HYACINTHOS 26516

[Le Viet An]:

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

Denote:

La, Lb, Lc = the Euler lines of A"B'C', B"C'A', C"A'B', resp.

1. La, Lb, Lc are concurrent.
2. The parallels to La, Lb, Lc through A", B", C", resp. are concurrent.
3. The NPC center of A"B"C" lies on the Euler line of ABC.

Points?

 

 

[César Lozada]:

1)       

Q = X(5)X(51) ∩ X(30)X(930)

= cos(B-C)*(4*(2*cos(A)+cos(3*A) )*cos(B-C)+2*cos(2*A)*cos(2*( B-C))-cos(4*A)+1/2) : : (trilinears)

= 3*X(549)-2*X(6150)

= on lines: {5, 51}, {30, 930}, {140, 1157}, {549, 6150}, {632, 10615}, {5432, 14102}, {7604, 11016}

= reflection of X(1157) in X(140)

= nine-points circle-inverse-of-X(13565)

= [ 2.722853867384805, -1.13111787658250, 3.167044303825427 ]

 

2)

Q’ = X(5)X(51) ∩ X(30)X(13512)

= cos(B-C)*(8*(2*cos(A)+cos(3*A) )*cos(B-C)+2*cos(2*A)*cos(2*( B-C))-3*cos(4*A)-1/2) : : (trilinears)

= 3*X(549)-2*X(1157)

= On lines: {5, 51}, {30, 13512}, {549, 1157} , {10615, 11539}

= [ 4.892835993134705, -1.94412908228801, 2.728368003583135 ]

 

3)

N’’ = reflection of X(4) in X(5501)

= (5*cos(2*A)+5*cos(4*A)+3/2)* cos(B-C)-2*(3*cos(A)+cos(3*A)) *cos(2*(B-C))+cos(2*A)*cos(3*( B-C))-cos(3*A)-cos(5*A)-7*cos( A) : : (trilinears)

= (5*R^2-2*SW)^2*X(3)-(9*R^4-8* SW*R^2-2*S^2+2*SW^2)*X(5)

= on line {2, 3}

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (964, 7475, 13628), (1370, 7475, 11331), (3090, 7460, 7487), (3533, 6990, 7435), (4244, 7715, 3850), (7442, 11345, 6955), (7473, 10223, 462)

= [ -0.698768022083452, -1.56644457729516, 5.047634430227237 ]

 

César Lozada