. Let $A B C$ be a triangle in which $\angle A B C=60^{\circ}$. Let $I$ and $O$ be the incentre and circumcentre of $A B C$, respectively. Let $M$ be the midpoint of the arc $B C$ of the circumcircle of $A B C$, which does not contain the point $A$. Determine $\angle B A C$ given that $M B=O I$.

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Since $\angle A B C=60^{\circ}$, we have $\angle A I C=\angle A O C=120^{\circ}$. Let $I^{\prime}, O^{\prime}$ be the points symmetric to $I, O$ with respect to $A C$, respectively. Then $I^{\prime}$ and $O^{\prime}$ lie on the circumcircle of $A B C$. Since $O^{\prime} I^{\prime}=O I=M B$, the angles determined by arcs $O^{\prime} I^{\prime}$ and $M B$ are equal. It follows that $\angle M A B=\angle I^{\prime} A O^{\prime}$.

Now, denoting $\angle B A C=\alpha$, we have

$$
\angle I^{\prime} A O^{\prime}=\angle I A O=|\angle I A C-\angle O A C|=\left|\frac{\alpha}{2}-30^{\circ}\right|
$$

It follows that $\frac{\alpha}{2}=\angle M A B=\angle I^{\prime} A O^{\prime}=\left|\frac{\alpha}{2}-30^{\circ}\right|$, i.e. $\alpha=30^{\circ}$.

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