HYACINTHOS 25346

[Le Viet An]:


Let ABC be a triangle and Fe, Fa, Fb, Fc the Feuerbach points.

Denote:

PaPbPc = the pedal triangle of I.

Qa, Qb, Qc  = the orthogonal projections of the excenters Ia,Ib, Ic on BC, CA, AB, resp.

The lines through Pa, Pb, Pc and parallels to FaQa, FbQb, FcQc, resp. bound a triangle A'B'C'.

The NPC of A'B'C' passes through the Feuerbach point Fe.

Which point is its center?

Le Viet An

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Note: Let A"B"C" be the triangle bounded by FaQa, FbQb, FcQc.

The homothetic center of A'B'C', A"B"C" is the Feuerbach point Fe.

APH

 

[Peter Moses]:


Hi Antreas,


FaFbFc = Feuerbach triangle.
PaPbPc = intouch triangle.
QaQbQc = extouch triangle.

A' = {-4 a^3+6 a^2 b+a b^2-3 b^3+6 a^2 c-2 a b c+3 b^2 c+a c^2+3 b c^2-3 c^3,(a-c) (a-2 b-c) (a-b+c),(a-b) (a-b-2 c) (a+b-c)}.

>The NPC of A'B'C' passes through the Feuerbach point Fe.

>Which point is its center?

2 a^6 b-a^5 b^2-5 a^4 b^3+2 a^3 b^4+4 a^2 b^5-a b^6-b^7+2 a^6 c+6 a^5 b c-5 a^4 b^2 c-7 a^3 b^3 c+2 a^2 b^4 c+a b^5 c+b^6 c-a^5 c^2-5 a^4 b c^2-10 a^3 b^2 c^2-6 a^2 b^3 c^2+a b^4 c^2+3 b^5 c^2-5 a^4 c^3-7 a^3 b c^3-6 a^2 b^2 c^3-2 a b^3 c^3-3 b^4 c^3+2 a^3 c^4+2 a^2 b c^4+a b^2 c^4-3 b^3 c^4+4 a^2 c^5+a b c^5+3 b^2 c^5-a c^6+b c^6-c^7::

on lines {{1, 30}, {442, 3833}, {758, 12267},...}.
(OI^2 - 8 R^2)  X[1]+ OI^2 X[79].

Best regards,
Peter Moses.