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\begin{document}
\begin{center}
{\large\bf On isogonal conjugacy, Miquel points,
(Newton-)Gauss lines, etc. Solutions.}

{\bf N.Beluhov, A.Zaslavsky, P.Kozhevnikov}
\end{center}

\section{Introductory problems}

\zd \pp Follows from sine Ceva theorem.

\pp Proof by counting angles.

\pp Let $P$ be a point of a given circle. Then measure of angle
$APB$ is fixed. Hence the sum of measures of angles $CAP$ and
$CBP$ is fixed. Therefore, measure of angle $AP'B$ is fixed, hence
$P'$ lies on a fixed circle passing through $A$ and $B$.

\pp By $P_A$, $P_B$, $P_C$ denote points symmetric to $P$ in $BC$,
$CA$, $AB$, respectively. Since $P_AC=PC=P_BC$, the perpendicular
bisector of the segment $P_AP_B$ passes through $C$, and hence it
is the bisector $CP'$ of angle $P_ACP_B$. Hence $P'$ is the
circumcenter of triangle $P_AP_BP_C$. By homothety with center $P$
and ratio $1/2$, the midpoint $T$ of $PP'$ is the center of the
circle passing through projections of $P$ to the sidelines.
Similarly, $T$ is the center of the circle passing through
projections of $P'$. These two circles coincide since $T$ is
equidistant from projections of $P$ and $P'$ to a centain line.

In the case when $P'$ is a point at infinity we obtain the Theorem
on Simson line. for a point lying on the circumcircle, its
projections of to the sidelines are collinear.

\pp Let the common point of $XY'$ and $YX'$ be $P$ and the common point of $XY$ and $X'Y'$ be $Q$. Let $XA \cap YX' = X_A$, and $Y'A \cap YX' = Y_A$. Also call as $Q_A$ intersection point of $QA$ and $YX'$. Prove that lines $AQ$ and $AP$ are symmetric with respect to the bisector of angle $A$. Since this is true for all angles points $P$ and $Q$ are isogonaly conjugated.

Consider cross-ratio $(Y,Q_A,X_A,X')$. Project these points from $Q$ to line $AX'$. The map of $Y$ is $X'$, the map of $Q_A$ is $A$, $X_A$ transforms to itself and $X'$ transforms to the common point of lines $X'Y'$ and $XA$. Now project obtained ratio from $Y'$ to line $YX'$. The map of $A$ is $Y_A$, the map of $X$ is $P$, $X_A$ transforms to itself and the map of $X'$ transforms back to $X'$. Thus $(Y,Q_A,X_A,X') = (P,Y_A,X_A,X') = (Y_A,P,X',X_A)$. It means that $AQ$ is the reflection of $AP$ in the bisector of angle $A$.

Another proof see also in the book of A.Akopyan and A.Zaslavsky "Geometry of conics"\, AMS, 2007.

\pp By $P'$ denote the point of intersection. Let $AB$ and $CD$
meet at $K$. For one of triangles $KBC$ and $KDA$, $P'$ is
isogonal conjugate to $P$. Hence, $P$ lies on the line symmetric
to $KP$ in the bisector of angle $K$. For the other of two
triangles $KBC$ and $KDA$, $P'$ is also isogonal conjugate to $P$.
We obtain that $P$ lies on all four lines symmetric to $AP$, $BP$,
$CP$, and $DP$, in the bisectors of corresponding angles.

\pp Suppose that projections of $P$ are concyclic. Then, similarly
to p.d), obtain that point $P'$ symmetric to $P$ in the center of
the circle is isogonal conjugate to $P$. The converse statement is
proved anagolously.

\pp See the book mentioned above.

\pp By the statement converse to the Theorem on Simson line, it is
sufficient to prove that the projections of parabola focus to the
sidelines are collinear. Since a point symmetric to the focus in
any tangent lies on a directrix, the required statement follows.

\vspace{1mm} \zd \pp Let the circumcircles of $ABF$ and $CDF$ meet
for the second time at $M$. We have $\angle (AM,\,MD) = \angle
(AM,\,MF) + \angle (MF,\,MD) = \angle (BA,\,BF) + \angle (CF,\,CD)
= \angle (BA,\,CD) = \angle (AE,\,ED)$, hence $M$ lies on the
circumcircle of $ADE$. Similarly for the other triangle.

\pp It is sufficient to prove that triangles $MBE$ and $MFD$ are
similar. We have $\angle (EB,\,BM) = \angle (CE,\,CM) = \angle
(CD,\,CM) = \angle (FD,\,FM)$. Angles $MEB$ and $MDF$ are equal by
the same reasoning.

\pp Let $B'$ and $C'$ are the positions of bugs at some moment.
The intersection point $P$ of the circumcircles of triangles $ABC$
and $AB'C'$ is the center of spiral similarity taking line $BB'$
to $CC'$ (ratio of this similarity is equal to the ratio of the
speeds of the bugs). Hence $P$ is Miquel point for quadruple of
lines $B'B$, $BC$, $CC'$, $C'B'$.

\pp Suppose $E$ moves from $B$ to $C$ at some constant speed,
while $F$ moves from $D$ to $A$ at the same speed. The condition
holds at any moment. We show that each of points $Q$ and $R$ moves
at a constant speed,
 so statement of the problem follows from the previous problem.
 Note that $\angle EQB = \angle FQD$,
$EB = FD$. Hence the circumcircles of triangles $EBQ$ $FDQ$ are
equal, and ratio $EQ/QF$ is constant. Therefore $Q$ moves at a
constant speed. Similarly, for $R$.

\pp Miquel point is the focus of a parabola inscribed ro the
quadrilateral.

\pp By Therem on Simson line, eah three of four projections are
collinear. Hence all four projections are collinear. It is known
that homothety with center at some point of the circumcircle and
ratio 2 takes Simson line to the line passing through the
orthocenter. Therefore, by homothety with center at Miquel point
and ratio $2$ we obtain the line $l$ passing through orthocenters
of four triangles. The orthocenters have equal powers with respect
to three circles constructed on the diagonals of quadrilateral,
hence $l$ is the radical axis for these three circles that is
perpendicular to Gauss line (passing through the centers of the
circles). For the inscribed parabola, the focus lies on the
circumcircles of the triangles. Hence, Miquel point is the focus.
The projections of Miquel point lie on a tangent to the parabola
at its vertex. The line passing through the orthocenters is the
directrix of the parabola. Hence, Gauss line is parallel to the
axis od parabola.

\vspace{1mm} \zd \pp By $M$ denote the midpoint of $AC$, $N$ ---
midpoint of $BD$, $T$ --- midpoint of $EF$. Let $F'$, $A'$ and
$B'$ are the midpoints of the sides of $ABF$. Note that $M$ lies
on $F'B'$, $N$ lies on $F'A'$, $T$ lies on $A'B'$. By homotheties
with centers at vertices of $ABF$ and ratio 2,
$\frac{\overrightarrow{F'M}}{\overrightarrow{MB'}} =
\frac{\overrightarrow{BC}}{\overrightarrow{CF}}$,
$\frac{\overrightarrow{B'T}}{\overrightarrow{TA'}} =
\frac{\overrightarrow{AE}}{\overrightarrow{EC}}$,
$\frac{\overrightarrow{A'N}}{\overrightarrow{NF'}} =
\frac{\overrightarrow{FD}}{\overrightarrow{DA}}$. Multiplying
these three equalities, and applying Menelaus theorem, we obtain
the required.

\pp Hint. Gauss line of quadrilateral $ABCD$ is the locus of
points $X$ such that $S_{XAB}+S_{XCD}=S_{XBC}+S_{XDA}$ (here the
areas are oriented).

\pp Follows from 2f.

\pp This is reformulation of b).

\eject

\section{Three Miquels for a Quartet.}


\zd From Problems 13, 15 it follows that $AX$, $BY$, $CZ$
pass through $D'$ (isogonal conjugate to $D$ with respect to triangle $XYZ$).

\vspace{1mm}
\zd Follows directly from Problem 13.

\vspace{1mm}
\zd Follows directly from Problems 13, 15.

\vspace{1mm}
\zd By problem 11, $\triangle XDZ \sim \triangle XYA$, $\triangle  XD'Z \sim \triangle XYA'$.
Hence $XA:XD'=(XA:XZ)(XZ:XD')=(XY:XD)(XA':XY)=XA':XD$, that equivalent to the
statement of the problem.

\vspace{1mm}
\zd Follows from the previous problem and Theorem on three homotheties
applied to the segments $AA'$, $DD'$, and $B'B$.
Indeed, $Z$ is the center of homothety that takes $A$ to $B'$, $A'$ to $B$, etc.

Alternative solution could be easily derived from Problem 1e).

\vspace{1mm}
\zd \pp $X$ is a center of a spiral similarity that takes $C$ to $D$, $A$ to $B$.
Since $K_Y$ is the image of $L_Y$ under this spiral similarity, angle $K_YXL_Y$
equals to $\angle (AC, BD)$. Hence $X$ lies on the circle $P_YK_YL_Y$.
Similarly obtain that $Z$ lies on the same circle.
\pp Since $X$ lies on the circle $AP_YB$, we have $\angle XP_YB=\angle XAB$.
Similarly, $\angle BP_YZ=\angle BCZ$. From these and four analogous equalities
it follows that $\angle XP_YZ+\angle ZP_XY+\angle YP_ZX=\pi$,
that implies the statement of the problem.
\pp From problem 15 it is clear that $\psi_X(P_Y)=P_Z$, etc.
This means that $P_X$, $P_Y$, $P_Z$ belong to a certain quartet, so
$\psi_X(P_X)=\psi_Y(P_Y)=\psi_Z(P_Z)$. We need to show that
$\psi_Z(P_Z)$ is isogonal conjugate to $T$, or equivalently, $\psi_Z(T)$
is isogonal conjugate to $P_Z$. Note that $\psi_Z$ takes circles  $ZXP_Y$ and $ZYP_X$
(passing through $T$) to lines $YP_X$ and $XP_Y$, hence $\psi_Z(T) = YP_X\cap  XP_Y$.
From equalities $\angle P_XYX=\angle ZYP_Z$, $\angle P_YXY=\angle ZXP_Z$
obtain the required statement.

\section{Quartets for three Miquels.}

\zd a)-b) Follows directly from the definition.

c) Each of the transformations $\psi_X$, $\psi_Y$, $\psi_Z$ is {\it circular}
(i.e. takes a circle either to a circle or to a line)
and preserves the orientation. Hence any product of $\psi_Z $, $\psi_Y$, $ \psi_X$
is a circular transformation preserving the orientation.
Moreover, from a)-b) follows $X$, $Y$, $Z$ are invariant points for $\psi_Z \psi_Y \psi_X$.
Note that a circular transformation preserving the orientation is uniquely defined
by the images of three points.
The statement of the problems is
independent of the order of $\psi_X$, $\psi_Y$, $\psi_Z$ in its product.
Hence $\psi_X$, $\psi_Y$, $\psi_Z$ commute to each other.
\zd By definition of $\psi_X$,  $\angle ZXD=\angle AXY$ and $XD\cdot XA=XY\cdot XZ$,
that implies the first similarity. The second similarity is proved analogously.
\zd From Problem 10 it follows, in particular, that
$\psi_X\circ\psi_Y=\psi_Z^{-1}=\psi_Z$. Therefore,
$\psi_Y(A)=\psi_Y(\psi_X(D))=\psi_Z(D)=C$, so $\psi_Y$ interchanges the points in the pairs
$(A, C)$, $(B, D)$. Similarly we obtain that $\psi_X$ interchanges the points in the pairs
$(A, D)$, $(B, C)$, while $\psi_Z$ interchanges the points in the pairs
$(A, B)$, $(C, D)$.
\zd Let $D'$, $A'$ be isogonal conjugates to  $D$, $A$ respectively.
Then $A'$ lies on $XD$, $D'$ lies on $XA$. Moreover,
$\angle XD'Z=\pi-\angle ZXD'-\angle D'ZX=\pi-\angle DXY-\angle YZD= \angle ZDX+\angle XYZ-\pi$.
By Problem 11 we have $\angle ZDX=\angle AYX$, i.e. $\angle XD'Z=\angle XYA'$.
Hence $\triangle XD'Z \sim \triangle XA'Y$, and $A'=\psi_X(D')$.
\zd
\pp The incenter and the excenters of $XYZ$.
\pp Point $O$ and three points symmetric to $X$, $Y$, $Z$ in the opposite sidelines of $XYZ$.
\pp From the definition of $\psi_Z$ it follows that its invariant point
lie on bisector of angle $XZY$, and on the circle with center $Z$ and radius
$\sqrt{ZX\cdot ZY}$. There are two such points $U$ and $V$.
From Problem 10 we obtain that $\psi_X(U)=\psi_Y(\psi_Z(U))=\psi_Y(U)=\psi_Z(\psi_X(U))$,
i.e. $\psi_X(U)$ is also an invariant point for $\psi_Z$. It is clear that $\psi_X(U)\not=U$,
therefore, $\psi_X(U)=\psi_Y(U)=V$, and a required quartet (both for $U$ and for $V$)
is $U$, $U$, $V$, $V$. Further, the isogonal conjugates to $U$, $V$
are also invariant under $\psi_Z$. Hence $U$ and $V$ are isogonal conjugates to each other,
and quartet $U$, $U$, $V$, $V$ coincides to conjugate quartet.
\zd \pp From the definition of $\psi_X$ and the Problem 12 it follows that $X$
is the center of spiral similarity taking $A$ to $B$, and $C$ to $D$.
By Problem 2b) this center is Miquel point.
\pp $Y$ is Miquel point for the quadruple of lines $AB$, $BC$,
$AD$, $CD$; $Z$ is Miquel point for the quadruple $AD$, $AC$,
$BD$, $BC$.
\pp $X$ is Miquel point, hence (in particular) $\triangle XBD \sim \triangle XAC$.
Let $P_X = AD\cap BC$, $P_Y= AC\cap BD$, $P_Z = AB\cap CD$.
We have $XA\cdot XD=XB\cdot XC=XP_Y\cdot XP_Z=R_a^2$, and the angles
$AXD$, $BXC$, $P_YXP_Z$ have a common bisector $l$.
The inversion with center $X$ and radius $R_a$ followed by the symmetry in $l$
takes triangles $ADP_Y$ and $BCP_Y$ to $DAP_Z$  and $CBP_Z$, respectively.
Therefore it tales intersection point $Y$ of the circles $ADP_Y$ and $BCP_Y$
to $Z$.
Hence the product of an inversion and a symmetry defined above is $\psi_X$.
\zd From Problem 13 it follows that the product of $\psi_X$ and the isogonal conjugacy
(in any order)
takes $D$ to $A'$.

\end{document}
