30. In a convex quadrilateral $ABCD$, $AB=a, BC=b, CD=c, DA=d, AC=e, BD=f$, and $\max \{a, b, c, d, e, f\}=1$. Find the maximum value of $abcd$.
(45th IMO Chinese National Team Test Question)
30. Discuss the following two cases: (1) When $e=f=1$, let $A C$ and $B D$ intersect at point $E$, and denote $A E=m, C E=n, B E=p, D E=q$, then $m+n=p+q=1$. By symmetry, assume without loss of generality that $p \geqslant m \geqslant n \geqslant q$, it is easy to see that $m n \geqslant p q$. When $p \leqslant \frac{\sqrt{3}}{2}$, by Stewart's Theorem, we have $p^{2}+m n=a^{2} n+b^{2} m \geqslant 2 a b \sqrt{m n}, q^{2}+m n=c^{2} m+d^{2} n \geqslant 2 c d \sqrt{m n}$. Therefore, $4 a b c d m n \leqslant\left(p^{2}+m n\right)$ $\left(q^{2}+m n\right)=p^{2} q^{2}+m n\left(p^{2}+q^{2}\right)+m^{2} n^{2}=m n+(m n-p q)^{2}=m n$. $\left[1+\left(\sqrt{m n}-\frac{p q}{\sqrt{m n}}\right)\right]$. Since $m n \leqslant \frac{(m+n)^{2}}{4}=\frac{1}{4}$, and when $x$ is closer to $1-x$, $x(1-x)\left(0 < x < \frac{1}{2}\right)$ is larger. When $p > \frac{\sqrt{3}}{2}$, by $q^{2}+m n=c^{2} m+d^{2} n \geqslant 2 c d \sqrt{m n}$ and $\sqrt{m n} \geqslant \sqrt{p q}>q$, we get $2 c d \leqslant \frac{q^{2}}{\sqrt{m n}}+$ $\sqrt{m n} \leqslant 2 q^{2}+\frac{1}{2}\left(f(x)=\frac{q^{2}}{x}+x\right.$ is increasing on $\left(q, \frac{1}{2}\right]$ $) \leqslant 2\left(1-\frac{3}{2}\right)^{2}+\frac{1}{2}=4$ $-2 \sqrt{3}$, so $c d \leqslant 2-\sqrt{3}$. Also, $0A B, C B^{\prime}>C B, B^{\prime} D>B D$, take point $B^{\prime}$ such that $\max \left\{A B^{\prime}, C B^{\prime}, D B^{\prime} \mid=1\right.$. At this point, if $D B^{\prime}=1$, then by (1), we get $a b c d \leqslant A B^{\prime} \cdot B^{\prime} C \cdot c d \leqslant 2-\sqrt{3}$. Otherwise, if $A B^{\prime}=1$ or $C B^{\prime}=1$, the problem reduces to the following (ii). (ii) If at least one of $a, b, c, d$ is 1, assume without loss of generality that $C D=c=1$. Draw arcs with centers at $C$ and $D$ and radius 1, intersecting at point $P$ (on the same side of $C D$ as $A$ and $B$), then $A$ and $B$ are inside or on the boundary of this region. If point $B$ is inside the region, draw the extension of the angle bisector of $\angle A B C$ to intersect the arc at point $B^{\prime}$, if point $A$ is inside the region, draw the extension of the angle bisector of $\angle B^{\prime} A D$ to intersect the arc at point $A^{\prime}$ (if points $A$ and $B$ are already on the arc, no further action is needed, but $A$ and $B$ cannot be on the arc simultaneously). If $B^{\prime}$ and $A^{\prime}$ are on arcs $\overparen{C P}$ and $\overparen{D P}$ respectively. Since $A B \leqslant A B^{\prime} \leqslant A^{\prime} B^{\prime}$, $B C \leqslant B^{\prime} C, A D \leqslant A^{\prime} D$, and the equalities cannot hold simultaneously, and $A^{\prime} C=B^{\prime} D=1$, then by (1) we know $a b c d<A^{\prime} B^{\prime} \cdot B^{\prime} C \cdot C D \cdot D A^{\prime} \leqslant 2-\sqrt{3}$. If $B^{\prime}$ and $A^{\prime}$ are on the same arc, say on $\overparen{D P}$, then $A^{\prime} B^{\prime}<A^{\prime} P$. Adjust $B^{\prime}$ to $P$, by (1), we get $a b c d<A^{\prime} P \cdot P C \cdot C D \cdot D A^{\prime}$ $\leqslant 2-\sqrt{3}$. In summary, the maximum value of $a b c d$ is $2-\sqrt{3}$.