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(2), X(115), X(125), X(523), X(525), X(542), X(647), X(1640), X(6103), X(14357), X(15526), X(18312), X(23967), X(34156), X(39170), X(42426), X(56399)

X(65722) → X(65736)

pole Ω = X(65724) = X(125) x X(542), on the lines {115,523}, {125,647}

midpoints of ABC

Geometric properties :

K1369 is the complement of K1368.

K1369 and K1353 share the same points at infinity. They meet again at six finite points on the circum-conic passing through X(2), X(1640), X(6103).

K1369 meets the nine point circle at X(115), X(125), X(42426) and the midpoints of ABC.

K1369 meets the Steiner inellipse at X(115), X(15526), X(23967) and the midpoints of ABC.

***

For every Ω on K1369, there is a point P on K1368 such that pK(Ω, P) meets the line at infinity at X(523), X(525), X(542).

This is obviously the case of the cubics K1353, K1368, K1369 with the respective pairs {Ω, P} = {X6103, X648}, {X2, X99 x X41254}, {X125 x X542, X2}.

Other examples of such pairs : {X115, X54395}, {X125, X542}, {X542, X110}, {X647, X62307}, {X1640, X525}, {X69 x X542, X99}.

***

Generalization

Let P1, P2, P3 be three distinct points on the line at infinity. Every pK(Ω, P) passing through these points must have

• its pole Ω on K(Ω) = psK(P1 x P2 x P3, X2, P1^2) passing through the barycentic squares P1^2, P2^2, P3^2 (on the Steiner inellipse) and the barycentic products P1 x P2, P2 x P3, P3 x P1.

• its pivot P on the complement K(P) of psK(P1 x P2 x P3, X2, P1) which passes though P1, P2, P3 and the vertices of the antimedial triangle. Hence, K(P) passes through A, B, C, P1, P2, P3 and the midpoints of ABC.

These two cubics become pKs when they contain X(2). Hence, if P1 and P2 are given, there is a unique corresponding point P3. In this case, K(Ω) also contains P1, P2, P3.

P3 is the infinite point of the line passing through G and the barycentric product of the infinite points of the trilinear polars of the isotomic conjugates of P1 and P2.

With P1 = {p,q,r} and P2 = {u,v,w}, we have P3 = F(P1, P2) = {-2 p u+q u+r u+p v+q v-2 r v+p w-2 q w+r w , p u+q u-2 r u+p v-2 q v+r v-2 p w+q w+r w , p u-2 q u+r u-2 p v+q v+r v+p w+q w-2 r w}.

Note that F(P1, P3) = P2 and F(P2,P3) = P1.

See another example in K1370 and K1371 when P1 = X(30), P2 = X(523) and then P3 = X(2799). See also K1372.