HYACINTHOS 27103

[Le Viet An]:

Consider a triangle ABC. Equilateral triangles BCA', CAB', ABC' are constructed on three sides BC, CA, AB, resp. having the same orientation (outer or inner orientation). 

The circle with center A'  passing  through  B, C  meets again AC, AB at Ac, Ab, resp.
Similarly  Bc, Ba and Ca, Cb .
The lines AbAc, BcBa, CaCb bound a triangle A"B"C" .
Let A_B, A_C be the orthogonal projections of of A" on AC, AB, resp.
Similarly B_C, B_A and C_A, C_B

Then:
1. The Euler lines of triangles  A"BC, B"CA, C"AB and  ABC are concurrent.
2. The Euler lines of triangles  A"A_BA_C, B"B_CB_A, C"C_AC_B are concurrent.
3. The lines parallel to the Euler lines of triangles  A"A_BA_C, B"B_CB_A, C"C_AC_B through A,B,C , resp. are concurrent.

Which are the points of concurrences ?



[César Lozada]:



1)  For outer A’: X(473)

For inner A’: X(472)

 
2) For outer A’:

Qo = X(5)X(51) ∩ X(15)X(54)

= cos(B-C)*(cos(3*A+1/3*Pi)*cos(B-C)-cos(A+1/6*Pi)*sin(3*A)-1/2) : : (trilinears)

= (SB+SC)*((8*R^2-3*SW)*S^2-sqrt(3)*(3*SA^2-S^2)*S-(4*R^2-SW)*SA^2)*(S^2+SB*SC) : : (barycentrics)

= (4*R^2+3*sqrt(3)*S-SW)*X(5)-3*(sqrt(3)*S+3*R^2-SW)*X(51)

= on lines: {5, 51}, {15, 54}

= reflection of Qi in X(973)

= [ 12.8127146085907700, -4.9114116169162860, 1.1273119358076080 ]

 

For inner A’:

Qi = X(5)X(51) ∩ X(16)X(54)

= cos(B-C)*(sin(3*A+1/6*Pi)*cos(B-C)+sin(A+1/3*Pi)*sin(3*A)-1/2) : : (trilinears)

= (SB+SC)*((8*R^2-3*SW)*S^2+sqrt(3)*(3*SA^2-S^2)*S-(4*R^2-SW)*SA^2)*(S^2+SB*SC) : : (barycentrics)

= (4*R^2-3*sqrt(3)*S-SW)*X(5)-3*(-sqrt(3)*S+3*R^2-SW)*X(51)

= on lines: {5, 51}, {16, 54}

= reflection of Qo in X(973)

= [ -6.5186486287706510, 2.3313277342694150, 5.0352754176150830 ]


3)      X(54) ( For outer- & inner- A’)

 
César Lozada