# 2012 AMC 10A Problems/Problem 18

The following problem is from both the 2012 AMC 12A #14 and 2012 AMC 10A #18, so both problems redirect to this page.

## Problem 14

The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2\pi}{3}$, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve? $[asy] defaultpen(fontsize(6pt)); dotfactor=4; label("\circ",(0,1)); label("\circ",(0.865,0.5)); label("\circ",(-0.865,0.5)); label("\circ",(0.865,-0.5)); label("\circ",(-0.865,-0.5)); label("\circ",(0,-1)); dot((0,1.5)); dot((-0.4325,0.75)); dot((0.4325,0.75)); dot((-0.4325,-0.75)); dot((0.4325,-0.75)); dot((-0.865,0)); dot((0.865,0)); dot((-1.2975,-0.75)); dot((1.2975,-0.75)); draw(Arc((0,1),0.5,210,-30)); draw(Arc((0.865,0.5),0.5,150,270)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0,-1),0.5,30,150)); draw(Arc((-0.865,-0.5),0.5,330,90)); draw(Arc((-0.865,0.5),0.5,-90,30)); [/asy]$ $\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt{3}\qquad\textbf{(C)}\ 3\pi+4\qquad\textbf{(D)}\ 2\pi+3\sqrt{3}+2\qquad\textbf{(E)}\ \pi+6\sqrt{3}$

## Solution 1

Draw the hexagon between the centers of the circles, and compute its area $(6)(0.5)(2\sqrt{3})=6\sqrt{3}$. Then add the areas of the three sectors outside the hexagon ( $2\pi$) and subtract the areas of the three sectors inside the hexagon but outside the figure( $\pi$) to get the area enclosed in the curved figure $(\pi+6\sqrt{3})$, which is $\boxed{\textbf{(E)}\ \pi+6\sqrt{3}}$.

## Solution 2 (Looking at the answer choices)

We see that after forming the hexagon using the sectors outside the hexagon, there will be three sectors left. Each sector has an area of $\frac{\pi}{3},$ so the three combined make $\pi.$ Since the side length of the hexagon is $2,$ it's area doesn't have $\pi$ in it, so we know that the final answer will be $\pi + \text{(area of hexagon)}.$ Looking at the answer choices, the only answer with only one $\pi$ is $\boxed{\textbf{(E)}}.$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 