Question 7
(NET 1) (IMO 4) Given a tetrahedron $ABCD$ whose all faces are acute-angled triangles, set
$$\sigma=\angle DAB+\angle BCD-\angle ABC-\angle CDA.$$
Consider all closed broken lines $XYZTX$ whose vertices $X,Y,Z,T$ lie in the interior of segments $AB,BC,CD,DA$ respectively. Prove that:
(a) if $\sigma\ne0$, then there is no broken line $XYZT$ of minimal length;
(b) if $\sigma=0$, then there are infinitely many such broken lines of minimal length. That length equals $2AC\sin(\alpha/2)$, where
$$\alpha=\angle BAC+\angle CAD+\angle DAB.$$
$$\sigma=\angle DAB+\angle BCD-\angle ABC-\angle CDA.$$
Consider all closed broken lines $XYZTX$ whose vertices $X,Y,Z,T$ lie in the interior of segments $AB,BC,CD,DA$ respectively. Prove that:
(a) if $\sigma\ne0$, then there is no broken line $XYZT$ of minimal length;
(b) if $\sigma=0$, then there are infinitely many such broken lines of minimal length. That length equals $2AC\sin(\alpha/2)$, where
$$\alpha=\angle BAC+\angle CAD+\angle DAB.$$