74. $A B C D$ is a regular tetrahedron with edge $a$. Let $M$ be the center of the face $A D C$, and $N$ be the midpoint of the edge $B C$. Find the radius of the sphere inscribed in the trihedral angle $A$ and tangent to the line $M N$.
74. When the sphere is intersected by the plane $A M N$, we obtain a circle inscribed in the triangle $A M N$. In this triangle, $|A N|=a \frac{\sqrt{\overline{3}}}{2}, \quad|A M|=a \frac{\sqrt{\overline{3}}}{3}, \quad|M N|=\frac{a}{2} \quad$ (determined from $\triangle C M N$). Therefore, if $L$ is the point of tangency of the inscribed sphere with $A M$, then

$$
|A L|=\frac{|A N|+|A M|-|M N|}{2}=\left(\frac{5}{12} \sqrt{\overline{3}}-\frac{1}{4}\right) a
$$

The sphere inscribed in $A B C D$ has a radius $r=\frac{1}{4} \sqrt{\frac{2}{3}} a$ and touches the plane $A C D$ at point $M$.

Thus, if $x$ is the radius of the desired sphere, then

$$
\frac{x}{r}=\frac{|A L|}{|A M|}=\frac{5-\sqrt{3}}{4}
$$

From this, $x=\frac{5 \sqrt{\overline{6}}-3 \sqrt{2}}{48} a$.