A point $P$ is chosen at random inside rectangle $ABCD$ with $AB=8$ and $BC=4$. What is the probability that $\overline{AB}$ is neither the shortest nor the longest side of $\triangle APB$?

1. Put $A=(0,0)$, $B=(8,0)$, so $P=(x,y)$ with $0\le y\le4$. Since $AB=8$, the side $AB$ is the middle one exactly when $AP<8
2. If $BP>8$, then $(8-x)^2>64-y^2\ge48$, so $8-x>4\sqrt3$ and $x<8-4\sqrt3<2$; hence $AP^2<4+16<64$ and $AP<8$ automatically. So the region $AP<8
3. This circle passes through $A$ and meets $\overline{CD}$ at a point $E$ with $BE=8$, $BC=4$, so $\angle EBC=60^\circ$, $\angle ABE=30^\circ$ and $EC=4\sqrt3$. The part of the rectangle inside the circle consists of the $30^\circ$ sector $BAE$ (area $\frac{30}{360}\pi\cdot8^2=\frac{16\pi}3$) and the right triangle $BCE$ (area $\frac12\cdot4\cdot4\sqrt3=8\sqrt3$).
4. So $[\,AP<8$$\frac{2\left(32-8\sqrt3-\frac{16\pi}3\right)}{32}=2-\frac{\sqrt3}2-\frac\pi3=\boxed{\dfrac{12-3\sqrt3-2\pi}6}.$$