Difference between revisions of "2005 AMC 10B Problems/Problem 7"
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== Problem == | == 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 | + | 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>\mathrm{(A)} \frac{\pi}{16} \qquad \mathrm{(B)} \frac{\pi}{8} \qquad \mathrm{(C)} \frac{3\pi}{16} \qquad \mathrm{(D)} \frac{\pi}{4} \qquad \mathrm{(E)} \frac{\pi}{2} </math> | <math>\mathrm{(A)} \frac{\pi}{16} \qquad \mathrm{(B)} \frac{\pi}{8} \qquad \mathrm{(C)} \frac{3\pi}{16} \qquad \mathrm{(D)} \frac{\pi}{4} \qquad \mathrm{(E)} \frac{\pi}{2} </math> | ||
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== Solution 1 == | == 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 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}\implies \boxed{\mathrm{(B)}\ \dfrac{\pi}{8}}</math>. | 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}\implies \boxed{\mathrm{(B)}\ \dfrac{\pi}{8}}</math>. |
Revision as of 21:11, 31 May 2021
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 diagonal of the square inscribed 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 smaller 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 larger square has sides of length . The ratio of the area of the smaller circle to the area of the larger square is therefore
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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