By convention, let $\angle A, \angle B, \angle C$ represent $\angle BAC, \angle ABC, \angle ACB$ respectively. If $O, H, I$ have two points coinciding, it is easy to see that $\triangle ABC$ is an equilateral triangle, and clearly $BH \neq OI$. Therefore, $O, H, I$ are pairwise distinct. Using the given conditions, $\angle A = 60^\circ$. Since $\triangle ABC$ is an acute triangle, the points $H, O, I$ are all inside $\triangle ABC$. It is easy to see that $\angle BHC = \angle BIC = \angle BOC = 120^\circ$.

Let the circumcircle of $\triangle BOC$ be $\Gamma$ with center $O^*$. From the above, points $H, I$ are on $\Gamma$. Using $BO = CO$, triangles $\triangle BOO^*$ and $\triangle COO^*$ have three pairs of corresponding sides equal and are therefore congruent. Thus, $\angle BOO^* = \angle COO^* = \frac{1}{2} \angle BOC = 60^\circ$. Hence, $\triangle BOO^*$ and $\triangle COO^*$ are equilateral triangles. Therefore, the radius of the circumcircle $\Gamma$ of $\triangle BOC$ is the circumradius $R$ of $\triangle ABC$.

It is easy to see that $AH = 2R \cos A = R$. In $\triangle AHI$ and $\triangle AOI$, $\angle IAH = |\angle BAI - \angle BAH| = | \frac{1}{2} \angle A - (90^\circ - \angle B)| = | \angle B - 60^\circ|$, $\angle IAO = | \frac{1}{2} \angle A - \angle OAC| = | \frac{1}{2} \angle A - \frac{1}{2}(180^\circ - \angle AOC)| = | \frac{1}{2} \angle A - 90^\circ + \angle B| = | \angle B - 60^\circ|$. Therefore, $\angle IAH = \angle IAO$. Thus, $\triangle AHI \cong \triangle AOI$. Therefore, $HI = OI$. Using $BH = OI$, we get $BH = HI$. Since $\triangle OBC$ is an isosceles triangle, $\angle BCO = \frac{1}{2}(180^\circ - \angle BOC) = 30^\circ$. On circle $\Gamma$, arcs $\text{BH , HI , OI}$ are equal. Thus, $\angle BCH = \frac{1}{3} \angle BCO = 10^\circ$. Therefore, $\angle B = 90^\circ - \angle BCH = 80^\circ$, $\angle C = 180^\circ - \angle A - \angle B = 40^\circ$.