HYACINTHOS 29000

[Angel Montesdeoca]:

 

 

Let ABC be a triangle, P a point and D, E, F the traces of its trilinear polar on the

sidelines of ABC.

ta, tb, tc are the  the tangents to the circumcircle  at A, B, C, resp.

A'=ta/\PD,  B'=tb/\PE, C'=tc/\PF

ABC is parallellogic with respect to a triangle A'B'C'   if and only if P lies on the sextic of barycentric equation:

3(a^2-b^2)(a^2-c^2)(b^2-c^2)x^2y^2z^2 +
Cyclic Sum[yz(-a^4(b^2-c^2)y^2z^2-b^2c^2x^3((-a^2+b^2-c^2)y+(a^2+b^2-c^2)z))]  = 0.

Points on the curve  A, B, C (which are triple), X(2), X(1988).

If P=X(1988):
*** The   parallelogic center of ABC with respect to A'B'C' is

U = ( a^2 (b^2 c^2 (b^2 - c^2)^6 - 10 a^10 (b^2 - c^2)^2 (b^2 + c^2) - 2 a^2 (b^2 - c^2)^4 (b^2 + c^2)^3 + a^12 (2 b^4 - 5 b^2 c^2 + 2 c^4) + 5 a^8 (b^2 - c^2)^2 (4 b^4 + 7 b^2 c^2 + 4 c^4) - 4 a^6 (b^2 - c^2)^2 (5 b^6 + 9 b^4 c^2 + 9 b^2 c^4 + 5 c^6) + a^4 (b^2 - c^2)^2 (10 b^8 + 13 b^6 c^2 + 18 b^4 c^4 + 13 b^2 c^6 + 10 c^8)) : ... : ...)

U lies on these  lines: {4,19209}, {20,2979}, {51,3087}, {97,6759}, {154,160}, {217,3172}, {512,23613}, {3198,11190}.

*** The   parallelogic center of A'B'C' with respect to ABC is reflection of  X(17434) in X(647)

U' = ( a^4 (a^2 - b^2 - c^2)^3 (b^2 - c^2) : ... : ...)

U' lies on these  lines:    {6,2430}, {112,6080}, {323,401}, {394,3265}, {416,23090}, {418,23613}, {450,2451}, {520,647}, {651,2639}, {1640,23128}, {3569,6753}, {9033,12077}.


Angel Montesdeoca    

https://amontes.webs.ull.es/otrashtm/HGT2019.htm#HG020519