2012 AMC 10A Problems/Problem 18
Problem 18
The closed curve in the figure is made up of 9 congruent circular arcs each of length 
, 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]](http://latex.artofproblemsolving.com/c/2/5/c25ea7245d49ffc173e58d29fdca0f31dda4f437.png)
Solution
Draw the hexagon between the centers of the circles, and compute its area (
). Then add the areas of the three sectors outside the hexagon (
) and subtract the areas of the three sectors inside the hexagon (
) to get the area enclosed in the curved figure (
), which is 
.
See Also
| 2012 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
