Dear Darij,

> >> > [DG] 2. Let Ma, Mb, Mc be the centers of the circles AB'C',

> >> > BC'A', CA'B' [it is well-known that triangles MaMbMc and

> >> > ABC are similar], let M' be the circumcenter of triangle

> >> > MaMbMc, and M the circumcenter of triangle ABC. Then

> >> > M'M = M'P.

> >> >

> >> > [This is a theorem by Peter Baum; proposed in the little

> >> > German periodical "Die Wurzel" (see the website

> >> > wurzel.org), in the 12/98 issue. Solved by Sefket

> >> > Arslanagic in the 5/99 issue; the solution was quite

> >> > involved.]

> >>

> >> [JPE] It is well known (Steiner 1827) that Ma,Mb,Mc,M,P are

> >> concyclic (Miquel cirle of the quadrilateral). Hence, this

> >> theorem is not a recent one.

>

> If I understand correctly, you assume that A', B', C' are

collinear.

> But they need not be! Therefore, the Baum theorem is a partial

> generalization of the Miquel circle in a quadrilateral, and not a

> corollary.

>

> Sorry for the possible-to-misunderstand use of the term "Miquel

> point".

Yes, you are perfectly right; I thought that you were talking about

the Miquel point of a complete quadrilateral.

With my apologizes for the misunderstanding. Jean-Pierre