Difference between revisions of "2002 AMC 8 Problems/Problem 23"
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| + | ==Problem== | ||
A portion of 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? | A portion of 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? | ||
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<math> \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 </math> | <math> \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 </math> | ||
| + | |||
| + | ==Solution== | ||
| + | The same pattern is repeated for every <math>6 \times 6</math> tile. Looking closer, there is also symmetry of the top <math>3 \times 3</math> 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 <math>4</math> dark tiles, and <math>9</math> total tiles, giving a fraction of <math>\boxed{\text{(B)}\ \frac49}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2002|num-b=22|num-a=24}} | ||
Revision as of 19:30, 23 December 2012
Problem
A portion of 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]](http://latex.artofproblemsolving.com/9/1/0/91026905d520b2d9e85b96a8902123543951d953.png)
Solution
The same pattern is repeated for every 
tile. Looking closer, there is also symmetry of the top 
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 
dark tiles, and 
total tiles, giving a fraction of 
.
See Also
| 2002 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
