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 3 of 5: Relate parallelism to the angle condition
Previous stepNext step
CD∥C′D′⟺∠DAB+∠BCD=∠CDA+∠ABCCD\parallel C'D'\Longleftrightarrow \angle DAB+\angle BCD=\angle CDA+\angle ABC
Detailed analysis

In the development, choose points M,N,PM,N,P so that CD∥MB∥AN∥PD′CD\parallel MB\parallel AN\parallel PD'. Comparing the decompositions of the four angles along these parallel lines reduces the stated angle equality to ∠C′D′P=0\angle C'D'P=0. Thus the equality holds exactly when CD∥C′D′CD\parallel C'D'.