the tangent at O passing through X(49) and the intersection Z = X(3292) of the real asymptotes. The polar conic of O is the Jerabek hyperbola. The polar cubic of O is the Orthocubic. The tangents at A, B, C, O are common to both curves. 

Q002 and the isogonal conjugates of the CPCC points These points are described in Table 11. They are the dark green points on the figure. They lie on the Darboux cubic, K172 = pK(X32, X3) and several other curves. Note that K172 meets the circumcircle at the same points as Q002 and the Thomson cubic. These are the vertices of the Thomson triangle. The tangents to Q002 at these points and at X(6) concur at E(227) = X(2)X(3) /\ X(6)X(373). 

Q002 and the isogonal conjugates of the Ixanticevian points These points are described in Table 23. They are the blue points on the figure. They lie on the Napoleon cubic, K073 = pK(X50, X3) and several other curves. Recall that K073 is a circular cubic passing through the isodynamic points. 

Q002 and the ellipse K Q002 contains the foci of the ellipse K i.e. the inconic (K) with center K. It is an ellipse when the triangle ABC is acute angle. They are the red points on the figure. These four foci also lie on the Pelletier strophoid K040 and many other curves. Note that Q002 and K040 meet at 12 known points since both curves are circular and contain X(1155) on the line IO. The vertices of the yellow triangle are the extraversions of X(1155) which lie on Q002. 

Q002 and the cubics of the Euler pencil The Euler pencil is formed by the isogonal pivotal cubics with pivot P on the Euler line. See Table 27. The frequent occurence of these cubics in the table above can be explained (and extended) by the following decomposition of Q002. Let K(P) be the cubic pK(X6, P) and H(P) be the rectangular hyperbola passing through O, K, P, X(2574), X(2575) hence having its asymptotes parallel to those of the Jerabek and Stammler hyperbolas which are two members of the pencil. K(P) and H(P) meet at six points namely O, P and four other points on Q002. Indeed, an easy computation shows that : Q002 = (a^2 + b^2 + c^2)(x + y + z) K(P) + (a^2 y z + b^2 z x + c^2 x y) H(P). Recall that Q002 and K(P) have 8 fixed common points A, B, C, X(1), X(3), the excenters, and then 4 more depending of P which must lie on H(P). Special cases : • P = O : H(P) is the Stammler hyperbola (tangent at O to the Euler line) and K(P) is the McCay cubic K003. The four points are the in/excenters and the tangents at these points are common to Q002 and K003, passing through O. • P = H : H(P) is the Jerabek hyperbola and K(P) is the Orthocubic K006. The four points are A, B, C, O and the tangents at these points are common to Q002 and K006, also passing through O. • P = X(20) : K(P) is the Darboux cubic K004 and the four points are the isogonal conjugates of the CPCC points. • P = X(5) : K(P) is the Napoleon cubic K005 and the four points are the isogonal conjugates of the Ixanticevian points. • P = G : K(P) is the Thomson cubic K002 and the four points are X(6) and the vertices of the Thomson triangle. • P = X(30) : K(P) is the Neuberg cubic K001 and the four points are the isodynamic points X(15), X(16) and the circular points at infinity. This property is generalized below. See also Q112 for a similar curve and another generalization. 

Q002 and the circular pivotal cubics 

Any isogonal circular pK must have its pivot P at infinity and its isopivot P* on the circumcircle (O). Since it already has 9 common points with Q002 (namely A, B, C, in/excenters, circular points at infinity), it must meet Q002 again at three other points Q1, Q2, Q3 which lie on a same line (L) passing through the Lemoine point K. This line (L) is actually the trilinear polar of the antipode Q of P* on (O). For example, with P = X(30), the cubic is K001 and (L) is the Brocard axis. The three points are X(3), X(15), X(16) as said above. Conversely, a line (L) passing through K has its trilinear pole Q on (O) and meets Q002 again at three points which lie on the isogonal circular pK whose pivot P is the isogonal conjugate of the antipode of Q. 
