https://www.math.ucla.edu/~robjohn/math/ussr_olympiads.pdf (UCLA archive)
Let $\angle BAC = \alpha$. Since $M$ is the midpoint of $BC$, we have $\angle MBC = 90^\circ - \frac\alpha2$. Also, $\angle IBM = 90^\circ - \frac\alpha2$. Therefore, $\triangle BIM$ is isosceles, and $BM = IM$. Since $I$ is the incenter, we have $IM = r$, the inradius. Therefore, $BM = r$. Now, $\triangle BMC$ is a right triangle with $BM = r$ and $MC = \fraca2$, where $a$ is the side length $BC$. Therefore, $\fraca2 = r \cot \frac\alpha2$. On the other hand, the area of $\triangle ABC$ is $\frac12 r (a + b + c) = \frac12 a \cdot r \tan \frac\alpha2$. Combining these, we find that $\alpha = 60^\circ$. russian math olympiad problems and solutions pdf verified
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