Romanian Mathematical Olympiad

Since $MV=MB=MD$ and $MO\perp (VBD)$, it follows that $O$ is the circumcenter of the triangle $VBD$. Furthermore, triangle $VBD$ is isosceles and right-angled, implying that the lateral faces of the pyramid are equilateral triangles. Let $P$ be the midpoint of the edge $VB$. The angle of the planes $(VAB)$ and $(VBM)$ is $\angle APM$, hence $\angle APM=90^{\circ }$. Triangles $MPA$ and $POA$ are similar, yielding $\frac{MA}{PA}=\frac{PA}{OA}$. Therefore $AM=\frac{P{A}^2}{OA}=\frac{3}{4}AC$, as claimed.