Sorry, an error occurred while loading the content.

## 6262Re: Some theorems on Miquel points

Expand Messages
• Jan 4, 2003
Dear Darij,
I've tried to find an easy synthetic proof of Braun theorem
> The Miquel point configuration is defined as follows:
>
> On the sidelines BC, CA, AB of a triangle ABC, points A', B', C'
are
> chosen and the circles AB'C', BC'A', CA'B' are drawn. Then, as we
> know, these circles have a common point P (the Miquel point).
>
> Now the two theorems:
>
> 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.]

PaPbPc is the pedal triangle of P.
A direct similitude with center P, angle T, ratio 1/cos(T) maps
PaPbPc to A'B'C'.
As the circumcenter of PPbPc is the midpoint A'' of AP, the
similitude maps this midpoint to Ma; hence the similitude maps the
midpoint I of PM (circumcenter of A''B''C'') to M'(circumcenter of
MaMbMc).
As PM'/PI = cos(T) and <IPM' = T, it follows that M' lies on the
perpendicular bisector of PM.
Friendly. Jean-Pierre
• Show all 8 messages in this topic