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

2005 AMC 10B Problems/Problem 7

Problem

A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smallest circle to the area of the largest square?

$\textbf{(A) } \frac{\pi}{16} \qquad \textbf{(B) } \frac{\pi}{8} \qquad \textbf{(C) } \frac{3\pi}{16} \qquad \textbf{(D) } \frac{\pi}{4} \qquad \textbf{(E) } \frac{\pi}{2}$

Solution 1

Let the side of the largest square be $x$. It follows that the diameter of the inscribed circle is also $x$. Therefore, the diagonal of the square inscribed inscribed in the circle is $x$. The side length of the smaller square is $\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}$. Similarly, the diameter of the smaller inscribed circle is $\dfrac{x\sqrt{2}}{2}$. Hence, its radius is $\dfrac{x\sqrt{2}}{4}$. The area of this circle is $\left(\dfrac{x\sqrt{2}}{4}\right)^2\pi=\dfrac{2\pi x^2}{16}=\dfrac{x^2\pi}{8}$, and the area of the largest square is $x^2$. The ratio of the areas is $\dfrac{\dfrac{x^2\pi}{8}}{x^2}=\frac{\cancel{x^2}\pi}{8}\cdot\frac{1}{\cancel{x^2}}=\boxed{\textbf{(B) }\frac{\pi}{8}}$.

Solution 2

Let the radius of the smallest circle be $r$. Then the side length of the smaller square is $2r$. The radius of the larger circle is half the length of the diagonal of the smaller square, so it is $\sqrt{2}r$. Hence the largest square has sides of length $2\sqrt{2}r$. The ratio of the area of the smallest circle to the area of the largest square is therefore $\frac{\pi r^2}{\left(2\sqrt{2}r\right)^2} =\boxed{\textbf{(B) }\frac{\pi}{8}}.$

[asy] draw(Circle((0,0),10),linewidth(0.7)); draw(Circle((0,0),14.1),linewidth(0.7)); draw((0,14.1)--(14.1,0)--(0,-14.1)--(-14.1,0)--cycle,linewidth(0.7)); draw((-14.1,14.1)--(14.1,14.1)--(14.1,-14.1)--(-14.1,-14.1)--cycle,linewidth(0.7)); draw((0,0)--(-14.1,0),linewidth(0.7)); draw((-7.1,7.1)--(0,0),linewidth(0.7)); label("$\sqrt{2}r$",(-6,0),S); label("$r$",(-3.5,3.5),NE); label("$2r$",(-7.1,7.1),W); label("$2\sqrt{2}r$",(0,14.1),N); [/asy] When facing a geometry problem, it is very helpful to draw a diagram.

See Also

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems/Problem_7&oldid=167730"