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(72674) = X(934)X(23865)∩X(8706)X(67038)

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

Antreas Hatzpolakis and Ercole Suppa, euclid 9755.

X(72674) lies on the circumcircle and these lines: {934, 23865}, {8706, 67038}

X(72674) = lies on all circumconics with perspector on the line {6, 62754}
X(72674) = intersection, other than A, B, C, of the circumconics: {{A,B,C,X(74),X(98)}, {A,B,C,X(36146),X(63192)}, {A,B,C,X(37139),X(61373)}}
X(72674) = trilinear pole of line {X(6), X(62754)}

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X(72675) = X(99)X(23864)∩X(101)X(4998)

Barycentrics    (a-b)*(a-c)*(a+b-c)*(a-b+c)*(a^3*b-2*a^2*b^2+a*b^3-a^3*c-b^3*c+a^2*c^2+b^2*c^2)*(a^3*b-a^2*b^2-a^3*c+2*a^2*c^2-b^2*c^2-a*c^3+b*c^3) : :  

    Antreas Hatzpolakis and Ercole Suppa, euclid 9755.

X(72675) lies on the circumcircle and these lines: {99, 23864}, {100, 67038}, {101, 4998}, {105, 34084}, {109, 1275}, {110, 4620}, {805, 17933}, {919, 39293}, {1447, 9082}, {2291, 35157}, {9436, 59019}, {46135, 53284}

X(72675) = lies on circumconic with center X(10001)
X(72675) = lies on all circumconics with perspector on the line {6, 664}
X(72675) = intersection, other than A, B, C, of the circumconics: {{A,B,C,X(74),X(98)}, {A,B,C,X(333),X(14727)}, {A,B,C,X(666),X(40419)}, {A,B,C,X(804),X(17992)}, {A,B,C,X(1275),X(4620)}, {A,B,C,X(3676),X(31286)}, {A,B,C,X(21453),X(46135)}}
X(72675) = trilinear pole of line {X(6), X(664)}
X(72675) = Collings transform of X(10001)
X(72675) = barycentric product X(i)*X(j) for these (i,j): {100, 34084}, {664, 60014}, {4569, 30627}, {4572, 59020}
X(72675) = barycentric quotient X(i)/X(j) for these (i,j): {109, 69030}, {651, 69031}, {664, 46180}, {1415, 69018}, {4554, 69032}, {30627, 3900}, {34084, 693}, {59020, 663}, {60014, 522}
X(72675) = trilinear product X(i)*X(j) for these (i,j): {101, 34084}, {651, 60014}, {658, 30627}, {4554, 59020}
X(72675) = trilinear quotient X(i)/X(j) for these (i,j): {109, 69018}, {514, 34084}, {650, 60014}, {651, 69030}, {657, 30627}, {664, 69031}, {3063, 59020}, {4554, 46180}, {4572, 69032}

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X(72676) = X(98)X(69926)∩X(22089)X(53886)

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

    Antreas Hatzpolakis and Ercole Suppa, euclid 9755.

X(72676) lies on the circumcircle and these lines: {98, 69926}, {22089, 53886}

X(72676) = lies on circumconic with center X(52585)
X(72676) = intersection, other than A, B, C, of the circumconics: {{A,B,C,X(74),X(98)}, {A,B,C,X(22089),X(42658)}}
X(72676) = Collings transform of X(52585)

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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.