IMO 1985 SL 2

Origin: BRA

Problem

A polyhedron has 12 faces and is such that: (i) all faces are isosceles triangles, (ii) all edges have length either x or y, (iii) at each vertex either 3 or 6 edges meet, and (iv) all dihedral angles are equal. Find the ratio x/y.

Solution

The polyhedron has 3 \cdot 12/2 = 18 edges, and by Euler’s formula, 8 vertices. Let v1 and v2 be the numbers of vertices at which respectively 3 and 6 edges meet. Then v1 + v2 = 8 and 3v1 + 6v2 = 2 \cdot 18, implying that v1 = 4. Let A, B, C, D be the vertices at which three edges meet. Since the dihedral angles are equal, all the edges meeting at A, say AE, AF, AG, must have equal length, say x. (If x = AE = AF ̸= AG = y, and AEF, AFG, and AGE are isosceles, \angleEAF ̸= \angleFAG, in contradiction to the equality of the dihedral angles.) It is easy to see that at E, F, and G six edges meet. One proceeds to conclude that if H is the fourth vertex of this kind, EFGH must be a regular tetrahedron of edge length y, and the other vertices A, B, C, and D are tops of isosceles pyramids based on EFG, EFH, FGH, and GEH. Let the plane through A, B, C meet EF, HF, and GF, at E′, H′, and G′. Then AE′BH′CG′ is a regular hexagon, and since x = FA = FE′, we have E′G′ = x and AE′ = x/ \sqrt