HMMT February

Solution:

Let $I$ and $I_{A}$ be the incenter and the $A$-excenter of $\triangle ABC$. The key observation is that $X$ is the circumcenter of $\triangle BDI_{A}$. To see why this is true, note that
$$\angle BXD = \angle C = 2\angle ICB = 2\angle II_{A}B = 2\angle DI_{A}B.$$
Analogously, $Y$ is the circumcenter of $\triangle CDI_{A}$. Hence, $XY$ is the perpendicular bisector of $DI_{A}$, which is clearly perpendicular to $AD$.

Solution:

Denote $\omega_{B}$ and $\omega_{C}$ as the circumcircle of $\triangle BXD$ and $\triangle CYD$. Also, let $AD$ intersect the circumcircle of $\triangle ABC$ at $M$. Since $\angle BXD = \angle ACB = \angle AMB$, we get that $M \in \omega_{B}$. Similarly, $M \in \omega_{C}$. From here, there are two ways to finish.
- Note by Law of Sine that the radius of $\omega_{B}$ and $\omega_{C}$ are $DB/(2\sin \angle MDB)$ and $DB/(2\sin \angle MDC)$, so they are actually equal. Thus, if $O_{B}$ and $O_{C}$ are the centers of $\omega_{B}$ and $\omega_{C}$, then $XO_{B} = YO_{C}$. Moreover, $XO_{B}$ and $YO_{C}$ are both clearly perpendicular to $BC$, so $XO_{B}O_{C}Y$ is a parallelogram, implying that $XY \parallel O_{B}O_{C} \perp DM$.
- Observe that
$$\angle MXD = \angle MBD = \frac{\angle A}{2} = 90^{\circ} - \angle XDY$$
so $XM \perp DY$. Similarly, $YM \perp XD$, so $M$ is the orthocenter of $\triangle DXY$, implying the result.

Solution:

Let $DX$ intersect $AC$ at $P$ and $DY$ intersect $AB$ at $Q$. Observe that $\angle BCP = \angle BXP$, so $B, C, P, X$ are concyclic. This implies that $CD = CP$ and that $DB \cdot DC = DX \cdot DP$.
Similarly, we have $BD = BQ$ and that $DB \cdot DC = DY \cdot DQ$. Thus, we actually have $DX \cdot DP = DY \cdot DQ$, implying that $X, Y, P, Q$ are concyclic.
Now, let $I$ be the incenter of $\triangle ABC$. Since $BD = BQ$, it follows that $BI$ is the perpendicular bisector of $DQ$, so $ID = IQ$. Similarly, $ID = IP$, so $I$ is actually the circumcenter of $\triangle DPQ$.
We then finish by angle chasing:
$$\angle XDI = 180^{\circ} - \angle PDI = 90^{\circ} + \angle DQP = 90^{\circ} + \angle DXY$$
implying the result.