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 2 of 5: Nonparallel copies give no minimum
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CD∦C′D′  ⟹  inf⁡ZZ′=DD′, unattained for Z∈(CD)CD\not\parallel C'D'\implies \inf ZZ'=DD'\text{, unattained for }Z\in(CD)
Detailed analysis

Use the alternative unfolding around BCBC, ABAB, and AD′AD'; the path becomes a segment ZZ′ZZ' joining Z∈(CD)Z\in(CD) to its image Z′∈(C′D′)Z'\in(C'D'). When CDCD and C′D′C'D' are not parallel, the shortest segment between the closed supporting segments occurs at the endpoint D,D′D,D'. Because ZZ and Z′Z' must be interior points, this is only a limiting value as Z→DZ\to D, so no admissible path attains a minimum.