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 $\angle LCA=60^{\circ }$, $\angle ACB=15^{\circ }$.