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

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==Problem==
 
==Problem==
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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
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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>
  
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===Solution 2===
 
===Solution 2===
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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 23:36, 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 $8$ 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 $\textbf{(A)}\ 396 \qquad \textbf{(B)}\ 404 \qquad \textbf{(C)}\ 800 \qquad \textbf{(D)}\ 10,\!000 \qquad \textbf{(E)}\ 10,\!404$

Solution

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

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.

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

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