Difference between revisions of "2020 AMC 8 Problems/Problem 24"
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+ | ==Problem== | ||
A large square region is paved with <math>n^2</math> gray square tiles, each measuring <math>s</math> inches on a side. A border <math>d</math> inches wide surrounds each tile. The figure below shows the case for <math>n=3</math>. When <math>n=24</math>, the <math>576</math> gray tiles cover <math>64\%</math> of the area of the large square region. What is the ratio <math>\frac{d}{s}</math> for this larger value of <math>n?</math> | A large square region is paved with <math>n^2</math> gray square tiles, each measuring <math>s</math> inches on a side. A border <math>d</math> inches wide surrounds each tile. The figure below shows the case for <math>n=3</math>. When <math>n=24</math>, the <math>576</math> gray tiles cover <math>64\%</math> of the area of the large square region. What is the ratio <math>\frac{d}{s}</math> for this larger value of <math>n?</math> | ||
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==Solution 1== | ==Solution 1== | ||
− | + | The area of the shaded region is <math>(24s)^2</math>. To find the area of the large square, we note that there is a <math>d</math>-inch border between each of the <math>23</math> pairs of consecutive squares, as well as from between first/last squares and the large square, for a total of <math>23+2 = 25</math> times the length of the border, i.e. <math>25d</math>. Adding this to the total length of the consecutive squares, which is <math>24s</math>, the side length of the large square is <math>(24s+25d)</math>, yielding the equation <math>\frac{(24s)^2}{(24s+25d)^2}=\frac{64}{100}</math>. Taking the square root of both sides (and using the fact that lengths are non-negative) gives <math>\frac{24s}{24s+25d}=\frac{8}{10} = \frac{4}{5}</math>, and cross-multiplying now gives <math>120s = 96s + 100d \Rightarrow 24s = 100d \Rightarrow \frac{d}{s} = \frac{24}{100} = \boxed{\textbf{(A) }\frac{6}{25}}</math>. | |
==Solution 2== | ==Solution 2== | ||
− | + | Without loss of generality, we may let <math>s=1</math> (since <math>d</math> will be determined by the scale of <math>s</math>, and we are only interested in the ratio <math>\frac{d}{s}</math>). Then, as the total area of the <math>576</math> gray tiles is simply <math>576</math>, the large square has area <math>\frac{576}{0.64} = 900</math>, making the side of the large square <math>\sqrt{900}=30</math>. As in Solution 1, the the side length of the large square consists of the total length of the gray tiles and <math>25</math> lots of the border, so the length of the border is <math>d = \frac{30-24}{25} = \frac{6}{25}</math>. Since <math>\frac{d}{s}=d</math> if <math>s=1</math>, the answer is <math>\boxed{\textbf{(A) }\frac{6}{25}}</math>. | |
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==Solution 3== | ==Solution 3== | ||
− | + | As in Solution 2, we let <math>s = 1</math> without loss of generality. For sufficiently large <math>n</math>, we can approximate the percentage of the area covered by the gray tiles by subdividing most of the region into congruent squares, as shown: | |
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<asy> | <asy> | ||
draw((0,0)--(13,0)--(13,13)--(0,13)--cycle); | draw((0,0)--(13,0)--(13,13)--(0,13)--cycle); | ||
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} | } | ||
</asy> | </asy> | ||
− | Each red square has side length <math>1+d</math>, so by solving <math>\frac{1^2}{(1+d)^2} = \frac{64}{100} \iff \frac{1}{1+d} = \frac{4}{5}</math>, we obtain <math>d = \frac{1}{4}</math>. The actual fraction of area covered by the gray tiles | + | Each red square has side length <math>(1+d)</math>, so by solving <math>\frac{1^2}{(1+d)^2} = \frac{64}{100} \iff \frac{1}{1+d} = \frac{4}{5}</math>, we obtain <math>d = \frac{1}{4}</math>. The actual fraction of the total area covered by the gray tiles will be slightly less than <math>\frac{1}{(1+d)^2}</math>, which implies <math>\frac{1}{(1+d)^2} > \frac{64}{100} \iff \frac{1}{1+d} > \frac{4}{5} \iff d < \frac{1}{4}</math>. Hence <math>d</math> (and thus <math>\frac{d}{s}</math>, since we are assuming <math>s=1</math>) is less than <math>\frac{1}{4}</math>, and the only choice that satisfies this is <math>\boxed{\textbf{(A) }\frac{6}{25}}</math>. |
==See also== | ==See also== |
Revision as of 10:51, 20 November 2020
Problem
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 .
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 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
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 .
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 | ||
All AJHSME/AMC 8 Problems and Solutions |
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