We will prove that $OA$ and $OK$ are tangents to $\odot(AIK)$ from which the desired conclusion trivially follows. Let $E$ and $F$ be the points such that $\omega_2$ is tangent to $AB$ and $AC$ at $E$ and $F$ respectively It is well known that $BEIK$ and $CFIK$ are cyclic (Refer EGMO Lemma 4.40 part (f)) and that $AI$ is the perp bisector of $EF$ with $E-I-F$ This gives us $$\angle AIK = 90^{\circ}+\angle EIK = 90^{\circ} + (180^{\circ} - \angle ABK) = 90^{\circ} + \angle ACK = 90^{\circ} +\frac{\angle AOK}{2} = 180^{\circ} - \angle KAO$$which means $\angle AKI = \angle IAO$ which means $OA$ is tangent to $\odot(AIK)$ and because $AO=KO$ the same holds true for $OK$ thus finishing the proof $\blacksquare$

Here's what I submitted in the contest..

Attachments:
Sharygin_Geometry_Olympiad P-10.pdf (112kb)

Say $M$ is the midpoint of $\widehat{BC}$ and $N$ is the midpoint of $\widehat{BAC}$.Clearly $\{M,O,N\},\{M,I,A\},\{N,I,K\}$ are collinear triplets. Say $P$ andd $Q$ are the midpoints of $AM$ and $NK$.Clearly $OP \perp AI,OQ \perp KI$.Now note that $OP=OM\sin \angle OMP=OM\sin \angle AKI$ and similarly $OQ=ON \sin \angle IAK \implies \frac{OP}{OQ}=\frac{\sin\angle IKA}{\sin\angle IAK}=\frac{IA}{IK}$ done! $\blacksquare$ Its known that for a point lying on the symmedian,the ratios of its projection on the 2 sides is same as the ratio of 2 sides.

It's well known that $K,I,M$ are collinear, where $M$ is the midpoint of $\overset{\LARGE\frown}{BAC}$ in $(ABC)$. Let $S$ be the midpoint of arc $BC$ in $\omega_{1}$, containing point $K$. Then obviously $A,I,S$ are collinear since $\angle BAI=\angle CAI$ and $\angle SAB=\angle SAC$. Note that points $M,S,O$ lie on the perpendicular bisector of $BC$, so $M,O,S$ are collinear. Since $\angle IMS=\angle KMS=\angle KAS=\angle KAI$ and $\angle MIS=\angle AIK$ it follows that $\triangle AIK\sim \triangle MIS$. Let $OI\cap AK=X$ and let $Y$ be the midpoint of $AK$. Because $\triangle AIK\sim \triangle MIS$ it follows that $\angle YIA=\angle MIO=\angle KIX\Longrightarrow IX$ is the symmedian in $\triangle AIK$, but $X\in OI\Longrightarrow$ line $OI$ contains the symmedian of $\triangle AIK$ ($XI$) and we've solved the problem.

Use this: https://artofproblemsolving.com/community/c6t48f6h2755824_a_geometry_ratio And this finish the problem

Indeed beautiful problem. Let $M,N$ be touch points of $\omega_2$ with $AB,AC;$ it's known that $I$ is the midpoint of $MN.$ It's also known, that $\omega_1$ meet $KI,KM,KM$ again by midpoints of arcs $BAC,AB,AC$ respectively. Hence (oriented arcs and angles) $$\angle AOK=2\angle ABK=\widehat{AC}+\widehat{CK}=\widehat{CK}+\frac{1}{2}(\widehat{AC}+\widehat{CB}+\widehat{BA})=2\angle KIN=2\angle KIA-\pi.$$But $O$ lies on perpendicular bisector of $AK,$ so the conclusion follows easily.

it sufices to show that $OA,OK$ are tangents to the circle of $AKI$ it s easy look at diagram below RH HAS

Attachments:

It is well known that $K,I,N$ are collinear where $N$ is the midpoint of arc $BC$ containing $A$.Let $M$ be the midpoint of minor arc $BC$. $AIK$ is similar to $NIM$ and $IO$ is median in $NIM$ so symedian in $AIK$

Let $M,N$ be the midpoints of arc $BC$ and $BAC$ respectively. Recall that $N,I,K$ are collinear. $$OI \text{ is a symmedian of } \Delta AIK \iff \frac{\sin \angle KAI}{\sin \angle AKI}=\frac{\sin \angle AIO}{\sin \angle KIO} \iff \frac{\sin \angle INM}{\sin \angle IMN}=\frac{\sin \angle MIO}{\sin \angle NIO}$$which is true since $O$ is the midpoint of $MN$.