Difference between revisions of "1992 AJHSME Problems/Problem 24"
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==See Also== | ==See Also== | ||
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Latest revision as of 23:10, 4 July 2013
Problem
Four circles of radius 
are arranged as shown. Their centers are the vertices of a square. The area of the shaded region is closest to
![[asy] fill((3,3)--(3,-3)--(-3,-3)--(-3,3)--cycle,lightgray); fill(arc((3,3),(0,3),(3,0),CCW)--(3,3)--cycle,white); fill(arc((3,-3),(3,0),(0,-3),CCW)--(3,-3)--cycle,white); fill(arc((-3,-3),(0,-3),(-3,0),CCW)--(-3,-3)--cycle,white); fill(arc((-3,3),(-3,0),(0,3),CCW)--(-3,3)--cycle,white); draw(circle((3,3),3)); draw(circle((3,-3),3)); draw(circle((-3,-3),3)); draw(circle((-3,3),3)); draw((3,3)--(3,-3)--(-3,-3)--(-3,3)--cycle); [/asy]](http://latex.artofproblemsolving.com/2/9/9/2993624e51e3a6603dd60630eadad79dacead787.png)
Solution
The side of the square of made up of two radii, and the area of the square is 
. The unshaded region of the square is composed of four quarter circles, and has the area of one circle, which is 
. The area of the shaded region is 
which less than 
, closest to 
.
See Also
| 1992 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
