Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2010 AMC 8 Problems/Problem 23: Difference between revisions

← Older editNewer edit →
Debussy (talk | contribs)
45 edits
Mwil (talk | contribs)
15 edits
Line 8: Line 8:


==Solution==
==Solution==
By the Pythagorean Theorem, the radius of the larger circle turns out to be <math>1^2 + 1^2 = \sqrt{2}</math>. Therefore, the area of the larger circle is <math>(\sqrt{2})^2\pi = 2\pi </math>. 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 <math>1^2\pi=\pi</math>. Finally, the ratio of the combined areas of the two semicircles to the area of circle <math>O</math> is <math>\boxed{\textbf{(B)}\ \frac{1}{2}}</math>.
By the Pythagorean Theorem, the radius of the larger circle turns out to be <math>\sqrt{1^2 + 1^2} = \sqrt{2}</math>. Therefore, the area of the larger circle is <math>(\sqrt{2})^2\pi = 2\pi </math>. 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 <math>1^2\pi=\pi</math>. Finally, the ratio of the combined areas of the two semicircles to the area of circle <math>O</math> is <math>\boxed{\textbf{(B)}\ \frac{1}{2}}</math>.


==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}}

Revision as of 16:59, 12 October 2017

Problem

Semicircles $POQ$ and $ROS$ pass through the center $O$. What is the ratio of the combined areas of the two semicircles to the area of circle $O$? [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] $\textbf{(A)}\ \frac{\sqrt 2}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{\pi}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{\sqrt 2}{2}$

Solution

By the Pythagorean Theorem, the radius of the larger circle turns out to be $\sqrt{1^2 + 1^2} = \sqrt{2}$. Therefore, the area of the larger circle is $(\sqrt{2})^2\pi = 2\pi$. 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 $1^2\pi=\pi$. Finally, the ratio of the combined areas of the two semicircles to the area of circle $O$ is $\boxed{\textbf{(B)}\ \frac{1}{2}}$.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_23&oldid=87780"