Difference between revisions of "2005 AMC 10B Problems/Problem 7"
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− | Let the radius of the | + | 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 <cmath>\frac{\pi r^2}{\left(2\sqrt{2}r\right)^2} =\frac{\pi}{8}\implies \boxed{\mathrm B}.</cmath> |
<asy> | <asy> |
Revision as of 21:12, 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 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
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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