Question

A point $P$ is selected at random from the interior of the pentagon with vertices $A = (0, 2),$ $B = (4, 0),$ $C = (2\pi + 1, 0),$ $D = (2\pi + 1, 4),$ and $E = (0, 4).$ What is the probability that $\angle APB$ is obtuse?

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{5}{16}$

$\dfrac{3}{8}$

$\dfrac{1}{2}$

Solution:

$\angle APB = 90^\circ$ 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 $\dfrac{5\pi/2}{8\pi} = \dfrac{5}{16}.$

Thus, the correct answer is C.

2001 AMC 12 Problem 17

Solution:

Problem 17 in Other Years