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X(3), X(4), X(8), X(76), X(847), X(3557), X(3558), X(3730), X(8743), X(10571), X(14246), X(14247), X(14248), X(14249), X(14250), X(14251), X(14252), X(14253), X(14254), X(14255), X(14256), X(14257), X(14258), X(14259), X(14260), X(14261), X(14262), X(14263), X(14264), X(14265), X(14266), X(14267), X(14268), X(25043), X(34243), X(34756), X(38539), X(38936), X(38937), X(38938), X(38939), X(39263), X(39264), X(39265), X(39267), X(39268), X(39269), X(39270)

four foci of the MacBeath inconic : X(3), X(4) and two imaginary

vertices of the cevian triangle of X(264)

extraversions of the Nagel point X(8) = vertices of the 2nd Conway triangle, see ETC X(9776)

infinite points of the McCay cubic, points on the circumcircle and the orthocubic

isogonal conjugates of Ua, Ub, Uc mentioned in the Neuberg cubic page. See also table 16 and table 18.

other points, details and figures below and in the page K027

 

This cubic is a nodal K60+ (or a stelloid) with a node at H where the tangents are parallel to the asymptotes of the Jerabek hyperbola. The tangents at A, B, C are the symmedians.

The three asymptotes concur at X(381), the midpoint of X(2)X(4), and are parallel to those of the McCay cubic.

It is (partially) described in Musselman's paper (see bibliography) in the following manner. Denote by A2, B2, C2 the reflections of P in the sidelines BC, CA, AB and by A3, B3, C3 the reflections of H in the lines AP, BP, CP. The three circles AB2C2, BC2A2, CA2B2 have a common point N on the circumcircle. The three circles PAA3, PBB3, PCC3 have P in common and another point L also on the circumcircle. These points N, L coincide if and only if P lies on the Musselman (second) cubic K027 and they are antipodes if and only if P lies on the Musselman (third) cubic K028 (Musselman, Some loci connected with a triangle. Monthly, p.354-361, June-July 1940).

K028 is also the isogonal transform of the Lemoine cubic K009, the isotomic transform of K257 and the anticomplement of K258. Note that K009 and K028 define a pencil of cubics that also contains K005 and K187. The X(1989)-isoconjugate of K028 is the nodal stelloid K724.

Another description is the following : let P be a point and Pa, Pb, Pc its projections on the perpendicular bisectors of ABC. Triangles ABC and PaPbPc are perspective if and only if P lies on the Stammler hyperbola. The locus of the perspector is the Musselman (third) cubic.

K028 is the Phi-transform of the circumcircle and the Psi-transform of the line at infinity. See CL037. In other words, K028 is the locus of the intersection of the line OP and the Steiner line of P when P traverses the circumcircle. See Q011 for an analogous property with the Simson line.

K028 is also psK(X4, X264, X3) in ABC and actually a psK in infinitely many triangles (see below) as in Pseudo-Pivotal Cubics and Poristic Triangles and spK(X3, X5) as in CL055. See a generalization and other related cubics in Table 43. See also CL067 and CL071.

 

Most of the following properties are easily generalized for other nodal circum-stelloids such as K054, K230, K594, K595, K596, K597, K724.

K028a

Groups of pivots and inscribed equilateral triangles

K028 is a stelloid or also a harmonic curve that solves Laplace's equation.

One of the groups of pivots on K028 consists in H (counted twice) and O. In other words, for any point M on K028 and for any arbitrary line L, 2(HM,L)+(OM,L) = constant (mod.π).

It follows that, for any points M and N on K028, we have 2(HM,HN)+(OM,ON) = 0 (mod.π).

Hence, any circle passing through O and H meets the cubic at three other points which are the vertices of an equilateral triangle inscribed in K028.

Indeed, if (HM,HN) = π/3 (mod.π), the equality above yields (OM,ON) = π/3 (mod.π) hence O, H, M, N are concyclic on a circle which must meet K028 again at a sixth point P such that (PM,PN) = π/3 (mod.π).

If M1, M2, M3 are the vertices of an equilateral triangle inscribed in the circumcircle of ABC then the Steiner lines of these points meet the corresponding lines passing through O at P1, P2, P3 and P1P2P3 is an equilateral triangle inscribed in K028 whose circumcircle passes through O and H.

Now, if two perpendiculars at H meet K028 again at M and N, then (HM, HN) = π/2 (mod.π) hence (OM, ON) = 0 (mod.π) and the line MN passes through O.

It follows that the line OP1 meets K028 at a third point Q1which is on the perpendicular at H to HP1. Q2 and Q3 are defined likewise and the triangle Q1Q2Q3 is also equilateral. Its circumcircle passes through O, H and is orthogonal to the circumcircle of P1P2P3.

When one perpendicular is the altitude AH, we obtain another point on K028 as the intersection of the line OA and the parallel at H to the sideline BC. Two other points on the lines OB and OC are defined similarly.

K028pivots

There are infinitely many groups of pivots {P0, P1, P2} on K028 and the circumcircle of P0P1P2 always meets K028 again at the vertices of an equilateral triangle inscribed in K028.

Let P0 be any point on K028 and let Q0 be the homothetic of P0 under h(X381, -1/2). Recall that X381 is the radial center of K028 i.e. the intersection of its asymptotes and then, any line passing through X381 meets K028 at three points whose isobarycenter is X381. This is obviously the case when the line is the Euler line.

The ellipse with foci G, H passing through Q0 is tangent at Q0 to the external bisector at Q0 of triangle GHQ0. The tangents drawn from P0 to this ellipse meet this bisector at two points P1, P2 which are the pivots associated to P0. Note that the ellipse is the Steiner inellipse of triangle P0P1P2.

It follows that, for any two points M, N on K028, we have (P0M,P0N)+(P1M,P1N)+(P2M,P2N) = 0 (mod. π).

A square inscribed in K028

The parallels at O to the nodal tangents (i.e. to the asymptotes of the Jerabek hyperbola) meet the circumcircle at four points M1, M2, M3, M4 and K028 at four points P1, P2, P3, P4 which are the vertices of a square inscribed in K028. The cicumcircle of this square is the circle with center O passing through H. See below for other circles with center O.

Naturally, each point Pi lies on the Steiner line of Mi.

A regular pentagon inscribed in K028

The circle with center H passing through O meets K028 at O and five other points R1, R2, R3, R4, R5 which are the vertices of a regular pentagon.

Indeed, since (HR1, HR2) = 2 (OR1, OR2), we find 5 (OR1, OR2) = 0 (mod.π) hence (HR1, HR2) = 2π/5 (mod.π).

K028b K028pentagone
K028sat

Points on the asymptotes

K028 meets its three asymptotes at three collinear points lying on the satellite of the line at infinity. This latter line (S) is parallel to the tangent at O to K028, the line through X(74), X(110), etc.

It meets the Euler line at the homothetic of X(381) under h(H, 1/3), now X(14269) in ETC (2017-09-06).

Parallels to the asymptotes

K028McCaytriangles

Let T be an equilateral triangle with center X(381) – the radial center of K028 – and sidelines parallel to the asymptotes of K028.

K028 meets these sidelines at six finite points which lie on a same circle with center Ω, a point on the line passing through X(381) and X(526) when T varies.

K028parallels

Three parallels at a point P to the asymptotes of K028 meets K028 again at six points which lie on a same rectangular hyperbola H(P).

H(P) is homothetic to the polar conic of P in K028, a rectangular hyperbola passing through H.

When P lies on the Euler line, these hyperbolas are homothetic to the Jerabek hyperbola.

If P = X(381), H(P) splits into the line at infinity and the satellite line (S) mentioned above.

The figure opposite shows H(O) passing through X(3), X(5), X(54), X(575), X(2574), X(2575), etc, with center X(140). H(O) meets the three parallels at O (counting for three) and three remaining points, the vertices of an equilateral triangle with center X(5).

 

K028K006

Points on the circumcircle

K028 and K006 meet the circumcircle at the same points namely A, B, C and A', B', C'.

The tangents at A', B', C' to K028 are concurrent at X lying on the lines X(3)-X(1495), X(5)-X(64), X(30)-X(599), X(74)-X(1995), X(110)-X(378), X(125)-X(381), etc. X is X(11472) in ETC.

These points also lie on the rectangular hyperbola through X(3), X(4), X(110), X(155), X(1351), X(1352), X(2574), X(2575).

Recall that the tangents to K006 at these points are concurrent at X(25).

K028med

Points on the perpendicular bisectors

K028 meets the perpendicular bisector of BC at O and two other points A1, A2 that lies on the circle HBC. Four other points B1, B2, C1, C2 are defined likewise.

K028 meets the line AO again at A' that lies on the parallel to BC at H. B' and C' are defined likewise.

K028Fuhrmann

Fuhrmann triangle and Fuhrmann circle

K028 passes through X(4), X(8), the endpoints of a diameter of the Fuhrmann circle.

Their three remaining common points are the vertices of the Fuhrmann triangle FaFbFc.

These points are the reflections in the sidelines of ABC of the vertices A1, B1, C1 of the circumcevian triangle of the incenter X(1).

 

Hessian, prehessian and flexes of K028

K028hess

Hessian

The hessian of K028 is a strophoid with node H and singular focus X(381).

Since K028 is a crunodal cubic, it has only three flexes and one only is real, namely F on the figure. Naturally, these flexes also lie on the hessian.

F lies on the perpendicular L at G to the Euler line.

The rectangular hyperbola with center X(110), with asymptotes parallel to those of the Jerabek hyperbola (therefore to the nodal tangents) and passing through X(895) meets the circumcircle at X(74) and three other points, one only being real. The corresponding points on K028 are the flexes.

This strophoid is the inverse in the polar circle of the rectangular hyperbola (H) with center X(468), with asymptotes parallel to those of the Jerabek hyperbola passing through H and X(67). The inverse of L is the circle with diameter HX(468) meeting (H) at the inverse of F.

K028prehess

Prehessian

K028 has only one prehessian. In other words, there is one and only one cubic whose hessian is K028.

This prehessian is also a nodal cubic with node H, passing through X(1656) and obviously the common flexes of K028 and its hessian.

 

Intersections of K028 with circles centered at O and psK cubics

Let (C) be a circle with center O, passing through M1 and M2 on the tangent at O to K028, a line passing through X(74) and X(110).

Let (H1), (H2) be the rectangular hyperbolas passing through M1, M2 and the fixed points O, H, X(2574), X(2575). These belong to the pencil generated by the Jerabek hyperbola and the union of the Euler line and the line at infinity. Each one is the reflection of the other about X(5). Their respective centers O1, O2 are the midpoints of HM1, HM2 and are obviously two points on the line X(5)X(125) symmetric about X(5).

K028cerclesOa

(H1) meets (C) at M1 and three other points Q1, Q2, Q3 on K028.

(H2) meets (C) at M2 and three other points R1, R2, R3 also on K028.

These two triangles Q1Q2Q3 and R1R2R3 have the same circumcenter O and the same orthocenter H.

The tangents to K028 at Q1, Q2, Q3 concur at K1 which is the Lemoine point of triangle Q1Q2Q3. It follows that K028 is a psK in this triangle hence K028 meets its sidelines again at three points, vertices of a triangle perspective to Q1Q2Q3.

The same results apply with triangle R1R2R3 and its Lemoine point K2.

K1 and K2 belong to a same rectangular hyperbola passing through X(3), X(6), X(381), X(599), X(2574), X(2575), etc, hence homothetic to the Jerabek hyperbola.

See two more detailed figures below.

K028cerclesOb K028cerclesOc