Saudi Arabia booklet, 2024

Let circle $(\omega)$ cuts $(O)$ again at $D$ and denote $O_1$ as center of $(\omega)$. Since $AD$ is the radical axis of $(\omega)$, $(O)$ then $OO_1 \perp AD$. But $OO_1 \perp OM$ since $OM$ is tangent to $(\omega)$ leads to $AD \parallel OM$ and then $AD \perp BC$.

This means that $AD$, $AO$ are isogonal in $\angle EAF$ then $OD \parallel EF$, which implies that $EFOD$ is an isosceles trapezoid. Note that $OP \perp AE$, $ON \perp AF$ and $O \in (AEF)$ so by Simson line property, $K$ is the projection of $O$ on $EF$. Denote $L$ as midpoint of $OD$ then $IL \perp OD$ so $OKIL$ is a rectangle. Thus
$$IK = OL = \frac{1}{2}OD = \frac{R}{2}.$$

Now consider the spiral similarity namely $\Omega$ of center $D$, maps $E \to B$ and $F \to C$ then $EF \to BC$ and
$$\Omega : \triangle DEF \to \triangle DBC.$$
Since $O_1$, $O$ are centers of $(DEF)$, $(DBC)$ so $\Omega : O_1 \to O$. Thus two triangles $DO_1O$ and $DIM$ are similar, but $O_1O = O_1D$ then $ID = IM$. By the isosceles trapezoid, one can check that $ID = IO$ also, which implies that $IM = IO$ then $IMO$ is an isosceles triangle.