HYACINTHOS 26370

[Le Viet An]:

Let ABC be a triangle.

Denote:

J = the reflection of I in the Feuerbach point Fe 

[ J = X(80)].

Then the cevian circle of J passes through Fe.

Which point is its center (lying on the OI line)?

 

 

[Angel Montesdeoca]:

The center of cevian circle of X(80) is W = (9r(r+R)-s^2) X(1) - 9r(2r-R) X(3).

W = ( a (4 a^6
               -a^5 (b+c)
                -a^4 (13 b^2-8 b c+13 c^2)
                +a^3 (2 b^3+11 b^2 c+11 b c^2+2 c^3)
               +2 a^2 (7 b^4-8 b^3 c-3 b^2 c^2-8 b c^3+7 c^4)
               -a (b-c)^2 (b^3+12 b^2 c+12 b c^2+c^3)
               -(b^2-c^2)^2 (5 b^2-8 b c+5 c^2) ) : ... : ... ).
               
(6 - 9 - 13) - search numbers  of W: (81.2652311218832, 67.3974022848629, -80.5261053100206).


The cevian circle of X(80) intersects the incircle in X(11) and X(3025).

Angel Montesdeoca