HMMT February

Solution:

Let lines $AB$ and $CD$ meet at $T$. Notice that
$$\begin{aligned} & \angle ATD = 180^{\circ} - \angle ABC - \angle DBC = 180^{\circ} - 2\theta \\ & \angle AXD = 180^{\circ} - 2(90^{\circ} - \theta) = 2\theta \end{aligned}$$
Therefore, $A$, $T$, $X$, and $D$ are concyclic. In particular, this implies that $\angle XTA = 90^{\circ} - \theta = \angle XTD$. Thus, $XT$ bisects $\angle BTC$. However, notice that $\triangle TBC$ is isosceles, so $XT$ is actually the perpendicular bisector of $BC$, implying that $BX = XC$.

Solution:

Without loss of generality, let $AB > CD$. Draw the circle $\gamma$ centered at $X$ and passing through $A$ and $D$. Let this circle intersect $CD$ again at point $P \neq D$. Then, notice that
$$\angle APD = \frac{\angle AXD}{2} = \theta$$
implying that $AP \parallel BC$. Combining with $\angle ABC = \angle BCP$, we get that quadrilateral $APCB$ is isosceles trapezoid. Since $X \in \gamma$, we have $X$ lies on the perpendicular bisector of $AP$, which is the same as the perpendicular bisector of $BC$, so we are done.

Solution:

Let the perpendicular bisector of $AD$ intersect $BC$ at point $K$. Notice that
$$\begin{aligned} \angle AXK = 90^{\circ} + \angle XAD = 180^{\circ} - \theta = \angle ABK &\Longrightarrow A, B, X, K \text{ are concyclic.} \\ &\Longrightarrow \angle XBC = \angle XAK \end{aligned}$$
Similarly, we get that $\angle XCB = \angle XDK$. However, since both $X$ and $K$ lie on the perpendicular bisector of $AD$, implying that $\angle XBC = \angle XCB$.

Solution:
Fix $\theta$ and points $A$, $B$, and $C$. Animate point $D$ along the fixed line through $C$. Since $\triangle XAD$ has a fixed shape, it follows that $X$ moves linearly along a fixed line. Since we want to show that $X$ lies on the perpendicular bisector of $BC$, which is fixed, it suffices to prove this for only two locations of $D$.
- When $AD \parallel BC$, it follows that $ABCD$ is an isosceles trapezoid, implying the result.
- When $D = C$, we notice that
$$\angle AXC = 2\theta = 2\angle ABC,$$
implying that $X$ is the circumcenter of $\triangle ABC$, and the result follows.