Školjka

%V0 All faces of the tetrahedron $ABCD$ are acute-angled. Take a point $X$ in the interior of the segment $AB$, and similarly $Y$ in $BC, Z$ in $CD$ and $T$ in $AD$. a.) If $\angle DAB+\angle BCD\ne\angle CDA+\angle ABC$, then prove none of the closed paths $XYZTX$ has minimal length; b.) If $\angle DAB+\angle BCD=\angle CDA+\angle ABC$, then there are infinitely many shortest paths $XYZTX$, each with length $2AC\sin k$, where $2k=\angle BAC+\angle CAD+\angle DAB$.