2001 AMC 12 Problem 17

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}

\angle APB

is obtuse exactly when

P

lies inside the circle having

AB

as diameter

That semicircle lies entirely inside the pentagon, so compare its area to the pentagon’s area

\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.

Problem 17 in other years