Δευτέρα 20 Μαρτίου 2023

POINTS ON THE McCAY CUBIC (K003) - 3

[APH = Antreas P. Hatzipolakis]

Let ABC be a triangle and P a point.

Denote 

 

Oa, Ob, Oc = the circumcenters of PBC, PCA, PAB, resp.

 

The perpendicular to POa at P intersects AB, AC at Ac, Ab, resp.
Similarly Bc, Ba and Ca, Cb

Which is the locus of P such that the six points lie on a conic?
And for whish P's the conic is a circle?

APH

-----------------------------------------------------------------

 

Which is the locus of P such that the six points lie on a conic?
The entire plane.

For P=x:y:z (barys), the center of the conic C(P) is:

O(P) = 2*(a^4*y*z+2*S^2*x^2)*y*z+x^2*((3*a^2-b^2+c^2)*b^2*z^2+(3*a^2+b^2-c^2)*c^2*y^2)+2*((2*y+z)*a^2*b^2*z^2+(y+2*z)*a^2*c^2*y^2+(y+z)*b^2*c^2*x^2)*x : :

 

ETC pairs (P,O(P)) for finite P: {1, 1}, {2, 42849}, {3, 182}, {4, 6}, {6, 42852}, {55, 42869}, {110, 34291}

 

O(X(5)) [= ETC X46775)] = X(6)X(17) ∩ X(30259)X(45971)

= 2*a^12-4*(b^2+c^2)*a^10-(5*b^4+6*b^2*c^2+5*c^4)*a^8+4*(b^2+c^2)*(5*b^4-2*b^2*c^2+5*c^4)*a^6-(20*b^8+20*c^8-(17*b^4+12*b^2*c^2+17*c^4)*b^2*c^2)*a^4+(b^4-c^4)*(b^2-c^2)*(8*b^4-17*b^2*c^2+8*c^4)*a^2-(b^2-c^2)^6 : :

= lies on these lines: {6, 17}, {30259, 45971}

= [ 8.4524873403652790, -3.7393299250273950, 2.3282833498347550 ]

 

O(X(20)) [= ETC 46776] = X(4)X(6) ∩ X(648)X(5895)

= 7*a^12-11*(b^2+c^2)*a^10-4*(b^4-9*b^2*c^2+c^4)*a^8+2*(b^2+c^2)*(5*b^4-14*b^2*c^2+5*c^4)*a^6-(b^2-c^2)^4*a^4+(b^4-c^4)*(b^2-c^2)*(b^4-10*b^2*c^2+c^4)*a^2-2*(b^4+4*b^2*c^2+c^4)*(b^2-c^2)^4 : :

= lies on these lines: {4, 6}, {648, 5895}, {8567, 44134}, {30549, 44247}

= [ 18.1059327021450400, 22.0688512504576200, -19.9943553232455300 ]

 

O(X(98)) [= ETC 46777] = X(2)X(6) ∩ X(98)X(804)

= (b^2+c^2)*a^10-3*(b^4+c^4)*a^8-2*(b^2-c^2)^2*b^4*c^4+3*(b^6+c^6)*a^6-(b^4-c^4)*(b^2-c^2)*a^2*b^2*c^2-(b^8+c^8-4*(b^2-c^2)^2*b^2*c^2)*a^4

= lies on these lines: {2, 6}, {98, 804}, {237, 36822}, {3053, 10684}, {4108, 46589}, {6785, 13240}, {9753, 36183}, {9755, 15920}

= crossdifference of every pair of points on line {X(512), X(11672)}

= crosssum of X(511) and X(34383)

= X(98)-daleth conjugate of-X(804)

= perspector of the circumconic {{A, B, C, X(99), X(34536)}}

= intersection, other than A, B, C, of circumconics {{A, B, C, X(98), X(2421)}} and {{A, B, C, X(325), X(43665)}}

= {X(2), X(385)}-harmonic conjugate of X(2421)

= [ 94.7477406753966700, 16.5884555656480700, -51.5733773752627400 ]

 

O(X(99)) [= ETC 46778] = X(6)X(523) ∩ X(99)X(670)

= (a^8+(b^4-3*b^2*c^2+c^4)*a^4-(b^2+c^2)*b^2*c^2*a^2+2*b^4*c^4)*(b^2-c^2) : :

= lies on these lines: {6, 523}, {99, 670}, {183, 669}, {308, 18105}, {512, 3734}, {599, 25423}, {888, 33755}, {1975, 14824}, {3098, 32472}, {3314, 44445}, {7610, 45317}, {7770, 23099}, {7778, 23301}, {8266, 21006}, {33799, 39292}, {37637, 44451}

= crossdifference of every pair of points on line {X(511), X(1084)}

= crosssum of X(512) and X(34383)

= X(99)-daleth conjugate of-X(804)

= perspector of the circumconic {{A, B, C, X(98), X(34537)}}

= inverse of X(6) in Kiepert parabola

= [ 3.9565698998036010, -1.4967506424783510, 2.8507672806369600 ]

 

And for whish P's the conic is a circle?

Conjecture: There is an unique center P0 such that C(P0) is a circle.

P0 = K003Q018Q098Q157

= [0.73551064390386005954,0.80676214172110474175,2.7426703942987557929] (6-9-13-ETC-search numbers)

 

There exist other points P for C(P) to be a circle, but they are either on the sidelines of ABC or in the infinity, which do not make sense for this construction.

 

César Lozada

 

************************************

[Bernard Gibert Q175]

Q175 passes through a point Z introduced in Euclid #4073 by César Lozada as follows.

Let P be a point and denote by Oa, Ob, Oc the circumcenters of PBC, PCA, PAB respectively. The perpendicular to POa at P intersects AB, AC at Ac, Ab respectively. Similarly define Bc, Ba and Ca, Cb. These six points lie on a conic for every P.

This conic is a proper circle for one and only one point Z that lies on K003Q018Q098Q157 and Q175.

Indeed, P must lie on three quintics QA, QB, QC of a same pencil. Each quintic (Q) meets the line at infinity as K024 and two other points on a rectangular circum-hyperbola (H). Q175 is the quintic obtained when (H) is the Jerabek hyperbola.

Note that (Q) and K024 ∪ (H) meet at 25 points on (L∞), on the sidelines of ABC and on a line passing through O which meets (H) on K003.

See the analogous Q177 also passing through Z.

[Bernard Gibert Q177]

 

Δεν υπάρχουν σχόλια:

Δημοσίευση σχολίου

Another relationship between Napoleon cubic and Neuberg cubic

Another relationship between Napoleon cubic K005 and Neuberg cubic K001 The world of Triangle Geometry is very intrincate. There are many...