HYACINTHOS 27098

[Antreas P. Hatzipolakis]:

Let ABC be a triangle and A'B'C' the antipedal triangle of O (tangential triangle)

Denote:

I1, I2, I3 = the excenters of A'B'C'

Aa, Ab, Ac = the orthogonal projections of I1 on OA, OB, OC, resp.

Ba, Bb, Bc = the orthogonal projections of I2 on OA, OB, OC, resp

Ca, Cb, Cc = the orthogonal projections of I3 on OA, OB, OC, resp

La, Lb, Lc = the Euler lines of AaAbAc, BaBbBc, CaCbCc, resp.

1. La, Lb, Lc are concurrent.
2. The parallels to La, Lb, Lc through I1, I2, I3, resp. are concurrent.

 

 

[César Lozada]:

 

 

1)

X(3)X(128) ∩ X(25)X(137)

= -2*(cos(4*A)-cos(6*A)+1)*cos( B-C)+2*(cos(3*A)-cos(5*A))* cos(2*(B-C))+2*cos(4*A)*cos(3* (B-C))+cos(5*A)-cos(7*A)+2* cos(A)-2*cos(3*A) : : (trilinears)

= on the tangential circle and these lines: {2, 14652}, {3, 128}, {4, 3432}, {22, 930}, {23, 11671}, {24, 1141}, {25, 137}, {68, 1658}, {110, 13504}, {1614, 13505}, {2937, 13512}, {6592, 7525}, {7502, 14072}, {12026, 12106}, {14674, 14703}

= X(110) of tangential triangle

= X(137) of Ara triangle

= [ 3.1383166334147800, 0.1339291913706113, 2.0994904416902110 ]

 

2)

X(3)X(128) ∩ X(25)X(1141) =

= (6*cos(2*A)+5*cos(4*A)-cos(6* A)+6)*cos(B-C)-(4*cos(A)+5* cos(3*A)-cos(5*A))*cos(2*(B-C) )+(2*cos(2*A)-cos(4*A)+1)*cos(3*(B-C))-7*cos(A)-cos(3*A)+( cos(7*A)-5*cos(5*A))/2 : : (trilinears)

= on lines: {3, 128}, {25, 1141}, {26, 9920}, {137, 1598}, {157, 381}, {930, 11414}, {1263, 7530} , {7159, 10833}, {7503, 14652} , {11441, 13504}, {11456, 13505}, {12026, 13861}, {12083, 13512}

= [ -0.5057296273667857, -5.6270925439901790, 7.7697576096084040 ]

 

César Lozada