HYACINTHOS 29551

[Antreas P. Hatzipolakis]:

Let ABC be a triangle.

Denote:

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

N1, N2, N3 = the NPC centers of NaBC, NbCA, NcAB, resp.,

NaN1, NbN2, NcN3 concur at X(30)
(ie they are parallels to Euler line)

La = the reflection of NaN1 in AI
Lb = the reflection of NbN2 in BI
Lc = the reflection of NcN3 in CI

A*B*C* = the triangle bounded by La, Lb, Lc

Then
ABC, A*B*C* are circumpaparallelogic.
The parallelogic center (ABC, A*B*C*) lies on the circumcircle of ABC and is X(74) of ABC
The parallelogic center of (A*B*C*, ABC) lies on the circumncircle of A*B*C* and is X(74) of A*B*C*
Which point wrt triangle ABC is it?

PS A question about circumparallelogic triangles:
Let ABC, A'B'C' be two circumparallelogic triangles.

If:
Parallelogic center (ABC, A'B'C') wrt ABC = U wrt triangle ABC (on the circumcircle of ABC)
Then:
Parallelogic center (A'B'C', ABC) = U wrt triangle A'B'C' (on the circumcircle of A'B'C')
(in the above U = X(74))

[Ercole Suppa]:

 

 

Hi Antreas

 

The parallelogic center (A*B*C*, ABC) is the point:

 

Q = a (2 a^9-3 a^8 (b+c)-2 a^7 (b^2-b c+c^2)+6 a^6 (b^3+c^3)+a^4 b c (8 b^3-11 b^2 c-11 b c^2+8 c^3)-2 a (b^2-c^2)^2 (2 b^4-b^3 c+5 b^2 c^2-b c^3+2 c^4)+(b-c)^2 (b+c)^3 (3 b^4-4 b^3 c+8 b^2 c^2-4 b c^3+3 c^4)-2 a^5 (3 b^4+b^3 c-7 b^2 c^2+b c^3+3 c^4)-a^2 (b-c)^2 (6 b^5+16 b^4 c+13 b^3 c^2+13 b^2 c^3+16 b c^4+6 c^5)+2 a^3 (5 b^6-b^5 c-5 b^4 c^2+3 b^3 c^3-5 b^2 c^4-b c^5+5 c^6)) : : (barys)

 

= lies on this line: {758,3579}

 

= ETC 6-9-13 search numbers: {-4.7908268204445384623, 2.2146125399012909624, 4.3186220252578877272}

 

Best regards

Ercole Suppa