Taiwan IMO Selection Camp

Extend the semicircle to a full circle $\Gamma$. Let line $LK$ meet $\Gamma$ at two points $P, Q$. On ray $OM$, find a point $W$ satisfying $OW \cdot OM = OA \cdot OB$. From this, we know that $P, Q, W, O$ are concyclic; let this circle be $\gamma$.

The radical center of the three circles $\Gamma, \gamma$, and the circumcircle of $XKOZ$ is the point $L$, because line $XZ$ and $PQ$ are radical axes. Therefore line $YO$ is the radical axis of $\Gamma$ and $\gamma$.
Let $T$ be the intersection of $XZ$ and $AB$. Since
$$KO \cdot KT = KA \cdot KB = KP \cdot KQ,$$

the point $T$ also lies on circle $\gamma$. Moreover, from
$$TA \cdot TB = TK \cdot TZ = TK \cdot TO,$$
we know that $\overline{TY}$ is tangent to $\Gamma$.
Let $S$ be the midpoint of segment $\overline{YW}$. Considering the homothety centered at point $W$ with ratio $2$, we know that the line through $S$ and the midpoint of $\overline{WT}$ is perpendicular to $\overline{YO}$. Moreover, $S$ lies on the perpendicular bisector of $\overline{KO}$. Therefore, $S$ is the center of the circumcircle of quadrilateral $XKOZ$.
Finally, since $Y, S, W$ are collinear, we have
$$OS \cdot OI = OW \cdot OM = OY^2,$$
from which we know $\angle OIM = \angle OWS = \angle OWY = \angle OYM$. Hence quadrilateral $YOMI$ is cyclic. Q.E.D.