HYACINTHOS 20787

Dear geometers:


Please, Could someone check and confirm this?


Let AH, BH and CH be the orthic triangle of ABC with orthocenter H.

Let A1 be the point where AAH cuts BHCH. Construct B1 and C1 cyclically.

Then:

1) the triangle formed with the perpendicular bisectors of AA1, BB1 and CC1
is in perspective with ABC and the perspector is

Barycentrics: f(A,B,C) : f(B,C,A) : f(C,A,B) where f(A,B,C)=
(cos(A)+4*cos(B)*cos(C))/(cos(A)+2*cos(B)*cos(C)) (Bolivar Point 1)


2) the triangle formed with the perpendicular bisectors of HA1, HB1 and HC1
is in perspective with ABC and the perspector is

Barycentrics: f(A,B,C) : f(B,C,A) : f(C,A,B) where f(A,B,C)=
(3*cos(A)+4*cos(B)*cos(C))/(cos(A)*(cos(B)*cos(C)+cos(A))*(cos(A)+2*cos(B)*c
os(C))) (Bolivar Point 2)


I’d like to name these centers as “Bolivar Points” after our great liberator
Simon Bolivar.


Note:

The triangle formed with the perpendicular bisectors of A1AH, B1BH and C1CH
is also in perspective with ABC and the perspector is X(275)


Best regards

Cesar Lozada

Venezuela