HYACINTHOS 27104

[Le Viet An]:

 

Let ABC be a triangle and IaIbIc the antipedal triangle of I.

Denote:

Ja, Jb, Jc = the incenters of IaBC, IbCA, IcAB, resp.

 

JbJc /\ IaIb = Kab
JbJc /\ IaIc = Kac
Similarly Kbc, Kba and Kca, Kcb

 

The lines KbaKca, KcbKab, KacKbc bound a triangle A'B'C'.

Ab, Ac = the orthogonal projections of Ia on JaJc, JaJb, resp.

 

Similarly Bc, Ba and Ca, Cb.

A'b, A'c = the orthogonal projections of A' on JaJc, JaJb, resp.

 Similarly B'c, B'a and C'a, C'b

 

Then:

1. The Euler lines of triangles A'JbJc, B'JcJa, C'JaJb  are concurrent.

2. The Euler lines of triangles IaAbAc, IbBcBa, IcCaCb are concurrent.

 

3. The Euler lines of triangles A'A'bA'c, B'B'cB'a, C'C'aC'b are concurrent.

Which are the concurrence points ?.


[César Lozada]:

 

1)      Long coordinates :

P1 = X(40)X(164) ∩ X(1128)X(2089) =

= F(a,b,c)*sin(A/2)+G(a,b,c)*sin(B/2)-G(a,c,b)*sin(C/2)+H(a,b,c) : :  (trilinears), where

F(a,b,c)  =  4*a*b*c*(b-c)*(2*a^5-3*(b+c)*a^4+4*a^3*b*c+(2*b-c)*(b-2*c)*(b+c)*a^2-2*(b^4-3*b^2*c^2+c^4)*a+(b^2-c^2)*(b-c)*(b^2+3*b*c+c^2))

G(a,b,c) = 2*c*(a^8+a^7*b-(6*b^2-2*b*c+3*c^2)*a^6+b*(7*b^2-4*c^2)*a^5-c*(2*b^3-5*b^2*c-3*c^3)*a^4-(9*b^4-c^4-4*b*c*(2*b^2+b*c-2*c^2))*b*a^3+(b^2-c^2)*(6*b^4+c^4-b*c*(8*b^2-b*c-2*c^2))*a^2+(b^2-c^2)^2*b*(b^2+2*c^2)*a-(b^2-c^2)^3*b^2)

H(a,b,c) = (b-c)*((b+c)*a^7-(b+c)^2*a^6-(b+c)*(3*b^2-5*b*c+3*c^2)*a^5+(3*b^4+3*c^4-b*c*(3*b^2+4*b*c+3*c^2))*a^4+(b+c)*(b^2+c^2)*(3*b^2-2*b*c+3*c^2)*a^3-(3*b^6+3*c^6-(4*b^4+4*c^4-b*c*(b^2+8*b*c+c^2))*b*c)*a^2-(b^2-c^2)*(b-c)*(b^4+c^4+b*c*(5*b^2+4*b*c+5*c^2))*a+(b^2-c^2)^2*(b+c)*(b^3+c^3))

= on lines: {40, 164}, {1128, 2089}, {10234, 12694}

= [ 7.1407056283277570, 5.7199996825000860, -3.6150455882823380 ]

 

César Lozada