Κυριακή, 29 Ιουνίου 2014

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

**********************

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


Πέμπτη, 5 Ιουνίου 2014

A PROBLEM

Source: Miguel Ochoa Sanchez in Facebook Group PERU GEOMETRICO

It is an old problem of unknown (to me) origin. It was published also in the Greek mathematical magazine EUCLID (June 1973)

The problem:

And the solution:


Τετάρτη, 4 Ιουνίου 2014

CONCYCLIC CIRCUMCENTERS - 2

Let ABC be a triangle and A'B'C' the cevian triangle of I.

Denote:

Ab,Ac = the reflections of A' in CC',BB', resp. (lying on AC, AB, resp.)

Similarly Bc,Ba = the reflections of B' in AA',CC', resp. and Ca,Cb = the reflections of C' in BB',AA', resp.

A1B1C1 = the antipedal triangle of I wrt triangle A'AbAc

Similarly A2B2C2 = the antipedal triangle of I wrt triangle B'BcBa and A3B3C3 = the antipedal triangle of I wrt triangle C'CaCb,

Oa,Ob,Oc = the circumcenters of the triangles A1B1C1, A2B2C2, A3B3C3, resp.

The circumcenter O of ABC and the circumcenters Oa,Ob,Oc of A1B1C1,A2B2C2,A3B3C3, resp. are concyclic.

Which point is the center X of the circle?

Antreas P. Hatzipolakis, 4 June 2014

X is X(5495)

Peter Moses 5 June 2014


 

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