Bulgarian National Olympiad - Final Round

Let $W$ be the midpoint of the major arc $BAC$, $A^{\prime }\in (ABC)$ be such that $A{A}^{\prime }\parallel BC$, $T$ be the intersection of the circle with diameter $AI$ and $(ABC)$, and let $\omega$ touch $AC$, $AB$ at $E$, $F$. We claim the two circles touch at $T$.

Firstly, observe that $Z\in (AEF)$, so

$$\begin{array}{rl}\angle ATZ& =\angle AIZ=90^{\circ }-\left(\frac{\alpha }{2}+\gamma \right)\\ & =\frac{\beta -\gamma }{2}=\beta -\left(90^{\circ }-\frac{\alpha }{2}\right)\\ & =\angle ABC-\angle WBC=\angle ABW=\angle ATW,\end{array}$$

hence $T$, $Z$, $W$ are collinear. Let $TW\cap BC=P$. By spiral similarity and angle bisector theorem we have $\frac{PB}{PC}=\frac{TB}{TC}=\frac{BF}{CE}$, so by converse of Menelaus theorem for $△ABC$ we obtain that $EF$, $TZ$, $BC$ are concurrent at $P$. Thus, by power of point at $P$, we obtain that $PX\cdot PY=PE\cdot PF=PZ\cdot PT$, so $XYZT$ is cyclic.

Finally, observe that the center of $(ZXY)$ lies on $IZ$, so $(ZXY)$ touches $A{A}^{\prime }$ at $Z$. Hence, by shooting lemma, since $W$ is the midpoint of the minor arc $A{A}^{\prime }$, we obtain that $(ZXY)$ touches $(ABC)$ at $T$ and we are done. $\square$