Difference between revisions of "2020 AMC 8 Problems/Problem 24"
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==Solution 3== | ==Solution 3== | ||
− | The area of the shaded region is <math>(24s)^2</math>. The total area of the square is <math>(24s+25d)^2</math> because for each side of the square there one extra row/column of the | + | The area of the shaded region is <math>(24s)^2</math>. The total area of the square is <math>(24s+25d)^2</math> because for each side of the square there one extra row/column of the border. Our equation is <math>\frac{(24s)^2}{(24x+25d)^2}=\frac{64}{100}</math>. Taking the square root of both sides gives <math>\frac{24x}{24x+25d}=\frac 45</math>. Cross multiplying and rearranging gives <math>\frac ds=\textbf{(A)} \frac{6}{25}</math>. |
-franzliszt | -franzliszt | ||
+ | |||
+ | ==Solution 4== | ||
+ | WLOG <math>s = 1</math>. For large enough <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: | ||
+ | <asy> | ||
+ | draw((0,0)--(13,0)--(13,13)--(0,13)--cycle); | ||
+ | filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray); | ||
+ | filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray); | ||
+ | filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray); | ||
+ | filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray); | ||
+ | filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray); | ||
+ | filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray); | ||
+ | filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray); | ||
+ | filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray); | ||
+ | filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray); | ||
+ | |||
+ | for(int i = 1; i <= 13; i += 4){ | ||
+ | draw((1,i)--(13,i), red); | ||
+ | draw((i,1)--(i,13), red); | ||
+ | } | ||
+ | </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 is 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 <math>\frac{d}{s}</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>. ~scrabbler94 | ||
==See also== | ==See also== |
Revision as of 17:34, 19 November 2020
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
WLOG, let . Then, the total area of the squares of side is , of the area of the large square, which would be , making the side of the large square . Then, borders have a total length of . Since if is the value we're asked to find, the answer is .
Solution 2
When , we see that the total height of the large square is . Similarly, when , the total height of the large square is . The total area of the gray tiles is and the area of the large white square is . We are given that the ratio of the gray area to the area of the large square is . Thus, our equation becomes . Square rooting both sides, we get . Cross multiplying, we get . Combining like terms, we get , which implies that .
~junaidmansuri
Solution 3
The area of the shaded region is . The total area of the square is because for each side of the square there one extra row/column of the border. Our equation is . Taking the square root of both sides gives . Cross multiplying and rearranging gives .
-franzliszt
Solution 4
WLOG . For large enough , 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 area covered by the gray tiles is slightly less than , which implies . Hence (and ) is less than , and the only choice that satisfies this is . ~scrabbler94
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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All AJHSME/AMC 8 Problems and Solutions |
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