## Some theorems on Miquel points

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• Dear Floor van Lamoen, and Hyacinthos, Since we have started to investigate Miquel points, here are two forgotten theorems on them. The Miquel point
Message 1 of 8 , Jan 1, 2003
Dear Floor van Lamoen, and Hyacinthos,

Since we have started to investigate Miquel points, here are
two "forgotten" theorems on them.

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:

1. If X is an arbitrary point in the plane, then the second points of
intersection of the lines XA, XB, XC with the circles AB'C', BC'A',
CA'B' (the first intersections being A, B, C) are concyclic with P
and X.

[Jacques Hadamard: Lecons de Geometrie Elementaire, exercise 344. I
don't know of a proof.]

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.]

Darij Grinberg
• Dear Darij ... CA B ... M ... It is well known (Steiner 1827) that Ma,Mb,Mc,M,P are concyclic (Miquel cirle of the quadrilateral). Hence, this theorem is not
Message 2 of 8 , Jan 1, 2003
Dear Darij

> 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.]

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.
Happy new year to all Hyacinthists. Jean-Pierre
• Dear Darij ... are ... of ... BC A , ... I ... Using oriented angles of lines (modulo Pi)
Message 3 of 8 , Jan 1, 2003
Dear Darij

> Since we have started to investigate Miquel points, here are
> two "forgotten" theorems on them.
>
> 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:
>
> 1. If X is an arbitrary point in the plane, then the second points
of
> intersection of the lines XA, XB, XC with the circles AB'C',
BC'A',
> CA'B' (the first intersections being A, B, C) are concyclic with P
> and X.
>
> [Jacques Hadamard: Lecons de Geometrie Elementaire, exercise 344.
I
> don't know of a proof.]

Using oriented angles of lines (modulo Pi)
<PXaX = <PXaA = <PC'A = <PC'B = <PXbB = <PXbX and Xa,Xb,P,X are
concyclic...
Happy New Year to all of you. Jean-Pierre
• (Answer to Hyacinthos message #6240) ... If I understand correctly, you assume that A , B , C are collinear. But they need not be! Therefore, the Baum theorem
Message 4 of 8 , Jan 3, 2003

In message #6240, Jean-Pierre Ehrmann wrote:

>> > [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".

Sincerely,
Darij Grinberg
• Dear Darij, ... collinear. ... Yes, you are perfectly right; I thought that you were talking about the Miquel point of a complete quadrilateral. With my
Message 5 of 8 , Jan 3, 2003
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
• Dear Darij, I ve tried to find an easy synthetic proof of Braun theorem ... are ... CA B ... M ... PaPbPc is the pedal triangle of P. A direct similitude
Message 6 of 8 , 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
• Dear Darij, I wrote ... Of course, I was meaning Baum theorem. Friendly. Jean-Pierre
Message 7 of 8 , Jan 4, 2003
Dear Darij, I wrote

> I've tried to find an easy synthetic proof of Braun theorem
Of course, I was meaning Baum theorem.
Friendly. Jean-Pierre
• We have PM /PI = 1/cos(T) and not PM /PI. With my apologizes. Jean-Pierre
Message 8 of 8 , Jan 4, 2003
We have PM'/PI = 1/cos(T) and not PM'/PI.
With my apologizes. Jean-Pierre
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