Difference between revisions of "2005 AMC 10B Problems/Problem 7"
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== Problem == | == Problem == | ||
− | == Solution == | + | 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? |
+ | |||
+ | <math>\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} </math> | ||
+ | |||
+ | == Solution 1 == | ||
+ | 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 == | ||
+ | Let the radius of the smallest circle be <math>r</math>. Then the side length of the smaller square is <math>2r</math>. The radius of the larger circle is half the length of the diagonal of the smaller square, so it is <math>\sqrt{2}r</math>. Hence the largest square has sides of length <math>2\sqrt{2}r</math>. The ratio of the area of the smallest circle to the area of the largest square is therefore <math>\frac{\pi r^2}{\left(2\sqrt{2}r\right)^2} =\boxed{\textbf{(B) }\frac{\pi}{8}}.</math> | ||
+ | |||
+ | <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 == | == See Also == | ||
− | + | {{AMC10 box|year=2005|ab=B|num-b=6|num-a=8}} | |
+ | |||
+ | [[Category: Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 01:54, 11 November 2024
Contents
[hide]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?
Solution 1
Let the side of the largest square be . It follows that the diameter of the inscribed circle is also . Therefore, the square's diagonal inscribed in the circle is . The side length of the smaller square is . Similarly, the diameter of the smaller inscribed circle is . Hence, its radius is . The area of this circle is , and the area of the largest square is . The ratio of the areas is .
Solution 2
Let the radius of the smallest circle be . Then the side length of the smaller square is . The radius of the larger circle is half the length of the diagonal of the smaller square, so it is . Hence the largest square has sides of length . The ratio of the area of the smallest circle to the area of the largest square is therefore
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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