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(72655) = X(7607)X(61972)∩X(7608)X(62032)

Barycentrics    (77*a^2+77*b^2-67*c^2)*(77*a^2-67*b^2+77*c^2) : :

Antreas Hatzipolakis and Ercole Suppa, euclid 9681.

X(72655) lies on the Kiepert circumhyperbola and these lines: {7607, 61972}, {7608, 62032}, {10155, 15640}, {10185, 69402}, {14494, 62018}, {15683, 53098}, {17578, 60332}, {50687, 60330}, {50689, 60334}, {50693, 60144}, {52519, 62002}, {53100, 61992}, {53103, 61966}, {54637, 63061}, {60123, 61954}, {60142, 62005}, {60322, 61989}, {60337, 61985}

X(72655) = isogonal conjugate of X(72656)
X(72655) = intersection, other than A, B, C, of the circumconics: {{A,B,C,X(2),X(4)}, {A,B,C,X(52281),X(62032)}, {A,B,C,X(52282),X(61972)}, {A,B,C,X(62018),X(62950)}, {A,B,C,X(63061),X(63064)}}

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X(72656) = X(3)X(6)∩X(3054)X(61806)

Barycentrics    a^2*(67*a^2-77*b^2-77*c^2) : :

Antreas Hatzipolakis and Ercole Suppa, euclid 9681.

X(72656) lies on these lines: {3, 6}, {3054, 61806}, {3055, 62120}, {3815, 62072}, {3832, 11742}, {11812, 43619}, {15702, 44541}, {18424, 61847}, {18584, 62155}, {31415, 62106}, {31489, 62094}, {37637, 61796}, {41196, 41197}, {41983, 44526}, {43457, 62140}, {43618, 62089}, {44519, 61816}, {53418, 62096}, {53419, 61817}, {62081, 62993}

X(72656) = isogonal conjugate of Χ(72655)
X(72656) = pole of the line {140, 63027} with respect to Terzić 1st hyperbola
X(72656) = {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): {3, 15602, 5210}, {6433, 6434, 55711}, {8588, 15815, 6}, {8589, 15655, 53095}, {11480, 11481, 55701}

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X(72657) = X(43537)X(62024)∩X(53859)X(62136)

Barycentrics    (77*a^2+77*b^2-97*c^2)*(77*a^2-97*b^2+77*c^2) : :

Antreas Hatzipolakis and Ercole Suppa, euclid 9681.

X(72657) lies on the Kiepert circumhyperbola and these lines: {43537, 62024}, {53859, 62136}

X(72657) = isogonal conjugate of X(72658)
X(72657) = intersection, other than A, B, C, of the circumconics: {{A,B,C,X(2),X(4)}, {A,B,C,X(62024),X(62955)}}

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X(72658) = X(3)X(6)∩X(3054)X(49133)

Barycentrics    a^2*(97*a^2-77*b^2-77*c^2) : :

Antreas Hatzipolakis and Ercole Suppa, euclid 9681.

X(72658) lies on these lines: {3, 6}, {3054, 49133}, {5355, 58189}, {15484, 61796}, {18584, 61862}, {41196, 41197}, {43291, 62096}, {43448, 62089}, {43618, 61878}, {43620, 62140}, {53418, 61847}, {62106, 62992}

X(72658) = isogonal conjugate of Χ(72657)

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X(72659) = (name pending)

Barycentrics    (77*a^2+77*b^2-82*c^2)*(77*a^2-82*b^2+77*c^2) : :

Antreas Hatzipolakis and Ercole Suppa, euclid 9681.

X(72659) lies on the Kiepert circumhyperbola and these lines: { }

X(72659) = isogonal conjugate of X(72660)
X(72659) = intersection, other than A, B, C, of the circumconics: {{A,B,C,X(2),X(4)}, {A,B,C,X(11588),X(43713)}}

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X(72660) = X(3)X(6)∩X(3543)X(11614)

Barycentrics    a^2*(82*a^2-77*b^2-77*c^2) : :

Antreas Hatzipolakis and Ercole Suppa, euclid 9681.

X(72660) lies on these lines: {3, 6}, {3543, 11614}, {7603, 61796}, {18424, 62096}, {41196, 41197}, {46332, 53419}

X(72660) = isogonal conjugate of Χ(72659)
X(72660) = {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): {187, 15602, 5041}, {5007, 15603, 187}

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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.
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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.