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%\greenit{\bf Предлагаю дать Лемму 0 и Лемму на стр. 2 в качестве задач 4.1 и 4.2, выдаваемых после промежуточного финиша. Это будет подсказкой к задачам разделов 0-3. Тогда все ссылки на эти леммы в решениях надо заменить на ссылки на задачи 4.1 и 4.2, соответственно, а доказательства лемм назвать решениями задач 4.1 и 4.2.}
%\bigskip
\title{Webs from lines and circles}
\author{Alexey Zaslavskiy,
Fedor Nilov,
Alexander Polyanskiy,
Mikhail Skopenkov}
\date{}
%\author{Присоединяйтесь!}
%\ead {zaslavsky AT mccme.ru, alexander.polyanskii AT yandex.ru, nilovfk AT mail.ru, skopenkov AT rambler.ru}
%\address [add2] {Institute for information transmission problems of the Russian Academy of Sciences}
%\begin{abstract}
%Даешь проект по элементарной геометрии с научным содержанием!
%\end{abstract}
%\end{frontmatter}
%\begin{figure}[hb]
%\includegraphics[width=16cm]{networks-fig0.png}
%\caption{Lattices of dimensions $1$, $2$, and $3$.}
%\label{fig0}
%\end{figure}
\maketitle
\footnotetext{Summer conference of the International mathematical Tournament of towns, August 2--10, 2012}
\vspace{-0.8cm}
\begin{center}
\includegraphics[width=0.4\textwidth]{simson.jpg}
\end{center}
\section*{Solutions of the problems}
Next lemma allows to solve problems 0.1, 0.5, 0.6.
\textit{Lemma 0.} Let $A'B'C'$ be a cevian triangle of some point wrt triangle $ABC$ (i.e the lines $AA',BB', CC'$ concur). The line passing through an arbitrary point $M_1$ of side $AC$ and parallel to $A'B'$ intersect $BC$ in point $M_2$; the line padding through $M_2$ and parallel to $A'C'$ intersect $AB$ in point $M_3$ etc. Then $M_1=M_7$.
\verb"Proof."
If $M_1$ coincide with $B_1$, then $M_4$ and $M_7$ also coincide with $B'$.
Else by Ceva theorem $\frac{\ov{AB'}}{\ov{B'C}}\frac{\ov{CA'}}{\ov{A'B}}\frac{\ov{BC'}}{\ov{C'A}}=1$. And by Thales theorem $\frac{\ov{B'C}}{\ov{CA'}}=\frac{\ov{B'M_1}}{\ov{M_2A'}}$, $\frac{\ov{A'B}}{\ov{BC'}}=\frac{\ov{M_2A'}}{\ov{C'M_3}}$, $\frac{\ov{C'A}}{\ov{AB'}}=\frac{\ov{C'M_3}}{\ov{M_4B'}}$. Placing three last equalities into the first one we obtain that: $\frac{\ov{B'M_1}}{\ov{M_4B'}}=1$. Thus $M_1$ and $M_4$ are symmetric wrt $B'$. Similarly $M_4$ and $M_7$ are symmetric wrt $B'$. Therefore $M_1=M_7$.
\\
\texttt{Solutions of problems of part 0.}
\textit{Solution of problem 0.1}
\emph{First solution}. Let $a,b,c$ be the side lengths of $BC$, $CA$, and $AB$, respectively. Let $x$ be the \emph{signed} length of $AA_1$ (i.e., the length of $AA_1$ taken with positive sign, if the vectors $\overrightarrow{AA_1}$ and $\overrightarrow{AB}$ have the same orientation, and with negative sign, if they have opposite orientation). Since $A_1A_2$ is orthogonal to the the bissector of the angle $BAC$ it follows that $AA_2=AA_1=x$ (with signs). Analogously we find consecutively $CA_3=CA_2=b-x$,
$BA_4=a-b+x$, $AA_5=c-a+b-x$, $CA_6=a-c+x$, $AA_7=x$ (with signs). Since $AA_7=x=AA_1$ (with signs) it follows that $A_7=A_1$.
\emph{Second solution}. Use lemma 0 to Gergonne triangle.
\smallskip
\textit{Solution of problem 0.2}
In problems 0.2, 0.3 and 0.4 we consider the angles between the directions (vectors).
Let $O$ be the common point of $l_1, l_2, l_3$. Let $(l_1,l_2)=\ff$,$(l_2,l_3)=\fff$,
$(OA_1,l_1)=\fa$. Since the lengths of segments $OA_i$ are equal (the symmetry conserve the length), it is sufficiently to prove that $(OA_7,l_1)=(OA_1,l_1)=\fa$.
Since $(OA_2,l_1)=-(OA_1,l_1)=-\fa$, then $(OA_2,l_2)=-\fa+\ff$.
Thus $(OA_3,l_2)=\fa-\ff$. From this we obtain that $(OA_3,l_3)=\fa-\ff+\fff$.
Therefore $(OA_4,l_3)=-\fa+\ff-\fff$. This yields that $(OA_4,l_1)=-\fa+\ff-\fff-\ff-\fff=-\fa-2\fff$.
Similarly we obtain that $(OA_7,l_1)=-(OA_4,l_1)-2\fff=\fa$.
\smallskip
\textit{Solution of problem 0.3}
Let $(l_1,l_2)=\ff$,$(l_2,l_3)=\fff$ and $(l_{(1,2)},l_1)=\fa$, where $l_{(i,i+1)}$ ix the vector from $O$ to the projection of $O$ to the line $A_iA_{i+1}$. Since the distances from $O$ to all lines $A_iA_{i+1}$ are equal (the symmetry conserve the length), it is sufficiently to prove that $(l_{(7,8)},l_1)=(l_{(1,2)},l_1)=\fa$.
Since $(l_{(2,3)},l_1)=-\fa$ then $(l_{(2,3)},l_2)=(l_{(2,3)},l_1)+(l_1,l_2)=-\fa+\ff$.
Similarly $(l_{(3,4)},l_2)=\fa-\ff$. Thus $(l_{(3,4)},l_3)=(l_{(3,4)},l_2)+(l_2,l_3)=\fa-\ff+\fff$.
Also $(l_{(4,5)},l_3)=-\fa+\ff-\fff$. Therefore $(l_{(4,5)},l_1)=(l_{(4,5)},l_3)+(l_1,l_3)=-\fa+\ff-\fff+(-\ff-\fff)=-\fa-2\fff$.
Similarly we obtain that $(l_{(7,8)},l_1)=-(l_{(4,5)},l_1)-2\fff=\fa$.
\smallskip
\textit{Solution of problem 0.4.}
Let $(l_1,l_2)=\ff$,$(l_2,l_3)=\fff$,
$(A_1A_2,l_1)=\fa$. Since the circumradii of triangles $OA_i A_{i+1}$ are equal (by the sinus theorem), it is sufficiently to prove that $(A_7A_6,l_3)=-(A_1A_2,l_2)=-\fa-\ff$ (by the inverse sinus theorem).
Since $(A_3A_2,l_3)=-(A_1A_2,l_1)=-\fa$ then $(A_3A_2, l_2)=-\fa-\fff$.
Since $(A_3A_4,l_1)=-(A_3A_2,l_2)=\fa+\fff$ then $(A_3A_4,l_3)=\fa+\ff+\fff+\fff$.
Since $(A_5A_4,l_2)=-(A_3A_4,l_3)=-\fa-\ff-2\fff$ then $(A_5A_4,l_1)=-\fa-\ff-2\fff-\ff$.
Since $(A_5A_6,l_3)=-(A_5A_4,l_1)=\fa+2\ff+2\fff$ then $(A_5A_6,l_2)=\fa+2\ff+2\fff-\fff$.
Since $(A_7A_6,l_1)=-(A_5A_6,l_2)=-\fa-2\ff-\fff$ then $(A_7A_6,l_3)=-\fa-2\ff-\fff+\ff+\fff=-\fa-\ff.$
\smallskip
\textit{Solution of problem 0.5.}
Use lemma 0 to the orthotriangle.
\textsc{Note.} The problem can be reformulated in the next way.
Let point $A_1$ lies on $AB$. The circumcircle of triangle $A_1AC$ secondary meets $BC$ in point $A_2$. The circumcircle of triangle
$A_2BA$ secondary meets $CA$ in point $A_3$ etc. Prove that $A_1=A_7$.
\smallskip
\textit{Solution of problem 0.6}
Use lemma 0 to the medial triangle.
\smallskip
\textit{Solution of problem 0.7}
a) The Pappus theorem in an equivalent statement is proved in the book
\cite[Chapter 1]{AkopZasl}. %[Прасолов], Problem 5.78.
%(how reduce this formulating to the given one?).
b) Consider the hexagon $A_1A_2A_3A_6A_5A_4$: the lines $A_1A_2, A_3A_6, A_5A_4$ concur in point $R$, and the lines $A_4A_1, A_2A_3, A_6A_5$ concur in point $G$. Therefore "the diagonals"\, $A_2A_5,$ $A_3A_4$ (these two lines pass through $B$) and $A_6A_1$ concur (in point $B$). Thus $A_7=A_1$.
\smallskip
\textit{Solution of problem 0.8}
a) The Brianchon theorem is proved in the book \cite[Chapter 1]{AkopZasl}.
%%[Прасолов], Problem 3.73
b) It is clear that the lines $A_4A_5$, $A_5A_6$, $A_6A_7$ are the reflections of $A_4A_3$, $A_3A_2$, $A_2A_1$ in $OI$. Therefore $A_1=A_7$.
%\greenit{\bf !!! Why $A_5A_6$ is a reflection of $A_3A_2$ ???}
\smallskip
\textit{Solution of problem 0.9}
\textit{First solution (D.~Yakutov)}.
Let us compute the angle
$\angle GA_{4}R$:
\begin{align*}
\angle GA_{4}R &= \pi - \angle GA_{4}B - \angle BA_{4}R \\
&= \pi - \angle GOB - \angle BA_{5}R \\
&= \pi - \angle GOB - (\pi - \angle GA_{5}R - \angle GA_{5}B) \\
&= \angle GA_{5}R + \angle GA_{5}B - \angle GOB \\
&= \angle GOR - \angle GOB + \angle GA_{6}B \\
&= \angle GOR - \angle GOB + (\pi - \angle GA_{6}R - \angle BA_{6}R) \\
&= \angle GOR - \angle GOB + (\pi - \angle GA_{7}R - \angle BOR) \\
&= (\pi - \angle BOR - \angle GOB + \angle GOR) - \angle GA_{7}R.
\end{align*}
Thus $\angle GA_{4}R + \angle GA_{7}R = \pi - \angle BOR - \angle GOB + \angle GOR$.
Similarly $\angle GA_{1}R + \angle GA_{4}R = \pi - \angle BOR - \angle GOB + \angle GOR$. Thus $\angle GA_{1}R = \angle GA_{7}R$. Also $\angle GA_{1}B = \angle GOB = \angle GA_{7}B$.
Hence the points $G, B, A_{1}, A_{7}$ belong to one circle and the points $G, R, A_{1}, A_{7}$ also belong to one circle. But these two points have at most two common point, one of which is $G$. Since $A_{1} \ne G$ and $A_{7} \ne G$ it follows that $A_{1}=A_{7}$.
\textit{Second solution}.
Consider an arbitrary inversion with center $O$. It transforms the red, green and blue circles to three lines. Let the image of the red point be $C$, the image of the blue point be $A$, and the image of the green point be $B$. Now look after points $A_i$. Through point $A_1$ ($\in AB$) we take a green (passing through $A$ and $C$) circle, which secondary meets in $A_2$ the blue "circle"\, --- line $BC$. Through $A_2$ we take a red circle secondary meeting in $A_3$ the green "circle"\, --- line $AC$. Through $A_3$ we take a blue circle secondary meeting in $A_4$ the red "circle"\, --- line $AB$ etc.
Thus we obtained the reformulating of problem 0.5 given in the note to this problem. Therefore $A_1=A_7$.
\\
%Now formulate a lemma which allows to solve some problems of parts 1,2,3. Next time we shall call the circles and the lines as \emph{general circles}.
%\textbf{Lemma.}
%Let next conditions be given:
%\begin{itemize}
%\item $\AAA$ --- некоторое семейство взаимно-однозначных преобразований плоскости в себя (переводящих окружности в окружности). Для любой точки $A$ плоскости (возможно, кроме одной) множество точек $\{f(A), \text{ где } f\in\AAA\}$ представляет собой окружность (возможно, без одной точки). Каждую такую окружность назовём красной. Для любых двух точек $A$ и $B$ красной окружности существует единственное $f\in \AAA$: $f(A)=B$.
%\item For each $t\in\R$ a map $R_t$ of the plane is defined.
%This map transforms all general circles to general circles.
%Suppose that for arbitrary $t,s\in\R$ and an arbitrary point $A$ $R_t(R_s(A))=R_{t+s}(A)$.
% Also for any point $A$ the set $\gamma_A=\{R_t(A):t\in\R\}$ is an arc of a general circle.
%\item Let $\gamma_1,\gamma_2$ be two different arcs of general circles passing through some point.
%For any point $A\in\gamma_1$ paint the arc $\gamma_A$ red.
%Paint the arcs $R_t(\gamma_1)$, where $t\in\R$ and the arcs $R_t(\gamma_2)$, where $t\in\R$ green and blue respectively.
%It is known that any two colored arcs have at most one common point.
% \item There exists a disc $\om$, such that exactly one arc of each color passes through each its point.
%\end{itemize}
%Then the red, green and blue arcs form a web.
%\verb"Proof."
%The foliation condition is true by the third condition of the lemma. Show that the closure condition also is true.
%Take an arbitrary point $O$ inside the disc.
%Draw trough it the red ($w_1$), green ($w_2$) and blue ($w_3$) arcs of general circles.
%Let all constructed points $A_i$ lie inside $\om$.
%Let $A_1\in w_1$ and $t\in\R$ is such that $R_t(O)=A_1$
%(such $t$ exists by the first condition of the lemma: if $w_1=\gamma_X$, where $X\in\gamma_1$, then there exist such $y,z\in \R$, that $R_y(X)=O$ and $R_z(X)=A_1$, thus $R_{z-y}(O)=R_{z-y}(R_y(X))=R_{z}(X)=A_1$.
%Therefore $t=z-y$.
%If $R_t(R_s(A))=R_{ts}(A)$ the sought $t$ can be found similarly.
%\greenit{\bf !!! Убрать сноску и написать этот текст прямо в тексте доказательства !!!}}
%).
%Draw through $A_1$ the green arc $w_2'$. Let $A_2$ be the common point of $w_2'$ and $w_3$. Draw through $A_2$ the red arc $w_1'$. Let $A_3$ be the common point of $w_1'$ and $w_2$. Draw through $A_3$ the blue arc $w_3'$. Let $A_4$ be the common point of $w_3'$ and $w_1$ etc.
%Now show that $R_t(O)=A_7$, this yields that $A_1=A_7$.
%We know that $R_t(O)=A_1$ , thus $R_t(w_2)=w'_2$ (it is true because $R_t(w_2)$ is the red arc passing through $A_1$, and the unique such arc is $w_2'$), therefore $R_t(A_3)\in w_2'\cap w_1'=A_2$. Since $R_t(A_3)=A_2$, then $R_t(A_4)=O$ (similarly). Since $R_t(A_4)=O$, then $R_t(A_5)=A_6$. Since $R_t(A_5)=A_6$, then $R_t(O)=A_7$.
%The lemma is proved.
%What is $f$?
%\\
\texttt{Solutions of problems of part 1.}
Most of the solutions of problems from sections 1--3 are based on Problem~4.4.
\\
\textit{Solution of problem 1.1.}
Use problem 4.4.
\begin{itemize}
\item For each $t\in\R$ take a homothety $H_O^{2^t}$ with center $O$ and coefficient $2^t$.
It is clear that for any point $A$ $H_O^{2^{t+s}}(A)=H_O^{2^{t}}(H_O^{2^s}(A))$.
The sets $\gamma_A$ are a rays with origin $O$.
\item Draw through point $A(1,1)$ the lines $y=1$ (this is $\gamma_1$) and $x=1$ (this is $\gamma_2$).
Draw through an arbitrary point $B\in\gamma_1$ a ray $\gamma_B$.
Paint all such rays red. Now paint green and blue respectively the lines $H_O^{2^t}(\gamma_1)$ ans $H_O^{2^t}(\gamma_2)$,
i.e the lines parallel to $Ox$ and $Oy$. It is evident that the rays or the lines of each color don't intersect.
\item Consider a disc with radius 1 and center (1;1).
It is clear that exactly one ray or line of each color passes through each point of this disc.
\end{itemize}
Therefore by problem 4.4 the constructed rays and lines form a web.
It is clear that the lines given in the problem also form a web.
\\
\textit{Solution of problem 1.2.}
\emph{First solution}. Follows from the assertion of problem 0.7b.
\emph{Second solution}. This follows from problem 1.1 by a projective transformation taking the line through $2$ points from the $3$ given ones to the infinitely distant line.
\textit{Solution of problem 1.3.}
Follows from the assertion of problem 0.8b).
%\greenit{\emph{Second solution}. This follows from the lemma applied to all the rotations $R_t$ through the angles $t$ around the origin $O$.}
\textit{Solution of problem 1.4.}
Follows from the assertion of problem 0.8a).
\textit{Solution of problem 1.5.}
These lines don't form a web.
\textit{Solution of problem 1.6.}
These lines don't form a web.
\\
\texttt{Solutions of problems of part 2.}
\textit{Solution of problem 2.1.}
Apply for example an arbitrary inversion to the web formed by the lines parallel to the sidelines of some triangle.
\textit{Solution of problem 2.2.}
\emph{First solution (E. Streltsova)}. Let us prove that these lines and circles form a wweb; see figure below. Take a disc in the first coordinate quarter above the line $y=1$ so that it has no common points with the unit disk. Let the radius of the disk be $<1$. Through each point of the disk there is exactly one red and one green line because there is exactly one tangent line of each color from each point. Through each point $T$ there is exactly one line with the center at the origin $(Z)$ because the radius $(ZT)$ and the center
$(Z)$ uniquely determine a circle. The circles with the center $Z$ cannot coincide with the tangents to the unit circle. And the green and the red line cannot coincide because the disk is above the line $y=1$.
Concentric circles cannot touch each other. And the green and the red lines cannot touch the circles with the center $Z$ because the latter have radius $>1$ and our disk have no common points with the unit disk. The tangents intersect the blue circles because they contain points inside the circles. Thus the foliation condition holds.
\smallskip
\input{liza.tex}
\smallskip
The green $(a)$ and the red $(b)$ lines through the point $O$ are symmetric with respect to the line $ZO$. Thus $A_3=S_{ZO}(A_4)$. Thus the red line through $A_3$ is symmetric to the green line through $A_4$
with respect to the line $ZO$. Hence $A_2=S_{ZO}(A_5)$. Hence the green line through $A_2$ $(b')$ is symmetric to the red line through $A_5$ $(a')$.
Further,
$a=S_{ZO}(a'), b=S_{ZO}(b')$. Thus $A_6=a\cap a'=S_{ZO}(b\cap b')=S_{ZO}(A_1)$. Thus
$A_6=S_{ZO}(A_1)$. Hence $ZA_6=S_{ZO}(ZA_1)$. Therefore $ZA_6=ZA_1$, hence
the blue circle through the point $A_6$ passes through $A_1$.
The closure condition has been checked.
\smallskip
\emph{Second solution}.
Use problem 4.4.
\begin{itemize}
\item For each $t\in\R$ take a rotation $R_O^{\pi t}$ around the origin $O$ to the angle $\pi t$.
It is clear that for any point $A$ $R_O^{\pi(t+s)}(A)=R_O^{\pi t}(R_O^{\pi s}(A))$,
if $t,s\in\R$. The sets $\gamma_A$ are an arcs of the circles with center $O$.
\item Draw through the point $A(0,2)$ the rays $y=\sqrt{3}x+2, x>-\sqrt{3}/2 $ (this is $\gamma_1$)
and $y=-\sqrt{3}x+2,x<\sqrt{3}/2$ (this is $\gamma_2$).
These rays touche the unit semicircle.
Draw through each point $B\in \gamma_1$ an arc $\gamma_B$ of the circle with the center in the origin.
Paint all such arcs the red.
Now paint the rays $R_O^{\pi t}(\gamma_1)$ and $R_O^{\pi t}(\gamma_2)$
green and blue respectively, these rays touche the unit semicercles.
\item Consider a disc with radius $1/2$ and center $(0;2)$.
It is clear that exactly one arc or ray of each color passes through each point of this disc.
\end{itemize}
Therefore by problem 4.4 the constructed arcs and rays form a web.
It is clear that the lines and the circles given in the problem also form a web.
\\
\textit{Solution of problem 2.3.}
Use problem 4.4.
\begin{itemize}
\item For each $t\in\R$ take a rotation $R_O^{\pi t}$ around the origin $O$ to angle $\pi t$.
It is clear that for any point $A$ $R_O^{\pi(t+s)}(A)=R_O^{\pi t}(R_O^{\pi s}(A))$, if $t,s\in\R$.
The sets $\gamma_A$ are an arcs of the circles with center $O$.
\item Draw through the point $A(0,2)$ the rays $y=\sqrt{3}x+2,x>-\sqrt{3}/2$ (this is $\gamma_1$)
and $x=0,y>0$ (this is $\gamma_2$).
One of these rays touches the unit semicircle and the other passes through the origin.
Draw through each point $B\in\gamma_1$ an arc $\gamma_B$ of the circle with center $O$.
Paint all such arcs the red.
Now paint the rays $R_O^{\pi t}(\gamma_1)$ and $R_O^{\pi t} (\gamma_2)$ green and blue respectively.
\item Consider a disc with radius $1/2$ and center $A(0;2)$.
It is clear that exactly one line of each color passes through each point of this disc.
\end{itemize}
Therefore by problem 4.4 the constructed rays and arcs form a web. It is clear that the lines and the circles given in the problem also form a web.
\\
\textit{Solution of problem 2.4.}
Use problem 4.4.
\begin{itemize}
\item For each $t\in\R$ take a translation $T_{(0,t)}$ to the vector $(0,t)$.
It is clear that for any point $A$ $T_{(0,t+s)}(A)=T_{(0,t)}(T_{(0,s)}(A))$,
if $t,s\in\R$. The sets $\gamma_A$ are lines parallel to $Oy$.
\item Draw through the point $A(1/2+1/\sqrt{8},1/2+1/\sqrt{8})$ the arc $(x-1/2)^2+(y-1/2)^2=1/4,x>1/2,y>1/2$
(this is $\gamma_1$) and the line $y=1/2+1/\sqrt{2}$
(this is $\gamma_2$). Through each point $B\in\gamma_1$ draw a line $\gamma_B$ parallel to
$Oy$. Paint all such lines red. Now paint the arcs $T_{(0,t)}(\gamma_1)$ and the lines $T_{(0,t)}(\gamma_2)$
green and blue respectively.
\item Consider a disc with radius $1/2-1/\sqrt{8}$ and center $A(1/2+1/\sqrt{8};1/2+1/\sqrt{8})$.
It is clear that exactly one arc of each color passes through each point of this disc.
\end{itemize}
Therefore by problem 4.4 the constructed arcs form a web. Then the generalized circles considered in the problem also form a web.
It is clear that the lines and the circles given in the problem also form a web.
\textit{Solution of problem 2.5.}
Use problem 4.4.
Consider as the maps the homotheties with center $O$. Then the red lines pass through the origin. The green circles touch both segments of the firs pair. The blue circles touch both segment of the second pair. By Problem 4.4 it follows that certain arcs of these circles form a web. Then the generalized circles considered in the problem also form a web.
By problem 4.4 these lines and circles form a web.
\textit{Solution of problem 2.6.}
Use problem 4.4.
Consider as the maps the homotheties with center $O$. Then the red lines pass through the origin. The green circles are the circles with center $O$. The blue circles touche two given segments.
By problem 4.4 these lines and circles form a web.
\\
\texttt{Solutions of problems of part 3.}
The assertions the analogous of the problem 4.4 are true for the torus and the hyperboloid of revolution.
\textit{Solution of problem 3.1.}
Use problem 4.4.
Consider as the maps the rotations around the axis of the torus.
%\greenit{\bf !!! В пространстве надо указывать не центр, а ось поворота!!!}
Then the parallels are the red circles. The Villarceau circles are the green and blue circles.
By problem 4.4 these circles form a web.
\textit{Solution of problem 3.2.}
Take an arbitrary point $O$ of the torus. Draw through it the meridean $\gamma_1$ and the Villarceau circles $\gamma_2, \gamma_3$.
Paint the meridians red, paint the Villarceau circles obtained from the circle $\gamma_2$ by rotation green,
paint the Villarceau circles obtained by rotation the circles $\gamma_2$ blue. Take a sphere with center $O$ and radius $R=\frac{r}{100}$ ($r$ is the distance between $\gamma_1$
and the axis of the torus). It is clear that any
two the Villarceu circles have at most one common point inside the sphere.
By $\Omega$ denote the intersection of the sphere and the torus.
Consider an arbitrary point $O'$ inside $\om$. Draw through it the red $w_1$, green $w_2$, and blue $w_3$ circles. Let all constructed points
$A_i$ lie inside $\Omega$. Let $A_1\in w_1$. Draw through $A_1$ the green circle $w_2'$. Let $A_2$ be the common point of $w_2'$ and $w_3$.
Draw through $A_2$ the red circle $w_1'$. Let $A_3$ be the common point of $w_1'$ and $w_2$.
Draw through $A_3$ the blue circle $w_3'$. Let $A_4$ be the common point of $w_3'$ and $w_1$.
Draw through $A_4$ the green circle $w_2''$. Let $A_5$ be the common point of $w_2''$ and $w_3$.
Draw through $A_5$ the red circle $w_1'$. Let $A_6$ be the common point of $w_1''$ and $w_2$.
Draw through $A_6$ the blue circle $w_3'$. Let $A_7$ be the common point of $w_3''$ and $w_1$.
Let $\alpha$ be a plane such that $w_1$ lie in. It is clear that the circles $w_3'$,$w_1'$, $w_2'$ are the reflections of $w_2''$, $w_1''$,
$w_3''$ in $\alpha$. Therefore $A_1=A_7$.
\textit{Solution of problem 3.3.}
%\greenit{\bf !!! В пространстве надо указывать не центр, а ось поворота!!!}
Use problem 4.4.
Consider as the maps the rotations around the axis of the torus. Then the parallels are the red circles. The line lying on the hyperboloid are the green and blue lines. By problem 4.4 these circles form a web.
\bigskip
\texttt{Hints and solutions of problems of part 4.}
\textit{Hint to problem 4.1.}
Let us list a few possible examples of the sets of blue general circles:
\begin{itemize}
\item (B) an arbitrary pencil of lines (Problems~1.1 and~1.2);
\item (B) circles with a common center;
\item (B) circular arcs obtained from a given one by parallel translations along either the $Ox$ or the $Oy$ axis (Problem~4.3);
\end{itemize}
By the Graf--Sauer theorem (Problem 4.12) there are no other examples of sets of blue lines.
By the Shelekhov classification of all webs from pencils of general circles \cite[Theorem~0.1]{shelekhov-2007-cw} it follows that there are no other examples in which the set of blue circles is
a pencil. Description of all possible examples, not necessarily pencils, is an open problem.
\textit{Hint to problem 4.2.}
Let us list a few possible examples of the sets of blue general circles:
\begin{itemize}
\item (B) an arbitrary pencil of lines (Problems~1.1 and~1.2);
\item (B) pencil of circles with a limit point at the origin $O$ and the common radical axis parallel to the $Ox$ axis;
\item (B) circular arcs obtained from a given one by homotheties with center at the origin (Problem~4.3).
\end{itemize}
By the Graf--Sauer theorem (Problem 4.12) there are no other examples of sets of blue lines.
By the Shelekhov classification of all webs from pencils of general circles \cite[Theorem~0.1]{shelekhov-2007-cw} it follows that there are no other examples in which the set of blue circles is
a pencil. Description of all possible examples, not necessarily pencils, is an open problem.
\\
\textit{Hint to problem 4.3.}
Perform an inversion with the center at one of the limit points.
The obtained pencils of general circles form a web by Problem~4.4.
\\
\textit{Solution of problem 4.4.}
The foliation condition is true by the third condition of the problem. Let us show that the closure condition also is true.
Take an arbitrary point $O$ inside the disc.
Draw trough it the red ($w_1$), green ($w_2$) and blue ($w_3$) arcs of general circles.
Let all constructed points $A_i$ lie inside $\om$.
Let $A_1\in w_1$ and $t\in\R$ is such that $R_t(O)=A_1$
(such $t$ exists by the first condition of the lemma: if $w_1=\gamma_X$, where $X\in\gamma_1$,
then there exist such $y,z\in \R$, that $R_y(X)=O$ and $R_z(X)=A_1$, thus $R_{z-y}(O)=R_{z-y}(R_y(X))=R_{z}(X)=A_1$.
Therefore $t=z-y$.
%If $R_t(R_s(A))=R_{ts}(A)$ the sought $t$ can be found similarly.
%\greenit{\bf !!! Убрать сноску и написать этот текст прямо в тексте доказательства !!!}}
).
Draw through $A_1$ the green arc $w_2'$. Let $A_2$ be the common point of $w_2'$ and $w_3$. Draw through $A_2$ the red arc $w_1'$. Let $A_3$ be the common point of $w_1'$ and $w_2$. Draw through $A_3$ the blue arc $w_3'$. Let $A_4$ be the common point of $w_3'$ and $w_1$ etc.
Now let us show that $R_t(O)=A_7$, this yields that $A_1=A_7$.
We know that $R_t(O)=A_1$ , thus $R_t(w_2)=w'_2$ (it is true because $R_t(w_2)$ is the red arc passing through $A_1$, and the unique such arc is $w_2'$), therefore $R_t(A_3)\in w_2'\cap w_1'=A_2$. Since $R_t(A_3)=A_2$, then $R_t(A_4)=O$ (similarly). Since $R_t(A_4)=O$, then $R_t(A_5)=A_6$. Since $R_t(A_5)=A_6$, then $R_t(O)=A_7$.
The assertion is proved.
%What is $f$?
\\
\textit{Hints to problems 4.5--4.6.}
These problems are discussed in \cite{tabachnikov:1988}.
\\
\textit{Hints to problem 4.7.}
The solution of this problem is given in the book by Prasolov and Solov'ev \cite{prasolov-soloviev}.
\emph{Hint}. Let the equations of red lines be $a_1x+b_1y-1=0$, $a_2x+b_2y-1=0$, $a_3x+b_3y-1=0$, and let the equation of blue ones be $c_1x+d_1y-1=0$, $c_2x+d_2y-1=0$, $c_3x+d_3y-1=0$. Prove that the equation of the curve has the form
$$p(a_1x+b_1y-1)(a_2x+b_2y-1)(a_3x+b_3y-1)+q(c_1x+d_1y-1)(c_2x+d_2y-1)(c_3x+d_3y-1)=0$$
for some real numbers $p$ and $q$.
\\
\textit{Hints to problem 4.8.}
This problem is obtained from the previous one by the projective duality.
\textit{Hints to problem 4.9.}
Use Problem 4.8.
\textit{Hints to problem 4.10--4.11.}
Use Problem 4.9. A figure to Problem~4.10 by
A. Ghaneiyan Sebdani and E. Ashourioun is shown at the first page of this document.
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