Ukrajina, 2008

Let $AC = BM = b$, $BH = MC = h$. We can easily show that point $H$ can not be on the segment $AM$. Let's consider two cases.

1) Point $H$ is on the ray $MA$ (fig.1).
Let's draw perpendicular $LC \perp AM$, point $L$ is on the line $AM$. As one of the angles of the right-angled triangle $ALC$ is $30^\circ$, $LC = \frac{1}{2}b$. Therefore, $\angle MCL = \angle MBH \Rightarrow$

Fig.1

$\cos \angle MCL = \frac{LC}{MC} = \frac{b}{2h} = \cos \angle MBH = \frac{BH}{BM} = \frac{h}{b} \Rightarrow \frac{b}{2h} = \frac{h}{b}$ and $b^2 = 2h^2$, which means that $\cos \angle MCL = \frac{b}{2h} = \frac{1}{\sqrt{2}}$ and $\angle MCL = 45^\circ$. Since $\angle LCA = 60^\circ$, $\angle ACB = 15^\circ$.

2) The point $H$ is on the ray $AM$ (fig.2). Let's draw perpendicular $LC \perp AM$, the point $L$ is on the line $AM$. Further we solve the problem as in the first case, and find that $\angle MCL = 45^\circ$. Since $\angle LCA = 60^\circ$, $\angle ACB = 15^\circ$.