HYACINTHOS 29031

[Vu Thanh Tung] :

Let

_  ABC: a triangle on the plane
_  A_1, B_1, C_1:  three excenters of triangle ABC
_  P: a point on the plane
_  A', B', C': the incenters of triangles PBC, PCA, PAB, resp.

 Then  A_1A', B_1B', C_1C' are concurrent.

 

Is this property well known before ?

 

 [César Lozada]:

 

Dear Mr Tung,

 

Your lines concur always.

 

Let P = U : V : W (exact trilinear coordinates) and let ρ_a, ρ_b , ρ_c be the sidelengths of the pedal triangle of P w/r to ABC, given by ρ_a = sqrt( V^2 + W^2 + 2*V*W*cos(A) ),…

 

Then the point of concurrence is:

Q(P) =3*U^2 + (cos(C)+2)*U*V + (cos(B)+2)*U*W - cos(A)*V*W + 2*( ρ_b + ρ_c)*U + (V+W)*(- ρ_a + ρ_b + ρ_c) +  ρ_b*ρ_c : : (trilinears)

 

Some related centers:

 

Q( X(1) ) = X(3645) (see this point in ETC)

 

Q( X(3) ) =  X(1)X(372) ∩ X(3)X(9)

= a*( a*(-a^2+b^2+c^2)+(-a+b+c)*S) : :  (barys)

= lies on the cubic K414 and these lines: {1, 372}, {2, 31540}, {3, 9}, {7, 31541}, {20, 30412}, {37, 1152}, {40, 30556}, {41, 30400}, {43, 1685}, {44, 1151}, {45, 6410}, {48, 19067}, {57, 16432}, {63, 16440}, {101, 31564}, {165, 6212}, {169, 31563}, {223, 13388}, {371, 1743}, {487, 4416}, {488, 3912}, {515, 7090}, {573, 31438}, {1100, 3594}, {1123, 31533}, {1124, 7074}, {1449, 3312}, {1587, 5393}, {2183, 19068}, {2951, 31545}, {3218, 21568}, {3219, 21567}, {3247, 6398}, {3305, 16441}, {3306, 21492}, {3311, 16670}, {3523, 30413}, {3576, 30557}, {3592, 16669}, {3723, 6438}, {3731, 6396}, {3911, 8957}, {3928, 21561}, {3929, 21560}, {3973, 6200}, {4292, 30324}, {4297, 31595}, {4357, 11291}, {5405, 13935}, {5437, 21548}, {6204, 15803}, {6351, 6460}, {6409, 16885}, {6411, 15492}, {6412, 16814}, {6420, 16667}, {6426, 16777}, {6430, 16672}, {6431, 16671}, {6432, 16666}, {6434, 16677}, {6450, 16676}, {6454, 16673}, {6469, 16674}, {6684, 14121}, {7308, 16433}, {9583, 13332}, {10164, 31594}, {11292, 17353}, {13411, 30325}, {21566, 27065}, {21569, 27003}

= (excentral)-isogonal conjugate of-X(6213)

= X(485)-of-excentral triangle

= X(488)-of-1st circumperp triangle

= X(641)-of-6th mixtilinear triangle

= X(6289)-of-hexyl triangle

= X(12257)-of-2nd circumperp triangle

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (20, 30412, 31562), (6502, 7133, 1)

= [ 5..0864295766909220, 3.9602012200478910, -1.4485961673676640 ]

 

César Lozada