HYACINTHOS 26425

[Le Viet An]:

Let ABC be a triangle and A'B'C' the antipedal triangle of N.

Denote:

Ea, Eb, Ec = the Euler lines of the triangles A'BC, B'CA, C'AB, resp.

1. Ea, Eb,Ec are concurrent.

 

2. the parallels to Ea, Eb, Ec through A,B, C, resp. concur on the circumcircle.

 

3. the parallels to Ea, Rb, Ec through A', B', C' are concurrent.

Points?

 

[César Lozada]:

 

1)      Z1 = reflection of X(5) in X(128)

= (S^2+SB*SC)*(4*SA^2+(6*R^2-4* SW)*SA-22*R^2*SW+4*SW^2+27*R^ 4) : : (barycentrics)

= 3*X(5)-2*X(137) = 3*X(128)-X(137) = 4*X(128)-X(1263) = 4*X(137)-3*X(1263) = 3*X(381)-X(11671)

= on cubics K465, K725 and these lines: {2,12026}, {3,2888}, {4,13512}, {5,128}, {30,930}, {140,1141}, {381,11671}, {495,3327}, {496,7159}, {539,6150}, {549,13372}, {6069,14050}

= midpoint of X(i) and X(j) for these {i,j}: {4,13512}, {6069,14050}

= reflection of X(i) in X(j) for these (i,j): (3,6592), (5,128), (1141,140), (1263,5)

= anticomplement of X(12026)

= [ 7.910112378714122, 4.04245651581737, -2.808780357680317 ]

 

2)      Z2=X(1141)

 

3)      Z3 = reflection of X(1263) in X(128)

= (S^2+SB*SC)*(8*SA^2+(12*R^2-8* SW)*SA+45*R^4+2*S^2+6*SW^2-36* R^2*SW) : : (barycentrics)

= 3*X(5)-4*X(128) = 5*X(5)-4*X(137) = 3*X(5)-2*X(1263) = 5*X(128)-3*X(137) = 6*X(137)-5*X(1263) = 3*X(549)-2*X(1141) = 3*X(549)-4*X(6592) = 5*X(632)-4*X(12026)

= On lines: {5,128}, {30,13512}, {74,550}, {546,11671}, {549,1141}, {632,12026}, {2888,10205}

= reflection of X(i) in X(j) for these (i,j): (550,930), (1141,6592), (1263,128), (11671,546)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (128, 1263, 5), (1141, 6592, 549)

= [ 15.267353015793340, 8.40301970251172, -9.223281319428352 ]

 

César Lozada