VMO

a) Let $(I)$ touch $CA$, $AB$ at $U,V$; $I_b,I_c$ be the excenters of vertices $B,C$ in triangle $ABC$. Clearly, $I_{a}A$, $I_{b}B$ and $I_{c}C$ are three altitudes of triangle $I_a{I}_b{I}_c$. On the other hand, notice that $IA\perp UV$, $IB\perp VD$, $IC\perp DU$ so two triangles $DUV$ and $I_a{I}_b{I}_c$ have corresponding parallel sides. So there exists a homothety $\mathcal{H}$ that turns $D$ into $I_a$, $U$ into $I_b$ and $V$ into $I_c$.

Through $\mathcal{H}$, $I$ becomes $O^{\prime }$ which is the circumcenter of triangle $I_a{I}_b{I}_c$. However, in triangle $I_a{I}_b{I}_c$, $I$ is the orthocenter and $(ABC)$ is the Euler circle, so $O$ belongs to $I{O}^{\prime }$, which implies the center of $H$ belongs to $OI$. On the other hand, notice that the center of $H$ also lies on $D{I}_a$; $D{I}_a$ and $OI$ intersect at $L$ so we infer that $L$ is the center of $H$.



We have $L,E,F$ are collinear and $DE\parallel I_{a}F$ so $F$ is the image of $E$ via the above homothety, so it lies on $(I_a{I}_b{I}_c)$. Furthermore, notice that triangle $I_a{I}_b{I}_c$ has $I$ as the orthocenter and $I_{a}F$ as the altitude, so we infer that $F$ and $I$ are symmetric through $I_c{I}_b$, hence, $AF=AI$.

b) It is easy to see that $J$ is the midpoint of the minor arc $BC$ of $(O)$, which is a fixed point. Let $K$ be the intersection of $MN$ with the perpendicular bisector of $BC$, we will prove that $K$ is a fixed point, it follows that $T$ always moves on a circle of diameter $JK$, which is a fixed circle. Let $S$ be the midpoint of $BC$, then because $BC$ and $I_c{I}_b$ are anti-parallel, then $I_{a}S$ is the symmedian of triangle $I_a{I}_b{I}_c$. Since the tangents at $U$ and $V$ of $(I)$ intersect at $A$, $DA$ is the symmedian of the triangle $DUV$. Moreover, the homothety $H$ turns triangle $DUV$ into triangle $I_a{I}_b{I}_c$ so we infer that the image of $DA$ through this homothety is $I_{a}S$. So $I_{a}S\parallel AD$, infer that $I_a,S$ and $M$ are collinear.



The tangent lines at $B$ and $C$ of $(J)$ intersect at $Q$, we will prove that $K$ is fixed by showing that $K$ is the midpoint of $SQ$.



Let $H$ be the projection of $I_a$ on $BC$. We have $I_{a}H$ and $I_{a}J$ are isogonal with respect to angle $B{I}_{a}C$, so it is easy to see $BH=DC$, i.e. $H$ and $D$ are symmetric with respect to $S$. Let $I_{a}Q$ meet $(J)$ again at $X$ and meet $BC$ at $Y$. We have $I_{a}Q$ is the symmedian of triangle $I_{a}BC$, so $X$ and $M$ are symmetric through $BC$, which implies $XH$ passes through $K$.

On the other hand, we have $(QY,X{I}_a)=-1$ so $H(QY,X{I}_a)=-1$. Since $QS$ is parallel to $H{I}_a$ then $K$ is the midpoint of $SQ$. $\square$