\documentclass[12pt]{article}
\usepackage[cp1251]{inputenc}
\usepackage[T2A]{fontenc}
\usepackage[russian]{babel}
\usepackage{graphicx}
\usepackage{amsmath, amsfonts, amssymb}
\usepackage{epsfig}
\textheight=24cm
\textwidth=17cm
\oddsidemargin=-1cm
\evensidemargin=-1cm
\topmargin=-1.5cm
\newcounter{zzz}
\newcounter{ppp}
\newcommand{\zd}{\par\addtocounter{zzz}{1}%
\arabic{zzz}.
\setcounter{ppp}{0}}
\newcommand{\pp}{\par\addtocounter{ppp}{1}
\alph{ppp}) }
\newcounter{ris}
\renewcommand{\r}{\refstepcounter{ris}%
fig.\arabic{ris}}
\begin{document}
%\vglue1cm
\def\tg{{\rm tg}}
\vskip 12pt
\begin{center}
{\bf\large On the Poncelet theorem}
{\bf Solutions.}
\end{center}
\section{Poncelet theorem for $n=3, 4$}
\ \par\vskip-\baselineskip
\zd Let the line joining $I$ with vertex $C$ of triangle $ABC$ meet for the second time the circumcircle of $ABC$ at point $C'$ (\r\label{formula1}).
\begin{center}
\includegraphics[width=0.9\textwidth]{formula1.eps}
Fig.\ref{formula1}
\end{center}
Since $C'A=C'B$, $\angle AIB=\pi-(\angle A+\angle B)/2=(\pi+\angle C)/2$ and $\angle AC'B=\pi-\angle C$, we obtain that $C'$ is the circumcenter of triangle $AIB$. Thus $IC'=C'B=2R\sin\frac{\angle C}2$. On the other hand $IC=r/\sin\frac{\angle C}2$, therefore
$$
R^2-OI^2=CI\cdot C'I=2Rr.
$$
\zd Let a triangle $ABC$ be given. Take the tangents from an arbitrary point $C'$ of its circumcircle to its incircle and find their common points $A'$, $B'$ with the circumcircle distinct from $C'$. We have to prove that line $A'B'$ also touches the incircle.
Suppose the opposite. For example let $A'B'$ do not intersect the incircle of $ABC$. Increase angle $A'C'B'$ in such a way that line $C'I$ stays its bisector. Then the distances from $I$ to lines $C'A'$ and $C'B'$ will increase, and the distance from $I$ to $A'B'$ will decrease, thus sometimes a circle with center $I$ and radius $r'>r$ will be the incircle of triangle $A'B'C'$. But by Euler formula the inradii of triangles $ABC$ and $A'B'C'$ are equal --- contradiction. The case when $A'B'$ intersect the incircle can be considered similarly.
\zd By the {\bf Feuerbach theorem}, the Euler circle passing through the midpoints $A_0$, $B_0$, $C_0$ of the sides of triangle $ABC$ touches its incircle. From this the center of this circle coinciding with the midpoint of $OH$ lie on the circle with center $I$ and radius $R/2-r$. Thus the trajectories of points $M$ and $H$ are homothetic to this circle with center $O$ and coefficients $2/3$ and 2 respectively.
The trajectory of the Gergonne point is a circle coaxial with the circumcircle and the incircle. Synthetic proof of this fact is unknown.
The trajectory of the Lemoine point is an ellipse with the minor axis lying on $OI$. Synthetic proof of this fact is also unknown.
\zd Let $A''$, $B''$, $C''$ be the second common points of the altitudes of $A'B'C'$ with the incircle of $ABC$. Then $A''A'$, $B''B'$, $C''C'$ are the bisectors of triangle $A''B''C''$, i.e. the orthocenter $H'$ of triangle $A'B'C'$ coincides with the incenter of $A''B''C''$. Also it is easy to see that the sidelines of $A''B''C''$ are parallel to the corresponding sidelines of $ABC$, thus these triangles are homothetic. This homothety maps $O$ and $I$ to $I$ to $H'$ respectively, therefore $H'$ lies on $OI$ and $IH'/OI=r/R$. When the triangle rotates $H'$ and the centroid of $A'B'C'$ dividing $H'I$ in ratio $2:1$ are immobile.
\zd {\bf Answer.} Let $P'$ be the point inverse to $P$ wrt the circumcircle. Consider a rotational homothety with center $P'$, transforming $P$ to $I$ and find the image $Q$ of $I$. The desired trajectory is a circle with center $Q$ (\r\label{ponsisog}). This can be proved using the complex numbers. Synthetic proof is unknown.
\begin{center}
\includegraphics[width=0.9\textwidth]{ponsisog.eps}
Fig.\ref{ponsisog}
\end{center}
\zd {\bf Hint.} The Simson line bisects the segment between the corresponding point and the orthocenter $H$ òðåóãîëüíèêà. The parallel line passing through $H$, meets for the second time the trajectory of $H$ at a fixed point (\r\label{ponsims}).
\begin{center}
\includegraphics[width=0.9\textwidth]{ponsims.eps}
Fig.\ref{ponsims}
\end{center}
\zd \pp Prove general assertion.
Let two circles one lying inside the other be given. From an arbitrary point $X$ of the external circle draw the tangents to the internal one and inscribe to the obtained angle a circle touching the external circle. Then the locus of the centers of such circles is an ellipse having the center of the external circle as a focus.
{\bf Proof.} Let $O$, $I$ be the centers of the given circles, and let $Y$ be a touching point. Then all chords $XY$ meet $OI$ at the same point --- the homothety center $H$ of the given circles. Since the center $Z$ of the semi-inscribed circle is the common point of lines $OY$ and $IX$, its trajectory is an ellipse. To prove that $O$ is its focus consider a polar map wrt the circle with center $O$. It transforms $I$, $H$ to a parallel lines $i$, $h$, also it transforms $X$, $Y$ to the tangents $x$, $y$ meeting at some point $P$ of line $h$. Let $U$, $V$ be the common point of $x$ and $y$ with $i$, and let $Q$ be the vertex of parallelogram $PUQV$. Then $Q$ lies on the line symmetric $h$ about $i$. Since $PQ$ is the fourth harmonic line to $x$, $y$, $h$, it meet the perpendicular from $O$ to $h$ at the fixed point --- the pole of $h$. Using the homothety with the center in this point we obtain that all lines $QU$, $QV$ (i.e. the polars of points $Z$) touches the same circle.
{\bf Alternative solution.} It is sufficient to prove that all semiinscribed circles have common radical center $L$. Suppose this is true, and all semiinscribed circles
have power equal to $p$ wrt $L$. Then they all touch the image of $\Gamma$ under inversion with center $L$ and radius $\sqrt{|p|}$ followed by the symmetry in $L$, for $p<0$.
Let us fix the outer circle $\Gamma$ (let $O$ be its center, and $R$ be its radius), and the center $I$ of the inner circle $\gamma $ (let $r$ be its radius, further $r$ could vary).
Let $Z, A, B, C, D \in \Gamma$ so that $ZA$, $AB$, $BC$, $CD$ touch $\gamma$. Let $\omega_A$, $\omega_B$, $\omega_C$ be semiinscribed circles of triangles $ZAB$, $ABC$, $BCD$
opposite to $A$, $B$, $C$, and touching $\Gamma$ at $T_A$, $T_B$, $T_C$, respactively. We will prove that the radical center $L$ of $\omega_A$, $\omega_B$, $\omega_C$
belongs to $IO$ (we use the proof of Stoyan Boyev, he applied it for the case of three semiinscribed circles of a triangle). By the jump argument it is sufficient
to conclude that all the circles from the family of semiinscribed circles have common radical center $L$.
Firstly, we will consider the homothety $h$ taking the $\gamma$ to $\Gamma$. According to the Monge's theorem we can easily find that lines
$AT_A$, $BT_B$ and $CT_C$ are concurent and their intersection point $T$ is the center of homothety $h$. Hence $T$, $I$ and $O$ are collinear points.
Let $Q_{AB} = AB\cap T_AT_B$, $Q_{BC} = BC\cap T_BT_C$.
Note that $Q_{AB}Q_{BC}$ is the polar line of point $T$ wrt $\Gamma$.
Since $T\in IO$, we have $Q_{AB}Q_{BC}\perp IO$.
Let $M_{AB}$, $M_{BC}$ be the midpoints of arcs $AB$, $BC$ (take arcs opposite to $I$).
Let $P_{AB}$ is the common point of $T_AT_B$ and the tangent to $\Gamma$ at $M_{AB}$. $P_{BC}$ defined similarly.
The radical axis $l(\omega_A,\omega_B)$ passes through $M_{AB}$ and the common point of tangents to $\Gamma$ through $T_A$ and $T_B$.
Hence $P_{AB}$ is the pole of $l(\omega_A,\omega_B)$ wrt $\Gamma$. Similarly, and $P_{BC}$ is the pole of the radical axis $l(\omega_B,\omega_C)$ wrt $\Gamma$.
Thus $P_{AB}P_{BC}$ is the polar of the radical center $L$ wrt $\Gamma$. To prove that $L\in IO$ it remains to prove that $P_AP_C\parallel Q_AQ_C$.
Let $X_{AB}$, $X_{BC}$ be touching points of $\omega_B$ with $AB$ and $BC$. Note that $T_B, X{AB}, M_{AB}$ are collinear,
$T_B, X{BC}, M_{BC}$ are collinear, and $\frac{T_BX_{BC}}{T_BM_{BC}}=\frac{T_BX_{AB}}{T_BM_{AB}}$ from homothety with center $T_B$ taking $\omega_B$ to $\Gamma$.
We have
$\frac{T_BQ_{BC}}{T_BP_{BC}}=\frac{T_BX_{BC}}{T_BM_{BC}}=\frac{T_BX_{AB}}{T_BM_{AB}}=\frac{T_BQ_{AB}}{T_BP_{AB}}$,
and this completes the proof.
%We have made a jump. From this solution follows similarly to Problem 9.
Considering a particular case with symmetry in $OI$, we obtain $\vec{OL} = \frac{2Rr}{R^2-d^2-r^2}\vec{OI}$, where $d=OI$.
\pp From previous result we obtain that all semi-inscribed circles touche the circle with center $O$ and another circle with the center in the remaining focus of the ellipse.
\pp {\bf Hint.} The locus of the centers of such circles is an ellipse with foci $O$ and $I$.
\zd {\bf Hint.} The trypolar passes through the Lemoine point $L$ and meets for the second time the trajectory of $L$ at a fixed point (\r\label{ponstryp}).
\begin{center}
\includegraphics[width=0.9\textwidth]{ponstryp.eps}
Fig.\ref{ponstryp}
\end{center}
\zd Let $R$, $r$ be the radii of the given circles, let $O$ be the center of the external circle, and $O'$ be the circumcenter of triangle $ABI$. We have that $O$ and $O'$ lie on the perpendicular bisector to segment $AB$.
Use the cosines theorem to triangles $AO'O$ and $OO'I$:
$$
R^2=O'A^2+O'O^2-2O'A\cdot O'O\cos\angle AO'O
$$
$$
OI^2=O'I^2+O'O^2-2O'I\cdot O'O\cos\angle IO'O.
$$
Substracting the second equality from the first one we obtain:
$$
R^2-OI^2=2O'O(O'A\cos\angle AO'O-O'I\cos\angle IO'O=2O'O\cdot r.
$$
Therefore the desired locus is the circle with center $O$ and radius $(R^2-OI^2)/2r$.
{\bf Alternative solution.} Let us fix the outer circle $\Gamma$ (let $O$ be its center, and $R$ be its radius), and the center $I$ of the inner circle $\gamma $ (let $r$ be its radius, further $r$ could vary).
Let $A, B, C \in \Gamma$ so that $AB$ and $BC$ touch $\gamma$. Let $S$ and $R$ be circumcenters of triangles $AIB$, $BIC$, respactively.
We have $OS\perp AB$, $OR\perp BC$, and $SR\perp IB$ (since $SR$ is the perpendicular bisector of $IB$). Since $\angle (AB, BI) = \angle (IB, BC)$, we have
$\angle (OS, SR) = \angle (SR, RO)$, hence $OS=OR$.
Let $X, Y\in \Gamma$, and $XY$ touches $\gamma $ so that $\angle (\vec{IX}, \vec{IY}) <\pi$. Define the map $g_r: \Gamma \to \Gamma$ such that $g_r(X)=Y$.
Let $S_r(X)$ be the circumcenter of triangle $XIg_r(X)$, and $OS_r(X) = \rho _r(X)$. Note that $\rho _r(X)$ is continous over $(X, r)$.
We have proved that $\rho _r(X) = \rho _r(g_r(X))$ (we made a jump).
From this jump we will derive that $\rho _r(X)$ is independent of $X$.
Let us call $r$ {\it regular} if the orbit $\{ X, g_r(X), g_r^2(X), \ldots \}$ is dence in $\Gamma$.
For each regular $r$ we obtain that $\rho _r(X)$ is independent of $X$.
If $r_0$ is not regular (there are closed Ponselet trajectories), then take a limit $\rho _{r_0}(X) = \lim \limits_{r\to r_0}\rho _{r}(X)$ over
regular $r$.
Thus we get that $\rho _r(X)$ is independent of $X$ for all $r$.
Considering a particular case with symmetry in $OI$, one can easily obtain $\rho _r(X) = \frac{R^2-d^2}{2r}$, where $d=OI$.
\zd Let $O$, $D$ be the centers of the given circles, and let $R$, $r$ be their radii; let $A'$, $B'$, $C'$ be the midpoints of arcs $BC$, $CA$, $AB$, and $I$ be the incenter of $ABC$. Then $I$ is the orthocenter of triangle $A'B'C'$, thus $\vec{OI}=\vec{OA'}+\vec{OB'}+\vec{OC'}$. Therefore vector $C'I$ lies on $CD$, and its length is equal to $2R\sin\angle OA'B'=2Rr/CD$. On the other hand $C'D=(R^2-OD^2)/CD$, i.e the ratio $C'I/C'D$ do not depend on $C$. Thus $DI/DC'$ also do not depend on $C$, i.e. $I$ lies on the circle homothetic to the given external circle with center $D$.
{\bf Alternative solution.} Let us fix the outer circle $\Gamma$ (let $O$ be its center, and $R$ be its radius), and the center $I$ of the inner circle $\gamma $ (let $r$ be its radius, further $r$ could vary).
Let $A, B, C \in \Gamma$ so that $AB$, $AC$, $BD$ touch $\gamma$. Let $AI$ and $BI$ intersect $\Gamma$ for the second time at $A'$ and $B'$, respectively.
Let $S$ and $R$ be incenters triangles $ABC$, $ABD$, respectively. We have $B'R/B'I = B'A/B'I = \lambda = A'B/A'I = A'S/A'I$. Hence $S$ and $R$ lie on the image of $\Gamma$
under homothety with center $I$ and ratio $1-\lambda$.
Considering special case one can calculate $\lambda$.
We have made a jump. From this solution follows similarly to Problem 9.
\zd Let $O$ be the center of the circle containing point $C$, and $O'$ be the center of the remaining circle. Since $OO'=\sqrt 3$, we obtain that $A'B'$
touches the second circle at some point $C'$. Therefore
$\angle A'O'A=\angle AO'C'+\frac12\angle C'O'B=2\angle ABC'+\angle C'AB=
\angle CB'A'+\frac12\angle CA'B'$, $\angle O'A'O=\angle O'A'B'+\angle B'A'O=
\frac{\pi}2-\angle C'O'A'+\frac{\pi}2-\angle BCA=
\pi-\angle BCA-\frac12\angle CA'B'=\angle CB'A'+\frac12\angle CA'B'$. Since
$O'A=OA'$, we obtain that $AO'A'O$ is an isosceles trapezoid, and $AA'=OO'=\sqrt 3$.
\zd Let $C$ be the fourth vertex of rectangle $PACB$. Since $OP^2+OC^2=OA^2+OB^2$, $C$ lies on the circle with center $O$. Thus the midpoint of $AB$ lies on the circle having the center at the midpoint of $OP$, and the inverse common point of the tangents also lies on a circle.
\zd {\bf Hint.} Prove that the segments joining the touching points of the incircle with the opposite sides are perpendicular and use the previous problem.
\zd {\bf Answer.}
$$
\frac{1}{r^2}=\frac{1}{(R+d)^2}+\frac{1}{(R-d)^2}.
$$
\zd {\bf Answer.}
$$
\frac{R+OP}{R-OP}=\frac{(R+d)^2}{(R-d)^2}.
$$
\zd Using the formula of the previous problem we obtain that $P$ is the limit point of the pencil containing the incircle and the circumcircle. Thus for any point $X$ of the circumcircle the ratio of distance $XP$ and the length of the tangent from $X$ to the incircle is the same. From this fact using that the lines joining the opposite touching points pass through $P$ we obtain the assertion of the problem.
\zd Point $M$ is the midpoint of the segment between the midpoints of the diagonals. Since the midpoints of the diagonals and the incenter are collinear it is easy to see that the trajectory of $M$ is a circle.
\zd \pp Let $U$, $V$ be the midpoints of the diagonals. Since $U$, $V$ lie on the circle with diameter $OP$, and line $UV$ passes through $I$, we have
$$
\tg\angle UPO\tg\angle VPO=\frac{UO\cdot VO}{UP\cdot VP}=\frac{S_{OUV}}{S_{PUV}}=\frac{OI}{IP}=\sqrt{\frac{R^2}{R^2-OP^2}}.
$$
\pp {\bf Hint.} The lengths of the diagonals are $2\sqrt{R^2-OP^2\sin^2\angle UPO}$ and $2\sqrt{R^2-OP^2\sin^2\angle VPO}$. From this and previous equality we obtain that the product of the diagonals is equal to $4R\sqrt{R^2-OP^2}$.
\eject
\section{An algebraic view on the Poncelet theorem}
\ \par\vskip-\baselineskip
\zd \pp Write an equation of line $A_0A_1$ and use that the distance from point $(d,0)$ to this line is equal to $r$. We obtain
$$
((R+d)^2-r^2)t_0^2t_1^2-r^2(t_0^2+t_1^2)+2(R^2-d^2)t_0t_1+((R-d)^2-r^2)=0
$$
\pp {\bf Hint.} Denote the above polynomial as $P_1$. Excluding $t_1$ from the system $P_1(t_0,t_1)=0$, $P_1(t_1,t_2)=0$, we obtain the polynomial from $t_0$, $t_2$, having degree 4 on each variable. Dividing it to $(t_0-t_2)^2$, we obtain the desired relation.
\pp The desired assertion can be obtained by induction similarly as the previous result.
\zd Let starting from some point $t_0$ we return to it through $n$ steps. Then we have: $P_n(t_0,t_n)=0$, $t_0=t_n$. Substituting the second equation to the first one we obtain the equation of degree 4 such that $t_0$ is its root. It is evident that $t_1, t_2, \ldots$ are also its roots. Since the number of roots of this equation is greater than its degree we obtain that it is an identity when $n \ge 5$. That proves the Poncelet theorem.
\zd {\bf Hint.} Let chords $A_0A_1$ and $A_1A_2$ touche two circles of the given pencil. Writing the corresponding equations and excluding $t_1$ we obtain a polynomial from $t_0$, $t_2$ having degree 4 on each variable. This polynomial can be decomposed to two multipliers each of them corresponding to some circle of the pencil. Therefore $A_0A_2$ touches one of these two circle depending on the choice of the tangents. Now we can reason similarly to the previous problem.
\zd Denote $x_i=r_i/(R+d_i)$, $y_i=r_i/(R-d_i)$, where $r_1$, $r_2$ are the radii of the circles touching the sides and the diagonals respectively, and $d_1$, $d_2$ are the distances from their centers to the circumcenter. Considering the symmetric polygons we obtain
$$
x_1=(x_2^2+y_2^2-1)/(1-x_2^2+y_2^2), \quad
y_1=(x_2^2+y_2^2-1)/(1+x_2^2-y_2^2) \eqno(1).
$$
resolving this system wrt $x_2$, $y_2$ we obtain:
$$
x_2=\sqrt{(x_1(1+y_1)/(x_1+y_1))}, \quad y_2=\sqrt{(y_1(1+x_1)/(x_1+y_1))}
\eqno(2)
$$
\zd
\pp For $n=3$ we have the Euler formula: $1/r=1/(R+d)+1/(R-d)$ i.e. $x+y=1$ or $x=\sin^2t$, $y=\cos^2t$,
$0