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2004 AMC 12A Problems/Problem 24

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Problem 24

A plane contains points and with . Let be the union of all disks of radius in the plane that cover . What is the area of ?

\text {(A)} 2\pi + \sqrt3 \qquad \text {(B)} \frac {8\pi}{3} \qquad \text {(C)} 3\pi - \frac {\sqrt3}{2} \qquad \text {(D)} \frac {10\pi}{3} - \sqrt3 \qquad \text {(E)}4\pi - 2\sqrt3


Solution

[Asy_image]

As the red circles move about segment , they cover the area we are looking for. On the left side, the circle must move around pivoted on . On the right side, the circle must move pivoted on However, at the top and bottom, the circle must lie on both A and B, giving us our upper and lower bounds.

This egg-like shape is .

[Asy_image]

The area of the region can be found by dividing it into several sectors, namely

\begin{align*}A &= 2(\mathrm{Blue\ Sector}) + 2(\mathrm{Red\ Sector}) - 2(\mathrm{Equilateral\ Triangle}) \\A &= 2\left(\frac{120^\circ}{360^\circ} \cdot \pi (2)^2\right) + 2\left(\frac{60^\circ}{360^\circ} \cdot \pi (1)^2\right) - 2\left(\frac{(1)^2\sqrt{3}}{4}\right) \\A &= \frac{8\pi}{3} + \frac{\pi}{3} - \frac{\sqrt{3}}{2} \\A &= 3\pi - \frac{\sqrt{3}}{2} \Longrightarrow \textbf {(C)}\end{align*}

See also

2004 AMC 12A (Problems)
Preceded by
Problem 23
Followed by
Problem 25
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