Difference between revisions of "2001 AMC 10 Problems/Problem 11"
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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. | ||
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| + | == See Also == | ||
| + | |||
| + | {{AMC10 box|year=2001|num-b=10|num-a=12}} | ||
Revision as of 15:21, 16 March 2011
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
Solutions
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 (Alternate Solution)
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 | ||
