Bulgarian Spring Tournament

Let $A{A}_1$ be the height from point $A$ in $△ABC$. Note that it lies on the continuation of $BC$, because this triangle is obtuse. The triangle $△A{A}_{1}B$ is right-angled with angle $30^{\circ }$. Therefore, $A{A}_1=AM=A_{1}M=BM=x$. Consequently, we directly find $\angle CMB=45^{\circ }$, $\angle A_{1}MA=60^{\circ }$, $\angle A_{1}MC=75^{\circ }$, $\angle MC{A}_1=75^{\circ }$. So $△MC{A}_1$ is isosceles. Then $C{A}_1=x$ and $△A{A}_{1}C$ is isosceles and right-angled, therefore $\angle AC{A}_1=45^{\circ }$ and therefore $\angle ACM=30^{\circ }$. We will calculate the area of $△ACM$ in two different ways.

$S_{ACM}=\frac{AC\cdot CM}{4}$ because $\angle ACM=30^{\circ }$ i.e. the height to $AC$ is equal to $\frac{MC}{2}$. But also $M$ is the middle of $AB$. So $S_{ACM}=\frac{S_{ABC}}{2}=\frac{BC\cdot A{A}_1}{4}=\frac{BC\cdot BM}{4}$ and we get what we are looking for. $\square$