2001 AMC 12 Problem 17

$\angle APB = 90^\circ$ when $P$ is on the circle with diameter $AB,$ centered at $(2, 1)$ with radius $\dfrac{|AB|}{2} = \dfrac{\sqrt{20}}{2} = \sqrt{5}.$ The angle is obtuse when $P$ is inside this circle. The relevant half-disk lies wholly within the pentagon, with area $\dfrac{1}{2}\pi(\sqrt{5})^2 = \dfrac{5\pi}{2}.$ The pentagon is the rectangle with corners $(0,0),$ $C,$ $D,$ $E$ minus triangle $OAB,$ so its area is $$4(2\pi + 1) - \dfrac{1}{2}(2)(4) = 8\pi.$$ The probability is $\dfrac{\frac{5\pi}{2}}{8\pi} = \dfrac{5}{16}.$ Thus, the correct answer is C.