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Difference between revisions of "1996 AHSME Problems/Problem 11"
 
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==Problem==
 
==Problem==
  
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Given a circle of raidus <math>2</math>, there are many line segments of length <math>2</math> that are tangent to the circle at their midpoints. Find the area of the region consisting of all such line segments.
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Given a [[circle]] of [[radius]] <math>2</math>, there are many line segments of length <math>2</math> that are [[Tangent (geometry)|tangent]] to the circle at their [[midpoint]]s. Find the area of the region consisting of all such line segments.
 
   
 
   
 
<math> \text{(A)}\ \frac{\pi} 4\qquad\text{(B)}\ 4-\pi\qquad\text{(C)}\ \frac{\pi} 2\qquad\text{(D)}\ \pi\qquad\text{(E)}\ 2\pi </math>
 
<math> \text{(A)}\ \frac{\pi} 4\qquad\text{(B)}\ 4-\pi\qquad\text{(C)}\ \frac{\pi} 2\qquad\text{(D)}\ \pi\qquad\text{(E)}\ 2\pi </math>
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<math>\triangle OAP</math> is a right triangle with right angle at <math>P</math>, because <math>AB</math> is tangent to circle <math>O</math> at point <math>P</math>, and <math>OP</math> is a radius.
 
<math>\triangle OAP</math> is a right triangle with right angle at <math>P</math>, because <math>AB</math> is tangent to circle <math>O</math> at point <math>P</math>, and <math>OP</math> is a radius.
  
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Since <math>AP^2 + OP^2 = OA^2</math>, we can find that <math>OA = \sqrt{1^2 + 2^2} = \sqrt{5}</math>.
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Since <math>AP^2 + OP^2 = OA^2</math> by the [[Pythagorean Theorem]], we can find that <math>OA = \sqrt{1^2 + 2^2} = \sqrt{5}</math>. Similarly, <math>OB = \sqrt{5}</math> also.
  
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Similarly, <math>OB = \sqrt{5}</math> also.
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Line segment <math>APB</math> can rotate around the circle.  The closest distance of this segment to the center will always be <math>OP = 2</math>, and the longest distance of this segment will always be <math>PA = PB = \sqrt{5}</math>.  Thus, the region in question is the [[annulus]] of a circle with outer radius <math>\sqrt{5}</math> and inner radius <math>2</math>.  This area is <math>\pi \cdot 5 - \pi \cdot 4 = \pi</math>, and the answer is <math>\boxed{D}</math>.
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−
Line segment <math>APB</math> can rotate around the circle.  The closest distance of this segment to the center will always be <math>OP = 2</math>, and the longest distance of this segment will always be <math>PA = PB = \sqrt{5}</math>.  Thus, the region in question is the annulus of a circle with outer radius <math>\sqrt{5}</math> and inner radius <math>2</math>.  This area is <math>\pi \cdot 5 - \pi \cdot 4 = \pi</math>, and the answer is <math>\boxed{D}</math>.
 
  
 
==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=10|num-a=12}}
 
{{AHSME box|year=1996|num-b=10|num-a=12}}
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 +
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 13:07, 5 July 2013

Problem

Given a circle of radius $2$, there are many line segments of length $2$ that are tangent to the circle at their midpoints. Find the area of the region consisting of all such line segments.

$\text{(A)}\ \frac{\pi} 4\qquad\text{(B)}\ 4-\pi\qquad\text{(C)}\ \frac{\pi} 2\qquad\text{(D)}\ \pi\qquad\text{(E)}\ 2\pi$

Solution

Let line segment $AB = 2$, and let it be tangent to circle $O$ at point $P$, with radius $OP = 2$. Let $AP = PB = 1$, so that $P$ is the midpoint of $AB$.

$\triangle OAP$ is a right triangle with right angle at $P$, because $AB$ is tangent to circle $O$ at point $P$, and $OP$ is a radius.

Since $AP^2 + OP^2 = OA^2$ by the Pythagorean Theorem, we can find that $OA = \sqrt{1^2 + 2^2} = \sqrt{5}$. Similarly, $OB = \sqrt{5}$ also.

Line segment $APB$ can rotate around the circle. The closest distance of this segment to the center will always be $OP = 2$, and the longest distance of this segment will always be $PA = PB = \sqrt{5}$. Thus, the region in question is the annulus of a circle with outer radius $\sqrt{5}$ and inner radius $2$. This area is $\pi \cdot 5 - \pi \cdot 4 = \pi$, and the answer is $\boxed{D}$.

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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