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X(3) triple point

X(1145), X(1312), X(1313), X(1511), X(2028), X(2029), X(2446), X(2447), X(5976), X(10354), X(56761), X(56785), X(56787), X(56788), X(56792), X(56793), X(56794), X(56795), X(56796), X(56973), X(61501), X(61502), X(61503), X(61504), X(61505), details below.

extraversions of X(1145), X(2446), X(2447)

Ma, Mb, Mc : midpoints of ABC (nodes)

Oa, Ob, Oc : vertices of the cevian triangle of O

points at infinity of the McCay cubic

common points of the circumcircle and the Thomson cubic i.e. vertices of the Thomson triangle

Q011 is the locus of point P such that P, its symgonal Q and O are collinear (together with the line at infinity, although the symgonal of a point at infinity is itself). See a generalization below.

Q011 is a circular quintic with singular focus O which is a triple point on the curve. It has three real asymptotes which are those of the McCay cubic.

The tangents at A, B, C concur at X(25).

Bui Quang Tuan has found that Q011 is the locus of the intersection Q of the line OP and the Simson line S(P) of P when P traverses the circumcircle (see Hyacinthos #14759). Note that, if the Simson line is replaced by the Steiner line, the locus is K028, the Musselman (third) cubic.

Thus, for P = u:v:w on the circumcircle, this point Q is : b^2c^2u^4(c^2v^2–b^2w^2)(c^2SC v–b^2SB w) : : .

Now, if P' is the antipode of P on the circumcircle, the line L(P) = OPP' and the Simson line S(P') of P' meet at Q' obviously on Q011. S(P) and S(P') meet at X on the nine point circle and the circle C(P) with diameter QQ' is tangent at X to the nine point circle. The center Ω of C(P) lies on K258, the complement of K028.

The barycentric product QxQ' of Q and Q' lies on the Steiner inscribed ellipse and the points X(115), X, QxQ' are collinear.

We obtain the following special cases :





points Q, Q' on Q011

Euler line

nine points circle



X(1312), X(1313)

Brocard axis

Moses circle



X(2028), X(2029)

line OI




X(2446), X(2447)

lines O-excenters


extraversions of X(11)


extraversions of X(2446), X(2447)

line O X(8)




X(1145) and X(1145)'

line O X(74)




X(1511) and X(1511)'

Notes :

X(1145)' = (b - c)^2 (b + c - 2a) / [(b + c)(a + b - c)(a - b + c) - 2abc], SEARCH = 4.0046105891, on the lines X(3)X(8), X(11)X(513).

X(1511)' = a^2 (b^2 - c^2)^2 (2SA + b c)*(2SA - b c) / [a^2 (2SA - a^2) + (b^2 - c^2)^2)], SEARCH =3.9490387472, on the lines X(3)X(74), X(125)X(523).

These two points are X(56761), X(56792) in ETC.


Intersection of Q011 and pK(Ω, O)

Let pK = pK(Ω, O) be a pivotal isocubic with pole Ω, pivot O. pK and Q011 meet at 15 points namely A, B, C, O (triple), Oa, Ob, Oc and six other points depending of Ω.

When Ω lies on the Brocard axis OK, these six points lie on a same circle (C) with center O, although these points might not be all real nor distinct. See CL021 for pivotal cubics with pivot O and specially those with pivot on OK.

If k is a positive real number and kR the radius of the circle (R is the circumradius), the coordinates of the corresponding pole Ω are a^2(cos^2 A – k^2) : : .


Special cases

1. When k = 0, Ω is X(577), the barycentric square of O. The cubic decomposes into the cevian lines of O and the circle (C) reduces to O itself.

2. When k = 1, Ω is X(32), the third power point, and pK is K172. (C) is the circumcircle itself and the six points are A, B, C again and the points where K172 and the Thomson cubic meet the circumcircle.

3. When k is infinite, Ω is the Lemoine point X(6) and pK is the McCay cubic. (C) degenerates into the line at infinity counted twice.

4. When k = 1/2, Ω is X(50) and pK is K073 or Ki which is a circular cubic passing through X(1511) also on Q011. It follows that Q011 and K073 must have three other common points on the circle (C) with radius R/2. These three points are the midpoints of O and the three common points (apart A, B, C) of the circumcircle and the Napoleon cubic K005. See the figure opposite.


5. When k^2 = – cosA cosB cosC, Ω = O and the cubic is pK(X3, X3). (C) is real when ABC is obtuse angle and its radius is half that of the polar circle. The six common points are not necessarily all real.

The figure opposite shows four real points.


Generalization of Q011

Let P = p : q : r be a fixed point. The locus of the point M such that M, its symgonal N and P are collinear is in general a circular circum-quintic Q(P), together with the line at infinity (in which case N = M).

The singular focus F is the midpoint of OP. Q(P) has three other points at infinity which are those of the cubic pK(X6, P).

Q(P) passes through O, P, the midpoints Ma, Mb, Mc of ABC which are double points, the vertices Pa, Pb, Pc of the cevian triangle of P.

O is a double point with perpendicular nodal tangents when P lies on the Euler line. It is a triple point when P = O and then Q(P) is Q011 and F = O.

Q(P) meets the bicevian conic C(X2, P) at 10 points namely Ma, Mb, Mc (counting for 6), Pa, Pb, Pc and a last point Z whose symgonal is P.

Q(P) meets the nine-point circle at 10 points namely Ma, Mb, Mc (counting for 6), the circular points at infinity and then, two other points N1, N2 which lie on a line passing through H. This is the Steiner line of the antipode on (O) of the isogonal conjugate of the infinite point of the line HP when P ≠ H. See below when P = H.

The tangents at A, B, C concur at the barycentric quotient X(32) ÷ P.

Q(P) meets the perpendicular bisectors of ABC at O, one of the midpoints of ABC and, since both points are double, at a 5th point Ta, Tb, Tc respectively.

These points coincide at O when P = O. They are collinear (on the line at infinity) when P = X(20).

Special cases

• When P lies on the line at infinity, Q(P) is bicircular. Hence, Q(P) and pK(X6, P) meet at 15 points, namely A, B, C, Pa, Pb, Pc, 4 points at infinity (P with the same tangent/asymptote and the circular points) and 5 remaining points which lie on K258.

• When P lies on the circumcircle, Q(P) splits into psK(gtgP, X2, X3) and a rectangular hyperbola passing through O, Ma, Mb, Mc and the antipode of the complement of P on the nine-point circle.

Note that gtgP lies on the Lemoine axis. This psK is circular with singular focus on C(O, R/2), see CL068 for a list of such cubics. These circular cubics form a pencil which contains only one pK, namely K043 obtained when P = X(187). Other examples are K446, K567, K570.

• When P = H, Q(P) splits into K009 and the nine-point circle. Recall that the symgonal of every point on the nine-point circle is H.