HYACINTHOS 28769

[Antreas P. Hatzipolakis]:

 

Taking perpendiculars instead of parallels of the problem Hyacinthos 28766

Let ABC be a triangle and P a point.

The perpendicular from P to BC meets the circles (PAB), (PAC) at Bc, Cb, resp..

The perpendicular from P to CA meets the circles (PBC), (PBA) at Ca, Ac, resp.

The perpendicular from P to AB meets the circles (PCA), (PCB) at Ab, Ba, resp.

Which is the locus of P such that:

1. The perpendicular bisectors of BaCa, CbAb, AcBc are concurrent?

2. The perpendicular bisectors of AbAc, BcBa, CaCb  are concurrent?

3 The triangle bounded by BaCa, AbCb, BcAc is perspective to ABC ?

4 The triangle bounded by AbAc, BcBa, CaCb is perspective to ABC ?

 

 

[César Lozada]: 

 

 

1)      Locus = {sidelines} ∪ {Linf } ∪ {circumcircle} ∪ {McCay cubic K003}

If P lies on the circumcircle the Bc=Cb, Ca=Ac and Ab=Ba and they all lie on the circumcircle, then the perpendicular bisectors concur at O.

ETC pairs (P,Q1(P)=point of concurrence) : (1,1), (3,1147), (4,3)

Points of concurrence are rather complicated for other P on K003.

 

2)      Locus = {sidelines} ∪ {Linf } ∪ {circumcircle} ∪ {degree 11}

Two ETC pairs (P,Q2(P)): (3, 6759), (4,3)

 

3)      Locus = {sidelines} ∪ {Linf } ∪ {circumcircle} ∪ {q6=degree 6}. No ETC’s on q6.

 

4)      Locus = The entire plane.. For P=u:v:w (trilinears) the perspector is:

Z(P) = a*v*w*(u^2+2*u*w*cos(B)+w^2)*(u^2+2*u*v*cos(C)+v^2)*(cos(C)*v-cos(B)*w) : :

-          If P lies on the circumcircle then Z(P)=isogonal of the antipode of P

-          ETC pairs (P,Z(P)) for P in the infinity: (30,14380), (523,15329), (1154,10412)

-          Other ETC-pairs (P,Z(P)): (1,513), (4,523), (6,8599), (15,20578), (16,20579), (67,8599), (186,10412)

-          Note: If P lies on {McCay cubic K003} ∪ {circumcircle} then Z(P) is in the  infinity.

 

Z(X(2)) = X(1383)X(2433) ∩ X(1384)X(17414)

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

= 3*X(3049)-X(10097)

= on lines: {512, 2030}, {523, 18907}, {647, 9517}, {1383, 2433}, {1384, 17414}, {1499, 23287}, {2501, 8599}, {2715, 11636}, {3049, 10097}, {17979, 30209}

= isogonal conjugate of the polar conjugate of X(8599)

= barycentric product X(i)*X(j) for these {i, j}: {3, 8599}, {125, 11636}, {525, 1383}, {598, 647}, {895, 23287}

= barycentric quotient X(i)/X(j) for these (i, j): (3, 9146), (184, 9145), (228, 3908), (512, 5094), (525, 9464), (598, 6331), (647, 599), (669, 8541), (1383, 648)

= trilinear product X(i)*X(j) for these {i, j}: {48, 8599}, {598, 810}, {656, 1383}

= trilinear quotient X(i)/X(j) for these (i, j): (48, 9145), (63, 9146), (71, 3908), (598, 811), (656, 599), (661, 5094), (798, 8541), (810, 574), (1383, 162)

= [ -1.3273090966733800, 2.4590213142259980, 2.5508693089848640 ]

 

Z(X(32)) = TRILINEAR POLE OF THE LINE {338, 2086}

= b^2*c^2*(b^2-c^2)*(a^6+a^4*c^2+c^6-(b^2-c^2)*a^2*c^2)*(a^6+a^4*b^2+b^6+(b^2-c^2)*a^2*b^2)

= on line {850, 5027}

= trilinear pole of the line {338, 2086}

= [ 3.7073412413989920, -2.4448187105392280, 3.6221507085581560 ]

 

César Lozada