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 5 of 5: Compute the common length
Previous stepNext step
CC′=2ACsin⁡(α/2),α=∠BAC+∠CAD+∠DABCC'=2AC\sin(\alpha/2),\qquad \alpha=\angle BAC+\angle CAD+\angle DAB
Detailed analysis

In the development, A,C,C′A,C,C' form an isosceles triangle with AC=AC′AC=AC', and its vertex angle at AA is α=∠BAC+∠CAD+∠DAB\alpha=\angle BAC+\angle CAD+\angle DAB. The perpendicular from AA to the base CC′CC' bisects that base, so the half-base length is ACsin⁡(α/2)AC\sin(\alpha/2). Therefore every shortest path has length ZZ′=CC′=2ACsin⁡(α/2)ZZ'=CC'=2AC\sin(\alpha/2).