Properties in a New Configuration Regarding Quadrilaterals
by XmL, May 30, 2016, 5:07 AM
Generalized Configuration:
is a quadrilateral. Define two circles
such that
and
.
Note that the two circles do not necessarily have to intersect; we will discuss such cases at a later time. Also, when the notion of locus is mentioned below, interpret its "domain" as all circle pairs
that satisfy the conditions above.
Preliminary Notations: Let
. Let
denote the Miquel point of
. Define
the spiral similarity transformation centered at
that sends
. (I just realized I used
twice here so please excuse that mistake)
Proposition 1:
lies on the circle of similitude of
.
Proof: By the definition of
, we know
under
. Thus
lies on the circle of similitude.
.
Proposition 2: Let
denote the external and internal center of similitude of
respectively. Prove that the loci of
(for all satisfactory
are straightlines that are perpendicular.
Proof: Because
, therefore
is constant; generally the shape of
stays the same up to similarity. Let
be the perpendicular bisector of
and consider the the spiral similarity transformation
. Suppose
under the transformation; clearly
. Since the transformation is independent of
, therefore
is a fixed line and the locus of
. Similarly define
as the image of
under
, we know
is the locus of
. Note that
thus
. 
Let
, then the perpendicularity implies that
lies on the circle of similitude. Hence we have the following Corollary:
Corollary 1: The circle of similitude of
passes through two fixed points, namely
and
.
Proposition 3:
Proof: Let
denote the center of the circle of similitude. Again since
is invariant up to similarity, the locus of
is the perpendicular bisector of
which satisfies
where the latter follows from
tangent to
. Because
as
are concyclic, it follows that
.
.
Lemma 1: Let
be the reflection of
over the perpendicular bisector of
. Let
be the reflection of
over the perpendicular bisector of
. Prove that
and
is parallel to the midline of
(Here I will force the term midline to mean the Newton-Gauss line).
Proof (for now): Clearly
and
under the spiral similarity transformation centered at
such that
. It follows that
must pass through
.
For the second assertion, let
be the orthocenter of
respectively. It is not hard to show that
, thus
. Since
is also perpendicular to the midline of
, therefore
is parallel to the midline.
.
Proposition 4:
is parallel to the midline of
.
Proof: Suppose we reflect
across the perpendicular bisector of
to obtain
. It is not hard to see that
, thus
are collinear. By Lemma 1,
, thus
is parallel to the midline of
. 
Lemma 2:
is a quadrilateral. Define
as in the introduction. Let
be a point outside
such that
(note that they are inversely similar). Prove that the radical axis of
passes through
.
Proof: Let
. By the construction of
we have
. Because
, therefore
. Hence
is cyclic. It follows that
is the radical axis of the two circles. 
Main result 1: The radical axis of
passes through a fixed point. In particular, the fixed point is the intersection of
with the line through
parallel to the midline of
.
Proof: Let
such that
is parallel to the midline. already gave a characterization of the fixed point, so it suffices to prove that
. We first show that
are isogonal conjugates wrt
.
By properties of Miquel point it's clear that
are isogonals wrt
and consequently
. Define
on
such that
. By Lemma1 we know
. Since
, it follows that
are isogonals wrt
. Hence
are isogonal conjugates of
. Similarly we can show that
are isogonal conjugates.
It is a property of isogonal conjugates that
thus
. From the isogonals we can obtain
; it follows that
. 
Let
be the circle that passes through
and is coaxal with
. We will develop some more properties before proving the following main results:
Main result 2:
passes through another fixed point on 
Main result 3:
.
It seems the most natural to investigate
using inversion. Let
denote the inversion through
with radius
. Besides
, other existing points
get mapped to
.
It is clear that
. Also, the Miquel point of
is
. Now let the inversion transform
into
. We can check with angle properties of inversion that these two circles still satisfy the special conditions stated in the introduction. This prompts us to construct
, the analog of
wrt
in
, i.e.
such that
are isogonals wrt
.
From Main result 1 we know
lies on the radical axis of
. However,
is mapped to this radical axis under
; therefore
. Since the construction of
is independent of
, therefore it is the fixed point on
. We have just proven Main result 2.
Lemma 3:
is a quadrilateral, Let
and
the Miquel point of
. If
is the circumcenter of
and
is the line through
perpendicular to
. Prove that
.
Proof: Let
. Note that
, thus
. Let
, thus
. Hence
. 
Proposition 5: Let
. Prove that
.
Proof:
. Define
as in Lemma 3, then
. Thus by Lemma 3,
and we are done(Note that we can show
in an analogous fashion. 
Let
respectively.
Proposition 6:
Proof: Suppose
. Because
are concyclic, therefore
. Thus
(property of inversion). Since
are isogonals wrt
and
by Proposition 5, we have
We can similarly show that
.
.
It follows from Proposition 6 that
, thus there exists a spiral similarity transfomation about
that sends
. Explicitly, the transformaiton is
. Note that
, thus
. Since
, it follows that
is an isosceles trapezoid.
Now suppose
under the spiral similarity defined above; it is clear that
. In addition,
lies on the perpendicular bisector of
, and thus
. Therefore
and we have proven that
, which is Main result 3.
Ending Remarks: The motivation to study this configuration came from the content of the previous post. I suspected that it would contain many fixed points and was lucky enough to find them by experimenting on my drawing software. This post examines the significance of the point
, whose definition makes me want to name it "anti-Miquel point". The connection of
with the vertices of the quadrilateral as isogonal conjugates is really surprising. The circle
is of great interest in the second part of the post. I seldem use inversion, but I realized the equivalence between the fixed point of
and
and it made the use of inversion natural (it also supports the post's organization.) Main result 3 is truly beautiful and I wonder if any proof exists without developing
.
Acknowledgement: I want to thank User Telv Cohl for reviewing the content of this post.




Note that the two circles do not necessarily have to intersect; we will discuss such cases at a later time. Also, when the notion of locus is mentioned below, interpret its "domain" as all circle pairs

Preliminary Notations: Let







Proposition 1:


Proof: By the definition of





Proposition 2: Let




Proof: Because
















![\[\angle (s_1,s_2)=\angle (s_1,l_1)+\angle (l_1,s_2)=\angle (MS_1,MO_1)+\angle (MO_1, MS_2)=\angle (MS_1,MS_2)=90^{\circ}\]](http://latex.artofproblemsolving.com/6/1/f/61f57b46b6d457fe42218bdf4f2eb79a03c4249e.png)


Let


Corollary 1: The circle of similitude of



Proposition 3:

Proof: Let




![\[\angle (l_1,l_s)=\angle (MO_1,MO_S)=\angle (O_2O_1, MO_2)\]](http://latex.artofproblemsolving.com/b/1/0/b10ea4239590daadcecb51a5ffcb8b41c0720b86.png)


![\[\angle (O_2O_1, MO_2)=\angle (AD,MD)=\angle (FA,MF)\]](http://latex.artofproblemsolving.com/b/8/1/b81213c3ee6c1181231d1cac97ef9aeaa8762ad9.png)



Lemma 1: Let









Proof (for now): Clearly






For the second assertion, let








Proposition 4:


Proof: Suppose we reflect









Lemma 2:







Proof: Let








Main result 1: The radical axis of




Proof: Let





By properties of Miquel point it's clear that













It is a property of isogonal conjugates that
![\[\frac {LA}{LC}=\frac {\sin \angle MAF}{\sin \angle MCF}, \frac {LD}{LB}=\frac {\sin \angle FDM}{\sin \angle MBF}\]](http://latex.artofproblemsolving.com/5/b/6/5b6bb6ce6e12e0134ac434f988a76375eb5ea3f1.png)
![\[\frac {LA}{LC}=\frac {LD}{LB}\]](http://latex.artofproblemsolving.com/8/6/0/86096607521268c76562c6ec92322673b2796fea.png)



Let



Main result 2:


Main result 3:

It seems the most natural to investigate







It is clear that












From Main result 1 we know








Lemma 3:










Proof: Let

![\[\angle (PE,PM)=\angle (DE,DM)=\angle (DA,DM)=\angle (FA,FM)\]](http://latex.artofproblemsolving.com/6/6/1/661c4ba85d3db14af147e17783ff0904c81ade7e.png)



![\[\angle (EO,h)=\angle (P'E,P'O)=\angle (P'E,P'P)=\angle (ME,MP)=\angle (ME,MF)\]](http://latex.artofproblemsolving.com/b/1/c/b1c63f656c14d5b32ef234173550966625b85cb2.png)

Proposition 5: Let


Proof:






Let

Proposition 6:

Proof: Suppose







![\[\angle UMA=\angle A'LU'=\angle F'LG=\angle F'FG=\angle EFA\]](http://latex.artofproblemsolving.com/6/4/4/6448f4a0931fd02bb11ff2655f88a5a61c8873ea.png)


It follows from Proposition 6 that




![\[\angle (AB,UV)=\angle (MA,MU)=\angle (AB,EF)\]](http://latex.artofproblemsolving.com/8/0/1/8010eb6e8bfc01efc912b96018969742213296af.png)



Now suppose







Ending Remarks: The motivation to study this configuration came from the content of the previous post. I suspected that it would contain many fixed points and was lucky enough to find them by experimenting on my drawing software. This post examines the significance of the point






Acknowledgement: I want to thank User Telv Cohl for reviewing the content of this post.
This post has been edited 7 times. Last edited by XmL, Jun 7, 2016, 11:16 PM
Reason: Telv
Reason: Telv