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\begin{center}

{\large\bf Three parabolas}

{\large\bf Solutions}

{\bf F.Nilov, A.Zaslavsky}

\end{center}


\section{Introductory problems}

\begin{enumerate}

\item Let $X'$ be the projection of $X$ to the directrix. Then $FX=XX'$.
Suppose that the bisectrix $l$ of the angle $X'XF$ isn't the
tangent to parabola. Then $l$ intersect the parabola in some point
$Y$, distinct from $X$. Note that the triangles $FXY$ and $X'XY$
are equal and so $FY=YX'$. Let $Y'$ be the projection of $Y$ to
the directrix. Then $YY'=FY=YX'$. But it is impossible because
$YY'$ and $YX'$ are the hypothenuse and the cathetus of rectangle
triangle $YX'Y'$.

\item Let $l$ touche the parabola in the point $X$. Then $FX=XX'$.
By problem 1 $l$ is the bisectrix of isosceles triangle $FXX'$. So
$l$ is the medial perpendicular to the segment $FX'$, and the
points $X'$ and $F$ are symmetric wrt $l$.

\item By problem 2 the lines $PX$ and $PY$ are medial perpendiculars
of segments $FX'$ и $FY'$. So their common point $P$ is the
circumcenter of $FX'Y'$.

\item Consider the medial line of the trapezoid $XX'Y'Y$. It is
perpendicular to the directrix and so parallel to the axis of the
parabola. Also it is the median of $PXY$, because it pass through
the midpoint of segment $XY$ and (by problem 3) through $P$.

\item Consider the line $l$ passing through $P$ and parallel to
the axis of the parabola. We must prove that the angle $\phi$
between $l$ and $PX$ is equal to the angle $FPY$. Let $X''$ and
$Y''$ be the projections of $F$ to the lines $PX$ and $PY$. By
problem 2 $X''$ and $Y''$ are the midpoints of $FX'$ and $FY'$. So
the lines $X'Y'$ and $X''Y''$ are parallel. This yields that $l$
and $X''Y''$ are perpendicular. Now we have $\angle
\phi=90^{\circ}-\angle PX''Y''=90^{\circ}-\angle PFY''=\angle
FPY$, q.e.d.

\item Let $P$ and $Q$ be isogonal conjugated. Note as $P_c$ and $Q_c$,
$P_a$ and $Q_a$, $P_b$ and $Q_b$ the reflections of $P$ and $Q$ in
$AB$, $BC$ and $AC$. It is evident that $BP_c=BP=BP_a$ and $\angle
P_cBP_a = 2\angle B$. Note that $\angle QBC=\angle PBA=\angle
ABP_c$. So $\angle P_cBQ = \angle B = 1/2 \angle P_cBP_a=\angle
P_aBQ$, and $QP_c=QP_a$. Similarly $QP_a=QP_b$ и $PQ_c=PQ_a=PQ_b$.
Using the homothety with center $P$ and coefficient $1/2$ we
receive that the circle $\omega$ with center in the midpoint of
$PQ$ and the radius $QP_c/2$ is the pedal circle of $P$. Similarly
$\omega$ is the pedal circle of $Q$.


\item As $\Pi_a$ and $\Pi_b$ are inscribed in the angles $A$ and $B$
their common points lie inside or on the sidelines of the triangle
$ABC$. $C$ isn't the unique common point of parabolas because
their tangents in $C$ doesn't coincide. So there exists the common
point $C'$ distonct from $C$. It is clear that the sidelines of
$ABC$ doesn't contain the common points distinct from $C$. Suppose
that $\Pi_a$ and $\Pi_b$ have inside the triangle the common point
$C''$ distinct from $C'$. As $A$, $C$, $C'$ и $C''$ lie on
$\Pi_b$, they are the vertex of convex quadrilateral. Similarly
$B$, $C$, $C'$, and $C''$ are the vertex of convex quadrilateral.
But it is impossible. So $\Pi_a$ and $\Pi_b$ have exactly two
common points

\item  It is known that an arbitrary affine map transforms any parabola
to the parabola. Consider the map transforming the triangle $ABC$
to regular triangle. It is evident that in regular triangle the
common points of parabolas lie on the medians. So $AA'$, $BB'$ and
$CC'$ are the medians of $ABC$ and concur in its centroid $M$.

\item By problems 4 and 5 the lines $AM$ and $AF_a$ are symmetric wrt
the bisectrix of angle $A$. So $AF_a$, $BF_b$ and $CF_c$ concur in
the Lemoine point $L$ of $ABC$.

\item Let the directrix $d_a$ of $\Pi_a$ intersect $BC$ in the point $A'$.
Note the projections of $B$ and $C$ to $d_a$ as $B_1$ and $C_1$.
It is clear that $BB_1=BF_a$ and $CC_1=CF_a$. As the triangles
$A'BB_1$ and $A'CC_1$ are similar $\frac {A'B} {A'C}=\frac
{BB_1}{CC_1}=\frac {BF_a}{CF_a}$. By problems 1 and 5 $\angle
F_aBA=\angle F_aAC$ and $\angle F_aCA=\angle F_aAB$. Usung the
sinus theorem for the triangle $F_aBA$ we receive that $BF_a=\frac
{\sin (\angle F_aAB) \cdot AF_a} {\sin (\angle ABF_a)}$. Similarly
$CF_a=\frac {\sin (\angle F_aAC) \cdot AF_a} {\sin (\angle
ACF_a)}$. So $\frac {BF_a}{CF_a}=\frac {\sin^2 (\angle F_aAB)}
{\sin^2 (\angle F_aAC)}$.

Let $d_b$ and $d_c$ be the directrix of $\Pi_b$ and $\Pi_c$. Note
as $B'$, $C'$ the common points of $d_b$ and $AC$, $d_c$ and $AB$.
Then
$$
\frac {A'B \cdot B'C \cdot C'A} {A'C \cdot B'A \cdot C'B} =
\frac {BF_a \cdot CF_b \cdot AF_c} {CF_a \cdot AF_b \cdot BF_c} =
\frac {\sin^2 (\angle F_aAB) \cdot \sin^2 (\angle F_bBC) \cdot
\sin^2 (\angle F_cCB)} {\sin^2 (\angle F_aAC) \cdot \sin^2 (\angle
F_bBA) \cdot \sin^2 (\angle F_cCA)} = 1.
$$
The last equality follows from the Ceva theorem for the cevians
$AF_a$, $BF_b$ and $CF_c$. By Menelaus theorem $A'$, $B'$ and $C'$
are collinear. So by Desargues theorem the triangles are
perspective.

\eject

\begin{center}

{\large\bf Three parabolas}

{\large\bf Solutions}

{\bf F.Nilov, A.Zaslavsky}

\end{center}
\section{Basic problems}

\item Note the midpoint of $AB$ as $C_0$. We have
$\angle C_0CB=\angle F_cCA$. By problems 1, 4 and 5 $\angle
MCB=\angle F_cBC$ и $\angle MCA=\angle F_cAC$. So $\angle
AF_cC=180^{\circ}-\angle F_cAC-\angle F_cCA=180^{\circ}-\angle
C=180^{\circ}-\angle F_cCB-\angle F_cBC=\angle BF_cC$.

\item By problem 11 $\angle AF_cC=\angle BF_cC=180^{\circ}-\angle C$.
So $\angle AF_cB=360^{\circ}-2(180^{\circ}-\angle C)=2\angle C=
\angle AOB$. This follows that $A$, $B$, $F_c$ and $O$ lie on the
circle $\omega$.

\item Let $A_0$ и $B_0$ be the midpoints of $BC$ and $AC$.
The problem 5 yields next

{\it Lemma. The focus of the parabola inscribed in the triangle
lies on the circumcircle of this triangle}

Prove that the circumcircle of $A_0B_0C$ pass through $F_c$. Note
as $C'$ the second common point of $CF_c$ with the cirumcircle of
$ABC$. By problems 1 and 5 $\angle F_cBC=\angle F_cCA$ and $\angle
F_cAC=\angle F_cCB$. So $\angle C'BA=\angle F_cCA=\angle F_cBC$ и
$\angle BAC'=\angle F_cCB$. From this the triangles $F_cCB$ and
$C'AB$ are similar and $\frac {CF_c}{C'A}=\frac {BF}{BC'}$. By
problem 11 $\angle BF_cC'=\angle AF_cC'$. Also $\angle
C'BF_c=\angle ABC=\angle AC'F_c$. So the triangles $F_cBC'$ and
$F_cC'A$ are similar and $\frac {F_cC'}{C'A}=\frac {BF}{BC'}=\frac
{CF_c}{C'A}$. It follows that $F_cC=F_cC'$. The homothety with
center $C$ and coefficient $\frac {1}{2}$ transforms $B$, $A$ and
$C'$ to $A_0$, $B_0$ and $F_c$, q.e.d.

\item By problem 11 the points $A$, $B$, $F_c$ and $O$ lie on the
circle $\omega$. Let $O'$ be the second common point of $CF_c$ and
$\omega$. By problem 10 $\angle AF_cO'=\angle BF_cO'$. So $O'$ is
is the midpoint of an arc $AB$ of $\omega$ and the segment $OO'$
is the diameter of $\omega$. From this $\angle LF_cO=\angle
OF_cO'=90^{\circ}$. Similarly $\angle LF_aO=\angle
LF_bO=90^{\circ}$. So $F_a$, $F_b$ lie $F_c$ lie on the circle
with diameter $OL$.

\item Let $A_0$ and $B_0$ be the midpoints of $BC$ and $AC$.
Consider the point $F'_c$ symmetric to $F_c$ wrt $A_0B_0$. By
problem 13 $A_1B_1$ touches $\Pi_c$. So by problem 2 $F'$ lies on
the directrix of $\Pi_c$. By problem 13 the quadrilateral
$CA_1F_cB_1$ is cyclic. So $\angle A_1F'_cB_1 =\angle
A_1F_cB_1=180^{\circ}-\angle C$ and $F'_c$ lies on the Euler
circle of $ABC$. Note as $M'$ the midpoint of $A_0B_0$. By problem
13 $A_1B_1$ is the bisectrix of angle $CMF_c$. So $F'_c$ lies on
$CM'$. Thus the common point of the median and Euler circle lies
on the directrix of $\Pi_c$.

\item
Let $A''$, $B''$, $C''$ be the vertex of directrix triangle. Note
as $A'$, $B'$ and $C'$ the common points of medians of $ABC$ and
its Euler circle. By problem 15 these points lie on the sidelines
of $A''B''C''$. Note as $C_0$ the midpoint of $AB$. Let $G$ be the
centroid of $A''B''C''$. Note that
$$
\frac {\sin (\angle GC''B'')} {\sin (\angle GC''A'')}=\frac {\sin
(\angle B'')} {\sin (\angle A'')}=\frac {\sin (\angle C_0MA)}
{\sin (\angle C_0MB)}=\frac {MB}{MA}=\frac {MA'}{MB'}=\frac {\sin
(\angle MC''B'')} {\sin (\angle MC''A'')}.
$$
The first and the third equalities are correct because $C''G$ and
$MC_0$ are the medians of $A''B''C''$ and $AMB$. So $\angle
GC''B''=\angle MC''B''$ and $G$ lies on $C''M$. Similarly $G$ lies
on $A''M$ and $B''M$. this follows that $G$ and $M$ coincide.

\item By problem 4 $AM$ is perpendicular to $d_a$. Similarly
$BM$ is perpendicular to $d_b$ and $CM$ is perpendicular to $d_c$.
So $M$ is the orthology center of considering triangles. Let $A'$,
$B'$ и $C'$ be the common points of respective directrix. As the
medians of $ABC$ are perpendicular to the sidelines of $A'B'C'$ we
have :

(1) $(\overrightarrow {C'A'}+ \overrightarrow {A'B'})\cdot (\overrightarrow {AB}+\overrightarrow {AC})=0$.

(2) $\overrightarrow {A'B'} \cdot (2\overrightarrow {CA}+\overrightarrow {AB})=0$.

(3) $\overrightarrow {C'A'} \cdot (2\overrightarrow {AB}+\overrightarrow {CA})=0$.

Summing (1) and (3) and substracting from the result (2), we
receive that $\overrightarrow {CA} \cdot \overrightarrow {A'B'}=
\overrightarrow {C'A'} \cdot \overrightarrow {AB}$. From this and
(1) $(\overrightarrow {C'A'}- \overrightarrow {A'B'})\cdot
(\overrightarrow {AB} - \overrightarrow {AC})=0$. This follows
that $BC$ is perpendicular to the median $A'G$ of $A'B'C'$. By
problem 16 the points $G$ and $M$ coincide. So $A'M$ and $BC$ are
perpendicular. Similarly $B'M$ and $AC$, $C'M$ and $AB$ are
perpendicular. So $M$ is the common orthology center of $ABC$ and
$A'B'C'$.

Note that two orthological triangles with coinciding orthology
centers are perspective. So we receive another solution of problem
10.

\item As $d_a$ is perpendicular to the median $AA_0$, then by problem 15
$d_a$ pass through the common point $A''$ of an altitude $AA'$ and
the Euler circle. Similarly the common points of Euler circle
$B''$, $C''$ of Euler circle and the altitudes $BB'$, $CC'$ lie on
$d_b$ and $d_c$. Note that Euler circle of $ABC$ is the pedal
circle of $M$ wrt the directrix triangle. By problem 16 $M$ is the
centroid of directrix triangle. So the medians of $ABC$ are
parallel to the lines $A''L'$, $B''L'$ and $C''L'$ , where $L'$ is
the Lemoine point of the directrix triangle. It is known that the
triangles $A''B''C''$ and $ABC$ are homothetic with center $H$ and
coefficient $1/2$. So, as $L'$ is the centroid of $A''B''C''$ it
is the midpoint of $HM$ and lies on the Euler line of $ABC$.

\item Let $T$ be the directrix triangle. Consider the parabolas
$\Pi_a'$, $\Pi_b'$ и $\Pi_c'$, touching the sidelines of $T$ in
its vertex. Note as $T'$ the triangle formed by the directrix of
these parabolas. The respective sidelines of $T'$ and $ABC$ are
perpendicular to the medians of $T$. So $T'$ and $ABC$ are
homothetic and $M$ is the homothety center because by problem 16
the centroids of $T'$ and $ABC$ coinside. Note the Lemoine point
of $T'$ as $L'$. By th homothety of $T'$ and $ABC$ the points $M$,
$L$ and $L'$ are collinear. By problem 18 $L'$ Lies on the Euler
line of $T$. So the line passing through $L$, $M$ and $L'$
coincide with the Euler line of $T$.

\item Let $C'$ be the common point of $AB$ and $P_aP_b$.
The points $C'$, $P_c$, $A$  $B$ are harmonic because $\frac {C'A}
{C'B} = \frac {AP_c} {P_cB}$. By problem 11 $\frac {AP_c} {P_cB} =
\frac {AF_c} {F_cB}$. Let $C''$ be the common point of $d_c$ and
$AB$. By problem 10 $\frac {C''A}{C''B} = \frac {AF_c}{F_cB}$. So
$\frac {C'A} {C'B} = \frac {AF_c}{F_cB} =\frac {C''A}{C''B}$ and
the points $C'$ and $C''$ coincide. By problem 10  $A''$, $B''$
and $C''$ are collinear. So $A'$, $B'$ and $C'$ are also collinear
and by the Desargues theorem the triangles $ABC$ and $P_aP_bP_c$
are perspective.

\item By problem 20 the perspective axis of $ABC$, $P_aP_bP_c$
and the directrix triangle coincide. Now use next

{\it Lemma. If the perspective axis of three mutually perspective
triangles coincide then their pair perspective centers are
collinear.}

{\it Proof.} Consider the projective map transforming the commo
perspective axis to the infinite line. It transforms given
triangles to the homothetic triangles. Their homothtety centers
are collinear.

\item Let $A_1$, $B_1$ и $C_1$ be the midpoints of the sides of
$ABC$ and $A_2$, $B_2$ и $C_2$ be the feet of its altitudes. Note
as $C'''$ the common point of $CC_1$ and the Euler circle. By
problem 15 $C'''$ lies on $d_c$. Let $C'$ be the common point of
$d_c$ and $A'B'$, and $C''$ be the common point of $CC_1$ and
$A_2B_2$. Note that $B_2C_1=AC_1=BC_1=A_2C_1$. So $\angle
B_2C'''C_1=\angle A_2C'''C_1$, because the quadrilateral
$A_2C'''B_2C_1$ is cyclic. By problem 5 $d_c$ is perpendicular to
$C'''C''$. So $C'''C'$ is the external bisectrix of angle $C'''$,
and $\frac {C'A_2}{C'B_2}=\frac {C''A_2}{C''B_2}$. Similarly
$\frac {B'C_2}{B'A_2}=\frac {B''C_2}{B''A_2}$ and $\frac
{A'B_2}{A'C_2}=\frac {A''B_2}{A''C_2}$, where $A'$, $B'$ are the
common points of $d_a$ and $B_2C_2$, $d_b$ and $A_2C_2$; $A''$,
$B''$ are the common points of $AA_1$, $BB_1$ and the Euler
circle. Note that
$$
\frac {C'A_2\cdot B'C_2\cdot A'B_2}{C'B_2\cdot B'A_2\cdot
A'C_2}=\frac {C''A_2\cdot B''C_2\cdot A''B_2}{C''B_2\cdot
B''A_2\cdot A''C_2}=1.
$$
So by Menelaus theorem $A'$, $B'$, $C'$ are collinear and by
Desargues theorem the directrix triangle and the orthotriangle are
perspective .

\item Consider the triangle formed by the medians of $ABC$.
By problem 5 its angles are equal to the angles of directrix
triangle. Prove that the Brocard angles of $ABC$ and the triangles
formed by its medians are equal. Next formulaes are correct for
any triangle:
\smallskip

(1) $\ctg \phi = \ctg \alpha + \ctg \beta + \ctg \gamma$.

\smallskip

(2) $\ctg \alpha + \ctg \beta + \ctg \gamma= \frac {a^2+b^2+c^2}{4S}$.

\smallskip

(3) $S=\frac {3}{4} S_m$.

\smallskip

(4) $a^2+b^2+c^2=\frac {3}{4} (m_a^2+m_b^2+m_c^2)$.

\smallskip

there $\alpha$, $\beta$ and $\gamma$ are the angles of triangle,
$\phi$ is its Brocard angle; $S$ and $S_m$ are the areas of the
triangle and the triangle formed by its medians; $a$, $b$ and $c$
are the lengths of the sides; $m_a$, $m_b$ and $m_c$ are the
lengths of the medians.

\smallskip

Using these formulaes to $ABC$ we receive

$$\ctg \phi =\ctg \alpha +\ctg \beta +\ctg \gamma=\frac {a^2+b^2+c^2}{4S}=\frac {m_a^2+m_b^2+m_c^2}{4S_m}=\ctg \phi_m,$$

where $\phi_m$ is the Brocard angle of the triangle formed by the
medians. So, $\phi =\phi_m$, q.e.d.


\end{enumerate}

\end{document}
