International Mathematical Olympiad 1971 Problem 4

All the faces of tetrahedron $ABCD$ are acute-angled triangles. We consider all closed polygonal paths of the form $XYZTX$ defined as follows: $X$ is a point on edge $AB$ distinct from $A$ and $B$; similarly, $Y, Z, T$ are interior points of edges $BC$, $CD$, $DA$, respectively. Prove:

(a) If $\angle DAB + \angle BCD \neq \angle CDA + \angle ABC$, then among the polygonal paths, there is none of minimal length.

(b) If $\angle DAB + \angle BCD = \angle CDA + \angle ABC$, then there are infinitely many shortest polygonal paths, their common length being $2AC\sin(\alpha/2)$, where $\alpha = \angle BAC + \angle CAD + \angle DAB$.