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 1 of 5: Unfold the path into a planar segment
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XYZTX⟼XYZTX′XYZTX\longmapsto XYZTX'
Detailed analysis

Rotate the faces successively about BCBC, CDCD, and AD′AD' so that the relevant faces lie in one plane. The path XYZTXXYZTX becomes a connected planar path XYZTX′XYZTX' between X∈ABX\in AB and its image X′∈A′B′X'\in A'B'. For fixed endpoints, the shortest path is the straight segment.