2001 AMC 12 Problems/Problem 17

Problem

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?

$\text{(A) }\frac {1}{5} \qquad \text{(B) }\frac {1}{4} \qquad \text{(C) }\frac {5}{16} \qquad \text{(D) }\frac {3}{8} \qquad \text{(E) }\frac {1}{2}$

Solution

The angle $APB$ is obtuse if and only if $P$ lies inside the circle with diameter $AB$ . (This follows for example from the fact that the inscribed angle is half of the central angle for the same arc.)

$[asy] defaultpen(0.8); real pi=3.14159265359; pair A=(0,2), B=(4,0), C=(2*pi+1, 0), D=(2*pi+1,4), E=(0,4), F=(0,0); draw(A--B--C--D--E--cycle); draw(circle((A+B)/2,length(B-A)/2)); label("$A$",A,W); label("$B$",B,SE); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,SW); draw(A--F--B,dashed); [/asy]$

The area of $AFB$ is $[AFB] = \frac {AF\cdot FB}2 = 4$ , and the area of $ABCDE$ is $CD\cdot DE - [AFB] = 4\cdot (2\pi+1) - 4 = 8\pi$ .

From the Pythagorean theorem the length of $AB$ is $\sqrt{2^2 + 4^2} = 2\sqrt{5}$ , thus the radius of the circle is $\sqrt{5}$ , and the area of the half-circle that is inside $ABCDE$ is $\frac{ 5\pi }2$ .

Therefore the probability that $APB$ is obtuse is $\frac{ \frac{ 5\pi }2 }{ 8\pi } = \boxed{\frac 5{16}}$ .

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2001 AMC 12 Problems/Problem 17

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