2. (BRA 3) 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 $x / y$.
2. The polyhedron has $3 \cdot 12 / 2=18$ edges, and by Euler's formula, 8 vertices. Let $v_{1}$ and $v_{2}$ be the numbers of vertices at which respectively 3 and 6 edges meet. Then $v_{1}+v_{2}=8$ and $3 v_{1}+6 v_{2}=2 \cdot 18$, implying that $v_{1}=4$. 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 $A E, A F, A G$, must have equal length, say $x$. (If $x=A E=A F \neq A G=y$, and $A E F$, $A F G$, and $A G E$ are isosceles, $\angle E A F \neq \angle F A G$, 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, $E F G H$ 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 $E F G, E F H, F G H$, and $G E H$. Let the plane through $A, B, C$ meet $E F$, $H F$, and $G F$, at $E^{\prime}, H^{\prime}$, and $G^{\prime}$. Then $A E^{\prime} B H^{\prime} C G^{\prime}$ is a regular hexagon, and since $x=F A=F E^{\prime}$, we have $E^{\prime} G^{\prime}=x$ and $A E^{\prime}=x / \sqrt{3}$. From the isosceles triangles $A E F$ and $F A E^{\prime}$ we obtain finally, with $\measuredangle E F A=\alpha$, $\frac{y}{2 x}=\cos \alpha=1-2 \sin ^{2}(\alpha / 2), x /(2 x \sqrt{3})=\sin (\alpha / 2)$, and $y / x=5 / 3$.