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\begin{center}
{\bf {\LARGE
Excircles and Dozens of Points.}\\
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}
\end{center}
\vspace{2mm}
\begin{center}
(\bf{\large{Presented by V. Filimonov and A. Zaslavsky.}})
\end{center}
\vspace{2mm}
\begin{center}
{\bf On the Origin of this Series of Problems.}
\end{center}
The work on this series started from the problem posed by D. Tereshin.
\vspace{3mm}
{\bf Problem} (D. Tereshin). Consider triangle $ABC$ and its
excircles: one of them touches the side $AC$ at $K$ and touches the lines $AB$ and $BC$ at $L$ and $M$,
the other touches the side $AB$ at $P$ and touches the lines $AC$ and $BC$ at $Q$ and $R$.
Prove that the intersection point $X$ of the lines $LM$ and $QR$
lies on the altitude (passing through $A$) of the triangle $ABC$.
\vspace{3mm}
At once the geometrical solution was not found,
only calculations work. Some observations made this problem more exciting.
It appears that the intersection point $Y$ of
lines $KM$ and $PR$ lies on the altitude of triangle $ABC$, and the length
of the segments
$AY$ and $AX$ equal to the radii of incircle and excircle touching the side
$BC$. Some other results were obtained, and some connections with known problems from
olympiads were established.
In search for the geometrical explanation of these results
we tried to consider in details the touch points of the
sides with incircle and excircles, and the lines joining these touch points.
The incircle and the excircles have some different properties
(for example, the incircle is smaller than any of the excircles,
the incircle lies inside the triangle while the excircle lies outside the triangle).
Nevertheless, these four circles have deep common properties:
each of them touches the three sidelines
of the triangle, the center of each circle is the intersection of three angle bisectors
(either internal or external).
So as a rule, if one of the four circles has some property, then
the others have an analogous property.
That is why these four circles in some sense enjoy equal rights with respect to the
original triangle.
To understand some important geometrical results
we need to consider all the four circles simultaneously.
Thus we introduce some non-regular but
symmetrical notation (see below).
The sections A, B, C of the project were made by the author of the text jointly with
I. Bogdanov, the section D was added by A. Zaslavsky.
Also A.~Akopyan, D. Prokopenko, and V. Protassov
had made many useful notes and additions.
\vspace{5mm}
\makebox[40em][r]{P. Kozhevnikov}
\newpage
\begin{center}
{\bf {\LARGE
Excircles and Dozens of Points.}\\
}
\end{center}
\vspace{10mm}
\begin{center}
{\bf Notation.}
\end{center}
In a fixed non-equilateral triangle $ABC$ let us denote:
$R, p$ --- the radius of the circumcircle and semiperimeter;
$a, b, c$ --- lengths of $BC$, $CA$, $AB$;
$A'$, $B'$, $C'$ --- midpoints of $BC$, $CA$, $AB$;
$AH_a$, $BH_b$, $CH_c$ --- altitudes, $H$ --- orthocenter of triangle $ABC$;
$\Omega$ --- circumcircle, $O$ --- circumcenter;
$\omega_0$ --- incircle, $I_0$ --- incenter;
$\omega_1$, $\omega_2$, $\omega_3$ --- excircles (touching segments
$BC$, $CA$, $AB$, respectively),
$I_1$, $I_2$, $I_3$ --- centers of the excircles,
$r_i$ --- radii of $\omega _i$;
$I_0', I_1', I_2', I_3'$ --- centers of incircle and excircles of triangle $A'B'C'$.
The notation has the following symmetry:
Note that 6 lines $I_iI_j$ ($i\neq j$) are internal and external bisectors of triangle $ABC$.
Therefore the quadruple $I_0, I_1, I_2, I_3$ is orthocentric,
and $ABC$ is the orthotriangle
(that is the triangle having feet of altitudes as vertices) for each of triangles
$I_0I_1I_2$,
$I_1I_2I_3$, $I_2I_3I_0$, $I_3I_0I_1$).
To each of the points $A, B, C$ we put into correspondence
a partition of 4-element set $\{0, 1, 2, 3\}$ into two 2-element subsets:
$A= I_0I_1 \cap I_2I_3$, $B= I_0I_2 \cap I_1I_3$, $C= I_0I_3 \cap I_1I_2$.
Also see the further notation
\vspace{10mm}
\begin{center}
{\bf Series А: The First Dozen: Touch Points}
\end{center}
Let $A_i$, $B_i$, $C_i$ ($i=0, 1, 2, 3$) be touch points
of $\omega_i$ and lines $BC$, $CA$, $AB$, respectively (see 12 red points in Fig. A).
Prove the following statements.
\begin{enumerate}
\item[A1.]
$A_0$ и $A_1$ (and also $A_2$ and $A_3$) are symmetric with respect to $A'$, moreover,
$A_0A_3 = A_1A_2 = c$, $A_0A_2 = A_1A_3 = b$, $A'A_0=A'A_1 = \dfrac{|b-c|}{2}$,
$A'A_2=A'A_3 = \dfrac{b+c}{2}$. (Similarly there is symmetry with respect to $B'$ and $C'$.)
\item[A2.]
а) $AA_i$, $BB_i$, $CC_i$ are concurrent.
б) $AA_1$, $BB_2$, $CC_3$ are concurrent.
(Similarly, triples of lines $AA_0$, $BB_3$, $CC_2$;
$AA_2$, $BB_1$, $CC_0$; $AA_3$, $BB_0$, $CC_1$
are either concurrent or parallel.)
\item[A3.] Radical axis of pairs $\omega_i$ and
$\omega_j$ are internal and external bisectors of triangle $A'B'C'$.
(Find the radical centers of triples of circles
$\omega_0$, $\omega_1$, $\omega_2$, $\omega_3$.)
\item[A4.]
In the set of circles touching $\omega_1$,
$\omega_2$, $\omega_3$ there exist three circles passing through $I_0'$.
(Formulate and prove the similar statement for the other triples of circles.)
\item[A5.]
$AA_1 \parallel I_0A'$ (similarly $AA_0 \parallel I_1A'$, $AA_2 \parallel I_3A'$,
$AA_3 \parallel I_2A'$, etc.).
\item[A6.]
$I_0A_1$, $I_1A_0$, $I_2A_3$, $I_3A_2$ are concurrent.
Determine the intersection point of these lines.
\end{enumerate}
\newpage
\begin{center}
{\bf Series B: The Second Dozen: "Foci"}
\end{center}
Let us denote $B_{01}= B_{10}= A_0B_0 \cap A_1B_1$,
$B_{23}= B_{32} = A_2B_2 \cap A_3B_3$. (Here $A_0B_0 \cap A_1B_1$ is
$B_{01}$, and not $A_{01}$, since $A$ corresponds to the partition of
$\{0, 1, 2, 3\}$ into pairs $0, 1$ and $2, 3$.)
Similarly define all 12 points:
$A_{ij}$ with $i\in\{0, 1\}$,
$j\in\{2, 3\}$ (we put $A_{ij} = A_{ji}$);
$B_{ij}$ with $i\in\{0, 2\}$, $j\in\{1, 3\}$;
$C_{ij}$ with $i\in\{0, 3\}$, $j\in\{1, 2\}$.
(See 12 violet points in Fig. B1, B2.)
Prove the following statements.
\begin{enumerate}
\item[B0.]
$\angle B_2B_{23}B_3 = 90^{\circ}$
(similarly for the other angles).
\item[B1.]
$B_{23}$, $C_{23}$, $A_2$, $A_3$ are concyclic.
Find the center of the circle passing through these points.
(Similarly, $B_{01}$, $C_{01}$, $A_0$, $A_1$ are concyclic, etc.,
thus red and violet points belong to 6 circles.)
\item[B2.]
$A_{ij}$ lies on the midline $B'C'$
(similarly, $B_{ij}$ and $C_{ij}$ lie on midlines,
thus 12 violet points lie on 3 lines $A'B'$, $B'C'$, $C'A'$).
\item[B3.]
$A_{13}$ (the same for $A_{02}, B_{01}, B_{23}$) lie on the circle with diameter $AB$,
(moreover, $A_{02} B_{01} A_{13} B_{23}$ is a rectangle.)
(Thus 12 violet points lie on 3 circles
with diameters $BC$, $CA$, $AB$).
\item[B4.]
Find the lengths $A_{02}A_{03}$, etc., in terms of $a, b, c$.
\item[B5.]
$A_{ij}$ lies on $I_iI_j$, moreover,
$A_{ij}$ is the projection of $A$ to $I_iI_j$.
(thus 12 violet points belong to 6 bisectors of
triangle $ABC$).
\item[B6.]
$A_{02}$ and $C_{02}$ are {\it foci} of $\omega_0$ and $\omega_2$
({\it Foci} means that $A_{02}$ and $C_{02}$ is a pair of points inverse to each other with respect
to each of two circles).
(Thus 12 violet points are partitioned into 6 pairs of foci; in particular, from that it follows that
each of $\omega_i$ contains exactly 3 violet points).
\item[B7.]
Determine the radical centers of triples of circles from Problem B1
having distinct centers.
\item[B8.]
Six points $A_{03}, A_{02}, C_{02}, C_{23}, B_{23}, B_{03}$ lie on a circle
(also there exist 3 circles constructed in the same manner). Determine
the centers of these circles.
Find the radii of these circles in terms of $a$, $b$, $c$.
\item[B9.]
$A_{02}$ and $A_{13}$ are either the centers of incircle and excircle
or the centers of excircles, for the triangle $B'H_aH_b$.
\end{enumerate}
\vspace{5mm}
\begin{center}
{\bf Series C: The Third Dozen: Intersections on the Altitudes}
\end{center}
Let $A_{(3)} = A_0C_0\cap A_1B_1$ (Here $A_0C_0\cap A_1B_1$ is
$A_{(3)}$, and not $A_{(2)}$, since $C$ corresponds to
the partition of $\{0, 1, 2, 3\}$ into pairs $0, 3$ and $1, 2$,
here 3 belongs to the pair containing 0). Similarly,
$A_{(2)} = A_0B_0\cap A_1C_1$,
$A_{(0)} = A_2B_2\cap A_3C_3$,
$A_{(1)} = A_2C_2\cap A_3B_3$; in the same manner define $B_{(i)}$ and $C_{(i)}$ ---
totally 12 green points in Fig. C.
Prove the following statements.
\begin{enumerate}
\item[C1.]
Points $A_{(i)}$ lie on the line $AH_a$
(similarly for $B_{(i)}$ and $C_{(i)}$,
thus 12 green points lie on the altitudes of triangle
$ABC$).
\item[C2.]
The length of $AA_{(i)}$ is equal to $r_i$.
\item[C3.]
$A_{(i)}A_i$ is parallel to one of two bisectors of angle $A$.
\item[C4.]
Prove that $A_{(1)}A_1$, $B_{(2)}B_2$, and $C_{(3)}C_3$
are concurrent.
(Similarly, there exist three triples of concurrent lines:
$A_{(0)}A_0$, $B_{(3)}B_3$, $C_{(2)}C_2$; $A_{(3)}A_3$, $B_{(0)}B_0$, $C_{(1)}C_1$;
$A_{(2)}A_2$, $B_{(1)}B_1$, $C_{(0)}C_0$.)
\item[C5.]
Triangles $A_1B_2C_3$ and $A_{(0)}B_{(0)}C_{(0)}$ are symmetric (with respect to a point).
Define the center of symmetry.
(Similarly, pair of triangles
$A_0B_3C_2$ and $A_{(1)}B_{(1)}C_{(1)}$,
$A_3B_0C_1$ and $A_{(2)}B_{(2)}C_{(2)}$,
$A_2B_1C_0$ and $A_{(3)}B_{(3)}C_{(3)}$
are symmetric.)
\item[C6.]
Triangles $A_{(1)}B_{(2)}C_{(3)}$,
$A_{(0)}B_{(3)}C_{(2)}$, $A_{(3)}B_{(0)}C_{(1)}$, $A_{(2)}B_{(1)}C_{(0)}$
Have a common circumcenter
(thus green points lie on 4 concentric circles). % с центром $H$).
Define the common circumcenter.
\item[C7.]
Find $AH$, $BH$, $CH$ in terms of radii $r_i$.
\item[C8.]
Find the radius of the circumcircle of triangle $A_{(1)}B_{(2)}C_{(3)}$ in terms of
$R$ and $r_0$. (Similarly, find the radii of the circles from Problem C6.)
\item[C9.]
$I_iA'$ passes through $A_{(i)}$
(similarly, $I_iB'$ passes through $B_{(i)}$,
$I_iC'$ passes through $C_{(i)}$).
\item[C10.]
Let $l_a$ be a line passing through $A$ and parallel to $BC$.
$M = A_0C_0 \cap l_a$, $N = A_0B_0 \cap l_a$.
Prove that $A_{(0)}$ is the circumcircle of the triangle $A_0MN$.
Determine the orthocenter of the triangle $A_0MN$.
\end{enumerate}
\begin{center}
{\bf Series D: The Fourth Dozen.}
\end{center}
Let $C^*_0=A_0B_0\cap A_1B_2$, and similarly define 12 blue points
$A^*_i$, $B^*_i$, $C^*_i$ (see Fig. D).
(The description of these points is as follows:
Take one of the circles, for example $\omega_0$.
Take its two touch points, say $A_0$, $B_0$.
Take the touch points of these sides with two other
circles that are symmetric to $A_0$, $B_0$ with respect to the midpoints of the sides
--- $A_1$, $B_2$. Take the intersection points of the lines joining pairs of these points.) %, в данном случае $C^*_0=A_0B_0\cap A_1B_2$.
Prove the following statements.
\begin{enumerate}
\item[D1.]
The sidelines of triangle $A^*_iB^*_iC^*_i$ pass through the vertices of triangle $ABC$.
\item[D2.]
A line passing through $C^*_i$ intersects $BC$, $AC$ at $A''$, $B''$, respectively.
Show that $A''B^*_i$ and
$B''A^*_i$ intersect at some point $C''$ of the line $AB$.
\item[D3.]
$AA''$, $BB''$, $CC''$ have a common point that is isogonally conjugate
to some point of the line $OI_i$.
\item[D4.]
The circumcircle of triangle $A''B''C''$ passes through the Feuerbach point $F_i$.
%Наверное, есть еще какие-то свойства.
\item[D5.]
Four blue points denoted by the same letter lie on a sideline
of the orthotriangle.
\item[D6.]
a) Triangles $A^*_iB^*_iC^*_i$ and $ABC$ are perspective (i.e. the lines joining corresponding vertises of
these triangles are concurrent)
b)Try to find some relations between four centers of perspective.
\item[D7.] (The generalization of the problem D4)
Let $C^{**}$ be a point on line $H_a H_b$. An arbitrary line passing through $C^{**}$
intersects $BC$, $AC$ at $A''$, $B''$, respectively. Let $P$ be the point of intersection of lines $AA''$ and $BB''$,
and $C''$ be the point of intersection of lines $CP$ and $AB$. Then
the circumcircles of all triangles $A''B''C''$ have the common point.
\end{enumerate}
Tasks from series A, B, C marked $\#$ and also from series D were given to the paticipants after the intermediate finish.
\end{document}