90. In the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, point $M$ is taken on $A C$ and point $N$ is taken on the diagonal $B D_{1}$ of the cube such that $\widehat{N M C}=$ $=60^{\circ}, \widehat{M N B}=45^{\circ}$. In what ratio do points $M$ and $N$ divide the segments $A C$ and $B D_{1}$?
90. Let the edge of the cube be 1. Denote by $O$ the center of the face $A B C D$. From the fact that $\widehat{N M C}=60^{\circ}$ and $\widehat{N O C}=90^{\circ}$, it follows that $O$ is between $M$ and $C$. Denote $|O M|=x,|N B|=y$. Then $|M N|=$ $=2 x,|N O|=x \sqrt{3},|M B|=\sqrt{\frac{1}{2}+x^{2}}$. Applying the cosine theorem to triangles $M N B$ and $O N B$, we get

$$
\left\{\begin{array}{l}
\frac{1}{2}+x^{2}=4 x^{2}+y^{2}-2 x y \sqrt{2} \\
3 x^{2}=\frac{1}{2}+y^{2}-\frac{2}{\sqrt{3}} y
\end{array}\right.
$$

From here, we find: $x=\frac{1}{\sqrt{6}}, y=\frac{2}{\sqrt{3}}$.

Answer: $|A M|:|M C|=2-\sqrt{3},|B N|:\left|N D_{1}\right|=2$.