Denote $\angle BAC = \alpha$. Using that $A, C, O, I$ are cyclic, we get $MB= 2R \cdot \sin \frac{\alpha}{2}$ and $OI = 2R \cdot \sin |\frac{\alpha}{2} - 30|$, so $\alpha=30$ will be our answer. (I think)

This problem was proposed by Burii.

$\angle AOC=2\angle B= 120, \angle AIC=90+\angle B/2=120$ so $AOIC$ is cyclic and so $\angle AIO=\angle ACO=30$ $\angle IBM=\frac{ \angle A+\angle B}{2}= 90-\frac{\angle C}{2}= 180-(90+\frac{\angle C}{2})=180-\angle AIB=\angle MIB$ so $MB=IM=OI$ So $\angle IMO=15$ and so $\angle MOB=30$ But $\angle A=\angle MOB=30$

RagvaloD wrote: $\angle AOC=2\angle B= 120, \angle AIC=90+\angle B/2=120$ so $AOIC$ is cyclic and so $\angle AIO=\angle ACO=30$ $\angle IBM=\frac{ \angle A+\angle B}{2}= 90-\frac{\angle C}{2}= 180-(90+\frac{\angle C}{2})=180-\angle AIB=\angle MIB$ so $MB=IM=OI$ So $\angle IMO=15$ and so $\angle MOB=30$ But $\angle A=\angle MOB=30$ For the thought of the pre last line: you don't know that points A, O and B are colinear! You have to prove that!

MI=MC (fact 5) <MIC=60, so MIC - equilateral and CI=MC AOIC - cyclic (<AOC = <AIC = 120) so <OIC = <BMC (180 - <BAC) Given OI=BM, triangle BMC congruent with triangle OIC. So OC = BC. BOC - equilateral (BO=OC). O lay on AB. So <BAC = 30.

Attachments:

Solution 1.[Trigonometry.] Let $x=\angle BAC$. As $MB=OI$, we get that $OM$ is tangent to $(AOI)$, hence $r^2=OM^2=MI\cdot MA=MC\cdot MA$. By Law of Sines, $$\frac{MC}{\sin{\frac{x}{2}}}=\frac{MA}{\sin{120^\circ-\frac{x}{2}}}=\frac{AC}{\sin{60^\circ}}=\frac{2\sqrt{3}}{3}AC$$and $$AC=2r\cdot \sin{60^\circ}=\sqrt{3}r,$$hence we get that \begin{align*} 4\cdot \sin{(120^\circ-\frac{x}{2})}\cdot \sin{\frac{x}{2}}&=1\\ 4\cdot (\frac{\sin{\frac{x}{2}}}{2}+\frac{\sqrt{3}}{2}\cos{\frac{x}{2}})\cdot \sin{\frac{x}{2}}&=1\\ \sqrt{3}\sin{x}+2\sin^2{\frac{x}{2}} &=1\\ \sqrt{3}\sin{x}+\frac{\sin{x} \cdot \sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}&=1\\ \sqrt{3}+\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}&=\frac{1}{\sin{x}}\\ \sqrt{3}+\frac{\sin{x}}{2\cos^2{\frac{x}{2}}}&=\frac{1}{\sin{x}}\\ \sqrt{3}+\frac{\sin{x}}{\cos{x}+1}&=\frac{1}{\sin{x}}\\ \sqrt{3}(\cos{x}+1)\sin{x}+\sin^2{x}&=\cos{x}+1\\ \sqrt{3}\sin{x}+\sin^2{x}-1&=\cos{x}-\sqrt{3}\cos{x}\sin{x}\\ (\sqrt{3}\sin{x}+\sin^2{x}-1)^2&=(1-\sqrt{3}\sin{x})^2(1-\sin^2{x})\\ 4\sin^4{x}&=\sin^2{x}\\ \sin{x}&=\frac{1}{2}\\ x&=30^{\circ}.\\ \end{align*} Solution 2.[Synthetic.] As $\angle ABC=60^\circ$, we get that $\angle CMI=60^\circ$ and as $MB=MI=MC$, we get that $IMC$ is equilateral. Also problem statement yields that $I$ is the circumcentre of $\triangle OMC$. Let $\angle BAM=a$, then $\angle OIM=180^\circ-2a$, hence $\angle OCM=90^\circ-a$, however on the other hand, $\angle OCM=\angle OCB+\angle BCM=60^\circ+a$, therefore $a=15^\circ$. We conclude that $\angle BAC=30^\circ$.