MOBY LOCI PROBLEMS and REWARD (PRIZE)

Let ABC be a triangle, P a point and PaPbPc the pedal triangle of P.

Denote:

Pab, Pac = the orthogonal projections of Pa on OB,OC, resp.

(N1) = the NPC of PaPabPac. Similarly (N2),(N3)

Ra = the radical axis of (N2),(N3. Similarly Rb, Rc.

Sa = the parallel to Ra through A. Similarly Sb, Sc

1. Which is the locus of P such that Sa,Sb,Sc are concurrent? The Euler Line + + ??

2. Let P be a point on the Euler Line.

2.1. Which is the locus of the radical center P' of (N1),(N2),(N3) [point of concurrence of Ra,Rb,Rc] as P moves on the Euler line?

2.2. Which is the locus of the point of concurrence P" of Sa,Sb,Sc (if concur) as P moves on the Euler line?

If the loci are new I name them 1st MOBY Locus and 2nd MOBY Locus and the points P',P" as P'-Moby Point and P"-Moby point.

REWARD:

For a complete solution I offer the rare book of J. Neuberg, Sur les projections et contre-projections d' un triangle fixe, Bruxelles 1890.

Antreas P. Hatzipolakis, 4 September 2014

************************************

1: Euler line and this conic:

2 a^2 (a^2 - b^2 - c^2) (a^6 b^2 - 3 a^4 b^4 + 3 a^2 b^6 - b^8 + a^6 c^2 + 2 a^2 b^4 c^2 - 3 b^6 c^2 - 3 a^4 c^4 + 2 a^2 b^2 c^4 + 8 b^4 c^4 + 3 a^2 c^6 - 3 b^2 c^6 - c^8) x^2 + (3 a^12 - 11 a^10 b^2 + 15 a^8 b^4 - 10 a^6 b^6 + 5 a^4 b^8 - 3 a^2 b^10 + b^12 - 11 a^10 c^2 + 21 a^8 b^2 c^2 + 6 a^6 b^4 c^2 - 33 a^4 b^6 c^2 + 19 a^2 b^8 c^2 - 2 b^10 c^2 + 15 a^8 c^4 + 6 a^6 b^2 c^4 + 56 a^4 b^4 c^4 - 16 a^2 b^6 c^4 - b^8 c^4 - 10 a^6 c^6 - 33 a^4 b^2 c^6 - 16 a^2 b^4 c^6 + 4 b^6 c^6 + 5 a^4 c^8 + 19 a^2 b^2 c^8 - b^4 c^8 - 3 a^2 c^10 - 2 b^2 c^10 + c^12) y z + cyclic

2.1: A nasty cubic.

2.2: circumconic thru X(3519)

(b^2-c^2) (a^2-b^2-c^2) (a^8-3 a^6 b^2+4 a^4 b^4-3 a^2 b^6+b^8-3 a^6 c^2-11 a^4 b^2 c^2+3 a^2 b^4 c^2-4 b^6 c^2+4 a^4 c^4+3 a^2 b^2 c^4+6 b^4 c^4-3 a^2 c^6-4 b^2 c^6+c^8) y z + cyclic

Peter Moses 4 September 2014

Suppose we parameterize a point on the Euler lines as a^2 SA + k SB SC::, then the concurrence is

1 / (a^2 SA (S^2 + 5 SA^2) + k (3 S^2 - SA^2) SB SC)::

Thus:

1): L, k = -1

concurrence = 1/(a^2 SA (S^2+5 SA^2)-(3 S^2-SA^2) SB SC)::

2): O, k = 0

concurrence = 1/(a^2 SA (S^2+5 SA^2)):: on lines {{4,3521},{93,403},...}

3): G, k = 1

concurrence = 1/(a^2 SA (S^2+5 SA^2)+(3 S^2-SA^2) SB SC)::

4): N, k = 2

concurrence = 1/(a^2 SA (S^2+5 SA^2)+2 (3 S^2-SA^2) SB SC)::

5): H, k = Infinity

concurrence = 1/((3 S^2-SA^2) SB SC):: = X(3519).

6): Schiffler, k = R/(r+R)

concurrence = 1/(a^2 (r+R) SA (S^2+5 SA^2)+R (3 S^2-SA^2) SB SC)::

Peter Moses 5 September 2014

ADDENDUM:

1. = X(34223)
2. = X(15424)

EULER LINES OF TRIAGLES BOUNDED BY REFLECTED PARALLEL LINES

Dao Thanh Oai:

Let ABC be a triangle and L1,L2,L3 three parallel lines through A,B,C respectively. The reflections of L1,L2,L3 in BC,CA,AB, resp. bound a triangle A1B1C1. The Euler line of A1B1C1 passes through a fixed point (as the three lines L1,L2,L3 move around A,B,C, being parallel)

Francisco Javier García Capitán:

The point is (f(a,b,c):f(b,c,a):f(c,a,b)) where f(a,b,c) is

a^22 - 8 a^20 b^2 + 28 a^18 b^4 - 56 a^16 b^6 + 70 a^14 b^8 - 56 a^12 b^10 + 28 a^10 b^12 - 8 a^8 b^14 + a^6 b^16 - 8 a^20 c^2 + 42 a^18 b^2 c^2 - 92 a^16 b^4 c^2 + 106 a^14 b^6 c^2 - 62 a^12 b^8 c^2 + 7 a^10 b^10 c^2 + 13 a^8 b^12 c^2 - 8 a^6 b^14 c^2 + 4 a^4 b^16 c^2 - 3 a^2 b^18 c^2 + b^20 c^2 + 28 a^18 c^4 - 92 a^16 b^2 c^4 + 113 a^14 b^4 c^4 - 62 a^12 b^6 c^4 + 17 a^10 b^8 c^4 - 9 a^8 b^10 c^4 + 5 a^6 b^12 c^4 - 6 a^4 b^14 c^4 + 13 a^2 b^16 c^4 - 7 b^18 c^4 - 56 a^16 c^6 + 106 a^14 b^2 c^6 - 62 a^12 b^4 c^6 + 4 a^10 b^6 c^6 + 4 a^8 b^8 c^6 + 8 a^6 b^10 c^6 - 6 a^4 b^12 c^6 - 18 a^2 b^14 c^6 + 20 b^16 c^6 + 70 a^14 c^8 - 62 a^12 b^2 c^8 + 17 a^10 b^4 c^8 + 4 a^8 b^6 c^8 - 12 a^6 b^8 c^8 + 8 a^4 b^10 c^8 + 3 a^2 b^12 c^8 - 28 b^14 c^8 - 56 a^12 c^10 + 7 a^10 b^2 c^10 - 9 a^8 b^4 c^10 + 8 a^6 b^6 c^10 + 8 a^4 b^8 c^10 + 10 a^2 b^10 c^10 + 14 b^12 c^10 + 28 a^10 c^12 + 13 a^8 b^2 c^12 + 5 a^6 b^4 c^12 - 6 a^4 b^6 c^12 + 3 a^2 b^8 c^12 + 14 b^10 c^12 - 8 a^8 c^14 - 8 a^6 b^2 c^14 - 6 a^4 b^4 c^14 - 18 a^2 b^6 c^14 - 28 b^8 c^14 + a^6 c^16 + 4 a^4 b^2 c^16 + 13 a^2 b^4 c^16 + 20 b^6 c^16 - 3 a^2 b^2 c^18 - 7 b^4 c^18 + b^2 c^20

Reference: Facebook Group "Short Mathematical Idea"

Let ABC be a triangle and L1,L2,L3 three parallel lines through A,B,C, resp. and let A'B'C' be the triangle bounded by the reflections of L1,L2,L3, in BC,CA,AB, resp.

Where the lines L1,L2,L3 move around A,B,C, being parallel, the Euler line of A'B'C' passing through a point Q which we shall call the Parry-Pohoata point. Barycentric coordinates for Q, of degree 22 in a,b,c, were found by J. F. Garcia Captitán (Hyacinthos #15827, Nov. 19, 2007) and are included in Pohoata's article Cosmin Pohoata, "On the Parry reflection point," Forum Geometricorum 8 (2008), 43-48

Angel Montesdeoca, Hyacinthos #22171

Conjecture:

Let A'B'C' be the antimedial triangle of ABC. The reflections of the parallel lines L1,L2,L3 trough A,B,C, resp. in the sidelines of A'B'C' (instead of ABC) bound a triangle whose the Euler line passes through a fixed point.

In general:

Let ABC, A'B'C' be two homothetic triangles. Let L1,L2,L3 be three parallel lines through A,B,C, resp. The reflections of L1,L2,L3 in the sidelines B'C',C'A',A'B', resp. of A'B'C' bound a triangle whose the Euler line passes through a fixed point.

Antreas P. Hatzipolakis, 21 April 2014

Special case:

If A’B’C’ is the antimedial triangle of ABC, the fixed point on Euler lines has trilinear coordinates:

(2*f(6)+6*f(4)+14*f(2)+7)*g(1) -2*(2*f(5)+2*f(3)+5*f(1))*g(2) +(2*f(2)+2*f(4)+3)*g(3) -2*(f(3)+f(1))*g(4) -(f(7)+f(5)+6*f(3)+6*f(1)) : :

Where f(n)=cos(n*A) and g(n)=cos(n*(B-C))

This point is on line (1141,3484) and has ETC-6-9-13 numbers:

(4.923838044604385591130, 4.248841971282335390013, -1.573382134182338747412)

César Lozada, Hyacinthos #22165

Radical axes:

Let (Oa), (Na) be the circumcenter, NPC center, resp. of the triangle A1B1C1 (=the triangle bounded by the reflections in BC,CA,AB of the parallels through A,B,C, resp.).

Conjecture:

The radical axis of (Oa), (Na) passes through a fixed point.

Generalization:

Let ABC, A'B'C' be two homothetic triangles. Let L1,L2,L3 be three parallel lines through A,B,C, resp. The reflections of L1,L2,L3 in the sidelines B'C',C'A',A'B', resp. of A'B'C' bound a triangle A1B1C1

Let (Oa), (Na) be the circumcenter, NPC center of A1B1C1. The radical axis of (Oa),(Na) passes through a fixed point.

Antreas P. Hatzipolakis, 22 April 2014

The point Y is (f(a,b,c):f(b,c,a):f(c,a,b)) where f(a,b,c) is

a^10(b^2+c^2) -a^8(3b^4+4b^2c^2+3c^4) +a^6(2b^6+5b^4c^2+5b^2c^4+2c^6) +2a^4(b^8-3b^6c^2+b^4c^4-3b^2c^6+c^8) -a^2(b^2-c^2)^2(3b^6-4b^4c^2-4b^2c^4+3c^6) +(b^2-c^2)^6+2a^4(b^8-3b^6c^2+b^4c^4-3b^2c^6+c^8)

Angel Montesdeoca, Hyacinthos #22171 and in HG

CONJECTURES:

Let ABC be a triangle and L1,L2,L3 three parallel lines through A,B,C, resp. and A'B'C' the triangle bounded by the reflections of L1,L2,L3 in the sidelines BC,CA,AB of ABC, resp.

Conjecture 1.

Let P be a fixed point. As the three lines L1,L2,L3 move around A,B,C, being parallel, the point P wrt triangle A'B'C' moves on a fixed circle. Note: We have seen the cases of the Euler lines and the radical axes of (O) and (N). Conjecture 2.

Let L be a fixed line. As the lines L1,L2,L3 move around A,B,C, being parallel, the line line L wrt A'B'C' passes through a fixed point (the envelope of the lines is a degenerated circle).

Problem: Which are the envelopes of fixed circles of A'B'C' (circumcircle, NPC, incircle ...) ?

Antreas P. Hatzipolakis, 25 April 2014

RADICAL AXES

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

Denote:

(Oa) = the circumcircle of (PBC)

(O1) = the reflection of (Oa) in BC

(O'1) = the reflection of (O1) in PA'

Ra = the radical axis of the pedal circle of P and (Oa)

R1 = the radical axis of the pedal circle of P and (O1)

R'1 = the radical axis of the pedal circle of P and (O'1)

Similarly Rb,Rc, R2,R3, R'2,R'3

Sa = the radical axis of the antipedal circle of P and (Oa)

S1 = the radical axis of the antipedal circle of P and (O1)

S'1 = the radical axis of the antipedal circle of P and (O'1)

Similarly Sb,Sc, S2,S3, S'2,S'3

Ab, Ac =

1. the orthogonal projections of Oa on Rb,Rc, resp.

APH, Hyacinthos #22150

2. the orthogonal projections of Oa on R2,R3, resp.

APH, Hyacinthos #22153

3. the orthogonal projections of Oa on R'2,R'3, resp.

APH, Hyacinthos #22153

4. the orthogonal projections of Oa on Sb,Sc, resp.

5. the orthogonal projections of Oa on S2,S3, resp.

6. the orthogonal projections of Oa on S'2,S'3, resp.

Similarly Bc,Ba and Ca, Cb

Which is locus of P such that the perpendicular bisectors of the line segments BaCa, CbAb,AcBc are concurrent?

1.

The perpendicular bisectors of the line segments BaCa, CbAb, AcBc are concurrent for all P.

If P=(x:y:z), the perpendicular bisectors concurrent in Q with first barycentric coordinate:

a^6y^2z^2(4x^2+4y*z+3x(y+z)) + a^4x*y*z(c^2y(5x^2+2x*y+2(y-z)z) + b^2z(5x^2+2x z+2y(-y+z))) - a^2x(c^4y^2(z^2(y+z)+2x*z(3y+2z)+2x^2(y+3z)) + b^4z^2(y^2(y+z)+2x^2(3y+z)+2x*y(2y+3z)) + 2b^2c^2y z(4x^2(y+z)-y*z(y+z) + x(y^2+4y*z+z^2))) + (b^2-c^2)x^3(c^4y^2(2y-z)+3b^2c^2y(y-z)z + b^4(y-2z)z^2)

The unique pairs of points {P, Q}, both being in ETC are, {X(1),X(3)} and {X(4),X(546)}.

If P=(x:y:z) lies on the circumcircle or line at infinity, the construction does not make sense.

However, as the coordinates sum of Q is 4(a-b-c)(a+b-c)(a-b+c)(a+b+c)xyz(x+y+z)(c^2xy+b^2xz+a^2yz), then Q lies on the line at infinity.

In fact, if P is in the circumcircle then Q is the isogonal conjugate of P.

Angel Montesdeoca, Hyacinthos #22152

NPCs. SEQUENCE OF POINTS

Let ABC be a triangle and P a point.

Denote:

A1, B1, C1 = The NPC centers of PBC, PCA, PAB, resp.

r11,r12,r13 = the radical axes of the NPCs: ((B1),(C1)), ((C1),(A1)), ((A1),(B1)), resp.

R1 = the point of concurrence of r11,r12,r13 [Radical center of the circles. It is the Poncelet point of P wrt ABC, lying on the NPC of ABC]

f11, f12, f13 = the parallels to r11, r12, r13 through A,B,C, resp.

The lines f11, f12, f13 concur at a point F1.

F1 is the reflection of P in R1.

Locus:

As P moves on a line (the Euler line, for example) which is the locus of F1 ?

Sequences of Ri, Fi:

Denote:

A2,B2,C2 = the NPC centers of A1BC, B1CA, C1AB, resp.

r21,r22,r23 = the radical axes of the NPC: ((B2),(C2)), ((C2),(A2)), ((A2),(B2)), resp.

R2 = the point of concurrence of r21,r22,r23

f21, f22, f23 = the parallels to r21, r22, r23 through A,B,C, resp.

The lines f21, f22, f23 concur at a point F2.

Similarly R3,R4...., Rn and F3,F4,.... Fn.

Where are lying the points Ri, Fi ?

Special points P: P = N, I

Antreas P. Hatzipolakis, 9 May 2013

RADICAL AXES OF NPCs

Let ABC be a triangle and P,P* two isogonal conjugate points.

Denote:

Ra = the radical axis of the circles (NPC of PBC), (NPC of P*BC)

Rb = the radical axis of the circles (NPC of PCA), (NPC of P*CA)

Rc = the radical axis of the circles (NPC of PAB), (NPC of P*AB)

Ra, Rb, Rc are concurrent.

Antreas P. Hatzipolakis, 12 April 2013

CONCURRENT RADICAL AXES

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[3] + 2 X[4] + X[155] = 3 X[2] + 2 X[3] – X[68].

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[2] + X[68] = 3 X[2] - X[1147]

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[1] + (2 r - R) X[104] = (r - R) X[3] + R X[10] = (2 r + 3 R) X[21] + R X[84] = 4 r X[3] + R X[8] - R X[20]

2.2.

P2.2 = X(1147)

Peter J. C. Moses, 9 April 2013

***************************************************

ETC X5448, X5449, X5450

Radical Axes and Loci

Let ABC be a triangle, L a line and A',B',C' the orthogonal projections of A,B,C on L.

Denote:

R1 = the radical axis of (B, BB'), (C, CC')

[ie the radical axis of the circles centered at B,C with radii BB',CC', resp.]

R2 = the radical axis of (C,CC'), (A,AA')

R3 = the radical axis of (A,AA'), (B,BB')

R1, R2, R3 are concurrent at a point R(L) = the Radical Center of the Circles (A,AA'), (B,BB'), (C, CC').

Loci:

1. Let P be a point and L a line passing through P.

Which is the locus of R(L) points as L moves around P?

2. Let P be a point on the circumcircle and L the Simson line of P.

Which is the locus of R(L) as P moves on the circumcircle?

3. Let P be a point and La,Lb,Lc three lines passing trrough P ( i) parallels or (ii) perpendiculars to BC,CA,AB, resp.

Which is the locus of P such that the triangles ABC, R(La)R(Lb)R(Lc) are perspective?

Antreas P. Hatzipolakis, Hyacinthos #21815

-----------------------

1. It is an ellipse centered at the midpoint of segment OP. The ellipse degenerates when P is on NPC.

2. It is the Steiner deltoid of the medial triangle.

3 (i) A hyperbola similar to Jerabek hyperbola, but through O and and the nine point center. It intersects the Jerabek hyperbola at the isogonal conjugate of X3523.

(ii) It is the Darboux cubic of the medial triangle.

Francisco Javier, Hyacinthos #21819

The ratio squared of the homothety that carries Jerabek hyperbola into the hyperbola in 3(i) is

(p^2 - r^2 - 4 r R + 8 R^2)/(4 (p - r - 2 R) (p + r + 2 R)).

Francisco Javier, Hyacinthos #21826

Equation of hyperbola 3(i):

5 a^4 b^4 c^2 x^2 - 6 a^2 b^6 c^2 x^2 + b^8 c^2 x^2 - 5 a^4 b^2 c^4 x^2 - 3 b^6 c^4 x^2 + 6 a^2 b^2 c^6 x^2 + 3 b^4 c^6 x^2 - b^2 c^8 x^2 + a^8 c^2 x y + 2 a^6 b^2 c^2 x y - 2 a^2 b^6 c^2 x y - b^8 c^2 x y - 3 a^6 c^4 x y - 3 a^4 b^2 c^4 x y + 3 a^2 b^4 c^4 x y + 3 b^6 c^4 x y + 3 a^4 c^6 x y - 3 b^4 c^6 x y - a^2 c^8 x y + b^2 c^8 x y - a^8 c^2 y^2 + 6 a^6 b^2 c^2 y^2 - 5 a^4 b^4 c^2 y^2 + 3 a^6 c^4 y^2 + 5 a^2 b^4 c^4 y^2 - 3 a^4 c^6 y^2 - 6 a^2 b^2 c^6 y^2 + a^2 c^8 y^2 - a^8 b^2 x z + 3 a^6 b^4 x z - 3 a^4 b^6 x z + a^2 b^8 x z - 2 a^6 b^2 c^2 x z + 3 a^4 b^4 c^2 x z - b^8 c^2 x z - 3 a^2 b^4 c^4 x z + 3 b^6 c^4 x z + 2 a^2 b^2 c^6 x z - 3 b^4 c^6 x z + b^2 c^8 x z - a^8 b^2 y z + 3 a^6 b^4 y z - 3 a^4 b^6 y z + a^2 b^8 y z + a^8 c^2 y z - 3 a^4 b^4 c^2 y z + 2 a^2 b^6 c^2 y z - 3 a^6 c^4 y z + 3 a^4 b^2 c^4 y z + 3 a^4 c^6 y z - 2 a^2 b^2 c^6 y z - a^2 c^8 y z + a^8 b^2 z^2 - 3 a^6 b^4 z^2 + 3 a^4 b^6 z^2 - a^2 b^8 z^2 - 6 a^6 b^2 c^2 z^2 + 6 a^2 b^6 c^2 z^2 + 5 a^4 b^2 c^4 z^2 - 5 a^2 b^4 c^4 z^2

Francisco Javier

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CIRCUMCIRCLE

Let ABC be a triangle and P, Q two isogonal conjugate points and P1P2P3, Q1Q2Q3 the pedal triangles of P,Q, resp.

Denote:

R1 = the radical axis of the circles ((P1, P1Q)[=circle centered at P1 with radius P1Q],(Q1, Q1P))

R2 = the radical axis of the circles ((P2, P2Q),(Q2, Q2P))

R3 = the radical axis of the circles ((P3, P3Q),(Q3, Q3P))

The triangles ABC, Triangle A'B'C' bounded by (R1,R2,R3) are perspective at a point D on the circumcircle.

Antreas P. Hatzipolakis, Hyacinthos #21733

REFLECTING PEDAL CIRCLE

Let ABC be a triangle, Q a point, QaQbQc the pedal triangle of Q, (X) the circumcircle of QaQbQc (=pedal circle of Q), P a point on the OQ line and PaPbPc the pedal triangle of P.

Denote:

(X1), (X2), (X3) the reflections of (X) in OPa,OPb,OPc, resp.

A'B'C' = the triangle bounded by the radical axes of ((O),(X1)),((O),(X2)),((O),(X3))

The triangles A'B'C', X1X2X3 are perspective at a point Qp.

Antreas P. Hatzipolakis, 13 Feb. 2013

CONICS CENTERED AT O

Let ABC be a triangle and r1,r2,r3 three not equal line segments.

Denote:

a1 = the circle centered at A with radius r1. Similarly ....

a1b2 = the radical axis of the circles a1 and b2. Similarly .....

Six Radical centers:

(a1,b2,c3), (a1,b3,c2), (a2,b3,c1), (a2,b1,c3), (a3,b1,c2), (a3,b2,c1)

Six other points of concurrent radical axes:

(a1b2,b3c1,c2a3), (a1b3,b2c1,c3a2), (a2b3,b1c2,c3a1), (a2b1,b3c2,c1a3), (a3b1,b2c3,c1a2),(a3b2,b1c3,c2a1)

The 12gon has opposite sides parallel and equal. It is inscribed on a conic centered at the circumcenter O = radical center of (a1,b1,c1) and (a2,b2,c2) and (a3,b3,c3)

Antreas P. Hatzipolakis. 29 Jan. 2013