HYACINTHOS 25070

[Tran Quang Hung]:


Let ABC be a triangle 

 

Excircles (Ia),(Ib),(Ic) touch BC,CA,AB at A',B',C', resp.

 

Draw tangnets B'Ab,C'Ac to (Ia) with Ab,Ac lie on (Ia).

 

Similarly, we have Bc,Ba,Ca,Cb.

 

Then

 

- Lines AbAc,BcBa,CaAb bound a triangle which is perspective to A'B'C'.

 

- Perpendicular bisectors of AbAc,BcBa,CaCb are concurrent.

 

.[Angel Monesdeoca]:

Dear  Tran Quang Hung,


 *** Lines AbAc,BcBa,CaAb bound a triangle which is perspective to A'B'C'. Perspector W on X(517)X(938)
 
 W = ( (a-b-c)/(a^4-2 a^2 (b^2-b c+c^2)+b^4-2 b^3 c+10 b^2 c^2-2 b c^3+c^4) : ... : ....)

 with (6-9-13)-search numbers (2.73241028071413,  2.68000811504665,  0.524161811160948).
 
 
 ***  Perpendicular bisectors of AbAc,BcBa,CaCb are concurrent at X(4882).

 Best regards,
 Angel Montesdeoca