Difference between revisions of "2020 AMC 8 Problems/Problem 24"

(Solution 2)
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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 boarder. 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>.
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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
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 +
==Solution 4==
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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:
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<asy>
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draw((0,0)--(13,0)--(13,13)--(0,13)--cycle);
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filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray);
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filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray);
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filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray);
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filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray);
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filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray);
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filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray);
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filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray);
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filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray);
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filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray);
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 +
for(int i = 1; i <= 13; i += 4){
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draw((1,i)--(13,i), red);
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draw((i,1)--(i,13), red);
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}
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</asy>
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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 $n^2$ gray square tiles, each measuring $s$ inches on a side. A border $d$ inches wide surrounds each tile. The figure below shows the case for $n=3$. When $n=24$, the $576$ gray tiles cover $64\%$ of the area of the large square region. What is the ratio $\frac{d}{s}$ for this larger value of $n?$

[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); [/asy]

$\textbf{(A) }\frac{6}{25} \qquad \textbf{(B) }\frac{1}{4} \qquad \textbf{(C) }\frac{9}{25} \qquad \textbf{(D) }\frac{7}{16} \qquad \textbf{(E) }\frac{9}{16}$

Solution 1

WLOG, let $s=1$. Then, the total area of the squares of side $s$ is $576$, $64\%$ of the area of the large square, which would be $900$, making the side of the large square $30$. Then, $25$ borders have a total length of $30-24=6$. Since $\frac{d}{s}=d$ if $s=1$ is the value we're asked to find, the answer is $\boxed{\textbf{(A) }\frac{6}{25}}$.

Solution 2

When $n=3$, we see that the total height of the large square is $3s+4d$. Similarly, when $n=24$, the total height of the large square is $24s+25d$. The total area of the $576$ gray tiles is $576s^2$ and the area of the large white square is $(24s+25d)^2$. We are given that the ratio of the gray area to the area of the large square is $\frac{64}{100}=\frac{16}{25}$. Thus, our equation becomes $\frac{576s^2}{(24s+25d)^2}=\frac{16}{25}$. Square rooting both sides, we get $\frac{24s}{24s+25d}=\frac{4}{5}$. Cross multiplying, we get $120s=96s+100d$. Combining like terms, we get $24s=100d$, which implies that $\frac{d}{s}=\frac{24}{100}=\frac{6}{25}\implies\boxed{\textbf{(A) }\frac{6}{25}}$.
~junaidmansuri

Solution 3

The area of the shaded region is $(24s)^2$. The total area of the square is $(24s+25d)^2$ because for each side of the square there one extra row/column of the border. Our equation is $\frac{(24s)^2}{(24x+25d)^2}=\frac{64}{100}$. Taking the square root of both sides gives $\frac{24x}{24x+25d}=\frac 45$. Cross multiplying and rearranging gives $\frac ds=\textbf{(A)} \frac{6}{25}$.

-franzliszt

Solution 4

WLOG $s = 1$. For large enough $n$, 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 $1+d$, so by solving $\frac{1^2}{(1+d)^2} = \frac{64}{100} \iff \frac{1}{1+d} = \frac{4}{5}$, we obtain $d = \frac{1}{4}$. The actual fraction of area covered by the gray tiles is slightly less than $\frac{1}{(1+d)^2}$, which implies $\frac{1}{(1+d)^2} > \frac{64}{100} \iff \frac{1}{1+d} > \frac{4}{5} \iff d < \frac{1}{4}$. Hence $d$ (and $\frac{d}{s}$) is less than $\frac{1}{4}$, and the only choice that satisfies this is $\boxed{\textbf{(A) }\frac{6}{25}}$. ~scrabbler94

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
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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