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(72486) = X(3)X(1232)∩X(140)X(184)

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

Antreas Hatzipolakis and Ercole Suppa, euclid 9624.

X(72486) lies on these lines: {3, 1232}, {25, 44732}, {32, 1595}, {140, 184}, {2200, 21012}, {3147, 40352}, {3541, 54034}, {10547, 14786}, {11411, 42065}, {17714, 52153}

X(72486) = isogonal conjugate of X(64051)
X(72486) = lies on circumconics with center X(i) for i in {11792, 17423}
X(72486) = lies on all circumconics with perspector on the line {3049, 55280}
X(72486) = intersection, other than A, B, C, of the circumconics: {{A,B,C,X(2),X(1595)}, {A,B,C,X(3),X(25)}, {A,B,C,X(4),X(140)}, {A,B,C,X(5),X(66)}, {A,B,C,X(6),X(13336)}, {A,B,C,X(24),X(34436)}, {A,B,C,X(30),X(3147)}, {A,B,C,X(54),X(2980)}, {A,B,C,X(64),X(96)}, {A,B,C,X(67),X(847)}}
X(72486) = trilinear pole of line {X(3049), X(55280)}
X(72486) = barycentric quotient X(6)/X(64051)

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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.