HYACINTHOS 25126

 

[Tran Quang Hung]:

Let ABC be a triangle with incenter I.

A'B'C' is pedal triangle of I.

A1 is reflection of I though B'C'.
A2 is reflection of A1 through BC.

A3 is reflection of I though BC.
A4 is reflection of A3 through B'C'.

Line A2A4 is da. Similarly we have line db,dc.

Then lines da,db,dc are concurrent.

Point of concurrence?

 

[Antreas P. Hatzipolakis]:

 

TANGENTIAL TRIANGLE VERSION


Let ABC be a triangle.

Denote:

A'B'C' =  the antipedal triangle of O (tangential triangle).

A1 = the  reflection of O in BC
A2 = the reflection of A1 in B'C'.

A3 = the  reflection of O in B'C'.
A4 = the reflection of A3 in BC

da = A2A4. Similarly db,dc.

The da, db, dc are concurrent.

Point of concurrence?

 

[César Lozada]:

 

T = (SB+SC)*(3*SW*SA^2+2*(R^2-2* SW)*SW*SA-(18*R^2-7*SW)*S^2) : : (barycentrics)

= a*(2*(b^2+c^2)*a^8-(4*b^4+9*b^ 2*c^2+4*c^4)*a^6+3*b^2*c^2*(b^ 2+c^2)*a^4+(4*b^8+4*c^8-b^2*c^ 2*(b^2+c^2)^2)*a^2-(b^4-c^4)*( b^2-c^2)*(2*b^4-3*b^2*c^2+2*c^ 4)) : : (trilinears)

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

= On lines: {3,8705}, {52,2393}, {511,3146}, {576,1614}, {1352,9973}, {1843,3542}, {2854,6243}, {7999,11188}, {9714,11216}, {9971,10095}

= reflection of X(1352) in X(9973)

= [ -8.594412946370884, -14.64197361057260, 17.744067572167350 ]

 

César Lozada