Taiwan IMO Selection Camp

Clearly $B'$ is the foot of the perpendicular from $L$ to $AB$, so it lies within segment $AB$. Similarly, $C'$ lies within segment $AC$. Therefore $O$ must lie inside $\triangle AB'C'$, and hence $\angle COB < \angle C'OB'$.

Now let $\alpha = \angle CAB$. It is easy to see that $\angle COB = 2\angle CAB = 2\alpha$ and $2\angle C'OB' = 360^\circ - \angle C'LB'$. Also, $\angle C'LB' = 180^\circ - \angle C'AB' = 180^\circ - \alpha$. Combining the above, we get
$$2\alpha = \angle COB < \angle C'OB' = \frac{360^\circ - \angle C'LB'}{2} = \frac{360^\circ - (180^\circ - \alpha)}{2} = 90^\circ + \frac{\alpha}{2},$$
so $\alpha < 60^\circ$.

Finally, let $O'$ be the reflection of $O$ across $BC$. Then in the quadrilateral $ABO'C$, we have
$$\angle CO'B + \angle CAB = \angle COB + \angle CAB = 2\alpha + \alpha < 180^\circ,$$
so $O'$ must lie outside the circumcircle of $\triangle ABC$. Therefore, the points $O$ and $O'$ lie respectively inside and outside the circumcircle of $\triangle ABC$, so the circle $\omega$ must intersect the circumcircle of $\triangle ABC$ at two points.