- Oct 27, 2006Dear All,
It appears that orthologic triangles are the theme of the moment.
Suppose that X is a triangle centre associated with triangle ABC.
Let us say that ABC is X-logic with triangle A'B'C' if the lines
through A', B', C' parallel to AX, BX, CX (repectively) concur.
It is easy to see that ABC and A'B'C' are H-logic precisely when
they are orthologic. As ever, H denotes the orthocentre.
Write X' for the corresponding centre associated with A'B'C'.
In this notation, we have the well-known results :
(1) If ABC is H-logic with A'B'C', then A'B'C' is H'-logic with ABC.
(2) If ABC is H-logic with A'B'C' and with B'C'A', then it is also
H-logic with C'A'B'.
If (2) holds, then we say the triangles are triply H-logic. We then
have some less well-known results :
(3) Suppose that ABC and A'B'C' are triply H-logic with concurrences
at the points A",B",C". Then triangles A'B'C' and A"B"C"
(a) are triply perspective, with perspectors at infinity, and
(b) have the same centroid and Steiner ellipses.
What may be new is that the same results hold if we replace H by G,
where G is the centroid. This is a fairly easy Maple exercise.
I have tried Cabri sketches and am convinced that H may NoT be
replaced in (1) by any other centre X(n), N < 11.
An obvious question - are there other centres which can replace H?
On the other hand, Maple shows that (2), (3) hold for any centre X.
One other new(?) result :
If X is a Fermat Point and ABC is X-logic wiyh A'B'C', then the
point of concurrence is a Fermat Point of A'B'C'.
- << Previous post in topic Next post in topic >>