HYACINTHOS 28747

[Aris Pavlakis](*)

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

The reflections of IN in B'C', C'A', A'B' bound a triangle A*B*C* similar to ABC.

The circumcenter O* of A*B*C* lies on the NPC.

Point?.  

 

(*) Aris Pavlakis, Romantics of Geometry 2583

APH

 

[Peter Moses]:

Hi Antreas,

X(3259); (2*a - b - c)*(b - c)^2*(a^2*b - b^3 + a^2*c - 2*a*b*c + b^2*c + b*c^2 - c^3) : : 

 

[...]

[Antreas P. Hatzipolakis]:

X(3259) is updated

http://faculty.evansville.edu/ck6/encyclopedia/ETCPart3.html#X3259

Now, let Q be the point X(3259) of the construction above.

There are three more points Qa, Qb, Qc on the NPC corresponding to pedal triangles of the excenters Ia, Ib, Ic, resp. and the lines NIa, NIb, NIc, resp.

I am wondering with which triangles of the reference triangle ABC is the triangle QaQbQc perspective or orthologic or parallelogic.

PS: QaQbQc is an Euler triangle
(Euler triangles are named the central triangles which are inscribed in the NPC)

 

[Peter Moses]:


Hi Antreas,

{Qa,Qb,Qc} is Perspective to the 4th Euler triangle at

 

= X(5)X(113)∩X(10)X(27555)

 

(b+c) (-2 a^4 b^2-3 a^3 b^3+a^2 b^4+3 a b^5+b^6-a^3 b^2 c-2 a^2 b^3 c+a b^4 c+2 b^5 c-2 a^4 c^2-a^3 b c^2-2 a^2 b^2 c^2-4 a b^3 c^2-b^4 c^2-3 a^3 c^3-2 a^2 b c^3-4 a b^2 c^3-4 b^3 c^3+a^2 c^4+a b c^4-b^2 c^4+3 a c^5+2 b c^5+c^6) : :

= 5 X[1698] - X[2940].

= lies on these lines: {5,113}, {10,27555}, {12,8287}, {115,24443}, {429,1861}, {442,22798}, {542,3615}, {857,29610}, {1495,9958}, {1698,2940}, {1737,14873}, {2392,3142}, {2899,27704}, {3013,8614}, {3136,25972}, {5044,22076}, {5221,8818}, {5587,27685}, {6723,24904}, {6739,18357}, {7173,8286}, {9780,27554}, {10175,27687}

= X(26734)-complementary conjugate of X(3741).

Best regards,
Peter Moses.