Saudi Arabian Mathematical Competitions

Denote $M$ as the midpoint of the arc $BC$ not containing $A$ of $(O)$, we need to prove $MK = 2OM$. Denote $BD, CE$ as the internal bisectors of $ABC$ then $D, E, L$ are collinear. Note that $L(AI, DB) = -1$ and take the orthogonal projection from $I$ to get $(Ix, Iy, IK, IM) = -1$ in which
$$IM \perp LA, IK \perp LI, Iy \perp LD, Ix \perp LB$$
Suppose that $Iy$ cuts $MK$ at $O'$, combining with $Ix \parallel MK$, then $O'$ is the midpoint of $MK$. Thus it remains to prove $M$ is the midpoint of $OO'$.
Denote $T$ as the ex-center with respect to angle $A$ of triangle $ABC$ then $M$ is the midpoint of $IT$. Then $MO = MO'$ if and only if $IO' \parallel OT$, or $OT \perp DE$.
Now take $U, V$ as the ex-centers with respect to angles $B, C$ of triangle $ABC$ then $A, B, C$ are the foot of altitudes in triangle $TUV$ and $I, O$ are the orthocenter and nine-point center of that triangle, thus the reflection $S$ of $I$ over $O$ is the circumcenter. Note that $B, D, U$ and $C, E, V$ are collinear then
$$\mathcal{P}_{D/(O)} = \overline{DA} \cdot \overline{DC} = \overline{DU} \cdot \overline{DI} = \mathcal{P}_{D/(IUV)}$$
and similarly, $\mathcal{P}_{E/(O)} = \mathcal{P}_{E/(IUV)}$ so $DE$ is the radical axis of $(O)$ and $(IUV)$ which implies that $DE$ is perpendicular to the line joining centers of $(O)$ and $(IUV)$. We know that center $Z$ of $(IUV)$ is the reflection of $S$ over $UV$, and since $IT = 2NS$, it is easy to check that $Z \in TO$. Thus, we have $DE \perp OT$ which finishes the proof. $\square$