ADGEOM 2546 * ADGEOM 2547 * ADGEOM 2553

 

#2546

Dear Dr. Paul Yiu, Dear Dr. Kimberling, and Mister Dergiades and all member,

More than year ago, I found a problem "A generalization of Lester circle theorem associated with Neuberg cubic" and no solution, and I sent this problem to Crux Mathematicorum Journal, but yesterday the editoral board reply to me that they reject this problem because this problem is very difficult

 

My generalization the Lester circle theorem as follows:

Consider ABC be a triangle, let P be a point on the Neuberg cubic. Pa,Pb,Pc be the reflection of P in three sidelines BC,CA,AB respectively. Well-known that APa,BPb,CPc are concurrent, define this point is Q.

 

Problem: Two Fermat points, P,Q lie on a circle.

I hope that there are some one like this problem and publish with a solution in FG,

I am thank to You very much

Best regards

Sincerely

Dao Thanh Oai

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#2547

Dear Dao:

 

Let P be a point and Pa,Pb,Pc the reflections of P in the sidelines of ABC. Let Q = (BPb)∩(CPc)

 

What is the locus of P such that P,Q and Fermat points F1, F2 are concylic?

 

The answer is P on {Neuberg cubic } U { another cubic}

 

Then we can conclude that if P is on Nueberg cubic then ABC and PaPbPc are perspective (as we knew) at Q(P) and P,Q(P),F1,F2 are concyclic.

 

When P moves on Neuberg cubic there are two remarkable facts:

 

1)  The locus of Q(P) is K060

 

2)  The center of the circle {P,Q(P),F1,F2} moves on the unnamed line (L)=(115,125).  

 

2.1) (L) is the trilinear polar of X(523) = isogonal conjugate of X(110)

 

2.2) (L) pasess through ETC`s: X(115), X(125) (both on NPC), X(245), X(246), X(247), X(686), X(690), X(868), X(1116), X(1562),  X(1637), X(1640), X(1648), X(2081), X(2088), X(2610), X(3120), X(3124), X(3125),  X(3269), X(3569), X(4120), X(6388), X(6627), X(6791)

 

2.3) (L) is parallel to (74,98), (99,110), (113,114) and others

 

2.4) (L) is perpendicular to (2,98) at X(125), to (6,13) at X(115) and others

 

2.5) Circles and respective poles of (L) w/r to them:

    [BROCARD, 184], ["DAO-MOSES-TELV", 542], [INCIRCLE, 4934], [LESTER, 542], ["NAPOLEON INNER", 6770],

    ["NAPOLEON INNER", 6773], ["ORTHOCENTROIDAL", 6], ["ORTHOPTIC", 98], [POLAR, 648]

 

2.5) Pole of (L) w/r to the NPC of ABC:

    N’ = (b^2-c^2)^2*(a^4-(b^2+c^2)*a^2-b^2*c^2) : : (barycentrics)

       = (2*cos(A)-cos(3*A)+cos(B-C))*sin(B-C)^2 : : (trilinears)

       = Complementary conjugate of X(3005)

       = Complement of X(1634)

       = On lines: (2,1634), (5,542), (6,3613), (11,3141), (98,1576), (115,804), (125,526) and others

       = ( 2.976330735223314, 3.41693479525486, -0.098596869525899 )

 

2.6) Pole of (L) w/r to the circumcircle of ABC:

     O’ = (cos(B-C)+2*cos(A)*cos(2*(B-C))-2*cos(A)+cos(3*A))*sin(A)^2 :: (trilinears)

        = Reflection of: (6/1976)                                                            

        = On lines: (3,67), (6,157), (25,1989), (50,2393), (98,338) and others

        = =( -4.241771111510047, -11.09547381263624, 13.279886865198720 )

 

Regards

César Lozada

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#2553

Dear César,

Some additional properties of your new points:

N' = crossdifference of every pair of points on line X(1625)X(1634)
N' is the perspector of the side- and vertex-triangles of the tangential triangles of the medial and orthic triangles.

O' = trilinear pole, with respect to the tangential triangle, of the Euler line
O' = isogonal conjugate of isotomic conjugate of X(3448)

Also, the isogonal conjugate of (L) is the hyperbola {A,B,C,PU(2)}, whose center is the complement of the isotomic conjugate of X(3448).  This hyperbola also passes through X(249), X(250), X(687), X(691) (its circumcircle intercept other than A, B, C), and X(2966).

Best regards,
Randy Hutson