Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
 
(11 intermediate revisions by 7 users not shown)
Line 1: Line 1:
−
== Problem ==
+
==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 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
  
−
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
+
<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>
  
−
<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>
  
−
== Solutions ==
+
==Solution==
−
 
+
===Solution 1===
−
=== Solution 1 ===
 
  
 
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.
Line 13: Line 23:
 
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.
 
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===
 +
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.
 +
 
 +
===Solution 3 (Less Rigorous)===
 +
 
 +
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 ==
 +
 
 +
{{AMC10 box|year=2001|num-b=10|num-a=12}}
 +
{{MAA Notice}}

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_11&oldid=222678"