Difference between revisions of "1996 AHSME Problems/Problem 11"
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| + | ==Problem== | ||
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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. | ||
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| + | <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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| + | ==Solution== | ||
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| + | Let line segment <math>AB = 2</math>, and let it be tangent to circle <math>O</math> at point <math>P</math>, with radius <math>OP = 2</math>. Let <math>AP = PB = 1</math>, so that <math>P</math> is the midpoint of <math>AB</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. | ||
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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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| + | 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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==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]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:07, 5 July 2013
Problem
Given a circle of radius 
, there are many line segments of length that are tangent to the circle at their midpoints. Find the area of the region consisting of all such line segments.
Solution
Let line segment 
, and let it be tangent to circle 
at point 
, with radius 
. Let 
, so that is the midpoint of 
.
is a right triangle with right angle at , because is tangent to circle at point , and 
is a radius.
Since 
by the Pythagorean Theorem, we can find that 
. Similarly, 
also.
Line segment 
can rotate around the circle. The closest distance of this segment to the center will always be , and the longest distance of this segment will always be 
. Thus, the region in question is the annulus of a circle with outer radius 
and inner radius . This area is 
, and the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
