2010 AMC 8 Problems/Problem 23
Problem
Semicircles 
and 
pass through the center 
. What is the ratio of the combined areas of the two semicircles to the area of circle ?
![[asy] import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.27,xmax=10.01,ymin=-5.65,ymax=10.98; draw(circle((0,0),2)); draw((-3,0)--(3,0),EndArrow(6)); draw((0,-3)--(0,3),EndArrow(6)); draw(shift((0.01,1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,179.76,359.76)); draw(shift((-0.01,-1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,-0.38,179.62)); draw((-1.4,1.43)--(1.41,1.41)); draw((-1.42,-1.41)--(1.4,-1.42)); label("$ P(-1,1) $",(-2.57,2.17),SE*lsf); label("$ Q(1,1) $",(1.55,2.21),SE*lsf); label("$ R(-1,1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf); dot((0,0),ds); label("$O$",(-0.24,-0.35),NE*lsf); dot((1.41,1.41),ds); dot((-1.4,1.43),ds); dot((1.4,-1.42),ds); dot((-1.42,-1.41),ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]](http://latex.artofproblemsolving.com/c/2/4/c249e85286e9954a4b55cd1290e9ae9dbd749f05.png)
Solution
According to the pythagorean theorem, The radius of the larger circle is:
Therefore the area of the larger circle is:
Using the coordinate plane given we find that the radius of the two semicircles to be 1. Therefore the area of the two semicircles is
Finally the ratio of the combined areas of the two semicircles to the area of circle is 
.
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
