[Antreas P. Hatzipolakis]:
Let ABC be a triangle and P a point.
Denote:
A'B'C' = the midway triangle of P
(ie A', B', C' = the midpoints of AP, BP, CP, resp.)
A1, B1, C1 = the reflections of A, B, C in B'C', C'A', A'B', resp.
A2, B2, C2 = the reflections of A', B',C' in BC, CA, AB, resp.
For P = O, the triangles A1B1C1, A2B2C2 are bilogic (perspective and orthologic)
Which is the locus of P such that A1B1C1, A2B2C2 are:
1. perspective
2. orthologic
?
[César Lozada]:
1) Locus = {infinity} \/ {Napoleon-Feuerbach cubic K005}
For P in the infinity, perspector Z(P)=P.
For P on K005, ETC-pairs (P,Z(P)): (1,1155), (3,5609), (4,4), (5,30), (195,10610)
Z( X(54) ) = ((cos(2*A)+3/2)*cos(2*(B-C))+ 3*cos(2*A)-cos(4*A))*sec(B-C) : : (trilinears)
= X(54)+3*X(1157)
= on line {3,54}
= [ 1.912345813217654, 7.71956432415496, -2.586270425531050 ]
Z(X(17)) = X(17)+3*X(8172) = on line {3,13} = [ -4.835641975899586, -4.18539572071610, 8.770080892818408 ]
Z(X(18)) = X(18)+3*X(8173) = on line {3,14} = [ 0.958187112494429, 7.83998462253035, -2.229257385610982 ]
* * * * * * * * *
2) Locus = {infinity} \/ {Euler line} \/ { circle with center=X(382), radius=3*R, no ETC centers on it }
For P on the Euler line. The loci of the orthologic center Z1(P) = (A1->A2) and Z2(P) = (A2->A1) are two distinct cubics.
ETC pairs (P,Z1(P)) = (20,4)
ETC pairs (P,Z2(P)) = (20,3627), (382,3627), (10296,5609)
Z1(O) = (5*cos(2*A)-2*cos(4*A)-3)*cos( B-C)+(-cos(A)+2*cos(3*A))*cos( 2*(B-C))+cos(A)-2*cos(3*A) : : (trilinears
= 4*X(389)-3*X(8550) = X(6241)-9*X(6403) = X(6241)+9*X(9973) = 9*X(11188)-5*X(11444)
= On lines: {3,8705}, {5,11649}, {6,3518}, {52,2854}, {143,8584}, {156,576}, {235,1843}, {389,2393}, {511,3627}, {524,6243}, {575,5944}, {1192,8549}, {1503,6240}, {2781,11381}, {3517,11216}, {5449,8262}, {9019,10625}, {9781,9971}, {11188,11444}, {11441,11477}
= midpoint of X(i) and X(j) for these {i,j}: {3,11663}, {6403,9973}
= reflection of X(5480) in X(1843)
= [ -0.906025026087269, -4.37351134192060, 7.086645422969685 ]
Z2(O) = (3*cos(2*A)-2*cos(4*A)+7/4)* cos(B-C)+(-cos(A)+2*cos(3*A))* cos(2*(B-C))-4*cos(A)-cos(3*A) : : (trilinears)
= on lines: {3518,11935}, {3627,6243}
= [ -7.449132760990365, -3.93726752184831, 9.804526348259531 ]
Z1(H) = (9*cos(2*A)-2*cos(4*A)+13/2)* cos(B-C)+(-3*cos(A)+2*cos(3*A) )*cos(2*(B-C))-19/2*cos(A)-3* cos(3*A) : : (trilinears)
= on lines: {24,3431}, {7530,9716}
= [ -13.678623913551800, -10.15032511945669, 16.981023678555230 ]
Z2(H) = (5*cos(2*A)-2*cos(4*A)-3)*cos( B-C)+(-cos(A)+2*cos(3*A))*cos( 2*(B-C))+cos(A)-2*cos(3*A) :: (trilinears)
= 4*X(389)-3*X(8550) = X(6241)-9*X(6403) = X(6241)+9*X(9973) = 9*X(11188)-5*X(11444)
= On lines: {3,8705}, {5,11649}, {6,3518}, {52,2854}, {143,8584}, {156,576}, {235,1843}, {389,2393}, {511,3627}, {524,6243}, {575,5944}, {1192,8549}, {1503,6240}, {2781,11381}, {3517,11216}, {5449,8262}, {9019,10625}, {9781,9971}, {11188,11444}, {11441,11477}
= midpoint of X(i) and X(j) for these {i,j}: {3,11663}, {6403,9973}
= reflection of X(5480) in X(1843)
= [ -0.906025026087269, -4.37351134192060, 7.086645422969685 ]
César Lozada
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