Seventeenth ROMANIAN MASTER OF MATHEMATICS

Consider the composition $\iota$ of an inversion centered at $A$ and the reflection in the bisector $AM$ that swaps $B$ and $C$. Then $\iota$ swaps the circle $\Omega$ with the line $BC$, hence it swaps $O$ with the reflection $L$ of $A$ in $BC$. Hence $\iota (\Gamma )$ is the circle ${\Gamma }^{\ast }=(BCL)$,

i.e., the reflection of $\Omega$ in $BC$ which passes through $H$. Let $S$ and $Q$ be the centers of $\Gamma$ and ${\Gamma }^{\ast }$, respectively; then $AM$ is the angle bisector of $\angle QAS$.

Since $\iota$ swaps $\Gamma$ and ${\Gamma }^{\ast }$, they are seen from $A$ at the same angle, so there exists a rotational homothety $h$ centred at $A$ mapping $\Gamma$ to ${\Gamma }^{\ast }$; the angle of $h$ is $\angle SAQ$. Notice that the rays $AH$ and $AF$ are obtained from $AO$ by reflections in $AM$ and $AS$, respectively, so $\angle HAF=2\angle MAS=\angle QAS$. This easily yields that $H=h(F)$. Hence the triangles $AHF$ and $AQS$ are similar, and $\angle AHF=\angle AQS$.



Now, let the circle ($AHF$) meet $\Gamma$ again at $T$. Using directed angles, we get $\angle ATF=\angle AHF=\angle AQS$; next, since $FO\perp AS$, we have $$\angle FT{A}^{\prime }=\angle FO{A}^{\prime }=\pi /2-\angle OAS=\pi /2-\angle QAL=\pi /2-\angle AQS,$$ (here the equality $\angle OAS=\angle QAL$ holds because these angles are symmetric to each other with respect to $AM$), so $\angle AT{A}^{\prime }=\angle ATF+\angle FT{A}^{\prime }=\pi /2$, as desired.

Remark. Existence of the rotational homothety $h$ may be shown in various ways. E.g., one may notice that $\Omega$ is an Apollonius circle of the segment $QS$, so the ratio of the radii of $\Gamma$ and ${\Gamma }^{\ast }$ is $BS/BQ=AS/AQ$, which also yields that $h$ exists.

---

Alternative solution.

Let $S$ be the centre of $\Gamma$ and let the circle on diameter $A{A}^{\prime }$ cross $\Gamma$ again at $T$. It is sufficient to prove that $T$ lies on circle $AFH$.

The points $O$ and $F$ are reflections of one another in $AS$; hence $S$ lies on the internal angle bisectrix of $\angle FA{A}^{\prime }$. On the other hand, since $SF=S{A}^{\prime }$, it lies on the perpendicular bisectrix of $F{A}^{\prime }$; so $S$ is the midpoint of the arc $A^{\prime }F$ on circle $A{A}^{\prime }F$ not containing $A$. In particular, $AFS{A}^{\prime }$ is cyclic.

Let $D$ be the orthogonal projection of $A$ on $BC$. We will prove that $O,D,T$ are collinear. Invert from $O$ with radius $OB$. This fixes $B$ and $C$, so $\Gamma$ maps to line $BC$. It follows that $A$ maps to $A^{\ast }=A{A}^{\prime }\cap BC$. Note that $A$ is fixed under this inversion, as $OA=OB$, so the image of the circle on diameter $A{A}^{\prime }$ is a circle $\delta$ through $A$ and $A^{\ast }$ — and, in fact, $\delta$ is the circle of diameter $A{A}^{\ast }$, as $A{A}^{\prime }$ passes through $O$. Hence $T$ maps to one of the points where line $BC$ crosses $\delta$. As $T\ne A^{\prime }$, its image is $D$, so $O,D,T$ are indeed collinear.

Letting $L$ be the reflection of $A$ in $BC$, we now prove that $HTLO$ is cyclic. As circle $HBC$ is the reflection of $\Gamma$ in $BC$, the quadrangle $HLBC$ is cyclic, so $HD\cdot DL=BD\cdot DC=OD\cdot DT$, whence $HTLO$ is indeed cyclic.



We are to prove that $\angle ATF+\angle FT{A}^{\prime }=90^{\circ }$. To this end, note the equivalences: $\angle ATF+\angle FT{A}^{\prime }=90^{\circ }\iff \angle AHF+\angle FOA=90^{\circ }\iff \angle AHF=\angle OAS$ (as $AS\perp OF)\iff \angle AHF=\angle SAF\iff AS$ is tangent to circle $AFH\iff \frac{AH}{\sin \angle HAS}=\frac{AF}{\sin \angle FAS}\iff \frac{AH}{AO}=\frac{\sin \angle HAS}{\sin \angle SAO}$. $(\ast )$



To prove $(\ast )$, let $AB$ and $AC$ meet $\Gamma$ again at $B^{\prime }$ and $C^{\prime }$, respectively. An easy angle chase shows that $O$ is the orthocentre of triangle $A{B}^{\prime }{C}^{\prime }$.

As triangles $ABC$ and $A{C}^{\prime }{B}^{\prime }$ are similar, $AH$ passes through the centre $O^{\prime }$ of circle $A{B}^{\prime }{C}^{\prime }$; and as circles $A{B}^{\prime }{C}^{\prime }$ and $BOC{C}^{\prime }{B}^{\prime }$ are reflections of one another in $B^{\prime }{C}^{\prime }$ and $A{O}^{\prime }SO$ is a parallelogram, it follows that $$\frac{\sin \angle HAS}{\sin \angle SAO}=\frac{\sin \angle O^{\prime }AS}{\sin \angle AS{O}^{\prime }}=\frac{O^{\prime }S}{A{O}^{\prime }}=\frac{AO}{A{O}^{\prime }}.(\ast \ast )$$ Further on, as triangles $ABC$ and $A{C}^{\prime }{B}^{\prime }$ are similar, $(\ast \ast )$ implies equal corresponding length ratios, so $AO/A{O}^{\prime }=AH/AO$. This establishes $(\ast )$ and concludes the solution.

Remark. Relation $(\ast )$ is equivalent to $AS$ being the $A$-symmedian of triangle $AOH$. This might very well be known and can actually be proved in several different ways.