Problem 1017

177. Suppose that no plane angle of the given trihedral angle is a right angle. Let $S$ be the vertex of this angle. Translate the second trihedral angle parallel to itself so that its vertex coincides with some point $A$ on one of the edges of the given angle (Fig. 37). $A B, A C$ and $A D$ are parallel to the edges of the second dihedral angle. Points $B$ and $C$ are on the edges of the given angle or their extensions. But $A B$ is perpendicular to $S C$, $A C$ is perpendicular to $S B$, so the projections of $B S$ and $C S$ on the plane $A B C$ will be perpendicular to $A C$ and $A B$ respectively, i.e., $S$ projects to the point of intersection of the altitudes of $\triangle A B C$, which means $A S$ is perpendicular to $B C$. Thus, the edge $A D$ is parallel to $B C$, which means that all the edges of the second trihedral angle lie in one plane. If one plane angle of the given trihedral angle is a right angle, then all the edges of the second must lie in one face of the given (the one corresponding to the right plane angle). If exactly two plane angles of the given trihedral angle are right angles, then two edges of the second must coincide with one edge of the given. Thus, the second trihedral angle can be non-degenerate only if all the plane angles of the given are right.

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Fig. 37.