2002 AMC 8 Problems/Problem 23

Problem

A corner of a tiled floor is shown. If the entire floor is tiled in this way and each of the four corners looks like this one, then what fraction of the tiled floor is made of darker tiles?

[asy] /* AMC8 2002 #23 Problem */ fill((0,2)--(1,3)--(2,3)--(2,4)--(3,5)--(4,4)--(4,3)--(5,3)--(6,2)--(5,1)--(4,1)--(4,0)--(2,0)--(2,1)--(1,1)--cycle, mediumgrey); fill((7,1)--(6,2)--(7,3)--(8,3)--(8,4)--(9,5)--(10,4)--(7,0)--cycle, mediumgrey); fill((3,5)--(2,6)--(2,7)--(1,7)--(0,8)--(1,9)--(2,9)--(2,10)--(3,11)--(4,10)--(4,9)--(5,9)--(6,8)--(5,7)--(4,7)--(4,6)--cycle, mediumgrey); fill((6,8)--(7,9)--(8,9)--(8,10)--(9,11)--(10,10)--(10,9)--(11,9)--(11,7)--(10,7)--(10,6)--(9,5)--(8,6)--(8,7)--(7,7)--cycle, mediumgrey); draw((0,0)--(0,11)--(11,11)); for ( int x = 1; x < 11; ++x ) { draw((x,11)--(x,0), linetype("4 4")); } for ( int y = 1; y < 11; ++y ) { draw((0,y)--(11,y), linetype("4 4")); } clip((0,0)--(0,11)--(11,11)--(11,5)--(4,1)--cycle);[/asy]

$\textbf{(A)}\ \frac{1}3\qquad\textbf{(B)}\ \frac{4}9\qquad\textbf{(C)}\ \frac{1}2\qquad\textbf{(D)}\ \frac{5}9\qquad\textbf{(E)}\ \frac{5}8$

Solution

The same pattern is repeated for every $6 \times 6$ tile. Looking closer, there is also symmetry of the top $3 \times 3$ square, so the fraction of the entire floor in dark tiles is the same as the fraction in the square. Counting the tiles, there are $4$ dark tiles, and $9$ total tiles, giving a fraction of $\boxed{\text{(B)}\ \frac49}$.

See Also

2002 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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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