CYCLOLOGIC

Let ABC be a triangle

Denote

1. Oa, Ob, Oc = the circumcenters of HBC, HCA, HAB, resp.

ABC, OaObOc are cyclologic, since Oa, Ob, Oc are the reflections of O in BC,CA,AB, resp.
Cyclologic center (OaObOc, ABC) = antigonal conjugate of O = X(265)

2. Sa, Sb, Sc = the X(54) of HBC, HCA, HAB, resp.

ABC,SaSbSc are cyclologic

Cyclologic centers?

ETC

X(72661) = X(3)X(59164)∩X(54)X(14978)

Barycentrics    b^2*c^2*(a^12-3*a^10*b^2+2*a^8*b^4+2*a^4*b^8-3*a^2*b^10+b^12-5*a^10*c^2+5*a^8*b^2*c^2+5*a^2*b^8*c^2-5*b^10*c^2+10*a^8*c^4+4*a^6*b^2*c^4+5*a^4*b^4*c^4+4*a^2*b^6*c^4+10*b^8*c^4-10*a^6*c^6-12*a^4*b^2*c^6-12*a^2*b^4*c^6-10*b^6*c^6+5*a^4*c^8+7*a^2*b^2*c^8+5*b^4*c^8-a^2*c^10-b^2*c^10)*(-a^12+5*a^10*b^2-10*a^8*b^4+10*a^6*b^6-5*a^4*b^8+a^2*b^10+3*a^10*c^2-5*a^8*b^2*c^2-4*a^6*b^4*c^2+12*a^4*b^6*c^2-7*a^2*b^8*c^2+b^10*c^2-2*a^8*c^4-5*a^4*b^4*c^4+12*a^2*b^6*c^4-5*b^8*c^4-4*a^2*b^4*c^6+10*b^6*c^6-2*a^4*c^8-5*a^2*b^2*c^8-10*b^4*c^8+3*a^2*c^10+5*b^2*c^10-c^12) : :

Antreas Hatzipolakis and Ercole Suppa, euclid 9721.

X(72661) lies on the Jerabek circumhyperbola and these lines: {3, 59164}, {54, 14978}, {264, 59143}, {1173, 60828}

X(72661) = isogonal conjugate of X(72662)
X(72661) = intersection, other than A, B, C, of the circumconics: {{A,B,C,X(3),X(4)}, {A,B,C,X(5),X(46138)}, {A,B,C,X(49),X(1298)}, {A,B,C,X(140),X(57765)}, {A,B,C,X(264),X(14978)}, {A,B,C,X(1263),X(6662)}, {A,B,C,X(3481),X(34302)}, {A,B,C,X(15318),X(32535)}, {A,B,C,X(22268),X(34385)}, {A,B,C,X(44176),X(70013)}}

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X(72662) = X(2)X(3)∩X(51)X(52540)

Barycentrics    a^4*(a^10*b^2-5*a^8*b^4+10*a^6*b^6-10*a^4*b^8+5*a^2*b^10-b^12+a^10*c^2-7*a^8*b^2*c^2+12*a^6*b^4*c^2-4*a^4*b^6*c^2-5*a^2*b^8*c^2+3*b^10*c^2-5*a^8*c^4+12*a^6*b^2*c^4-5*a^4*b^4*c^4-2*b^8*c^4+10*a^6*c^6-4*a^4*b^2*c^6-10*a^4*c^8-5*a^2*b^2*c^8-2*b^4*c^8+5*a^2*c^10+3*b^2*c^10-c^12) : :

Antreas Hatzipolakis and Ercole Suppa, euclid 9721.

X(72662) lies on these lines: {2, 3}, {51, 52540}, {93, 32428}, {156, 23038}, {184, 20574}, {1157, 1614}, {1199, 23195}, {5000, 5001}, {6152, 46025}, {6242, 42441}, {6288, 34292}, {6344, 51888}, {8154, 32379}, {10263, 15345}, {11576, 58468}, {12325, 50947}, {13558, 34418}, {15033, 25042}, {18016, 68084}, {19210, 52417}, {26881, 45083}, {32340, 63816}, {41598, 71245}, {43808, 68741}, {51255, 61715}, {52681, 61139}, {59371, 64032}

X(72662) = isogonal conjugate of X(72661)
X(72662) = pole of the line {3, 59164} with respect to Stammler hyperbola
X(72662) = intersection, other than A, B, C, of the circumconics: {{A,B,C,X(5),X(20574)}, {A,B,C,X(140),X(46089)}, {A,B,C,X(184),X(3078)}}
X(72662) = {X(13564),X(15770)}-harmonic conjugate of X(3)

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X(72663) = X(2)X(10182)∩X(96)X(578)

Barycentrics    a^14*b^2-6*a^12*b^4+15*a^10*b^6-20*a^8*b^8+15*a^6*b^10-6*a^4*b^12+a^2*b^14+a^14*c^2-6*a^12*b^2*c^2+14*a^10*b^4*c^2-10*a^8*b^6*c^2-10*a^6*b^8*c^2+19*a^4*b^10*c^2-9*a^2*b^12*c^2+b^14*c^2-6*a^12*c^4+14*a^10*b^2*c^4-6*a^8*b^4*c^4-5*a^6*b^6*c^4-12*a^4*b^8*c^4+21*a^2*b^10*c^4-6*b^12*c^4+15*a^10*c^6-10*a^8*b^2*c^6-5*a^6*b^4*c^6-2*a^4*b^6*c^6-13*a^2*b^8*c^6+15*b^10*c^6-20*a^8*c^8-10*a^6*b^2*c^8-12*a^4*b^4*c^8-13*a^2*b^6*c^8-20*b^8*c^8+15*a^6*c^10+19*a^4*b^2*c^10+21*a^2*b^4*c^10+15*b^6*c^10-6*a^4*c^12-9*a^2*b^2*c^12-6*b^4*c^12+a^2*c^14+b^2*c^14 : :
X(72663) = X[4]+2*X[63816], 2*X[140]+X[31867]

Antreas Hatzipolakis and Ercole Suppa, euclid 9721.

X(72663) lies on these lines: {2, 10182}, {4, 63816}, {96, 578}, {140, 31867}, {5890, 9221}, {6143, 32904}, {35482, 43666}, {35717, 37119}

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G - Orthologic

Let ABC be a triangle, P = G = X(2) and Q a point on the Euler line.

Denote:

Bc, Cb = the orthogonal projections of B, C on GC, GB, resp.

Qa = same to Q point of the triangle ABcCb.
Similarly Qb, Qc.

ABC, QaQbQc are orthologic.

For Q = G = X(2)
Orthologic center (ABC, QaQbQc) = G* = ?
Orthologic center (QaQbQc, ABC) = G** = ?

For Q = X(3) = O:
Orthologic center (ABC, QaQbQc) = O* = X(36889)
Orthologic center (QaQbQc, ABC) = O** = X(1352)
Euclid 9541

Q = H = X(4)
Orthologic center (ABC, QaQbQc) = H* = X(3)= O
Orthologic center (QaQbQc, ABC) = H** = ?

Q = N = X(5)
Orthologic center (ABC, QaQbqc) = N* = ?
Orthologic center (QaQbQc, ABC) = N** = ?

The locus of the orthologic center (QaQbQc, ABC) = Q**, as Q moves on the Euler line, is a line.
(OQ/OH = O**Q**/O**H**)

Locus of the orthologic center (ABC, QaQbQc) ?

H - Orthologic

Let ABC be a triangle, P = H = X(4) and Q a point on the Euler line.

Denote:

Bc, Cb = the orthogonal projections of B, C on HC, HB, resp.

Qa = same to Q point of the triangle ABcCb.
Similarly Qb, Qc.

ABC, QaQbQc are orthologic.

For Q = G = X(2)
Orthologic center (ABC, QaQbQc) = G* = ?
Orthologic center (QaQbQc, ABC) = G** = G of orthic = X(51)

For Q = X(3) = O:
Orthologic centers = X(4) = H

Q = H = X(4)
Orthologic center (ABC, QaQbQc) = H* = X(3) = O
Orthologic center (QaQbQc, ABC) = H** = ?

For Q = N = X(5)
Orthologic center (ABC, QaQbQc) = N* = ?
Orthologic center (QaQbQc, ABC) = N** = ?

The locus of the orthologic center (QaQbQc, ABC) = Q**, as Q moves on the Euler line, is a line. (The line {4,51})
(OQ/OH = O**Q**/O**H**)

Locus of the orthologic center (ABC, QaQbQc) ?

O - Orthologic

Let ABC be a triangle, P = O = X(3) and Q a point on the Euler line.

Denote:

Bc, Cb = the orthogonal projections of B, C on OC, OB, resp.

Qa = same to Q point of the triangle ABcCb.
Similarly Qb, Qc.

ABC, QaQbQc are orthologic.

Orthologic center (QaQbQc, ABC) = Q

For Q = G = X(2)
Orthologic center (ABC, QaQbQc) = G* = ?

For Q = X(3) = O:
Orthologic center (ABC, QaQbQc) = O* = X(72422) = X(2)X(9291)∩X(4)X(290)

For Q = H = X(4)
Orthologic center (ABC, QaQbQc) = H* = ?

For Q = N = X(5)
Orthologic center (ABC, QaQbQc) = N* = ?

Locus:
The locus of the orthologic center (ABC, QaQbQc) = Q*, as Q moves on the Euler line, is a CIRCLE

LOCI

Let ABC be a triangle and P a point.

Denote:
Bc, Cb = the orthogonal projections of B, C on PC, PB, resp.

A' = the other than A intersection the circumcircles of ABC and ABcCb
Similarly B',C'

La, B, Lc = Euler lines of A'BC, B'CA, C'AB, resp.

1. Which is the locus of P such that ABC, A'B'C' are orthologic?
O lies on the locus
Orthologic center (ABC, A'B'C') = (3) = O
Orthologic center ( A'B'C', ABC) = Χ(20)

2. Which is the locus of P such that the parallels to La,Lb, Lc through A, B, C,resp, are concurrent?
O lies on the locus.
.

Circumcenters - Orthologic

Let ABC be a triangle and P a point.

Denote:

Bc, Cb = the orthogonal projections of B, C on PC, PB, resp.

Oa = the circumcenter of ABcCb.
Similarly Ob, Oc.

Which is the locus of P such that ABC, OaObOc are orthologic?

H, O, G lie on the locus.