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 $O{O}_1\perp AD$. But $O{O}_1\perp OM$ 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 $$\Omega :△DEF\to △DBC.$$ Since $O_1$, $O$ are centers of $(DEF)$, $(DBC)$ so $\Omega :O_1\to O$. Thus two triangles $D{O}_{1}O$ and $DIM$ are similar, but $O_{1}O=O_{1}D$ 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.