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\begin{center}

{\Large\bf Auxiliary conics}

{\bf A.Zaslavsky, A.Akopjan}

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\section*{Solutions}

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\zd Using the theorem  3 for four given lines and infinity line we receive
that tere exists the single parabola touching the given lines. By the theorem
1 its focus is the Michel point of the quadrilateral and the directrix is its
Aubert line. Also we can apply the dual theorem 10 to for given lines, the
infinity line and three medial lines of the diagonal triangle. It results
that the medial lines touche the same parabola, so we receive the proof of
p.c).

\zd Let tehe conic $ABCPQ$ be considered. The pole of $PQ$ with respect to
this conic is the sought points. This follows from the p.c) of the addition
and the theorem 4.

{\bf Addition}
\pp the incenter;

\pp the Lemoine point isogonally conjugated to the centroid;

\pp the infinite point with the direction perpendicular to $PQ$.

\zd Let $X'$ be the projection of $X$ to the directrix. Suppose that the
bissector of angle $FXX'$ don't coincide with the tangent. Then it intersect
the parabola in some point $Y$ distinct from X and $FY=YY'$, where $Y'$ is
the projection of $Y$ to the directrix. As $FX=XX'$, $\angle FXY=\angle X'XY$
the triangles $FXY$ and $X'XY$ are congruent, so $FY=YX'$. But $YY'<YX'$~---
contradiction.

\zd Using the notations of previous problem we receive that $F$ and $X'$ are
symmetric with respect to the tangent.

\zd Using the result of previous problem we receive that the sought locus is
the tangent to the parabola in its vertex.

\zd Let the pencil of conics passing through the incenter and the excenters of
$ABC$ be considered. By the theorem 8 all polars of $X$ with respect to the
conics of this pencil pass through $X'$. Let ${\cal C}$ be such conic that
the polar of $X$ with respest to ${\cal C}$ coincide with  $X'Y'$. Then the
polar of $Y'$ pass through $X$, and so coincide with $XY$. As $U$ is the
common point of polars of $X$ and $Y'$, $XY'$ is the polar of $U$. So the
polar of $U$ pass through $V$. Similarly we can find such conic that the polar
of $U$ coincide with $X'Y$. By the theorem 8 $V$ is isogonally conjugated to
$U$.

\zd By the Brianchon theorem the given triangles have cjmmjn inscribed conic.
Let some line passing through $A'$ be intersect $BC$ in the point $P$, and
the parallel line passing through $B'$ be intersect $AC$ in the point $Q$.
Using the Brianchon theorem for the hexagon $A'XQPYB'$ where $X$, $Y$ are the
the infinite points of $AC$ and $BC$ repectively, we obtain that the line $PQ$
also touches this conic. This follows the sought assertion.

\zd The common points of common external tangents are collinear. So we can use
the dual four conics theorem.

\zd By the theorem 7 $T$ is the center of some inscribed conic. As the
distances from $T$ to opposite sidelines are equal, these sideline are
parallel or symmetric with respect to some axis of this conic. If the both
pairs of opposite sideline are symmetric with respect to the same axis then
there exists the circle inscribed to given quadrilateral. If the symmetry
axis are distinct then the bissectors of two angles formed by the opposite
sideline are perpendicular and so the quadrilateral is cyclic.

\zd This is the reformulation of previous problem for the quadrilateral formed
by the lines $AX$, $BX$, $AY$, $BY$.

\zd By the theorems 1, 12 there exists a parabola with focus $F$ touching the
sidelines of the triangle and two perpendicular lines passing through its
orthocenter $H$. The midpoints of stiked segments are the circumcenters of
respective rectangular triangles, so they lie in medial perpendicular to $HF$.

\zd If the triangle $ABC$ is rectangular and isosceles ($AC=BC$) and $C^*$ is
the infinite point perpendicular to $AB$ then the locus of points $P$ is the
circumcenter of $ABC$. Using respective projective transformation we obtain
that in general case the locus of $P$ is the circumconic.

Now as this conic pass through the orthocenter it is the equilateral hyperbola
and by theorem 15 all circles $A'B'C'$ pass through its center.

\zd The orthocenter, the incenter the Gergonne point and the Nagel point are
isogonally conjugated to the circumcenter, the incenter and two homothety
centers of the incircle and the circumcircle respectively. So the isogonal
images of four points lie on the line passing through $O$. By the theorem 9
these points lie on the equilateral circumhiperbola and we can use the
assertion of previous problem.

\zd The pedal circles of two isogonal points coincide. So the assertion of
problem follows from the theorems 9, 15.

\zd The similarity ${\cal F}$ transforming one of given triangles to the
other is the composition of reflection in some line $l$ passing through common
orthocenter $H$ and homothety with center $H$. Let $\Gamma$ be an equilateral
circumhyperbola of given triangle $ABC$ such that one of its asymptotes is
parallel to $l$. Similarly $\Gamma'$ is the equilateral hyperbola of second
triangle $A'B'C'$ with the same directions of asymptotes. $\Gamma$ and
$\Gamma'$ have three real common points: $H$, and two infinite points. So
their fourth common point $P$ is also real. ${\cal F}$ define projective
tranformation from $\Gamma$ to $\Gamma'$ such that three common points of
hyperbolaes leave fixed. The single transformation with this property is the
projection from fourth common point $P$. So $P$ is the perspective center of
given triangles.

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