ETC

X(72955) = X(2)X(99)∩X(30)X(37746)

Barycentrics    (2*a^4-5*a^2*b^2-b^4-5*a^2*c^2+10*b^2*c^2-c^4)*(4*a^4-a^2*b^2+b^4-a^2*c^2-4*b^2*c^2+c^4) ; :
X(72955) = 3*X[2]-X[11162]

Antreas Hatzipolakis and Ercole Suppa, euclid 10179.

X(72955) lies on these lines: {2, 99}, {30, 37746}, {125, 11569}, {373, 33962}, {524, 34227}, {538, 52231}, {2793, 9135}, {3066, 11159}, {3363, 5512}, {3849, 62293}, {4563, 66458}, {5077, 23699}, {8367, 51535}, {11164, 60866}, {12036, 37745}, {17952, 17968}, {35955, 63408}, {47467, 62508}

X(72955) = reflection of X(72412) in the line X(523)X(597)
X(72955) = complement of X(11162)
X(72955) = X(61071)-Dao conjugate of X(43674)
X(72955) = X(i)-reciprocal conjugate of X(j) for these {i,j}: {2030, 52230},{2793, 43674},{18800, 63854},{52229, 5503},{68455, 46144}
X(72955) = inverse of X(9877) in orthoptic circle of Steiner inellipse
X(72955) = foot of the perpendicular from X(i) to the line X(j)X(k) for these {i,j,k}: {9135, 2, 99}, {2, 9135, 14327}
X(72955) = perspector of the circumconic through X(892) and X(17937)
X(72955) = intersection, other than A, B, C, of the circumconics: {{A,B,C,X(2),X(12036)}, {A,B,C,X(99),X(18800)}, {A,B,C,X(111),X(9135)}, {A,B,C,X(671),X(2793)}, {A,B,C,X(22329),X(37745)}, {A,B,C,X(34245),X(68455)}, {A,B,C,X(63853),X(68169)}}
X(72955) = pole of the line {2793, 9877} with respect to orthoptic circle of Steiner inellipse
X(72955) = pole of the line {5468, 68169} with respect to Kiepert parabola
X(72955) = pole of the line {187, 2709} with respect to Stammler hyperbola
X(72955) = pole of the line {524, 46144} with respect to Wallace hyperbola
X(72955) = pole of the line {37745, 53798} with respect to Thomson-Gibert-Moses hyperbola
X(72955) = barycentric product X(i)*X(j) for these (i,j): {2793, 68455}, {12036, 63853}, {17952, 37745}, {22329, 52229}
X(72955) = barycentric quotient X(i)/X(j) for these (i,j): {2030, 52230}, {2793, 43674}, {18800, 63854}, {52229, 5503}, {68455, 46144}
X(72955) = trilinear product X(17959)*X(37745)

---

X(72956) = X(2)X(5684)∩X(137)X(140)

Barycentrics    (a^12-5*a^10*b^2+10*a^8*b^4-10*a^6*b^6+5*a^4*b^8-a^2*b^10-5*a^10*c^2+16*a^8*b^2*c^2-17*a^6*b^4*c^2+8*a^4*b^6*c^2-4*a^2*b^8*c^2+2*b^10*c^2+10*a^8*c^4-17*a^6*b^2*c^4+a^4*b^4*c^4+5*a^2*b^6*c^4-8*b^8*c^4-10*a^6*c^6+8*a^4*b^2*c^6+5*a^2*b^4*c^6+12*b^6*c^6+5*a^4*c^8-4*a^2*b^2*c^8-8*b^4*c^8-a^2*c^10+2*b^2*c^10)*(2*a^16-11*a^14*b^2+25*a^12*b^4-31*a^10*b^6+25*a^8*b^8-17*a^6*b^10+11*a^4*b^12-5*a^2*b^14+b^16-11*a^14*c^2+34*a^12*b^2*c^2-31*a^10*b^4*c^2-4*a^8*b^6*c^2+31*a^6*b^8*c^2-38*a^4*b^10*c^2+27*a^2*b^12*c^2-8*b^14*c^2+25*a^12*c^4-31*a^10*b^2*c^4-5*a^6*b^6*c^4+34*a^4*b^8*c^4-51*a^2*b^10*c^4+28*b^12*c^4-31*a^10*c^6-4*a^8*b^2*c^6-5*a^6*b^4*c^6-14*a^4*b^6*c^6+29*a^2*b^8*c^6-56*b^10*c^6+25*a^8*c^8+31*a^6*b^2*c^8+34*a^4*b^4*c^8+29*a^2*b^6*c^8+70*b^8*c^8-17*a^6*c^10-38*a^4*b^2*c^10-51*a^2*b^4*c^10-56*b^6*c^10+11*a^4*c^12+27*a^2*b^2*c^12+28*b^4*c^12-5*a^2*c^14-8*b^2*c^14+c^16) ; :
X(72956) = 3*X[2]+X[5684]

Antreas Hatzipolakis and Ercole Suppa, euclid 10179.

X(72956) lies on these lines: {2, 5684}, {137, 140}

X(72956) = QAP1: Quadrangle Centroid of X(5684)

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X(72957) = X(2)X(15169)∩X(3)X(10121)

Barycentrics    (a^2-b^2) (a^2-c^2) (a^16-5 a^14 b^2+10 a^12 b^4-11 a^10 b^6+10 a^8 b^8-11 a^6 b^10+10 a^4 b^12-5 a^2 b^14+b^16-8 a^14 c^2+24 a^12 b^2 c^2-26 a^10 b^4 c^2+10 a^8 b^6 c^2+10 a^6 b^8 c^2-26 a^4 b^10 c^2+24 a^2 b^12 c^2-8 b^14 c^2+28 a^12 c^4-33 a^10 b^2 c^4+7 a^8 b^4 c^4+11 a^6 b^6 c^4+7 a^4 b^8 c^4-33 a^2 b^10 c^4+28 b^12 c^4-56 a^10 c^6-16 a^8 b^2 c^6-16 a^6 b^4 c^6-16 a^4 b^6 c^6-16 a^2 b^8 c^6-56 b^10 c^6+70 a^8 c^8+89 a^6 b^2 c^8+93 a^4 b^4 c^8+89 a^2 b^6 c^8+70 b^8 c^8-56 a^6 c^10-96 a^4 b^2 c^10-96 a^2 b^4 c^10-56 b^6 c^10+28 a^4 c^12+45 a^2 b^2 c^12+28 b^4 c^12-8 a^2 c^14-8 b^2 c^14+c^16) (a^16-8 a^14 b^2+28 a^12 b^4-56 a^10 b^6+70 a^8 b^8-56 a^6 b^10+28 a^4 b^12-8 a^2 b^14+b^16-5 a^14 c^2+24 a^12 b^2 c^2-33 a^10 b^4 c^2-16 a^8 b^6 c^2+89 a^6 b^8 c^2-96 a^4 b^10 c^2+45 a^2 b^12 c^2-8 b^14 c^2+10 a^12 c^4-26 a^10 b^2 c^4+7 a^8 b^4 c^4-16 a^6 b^6 c^4+93 a^4 b^8 c^4-96 a^2 b^10 c^4+28 b^12 c^4-11 a^10 c^6+10 a^8 b^2 c^6+11 a^6 b^4 c^6-16 a^4 b^6 c^6+89 a^2 b^8 c^6-56 b^10 c^6+10 a^8 c^8+10 a^6 b^2 c^8+7 a^4 b^4 c^8-16 a^2 b^6 c^8+70 b^8 c^8-11 a^6 c^10-26 a^4 b^2 c^10-33 a^2 b^4 c^10-56 b^6 c^10+10 a^4 c^12+24 a^2 b^2 c^12+28 b^4 c^12-5 a^2 c^14-8 b^2 c^14+c^16) ; :
X(72957) = 3*X[2]-2*X[15169], 2*X[3]-X[10121], 5*X[631]-4*X[10120]

Antreas Hatzipolakis and Ercole Suppa, euclid 10179.

X(72957) lies on the circumcircle and these lines: {2, 15169}, {3, 10121}, {631, 10120}, {6345, 14140}, {33643, 47608}

X(72957) = reflection of X(i) in X(j) for these {i,j}: {10121, 3}
X(72957) = anticomplement of X(15169)
X(72957) = circumperp conjugate of X(10121)
X(72957) = X(15169)-Dao conjugate of X(15169)

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X(72958) = X(2)X(32425)∩X(30)X(11568)

Barycentrics    a^2*(a^2-b^2)*(a^2-c^2)*(2*a^6-3*a^4*b^2+5*b^6-6*a^4*c^2+9*a^2*b^2*c^2-6*a^2*c^4-3*b^2*c^4+2*c^6)*(2*a^6-6*a^4*b^2-6*a^2*b^4+2*b^6-3*a^4*c^2+9*a^2*b^2*c^2-3*b^4*c^2+5*c^6) ; :
X(72958) = 2*X[3]-X[32425]

Antreas Hatzipolakis and Ercole Suppa, euclid 10179.

X(72958) lies on the circumcircle and these lines: {3, 32425}, {30, 11568}, {98, 62294}, {111, 8705}, {352, 6323}, {353, 843}, {523, 67731}, {524, 6325}, {1499, 6236}, {2770, 11628}, {9831, 59794}, {32583, 53613}

X(72958) = reflection of X(32425) in X(3)
X(72958) = reflection of X(i) in the line X(j)X(k) for these {i,j,k}: {6325, 3, 669}, {11568, 3, 523}, {67731, 2, 3}
X(72958) = circumperp conjugate of X(32425)
X(72958) = intersection, other than A, B, C, of the circumconics: {{A,B,C,X(6),X(32583)}, {A,B,C,X(74),X(98)}, {A,B,C,X(524),X(8705)}, {A,B,C,X(2421),X(62294)}, {A,B,C,X(9124),X(64218)}}
X(72958) = pole of the line {110, 32425} with respect to orthoptic circle of Stammler hyperbola
X(72958) = pole of the line {99, 32425} with respect to orthoptic circle of Wallace hyperbola

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A CYCLOLOGIC THEOREM RELATED TO EXCENTRAL TRIANGLE.

[APH]

Excentral version

Let ABC be a triangle, IaIbIc the excentral triangle and P a point..

Denote

Pa, Pb, Pc = same to P points of IaBC, IbCA, IcAB, resp.

ABC, PaPbPc are circumcyclologic

Cyclologic center (ABC, PaPbPc) = Q = ? (on the circumcircle of ABC)
Cyclologic center (PaPbPc, ABC) = Q* = ? (on the circumcircle of PaPbPc)

[Ercole Suppa]

1. P on the Euler Line:

Q = X(100)
Locus of Q* as P moves on the Euler line: K086 2. P on the Brocard axis:

Q = X(101)

[Bernard Gibert]

If P is on a line through X(3) and a strong point M then the locus seems to be a circular cK(#X1,R) with singular focus F.

When M = X2, you get K086.

When M = X6, you get K040.

[APH]

For the case of the cyclologic center (ABC, PaPbPc) M can be any point strong or not.
That is:
Let ABC be a triangle, IaIbIc the excentral triangle, M a fixed point and P a point on the line OM.

Denote:

Pa, Pb,Pc = same to P points of IaBC,IbCA,IcAB, resp.

The triangles ABC, PaPbPc are circumcyclologic.

As P moves on the line OM:
The cyclologic center (ABC, PaPbPc) is a fixed point Q on the circumcircle.
The locus of the cyclologic center (PaPbPc, ABC) is a cubic.

Let's see the Q's. The class of the cubics is a subject of Bernard Gibert.
1. P on the Euler line
Q = X(100) = Reflection point of IO = X(1)X(3) line = Reflection point of Euler line of INTOUCH triangle (pedal triangle of I).

2. P on the Brocard axis
Q = X(101) = Reflection point of X(1)X(7) line = Reflection point of Brocard axis of INTOUCH triangle (pedal triangle of I).

Generalization

Let M be a fixed Point and P be a point on the line OM = L
Denote: P' = the same to P point of the INTOUCH triangle.
L' = the same to L line of the INTOUCH triangle.

Then Q is the reflection point of the L' line = IP' line of ABC

Note:
Reflection point of a line L:
The reflections La, Lb, Lc of L in the siedelines BC, CA, AB, resp. bound a triangle A*B*C*.
ABC, A*B*C* are perspective.The perspector, lying on the circumcircle, is called "Reflection pont of the line L"
It is the incenter (or an excenter) of the triangle A*B*C*.

Case of OM with M = I= X(1)

1. P = X(1) = O of INTOUCH triangle.
Pa, Pb, Pc = X(1) of IaBC, IbCA, IcAB, resp.
Cyclologic center (ABC, PaPbPc) = Q

2. P = X(354) = X(2) of the INTOUCH triangle
Pa, Pb, Pc = X(2) of the intouch triangles of IaBC, IbCA, IcAB, resp.
Cyclologic center (ABC, PaPbPc) = Q1

3. P = X(942) = X(5) of the INTOUCH triangle
Pa, Pb, Pc = X(5) of the intouch triangles of IaBC, IbCA, IcAB, resp.
Cyclologic center (ABC, PaPbPc) = Q2

Q, Q1, Q2 coincide.

Q is the reflection point of the IO line of the INTOUCH triangle.
It is the line passing thrpugh the incenters of ABC and the intouch triangle.

A CYCLOLOGIC THEOREM RELATED TO ORTHIC TRIANGLE

[APH]

Orthic Version

Let ABC be a triangle, HaHbHc the orthic triangle and P a point.

Denote:

Pa, Pb, Pc = same to P points of AHbHc, BHcHa, CHaHb, resp.

HaHbHc, PaPbPc are circumcyclologic

Cyclologic center (HaHbHc, PaPbPc = ? (on the circumcircle of HaHbHc = NPC)
Cyclologic center (PaPbPc, HaHbHc) = ? (on the circumcircle of PaPbPc)

[Ercole Suppa]

cyclologic center (HaHbHc, PaPbPc) = Poncelet point(isogonal conjugate(P))

cyclologic center (PaPbPc, HaHbHc) = orthoassociate(isogonal conjugate(circumcircleInverse(P)))

Euclid 10026