HYACINTHOS 28698

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle.

Denote:

Na, Nb, Nc = the NPC centers of IBC, ICA, IAB, resp.

Aa, Ab, Ac = the orthogonal projections of A on INa, INb, INc, resp.

The Euler line of AaAbAc passes through the Feuerbach point Fe.

Which point is Fe wrt triangle AaAbAc?


 [Randy Hutson]:

Hi Antreas,

Triangles AaAbAc and BcBbBc, CaCbCc (constructed cyclically) are all similar to the excentral triangle. While Fe lies on the Euler lines of AaAbAc, BcBbBc, and CaCbCc, it does not occupy the same position relative to each triangle, therefore it is not a center of any of them.  There are some interesting results, however:

The similitude center of BcBbBc and CaCbCc is the A-vertex of the intouch triangle, and cyclically ...

Let Sa be the AaAbAc-to-excentral similarity image of Fe, and define Sb, Sc cyclically.

Sa, Sb, Sc lie on the Euler line of the excentral triangle (line X(1)X(3)), and the centroid of SaSbSc is X(165) (centroid of excentral triangle).

The lines ASa, BSb, CSc concur in:

= ISOGONAL CONJUGATE OF X(1768)

Trilinears 1/(a^5 - a^4 (b + c) - a^3 (2 b^2 - 5 b c + 2 c^2) + 2 a^2 (b - c)^2 (b + c) + a (b - c)^2 (b^2 - b c + c^2) - b^5 + b^4 c + b c^4 - c^5) : :

= lies on these lines:  {484, 1785},  {516, 5080},  {517, 1456},  {522, 1768},  {910, 5537}, {1325, 5538}, {5536, 22464} et al

 

= isogonal conjugate of X(1768)
= [X(4)-Ceva conjugate of X(110)]-of-excentral triangle

Search = [25.675244349994029, 20.896520660086082, -22.676270290457076]

Best regards,
Randy Hutson