LOCUS PROBLEM

Problem by Antreas Hatzipolakis
Solution by Francisco Javier García Capitán

ETC LISTING OF Q

X(72803)

ETC

X(72804) = X(4)X(94)∩X(32205)X(47117)

Barycentrics    a^2*(a^18*b^2-5*a^16*b^4+8*a^14*b^6-14*a^10*b^10+14*a^8*b^12-8*a^4*b^16+5*a^2*b^18-b^20+a^18*c^2-10*a^16*b^2*c^2+31*a^14*b^4*c^2-47*a^12*b^6*c^2+41*a^10*b^8*c^2-11*a^8*b^10*c^2-35*a^6*b^12*c^2+59*a^4*b^14*c^2-38*a^2*b^16*c^2+9*b^18*c^2-5*a^16*c^4+31*a^14*b^2*c^4-58*a^12*b^4*c^4+44*a^10*b^6*c^4-29*a^8*b^8*c^4+68*a^6*b^10*c^4-111*a^4*b^12*c^4+85*a^2*b^14*c^4-25*b^16*c^4+8*a^14*c^6-47*a^12*b^2*c^6+44*a^10*b^4*c^6-32*a^6*b^8*c^6+84*a^4*b^10*c^6-77*a^2*b^12*c^6+20*b^14*c^6+41*a^10*b^2*c^8-29*a^8*b^4*c^8-32*a^6*b^6*c^8-48*a^4*b^8*c^8+25*a^2*b^10*c^8+26*b^12*c^8-14*a^10*c^10-11*a^8*b^2*c^10+68*a^6*b^4*c^10+84*a^4*b^6*c^10+25*a^2*b^8*c^10-58*b^10*c^10+14*a^8*c^12-35*a^6*b^2*c^12-111*a^4*b^4*c^12-77*a^2*b^6*c^12+26*b^8*c^12+59*a^4*b^2*c^14+85*a^2*b^4*c^14+20*b^6*c^14-8*a^4*c^16-38*a^2*b^2*c^16-25*b^4*c^16+5*a^2*c^18+9*b^2*c^18-c^20) : :
X(72804) = 2*X[32205]-X[54073]

Antreas Hatzipolakis and Ercole Suppa, euclid 9899.

X(72804) lies on these lines: {4, 94}, {32205, 47117}

X(72804) = reflection of X(54073) in X(32205)

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X(72805) = X(4)X(94)∩X(54)X(125)

Barycentrics    (a^2+b^2-c^2)*(a^2-b^2+c^2)*(a^12-4*a^10*b^2+7*a^8*b^4-8*a^6*b^6+7*a^4*b^8-4*a^2*b^10+b^12-4*a^10*c^2+8*a^8*b^2*c^2-4*a^6*b^4*c^2-2*a^4*b^6*c^2+4*a^2*b^8*c^2-2*b^10*c^2+7*a^8*c^4-4*a^6*b^2*c^4-a^4*b^4*c^4-b^8*c^4-8*a^6*c^6-2*a^4*b^2*c^6+4*b^6*c^6+7*a^4*c^8+4*a^2*b^2*c^8-b^4*c^8-4*a^2*c^10-2*b^2*c^10+c^12) : :
X(72805) = 3*X[2]-2*X[54073], X[3448]+X[59493], 2*X[125]-X[58881]

Antreas Hatzipolakis and Ercole Suppa, euclid 9899.

X(72805) lies on these lines: {2, 54073}, {4, 94}, {54, 125}, {67, 45972}, {68, 5963}, {74, 21659}, {110, 5449}, {136, 68638}, {186, 32423}, {399, 16868}, {526, 562}, {542, 19128}, {1209, 27866}, {1511, 2888}, {1594, 2914}, {1899, 32607}, {2131, 5373}, {3269, 13527}, {3410, 14643}, {3520, 10264}, {6288, 11561}, {6699, 51033}, {7505, 14683}, {7547, 12165}, {7577, 15087}, {7731, 18381}, {9140, 15463}, {9927, 12270}, {10628, 25739}, {11264, 15089}, {11442, 68453}, {11457, 12244}, {11562, 58922}, {11565, 12041}, {11597, 34826}, {11801, 54001}, {12227, 15081}, {12281, 67903}, {12308, 35488}, {12902, 34797}, {13619, 17702}, {14448, 32365}, {14644, 18388}, {15027, 32136}, {16223, 41171}, {17506, 34153}, {19379, 34118}, {19481, 35482}, {19504, 52295}, {20304, 36153}, {22584, 72680}, {23306, 56292}, {25330, 39588}, {35471, 64183}

X(72805) = midpoint of X(3448) and X(59493)
X(72805) = reflection of X(58881) in X(125)
X(72805) = anticomplement of X(54073)
X(72805) = orthoassociate of X(143)
X(72805) = inverse of X(32140) in anticomplementary circle
X(72805) = inverse of X(18379) in Johnson triangle circumcircle
X(72805) = inverse of X(143) in polar circle
X(72805) = {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): {125, 3043, 6143}, {125, 10114, 54}, {146, 3448, 32140}, {265, 7722, 4}, {265, 7728, 18379}, {3448, 59493, 5663}, {33565, 43808, 125}

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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