NEUBERG CUBIC - PRIZE (REWARD)

Let ABC be a triangle and P = I the incenter.

Denote:

La,Lb,Lc = the Euler lines of PBC, PCA, PAB, resp. (concurrent at Schiffler point)

(Na), (Nb), (Nc) = the NPCs of PBC,PCA,PAB, resp. (concurrent at Poncelet Point of I)

The perpendicular to La at the NPC center Na intersects BC,CA,AB at Aa, Ab, Ac, resp.

A' = BAb /\ CAc. Similarly B', C'.

Conjecture:

The triangles ABC, A'B'C' are perspective and equivalently the points Aa, Bb, Cc are collinear.

General Conjecture: It is true for any P on the Neuberg Cubic (in this case the Euler lines of PBC,PCA, PAB are concurrent).

If the general conjecture is true: For a proof I offer the book (original edition):

Richard Heger: Elemente der analytischen Geometrie in homogenen Coordinaten. 1872

Note: It is available on-line at:

https://archive.org/details/elementederanal01hegegoog

Antreas P. Hatzipolakis, 29 June 2014

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The conjecture is TRUE, since rhe Neuberg cubic is part of the general locus of P such that The triangles ABC, A'B'C' are perspective !

Peter Moses won the book !

Peter Moses:

Hi Antreas,

Circumcircle + Infinity + Neuberg Cubic, cyclicsum[a^2 ((S^2 - 3 SA SB) y^2 z - (S^2 - 3 SC SA) y z^2)], + maybe 3 imaginary ellipses, (-a^2 + b^2 + c^2) y z + c^2 y^2 + b^2 z^2 = 0, centered on the vertices of ABC ?

Best regards,

Peter.

Hyacinthos #22483