Home page | Catalogue | Classes | Tables | Glossary | Notations | Links | Bibliography | Thanks | Downloads | Related Curves

too complicated to be written here. Click on the link to download a text file.

X(3), X(13), X(14), X(83), X(99), X(1379), X(1380), X(1625), X(2479), X(2480), X(14370), X(39681), X(39683), X(39685), X(51876), X(51878)

vertices of the reflected first Brocard triangle, see Table 32

projections of A, B, C on the corresponding perpendicular bisectors of ABC

infinite points of pK(X6, X14712)

other points below

Geometric properties :

K1159 is the isogonal transform of the Altintas cubic K1158.

K1159 is invariant under the Cundy-Parry like transformation (see CL037) described as follows. Denote by P* the image of P in the isoconjugation with pole X(4558) that swaps X(3) and X(99). Let Q be the intersection of lines X(3)P and X(99)P*. When P lies on K1159, Q also lies on K1159 and obviously, O, P, Q are collinear on K1159.

The six finite common points of K1159 and pK(X6, X14712) are A, B, C, X(14370) and two other points on the line through X(2), X(99) and the circumconic with perspector X(3005), center X(3124), passing through X(i) for i = 6, 39, 76, 141, 755, 882, 1843, 2353, etc.

The tangents at X(1379), X(1380), X(2479), X(2480) to K1159 concur at X(1625) since the polar conic of X(1625) passes through these four points. The vertices of their diagonal triangle must lie on the cubic and they are X(3), Z1 , Z2.

Z1 = X(1379), X(2480) /\ X(1380), X(2479),

Z2 = X(1379), X(2479) /\ X(1380), X(2480).

These two points lie on the cubic and on the line through X(287), X(401), X(511), X(1972), etc, which must meet the cubic at a third point Z3 = X(39683), the third point of K1159 on the line X(3), X(1625).

Other points on K1159 :

Z4 = X(83), X(1379) /\ X(99), X(1380) = X(51878)

Z5 = X(83), X(1380) /\ X(99), X(1379) = X(51876)

Z6 = X(83), X(2479) /\ X(99), X(2480)

Z7 = X(83), X(2480) /\ X(99), X(2479)

Z8 = Z4Z5 /\ Z6Z7, also on the lines {39, 1915}, {141, 384}

Q1 = X(3), X(83) /\ X(99), X(1625) = X(39681)

Q2 = X(3), X(99) /\ X(83), X(1625) = X(39685)

***

Peter Moses observes that the triangle formed by the centers of the osculating circles at A, B, C is perspective to ABC at X(54). It is also perspective to the Kosnita (see ETC X(1658)), the 3rd Hatzipolakis (see ETC X(17817)), the Hatzipolakis-Moses (see ETC X(6145)) triangles.