Difference between revisions of "2001 AMC 10 Problems/Problem 11"
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| − | == Problem == | + | ==Problem== |
Consider the dark square in an array of unit squares, part of which is shown. The first ring of squares around this center square contains <math> 8 </math> unit squares. The second ring contains <math>16</math> unit squares. If we continue this process, the number of unit squares in the <math>100^\text{th}</math> ring is | Consider the dark square in an array of unit squares, part of which is shown. The first ring of squares around this center square contains <math> 8 </math> unit squares. The second ring contains <math>16</math> unit squares. If we continue this process, the number of unit squares in the <math>100^\text{th}</math> ring is | ||
| − | <math> \textbf{(A)}\ 396 \qquad \textbf{(B)}\ 404 \qquad \textbf{(C)}\ 800 \qquad \textbf{(D)}\ 10,\!000 \qquad \textbf{(E)}\ 10,\!404 </math> | + | <math>\textbf{(A)}\ 396 \qquad \textbf{(B)}\ 404 \qquad \textbf{(C)}\ 800 \qquad \textbf{(D)}\ 10,\!000 \qquad \textbf{(E)}\ 10,\!404</math> |
| − | == | + | ==Solution== |
| − | + | ===Solution 1=== | |
| − | === Solution 1 === | ||
We can partition the <math> n^\text{th} </math> ring into <math> 4 </math> rectangles: two containing <math> 2n+1 </math> unit squares and two containing <math> 2n-1 </math> unit squares. | We can partition the <math> n^\text{th} </math> ring into <math> 4 </math> rectangles: two containing <math> 2n+1 </math> unit squares and two containing <math> 2n-1 </math> unit squares. | ||
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Thus, the <math>100^\text{th}</math> ring has <math> 8 \times 100 = \boxed{\textbf{(C)} 800} </math> unit squares. | Thus, the <math>100^\text{th}</math> ring has <math> 8 \times 100 = \boxed{\textbf{(C)} 800} </math> unit squares. | ||
| − | === Solution 2 | + | ===Solution 2=== |
We can make the <math> n^\text{th} </math> ring by removing a square of side length <math> 2n-1 </math> from a square of side length <math> 2n+1 </math>. | We can make the <math> n^\text{th} </math> ring by removing a square of side length <math> 2n-1 </math> from a square of side length <math> 2n+1 </math>. | ||
Revision as of 20:17, 16 March 2020
Problem
Consider the dark square in an array of unit squares, part of which is shown. The first ring of squares around this center square contains 
unit squares. The second ring contains 
unit squares. If we continue this process, the number of unit squares in the 
ring is
Solution
Solution 1
We can partition the 
ring into 
rectangles: two containing 
unit squares and two containing 
unit squares.
There are 
unit squares in the ring.
Thus, the ring has 
unit squares.
Solution 2
We can make the ring by removing a square of side length from a square of side length .
This ring contains 
unit squares.
Thus, the ring has 
unit squares.
See Also
| 2001 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
