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Difference between revisions of "2001 AMC 12 Problems/Problem 17"

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From the [[Pythagorean theorem]] the length of <math>AB</math> is <math>\sqrt{2^2 + 4^2} = 2\sqrt{5}</math>, thus the radius of the circle is <math>\sqrt{5}</math>, and the area of the half-circle that is inside <math>ABCDE</math> is <math>\frac{ 5\pi }2</math>.  
 
From the [[Pythagorean theorem]] the length of <math>AB</math> is <math>\sqrt{2^2 + 4^2} = 2\sqrt{5}</math>, thus the radius of the circle is <math>\sqrt{5}</math>, and the area of the half-circle that is inside <math>ABCDE</math> is <math>\frac{ 5\pi }2</math>.  
  
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Therefore the probability that <math>APB</math> is obtuse is <math>\frac{ \frac{ 5\pi }2 }{ 8\pi } = \boxed{\frac 5{16}}</math>. Answer choice (C)
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Therefore the probability that <math>APB</math> is obtuse is <math>\frac{ \frac{ 5\pi }2 }{ 8\pi } = \boxed{\frac 5{16}}</math>. Answer choice <math>(C)</math>
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2001|num-b=16|num-a=18}}
 
{{AMC12 box|year=2001|num-b=16|num-a=18}}

Revision as of 12:30, 18 February 2013

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 $\frac{ \frac{ 5\pi }2 }{ 8\pi } = \boxed{\frac 5{16}}$. Answer choice $(C)$

See Also

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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