(5) In a regular tetrahedron $ABCD$, $AO \perp$ plane $BCD$, with the foot of the perpendicular being $O$. Let $M$ be a point on the line segment $AO$ such that $\angle BMC=90^{\circ}$. Then $\frac{AM}{MO}=$ $\qquad$ .
(5) 1 Hint: As shown in the figure, connect $O B$. Let the edge length of the regular tetrahedron $A B C D$ be $a$, then
$$
O B=\frac{\sqrt{3}}{3} a, M B=\frac{\sqrt{2}}{2} a .
$$

So $M O=\sqrt{M B^{2}-O B^{2}}=\frac{\sqrt{6}}{6} a=$
$$
\begin{array}{r}
\frac{1}{2} A O=A M . \\
\text { Therefore } \frac{A M}{M O}=1 .
\end{array}
$$