104, For any set $S$ of 5 points in the plane, satisfying that no three points in $S$ are collinear, let $M(S)$ and $m(S)$ be the maximum and minimum areas of triangles formed by any 3 points in $S$, respectively. Find the minimum value of $\frac{M(S)}{m(S)}$ $\quad($ Problem 43 of the 1 M O Preliminary Selection $)$
104. When these 5 points are the vertices of a regular pentagon, it is easy to know that $\frac{M(S)}{m(S)}$ equals the golden ratio $\tau=\frac{\sqrt{5}+1}{2}$.

Let the 5 points in $S$ be $A, B, C, D, E$, and the area of $\triangle A B C$ is $M(S)$. We prove that there exists a triangle whose area is less than or equal to $\frac{M(S)}{\tau}$.

As shown in Figure (1), construct $\triangle A^{\prime} B^{\prime} C^{\prime}$ such that each side is parallel to the corresponding side of $\triangle A B C$, and $A, B, C$ are the midpoints of $B^{\prime} C^{\prime}, C^{\prime} A^{\prime}, A^{\prime} B^{\prime}$, respectively. Then points $D, E$ are inside or on the boundary of $\triangle A^{\prime} B^{\prime} C^{\prime}$.

Since at least one of $\triangle A^{\prime} B C, \triangle A B^{\prime} C, \triangle A B C^{\prime}$ does not contain $D$ or $E$, assume without loss of generality that $D, E$ are inside the quadrilateral $B C B^{\prime} C^{\prime}$.

As shown in Figure (2), since the ratio $\frac{M(S)}{m(S)}$ remains unchanged under affine transformation, assume that $A, B, C$ are the vertices of the regular pentagon $A P B C Q$. Because we can first use an affine transformation to meet the requirement, thus, $\angle A B P = \angle B A C = 36^{\circ}$, so, $P$ is on $B C^{\prime}$, and similarly, $Q$ is on $C B^{\prime}$.

If $D$ or $E$ is inside the regular pentagon $A P B C Q$, assume without loss of generality that $D$. Since $S_{\triangle A P B} = \frac{M(S)}{\tau}$, if $D$ is inside $\triangle A P B$, then $S_{\triangle D A B} \leqslant S_{\triangle A P B}$.

Similarly, if $D$ is inside $\triangle A Q C$, then $S_{\triangle D A C} \leqslant S_{\triangle Q A C} = S_{\triangle A P B}$. If $D$ is inside $\triangle A B C$, then at least one of $S_{\triangle D A B}, S_{\triangle D B C}, S_{\triangle D C A}$ does not exceed $\frac{M(S)}{3} < \frac{M(S)}{\tau}$.

If $D$ and $E$ are inside $\triangle A P C^{\prime}$ and $\triangle A Q B^{\prime}$, then $\max \{A E, A D\} \leqslant A P = A Q$. Since $0^{\circ} < \angle D A E \leqslant 36^{\circ}$ (if $D$ and $E$ are in the same triangle) or $108^{\circ} \leqslant \angle D A E < 180^{\circ}$ (if $D$ and $E$ are in the same triangle), thus, $S_{\triangle A D E} = \frac{1}{2} A D \cdot A E \cdot \sin \angle D A E \leqslant \frac{1}{2} A P \cdot A Q \cdot \sin 108^{\circ} = S_{\triangle A P Q} = \frac{M(S)}{\tau}$.

Therefore, the minimum value sought is $\tau = \frac{\sqrt{5}+1}{2}$.