Difference between revisions of "1996 AHSME Problems/Problem 11"
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==Problem== | ==Problem== | ||
− | Given a circle of | + | 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. | ||
− | Since <math>AP^2 + OP^2 = OA^2</math>, we can find that <math>OA = \sqrt{1^2 + 2^2} = \sqrt{5}</math>. | + | 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. |
− | + | 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]] |
Revision as of 12:11, 19 August 2011
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 |