Japan 2007

Take $\angle APB=\theta$ and let $S$ be the area of $APB$. Then $\frac{1}{2}\times AP\times BP\times \sin \theta =S=\frac{3\times 7}{2}=\frac{21}{2}$. Since $\sin \theta$ is positive, $AP\times BP$ takes minimum value when $\sin \theta$ takes maximum. Since $\frac{7}{2}>3$, we can take $P$ on the circle with diameter $AB$. Then $\sin \theta$ takes maximum value $1$ and $AP\times BP$ takes minimum value $21$.