Difference between revisions of "2005 AMC 10B Problems/Problem 7"
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− | Let the radius of the smaller 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 larger square has sides of length <math>2\sqrt{2}r</math>. The ratio of the area of the smaller circle to the area of the larger square is therefore | + | Let the radius of the smaller 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 larger square has sides of length <math>2\sqrt{2}r</math>. The ratio of the area of the smaller circle to the area of the larger square is therefore <cmath>\frac{\pi r^2}{\left(2\sqrt{2}r\right)^2} =\boxed{\frac{\pi}{8}}.</cmath> |
− | \frac{\pi r^2}{\left(2\sqrt{2}r\right)^2} =\boxed{\frac{\pi}{8}}. | ||
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[asy] | [asy] |
Revision as of 22:13, 26 April 2016
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 smaller circle to the area of the larger 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
[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("",(-6,0),S); label("",(-3.5,3.5),NE); label("",(-7.1,7.1),W); label("",(0,14.1),N); [/asy]
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.