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Difference between revisions of "2005 AMC 10B Problems/Problem 7"
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== Solution 1 ==
 
== Solution 1 ==
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Let the side of the largest square be <math>x</math>. It follows that the diameter of the inscribed circle is also <math>x</math>. Therefore, the diagonal of the square inscribed inscribed in the circle is <math>x</math>. The side length of the smaller square is <math>\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}</math>. Similarly, the diameter of the smaller inscribed circle is <math>\dfrac{x\sqrt{2}}{2}</math>. Hence, its radius is <math>\dfrac{x\sqrt{2}}{4}</math>. The area of this circle is <math>\left(\dfrac{x\sqrt{2}}{4}\right)^2\pi=\dfrac{2\pi x^2}{16}=\dfrac{x^2\pi}{8}</math>, and the area of the largest square is <math>x^2</math>. The ratio of the areas is <math>\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}}</math>.
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Let the side of the largest square be <math>x</math>. It follows that the diameter of the inscribed circle is also <math>x</math>. Therefore, the square's diagonal inscribed in the circle is <math>x</math>. The side length of the smaller square is <math>\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}</math>. Similarly, the diameter of the smaller inscribed circle is <math>\dfrac{x\sqrt{2}}{2}</math>. Hence, its radius is <math>\dfrac{x\sqrt{2}}{4}</math>. The area of this circle is <math>\left(\dfrac{x\sqrt{2}}{4}\right)^2\pi=\dfrac{2\pi x^2}{16}=\dfrac{x^2\pi}{8}</math>, and the area of the largest square is <math>x^2</math>. The ratio of the areas is <math>\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}}</math>.
  
 
== Solution 2 ==
 
== Solution 2 ==
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label("$2\sqrt{2}r$",(0,14.1),N);
 
label("$2\sqrt{2}r$",(0,14.1),N);
 
</asy>
 
</asy>
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When facing a geometry problem, it is very helpful to draw a diagram.
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Tip: When facing a geometry problem, it is very helpful to draw a diagram.
  
 
== See Also ==
 
== See Also ==
 
{{AMC10 box|year=2005|ab=B|num-b=6|num-a=8}}
 
{{AMC10 box|year=2005|ab=B|num-b=6|num-a=8}}
  
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[[Category:Introductory Geometry Problems]]
+
[[Category: Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:54, 11 November 2024

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 square's diagonal 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] Tip: 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

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