Difference between revisions of "2011 AMC 8 Problems/Problem 25"
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==Solution 1== | ==Solution 1== | ||
− | The area of the smaller square is | + | The area of the smaller square is one half of the product of its diagonals. Note that the distance from a corner of the smaller square to the center is equivalent to the circle's radius so the diagonal is equal to the diameter: <math>2 \cdot |
+ | 2 \cdot \frac{1}{2}=2.</math> | ||
The circle's shaded area is the area of the smaller square subtracted from the area of the circle: <math>\pi - 2.</math> | The circle's shaded area is the area of the smaller square subtracted from the area of the circle: <math>\pi - 2.</math> | ||
Line 24: | Line 25: | ||
Note that this solution is not rigorous, because we still should show that the ratio is less than <math>\frac{3}{4}</math>. | Note that this solution is not rigorous, because we still should show that the ratio is less than <math>\frac{3}{4}</math>. | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Set the side length of the bigger square to be <math>8</math>. | ||
+ | Then the area of the big square is <math>8^2 =64</math> and | ||
+ | the area of the small square <math>(4\sqrt{2})^2 = 32</math>. | ||
+ | The difference is <math>32</math>. | ||
+ | The area of the circle is <math>4^2</math> times <math>\pi</math> which is <math>16 \pi</math> or about <math>48</math>. | ||
+ | Knowing the area of the small square is <math>32</math>. <math>48-32</math> is <math>16</math>. | ||
+ | The area of the big square is <math>64</math>. | ||
+ | So <math>32/64</math> is <math>1/2</math>, or <math>\boxed{\textbf{(A)}\ \frac12}</math>. | ||
+ | |||
+ | |||
+ | - by goldenn | ||
+ | |||
+ | - (tex) updated by CasperYC on 26th July 2020 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=24|after=Last Problem}} | {{AMC8 box|year=2011|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:53, 1 December 2021
Problem
A circle with radius is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?
Solution 1
The area of the smaller square is one half of the product of its diagonals. Note that the distance from a corner of the smaller square to the center is equivalent to the circle's radius so the diagonal is equal to the diameter:
The circle's shaded area is the area of the smaller square subtracted from the area of the circle:
If you draw the diagonals of the smaller square, you will see that the larger square is split congruent half-shaded squares. The area between the squares is equal to the area of the smaller square:
Approximating to the ratio of the circle's shaded area to the area between the two squares is about
Solution 2
For the ratio of the circle's shaded area to the area between the squares to be they would have to be approximately the same size. For any ratio larger than that, the circle's shaded area must be greater. However, we can clearly see that the circle's shaded area is part of the area between the squares, and is approximately .
Note that this solution is not rigorous, because we still should show that the ratio is less than .
Solution 3
Set the side length of the bigger square to be . Then the area of the big square is and the area of the small square . The difference is . The area of the circle is times which is or about . Knowing the area of the small square is . is . The area of the big square is . So is , or .
- by goldenn
- (tex) updated by CasperYC on 26th July 2020
See Also
2011 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.