3. (CZS 3) ${ }^{\mathrm{MO} O 1}$ In a convex quadrangle with area $32 \mathrm{~cm}^{2}$, the sum of the lengths of two nonadjacent edges and of the length of one diagonal is equal to $16 \mathrm{~cm}$.
(a) What is the length of the other diagonal?
(b) What are the lengths of the edges of the quadrangle if the perimeter is a minimum?
(c) Is it possible to choose the edges in such a way that the perimeter is a maximum?
3. (a) Let $A B C D$ be a quadrangle with $16=d=A B+C D+A C$, and let $S$ be its area. Then $S \leq(A C \cdot A B+A C \cdot C D) / 2=A C(d-A C) / 2 \leq$ $d^{2} / 8=32$, where equality occurs if and only if $A B \perp A C \perp C D$ and $A C=A B+C D=8$. In this case $B D=8 \sqrt{2}$.
(b) Let $A^{\prime}$ be the point with $\overrightarrow{D A^{\prime}}=\overrightarrow{A C}$. The triangular inequality implies $A D+B C \geq A A^{\prime}=8 \sqrt{5}$. Thus the perimeter attains its minimum for $A B=C D=4$.
(c) Let us assume w.l.o.g. that $C D \leq A B$. Then $C$ lies inside $\triangle B D A^{\prime}$ and hence $B C+A D=B C+C A^{\prime}<B D+D A^{\prime}$. The maximal value $B D+D A^{\prime}$ of $B C+A D$ is attained when $C$ approaches $D$, making a degenerate quadrangle.