HYACINTHOS 26731

[Antreas P. Hatzipolakis]:

 

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

Denote:

MaMbMc = the midheight triangle
(ie Ma, Mb, Mc = the midpoints of AA', BB', CC', resp.)

Maa, Mab, Mac = the reflections of Ma in BC, CA, AB, resp.
Mba, Mbb, Mbc = the reflections of Mb in BC, CA, AB, resp.
Mca, Mcb, Mcc =  the reflections of Mc in BC, CA, AB, resp.

Oa, Ob, Oc = the circumcenters of MaaMabMac, MbaMbbMbc, McaMcbMcc, resp.

A'B'C', OaObOc are orthologic.
The orthologic center (OaObOc, A'B'C') is the O.
The other one (A'B'C', OaObOc) ?

 

Conjecture:

Let Pa, Pb, Pc be same points on the Euler lines of MaaMabMac, MbaMbbMbc, McaMcbMcc, resp
A'B'C', PaPbPc are orthologic,
Which are the loci of the orthologic centers as Pa, Pb, Pc move on the respective Euler lines being same?

 

 

[César Lozada]:

 

 

For P such that OP=t*OH, orthologic centers are:

 

Qa = (A’->P) =

= (-4*S^2*(18*a^10-21*(b^2+c^2)* a^8-4*(9*b^4-14*b^2*c^2+9*c^4) *a^6+54*(b^4-c^4)*(b^2-c^2)*a^ 4-2*(b^2-c^2)^2*(3*b^4-14*b^2* c^2+3*c^4)*a^2-3*(b^4-c^4)*(b^ 2-c^2)*(3*b^2+c^2)*(b^2+3*c^2) )*t+8*a^2*b^2*c^2*(3*a^8-4*(b^ 2+c^2)*a^6-6*(b^2-c^2)^2*a^4+ 12*(b^4-c^4)*(b^2-c^2)*a^2-(5* b^4+6*b^2*c^2+5*c^4)*(b^2-c^2) ^2))*(-4*S^2*(2*a^6+5*(b^2+c^ 2)*a^4-4*(b^2+c^2)^2*a^2-3*(b^ 4-c^4)*(b^2-c^2))*t+4*a^2*b^2* c^2*(3*a^4-2*(b^2+c^2)*a^2-(b^ 2-c^2)^2))*(a^2-b^2+c^2)*(a^2+ b^2-c^2) : :  (barycentrics)

= on this conic:

∑ [ ((3*a^12-42*(b^2+c^2)*a^10+39* (3*b^4+2*b^2*c^2+3*c^4)*a^8- 12*(b^2+c^2)*(9*b^4-2*b^2*c^2+ 9*c^4)*a^6-(3*b^8+3*c^8-2*b^2* c^2*(70*b^4-73*b^2*c^2+70*c^4) )*a^4+2*(b^4-c^4)*(b^2-c^2)*( 27*b^4-38*b^2*c^2+27*c^4)*a^2- (21*b^4+46*b^2*c^2+21*c^4)*(b^ 2-c^2)^4)*(-a^2+b^2+c^2)^2*x^ 2-(a^2+b^2-c^2)*(a^2-b^2+c^2)* (a^4+2*(b^2+c^2)*a^2-3*(b^2-c^ 2)^2)*(5*a^8-8*(b^2+c^2)*a^6- 2*(3*b^4-14*b^2*c^2+3*c^4)*a^ 4+16*(b^4-c^4)*(b^2-c^2)*a^2-( 7*b^4+18*b^2*c^2+7*c^4)*(b^2- c^2)^2)*z*y)*(b^2-c^2) ] = 0 (barys) (complicated center and perspector)

 

Qp = (P->A’) =

= a^2*(12*S^2*((b^2+c^2)*a^2-b^ 4-c^4)*t+4*a^2*b^2*c^2*(-a^2+ b^2+c^2)) : : (barys)

= on Brocard axis {X(3), X(6)}

= (4*R^2-3*SW*t)*X(3)+3*t*SW*X( 6)

 

OQp = (3*t*SW/(4*R^2))*OK

 

ETC pairs (P,Qa): (2, 13202)

 

ETC pairs (P,Qp): (2,389), (3,3), (381,52), (549,9729), (1597,11477), (3534,10625), (3830,6243), (5054,9730), (5055,568), (6904,574), (7514,575), (8703,13348), (9818,576), (11346,8399)

 

For P=O, we have:

 

Qa(X(3)) = X(4)X(64) ∩ X(20)X(1249)

= (3*a^4-2*(b^2+c^2)*a^2-(b^2-c^ 2)^2)*(3*a^8-4*(b^2+c^2)*a^6- 6*(b^2-c^2)^2*a^4+12*(b^4-c^4) *(b^2-c^2)*a^2-(5*b^4+6*b^2*c^ 2+5*c^4)*(b^2-c^2)^2)*(a^2-b^ 2+c^2)*(a^2+b^2-c^2) : : (barycentrics)

= On lines: {4, 64}, {20, 1249}, {122, 631}, {196, 950}, {516, 3176}, {1204, 6619}, {2883, 3079}, {3146, 14361}, {5894, 10002}

= Trilinear product of X(1192) and X(1895)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (4, 3183, 459), (5895, 6525, 4)

= [ 6.962159815999234, 6.93538643557720, -4.374061427030253 ]

 

César Lozada

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