Difference between revisions of "2001 AMC 10 Problems/Problem 11"

Revision as of 15:18, 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 $8$ unit squares. The second ring contains $16$ unit squares. If we continue this process, the number of unit squares in the $100^\text{th}$ ring is

$\textbf{(A)}\ 396 \qquad \textbf{(B)}\ 404 \qquad \textbf{(C)}\ 800 \qquad \textbf{(D)}\ 10,\!000 \qquad \textbf{(E)}\ 10,\!404$

Solutions

Solution 1

We can partition the $n^\text{th}$ ring into $4$ rectangles: two containing $2n+1$ unit squares and two containing $2n-1$ unit squares.

There are $2(2n+1)+2(2n-1)=4n+2+4n-2=8n$ unit squares in the $n^\text{th}$ ring.

Thus, the $100^\text{th}$ ring has $8 \times 100 = \boxed{\textbf{(C)} 800}$ unit squares.

Solution 2 (Alternate Solution)

We can make the $n^\text{th}$ ring by removing a square of side length $2n-1$ from a square of side length $2n+1$ .

This ring contains $(2n+1)^2-(2n-1)^2=(4n^2+4n+1)-(4n^2-4n+1)=8n$ unit squares.

Thus, the $100^\text{th}$ ring has $8 \times 100 = \boxed{\textbf{(C)}\ 800}$ unit squares.

@@ Line 13: / Line 13: @@
 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>.
+Thus, the <math>100^\text{th}</math> ring has <math> 8 \times 100 = \boxed{\textbf{(C)} 800} </math> unit squares.
+=== Solution 2 (Alternate Solution) ===
+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>.
+This ring contains <math> (2n+1)^2-(2n-1)^2=(4n^2+4n+1)-(4n^2-4n+1)=8n </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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