Difference between revisions of "2001 AMC 10 Problems/Problem 11"
Pidigits125 (talk | contribs) (Created page with '== 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 s…') |
Pidigits125 (talk | contribs) (→Solution 1) |
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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. | ||
− | There are <math> 2(2n+1)+2(2n-1)=4n+2+4n-2=8n </math> unit squares in the <math> n^\text{th} | + | There are <math> 2(2n+1)+2(2n-1)=4n+2+4n-2=8n </math> unit squares in the <math> n^\text{th} </math> ring. |
− | + | Thus, the <math>100^\text{th}</math> ring has <math> 8 \times 100 = \boxed{\textbf{(C)} 800} </math>. |
Revision as of 15:15, 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 .