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

Latest revision as of 20:41, 15 July 2024

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

$[asy] unitsize(3mm); defaultpen(linewidth(1pt)); fill((2,2)--(2,7)--(7,7)--(7,2)--cycle, mediumgray); fill((3,3)--(6,3)--(6,6)--(3,6)--cycle, gray); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle, black); for(real i=0; i<=9; ++i) { draw((i,0)--(i,9)); draw((0,i)--(9,i)); }[/asy]$

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

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.

Solution 3 (Less Rigorous)

Notice that the first ring around the center square contains $8$ unit squares, the second ring contains $16$ unit squares, the third contains $24$ unit squares, and so on. The number of squares in the $n^\text{th}$ ring is determined by the expression $8 \times n$ . Thus, the number of unit squares in the $100^\text{th}$ ring is equal to $8 \times 100$ , which equals $\boxed{\textbf{(C) }800}$ unit squares.

-Darth_Cadet

Video Solution by Daily Dose of Math

https://youtu.be/y52knpoCVYo?si=dYATo3Zxoj4obeMV

~Thesmartgreekmathdude

@@ Line 34: / Line 34: @@
 ===Solution 3 (Less Rigorous)===
-Notice that the first ring around the center square contains 8 unit squares, the second ring contains 16 unit squares, the third contains 24 unit squares, and so on. The number of squares in the <math> n^\text{th} </math> ring is determined by the expression <math> 8 \times n </math>. Thus, the number of unit squares in the <math>100^\text{th}</math> ring is equal to <math> 8 \times 100 </math>, which equals <math> \boxed{\textbf{(C) }800} </math> unit squares.
+Notice that the first ring around the center square contains <math> 8 </math> unit squares, the second ring contains <math> 16 </math> unit squares, the third contains <math> 24 </math> unit squares, and so on. The number of squares in the <math> n^\text{th} </math> ring is determined by the expression <math> 8 \times n </math>. Thus, the number of unit squares in the <math>100^\text{th}</math> ring is equal to <math> 8 \times 100 </math>, which equals <math> \boxed{\textbf{(C) }800} </math> unit squares.
 -Darth_Cadet
+==Video Solution by Daily Dose of Math==
+https://youtu.be/y52knpoCVYo?si=dYATo3Zxoj4obeMV
+~Thesmartgreekmathdude
 == 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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