[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
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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 = K003 ∩ Q018 ∩ Q098 ∩ Q157
= [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
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[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 K003, Q018, Q098, Q157 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]
