## 22180Conic

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• Apr 25, 2014
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X(37): Let {cA} be the circle with center on (BC) and tangent to (AB) and (AC). {cA} cuts (BC) at Ab, Ac. Build Ba, Bc, Ca, Cb cyclically. Points Ab, Ac, Ba, Bc, Ca, Cb lie on an ellipse with center X(37). [03-13-2014]

http://www.marveloustriangles.blogspot.gr/2014/04/centers-facts-2.html

The centers of the three circles are the traces on the sidelines of the
angle bisectors of the triangle.

We have seen here other conics constructed by these traces.
I recall this: Let A'B'C' be the cevian triangle of I and Ab, Ac the reflections
of A' in BB', CC', resp. (lying on AC, AB, resp.) and similarly Bc,Ba and Ca,Cb.
The six points Ab,Ac,Bc,Ba,Ca,Cb lie on a conic (somewhere in Anopolis archive !)

Here is another one I do not remember if we have discussed it.

Let A'B'C' be the cevian triangle of I, Ab, Ac the orthogonal projections of A
on BB', CC', resp. and A2,A3 the orthogonal proojections of Ab,Ac on AB,AC,
resp. Similarly B3,B1 and C1,C2. The six points A2,A3, B3,B1, C1,C2 lie on a conic (for the sides B1C1, C2A2, A3B3 are parallel to sidelines BC,CA,AB, resp.
of the triangle, where are lying the other three sides of the hejagon).

How about if A'B'C' is the cevian triangle of a point P?
Are the six points lying on a conic for all P's (or for which ones anyways ... :-)?

Antreas

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