HYACINTHOS 26369

[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)?


[Peter Moses]:


Hi Antreas,

The circumcircle of the cevian of a point P passes through X(11) for P on the Feuerbach circumhyperbola and on a (b-c) y^2 z^2 + cyclic = 0.

Specifically for P = X(80), the circle's center is

a (4 a^6-a^5 b-13 a^4 b^2+2 a^3 b^3+14 a^2 b^4-a b^5-5 b^6-a^5 c+8 a^4 b c+11 a^3 b^2 c-16 a^2 b^3 c-10 a b^4 c+8 b^5 c-13 a^4 c^2+11 a^3 b c^2-6 a^2 b^2 c^2+11 a b^3 c^2+5 b^4 c^2+2 a^3 c^3-16 a^2 b c^3+11 a b^2 c^3-16 b^3 c^3+14 a^2 c^4-10 a b c^4+5 b^2 c^4-a c^5+8 b c^5-5 c^6:: 

on lines {{1, 3}, {88, 12515}}.
(9 r (r + R) - s^2) X[1] + 9 r (R - 2 r) X[3].

Best regards,
Peter Moses.