1996 AHSME Problems/Problem 11

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 $\boxed{D}$ .

1996 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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1996 AHSME Problems/Problem 11

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