14954Re: Four Circum Equilateral Triangles
- Mar 3, 2007Dear friends
> Here is a more symetric way to draw this figure :the
> Ca is the circle going through A with center the common point of
> perpendicular bisector of BC with the line AX (The radius ofthis
> circle is OH/|2.cos A|). Similarly, we get Cb, Cc (may be, there isa
> special name for these three circles because they are closelythen
> connected with the Mac Cay circles)
> Three parallel lines going respectively through A, B, C intersect
> again Ca, Cb, Cc respectively at A'', B'', C''; A' = BC inter
> B''C'', ...
> I've checked by computation, that, if we consider three directly
> congruent non isosceles triangles inscribed in the circumcircle,
> Ta,Tb,Tc where A is a vertex of Ta and La the opposite sideline,
> the points BC inter La, CA inter Lb, AB inter Lc will lie on a sameIn fact, this is a bit tedious because the centers of the three
> line only in the case above.
circles Ca, Cb, Cc are homothetic in (O,-1/2) of the vertices of the
That's enough for me for today.
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