POLYA012: The Feuerbach point is the perspector (proposed by Tatiana Emelyanova, 11/18/01).
Let A0B0C0, A1B1C1, A2B2C2, A3B3C3 be the medial, the I-cevian, the
orthic, the intouch triangles of ABC respectively. In all following cases
the Feuerbach point of ABC is the perspector of triangles 1) and 2):
a) 1) the side-triangle of A0B0C0 and A1B1C1, 2) A2B2C2;
b) 1) the side-triangle of A1B1C1 and A2B2C2, 2) A0B0C0;
c) 1) the side-triangle of A2B2C2 and A0B0C0, 2) A1B1C1;
d) 1) the side-triangle of A3B3C3 and A0B0C0, 2) A3B3C3.
I can prove geometrically only case b).
(The side-triangle of A1B1C1 and A2B2C2 has vertices A1B1 /\ A2B2, etc.)
Discussion. [JPE 11/20/01]: Here are some general
ideas related to this sort of problems:
If T(P) is the cevian triangle of P, the triangles T(P)
and T(Q) are clearly self-polar wrt the circumconic C(P,Q) going through
P and Q. Hence the vertices of the side-triangle S(P,Q) of T(P) and T(Q)
are the poles wrt C(P,Q) of the side lines of ABC. Hence the side lines
of S(P,Q) are tangent at A, B, C to C(P,Q); which means that S(P,Q) is
the precevian triangle of the perspector K(P,Q) of C(P,Q) (note that S(P,Q)
depends only of the circumconic) This means that, for any point R, S(P,Q)
and T(R) are perspective at the cevian quotient R/K(P,Q). Now if we take
R = G, we get that S(P,Q) and the medial triangle are perspective at the
center of C(P,Q).
If we take P, Q on Feuerbach hyperbola (for instance
incenter, orthocenter, Nagel, Gergonne...), S(P,Q) and the medial triangle
are perspective at the Feuerbach point F. This is the assertion 2).
A geometric proof of the other ones seems much more difficult.
Here is something more general :
The side triangle of the medial triangle and T(P) is
the precevian triangle of the infinite point P1 of the trilinear polar
of P. Hence, if P moves on the circumconic with center W, the cevian product
W*P1 moves on the same conic and the line P, W*P1 goes through a fixed
point (P->W*P1 is an involution of the circumconic). The fixed point of
the line P,W*P1 is the cevian quotient K'/K where K = G/W is the perspector
of the
circumconic and K' the isotomic conjugate of K. This
means that for any pair of points P,Q lying on the circumconic with center
W and such as the line PQ goes through K'/K, the side triangle of T(P)
and the medial triangle is perspective at W with T(Q).
Your assertions seem to mean that, in the particular
case of Feuerbach hyperbola, the involution P->Q maps I to H and has the
Gergonne point and the Nagel point as fixed points. I'm unable to give
a geometric proof of this fact.
Bibliography.
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