[ Quadrilaterals (extreme properties) ]

The diagonals of a convex quadrilateral $A B C D$ intersect at point $O$. What is the smallest area that this quadrilateral can have if the area of triangle $A O B$ is 4 and the area of triangle $C O D$ is 9?
Since $S_{\mathrm{AOB}}: S_{\mathrm{BOC}}=A O: O C=S_{\mathrm{AOD}}: S_{\mathrm{DOC}}$, then $S_{\mathrm{BOC}} \cdot S_{\mathrm{AOD}}=S_{\mathrm{AOB}} \cdot S_{\mathrm{DOC}}=36$. Therefore, $S_{\mathrm{BOC}}+S_{\mathrm{AOD}}$ $\geq 2 \sqrt{S_{B O C} \cdot S_{A O D}}=12$, and equality is achieved if $S_{\mathrm{BOC}}=S_{\mathrm{AOD}}$, i.e., $S_{\mathrm{ABC}}=S_{\mathrm{ABD}}$, from which $A B \mid C D$. In this case, the area of the quadrilateral is $4+9+12=25$.

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