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
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