## Τρίτη, 9 Απριλίου 2013

1.1.

Let ABC be a triangle, A'B'C' the pedal triangle of H and A"B"C" the circumcevian triangle of H wrt A'B'C'.

Denote:

Ra = the radical axis of ((B', B'C"), (C', C'B"))

Rb = the radical axis of ((C', C'A"), (A', A'C"))

Rc = the radical axis of ((A', A'B"), (B', B'A"))

The Ra,Rb,Rc are concurrent.

Note: The radical axes of ((B", B"C'), (C", C"B')),((C", C"A'), (A", A"C')), ((A", A"B'), (B", B"A')) concur at H (all circles pass through H).

1.2.

Let ABC be a triangle, A'B'C' the pedal triangle of O and A"B"C" the circumcevian triangle of O wrt A'B'C'.

Denote:

Ra = the radical axis of ((B", B"C'), (C", C"B'))

Rb = the radical axis of ((C", C"A'), (A", A"C'))

Rc = the radical axis of ((A", A"B'), (B", B"A'))

The Ra,Rb,Rc are concurrent.

Note: The radical axes of ((B', B'C"), (C', C'B")),((C', C'A"), (A', A'C")), ((A', A'B"), (B', B'A")) concur at O (all circles pass through O).

LOCUS:

Let ABC be a triangle, P a point, A'B'C' the pedal triangle of P and A"B"C" the circumcevian triangle of P wrt A'B'C'.

Denote:

Ra = the radical axis of ((B', B'C"), (C', C'B"))

Rb = the radical axis of ((C', C'A"), (A', A'C"))

Rc = the radical axis of ((A', A'B"), (B', B'A"))

R'a = the radical axis of ((B", B"C'), (C", C"B'))

R'b = the radical axis of ((C", C"A'), (A", A"C'))

R'c = the radical axis of ((A", A"B'), (B", B"A'))

Which is the locus of P such that 1. Ra,Rb,Rc 2. R'a,R'b,R'c are concurrent?

2.1.

Let ABC be a triangle and A'B'C' the circumcevian triangle of I.

Denote:

Ra = the radical axis of ((B, BC'), (C, CB'))

Rb = the radical axis of ((C, CA'), (A, AC'))

Rc = the radical axis of ((A, AB'), (B, BA'))

The Ra, Rb, Rc are concurrent.

Note: The radical axes of ((B', B'C), (C', C'B)), ((C', C'A), (A', A'C)), ((A', A'B), (B', B'A)) concur at I (all circles pass through I)

2.2.

Let ABC be a triangle and A'B'C' the circumcevian triangle of H.

Denote:

Ra = the radical axis of ((B', B'C), (C', C'B))

Rb = the radical axis of ((C', C'A), (A', A'C))

Rc = the radical axis of ((A', A'B), (B', B'A))

The Ra, Rb, Rc are concurrent.

Note: The radical axes of ((B, BC'), (C, CB')), ((C, CA'), (A, AC')), ((A, AB'), (B, BA')) concur at H (all circles pass through H)

LOCUS:

Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.

Denote:

Ra = the radical axis of ((B, BC'), (C, CB'))

Rb = the radical axis of ((C, CA'), (A, AC'))

Rc = the radical axis of ((A, AB'), (B, BA'))

R'a = the radical axis of ((B', B'C), (C', C'B))

R'b = the radical axis of ((C', C'A), (A', A'C))

R'c = the radical axis of ((A', A'B), (B', B'A))

Which is the locus of P such that 1. Ra,Rb,Rc 2. R'a,R'b,R'c are concurrent?

Antreas P. Hatzipolakis, 9 April 2013

***Points of Concurrence*****

1.1.

P.1.1 = (-a^2+b^2+c^2) (2 a^6 b^2-3 a^4 b^4+b^8+2 a^6 c^2+4 a^4 b^2 c^2-4 b^6 c^2-3 a^4 c^4+6 b^4 c^4-4 b^2 c^6+c^8)::

Search = -1.2087990869888377496.

On lines {{3,1568},{4,110},{5,389},{52,403},{68,1173},{155,195},{185,2072},{541,3357},{1533,5073},{1614,3153},{3167,3843},{3546,4846},{3564,3850}}

Midpoint of X(4) and X(1147).

X + 2 X + X = 3 X + 2 X – X.

1.2.

P1.2 = (a^2-b^2-c^2) (a^4 b^4-2 a^2 b^6+b^8+2 a^2 b^4 c^2-4 b^6 c^2+a^4 c^4+2 a^2 b^2 c^4+6 b^4 c^4-2 a^2 c^6-4 b^2 c^6+c^8) :: = b^4 SB (SB^2-S^2)+c^4 SC (SC^2-S^2) ::

= complement X(1147) Search = 2.3145425702586469385

On lines {{2,54},{3,125},{5,389},{52,1594},{136,847},{155,1656},{156,542},{343,1216},{568,3574},{575,3564},{912,3812},{1614,3448},{1899,3549},{3167,5070}}

midpoint of P1.1 and P1.2 = X(5)

P1.2 = midpoint X(68) and X(1147)

P1.2 = 3 X + X = 3 X - X

2.1

P2.1 = a (a^6-a^5 b-2 a^4 b^2+2 a^3 b^3+a^2 b^4-a b^5-a^5 c+4 a^4 b c-a^3 b^2 c-3 a^2 b^3 c+2 a b^4 c-b^5 c-2 a^4 c^2-a^3 b c^2+4 a^2 b^2 c^2-a b^3 c^2+2 a^3 c^3-3 a^2 b c^3-a b^2 c^3+2 b^3 c^3+a^2 c^4+2 a b c^4-a c^5-b c^5)::

Search = -9.3788352311451100575

On lines {{1,104},{3,10},{4,36},{5,2829},{8,2077},{21,84},{30,3829},{35,944},{40,2975},{48,1765},{56,946},{318,1309},{411,5303},{631,5251},{995,3073},{999,3671},{1006,1490},{1071,2646},{1125,3560},{1210,1470},{1385,5248},{1457,1777},{1482,4084},{2096,3485},{3072,4257},{3149,5204},{4231,5345}}

P2.1 = Midpoint of X(1) and X(1158)

P2.1 = R X + (2 r - R) X = (r - R) X + R X = (2 r + 3 R) X + R X = 4 r X + R X - R X

2.2.

P2.2 = X(1147)

Peter J. C. Moses, 9 April 2013

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