VI OBM

The centers of the faces of a regular dodecahedron form a regular icosahedron. Let $P,P^{\prime }$ be two opposite vertices of a regular icosahedron. Then 5 of the remaining vertices are adjacent to $P$ and the other 5 are adjacent to $P^{\prime }$. So we may label the other pair of opposite vertices $Q$ and $Q^{\prime }$, where $Q$ is adjacent to $P$ and hence $Q^{\prime }$ is adjacent to $P^{\prime }$.



Let the side of the icosahedron be $k$. Then $PQ{P}^{\prime }{Q}^{\prime }$ is obviously a rectangle and one pair of sides has length $k$. The other side is the diagonal of the regular pentagon $XY{P}^{\prime }ZQ$ of side $k$ and hence has length $\left(\frac{1+\sqrt{5}}{2}\right)k$. $ABCD$ is similar to $PQ{P}^{\prime }{Q}^{\prime }$ and so has the same ratio $\frac{1+\sqrt{5}}{2}$ for its side lengths.