LOCUS PROBLEM

Problem by Antreas Hatzipolakis
Solution by Francisco Javier García Capitán

ETC LISTING OF Q

X(72803)

ETC

X(72803) = X(3)X(523)∩X(5)X(525)

Barycentrics    -((b^2-c^2)*(-a^8+3*a^6*b^2-2*a^4*b^4-a^2*b^6+b^8+3*a^6*c^2-5*a^4*b^2*c^2+3*a^2*b^4*c^2-b^6*c^2-2*a^4*c^4+3*a^2*b^2*c^4-a^2*c^6-b^2*c^6+c^8)) : :
X(72803) = 3*X[2]-X[5489], X[3]-3*X[5664], 5*X[3]-3*X[18556], 5*X[5664]-X[18556], X[4]-3*X[65754], X[20]+3*X[65714], 2*X[140]-3*X[45681], 5*X[631]-3*X[65723], X[684]+X[44427], 3*X[2394]-7*X[3090], X[16230]+X[41077], 5*X[3091]-3*X[42733], 5*X[3091]+3*X[63248], X[42733]+X[63248], X[3146]-3*X[58346], 7*X[3523]-3*X[53383], 4*X[3628]-3*X[14566], 4*X[3850]-3*X[39491], X[9409]+X[65871], 3*X[10278]-2*X[23105], 3*X[14223]+X[23235], X[38664]-3*X[42738], X[41079]-2*X[45259], 3*X[42731]-X[53345], 2*X[44818]-X[68791]

Antreas Hatzipolakis and Francisco Javier García Capitán, euclid 9884.

X(72803) lies on these lines: {2, 5489}, {3, 523}, {4, 65754}, {5, 525}, {20, 65714}, {30, 68412}, {140, 45681}, {194, 33294}, {389, 520}, {512, 46626}, {526, 25711}, {550, 62510}, {631, 65723}, {648, 54057}, {684, 44427}, {690, 38745}, {826, 32348}, {850, 26166}, {924, 68026}, {1075, 57065}, {2394, 3090}, {2435, 14542}, {2797, 16230}, {3091, 42733}, {3146, 58346}, {3265, 7763}, {3520, 22089}, {3523, 53383}, {3566, 5878}, {3628, 14566}, {3767, 6587}, {3850, 39491}, {3906, 3934}, {3926, 62555}, {5013, 62384}, {5649, 47293}, {8029, 35922}, {8057, 66762}, {9007, 63722}, {9033, 15774}, {9409, 65871}, {9815, 63249}, {10190, 55308}, {10278, 11007}, {11413, 53330}, {14223, 23235}, {18560, 59932}, {22467, 39201}, {23301, 52532}, {24904, 69318}, {30221, 51262}, {38664, 42738}, {39228, 43615}, {39265, 66077}, {41079, 45259}, {42731, 53345}, {44818, 68791}, {46371, 52624}, {58342, 63640}, {58757, 59424}, {59422, 66124}, {59744, 68470}

Χ(72803) = midpoint of X(i) and X(j) for these {i,j}: {684, 44427}, {9409, 65871}, {16230, 41077}, {42733, 63248}
Χ(72803) = reflection of X(i) in X(j) for these {i,j}: {41079, 45259}, {68791, 44818}
Χ(72803) = complement of X(5489)
Χ(72803) = reflection of X(i) in X(j)X(k) for these {i,j,k}}: {68412, 140, 523}
Χ(72803) = foot of the perpendicular from X(i) to the line X(j)X(k) for these {i,j,k}: {68412, 3, 5664}
Χ(72803) = perspector of the circumconic through X(1972) and X(2986)
Χ(72803) = intersection, other than A, B, C, of the circumconics: {{A,B,C,X(14542),X(51960)}, {A,B,C,X(15328),X(60036)}, {A,B,C,X(15454),X(47304)}, {A,B,C,X(40804),X(66078)}, {A,B,C,X(52772),X(56683)}}
Χ(72803) = center of circle {X(i),X(j),X(k)} for these {i,j,k}: {2, 376, 67222}, {4, 20, 185}, {107, 2693, 70067}, {110, 477, 3258}, {112, 132, 2697}, {113, 6033, 11562}, {114, 36471, 53737}, {125, 147, 1113}, {381, 3534, 67217}, {974, 12131, 67281}, {3184, 16177, 53757}, {6759, 18381, 41725}, {9409, 32119, 65871}, {9840, 35099, 37425}, {33813, 38613, 51872}
Χ(72803) = pole of line {2072, 38743} with respect to the Droz-Farny 1st circle
Χ(72803) = pole of line {1503, 2072} with respect to the nine-point circle
Χ(72803) = pole of line {403, 6761} with respect to the polar circle
Χ(72803) = pole of line {1503, 18859} with respect to the Stammler circles radical circle
Χ(72803) = pole of line {1503, 23236} with respect to the Steiner 1st circle
Χ(72803) = pole of line {1503, 5055} with respect to the Warren G-circle
Χ(72803) = pole of tripolar of X(44766) with respect to the Warren H-circle
Χ(72803) = pole of line {526, 41673} with respect to the Kiepert parabola
Χ(72803) = pole of line {15329, 39138} with respect to the Stammler hyperbola
Χ(72803) = pole of line {323, 3331} with respect to the Steiner circumellipse
Χ(72803) = pole of line {249, 297} with respect to the Steiner inellipse
Χ(72803) = pole of line {2071, 67093} with respect to the orthoptic circle of 1st DrozFarny circle
Χ(72803) = pole of line {1503, 37938} with respect to the orthoptic circle of nine-point circle
Χ(72803) = pole of line {1503, 18403} with respect to the orthoptic circle of MacBeath inconic
Χ(72803) = pole of line {230, 3284} with respect to the dual conic of DeLongchamps circle
Χ(72803) = pole of line {44888, 62338} with respect to the dual conic of polar circle
Χ(72803) = pole of line {6130, 32193} with respect to the dual conic of Wallace hyperbola
Χ(72803) = {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): {16230, 41077, 2797}

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FOUR POINTS ON THE EULER LINE

Problem by Antreas Hatzipolakis
Solution by Francisco Javier García Capitán
FOUR POINTS ON THE EULER LINE

ETC LISTINGS

X(72698)

X(72699)

X(72700)

X(72701)

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?

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