11. G5 (AUS) For any set \( S \) of five points in the plane, no three of which are collinear, let \( M(S) \) and \( m(S) \) denote the greatest and smallest areas, respectively, of triangles determined by three points from \( S \). What is the minimum possible value of \( M(S) / m(S) \) ?
11. When $S$ is the set of vertices of a regular pentagon, then it is easily verified that $\frac{M(S)}{m(S)}=\frac{1+\sqrt{5}}{2}=\alpha$. We claim that this is the best possible. Let $A, B, C, D, E$ be five arbitrary points, and assume that $\triangle A B C$ has the area $M(S)$. We claim that some triangle has area less than or equal to $M(S) / \alpha$. Construct a larger triangle $A^{\prime} B^{\prime} C^{\prime}$ with $C \in A^{\prime} B^{\prime}\left\|A B, A \in B^{\prime} C^{\prime}\right\| B C$, $B \in C^{\prime} A^{\prime} \| C A$. The point $D$, as well as $E$, must lie on the same side of $B^{\prime} C^{\prime}$ as $B C$, for otherwise $\triangle D B C$ would have greater area than $\triangle A B C$. A similar result holds for the other edges, and therefore $D, E$ lie inside the triangle $A^{\prime} B^{\prime} C^{\prime}$ or on its boundary. Moreover, at least one of the triangles $A^{\prime} B C, A B^{\prime} C, A B C^{\prime}$, say $A B C^{\prime}$, contains neither $D$ nor $E$. Hence we can assume that $D, E$ are contained inside the quadrilateral $A^{\prime} B^{\prime} A B$. An affine linear transformation does not change the ratios between areas. Thus if we apply such an affine transformation mapping $A, B, C$ into the vertices $A B M C N$ of a regular pentagon, we won't change $M(S) / m(S)$. If now $D$ or $E$ lies inside $A B M C N$, then we are done. Suppose that both $D$ and $E$ are inside the triangles $C M A^{\prime}, C N B^{\prime}$. Then $C D, C E \leq C M$ (because $C M=C N=C A^{\prime}=C B^{\prime}$ ) and $\angle D C E$ is either less than or equal to $36^{\circ}$ or greater than or equal to $108^{\circ}$, from which we obtain that the area of $\triangle C D E$ cannot exceed the area of $\triangle C M N=M(S) / \alpha$. This completes the proof.