HMMT February, 2025

Solution:
Let $M$ be the midpoint of $AB$. Let $O_{1}$ and $O_{2}$ be the circumcenters of $\triangle AXD$ and $\triangle BXC$, respectively. Since $O_{1}M$ is the perpendicular bisector of $AX$ and $O_{2}M$ is the perpendicular bisector of $BX$, we get that $\angle O_{1}MO_{2} = 90^{\circ}$.
Let $P_{1}$ and $P_{2}$ be the projections of $O_{1}$ and $O_{2}$ onto segment $AB$, respectively, and let $AB = 2x$. By the Pythagorean theorem, $P_{1}A = \sqrt{O_{1}A^{2} - O_{1}P_{1}^{2}} = \sqrt{15^{2} - 12^{2}} = 5$, so $MP_{1} = MA - P_{1}A = x - 5$. Likewise, $MP_{2} = MB - \sqrt{15^{2} - 12^{2}} = x - 9$. Since $\triangle MP_{1}O_{1} \sim \triangle O_{2}P_{2}M$, we know
$$(x - 5)(x - 9) = MP_{1} \cdot MP_{2} = O_{2}P_{2} \cdot P_{1}O_{1} = 12^{2}.$$
Solving this, we get $x = 7 + 2\sqrt{37}$, which implies that $AB = 2x = \boxed{14 + 4\sqrt{37}}$. (The condition that $\triangle AXD$ and $\triangle BXC$ are acute rules out $14 - 4\sqrt{37}$.)

Solution:

Let $P$ be the antipode of $A$ in $\odot (AXD)$ and $Q$ be the antipode of $B$ in $\odot (BXC)$. From $\angle PDA = \angle QCB = 90^{\circ}$, we get that $P$ and $Q$ lie on $CD$. Moreover, from $\angle PXA = 90^{\circ}$, we get that $P \in BX$, and similarly $Q \in AX$.
Being a diameter, $AP = 2 \cdot 13 = 26$, so by the Pythagorean theorem, $DP = \sqrt{26^{2} - 24^{2}} = 10$. Similarly, $BQ = 30$ and $CQ = \sqrt{30^{2} - 24^{2}} = 18$. Letting $AB = x$, we get $PQ = x - 28$. Quadrilateral $ABQP$ has perpendicular diagonals, so $AB^{2} + PQ^{2} = AP^{2} + BQ^{2}$, which means that $x^{2} + (x - 28)^{2} = 26^{2} + 30^{2}$.
Solving this quadratic gives $x = \boxed{14 + 4\sqrt{37}}$. (The condition that $\triangle AXD$ and $\triangle BXC$ are acute rules out $14 - 4\sqrt{37}$.)