2020 AMC 8 Problems/Problem 24
Contents
Problem 24
A large square region is paved with gray square tiles, each measuring inches on a side. A border inches wide surrounds each tile. The figure below shows the case for . When , the gray tiles cover of the area of the large square region. What is the ratio for this larger value of
Solution 1
The area of the shaded region is . To find the area of the large square, we note that there is a -inch border between each of the pairs of consecutive squares, as well as from between first/last squares and the large square, for a total of times the length of the border, i.e. . Adding this to the total length of the consecutive squares, which is , the side length of the large square is , yielding the equation . Taking the square root of both sides (and using the fact that lengths are non-negative) gives , and cross-multiplying now gives .
Note: Once we obtain to ease computation, we may take the reciprocal of both sides to yield so Multiplying both sides by yields the same answer as before. ~peace09
Solution 2
Without loss of generality, we may let (since will be determined by the scale of , and we are only interested in the ratio ). Then, as the total area of the gray tiles is simply , the large square has area , making the side of the large square . As in Solution 1, the side length of the large square consists of the total length of the gray tiles and lots of the border, so the length of the border is . Since if , the answer is .
Solution 3 (using answer choices)
As in Solution 2, we let without loss of generality. For sufficiently large , we can approximate the percentage of the area covered by the gray tiles by subdividing most of the region into congruent squares, as shown: Each red square has side length , so by solving , we obtain . The actual fraction of the total area covered by the gray tiles will be slightly less than , which implies . Hence (and thus , since we are assuming ) is less than , and the only choice that satisfies this is .
Video Solution
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Video Solution by OmegaLearn
https://youtu.be/UpCURw5Moig?t=31
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Video Solution by WhyMath
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Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=1515
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Video Solution by STEMbreezy
https://youtu.be/wq8EUCe5oQU?t=353
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See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 | ||
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