## 14954Re: Four Circum Equilateral Triangles

Expand Messages
• Mar 3, 2007
Dear friends
I wrote

> Here is a more symetric way to draw this figure :
> Ca is the circle going through A with center the common point of
the
> perpendicular bisector of BC with the line AX[74] (The radius of
this
> circle is OH/|2.cos A|). Similarly, we get Cb, Cc (may be, there is
a
> special name for these three circles because they are closely
> 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,
then
> the points BC inter La, CA inter Lb, AB inter Lc will lie on a same
> line only in the case above.

In fact, this is a bit tedious because the centers of the three
circles Ca, Cb, Cc are homothetic in (O,-1/2) of the vertices of the
tangential triangle.
That's enough for me for today.
Friendly. Jean-Pierre
• Show all 14 messages in this topic