Difference between revisions of "2004 AMC 12A Problems/Problem 24"

Problem 24

A plane contains points $A$ and $B$ with $AB = 1$. Let $S$ be the union of all disks of radius $1$ in the plane that cover $\overline{AB}$. What is the area of $S$?

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

Solution

$[asy] pair A=(-.5,0), B=(.5,0), C=(0,3**(.5)/2), D=(0,-3**(.5)/2); draw(arc(A,2,-60,60),blue); draw(arc(B,2,120,240),blue); draw(circle(C,1),red); draw(A--(.5,3^.5)); draw(B--(-.5,3^.5)); draw(A--(.5,-3^.5)); draw(B--(-.5,-3^.5)); draw(A--B); dot(A);dot(B);dot(C);dot(D); label("$$1$$",(0,0),N); label("$$1$$",A/2+D/2,W); label("$$1$$",A/2+C/2,W); label("$$1$$",B/2+D/2,E); label("$$1$$",B/2+C/2,E); label("$$1$$",A/2+3D/2,W); label("$$1$$",A/2+3C/2,W); label("$$1$$",B/2+3D/2,E); label("$$1$$",B/2+3C/2,E); label("$$A$$",A,W); label("$$B$$",B,E); label("$$C$$",C,W); label("$$D$$",D,E); [/asy]$

As the red circles move about segment $AB$, they cover the area we are looking for. On the left side, the circle must move around pivoted on $B$. On the right side, the circle must move pivoted on $A$ 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 $S$.

$[asy] pair A=(-.5,0), B=(.5,0), C=(0,3**(.5)/2), D=(0,-3**(.5)/2); draw(arc(A,2,-60,60),blue); draw(arc(B,2,120,240),blue); draw(arc(C,1,60,120),red); draw(arc(D,1,-120,-60),red); draw(A--(.5,3^.5)); draw(B--(-.5,3^.5)); draw(A--(.5,-3^.5)); draw(B--(-.5,-3^.5)); draw(A--B); dot(A);dot(B);dot(C);dot(D); label("$$A$$",A,W); label("$$B$$",B,E); label("$$C$$",C,W); label("$$D$$",D,E); label("$$1$$",(0,0),N); label("$$1$$",A/2+D/2,W); label("$$1$$",A/2+C/2,W); label("$$1$$",B/2+D/2,E); label("$$1$$",B/2+C/2,E); label("$$1$$",A/2+3D/2,W); label("$$1$$",A/2+3C/2,W); label("$$1$$",B/2+3D/2,E); label("$$1$$",B/2+3C/2,E); [/asy]$

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*}