HYACINTHOS 27510

[Antreas P. Hatzipolakis]:

 

 

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

Denote:

Bc, Cb = the reflections of B', C' in NC', NB', resp.

A1 = the orthogonal projection of N on BcCb
A2 = BcCb /\ NA'

We have A1 = A2 := A*

Similarly B*, C*

 

ABC, A*B*C* are perspective.

Perspector?

LOCI:

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

Denote:

Bc, Cb = the reflections of B', C' in PC', PB', resp.

A1 = the orthogonal projection of P on BcCb
A2 = BcCb /\ PA' 

Similarly B1, B2 and C1, C2

Which is the locus of P such that
1. ABC, A1B1C1 are perspective?

2. ABC, A2B2C2 are perspective?

 

 

 

 [César Lozada]:

 

[...] 

 

 

> LOCI:

> Which is the locus of P such that
> 1. ABC, A1B1C1 are perspective?

> 2. ABC, A2B2C2 are perspective?

A1 = A2 for any P

 

Locus: An excentral-circum-sixtic through ETC’s 1, 5, 54.

Perspectors: Q(X(1)) = X(1) & Q(X(54))= X(16030)

 

 

[César Lozada]:

 

 

CORRECTION:

Sorry. I confused A’ with A1 in A2 = BcCb /\ PA'.

> Which is the locus of P such that
> 2. ABC, A2B2C2 are perspective?

Locus = {l=Line X(5)X(523)} {q=rectangular circum-hyperbola with center=X(5099) and perspector=X(2492)}

 

ETC centers on l: 5, 523, 6757, 8151, 10287, 10288, 10412, 14254, 14356, 14566, 14592, 14670, 15475

ETC centers on q: 4, 23, 316, 842, 1383, 8744, 10422, 10561, 10630, 13485, 13574, 14246

 

l is the tripolar of X(94) and is perpendicular to the Euler line at X(5)=N

 

q is the isogonal conjugate of the line X(3)X(67), perpendicular to the Fermat axis X(6)X(13) at X(542), having tripole:

 

ETC-pairs (P,Q=perspector): (4,4), (523,4)

 

T = isogonal conjugate of X(2492)

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

= on MacBeath circumconic  and these lines: {67, 524}, {110, 525}, {249, 4576}, {287, 323}, {441, 14919}, {476, 2799}, {648, 850}, {671, 9141}, {805, 9076}, {2966, 3268}, {2987, 3580}, {3265, 4558}, {3455, 9888}, {5503, 10511}, {6390, 15900}

= isogonal conjugate of X(2492)

= isotomic conjugate of X(9979)

= trilinear pole of the line {3, 67}

= [ -5.4729673199657220, 2.8243990691198060, 4.2112961971163850 ]

 

One perspector:

Q(X(23)) = X(4)X(83) ∩ X(9138)X(9979)

= ((SB+SC)*(3*SA^2-3*R^2*(3*SA- 2*SW)-3*SB*SC-SW^2)-(4*SA+SB+ SC-12*R^2)*S^2)*(S^2+SA^2-4* SB*SC) : : (barys)

= on lines: {4, 83}, {9138, 9979}

= [ -2.2113879199533490, -2.8271461470550880, 6.6185600852317540 ]

 

César Lozada