Difference between revisions of "2010 AMC 8 Problems/Problem 23"
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| + | ==Video Solution by MathTalks== | ||
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| + | https://youtu.be/mSCQzmfdX-g | ||
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| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/yBpbVpoVUXI | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=22|num-a=24}} | {{AMC8 box|year=2010|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 23:29, 17 December 2023
Contents
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/6/f/e/6fef195a7bd352396d3561535ac6c4b7e2c67129.png)
Solution
By the Pythagorean Theorem, the radius of the larger circle turns out to be 
. Therefore, the area of the larger circle is 
. Using the coordinate plane given, we find that the radius of each of the two semicircles to be 1. So, the area of the two semicircles is 
. Finally, the ratio of the combined areas of the two semicircles to the area of circle is 
.
Video Solution by OmegaLearn
https://youtu.be/abSgjn4Qs34?t=903
~ pi_is_3.14
Video Solution by MathTalks
Video Solution by WhyMath
~savannahsolver
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. 
