MathLabs

Problem 4

All faces of tetrahedron ABCDABCD are acute-angled triangles. Consider closed polygonal paths XYZTXXYZTX, where XX is interior to ABAB, and Y,Z,TY,Z,T are interior to BC,CD,DABC,CD,DA, respectively. Prove: (a) if ∠DAB+∠BCD≠∠CDA+∠ABC\angle DAB+\angle BCD\ne\angle CDA+\angle ABC, no path has minimal length; (b) if ∠DAB+∠BCD=∠CDA+∠ABC\angle DAB+\angle BCD=\angle CDA+\angle ABC, infinitely many shortest paths exist, with common length 2ACsin⁡(α/2)2AC\sin(\alpha/2), where α=∠BAC+∠CAD+∠DAB\alpha=\angle BAC+\angle CAD+\angle DAB.
Step 4 of 5: Obtain infinitely many equal shortest paths
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CD∥C′D′  ⟹  CC′∥ZZ′∥DD′CD\parallel C'D'\implies CC'\parallel ZZ'\parallel DD'
Detailed analysis

If the copies are parallel, then CC′∥ZZ′∥DD′CC'\parallel ZZ'\parallel DD'. The acute-face hypothesis ensures that the parallel segment ZZ′ZZ' intersects the intervening edge segments CBCB, BABA, and AD′AD' for every ZZ in a nonempty interval of interior points of CDCD. Hence infinitely many admissible paths have the same minimum length.