42nd Balkan Mathematical Olympiad

Denote by $l$ the line through $A$ parallel to $BC$. Assume without loss of generality that $X$ is closer to $A$ than $Y$. First we show that points $X$ and $Y$ are uniquely defined. Suppose that $X^{\prime }$ and $Y^{\prime }$ are also points on $l$ such that $B{X}^{\prime }=C{Y}^{\prime }$ and that $AC$ is tangent to $(X^{\prime }D{Y}^{\prime })$. Assume also that $X^{\prime }$ is closer to $A$ than $Y^{\prime }$. Clearly $BCYX$ and $BC{Y}^{\prime }{X}^{\prime }$ are isosceles trapezoids, hence $XY$ and $X^{\prime }{Y}^{\prime }$ have the same midpoint - call it $N$. Then $A{N}^2-N{X}^{\prime 2}=A{X}^{\prime }\cdot A{Y}^{\prime }=A{D}^2=AX\cdot AY=A{N}^2-N{X}^2$. This implies $N{X}^{\prime 2}=N{X}^2$, which means $X=X^{\prime }$ and $Y=Y^{\prime }$. Let $M$ be the midpoint of the arc $BAC$. We claim that $M$ is the center of $(XDY)$. Let $F$ be a point on the ray $BA$ such that $FB=FC$. Then, by symmetry, $FM$ is the angle bisector of $\angle AFC$. It is also well-known that $AM$ is the external angle bisector of $\angle BAC$, meaning $\angle MAC=\angle MAF$. These two imply that $M$ is the incenter of $△FAC$. Denote the incircle of this triangle by $\omega$. Let $\omega$ touch $AF$ and $AC$ at $K$ and $D^{\prime }$, respectively. Then $A{D}^{\prime }=\frac{AF+AC-FC}{2}=\frac{AC-(FC-AF)}{2}=\frac{AC-(FB-AF)}{2}=\frac{AC-AB}{2}$, which means that $D^{\prime }=D$. If $\omega$ intersects $l$ at $U$ and $V$, we have that $BU=CV$ by symmetry and that $AD$ touches $(UDV)$, so $U$ and $V$ are in fact $X$ and $Y$ (by uniqueness shown earlier). Therefore, $(XDY)$ is $\omega$. Let $T$ be the midpoint of $BC$. Then, $MT$ is perpendicular to $BC$, hence points $K,D$ and $T$ belong to the Simson line of $M$. Denote $\angle CAB=\alpha$, $\angle ABC=\beta$ and $\angle BCA=\gamma$. Since $\angle KAD=180^{\circ }-\alpha$ and $AK=AD$, we have $\angle ADK=\frac{\alpha }{2}$, and thus $\angle TDC=\frac{\alpha }{2}$. We now have $\angle DTM=\angle DTC-\angle MTC=(180^{\circ }-\angle TDC-\angle DCT)-90^{\circ }=90^{\circ }-\frac{\alpha }{2}-\gamma =\frac{\beta -\gamma }{2}$.

On the other hand, $MADN$ is cyclic ($\angle MNA=\angle MDA=90^{\circ }$), so $\angle MDN=\angle MAN$, and $\angle MAN=\angle MAC-\angle NAC=\angle MBC-\angle ACB=90^{\circ }-\frac{\alpha }{2}-\gamma =\frac{\beta -\gamma }{2}$. We obtained $\angle MDN=\angle DTM$, meaning that $MD$ is tangent to $(NDT)$, which implies $M{D}^2=MN\cdot MT$. Since $MX=MD$, we obtain $M{X}^2=MN\cdot MT$. This together with $\angle MNX=90^{\circ }$ means that $\angle MXT=90^{\circ }$, so $TX$ is tangent to $\omega$. We similarly show that $TY$ is also tangent to $\omega$, so the tangents to $\omega$ at $X$ and $Y$ intersect at $T$, which finishes the proof.