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\begin{document}
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\begin{flushright}
{\bf A.Zaslavsky, A.Akopyan,\\ P.Kozhevnikov, A.Zaslavsky}
\end{flushright}
\begin{center}
{\bf\large Brocard points}
{\bf Solutions}
\end{center}
\vskip 12pt
\section{Brocard points in triangles}
\ \par\vskip-\baselineskip
\zd Since $\angle PAB=\angle PBC$ we have $\angle BPA=\pi-\angle B$, i.e. $P$ lies on the circle passing through $A$ and $B$ and touching $BC$. Since
$\angle PBC=\angle PCA$, $P$ lies on the circle passing through $B$ and $C$ and touching $CA$. Therefore $P$ is the common point of these circles distinct from $B$. Point $Q$ is constructed similarly.
\zd Answer $\ctg\phi=\ctg A+\ctg B+\ctg C$ follows rom the formula which will be proved later.
\zd Let $A'$, $B'$, $C'$ be the reflections of $P$ in $BC$, $CA$, $AB$. Since $CA'=CP=CB'$ and $\angle PCA=\angle QCB$, we obtain that $CQ$ is the perpendicular bisector to segment $A'B'$, i.e. $Q$ is the circumcenter of $A'B'C'$. Thus the midpoint of $PQ$ is the center of the circle passing through the projections of $P$ to the sidelines of $ABC$. Similarly the midpoint of $PQ$ is the center of the circle passing through the projections of $Q$. It is clear that the radii of these circles are equal.
\zd Let $AP$, $BP$, $CP$ meet for the second time the circumcircle of $ABC$ in points $A'$, $B'$, $C'$. Then the arcs $BA'$, $CB'$ and $AC'$ are equal,
i.e. triangle $B'C'A'$ is the rotation of $ABC$ around $O$ to angle $2\phi$. Then $P$ is the second Brocard point of $A'B'C'$ and this yields both assertions of the problem.
\zd Let $C'$ be a common point of lines $AP$ and $BQ$. Since the angle between these lines is equal to $2\phi$ we obtain by previous problem that $C'$ lie on the circle $OPQ$. Also it is evident that $OC'\perp AB$. Therefore it is sufficient to prove that $C'L\parallel AB$. Let points $A'$, $B'$ be defined similarly as $C'$. Since triangles $ABC'$, $BCA'$, and $CAB'$ are similar the distances from $A'$, $B'$, $C'$ to the correspondent sides of $ABC$ are proportional to the lengths of these sides. The ratio from the point of circle $OPQ$ opposite to $O$ to these sides are the same. Thus this point coincide with $L$.
\zd \pp, \pp {\bf Hint.} Consider the spiral similarities with center $P$ ($Q$), transforming $Q$ ($P$) to $O$.
\pp {\bf Answer.} The midpoint of $OL$, $\pi-2\phi$.
\zd Since $\angle PAC'=\angle QBC'=\phi$ and $\angle PC'A=\angle QC'B$, triangles $APC'$ and $BQC'$ are similar, i.e. $AC'/BC'=AP/BQ$. But from triangles $ACP$, $BCQ$ we have $AP/\sin\phi=AC\sin A$, $BQ/\sin\phi=BC/\sin B$. Therefore, $AC'/BC'=AC^2/BC^2$ and $CC'$ is a symmedian.
\zd This is a partial case of problem 27.
\eject
\section{Brocard point in quadrilaterals}
\ \par\vskip-\baselineskip
\zd The proof is the same as in problem 1.
\zd Since $\angle APB=\pi-\angle B$, $\angle BPC=\pi-\angle C$, we obtain using the sinus theorem to triangles $APB$ and $BPC$
$$
\frac{PB}{\sin\phi}=\frac{AB}{\sin B}, \quad
\frac{PB}{\sin(C-\phi)}=\frac{BC}{\sin C}.
$$
Dividing the first equation to the second one we have
$$
\ctg\phi=\frac{AB}{BC\sin B}+\ctg C.
$$
\zd The condition $\phi(ABCD)=\phi(DCBA)$ is equivalent to
$$
\frac{AB}{BC\sin B}-\ctg B=\frac{CD}{BC\sin C}-\ctg C.
$$
Adding the unit to the squares of both parts we obtain
$$
\frac{AB^2+BC^2-2AB\cdot BC\cos B}{\sin^2 B}=
\frac{CD^2+BC^2-2CD\cdot BC\cos C}{\sin^2 C},
$$
i.e. $\frac{AC}{\sin B}=\frac{BD}{\sin C}$, ÷.ò.ä.
\zd By the construction of $P_i$ we obtain that quadrilaterals $BCP_1P_2$,
$CDP_2P_3$, $DAP_3P_4$, $ABP_4P_1$ are cyclic. From this
$\angle P_1P_4P_3+\angle P_3P_2P_1=\angle A+\angle C=\pi$.
\zd Since $BCP_1P_2$ is a cyclic quadrilateral we obtain
$$
\frac{P_1P_2}{BC}=
\frac{\sin(\phi(ABCD)-\phi(BCDA)}{\sin(C+\phi(ABCD)-\phi(BCDA)}=
\frac{Q_1Q_2}{CD}.
$$
\zd See http://jcgeometry.org/Articles/Volume2/Belev\_Brocard\_points.pdf
\zd See http://jcgeometry.org/Articles/Volume2/Belev\_Brocard\_points.pdf
\zd See http://jcgeometry.org/Articles/Volume2/Belev\_Brocard\_points.pdf
\zd Since $\phi(ABCD)=\phi(DCBA)$, the sought equality is equivalent to
$$
\frac{AB}{BC\sin B}+\ctg C=\frac{AD}{DC\sin D}+\ctg C.
$$
Since $\sin B=\sin D$ we obtain the assertion of the problem.
\zd Since $ABCD$ is cyclic, $AB\cdot CD+AD\cdot BC=AC\cdot BD$, i.e.
$AB\cdot CD=AC\cdot BD/2$. Let $M$ be the midpoint of $AC$. Then $CM\cdot BD=
BC\cdot AD$, i.e. $BC/CM=BD/AD$. Since $\angle BCM=\angle BDA$,
triangles $BCM$ and $BDA$ are similar. Therefore, $\angle MBC=\angle ABD$ and
$BD$ is the symmedian of $ABC$, which yields a)-c).
For prove d) note that four arbitrary points can be transformed by an inversion to the vertices of a parallelogram. If the given points are concyclic this parallelogram will be a rectangle with the same ratio of the products of the opposite sides. Thus the vertices of a harmonic quadrilateral will be transformed to the vertices of a square.
For prove e) consider a central projection conserving the circumcircle of $ABCD$ and transforming the common point of its diagonals to the center. Then the image of the quadrilateral will be a rectangle. Since the tangents to the circumcircle in the opposite vertices of this rectangle are parallel to its diagonal the rectangle is a square.
\zd Since $\ctg\phi=\frac{AB}{BC\sin B}+\ctg C=\frac{BC}{AB\sin B}+\ctg A$,
$\ctg^2\phi-\ctg^2 A=\frac1{\sin^2 B}$ or
$$
\frac1{\sin^2\phi}=\frac1{\sin^2 A}+\frac1{\sin^2 B}.
$$
\zd The proof is the same as in problem 4.
\zd the proof is the same as in problem 5.
\section{Brocard points in polygons}
\ \par\vskip-\baselineskip
\zd {\bf Hint.} Prove that all lines $X_iX_{i+1}$ are the tangents to the same ellipse.
\zd {\bf Answer.}
$$
\frac{OL^2}{R^2}+\tg^2\phi\tg^2\frac{\pi}{n}=1.
$$
\zd {\bf Hint.} Consider the rotations around $O$ to $\pm\phi$.
\zd The proof is the same as for $n=4$.
\zd The proof is the same as for $n=3$.
\zd {\bf Hint.} $T_1$, $T_2$ are the limit points of the pencil containing the circumcircle of the polygon and the circumcircle of $OPQ$.
\end{document}